Algebra+Prelim+Solutions

Algebra Preliminary Examination Solutions

David Paige

Summer 2007

Author’s Note

Here you have almost 300 pages of math that I did to study for the AlgebraPrelim. It took a lot of work to do it all and now you might be using it tohelp you study. In other words, you may be benefitting from the sweat of mybrow. Therefore, when you read stuff from it, don’t say things like ‘Man thatDavid Paige is a huge loooooser! Why did he do all of this? He works toohard, plus I don’t think any of these solutions are right.’ Instead, you shouldsay stuff like ‘Dude that David Paige, what a nice guy!! I really appreciate allthe hard work he did. I hope that all of his sports teams win their respectivenational championships and that his wife has lots of sex with him. Next timeI see him I’m going to buy him a beer.’ All of that being said, if you do endup reading any of it and find any glaring typos or especially if you notice asolution that is incorrect, let me know. Have fun!!

Instructions

There were some cases where problems or (more commonly) arguments re-quired in problems were repeated. When this occurred, I made a link in mydocument to the place where the argument was given. Also there were timeswhen I found hints or solutions in a textbook and I made note of them withfoot notes. The textbooks I used were:

• Dummit, David S. and Foote, Richard M. Abstract Algebra 3rd edition.

• Lang, Serge. Algebra. 3rd edition.

• Milne, J.S. Group Theory. Available online here.

• Rotman, Joseph J. Advanced Modern Algebra.

Finally, at the end, there is an index of problem topics created by MarkNorfleet (and adjusted some by me). The topics are divided into five majorareas:

1. Fields and Galois Theory

2. Groups

3. Linear Algebra

4. Modules

5. Rings

Contents

I The University Texas 1

1 Spring 2007 3

2 Fall 2006 13

3 Spring 2006 23

4 Fall 2005 33

5 Spring 2005 41

6 Fall 2004 51

7 Spring 2004 63

8 Fall 2003 69

9 Spring 2003 85

10 Fall 2002 91

11 Spring 2002 97

12 Fall 2001 105

13 Spring 2001 111

14 Fall 2000 117

15 Spring 2000 131

16 Fall 1999 137

17 Fall 1998 145

18 Spring 1998 155

19 Fall 1997 165

20 Spring 1997 173

21 Fall 1996 183

22 Spring 1996 193

23 Fall 1995 199

24 Spring 1995 209

II Stanford University 217

25 Stanford Fall 2006 21925.1 Morning Session . . . . . . . . . . . . . . . . . . . . . . . . . . 21925.2 Afternoon Session . . . . . . . . . . . . . . . . . . . . . . . . . 228

26 Stanford Spring 2006 23326.1 Morning Session . . . . . . . . . . . . . . . . . . . . . . . . . . 23326.2 Afternoon Session . . . . . . . . . . . . . . . . . . . . . . . . . 241

27 Stanford Fall 2005 24727.1 Morning Session . . . . . . . . . . . . . . . . . . . . . . . . . . 24727.2 Afternoon Session . . . . . . . . . . . . . . . . . . . . . . . . . 252

28 Stanford Spring 2005 25928.1 Morning Session . . . . . . . . . . . . . . . . . . . . . . . . . . 25928.2 Afternoon Session . . . . . . . . . . . . . . . . . . . . . . . . . 266

Part I

The University Texas

Chapter 1

Spring 2007

1. (i) Let G be a finite group and H < G a proper subgroup. Provethat

G 6=⋃g∈G

gHg−1.

Since G acts transitively on S = gHg−1|g ∈ G by conjugation, we havethat |S| = |G/stab(H)| = |G|/|NG(H)|. Furthermore, we have H < NG(H)so that |G/NG(H)| ≤ |G/H|. But in each gHg−1, there are only |H| el-ements and these sets are not disjoint (they intersect at least at the iden-tity). Hence the total number of elements in ∪g∈GgHg−1 is strictly less than|G/H||H| = |G|. The statement follows.1

(ii) Let G be a group acting transitively on a finite set X with#X > 1. Prove that there is an element g ∈ G with no fixed pointsof its action on X.

First consider the case when the action is faithful. This gives an injectionG→ S(X), so that in particular, G is finite. Now since G acts transitively onX (and |X| > 1), we may find an element x0 ∈ X such that H = stab(x0) 6=G. Then we have an isomorphism of G-sets given by G/H → X given bygH 7→ g · x0. Hence it suffices to consider the case when G is a finite groupwhich acts (by left multiplication) on G/H where H is a proper subgroup of

1This is exercise #20 in Milne’s notes; the solution is given on p. 74

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G. In this case, we have, for x ∈ G,

stab(xH) = g ∈ G|gxH = xH= g ∈ G|x−1gxH = H= g ∈ G|x−1gx ∈ H= xHx−1.

By the above we may find an element g0 ∈ G which does not lie in xHx−1

for any x ∈ G. Thus for each xH ∈ G/H, we have g0 6∈ stab(xH) which isto say that g0 has no fixed points of its action on X.

In the general case, we have a homomorphism G → S(X). If N is thekernel of this homomorphism, we have an embedding G/N → S(X). Thisgives a faithful action of G/N on X which is still transitive (to send x ∈ X toy ∈ X, find an element in G which accomplishes it and then reduce moduloN). Hence we can find an element g ∈ G which has no fixed points in theaction of G/N on X. Then g ∈ G will have no fixed points on its action on X.

(iii) Let G = GL2(C) and H < G the subgroup of upper triangularmatrices. Prove

G =⋃g∈G

gHg−1.

Let A ∈ GL2(C). Since C is algebraically closed, it contains all theeigenvalues of A. Hence there is a matrix P ∈ GL2(C) such that PAP−1 isin Jordan Canonical Form. In particular, PAP−1 ∈ H so that A ∈ P−1HP .

2. Determine all groups with exactly four subgroups.

Let G be a group with exactly four subgroups. First suppose that Gis infinite. If x ∈ G has infinite order, then 〈x〉 is isomorphic to Z, whichhas infinitely-many subgroups, a contradiction. Hence every element of Gmust have finite order. Now let x1 ∈ G. Then 〈x1〉 is finite so we we canfind x2 ∈ G\ 〈x1〉. Likewise, we can find x3 ∈ G\ [〈x1〉 ∪ 〈x2〉]. Thus 1,

5

〈x1〉, 〈x2〉, 〈x3〉, and G are five distinct subgroups of G. Hence G cannot beinfinite.

Now suppose that |G| = pk, where p is a prime. Since G is a p-group,it has a normal subgroup, of order pi for each 0 ≤ i ≤ k. For k ≥ 4, thisis more than four subgroups. Hence we must have |G| = 1, p, p2, or p3.Certainly, G cannot be the trivial group as this gives only one subgroup.Likewise, if |G| = p, the only subgroups of G are 1 and G. Suppose thenthat |G| = p2. Since Z(G) is nontrivial, it must have order p or p2. In theformer case, |G/Z(G)| = p so that G/Z(G) is cyclic and G is abelian (acontradiction). In the latter case G is abelian. This implies that G ∼= Zp2 orG ∼= Zp × Zp. Zp2 has only three subgroups (Zp2 , the collection of elementsof order p, and 1). Zp × Zp has four subgroups: 0, Zp × 0, 0 × Zp

and Zp × Zp.Now let |G| = p3. By a similar argument as above, we have G is abelian

or |Z(G)| = p. If |Z(G)| = p then |G/Z(G)| = p2 so that G/Z(G) ∼= Zp2

or G/Z(G) ∼= Zp × Zp. This first case cannot happen as this implies thatG is abelian (so that |Z(G)| 6= p). In the latter case, we may apply theCorrespondence Theorem to conclude that G has four subgroups that containZ(G). Adding the subgroup 1, we have at least five subgroups and so thelatter case is also a contradiction. Hence, Z(G) cannot have order p and soG must be abelian. This gives G ∼= Zp3 , G ∼= Zp×Zp2 , or G ∼= ZP ×Zp×Zp.The first possibility has four subgroups, the second has six, and the thirdhas eight. Thus we must have G ∼= Zp3 in this case.

Next suppose that G is divisible by three distinct primes p, q, and r. Thenwe have 1, a Sylow p-subgroup, a Sylow q-subgroup, a Sylow r-subgroup,and G as subgroups of G. Since this is five subgroups, G cannot be divisibleby three distinct primes.

Finally we have restricted ourselves to the case that |G| = piqj for distinctprimes p and q. Since we have a Sylow p-subgroup, a Sylow q-subgroup, 1,and G, we must have unique Sylow subgroups so that (applying the DirectProduct Recognition Theorem), G ∼= P×Q, where P is the Sylow p-subgroupand Q is the Sylow q-subgroup. By the above (and considering the subgroupsof P and Q), we cannot have i > 3 or j > 3 as this would imply at least fivesubgroups of order divisible by p and or q respectively. Likewise, if i = 3, wecan apply the above to conclude that G has at least four subgroups of orderdivisible by p. Adding a Sylow q-subgroup, we have at least five subgroups sothat i 6= 3. Similarly, j 6= 3. Now suppose that |G| = piq2 (where i = 1, 2).Then P has at least two subgroups and (from above) Q has at least three.

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Noting that these two collections overlap only at the trivial subgroup andadding G itself, we have at least five subgroups which rules out this case.Likewise, we cannot have |G| = p2qj. Hence we are left with |G| = pq. Inthis case, we have G ∼= Zp×Zq, which does indeed have four subgroups (1,Zp × 1, 1 × Zq, and G).

Thus we have two possibilities: Zp×Zq, where p and q are (not necessarilydistinct) primes and Zp3 .

3. Let A be an n×n matrix with entries in a field K. Prove that Ais similar to is transpose, i.e., that there exists an invertible n× nmatrix U with entries in K such that

UAU−1 = AT .

Let L be the algebraic closure of K (if we don’t want to invoke Zorn’sLemma, we may just take L to be the splitting field of the characteristicpolynomial of A). Then over L, A has all its eigenvalues. We can make Kn

into a K[x]-module, VA via A. Then VA ∼= ⊕K[x]/(x − λi)di for some finite

(thought not necessarily distinct) λi ⊂ L. Pick a certain eigenvalue µ andlet ni be the number of occurrences of K[x]/(x−µ)i in the decomposition ofVA. A−µ annihilates any vector in K[x]/(x−µ) (a one-dimensional space),and a one-dimensional subspace in K[x]/(x − µ)r for r > 0 (namely thatgenerates by the equivalence class of (x− µ)r−1). Hence dim ker(A− µI) =∑ni (no vector in K[x]/(x − λ) for λ 6= µ will be annihilated by x − µ).

Likewise dim ker((A−µI)2) = n1+∑

i>1 2ni. In general, dim ker((A−µI)j) =

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∑i<j ini +

∑i≥j jni. This implies that

dim ker((A− µI)j+1)− dim ker((A− µI)j) =∑i<j+1

ini +∑i≥j+1

(j + 1)ni

−∑i<j

ini −∑i≥j

jni

=∑i≥j+1

ni.

Hence each ni is determined by dim ker(A− µI)i. Thus the elementarydivisors and therefore the invariant factors and Rational Canonical Form ofA over L is determined by dim ker((A− µI)i) for µ ∈ L.

Now dim ker((A−µI)i) is nminus the column rank for (A−µI)i. Likewise,dim ker((AT − µI)i) = dim ker([(A − µI)i]T ) is n minus the column rank of[(A − µI)i]T , that is, the row rank of (A − µI)i. But the column rankand the row rank of a matrix are the same so that dim ker((A − µI)i) =dim ker((AT − µI)i) for any µ ∈ L and i > 0. From the above this impliesthat A and AT and A have the same Rational Canonical Form over L. Nowby uniqueness, A has the same Rational Canonical Form over L as it doesover K (if these were distinct, the latter would also be a Rational CanonicalForm over L which violates uniqueness). Since the same is true of AT , wemay conclude that they have the same Rational Canonical Form over K.This implies that they are similar over K.

4. Find explicitly a polynomial f ∈ Q[x] such that its splitting fieldK satisfies:

(i) Gal(K/Q) is cyclic of order 4.

(ii) There exists a ∈ K such that a2 = 17.

Let ξ be a primitive 17th root of unity and consider the extension Q(ξ).We have Gal(Q(ξ)/Q) ∼= (Z/17Z)× ∼= Z/16Z. Consider the subgroup of(Z/17Z)× generated by 4. This is H1 = 1, 4, 13, 16 and so is a subgroup

8 Spring 2007

of order four. By the Fundamental Theorem of Galois Theory, the fixedfield of this subgroup will be an extension of degree four. Now, since this is acyclotomic extension, the fixed field ofH is Q(η1), where η1 = ξ+ξ4+ξ13+ξ16.Hence the minimal polynomial of η1 will be a polynomial whose Galois groupisomorphic to (Z/17Z)×/H ∼= Z/4Z, that is, cyclic of order four.

Likewise, a subgroup of (Z/17Z)× of order 8 isH2 = 1, 2, 4, 8, 9, 13, 15, 16(this is the subgroup generated by two). Note that H1 ⊂ H2. Hence the fixedfield of H1 contains that of H2. The fixed field of H2 is given by Q(λ), whereλ = ξ + ξ2 + ξ4 + ξ8 + ξ9 + ξ13 + ξ15 + ξ16. We can check that λ solves theequation x2 + x − 4. But, applying the quadratic formula, the roots of thispolynomial are

−1±√

17

2.

Hence we may conclude that√

17 ∈ Q(λ) ⊂ Q(η1). Hence the polynomialabove satisfies the two properties. Thus it suffices to calculate the polyno-mial.

To calculate the polynomial, we first find its other three roots. SinceQ(η1) is Galois over Q, we know that they are elements of Q(η1). By degreeconsiderations they must generate Q(η1) over Q. This leads us to apply Hto other primitive 17th roots of unity (we found η1 by applying it to ξ). Thisgives

η1 = ξ + ξ4 + ξ13 + ξ16

η2 = ξ2 + ξ8 + ξ9 + ξ15

η3 = ξ6 + ξ7 + ξ10 + ξ11

η4 = ξ3 + ξ5 + ξ12 + ξ14.

Constructing the polynomial with these roots, we arrive at

f(x) = x4 + x3 − 6x3 − x+ 1.

5. (i) Prove that an algebraically closed field is infinite.

9

There is a very simple proof of this, but we will need a little bit more onpart (ii). We give the simple proof first. Let F be a finite field and suppose itsorder is q. Then we have that αq−α = 0 for all α ∈ F. Hence the polynomialxq − x+ 1 has no root in F which shows that F is not algebraically closed.

To prove the stronger statement, we will first note a few statements wewill need. First, the polynomial x− 1 divides xd− 1 for any d ∈ N (certainlyx = 1 is a root of xd− 1). Evaluating at a ∈ N, we have a− 1 divides ad− 1.Taking a = pn for prime p and n ∈ N, we have pn− 1 divides pnd− 1. Hence,any pn − 1st root of unity is also a pnd − 1st root of unity. Thus over anyfield, xp

n−1 − 1 divides xpnd−1 − 1.

Now let F be a finite field of order pn. Then again F is the set of solutionsto xp

n − x. Now let F′ be the finite field of order pnd for some n ∈ N. ThenF′ is the set of solutions to xp

m − x = x(xpnd−1 − 1). By the above, xp

n − xdivides xp

nd − x. Since the latter has all its roots in F′ so must the former.But then the set of solutions (over F′) is a subfield of F′ isomorphic to F(applying uniqueness of splitting fields or of finite fields of a given order) andwe may then consider F to be a subfield of F′. Now by order considerations,we have [F′ : F] = d. Also by the Primitive Element Theorem, we have thatF′ = F(α) for some α ∈ F′. The irreducible polynomial of α over F mustthen have order d. Hence there exist (monic) irreducible polynomials over Fof every degree. In particular, F is not algebraically closed.

(ii) Let V be a vector space of dimension n > 2 over a finite fieldk. For a k-linear map f : V → V and a basis B of V let fB be theassociated n×n matrix with entries in k. Prove there exists f suchthat fB is not upper-triangular for any choice of basis B.

There are two proofs of this statement. By part (i), we can find anirreducible monic polynomial g(x) ∈ k[x] of degree n. Let A be the com-panion matrix of g(x), choose a basis for V , and let f be the linear trans-formation that corresponds to A with respect to this basis. To prove thestatement, then, it suffices to shows that PAP−1 is not upper triangular forany P ∈ GLn(k). That is, that A is not similar to any upper triangularmatrix. Suppose for a contradiction that A is similar to an upper triangu-lar matrix B with diagonal entries a1, a2, . . . , an ∈ k. Expanding about thetop row (and the top row of the resulting submatrix repeatedly), we havedet(xI −B) = (x− a1) · · · (x− an). In particular, the characteristic polyno-mial of B splits over k. But the characteristic polynomial of A is the same as

10 Spring 2007

that of B (as the product of the invariant factors). Hence we have that thecharacteristic polynomial of A splits. But the characteristic polynomial of Ais g(x) (expand about the last column), which by assumption is irreducibleover k. This is a contradiction and so completes the first proof.

The second proof is as follows. We have that G = GLn(k) is a finitegroup. Let H be the subgroup of G of upper triangular matrices. By #1,we may find a matrix A ∈ GLn(k) which is not conjugate to any elementof H. Choose a basis for V and let f be the linear transformation whichresults for A. To show that f is the claimed transformation, it suffices toshow that PAP−1 is not upper triangular for any P ∈ GLn(k). Suppose byfor a contradiction that we have PAP−1 = B where B is upper triangular.Them B = P−1AP ∈ GLn(k) so that B ∈ H. But this contradicts our choiceof A and so completes the second proof.

6. (i) Let R be a UFD. Prove that R[x1, x2, . . .] is a non-NoetherianUFD. (You may quote, without giving a proof, Gauss’s Lemma.)

First we will show that R[x] is a UFD. Let f(x) ∈ R[x] be a nonconstantpolynomial of content one. We will show that f(x) is irreducible over F [x](where F is the field of fractions of R, which exists since R is an integral do-main) if and only if it is irreducible over R[x]. By Gauss’s Lemma, reducibil-ity over F implies that over R. Conversely, suppose that f(x) = a(x)b(x)where a(x) and b(x) are irreducible. Since neither a(x) nor b(x) is a unit andno element of R that is not a unit divides the coefficients of f(x), we cannothave a(x) nor b(x) being an element of R. Hence f(x) = a(x)b(x) shows thatf(x) is reducible over F and gives the claim.

Now we will show that f(x) is a product of irreducibles. Since F [x] isa UFD (as a Euclidean Domain), f(x) has a factorization over F [x] intoirreducibles: f(x) = p1(x) · · · pr(x). Again by Gauss’s Lemma, we may finda1, . . . ar ∈ F such that aipi(x) ∈ R(x) and f(x) = a1p1(x) · · · arpr(x). Thecontent of aipi(x) must be one: if a prime p divided aipi(x), it would thendivide f(x). Thus by the above, aipi(x) is irreducible over R and we havewritten f(x) as a product of irreducibles. In the case that the content off(x) is d, write f(x) = dg(x) where the content of g(x) must be 1. Since

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R is a UFD, we may write d as a product of irreducibles over R (which bydegree considerations are also irreducible over R[x]). Applying the previousargument we may write g(x) as a product of irreducibles an we have writtenf(x) as a product of irreducibles.

Next suppose that f(x) = p1(x) · · · pr(x) = q1(x) · · · qs(x), where each piand qj is irreducible (over R[x]) . Similar to the above, it suffices to considerthe case when the the content of f(x) is one. Then each pi and qj musthave content one (in particular, they cannot be elements of R). Hence bythe above, the two factorizations are also such over F . Hence we have s = rand (after possible reordering) that pi is an associate over F of qi. It thensuffices to prove that they are associates over R. Write pi(x) = (a/b)qi(x)so that bpi(x) = aqi(x) (a, b ∈ R). The content of the left is b and that ofthe right is a. Hence a and b must be associates (over R) so that b = au forsome u ∈ R×. This gives pi(x) = uqi(x) and proves the claim.

By induction, R[x1, . . . , xn] ∼= R[x1, . . . , xn−1][xn] is a UFD for any n.Now let f ∈ R[x1, x2, . . .]. Since f only involves finitely many variables, letxn be the largest which appears in f (in any monomial of zero coefficient).Now, suppose g is a polynomial whose degree with respect to some xm ispositive. If m > n, g cannot divide f . To see this, suppose f = gh whichh ∈ R[x1, x2, . . .]. Viewing R[x1, x2, . . .] as a polynomial ring over the singlevariable xm, the degree of f is strictly less than that of gh, a contradiction.

Now assume f is nonconstant. Viewing f as an element of R[x1, . . . , xn].We can write f as product of irreducibles over R[x1, . . . , xn]. By the abovethese polynomials are also irreducible over R[x1, x2, . . .]. Furthermore if fhas two factorizations, we can again apply the above to conclude that all thefactors are elements of R[x1, . . . , xn] and the factorizations must be uniqueby the above (the units being the same for both rings). Hence R[x1, x2, . . .]is a UFD.

Finally consider the ideal I = (x1, x2, . . .). Suppose for a contradictionthat I = (f1, . . . , fn) for some fi ∈ R[x1, . . .]. Since each fi only involvesfinitely-many variables, we can find an xm which does not appear in anyof the fi. Since xm ∈ I, we can write xm = g1f1 + . . . + gnfn for someg1, . . . , gn ∈ R[x1, x2, . . .]. Evaluating this equality at xm = 1 with all theother variables zero, we have 1 = f0, where f0 is the constant term of thepolynomial on the right. Evaluating all variables at zero gives 0 = f0, acontradiction. Thus I is not finitely generated which shows R[x1, x2, . . .] is

12 Spring 2007

not Noetherian.2

(ii) Let R be a UFD such that for any a, b ∈ R the ideal (a, b) isprincipal. Prove R is a PID.

First we will show that if I is an ideal and a1, . . . , ak ∈ I then the greatestcommon divisor of ai lies in I (which exists since R is a UFD). We will proceedby induction on k. In this case we have the gcd of a1, a2, . . . , ak−1 is in I.Write it as d. Now, consider (d, ak). By assumption this is generated by anelement d′ which is then the gcd of d and ak and so a1, . . . , ak. But thend′ ∈ (d, ak) so that d′ = sd+ tak for some st ∈ R. This shows d′ ∈ I and wehave the result.

Again let I be an ideal in R. Since R = (1), it suffices to assume I 6= Rso that no element of I is a unit. Similarly, it suffices to assume I is nonzero.Select a nonzero element r ∈ I. Since r is not a unit, it can be writtenas a product of finitely many irreducibles. Suppose that p1, p2, . . . , pn arethe (distinct) irreducibles which appear in this factorization. Let Si be thecollection of powers of pi which appear in the factorizations of elements ofI. By the well ordering principal, we can find a minimum ai of each Si. Letd = pa1

1 · · · pann . Find elements ri ∈ I such that pi appears in the factorization

of ri exactly to the power of ai (which is possible by the definition of ai).The power of pj is then at least aj for each ri. Since no other irreduciblesappear in the factorization of a, we may conclude that the gcd of a, r1, . . . , rnis d. This shows that d ∈ I. Also by definition, d divides every element of I.Thus I = (d) and we have shown R is a PID.

2This proof is in Dummit and Foote, Section 9.3, starting on p. 302

Chapter 2

Fall 2006

1. Let k be a field and G = GL3(k). Let A ∈ G and let M be thek[x]-module k3 with x acting as A.

a) Show how the invariant factors of M can be obtained from theminimal and characteristic polynomials of A. Conclude that theminimal and characteristic polynomials of A determine the conju-gacy class of A in G.

The characteristic polynomial is of degree three (since we are workingin a vector space of dimension three). Since the minimal polynomial is anonconstant polynomial which divides the characteristic polynomial, we canconclude that it has degree one, two, or three. Note also that the character-istic polynomial is the product of the invariant factors of which the minimalpolynomial is the largest. Hence, if the minimal polynomial has degree three,it is equal to the characteristic polynomial and so is the unique invariant fac-tor. If it has degree two, the characteristic polynomial is the product of theminimal polynomial and and a one dimensional polynomial (which can thenbe determined by polynomial division). Hence these two polynomials are theinvariant factors. Finally, if the minimal polynomial has degree one, we knowthat all the other invariant factors divide it and so must equal it (as they arenonconstant). Since their product is degree three, the invariant factors arethen three copies of the minimal polynomial. Once we know the invariantfactors, we know the rational canonical form and therefore the conjugacyclass.

14 Fall 2006

b) Let k = F2, the field of two elements. List all possible conjugacyclasses of G and show that if the characteristic polynomial of A isirreducible then the conjugacy class of A has 24 elements.

First suppose that the minimal polynomial of A has degree one. Thenit can be x or x + 1 and the rational canonical form of A is three blocks ofthe corresponding companion matrix. The rational canonical form is then arepresentative of the conjugacy class of A. Next suppose that the minimalpolynomial of A has degree two. Then it can be x2, x2+1, x2+x, or x2+x+1.The remaining invariant factor must then divide this. Since x2 + x + 1 isirreducible, it cannot be the minimal polynomial. If the minimal polynomialis x2 or x2 + 1 = (x + 1)2, the remaining invariant factor must then x orx+ 1 respectively as these are the only degree one polynomials which dividethe corresponding minimal polynomial. Likewise if the minimal polynomialis x2 + x = x(x + 1), the other factor can be either x or x + 1. Again oncewe know the invariant factors of A we can determine its rational canonicalform. Lastly if the minimal polynomial has degree three, it can be x3, x3 +1,x3 + x, x3 + x+ 1, x3 + x2, x3 + x2 + 1, x3 + x2 + x or x3 + x2 + x+ 1. Allthese choices give the possibilities below (noting that −1 = 1 in F2). 0 0 0

0 0 00 0 0

1 0 00 1 00 0 1

0 0 00 0 00 1 0

1 0 00 0 10 1 0

0 0 0

0 0 00 1 1

1 0 00 0 00 1 1

0 0 01 0 00 1 0

0 0 11 0 00 1 0

0 0 0

1 0 10 1 0

0 0 11 0 10 1 0

0 0 01 0 00 1 1

0 0 11 0 00 1 1

0 0 0

1 0 10 1 1

0 0 11 0 10 1 1

Since we are under the restriction A ∈ GL3(F2), we then need to remove

all the matrices above which have determinant zero, i.e., are not invertible.

15

The remaining matrices will be a collection of unique representatives of theconjugacy classes in G. This gives the collection below. 1 0 0

0 1 00 0 1

1 0 00 0 10 1 0

0 0 11 0 00 1 0

0 0 11 0 10 1 0

0 0 1

1 0 00 1 1

0 0 11 0 10 1 1

Now x8 − x = x(x7 − 1) is the product of all the irreducible polynomials

over F2 of order dividing 3. Hence the matrices whose minimal polynomial isan irreducible third degree polynomial have order 7. The minimal polynomialof the matrices above are x + 1, x2 + 1, x3 + 1, x3 + x + 1, x3 + x2 + 1,x3 + x2 + x + 1 = (x + 1)3. The first three visibly have orders 1, 2, and 3,respectively. The next two have irreducible polynomials of degree three andso have order 7. We can check that the order of the last matrix is four. Hencethe matrices whose minimal polynomials are irreducible cubics are the onlymatrices of order seven. Let B be such a matrix. Now the order of the groupis 168: 7 choices for the first column (excluding the zero column), 6 for thesecond (excluding a linear combination of the first column), and 4 for thethird (excluding a linear combination of the first two columns). Since thecentralizer of B has order at least 7, its conjugacy class has order at most 24.Let n7 be the set of Sylow 7-subgroups. Then either n7 = 1 or n7 is at least8. In the latter cause we need at least 48 elements of order 7. Hence n7 iseither 1 or 8. In the latter case the two conjugacy classes would then have 24elements. Hence it suffice to show that there is more than one subgroup oforder 7. Taking powers of the fourth matrix above, we see that the subgroupit generates does not include the other conjugacy class’ rational canonicalform. Hence there is more than one subgroup of order seven which gives theresult.

2. Let F be a finite field with p elements, p a fixed prime. Let Fbe the algebraic closure of F . Consider the set L ⊂ F of elements

16 Fall 2006

of degree prime to p over F .

a) Show that L is a field. Show that every algebraic extension of Lhas p power degree.

Suppose that α, β ∈ L. Then [F (α) : F ] and [F (β) : F ] are not divisibleby p. F (α)/F is Galois as we are working over a finite field (it is the splittingfield of the minimal polynomial of α). Now by the lifting property of Galoisgroups, we have that Gal(F (α)F (β)/F (α)) ∼= Gal(F (β)/F (α) ∩ F (β)). ButF (α)∩F (β) ⊂ F (β) and so by the Fundamental Theorem of Galois Theory,the latter Galois group is a subgroup of Gal(F (β)/F ). In particular it cannothave order divisible by p as this would imply [F (β) : F ] were divisible by p.Hence [F (α)F (β) : F (α)] is not divisible by p. Thus we have

[F (α)F (β) : F ] = [F (α)F (β) : F (α)][F (α) : F ]

is not divisible by p. Now let γ ∈ F (α)F (β) and suppose that [F (γ) : F ]is divisible by p. Then [F (α)F (β) : F ] = [F (α)F (β) : F (γ)][F (γ) : F ] isdivisible by p, contradicting what we have shown. Hence [F (γ) : F ] is notdivisible by p and so F (α)F (β) ⊂ L. This gives α+ β, αβ, α−1,−α ∈ L andwe may conclude that L is a subfield of F .

Consider now a (monic) irreducible polynomial f(x) over F of degree pk,which exists since the finite field K of degree pk over F is simple. Let α bea root of this polynomial the F (α) = K. Then all the elements of F (α) areroots of the polynomial

xppk

− x,

which is the product of all the irreducible polynomials over F with degreesdividing pk. Since all these polynomials thus of degree a power of p, we mayconclude that β ∈ F (α), β 6∈ F implies that p|[F (β) : F ]. Hence L∩F (α) =F . Moreover, F (α)L = L(α) so that we may apply the lifting of Galoisextensions to conclude that Gal(L(α)/L) ∼= Gal(F (α)/F ). In particular, αhas degree pk over L. But L ⊂ L(α) ⊂ F and so

[F : L] = [F : L(α)][L(α) : L] ≥ pk.

Since this holds for all k, we may conclude that [F : L] = ∞. Since F isalgebraic over L (as it is algebraic over F ), we may conclude that not everyalgebraic extension of L is of degree a power of p. Thus the second part ofthe statement is incorrect.

17

Now suppose that K/L is a finite extension. Then K is generated over Lby a finite collection α1, α2, . . . , αn of elements. Let fi(x) be the irreduciblepolynomial of αi over L. Since the coefficients of each fi(x) are algebraicover F (by definition of L) and there are finitely many, they generate a finiteextension Fpk over F . Each αi is then algebraic over Fpk so that they generatea finite extension Fpn of Fpk . Since this Fpn is a finite field, we may concludethat it is Galois over F . We may then apply the lifting property of GaloisExtensions to conclude that Gal(LFpn/L) ∼= Gal(Fpn/L∩Fpn) < Gal(Fpn/F ).In particular, the Galois group of LFpk over L is finite. Since a finite grouphas finitely many subgroups, we may apply the Fundamental Theorem ofGalois Theory to conclude that there are finitely many subfields between Land LFpk . But K ⊂ LFpk by construction and so we may conclude that thereare finitely many subfields between L and K. Hence, K is simple over L.

Set K = L(α) and let f(x) ∈ L[x] be the irreducible polynomial of α overL. Let n = pkm (where m is prime to p) be the degree of f(x). Again thecoefficients of f(x) lie in L and so have degrees over F which are relativelyprime to p. If we let d be there least common multiple, we have that they areall roots of xp

d − x (since this is the product of all irreducible polynomialsover F of degree dividing d). Hence f(x) ∈ Fpd [x]. Since f(x) is irreducibleover L it is also irreducible over Fpd [x] and we have that Fpd(α) = Fpnd . The

elements of Fpnd are then the roots of xpnd − x. Since md is a divisor of nd,

we may find a root β of an irreducible polynomial over F of degree md inFpnd = F (α). The equation [F (α) : F ] = [F (α) : F (β)][F (β) : F ] then allowsus to conclude that [F (α) : F (β)] = pk. Since md is prime to p, β ∈ L.Then F (α)∩L is an intermediate field between F (α) and F (β) and so F (α)has degree pi (for some i ≤ k) over F (α) ∩ L (applying the FundamentalTheorem of Galois Theory and Lagrange’s Theorem). Again applying thelifting property of Galois groups, we may conclude that K = L(α) = F (α)Lhas degree pi over L, which gives the claim.

b) Let q be any prime distinct from p and ρ ∈ F a primitive qthroot of one. Let n be the order of p modulo q. Show that ρ ∈ L ifand only if n is prime to p.

Let ζ be a primitive qth root of unity in F . Since ζ is algebraic over F ,[F (ζ) : F ] is finite and so F (ζ) is a finite field. Hence F (ζ)× has order pk− 1for some k. But the order of ζ (as a member of this multiplicative group) is

18 Fall 2006

q and this must divide the order of the group. Hence q divides pk − 1 or pk

is equivalent to 1 modulo q. Conversely if pm − 1 is equivalent to 1 moduloq, Cauchy’s Theorem implies that there is a primitive qth root of unity inFpm which will then generate all the primitive qth roots of unity (since q isprime). Thus Fpn is the smallest extension of F which contains ζ. Hence ζhas degree n over F . By the definition of L, this says that ζ ∈ L if and onlyif n is prime to p.

c) If p has order prime to p modulo q, we call q a fine prime. If q isa fine prime and a ∈ L×, show that one can solve xq = a in L.

Suppose that xn − a, with n prime to p, has no root in L, that is, it hasno linear factor. We showed in part a) that the only irreducible polynomialsover L are either of degree pi for some i > 0 or linear. Since xn − a has nolinear factor, me may conclude that there is some finite collection of di > 0such that xn − a factors into polynomials of degree pdi . This implies that∑pdi = n so that n is divisible by p, a contradiction. Hence xn−a is solvable

in L if n is prime to p. In particular, xq−a has a root in L, since q is a primedistinct from p (p has no order modulo itself; it is zero).

d) Show that L× is a divisible group. That is, for any a ∈ L× andany integer n, show that xn = a is solvable in L.

First consider the case xpk − a. Now a ∈ L is algebraic over F . Hence

if d is its degree (which must then be prime to p), it lies in the finite fieldFpd . But then it also lies in Fpr as long as d divides r. Choose r such thatd divides it, r > k, and r is prime to p (taking r to be a sufficiently highpower of d would accomplish this). Then the Galois group of Fpr over F isgenerated by the Frobenius automorphism σ : α 7→ αp which has order r.Then σk : α 7→ αp

kis an automorphism of Fpr . In particular it is surjective

so that we may find an α ∈ Fpr such that αpk

= a. The degree of α over F

is a divisor of r and so is prime to p which shows that α ∈ L. Hence xpk − a

is solvable in L

Now consider the general case of xn − a. Write n = pkm for some mprime to p. Then by the previous argument, we can find an α ∈ L such thatαp

k= a. Then xn − a = (xm)p

k − αpk

= (xm − α)pk

(since we are working incharacteristic p). But from c), we can solve xm − α over L since m is prime

19

to p. A root of this polynomial is thus a root to xn − a.

3. Let R be a commutative ring with ideals I, J ⊂ R. Form the Rmodule M = R/I ⊗R R/J (so r(a⊗ b) = ra⊗ b = a⊗ rb).

(a) Show that there is a surjective R-module homomorphism

φ : M → R/(I + J).

Define a map Φ : R/I×R/J → R/(I+J) by (x+ I, y+J) 7→ xy+ I+J .To see that this is well defined, suppose that s ∈ I and t ∈ J . Then

Φ(x+ s+ I, y + t+ J) = (x+ s)(y + t) + I + J

= xy + sy + xt+ st+ I + J

= xy + I + J

= Φ(x+ I, y + J),

since sy, xt, st ∈ I + J . Next let x1, x2, x, y, r ∈ R. Then we have

Φ(x1 + I + x2 + I, y + J) = Φ(x1 + x2 + I, y + J)

= (x1 + x2)y + I + J

= x1y + I + J + x2y + I + J

= Φ(x1 + I, y + J) + Φ(x2 + I, y + J)

and

Φ(rx+ I, y + J) = rxy + I + J

= r(xy + I + J)

= rΦ(x+ I, y + J),

so that Φ is linear in the first coordinate. Similarly, Φ is linear in the sec-ond coordinate. Thus Φ is R-bilinear and so we may apply the universal

20 Fall 2006

property of the Tensor Product to conclude that there is and R-linear mapφ : M → R/(I + J) which satisfies φ((x + I) ⊗ (y + J)) = xy + I + J . Ifx+ I + J ∈ R/(I + J), we have φ((x+ I)⊗ (1 + J)) = x+ I + J so that φis surjective.

b) Show that the element α = (1 + I) ⊗ (1 + J) generates M as anR-module.

Let (x+ I)⊗ (y + J) ∈M . Then x, y ∈ R so that

xyα = x[y(1 + I)⊗ (1 + J)]

= x(1 + I)⊗ (y + J)

= (x+ I)⊗ (y + J),

which shows that (x+ I)⊗ (y + J) ∈ Rα and gives the result.

c) Show that I + J = r ∈ R|rα = 0. (Hint: φ(α) =?)

First suppose that r ∈ I + J so that r = x + y for some x ∈ I, y ∈ J .Then

rα = (x+ y)[(1 + I)⊗ (1 + J)]

= x[(1 + I)⊗ (1 + J)] + y[(1 + I)⊗ (1 + J)]

= (x+ I)⊗ (1 + J) + (1 + I)⊗ (y + J)

= 0.

Conversely, if rα = 0, we have φ(rα) = 0 so that rφ(α) = 0. But φ(α) =1 + I + J so r + I + J = 0 + I + J and r ∈ I + J .

d) Show that φ is an isomorphism.

Suppose m ∈ ker(φ). Since we have shown α generates M , we maywrite m = rα for some r ∈ R. This gives φ(rα) = 0 + I + J so thatr+ I + J = 0 + I + J . Hence r ∈ I + J which means that m = rα = 0. Thusφ is injective and so an isomorphism.

21

4. A discrete valuation ring (DVR) is a domain R with the prop-erty that R is a principal ideal domain and R has a single non-zeroprime ideal.

a) Show that if R is a DVR and K is the fraction field of R thenfor any non-zero element x ∈ K we have either x ∈ R or x−1 ∈ R.

Since R has a unique nonzero prime ideal, it has (up to multiplication byunits) a unique prime element p. Suppose a/b ∈ K is nonzero. Since R isa PID, it is also a UFD. Hence a can decomposed into a product of primes.Since p is the unique prime, we have a = upi for some i ≥ 0 and u ∈ R×.Likewise b = vpj for some j ≥ 0 and v ∈ R×. If i ≥ j, a/b = uv−1pi−j ∈ R.If j ≥ i, (a/b)−1 = u−1vpj−i ∈ R. This gives the result.

b) If R is a DVR and P denotes its maximal ideal, show that∩n≥1P

n = 0.

As above an element a ∈ R which is not zero can be written as a = upi

for some i ≥ 0 and u ∈ R×. Suppose then that a ∈ P i+1 = (p)i+1 = (pi+1).Then a = bpi+1 for some b ∈ R. This gives u = bp. So that u−1bp = 1 andp is a unit. Since p generates a prime ideal this is not the case. Thus givenany element of a ∈ R\0 we can find an n such that a 6∈ P n which impliesthat ∩n≥0P

n = 0.

c) Show that if R is a DVR, then R is integrally closed, that is,if a ∈ K is an element such that there exists a monic polynomialf(x) ∈ R[x] with the property that f(a) = 0, then necessarily a ∈ R.

Let a ∈ K\R. The from part a), we have that a = u/pk for someu ∈ R× and k > 0 (if k = 0, a = u ∈ R). Suppose that f(x) ∈ R[x] isf(x) = xn + an−1x

n−1 + · · ·+ a0. Then

f(a) =uk

pkn+un−1an−1

pkn−k+ · · ·+ a0

=uk + un−1an−1p

k + · · ·+ a0pkn

pkn.

If this were zero, we would have uk = −(un−1an−1pk+ · · ·+a0p

kn) ∈ P , whichcannot happen since uk is a unit. Hence we have shown that R is integrally

22 Fall 2006

closed.

d) Is the polynomial ring Q[x, y] a DVR? Explain your answer. (Qis the rational field).

Consider the ideal (x, y) ∈ Q[x, y] and suppose that is generated byf(x, y) ∈ Q[x, y]. Then x = g(x, y)f(x, y) for some g(x, y) ∈ Q[x, y]. View-ing both sides as polynomials over y, we can conclude that the degree in yof f(x, y) is zero. That is, f(x, y) is a polynomial in x. But there also existsh(x, y) ∈ Q[x, y] such that y = h(x, y)f(x, y) so that f(x, y) is a polynomialin y. From this we can conclude that f(x, y) = a ∈ Q. But then a ∈ (x, y)so we can find s(x, y), t(x, y) ∈ Q[x, y] such that a = s(x, y)x + t(x, y)y. Ifs(x, y) is nonzero, the degree on the right in x is at least one, whereas it iszero on the right. Hence s(x, y) = 0. Likewise, t(x, y) = 0 so that a = 0 and(x, y) = (0). Since x ∈ (x, y), this is not the case and so we may concludethat (x, y) is not principle. Thus Q[x, y] is not a PID and so, in particular,it is not a DVR.

e) Show that if R is a DVR, then its fraction field cannot be alge-braically closed.

Consider the polynomial x2 + p. Suppose that it has roots α, β ∈ K(K the fraction field of R). Then we may have x2 + p = (x − α)(x − β) =x2 − (α + β)x + αβ. Hence α and β multiply to p. Since α, β ∈ K, we mayapply part a) to find u, v ∈ R× and i, j ∈ Z such that α = upi and vpj. Thisgives uvpi+j = p. By unique factorization, uv = 1 and i + j = 1. But i andj must both be nonnegative from the fact that R is integrally closed. Hence(without loss of generality), i = 1 and j = 0. This shows that v is a rootof x2 + p. Hence v2 = −p ∈ P . But v2 is a unit and so it cannot be in P .This is a contradiction. Thus x2 + p does not split in K and so K is notalgebraically closed.

Chapter 3

Spring 2006

1. Let G be the group given by generators and relations as⟨x, y, z : xy−1[x, y] = yz−1[y, z] = zx−1[z, x] = 1G

⟩,

where [a, b] = aba−1b−1 denotes the commutator of elements a and b.Prove that G is an infinite group.

Let F denote the free group on generators S = x, y, z. Then considerthe map of sets S → Z given by x, y, z 7→ 1. By the universal property offree groups, we have an induced group homomorphism F → Z which mapsx, y, and z to 1. Note also that xy−1[x, y] = 1−1+1+1−1−1 = 0 (writingthe operation in F as multiplication and that in Z as addition). Likewise,yz−1[y, z] and zx−1[z, x] are mapped to zero. Hence our homomorphismfactors through F/R where R is the normal subgroup generated by thesethree expressions. That is, we have a homomorphism G→ Z which maps x,y, and z to 1. The image of this map is a subgroup containing 1 and so mustequal Z. Hence the map is surjective which shows that G is infinite.

2. Let E/K be a Galois extension of degree p2q where p and q areprimes, p > q and q does not divide p2 − 1.

24 Spring 2006

(i) Prove that there exist unique intermediate fields L and M suchthat

K ⊂ L ⊂ E, K ⊂M ⊂ E,

[L : K] = p2, and [M : K] = q.

Let G be the Galois group and set np and nq to be the number of Sylowp-subgroups and the number of Sylow q-subgroups in G, respectively. Fromthe Sylow Theorems, np divides the order of G and is equivalent modulo p to1. Hence it must be either 1 or q. Because p is greater then q, however, q isnot equivalent to one modulo p. Thus we have np = 1. Similarly, nq must be1, p, or p2. Since q does not divide p2− 1, p2 is not equivalent to one moduloq. Thus nq 6= p2. Since p2 − 1 = (p+ 1)(p− 1), q also cannot divide p− 1 sothat nq 6= p. Thus nq = 1. We may conclude that both the Sylow subgroupsare unique and therefore normal.

Now from the Fundamental Theorem of Galois Theory, a subgroup L asabove corresponds to a subgroup in G of index p2, that is, of order q. Thisis a Sylow q-subgroup. Since we have G has a unique such subgroup, wemay conclude that E has a unique subfield satisfying the given properties.Likewise M corresponds to the unique Sylow p-subgroup.

(ii) Prove that the fields L and M are each Galois over K.

We have shown previously that the subgroups to which L and M corre-spond are normal. This implies that L and M are Galois over K.

(iii) Prove that the Galois group Aut(E/K) is abelian.

Let P be a Sylow p-subgroup and Q as Sylow q-subgroup. We have seenthat both subgroups are normal. Furthermore, |P ||Q| = p2q = |G| andP ∩ Q = 1 (by order considerations). Hence we may apply the DirectProduct Recognition Theorem to conclude that G ∼= P × Q. Now Q is ofprime order and so is cyclic. Likewise, P has order p2 and so is abelian (theargument for this is given in Spring 2007 #2). Thus G is a direct product ofabelian groups and so abelian itself.

25

3. Let φ : ZN → ZN be a homomorphism of Z-modules.

(i) Let β1, β2, . . . , βN be a basis for ZN as a Z-module and let A bethe N ×N matrix with entries in Z determined by φ and this basis.Prove that | detA| is independent of the choice of basis and hence| detφ| is well defined.

Changing basis is equivalent to performing elementary row and columnoperations on A. That is, it is equivalent to taking PAQ, where P,Q ∈GLN(Z). Now, we may also consider P,A,Q as matrices over Q and inthis case the determinant is multiplicative. Hence the determinant is multi-plicative over Z. We have then that det(P ) det(P−1) = det(I) = 1. Hencedet(P ) = ±1. Likewise, det(Q) = ±1. Thus

| det(PAQ)| = | det(P )|| det(A)|| det(Q)| = | det(A)|,

as claimed.

(ii) Letχ = ZN/φ(z) : z ∈ ZN

denote the cokernel of φ. Prove that χ is finite if and only if| detφ| 6= 0.

First, suppose that b1, . . . , bN are a basis for ZN , i.e. ZN = Zb1⊕· · ·⊕ZbNand let a1, . . . , aN ∈ N. We will show that[

Zb1 ⊕ · · · ⊕ ZbN]/[

Za1b1 ⊕ · · · ⊕ ZaNbN]∼= (Z/a1Z)⊕ . . .⊕ (Z/aNZ)

as Z-modules (i.e. as groups). For this consider the map

ψ : Zb1 ⊕ · · · ⊕ ZbN → (Z/a1Z)⊕ . . .⊕ (Z/aNZ)

given by(c1, . . . , cN) 7→ (c1 mod(a1), . . . , cN mod(aN)).

This is a group homomorphism as a direct sum of quotient maps. Further-more, the (c1, . . . , cN) is mapped to zero if and only if ci ∈ (ai) for eachai. That is, if (c1, . . . , cN) ∈ Za1b1 ⊕ · · · ⊕ ZaNbN . The First IsomorphismTheorem then gives the claim.

26 Spring 2006

Now since Z is a Euclidean Domain, we may find a select a basis B =b1, . . . , bN of ZN with respect to which the matrix, B of φ is in SmithNormal Form. That is B is a diagonal matrix with entries a1, . . . , aN wherea1| · · · |aN . Since we are allowed to multiply by units, we may assume thatai ≥ 0. The image of the column vector [c1, . . . , cN ]T is thus [a1c1, . . . , aNcN ]T ,which shows that the image of φ is Za1b1 ⊕ · · · ⊕ ZaNbN . By the above wemay conclude that χ ∼= (Z/a1Z) ⊕ . . . ⊕ (Z/aNZ). Hence χ is finite if andonly if each ai is nonzero. But now we have | det(φ)| = | det(B)| = a1 · · · aN .This is nonzero if and only if each ai is nonzero, which gives the claim.

(iii) Assume that | detφ| 6= 0 and show that the cardinality of χ is| detφ|.

With notation as above, we have | detφ| = a1 . . . aN 6= 0 and χ ∼=(Z/a1Z) ⊕ · · · ⊕ (Z/aNZ). The order of the latter is a1 · · · aN , which givesthe result.

4. Let R = Z[√

2] and let N : R→ 0, 1, 2, . . . be defined by

N(a+ b√

2) = |a2 − 2b2|.

(i) Prove that N is a norm on the ring R and R is a Euclideandomain with respect to this norm.

Let a + b√

2, c + d√

2 ∈ Q[√

2] and extent the norm to Q[√

2] in the

27

obvious way. Then

N((a+ b

√2)(c+ d

√2))

= N(ac+ 2bd+ (ad+ bc)

√2)

=∣∣∣(ac+ 2bd)2 − 2(ad+ bc)2

∣∣∣=∣∣∣a2c2 + 4abcd+ 4b2d2 − 2a2d2 − 4abcd− 2b2c2

∣∣∣=∣∣∣a2c2 − 2a2d2 − 2b2c2 + 4b2d2

∣∣∣=∣∣∣(a2 − 2b2)(c2 − 2d2)

∣∣∣= N(a+ b

√2)(c+ d

√2).

Thus this norm preserves multiplication in Q[√

2]. In particular, it does soin R.

Now to see that R is a Euclidean domain under this norm, let a, b ∈ Rwith b 6= 0. Since Q[

√2] is a field (

√2 is degree two over Q), we may find

r, s ∈ Q such that a/b = r + s√

2. Let p and q be the closest integers to rand s, respectively (if either r or s is equally between two integers, i.e., ishalf of an odd integer, pick either integer). Then we have |p− r| ≤ 1/2 and|q − s| ≤ (1/2). This gives

N((p+ q

√2)b− a

)= N

((r + s

√2)b+ ([p− r] + [q − s]

√2)b− a

)= N

(([p− r] + [q − s]

√2)b)

= N([p− r] + [q − s]

√2)N(b)

≤ (|p− r|2 + 2|q − s|2)N(b)

≤ (3/4)N(b)

< N(b).

This gives the result.

(ii) Prove that the element 1 +√

2 in R is a unit of infinite order.

First we will show that an element a ∈ R is a unit if and only if N(a) = 1.If a is a unit, we have some u ∈ R such that au = 1. Hence N(a)N(u) =N(au) = N(1) = 1. Thus N(a) is a nonnegative integer which divides 1 and

28 Spring 2006

so must one. Conversely if N(a) = 1, we have that a 6= 0, as N(0) = 0.Hence we may find r, u ∈ R such that 1 = au + r and N(r) < N(a). Thisimplies that N(r) = 0. Suppose r = c + d

√2. Then we have c2 = 2d2.

Supposing that d 6= 0 gives (c/d)2 = 2, which is impossible for c, d ∈ Z.Hence d = 0 and so c = 0. Thus r = 0 in R and we have shown 1 = au sothat a is a unit.

Now 1 +√

2 is a unit as it has norm one (also [1 +√

2][−1 +√

2] = 1).Now suppose that a+ b

√2 ∈ R satisfies a, b > 0. Then (a+ b

√2)(1 +

√2) =

(a + 2b) + (a + b)√

2. Since a + 2b > 0, a + b > 0 we see that this elementsatisfies the original condition. Hence we may apply induction to concludethat the coefficients on (1+

√2)n on both 1 and

√2 are positive for all n. In

particular, (1 +√

2)n 6= 1 (as this would require a coefficient of zero on√

2).Hence 1 +

√2 has infinite order.

(iii) Explain why the identities

7 = (5 + 3√

2)(5− 3√

2)

= (27 + 19√

2)(27− 19√

2)

= (75 + 53√

2)(75− 53√

2)

do not violate unique factorization in R.

First suppose that x1x2 = y1y2 in R with x1 an associate of y1. Thenwrite y1 = x1u where u ∈ R×. This gives x1x2 = ux1y2 so that (since R is anintegral domain) x2 = uy2. Thus x2 and y2 are also associates under theseconditions.

Note that neither 5 + 3√

2 nor 5 − 3√

3 is a unit since they both havenorm 7. Hence we may conclude that seven is not irreducible. Now noticethat 27+19

√2 = (5+3

√2)(1+

√2)2. Since (1+

√2)2 is a unit from (i) (as a

product of units), we have that 5+3√

2 is an associate of 27+19√

2. By theabove then, 5− 3

√2 is an associate of 27− 19

√2. Hence the equality of first

two products above does not violate unique factorization. Likewise, we have75− 53

√2 = (5 + 3

√2)(99− 70

√2) and 99− 70

√2 is a unit as it has norm

one. Hence, as in the previous case, we have that equality between the firstand third products does not violate unique factorization. The transitivity ofthe associates relation shows that equality of the second and third productdoes not violate unique factorization.

29

(iv) Determine a factorization of 7 in prime elements of R.

First suppose that a ∈ R has norm seven and that a = bc for someb, c ∈ R. Then 7 = N(a) = N(b)N(c) so that either N(b) or N(c) has norm1. Hence either b or c is a unit so that a is irreducible. Since all the numbersabove (aside from 7) have norm 7, this shows that they are all irreducible.Hence any of the expressions given in (iii) is a factorization of 7 into primeelements of R.

5. Let K = F1024 be the finite field with 1024 = 29 elements, and letL/K be a field extension of degree 2.

(i) Prove there is a unique automorphism σ : L → L which fixes Kand has order 2.

First note that L/K is a finite extension of a finite field (which is thereforeperfect) and so is a separable extension. Furthermore, let α ∈ L\K. Thenα is degree two over K and so satisfies a polynomial of the form x2 + ax+ bwhich is irreducible over K. If β is another root of this polynomial (in somefixed algebraic closure of L) then αβ = b so that β = b/α ∈ L. This showsthat L is the splitting field of x2 + ax+ b and so is a normal extension of K.Hence L/K is Galois. The Galois groups must then consist of the identityand an a unique element, σ, of order two. This is the claimed automorphism.

(ii) Determine the number of elements x in the multiplicative groupL× such that σ(x) = x−1.

First we will show that every quadratic (monic) polynomial in K[x] splitsover L. Suppose for a contradiction that f(x) = x2 + ax + b ∈ K[x] isirreducible over L. Let β, β′ be a roots of this polynomial in some fixedalgebraic closure of L. Let α ∈ L\K. Then we have that [K(α, β) : L] = 2.Hence [K(α, β) : K] = [K(α, β) : L][L : K] = 4 so that the Galois grouphas order 4 (this is a Galois extension as the splitting field of the product ofthe irreducible polynomials of α and β over K, a perfect field). Let α′ be

30 Spring 2006

the other root of the irreducible polynomial of α over K. Then any elementτ ∈ Gal(K(α, β)/K) either preserves α or maps it to α′. In the latter caseit must then map α′ to α. This shows that in either case τ 2 preserves α.Likewise, τ 2 preserves β. Hence me may conclude that every element of theGalois groups has order dividing 2. Hence the Galois groups is the Klein4-group (alternatively, K(α, β) = K(α)K(β) and K(α) ∩K(β) = K so thatthe Galois group of the composite is the direct product of the two smallerGalois groups). K(α, β), however is also an extension of F2 and so we havethat its Galois group over F2 is cyclic. But then K is an intermediate fieldso that Gal(K(α, β)/K) is isomorphic to a subgroup of a cyclic group andso cyclic itself. This is a contradiction.

Consider now a polynomial of the form f(x) = x2 + bx + 1 with b ∈ K.Suppose α and β are the roots of f(x). Then we have f(x) = (x+α)(x+β) =x2 + (α + β)x + αβ (recall we are in characteristic 2). Hence α and β areinverses and add to b. Thus given an element α ∈ L and its inverse suchthat α + α−1 ∈ K, we get such a polynomial and vice-versa. Furthermore,note that α = α−1 implies α is a root of x2 − 1 = (x − 1)2 so that α =1. Excluding one and zero, we get (29 − 2)/2 = 28 − 1 polynomials fromelements in K (each polynomial corresponds to two distinct elements in thiscase). Adding the polynomial corresponding to 1 (which is x2 − 1), we get28 polynomials which come from elements in K. The remaining 29−28 mustthen be irreducible over K. By the above, these must have roots in L so thatwe have 2(29 − 28) = 210 − 29 elements of L\K which satisfy the propertymentioned.

Now suppose α ∈ L\K with α + α−1 ∈ K. Then from the above, α andα−1 have the same irreducible polynomial over K so that σ(α) = α−1 (σcannot preserve α as this would imply that α was in the fixed field of theGalois groups, i.e., K). Conversely, if σ(α) = α−1, we may conclude that αand α−1 have the same irreducible polynomial over K so that α+ α−1 ∈ K.Hence, we have 210 − 29 elements of L\K which satisfy σ(x) = x−1. If anelement a ∈ K satisfies this property, we have x = x−1. We have shownpreviously that this is true only of 1. Hence we have 210− 29 + 1 elements ofL× which satisfy the given property.

31

6. Let V and W be finite dimensional vector spaces over the fieldk. Let V ∗ be the dual vector space of all linear maps from V intok. Define a map

φ : V ∗ ⊗W → Hom(V,W )

byφ(v∗ ⊗ w)(v) = v∗(v)w.

(i) Prove that φ is well defined and a homomorphism.

Consider the map Φ : V ∗ × W → Hom(V,W ) given by Φ(v∗, w)(v) =v∗(v)w. First we check that this is well defined. Fix v∗ ∈ V ∗, w ∈ W ,v1, v2 ∈ V , and r ∈ k. Then

Φ(v∗, w)(v1 + v2) = v∗(v1 + v2)w

= v∗(v1)w + v∗(v2)w

= Φ(v∗, w)(v1) + Φ(v∗, w)(v2)

and

Φ(v∗, w)(rv1) = v∗(rv1)w

= rv∗(v1)w

= rΦ(v∗, w)(v1),

so that Φ is well defined.Next we will show that Φ is k-bilinear. For this, let v ∈ V , v∗1, v

∗2, v

∗ ∈ V ∗,w1, w2, w ∈ W , and r ∈ k. Then we have

Φ(v∗1 + v∗2, w)(v) = (v∗1 + v∗2)(v)w

= v∗1(v)w + v∗2(v)w

= Φ(v∗1, w)(v) + Φ(v∗2, w)(v),

which shows that Φ(v∗1, w) + Φ(v∗2, w) = Φ(v∗1 + v∗2, w). Likewise,

Φ(v∗, w1 + w2)(v) = v∗(v)(w1 + w2)

= v∗(v)w1 + v∗(v)w2

= Φ(v∗, w1)(v) + Φ(v∗, w2)(v)

32 Spring 2006

so that Φ(v∗, w1) + Φ(v∗, w2) = Φ(v∗, w1 + w2). Finally we have

Φ(rv∗, w)(v) = (rv∗)(v)w

= r[v∗(v)w]

= rΦ(v∗, w)(v)

= rv∗(v)w

= v∗(v)(rw)

= Φ(v∗, rw)(v),

which shows that Φ(rv∗, w) = rΦ(v∗, w) = Φ(v∗, rw). Since Φ is k-bilinear,we can apply the universal property of tensor products to conclude that thereis a k-linear map φ : V ∗ ⊗ V → Hom(V,W ) which satisfies φ(v∗ ⊗ (w))(v) =v∗(v)w. This shows that φ is a well-defined homomorphism.

(ii) Prove that φ is injective.

Choose a basis a1, . . . , an of V and a basis b1, . . . , bm of W . Then a basisfor V ∗ is v∗1, . . . , v

∗n, where v∗i (aj) = δij. This implies that a basis of V ∗⊗W is

v∗i ⊗bj. Suppose then that φ maps∑

i

∑j cijv

∗i ⊗bj to zero (cij ∈ k). Then

for each k, we have∑

i

∑j cijv

∗i (ek)bj = 0. That is,

∑j ckjbj = 0. Since bj

is linearly independent, we may conclude that ckj = 0. Thus we have shownthat the kernel of φ is zero as desired.

(iii) Prove that φ is surjective.

Fix the basis as above and let f ∈ Hom(V,W ). Write f(ai) =∑

j cijbj.Consider

∑i

∑j cijv

∗i ⊗ bj ∈ V ∗ ⊗W . We have[∑

i

∑j

cijv∗i ⊗ bj

](ek) =

∑i

∑j

cijv∗i (ek)bj

=∑j

ckjbj

= f(ek).

Hence the image of∑

i

∑j cijv

∗i ⊗ bj and f agree on a basis of V which

implies that they are equal. This shows that f is in the image of φ and so φis surjective.

Chapter 4

Fall 2005

1. Let G be a finite group, let Z ⊂ G be the center of G and supposethat K ⊂ G is a normal subgroup.

(i) Let p be a prime number and assume that p divides |Z| but pdoes not divide |Z ∩K|; (here |Z| is the cardinality of Z and |Z ∩K|is the cardinality of Z ∩K). Show that p divides the index [G : K].

Since Z < NG(K), we have, by Second Isomorphism Theorem, thatZK/K ∼= Z/Z ∩ K. In particular, |ZK/K| = |Z|/|Z ∩ K|. Since thefraction on the right is an integer, p divides the top, and p does not dividethe bottom, it is divisible by p. Thus p divides the order of ZK/K. Now,we have K < ZK < G so that [G : K] = [G : ZK][ZK : K] is divisible by p.

(ii) Let G′ ⊂ G be the commutator subgroup of G, p a prime num-ber, and assume the p divides |Z| by p does not divide |Z ∩ G′|.Show that G has a subgroup of index p.

First suppose A is an abelian group and n divides |A|. We will show Ahas a subgroup of order n. By the structure theorem for abelian groups, wehave A ∼= A1 × · · · × Ar where |Ai| = pαi

i and |A| = pα11 × · · · × pαr

r is thefactorization of |A| into primes. Since n divides |A| its factorization is givenby n = pβ1

1 · · · pβrr where βi ≤ αi (and possibly zero). Now, since Ai is a

p-group, it contains a subgroup, Hi, of order pβi

i . Then H1 × · · · × Hn is asubgroup of A of order n.

By (i), p divides |G/G′|. Writing |G/G′| = np, we may apply the

34 Fall 2005

above to conclude that G/G′ has a subgroup K of order n (since G/G′ isabelian). Now by the Correspondence Theorem we have a subgroup of G,H, which contains G′ such that K = H/G′. We have G′ < H < G so that[G : G′] = [G : H][H : G′], from which we may conclude that [G : H] = p.

(iii) Prove that

Sp(K) = P ∩K : P ∈ Sp(G),

where Sp(K) is the set of Sylow p-subgroups of K and Sp(G) is theset of Sylow p-subgroups of G.

Let PK be a Sylow p-subgroup of K. Since PK is then a p-subgroup of G,we have a Sylow p-subgroup ofG, PG such that PK ⊂ PG. Then PK ⊂ K∩PG.But then the latter is a p-group contained in K which contains a maximalp-group (in K) so we must have equality, that is, PK = PG ∩K. This givesone containment.

Now let QG be a Sylow p-subgroup of G. We have already seen that thereexists a Sylow p-subgroup, PG, of G such that PK = PG ∩K, where PK is aSylow p-subgroup of K. By the Sylow Theorems, there is a g ∈ G such thatQG = gPGg

−1. Consider gPKg−1, a subgroup of QG. By normality, it is con-

tained in K. Since conjugation preserves the order of a subgroup, it is also aSylow p-subgroup of K. Lastly, suppose that h ∈ QG ∩K then g−1hg ∈ PGand g−1hg ∈ K (by normality). This gives h ∈ gPG ∩ Kg−1 = gPKg

−1.Hence we have QG ∩K = gPKg

−1, a Sylow p-subgroup of K. This gives theother containment.

(iv) Prove that |Sp(K)| divides |Sp(G)|.

By the Sylow Theorems, G acts transitively on the set of Sylow p-subgroupsof G. By normality, and the fact that conjugation preservers order, it actson the set of Sylow p-subgroups of K. Since the action of K is a re-striction of this action and K acts transitively, G acts transitively. Hence|Sp(G)| = |G|/|NG(PG)| and |Sp(K)| = |G|/|NG(PK)| where PG is a Sylowp-subgroup of G and PK is that of K. Since neither equation depends on thechoice of subgroup, we may assume (by the above) that PK = K ∩ PG.Suppose than that g ∈ NG(PG). Then gPGg

−1 = PG (by assumption)and gKg−1 = K (by normality). Hence g(PG ∩ K)g−1 = PG ∩ K so that

35

g ∈ NG(PK). Hence NG(PG) < NG(PK) so that |NG(PG)| divides |NG(PK)|.This gives the statement.

2. Let R be a principal ideal domain so that Rn is a free R-moduleof rank n.

(i) Prove that if M ⊂ Rn is an R-submodule, then M is free of rankat most n.

For each 1 ≤ k ≤ m, let Mk = M ∩Rk (where Rk is the submodule of Rn

generated by the first k basis elements). We will show by induction that Mk

is free of rank less than or equal to k. First consider M1 is a submodule ofR. Hence it is an ideal in R and so equal to (a1) for some a1 ∈ R. If a1 = 0,we have M1 = 0. Otherwise, M1

∼= R via the map ra1 7→ r.

Now assume that Mk is free of rank less than or equal to k. Let I ⊂ Rbe the set of elements a ∈ R for which there is some x ∈ Mk+1 withx = (b1, b2, . . . , bk, a, 0, . . . , 0). Then I is clearly and ideal and so can bewritten as I = (ak+1) for some ak+1 ∈ R. If ak+1 = 0, we have Mk = Mk+1

which completes the induction step. Otherwise, let w ∈Mk+1 be an elementwhose r + 1st coordinate is ak+1. If y ∈ Mk+1, the k + 1st coordinate ofy is divisible by ak+1 so we may find c ∈ R such that y − cw ∈ Mr. Thisshows that Mk+1 = Mk + Rw. Lastly suppose that y ∈ Mk+1 ∩ Rw. Thenwe may find r ∈ R such that y = rw so that the k + 1st coordinate of y israr+1. But y ∈ Mk so that the k + 1st coordinate is zero. Since ar+1 6= 0,we have r = 0 and so y = 0. Hence we have Mk+1 = Mk ⊕ Rw ∼= Mk ⊕ R.This shows that Mk+1 is free of rank less than or equal to k + 1 and givethe claim by induction. Applying this to the case k = n gives the statement.1

(ii) Let M be a finitely generated R-module. Show that there existnonnegative integers m ≤ n and suitable maps so that

0 → Rm → Rn →M → 0

1This proof is in Lang, p.146

36 Fall 2005

is an exact sequence of R-modules.

Find a set a1, . . . , an which generates M over R. Let Rn have basisb1 . . . , bn and consider the map φ : Rn → M given by bi 7→ ai (whichexists and is well-defined by the universal property of free modules). Thensince every x ∈ M can be written as r1a1 + · · · + rnan for some elementsr1, . . . , rn of R, we have that φ is onto. Let K ⊂ M be the kernel of φ. By(i), we have that K is isomorphic to Rm for some m ≤ n. Hence we have anisomorphism Rm → K. Composing this with the inclusion map K → Rn, weget an injective map ψ : Rm → Rn whose image is K, the kernel of φ. Hencewe these maps the sequence given is exact.

(iii) Let M be a finitely generated R-module. Prove that everyR-submodule of M is again finitely generated.

Let φ : Rn → M be the map above and let N ⊂ M be a submodule.Then φ−1(N) is a submodule of Rn and so is isomorphic to Rk for some k. Inparticular, it is finitely generated by some β1, . . . , βk ∈ Rn (k ≤ n). Considerthe submodule, L, of M generated by φ(β1), . . . , φ(βk). By definition, allthese elements are contained in N and so that L ⊂ N . Let x ∈ N . Then bysurjectivity, we may find a a ∈ Rn such that φ(x) = a. Then x ∈ φ−1(N) sothat we may find some r1, . . . , rk ∈ R such that x = r1β1 + · · ·+ rkβk. Thisgives n = φ(x) = r1φ(β1) + . . . + rkφ(βk) ∈ L. Hence we have shown the Nis finitely generated.

3. Let V be a finite dimensional vector space over a field k, andM : V → V an endomorphism such that M2 = M .

(i) Show that 1−M is also an idempotent, that is, (1−M)2 = 1−M ,where 1 : V → V is the identity map.

Recall that EndR(V ) is a ring, where addition is addition of functions andmultiplication is composition. We have (1−M)2 = 1−M−M+M2 = 1−Msince M2 −M is zero by assumption.

37

(ii) Write

R(M) = Mv : v ∈ V and N(M) = v ∈ V : Mv = 0

for the range an null space of M , respectively. Define R(1−M) andN(1−M) similarly. Prove that V = R(M)⊕ R(N).

Let v ∈ V and consider w = v−Mv. We have M(v−Mv) = Mv−M2v =Mv −Mv = 0. Hence w = v −Mv ∈ N(M). v = Mv + w then shows thatM = N(M) + R(M). To see that the sum is direct, suppose that v liesin the intersection. Then we have v = Mw for some w ∈ V . We have0 = Mv = M2w = Mw = v. Hence the sum is direct.

(iii) Show that an eigenvalue of M is either 0 or 1.

We have that the characteristic polynomial of M is the product of its in-variant factors of which the minimal polynomial is the largest. Since all theinvariant factors divide the minimal polynomial, the characteristic polyno-mial of M divides some power of the minimal polynomial of M . In particularthey have the same roots so that the eigenvalues (the roots of the character-istic polynomial) must be roots of the minimal polynomial. Since M satisfiesx2 − x, its minimal polynomial divides x(x− 1). Hence the minimal polyno-mial can only have one or zero as a root. This gives the statement.

(iv) Show that V has a basis such that each basis is an eigenvectorM .

Since the minimal polynomial of M has only the factors x and x− 1 (notnecessarily both), all the invariant factors (by divisibility) must be eitherx− 1, x, or x(x− 1). In particular the elementary divisors of M only consistof x − 1 or x. Hence the Jordan Canonical Form of M is is a diagonalmatrix, D, with ones and zeros on the diagonal. Hence we may find a basis,e1, . . . , en of V , with respect to which M has matrix D. If D has an a(either zero or one) in the ith column, the image of ei under M is aei. Henceevery basis vector is an eigenvector.

38 Fall 2005

4. Let k be a field and f(x) be a separable, irreducible monicpolynomial in k[x] of degree p, where p is a prime number. Inthis problem, we assume that K is a splitting field for f(x) over k,G = Aut(K/k) is the Galois group of all automorphisms of K thatfix k, P ⊂ G is a Sylow p-subgroup of G, N ⊂ G is the normalizer ofP in G, and N = P .

(i) Prove that |G| = ps where s is an integer not divisible by p.

An element σ ∈ G permutes the roots of f(x) (of which there are p).Furthermore, K is generated over k be these roots. Hence, if σ acts triv-ially on the set of roots it must be the identity. Thus we have an injectivehomomorphism G → Sp so that |G| divides p!. If α is a root of f(x), wehave [k(α) : k] = p. Since [K : k] = [K : k(α)][k(α) : k], this implies that pdivides [K : k], which by the Fundamental Theorem of Galois Theory equals|G|. Hence |G| = ps for some s. By the above, s divides (p − 1)!. If p wereto divide s, we would have p dividing (p − 1)!, which is not the case (since(p− 1)! is a product of numbers less than p). This gives the statement.

(ii) With |G| = ps, show that P has exactly s distinct conjugates inG.

By the Sylow Theorems, G acts transitively by conjugation on the set ofSylow p-subgroups. Hence the number of conjugates will be |G| divided bythe stabilizer of any element of this action. That is, |G|/|N | = |G|/|P | =ps/p = s (applying (i) to find the order of P ).

(iii) With |G| = ps, show that there are exactly s(p − 1) distinctelements in G of order p.

Every Sylow p-subgroup had order p and so contains p−1 elements of or-der p (excluding the identity and applying Lagrange’s Theorem). Since eachelement in such a group generates the group, distinct Sylow p-subgroups can-not overlap. By (ii), there are s subgroups which gives s(p− 1) elements oforder p.

(iv) Let α and β be distinct roots of f(x) in K. Write Hα ⊂ G forthe Galois group of K over k(α) and write Hβ ⊂ G for the Galois

39

group of K over k(β). Prove that |Hα| = s.

By the Fundamental Theorem of Galois Theory, p = [k(α) : k] = [G : Hα].Hence |Hα| = s.

(v) Show that

G = Hα ∪ σ ∈ G : ordG(σ) = p

and show that this is a disjoint union.

No element of Hα can have order p by Lagrange’s Theorem. This provesthat the union is disjoint. By (iii), the order of the set on the right is ps− s.By (iv), the order of the set on the left is s. Hence the order of their unionis ps which shows that it is the whole group.

(vi) Prove that Hα = Hβ.

By (v), Hα = G\σ ∈ G : ordG(σ) = p = Hβ.

(vii) Prove that K = k(α) and that P = G.

By the Fundamental Theorem of Galois Theory, the subgroups corre-sponding to two fields are equal only if the two fields are equal. HenceF (α) = F (β) by (vi). In particular, β ∈ F (α). This applies to any root off(x) so that F (α) contains all the roots. Hence, K = F (α). But then G hasorder p so we must have G = P .

40 Fall 2005

Chapter 5

Spring 2005

1. Let L/F be a finite extension of fields. Let f(x) ∈ F [x] be apolynomial.

a) Show that F [x]/f(x)F [x]⊗F L ∼= L[x]/f(x)L[x].

Consider the map Φ : F [x]/(f(x)) × L → L[x]/(f(x)) given by (g(x) +(f(x)), α) 7→ αg(x) + (f(x)). For g(x), h(x) ∈ F [x], α ∈ L, we have that

Φ(g(x) + h(x)f(x) + (f(x)), α) = αg(x) + αh(x)f(x) + (f(x))

= αg(x) + (f(x))

= Φ(g(x) + (f(x)), α),

so that Φ is well-defined. Now suppose additionally that β ∈ L, and a, b ∈ F .Then

Φ(g(x) + (f(x)), aα+ bβ) = (aα+ bβ)g(x) + (f(x))

= a[αg(x) + (f(x))] + b[βg(x) + (f(x))]

= aΦ(g(x) + (f(x)), α) + bΦ(g(x) + (f(x)), β),

so that Φ is F -linear in the second coordinate. It is similarly F -linear inthe first coordinate and so F -bilinear. Hence we may apply the universalproperty of the tensor product to conclude that there exists a F -modulehomomorphism φ : F [x]/(f(x)) ⊗ L → L/(f(x)) which satisfies [g(x) +(f(x))] ⊗ α 7→ αg(x) + f(x). In fact, it is clear this map is even L-linear(F [x]/(f(x)) ⊗ L has an L-module structure by multiplying on the secondcoordinate).

42 Spring 2005

Likewise, consider the map Ψ : L[x] → F [x]/(f(x))⊗ L given by

αnxn + . . .+ α0 7→ [xn + (f(x))]⊗ αn + . . .+ [1 + (f(x))]⊗ α0).

It is routine to check that Ψ is L-linear. We can also see that Ψ preservesmultiplication: let h(x) = βmx

m + . . .+ β + 0, g(x) = αxn + . . .+ α0 ∈ L[x].Then we have

Ψ(g(x)h(x)

)= Ψ

(m+n∑k=0

k∑i=0

αiβk−ixk

)

=m+n∑k=0

k∑i=0

Ψ(αiβk−ixk)

=m+n∑k=0

k∑i=0

[xk + (f(x))]⊗ (αiβk−1)

=m+n∑k=0

k∑i=0

([xi + (f(x))]⊗ αi

)([xj + (f(x))]⊗ βk−i

)=

(n∑i=0

[xi + (f(x))]⊗ αi

)(m∑j=0

[xj + (f(x))]⊗ βi

)= Φ

(g(x)

)Ψ(g(x)

).

Note also that f(x) is mapped to zero. Hence we get an induced F -algebrahomomorphism ψ : L[x]/(f(x)) → F [x]/(f(x)) ⊗ L which satisfies α 7→[1 + (f(x))] ⊗ α and x 7→ [x + f(x)] ⊗ 1. Now L[x]/(f(x)) is generated asan F -algebra by x and α ∈ L and ψ and φ act as inverses on this set. Like-wise, F [x]/(f(x))⊗ L is generated as an F -algebra by [x+ (f(x))]⊗ 1) and[1 + (f(x)) ⊗ α] for α ∈ L and ψ and φ act as inverses on this set. Henceψ and φ are inverses and F [x]/)f(x))⊗L and L[x]/(f(x)) are isomorphic asL-algebras.

b) Suppose L,L′ are both finite extensions of F . If L′/F is separa-ble, use a) to show that L⊗ L′ is a direct sum of fields.

Since L′/F is separable, we can consider the Galois closure K of L′ withrespect to F . The extension K/F will then be a finite Galois extension.

43

Since the Galois group (as a finite groups) has finitely many subgroups, wemay apply the Fundamental Theorem of Galois Theory to conclude thatthere are finitely many intermediate fields F ⊂ F ′ ⊂ L. This implies thatthere are finitely many intermediate field F ⊂ F ′ ⊂ L′. Hence L′/F is asimple extensions and we may find α ∈ L′ such that L′ = F (α). Lettingf(x) ∈ F [x] be the minimal polynomial of α, we have L′ ∼= F [x]/(f(x)).This gives L⊗ L′ ∼= L′ ⊗ L ∼= F [x]/(f(x))⊗ L ∼= L[x]/(f(x)).

Now over L, we can factor f(x) into irreducibles:

f(x) = f1(x)f2(x) · · · fr(x).

Since L′/F is separable, f(x) is separable (as the minimal polynomial ofα ∈ L′). Hence the fi’s are distinct. This is implies that for i 6= j, wehave (fi(x)) and (fj(x)) are comaximal ideals in L[x]. Moreover, we have(f1(x)) · · · (fr(x)) = (f(x)). Hence we may apply the Chinese RemainderTheorem to conclude that L[x]/(f(x)) ∼= L[x]/(f1(x)) ⊕ · · · ⊕ L[x]/(fr(x))as L-algebras (the statement of the Chinese Remainder Theorem is only forrings, but the map is actually an L-algebra isomorphism in this case). Butsince fi(x) is irreducible, (fi(x)) is a maximal ideal in L[x] so that L[x]/(fi(x))is a field. Hence we have shown that L⊗ L′ is a direct sum of fields.

c) If L/F is Galois, show that L⊗F L is a direct sum of copies of L.

Applying the same argument as above, we have L = F (α) for some α ∈ Land if f(x) is the minimal polynomial of α over F we can conclude thatL ⊗ L ∼= L[x]/(f1(x)) ⊕ L[x]/(f2(x)) (as L-algebras). But L/F is a normalextension and f(x), an irreducible over F has a root in L. Hence f(x) splitsinto linear factors over L so that fi(x) = x − αi for some αi ∈ L. Now,L[x]/(x−αi) ∼= L as L-algebras. Hence L⊗L is a direct sum of copies of L.

d) If F has nonzero characteristic p, and L = F (a1/p), use nilpotenceto show that L⊗F L is not a direct sum of fields.

First we need to add the assumption that a does not have a pth rootin F , otherwise we have L = F (a1/p) = F so that L/F is trivially separa-ble. We may then apply b) to conclude that L⊗ L is a direct sum of fields.Hence we must assume that xp− a has no root over F . Over L, however, wehave (x − a1/p)p = xp − a. Hence the minimal polynomial of a1/p over F is

44 Spring 2005

(x − a1/p)k for some k > 1 (since a1/p 6∈ F ). Hence L ∼= F [x]/((x − a1/p)k)which by a) gives L ⊗ L ∼= L[x]/((x − a1/p)k). Now in L[x]/((x − a1/p)k),x − a1/p is nilpotent. Hence L ⊗ L has nonzero nilpotent elements. Since adirect sum of fields has no nilpotent elements, we may conclude that L⊗ Lis not a direct sum of fields.

e) Suppose that L/F is finite and L ⊗F L is a direct sum of copiesof L. Show that L/F is Galois.

Before we prove the statement, we will need a small lemma. Suppose Kis a field and Si and Tj are two collections of K-algebras that are alsofields. Suppose further that

S1 ⊕ · · ·Sn ∼= T1 ⊕ · · ·Tm,

as K-algebras. Then m = n and after possible reordering, Si ∼= Ti as K-algebras. We proceed by induction on n. Sn is an ideal in S1 ⊕ Sn which isalso a field. Hence it must correspond under the isomorphism to an ideal inT1 ⊕ · · ·Tm which is a field. But the only ideals in a direct sum are a directsum of ideals in the individual factors. In this case, the individual factors arefields and so have no non trivial ideal. Hence Sn corresponds to a direct sumof the Tj. But if the direct sum has more then one nonzero factor, it is nota field (i.e., something like (1, 0, 0, . . . , 0)) is nonzero and has no inverse inT1⊕T2⊕0⊕· · ·⊕0). Hence Sn corresponds to some Tj. After reordering,we may assume that it is Tm. But then

S1 ⊕ · · · ⊕ Sn−1∼=(S1 ⊕ · · · ⊕ Sn

)/Sn

∼=(T1 ⊕ · · · ⊕ Tm

)/Tm

∼= T1 ⊕ · · · ⊕ Tm−1,

(where the first isomorphism is induced by the map S1⊕Sn → S1⊕Sn givenby (s1, . . . , sn) 7→ (s1, . . . , sn−1, 0) and similarly for the last isomorphism).We may then apply the induction hypothesis to conclude the statement.

Suppose that L/F is inseparable. In particular, F must have nonzerocharacteristic p. Then we may find α ∈ L such that the minimal polynomialof α over F , f(x), is inseparable. Hence f(x) = g(xp) for some g(x) ∈ F [x].Now F (α) ∼= F [x]/(f(x)). Over F (α), we have g(αp) = f(α) = 0 so that αp

45

is a root of g(x). Hence g(x) = (x − αp)h(x) for some h(x) ∈ L[x]. Hencef(x) = g(xp) = (xp−αp)h(xp) = (x−α)ph(xp). This shows that (x−α)h(xp)is nilpotent in L[x]/(f(x)) so that F (α)⊗F L has nonzero nilpotent elements.But F (α) ⊂ L and L is a flat F -module (as it is free). Hence F (α)⊗L→ L⊗Lis injective and we may view F (α)⊗L as a submodule of L⊗L. This impliesthat L⊗L has nonzero nilpotent elements and so is not a direct sum of fields.

Hence we may assume that L/F is separable. Thus since it is finite, wemay apply the Galois closure argument above to conclude that L = F (α)for some α ∈ L. If f(x) ∈ F [x] is the minimal polynomial of α over F andf(x) = f1(x) · · · fr(x) is the factorization of f(x) into irreducibles over L, wehave

⊕L ∼= L⊗ L ∼= L[x]/(f(x)) ∼= L[x]/(f1(x))⊕ · · · ⊕ L[x]/(fr(x))

(as L-algebras). By the above, we may then conclude that L[x]/(fi(x)) ∼= Lfor each i as L-algebras. In particular, they are isomorphic as L-modules sothat the dimension of L[x]/(fi(x)) over L is one. But if the degree of fi(x) isn, a basis for L[x]/(fi(x)) is 1, x, . . . , xn−1 so that its dimension is n. Hencewe may conclude that the degree of each fi(x) is one and so f(x) has all itsroots over L. Certainly L is the smallest field extension of K which containsall the root of f(x), as it is generated over K by one of the roots. Hence L isthe splitting field of f(x) which shows that is a normal and therefor Galoisextension of F .

2. Let F be an algebraically closed field and Mn(F ) the ring ofn× n matrices over F . We say A ∈Mn(F ) is diagonalizable if thereis an invertible P ∈ Mn(F ) such that PAP−1 is a diagonal matrix.That is, all the off diagonal entries of PAP−1 are zero. p(x) ∈ F [x]is called the minimum polynomial of A if p(A) = 0 and p(x) is themonic polynomial of least degree with this property.

a) Show A is a diagonalizable matrix if and only if the minimumpolynomial of A has distinct roots.

46 Spring 2005

Since all the invariant factors of A divide the minimal polynomial, we mayinclude that all the invariant factors of A have distinct roots. Hence everyelementary divisor of A is linear. This implies that each the each Jordanblock of the Jordan Canonical Form of A is 1 × 1, that is, it is diagonal.Hence A is diagonalizable.

Conversely, suppose that A is diagonalizable and find P ∈ GLn(F ) suchthat B = PAP−1 is diagonal. Let λ1, λ2, . . . , λk ∈ F be the distinct diagonalentries of B. Then B − λiI is a diagonal matrix with a 0 in each place inwhich there was λi in B. Hence (B−λ1I) · · · (B−λkI) = 0 as each diagonalplace is zero in one of the factor matrices. Hence the minimal polynomial ofB divides (x − λ1) · · · (x − λk) (in fact this is the minimal polynomial) andso has distinct roots. Since the minimal polynomial of a matrix is preservedunder conjugation by an invertible matrix, we may conclude that that mini-mal polynomial of A has distinct roots.

b) Now suppose B ⊂Mn(F ) is a subalgebra which is simple as ring.Show that there is a d dividing n and invertible P ∈ Mn(F ) suchthat PBP−1 consists of all block diagonal matrices

A 0 0 · · · 00 A 0 · · · 00 0 A · · · 0...

......

. . ....

0 0 0 · · · A

for all d× d matrices A.

Since B is a subalgebra, it is in particular a k vector subspace of Mn(F )and so is a finite-dimensional vector space over k. If I1 ⊃ I2 . . . is a decreasingsequence of ideals in B, it is, in particular, a decreasing sequence of k vectorssubspaces of B which must stabilize by dimension considerations. HenceB is Artinian and simple. This implies, by Wedderburn’s Theorem, thatB ∼= Md(∆) for some division ring ∆ and d. But then ∆ is a division algebraover k and we may conclude by an argument given in Fall 1995 # 5 e that∆ = F .

Now let C be the centralizer of B in Mn(F ). Then from the Double Cen-tralizer Theorem, we may conclude that dimkMn(F ) = (dimk B)(dimk C).Hence d2 divides n2 and so d divides n. Now if D is the set above, we clearly

47

have D ∼= Mn(∆). This gives an isomorphism B → D. Hence we may applySkolem-Noether to find a P ∈ GLn(F ) such that D = PBP−1, as claimed.

c) Next suppose B ⊂ Mn(F ) has Jacobson radical 0. State andprove a version of b).

I never figured this one out before I took the prelim. Let me know if youfigure it out.

3. Let G be a finite group and F a field. Let F (G) be the field offractions of the polynomial ring F [xg|g ∈ G]. It is clear (and youdo not have to prove) that there is an action of G on F (G) withg(xg′) = xgg′. Let F (G)G be the fixed field. Let p be any prime. IfL/K is a field extension, we say K ′ is an intermediate field of degreep if K ⊂ K ′ ⊂ L and K ′/K has degree p.

a) Show that F (G)/F (G)G is Galois with group G.

Consider the polynomial

f(x) =∏g∈G

(x− xg).

For a ∈ G, we have

a · f(x) =∏

(x− xag) =∏

(x− xg).

Hence we have f(x) ∈ F (G)G. But then F (G) is generated over F (G)G ⊃ Fby the roots of f(x), a separable polynomial. Hence F (G) is Galois overF (G)G. Let K be the Galois group of F (G) over F (G)G. We may identifyeach element of G with the automorphism of F (G) that its action induces.Since G fixes F (G)G by assumption, we have that G is a subgroup of K. Butthen the fixed field of G is F (G)G so that we may apply the FundamentalTheorem of Galois Theory to conclude that G = K.

48 Spring 2005

b) Suppose that L ⊃ K ′ ⊃ K are field extensions such that L/K isGalois with group G and G is a p-group. If K ′/K has degree greaterthan p, show that K ′/K has an intermediate field of degree p.

Suppose |G| = pk and let H be the subgroup of G corresponding to K ′.Then [G : H] = [K ′ : K] = pl for some 2 ≤ l ≥ k. This implies that|H| = pk−l. Now, H is contained in a maximal subgroup M . Because G is ap-group, this subgroup has index p in G (and so doesn’t equal H). Let KM

be the fixed field of M . Then K ⊂ KM ⊂ K ′ and [KM : k] = [G : M ] = p.

c) Assume L/K is separable of degree p2, and K ′ is an intermediatefield of degree p. Write K ′ = K(θ) and L = K(α). Let g(x), f(x) ∈ K[x]be the monic minimal polynomials of θ and α respectively over K.Let K ′′ be the field generated by all the roots of g(x). Let G be theGalois group of K ′′/K and H that of K ′′/K ′. Choose gi such thatG is the disjoint union of the cosets giH. Show that f(x) has anirreducible factor, h(x), of degree p in K ′[x]. Using then gi, showthat f(x) is the product of p factors of degree p over K ′′[x].

We have [L : K] = [L : K ′][K ′ : K] so that [L : K ′] = p. Also wehave L = K(α) = K ′(α) since K ⊂ K ′ ⊂ L. Hence we may conclude thatthe (irreducible) minimal polynomial, h(x), of α over K ′ has degree p. Butf(α) = 0 so that h(x) must be a factor of f(x) in K ′[x]. This gives the firstclaim.

Now from the Galois correspondence, we have [G : H] = [K ′ : K] = p.Thus g1, . . . , gp is a collection of distinct coset representatives for H inG. Since f(x) ∈ K[x], gi preserves f(x). Thus gi maps h(x) to a differentfactor of f(x) in K ′′[x]. Suppose now that gih(x) = gjh(x). Then we havethat gig

−1j preserves h(x). Since f(x) is irreducible over K, h(x) have at

least one coefficient which does not like in K. Hence gig−1j fixes an element

γ ∈ K ′−K. Let F be the fixed field of gig−1j and consider F ∩K ′. We have

K ⊂ F ∩K ′ ⊂ K ′ so that p = [K ′ : K] = [K ′ ⊂ F ∩K ′][K ′ ∩ F : K]. But[K ′ ∩ F : K] 6= 1 and so we may conclude that [K ′ ⊂ F ∩K ′] = 1. That is,K ′ ⊂ F . This gives gig

−1j ∈ H and so we must have i = j. Hence gih(x) is

a collection of p distinct factors of f(x) of degree p. Since the degree of f(x)is p2 the claim follows.

49

d) Suppose L′ is the extension generated by all the roots of f(x).Show that L′/K has degree no more than (p!)p+1.

Note that since g(x) has degree p over K, L has degree at most p! overK. Now L′ is a splitting field over K and so any irreducible polynomial overK which has a root in L′ has all its roots in L. Hence θ ∈ K(θ) ⊂ K(α) ⊂ L′

implies that g(x) splits over L′. That is, K ′′ ⊂ L′. We have seen that f(x)splits into p irreducible factors of degree p over K ′′. Let fi(x) be one of thesefactors and K ′′ ⊂ Ki ⊂ L′ be the field generated by its roots. Since fi(x)has degree p, we have [Ki : K ′′] ≤ p!. We have K1 · · ·Kp = L′ so that

[L′ : K ′′] ≤ [K1 : K ′′] · · · [Kp : K ′′] = (p!)p.

This gives [L′ : K] = [L′ : K ′′][K ′′ : K] ≤ (p!)p+1, as claimed.

e) Show that there is a field extension L/K of degree p2 with nointermediate field of degree p.

Consider a group G of order 12 and suppose that it has a subgroup Nof order 6. Then the index of N in G is 2, which is the smallest prime thatdivides the order of G By Spring 2003(i), this implies that N is normal inG. Now, from the Sylow Theorems, N has a unique subgroup H of order3. This implies that H is normal in G (conjugating H by an element of Ggives a subgroup of N of order 3, which then must be H). From the SylowTheorems, this implies that H is the only subgroup of order 3. Hence if agroup of order 12 has a subgroup of order 6, it has a unique subgroup oforder 3. Consider then, the group A4. This group does not have a uniquesubgroup of order 3 as 〈(123)〉 and 〈(124)〉 are two distinct subgroups of order3. Hence A4 has no subgroup of index 2.

By (a), we may find a Galois extension L/F with Galois group A4 Let Hbe a subgroup of order 3. By the Galois Correspondence, the fixed field ofH, K, is of degree 4 = 22 over K. However, if L/K had a field of degree 2,we would have a subgroup of A4 of index 2, which we have shown not to bethe case.

50 Spring 2005

Chapter 6

Fall 2004

1. Let F = Z/2Z be the field of two elements. In this problemabelian groups in which every element has order dividing 2 will beregarded as vector spaces over F . If a is a nonzero rational num-ber, then we will write a′ for its image in the group Q∗/(Q∗)2, whereQ is the rational field viewed as a subfield of the complex field C.Let a1, . . . , an ∈ Q∗ and let K = Q(

√a1, . . . ,

√an) ⊂ C. Let A be the

subgroup of Q∗/(Q∗)2 generated by the a′i.

a. Show that K/Q is Galois with abelian Galois group G, and thatevery element of G has order dividing 2. (Hint: adjoin the rootsone by one.)

Consider the polynomial f(x) = (x2− a1) · · · (x2− an) ∈ Q[x]. The rootsof this polynomial are ±√a1, . . . ,±

√an ∈ C. Hence this polynomial splits

in K and K is generated by the roots. Thus K is a normal extension of Q.Since Q is a perfect field (as a field of characteristic zero), the extension K/Qis separable. Thus K is a Galois extension of Q.

Let σ ∈ G. Now the minimal polynomial for√ai over Q is either x2− ai,

if√ai 6∈ Q, or x − √

ai if√ai ∈ Q. In either case σ must map

√ai to

±√ai since these are the only roots of either polynomial. If σ preserves√ai, then σ2 trivially preserves

√ai. Likewise, if σ(

√ai) = −√ai, we have

σ2(√ai) = σ(−√ai) =

√ai. Hence σ2 preserves

√ai. Since σ2 then preserves

Q and a generating set for K over Q, we may conclude that σ2 is the identityon K and so σ has order divisible by 2. Every element in G has either order2 or is the identity. In either case, each element is its own inverse. Hence it

52 Fall 2004

σ, τ ∈ G, we have στ = (στ)−1 = τ−1σ−1 = τσ. Thus G is abelian.b. Consider the pairing (, ) : A×G→ F defined be the condition

(a′, σ) =

0 if σ fixes

√a

1 if σ negates√a

for all a ∈ Q∗ with a′ ∈ A. Show that this pairing is well definedand bilinear.

Suppose that a, b ∈ Q with a′ = b′ in A. Then there is some c ∈ Q∗ withb = ac2. Then

√b = c

√a. Let σ ∈ G and suppose σ fixes

√a. Then σ(

√b) =

σ(c√a) = c

√a =

√b (since σ fixes c ∈ Q) so that σ fixes

√b. Conversely,

suppose that σ negates√a. Then σ(

√b) = σ(c

√a) = −c

√a = −

√b. Hence

the action of σ on√a is the same as that on

√b. This shows that the pairing

is well-defined.Now suppose that a′, b′ ∈ A and σ, τ ∈ G. Consider (a′b′, σ). If σ negates

both√a and

√b, it will fix

√a′b′ so that (a′b′, σ) = 0 = 1+1 = (a′, σ)+(b′, σ).

Likewise, if σ preserves only one of the individual roots, it will negate√a′b′

so that that (a′b′, σ) = 1 which will then be the sum of (a′, σ) and (b′, σ).Lastly, if σ preserves both individual roots, it will preserve

√a′b′ so that

(a′b′, σ) = (a′, σ) + (b′, σ) will be satisfied by all terms being zero. Hence thepairing preserves the group operation in the first coordinate. Similar reason-ing shows that (a′, σ + τ) = (a′, σ) + (a′, τ). Since both coordinates preservethe group operation, we automatically have that the pairing is F -bilinear asthe action of F is addition (modulo 2).

c. Prove that the pairing (, ) is perfect; i.e. show that if (a′, σ) = 0for all σ ∈ G then a′ is the identity of A and that if (a′, σ) = 0 for alla′ ∈ A then σ is the identity in G. Conclude that A and G have thesame dimensions as vector spaces over F .

Suppose that a′ ∈ A satisfies (a′, σ) = 0 for all σ ∈ G. Then σ(√a) =

√a

for all σ ∈ G. Hence√a is in the fixed field of G. By the Fundamental

Theorem of Galois Theory, this implies that√a ∈ Q. Hence a ∈ (Q∗)2 (a

is a square in Q and it cannot be a square of zero as this would imply thatit were zero). Hence a′ = 1 in A. Likewise, suppose that σ ∈ G satisfies(a′, σ = 0) for all a′ ∈ A. Then σ(

√a) =

√a for all a′ ∈ A. In particular, we

have σ(√ai) = 0 for each ai. Then σ acts as the identity on a generating set

of K over Q and so σ is the identity on K. This gives the first statement.

53

Now suppose that F is a field and V and W are finite dimensional vectorspaces over F. Suppose 〈, 〉 : U ×W → F is F-bilinear and perfect. We willshow that W and V have the same dimension. We get a map φ : W → V ∗

given by w 7→ 〈·, w〉. By bilinearity, this map is well defined and F-linear.Now suppose that w ∈ W is mapped to zero. Then (v, w) = 0 for all v ∈ V .This implies that w = 0 so that φ is injective. But, by picking a basis for V ,we have an isomorphism V ∗ → V . Composing with φ, we get an injectivemap W → V . This shows that dimFW ≤ dimF V . By symmetry we alsohave dimF V ≤ dimFW . This gives the claim. Since both G and A are finitedimensional over F (the former since it is the Galois group of an extensiongenerated by finitely many algebraic elements and the latter since it is afinitely generated abelian group where every element has finite order), wecan conclude that A and G have the same dimension over F .

d. Show that for every r there is a Galois K/Q such that the Galoisgroup of K/Q is (Z/Z)r.

Let p1, . . . , pr be distinct primes. Let A = 〈p′1, . . . , p′r〉 ⊂ Q/(Q)2. Thenevery element of A can be represented by an element in Q of the formpα2

1 · · · pαrr for αi = 0, 1. Now suppose two such expression represent the

same equivalence class. Then we have pα11 · · · pαr

r = pβ1

1 · · · pβrr (a2/b2) (where

at least one αi does not equal the corresponding β1) for some a, b ∈ Z whichwe may assume to be relatively prime. Now, dividing out the primes whichoccur in both expressions, neglecting the primes which are raised to the powerzero, we have q1 · · · qm = (a2/b2)r1 · · · rn for two collections, q1, . . . , qm andr1, . . . , rn, of distinct primes. Since the two collections are be different wemay assume (after possible reordering and taking the reciprocal of a2/b2)that q1 is not equal to any of the ri. This gives q1 · · · qmb2 = r1 · · · rna2 andso we may conclude that q1 divides r1 · · · rna2. Whenever a prime dividesa product, it must divide one of the factors. Hence we may conclude thatq1 divides a. Writing a = q1k, we have a2 = q2

1k2. Substituting this into

the above and dividing by q1 gives q2 . . . qmb2 = r1 · · · rnk2q1. But this im-

plies that q1 divides b and we have assumed a and b to be relatively prime.Hence we have a contradiction and so no two expressions represent the sameequivalence class. Hence the order of A is 2r.

Now a finite dimensional vector space over F is a finite abelian groupwhere every element has order dividing two. We can apply the ClassificationTheorem for Abelian Groups to conclude that any finite dimensional vector

54 Fall 2004

space over F is isomorphic as an F -vector space to (Z/2Z)k for some k(two vector spaces over F which are isomorphic as groups are automaticallyisomorphic as F -vector spaces). By order considerations, the dimension ofthe group will then be k and conversely any F -vector space of dimensionk must be isomorphic to (Z/2Z)k. Hence, by the above and applying c.,the Galois group of Q(

√p1, . . . ,

√pr)/Q is an F -vector space of dimension r.

Hence it is (Z/2Z)r.

2. In this problem you may use without proof the fact that thealternating groups An are simple for n ≥ 5.

a. Use Sylow’s theorems to show that the symmetric group S5 canbe made to act transitively on a set of size six.

Let n5 be the number of Sylow 5-subgroups of G. From the Sylow Theo-rems, n5 divides |S5| = 120 = 23 · 3 · 5. Since n5 must also be equivalent to1 modulo 5, it cannot have a factor of 5. This gives n5 = 1, 2, 4, 8, 3, 6, 12, 24are possibilities. Of these, only 1 and 6 are equivalent to 1 modulo 5. If n5

were 1, we would have a normal Sylow 5-subgroup. But the only elements oforder 5 in S5 are 5-cycles. Hence the normal subgroup would be a subgroupof A5. Since A5 is simple, this is a contradiction. Therefore, n5 = 6. Butby the Sylow Theorems, S5 acts transitively (by conjugation) on the set ofSylow 5-subgroups. This gives the result.

b. The symmetric group S6 contains at least two conjugacy classesof subgroups that are isomorphic to S5.

First we will show that the only nontrivial normal subgroup of S5 is A5.Let N C S5. Then N ∩ A5 is a normal subgroup of S5 as an intersection ofnormal subgroups. But then it is also a normal subgroup of A5.

By the simplicity of A5, we that have N ∩A5 is either 1 or A5. Considerthe first case and suppose that we have σ, τ ∈ N\1. Then σ and τ areodd and so their product is even. But then their product lies in A5 ∩ N .Hence τ = σ−1. Thus every element of N\1 is the inverse of every other

55

element (including itself). Since inverses are unique, we may conclude thatN = 1, σ for some σ ∈ S5. But then σ is its own conjugacy class, thatis, the only element with its cycle decomposition. Furthermore, σ has ordertwo and so must be a 2-cycle or a product of two disjoint 2-cycles. Neitherone of these cycles types gives a a unique element (for example, we have (12)and (23) and (12)(34) and (13)(24)). Hence we must have N = 1.

Now suppose that N ∩A5 = A5. Then A5 ⊂ N . Hence by the correspon-dence theorem, N corresponds to a subgroup of S5/A5

∼= Z2. Since the onlysubgroups of Z2 are 1 and Z2. We must have that N is A5 of S5. Thisgives the claim.

Now from the above we have a group homomorphism S5 → S6, wherethe corresponding action of S5 on 1, 2, 3, 4, 5, 6 is transitive. Let N bethe kernel of this map. If N = S5, the corresponding action is trivial andso cannot be transitive. If N = A5, each element of S5 acts in one of twoways: trivially, according to 0 ∈ Z2

∼= S5/A5 or nontrivially, according to1 ∈ Z2

∼= S5/A5. Hence the image of 1 under the action of an element of S5

is either itself or some fixed number k. But the action is transitive and so wemust be able to send 1 to any of 1, 2, 3, 4, 5, 6 by the action of an elementof S5. Hence we have a contradiction and so may conclude that N = 1.Thus we have an injective map S5 → S6, where the corresponding action istransitive on 1, 2, 3, 4, 5, 6. Hence S5 is a subgroup of S6 in this way.

Now we also we have S5 < S6 simply by taking the elements in S6 whichfix 6. But if conjugate this subgroup by any element of S6, the elements ofthe resulting subgroup will always have a fixed point in common (the numberwhich is mapped to six by the conjugating cycle). Hence any conjugate ofthis subgroup cannot act transitively on 1, 2, 3, 4, 5, 6 as does the subgroupabove. This gives the result.

c. If X and Y are sets, then a bijection between them inducesan isomorphism of the symmetric group of X with the symmetricgroup of Y .

Let f : X → Y be a bijection. Define a map φ : SY → SX by φ(σ) =f−1 σ f . First we will show that φ is well-defined. Let σ ∈ SY . Thenφ(σ)f−1 σf = f−1 σf f−1 σf is the identity map on X. Likewise,f−1 σ f φ(σ) is the identity map on X. Hence φ(σ) has an inverse andso is a bijection on X.

56 Fall 2004

Next suppose that σ, τ ∈ SY . We have

φ(σ τ) = f−1 σ τ f−1

= f−1 σ f f−1τ f−1

= φ(σ) φ(τ),

so that φ is a group homomorphism. Suppose next that σ is in the kernelof φ. Then f−1 σ f is the identity map. Applying this to f−1(x) andcomposing with f (on the left). We obtain that σ(x) = x, that is σ is theidentity map. Hence φ is injective. Lastly suppose that τ ∈ SX . Define amap on Y by f τ f−1. Similar to the above, this is actually a bijection on Y .Moreover, we have φ(fτf−1) = τ so that φ is surjective. This gives the result.

d. There is a nontrivial outer automorphism of S6-that is, and au-tomorphism that is not the conjugation map of any element of S6.

Before we give the automorphism, we will first prove a presentation forSn. Consider the group

Gn =⟨x1, x2, . . . , xn−1|x2

i = xix3i+1 = 1 and xixj = xjxi if |i− j| ≥ 2

⟩.

Let Fn be the free group on x1, . . . , xn−1. Then by universal property offree groups, we have a homomorphism Gn → Sn given by xi 7→ (i, i + 1)(the transposition). Furthermore, (i, i + 1)2 = 1, [(i, i + 1)(i + 1, i + 2)] =(i, i + 1, i + 2)3 = 1 and (i, i + 1)(j, j + 1) = (j, j + 1)(i, i + 1). Hence thishomomorphism factors through the normal subgroup, R, of Fn generated bythe relations. That is, we have a map φn : Gn → Sn given by xi 7→ (i, i+ 1).Since the simple transpositions generate Sn, this map is surjective.

Now consider a word in Gn. We will show that using the relations given,this word can be put in the form xi . . . xn−1y where 1 ≤ i ≤ n (i = n isinterpreted to mean xi · · ·xn−1 = 1) and y is a word that does not containxn−1 (that is the equivalence class in Fn/R of each word in can be representedby a word in this form). Find the left-most occurrence of xn−1 in the word (ifthere is no occurrence we are done with this step). Using the commutativityof xn−1 with all of the xi aside from xn−2, we can move our xn−1 to theleft until we hit a xn−2 (or reach the end of the word in which case we aredone). We can then move our xn−2xn−1 block to the left until we hit anxn−2, in which case we can apply x2

n−1 = 1 and have again only xn−1 (and

57

so reduce back to the first step); until we hit an xn−3 and then we will havea xn−3xn−2xn−1 block; or until we reach the left-most edge of the word. Inthe case of a xn−3xn−2xn−1 block, we can move the block until we hit oneof three things, xn−3, xn−2, or xn−1. In the first case, we simply cancel thexn−3 and go back to an block of xn−2xn−1. In the second we use the relationxn−2xn−3xn−2 = xn−3xn−2xn−3 (which follows from the order three relation)to first convert to xn−3xn−2xn−3xn−1 and then to xn−3xn−2xn−1xn−3 whichmeans our xn−3xn−2xn−1 block has moved one to the left. In the third case, wenow have a block of xn−4xn−3xn−2xn−1. Hence by applying these procedures(and their generalizations), we will eventually reach to the left-most point ofthe word with a block of the form xi · · ·xn−1.

Now by applying these same steps to the next instance of xn−1 (if thereisn’t one, we are done), we can reach our first block of xi · · ·xn−1 with asecond block of the form xj · · ·xn−1. Then we have that

xi · · ·xn−1xj · · ·xn = xi · · ·xn−2xjxn−1xj+1 · · ·xn−1

= . . .

= xi · · ·xn−2xj · · ·xn−3xn−1xn−2xn−1

= xi · · ·xn−2xj · · ·xn−3xn−2xn−1xn−2,

so that we have reduced the number of occurrences of xn−1 in our word byone (in the case j = n−1, we simply cancel two occurrences of xn−1). Hencewe may repeat this entire procedure until we have only one occurrence ofxn−1 and then move it, along with a block of the form xi · · ·xn−1, to theleft-most end of the word. This gives the claim.

Now we will show by induction that φn is injective. The base case istrivial. Then suppose that φn−1 is injective. We may view Sn−1 as a subgroupof Sn (the set of permutations which preserve n) and in this case φn−1 isjust the restriction of φn to Gn−1 (again viewing Gn−1 as a subgroup ofGn in the obvious way). Now suppose that a ∈ ker(φn). By the abovewe may write a = xi · · ·xn−1y for some i and y ∈ Gn−1. Then φn(a) =(i, i+1) · · · (n− 1, n)φn(y) = (i · · ·n)φn−1(y). Since φn−1(y) preservers n, wemay conclude that (i · · ·n)φn−1(y) maps n to i. Since this is the identity, wemust have i = n, that is, a = y. Hence a = y ∈ ker(φn−1) which we haveassumed to be trivial. This shows that φn is injective and gives the claim.

Now let t1, . . . , t5 ∈ S6 be given by

t1 = (12)(34)(56) t2 = (14)(25)(36) t3 = (13)(24)(56)t4 = (12)(36)(45) t5 = (14)(23)(56).

58 Fall 2004

Let T be subgroup of S6 generated by these elements. Applying the per-mutation of corresponding to t3, t4, or t5 to t1 results in t1 so that all theseelements commute with t1. Similarly titj = tjti if |j − i| ≥ 2. Furthermorewe have

t1t2 = (135)(264),

t2t3 = (162)(345),

t3t4 = (146)(235), and

t4t5 = (153)(254),

which all have order three. Since all the ti’s have order two, we see that the ti’ssatisfy the relations of xi’s above to that we get a surjective homomorphismG6 → T given by xi 7→ ti. Composing this with the isomorphism to S6 → G6

we get a surjective homomorphism ψ : S6 → T given by (i, i+ 1) 7→ ti. Nowthe kernel of ψ must be a normal subgroup and we have shown above thatthe only normal subgroups of S6 are 1, A6, and S6. If the kernel is either ofthe latter two, we would have |T | ≤ 2 which is certainly not the case. Henceker(ψ) = 1. Hence ψ : S6 → T is an isomorphism. We may conclude thatT = S6 (by order considerations) and so ψ is an automorphism of S6 whichdoes not preserve conjugacy class and so is not an inner automorphism.

3. Suppose N is a linear transformation of an n-dimensional vec-tors space over a field, and that Nd = 0 but Nd−1 6= 0.

a. What is the characteristic polynomial of N+I (I being the iden-tity map)?

First note that since N satisfies xd = 0, the minimal polynomial of Nmust divide xd. Hence it must be xi for some i ≤ d. But if i < d, wehave Nd−1 = N iNd−i−1 = 0, contradicting our assumption. Thus xd is theminimal polynomial of N . The characteristic polynomial of M divides somepower of the minimal polynomial, that is, it divides xk for some k. Sincethe characteristic polynomial must have degree n, we may conclude that thecharacteristic polynomial of N is xn. That is, det(xI−N) = xn. This implies

59

that

det(xI − [N + I]) = det(xI − I −N)

= det([x− 1]I −N)

= (x− 1)n.

Hence, the characteristic polynomial of N + I is (x− 1)n.

b. What is the minimal polynomial of N + I?

We have (N + I − I)d = Nd = 0 so that N + I satisfies (x − 1)d. If Nsatisfies (x−1)i for i < d, we have N i = (N+I−I)i = 0, a contradiction, bythe above. Hence we can conclude that (x − 1)d is the minimal polynomialof N .

c. Give an example where the minimal and characteristic polyno-mials are different.

Consider the matrices

N =

0 1 00 0 10 0 0

and N + I =

1 1 00 1 10 0 1

over Q. We have that

xI − (N + 1) =

x− 1 −1 00 x− 1 −10 0 x− 1

and, expanding about the first column, that det[xI − (N + 1)] = (x − 1)3.Furthermore N2 = 0 and (N + I − I)2 = 0 so that the minimal polynomialof N + I is not (x− 1)3 (in fact, it’s (x− 1)2).

60 Fall 2004

4. In this problem you may assume that for any ring R and anymodule M over R, R⊗RM ∼= M .

a. Let A and B be abelian groups, which are not assumed tobe finitely generated. Suppose in A ⊗Z B, we have the relation∑ai ⊗ bi = 0. Prove that there are finitely generated subgroups

A′ ⊂ A and B′ ⊂ B such that all ai ∈ A′, all bi ∈ B′, and∑ai ⊗ bi = 0

in A′ ⊗Z B′. (You may use the description of the tensor product in

terms of generators and relations).

We have that A ⊗ B is the free abelian group on the cartesian productA × B modulo the subgroup, N , generated by the elements (a1 + a2, b) −(a1, b) − (a2, b), (a, b1 + b2) − (a, b1) − (a, b2), and (na, b) − (a, nb) for anya, a1, a2 ∈ A, b, b1, b2 ∈ B, and n ∈ Z. Since

∑ai⊗ bi = 0 in A⊗B, we must

have that∑

(ai, bi) ∈ N . Hence∑

(ai, bi) is a (finite) sum of elements of the

above form. That is, we may find finite sets a(1)i , a

(2)i , αj, α′k ⊂ A,

b(1)j , b

(2)j , βi, β′k ⊂ B, and nk ∈ Z such that∑

(ai, bi) =∑[

(a(1)i + a

(2)i , βi)− (a

(1)i , βi)− (a

(2)i , βi)

]+∑[

(αj, b(1)j + b

(2)j )− (αj, b

(1))− (αj, b(2)j )]

+∑[

(nkα′k, β

′k)− (α′k, nkβ

′k)].

Let then A′ ⊂ A be group generated by ai, a(1)i , a

(2)i , αj, and α′k.

Similarly let B′ ⊂ B be the group generated by bi, βi, b(1)j , b(2)j ,

and β′k. Then A′ and B′ are finitely generated by definition. Furthermore,∑ai⊗ bi is still a sum of relations in A′⊗B′ and so

∑ai⊗ bi = 0 in A′⊗B′,

as required.

b. Consider the rational field Q as an abelian group. If A′ ⊂ Q is afinite generated nonzero subgroup, show that A′ ∼= Z.

Of course, we must through out the trivial case A′ = 0 for in this casethe statement is false. Now since A′ is finitely generated, we may apply thestructure theorem for abelian groups to conclude that A′ ∼= (A′)tor ⊕ Zr forsome r (where (A′)tor is the torsion submodule of A′). But for x ∈ (A′)tor,we have some n ∈ Z such that nx = 0. Since Q is a field (and in particular

61

an integral domain) this implies that x = 0. Hence A′ ∼= Zr. Now, supposer ≥ 2. Then we may find two elements a/b, c/d ∈ Q which are linearly in-dependent over Z. But we have bc(a/b) − ad(c/d) = 0, contradicting linearindependence. Hence r = 1 and so A′ ∼= Z.

c. Consider C = Q⊗Z Q. Show that if α ∈ C and nα = 0 for integern 6= 0 then α = 0.

By a., we may find finitely generated abelian groups A′, B′ ⊂ Q such thatα ∈ A′ ⊗Z B

′ and nα = 0 still holds in A′ ⊗Z B′. But by b., A′ ∼= B′ ∼= Z (if

either A′ or B′ is 0, the statement is already clear) so we have α ∈ Z⊗ Zwith nα = 0 (viewing the isomorphic groups as identical). But Z⊗Z Z ∼= Zso that nα = 0 in Z. Hence α = 0 as Z is an integral domain.

d. Show that Q⊗Z Q ∼= Q⊗Q Q ∼= Q.

Let A be the free abelian group on the cartesian product Q × Q. Thenwe have an epimorphism A → A induced by the identity map on Q × Q.Composing this with the quotient map A→ Q⊗Z Q, we get an epimorphismA→ Q⊗Q Q. Since the relations on Q⊗Z Q are a subset of those on Q⊗Q Q(the latter allow elements of Q to move from one factor to another whereasthe former only allow elements of Z), this last map factors through Q⊗Z Qand we get an epimorphism φ : Q⊗Z Q → Q⊗Q Q given by a⊗Z b 7→ a⊗Q b.

Now consider the map Ψ : Q × Q → Q ⊗Z Q given by (a, b) 7→ a ⊗Z b.This map trivially preserves addition in each coordinate. Now suppose thata1/b1, a2/b2, a3/b3 ∈ Q. Then we have

b3

[Ψ

(a3

b3· a1

b1,a2

b2

)−Ψ

(a1

b1,a3

b3· a2

b2

)]=

[b3a3

b3· a1

b1

]⊗Z

a2

b2

− ·a1

b1⊗Z

[b3a3

b3· a2

b2

]=

[a3 ·

a1

b1

]⊗Z

a2

b2

− ·a1

b1⊗Z

[a3 ·

a2

b2

]= 0,

62 Fall 2004

since a3 ∈ Z. By c., we can conclude that

Ψ

(a3

b3· a1

b1,a2

b2

)= Ψ

(a1

b1,a3

b3· a2

b2

)and essentially from the definition of Q⊗Z Q as a Q-module, we have

Ψ

(a3

b3· a1

b1,a2

b2

)=a3

b3·Ψ(a1

b1,a2

b2

)Hence the map is Q-bilinear and we may apply the universal property of thetensor product to conclude that we have a Q-linear map ψ : Q⊗QQ → Q⊗ZQgiven by a⊗Q b 7→ a⊗Z b. This ψ is clearly an inverse to φ above and so themay conclude the first isomorphism. The second isomorphism follows fromthe statement at the beginning of the problem.

Chapter 7

Spring 2004

1. Let Q be the rational field and set F = Q(√

2). Consider thefollowing three elements of F :

α1 = 2 +√

2

α2 = 3 +√

2

α3 = −3 + 2√

2

Let Ki = F (√αi) for i = 1, 2, 3.

a) Show that [Ki : Q] = 4 for i = 1, 2, 3.

We can see that α1 a root of f1(x) = x4 − 4x2 + 2. This polynomialis Eisenstein at p = 2 and so is irreducible. Hence f1(x) is the minimalpolynomial of α1 over Q and we may conclude that [K1 : Q] = 4. Note that

f1(x) = (x2 − 2−√

2)(x2 − 2 +√

2)

so that the roots of f1(x) are ±√

2 +√

2 and ±√

2−√

2.Likewise, α2 is a root of f2(x) = x4 − 6x2 + 7. The roots of f2(x) are

±√

3 +√

2, and ±√

3−√

2. Multiplying them pairwise, we never get a ra-tional number. Hence f2(x) cannot be divisible by any quadratic over Q.Furthermore, f2(x) has no rational roots and so we may conclude that it isirreducible. This implies that [K2 : Q] = 4.

Lastly, α3 satisfies f3(x) = x4 + 6x2 + 1, which has roots ±√−3 + 2

√3

and ±√−3− 2

√2. Multiplying the first two doesn’t give a rational and nei-

ther does multiplying the second two. Hence if f3(x) is going to split into two

64 Spring 2004

rational polynomials over Q, we must have√−3 + 2

√2±

√−3− 2

√2 ∈ Q,

a contradiction. Thus as above f3(x) is irreducible and [K3 : Q] = 4.

b) Determine whether Ki/Q is:

i) non-Galois,

ii) Galois with cyclic Galois group, or

iii) Galois with noncyclic Galois group.

Note that√

2 +√

2√

2−√

2 =√

2. Hence we may conclude that√2−

√2 ∈ K1

(since of course√

2 ∈ K1). Hence K1 is the splitting field of f1(x) overQ and so is normal and therefore Galois over Q (since Q has characteristiczero). Set G1 to be the Galois group of K1 which of course the same as theGalois group of f1(x) and view G1 as a subgroup of S4 in the usual way.Since we know all the roots of f1(x), we can compute that its discriminantis 2048 = 211. Since this is not a square in Q, G1 is not contained in A4.

SupposeG1 does not contain a 4-cycle. Then sinceG1 6⊂ A4, we must havea 2-cycle in G1. We assume without loss of generality that (12) ∈ G1. Letσ ∈ G1 be another element which is not the identity. By order considerationsand our assumption that G1 does not contain a 4-cycle, σ must be either aproduct of disjoint 2-cycles or another 2-cycle. In the latter case, if σ doesnot fix 1 and 2, its product with (12) is a 3-cycle, contradicting the fact that|G1| = 4. Thus σ = (34). But then G1 = 1, (12), (34), (12)(34). Thissubgroup does not act active transitively (1 is never sent to 3 for example),a contradiction. Hence σ must be a product of disjoint 2-cycles. Again, ifσ 6= (12)(34), (12)σ is 4-cycle, again a contradiction. Hence σ = (12)(34), butthat gives the same subgroup above and so we have the same contradiction.Thus we may conclude that G1 contains an element of order 4 and so theGalois group of K1 satisfies ii).

Before we consider K2, we need to prove two lemmas: First we will showthat

√7 6∈ Q(

√2). The contrary would imply that we had a, b ∈ Q such

that√

7 = a+ b√

2. Squaring both sides gives 7 = a2 + 2b2 + 2ab√

2. Hencewe have either a = 0 or b = 0 which gives b2 = 7/2 or a2 = 7, respectively,

65

both are contradictions. Hence [Q(√

2,√

7) : Q] = 4. Next we will showthat α2 6∈ Q(

√2,√

7). This will imply that Q(√

7, α2) = Q(√

2,√

7, α2) hasdegree greater than four over Q so that

√7 6∈ K2. Suppose the contrary.

Since 1,√

2 is a basis for Q(√

2) over Q and 1,√

7 is that for Q(√

7), wehave that a basis for Q(

√2,√

7) over Q is 1,√

2,√

7,√

14.Now if α2 ∈ Q(

√2,√

7), we have a, b, c, d ∈ Q such that α2 = a+ b√

2 +c√

7 + d√

14. Squaring both sides gives

3 +√

2 = (a2 + 2b2 + 7c2 + 14d2) + (2ab+ 14cd)√

2

+ (2ac+ 4bd)√

7 + (2ad+ 2bc)√

14.

By linear independence, we have ac = −2bd and ad = −bc. Suppose thata 6= 0 and c 6= 0. Then we have d = −(bc)/a so that ac = (2b2c)/a whichimplies (1/2) = b2/a2, a contradiction. Next suppose that a = 0 and c =0. Then 1 = 2ab + 14cd = 0, a contradiction. Hence we must have aor c zero, but not both. The former case gives −bc = ad = 0 so thatb = 0. Then 14cd = 1 and 7c2 + 14d2 = 3. Then d = 1/(14c) and so7c2+1/(14c2) = 3. Multiplying through by 14c2, we have 98c4−42c2+1 = 0.Since 98 = 72 · 2, the Rational Roots Test gives the possible values for c as±1,±7,±49,±2,±14,±98. Clearly, we need only check the positive ones,none of which give zero. Hence we have a contradiction. Finally we musthave c = 0. This gives ad = 0 so that d = 0. But then 1 = 2ab+ 14cd = 0, acontradiction. Hence we have shown the claim:

√7 6∈ K2.

Finally we have√

3 +√

2√

3−√

2 =√

7. Hence√

3−√

2 6∈ K2. Thusf2(x) is irreducible over Q and has a root in K2 but does not split over K2.This shows that K2 is not normal over Q and so K2 satisfies i).

Lastly, we have√−3 + 2

√2√−3− 2

√2 = 1 so that −3−2

√2 ∈ K3 and

so K3 is the splitting field for f3(x). This shows that K3 is Galois over Q.Let G3 be the Galois group, viewed as a subgroup of S4 in the usual way.Now the discriminant of f3(x) is 16384 = 214, which is a square in Q. HenceG3 ⊂ A4. In particular, G3 cannot contain a 4-cycle and so no element of G4

has order 4. Thus K3 satisfies iii).

2. Let R be a commutative integral domain with unit. Let I, J ⊂ Rbe two ideals. Suppose I, J, and I + J are principal ideas. Provethat I ∩ J is principal by finding a generator.

66 Spring 2004

Let I = (a), J = (b), and I + J = (c). Now a ∈ I ⊂ I + J so we mayfind a′ ∈ R such that a = a′c. Likewise, we may find b′ ∈ R with b = b′c.Furthermore, c ∈ I + J , so we may find s, t ∈ R such that c = sa+ tb. Thisimplies that c = sa′c + tb′c = (sa′ + tb′)c. By the cancelation property of adomain, we have 1 = sa′ + tb′ (since R has an identity).

We claim that I ∩J = (a′b′c). Suppose that z ∈ (a′b′c). Find x such thatz = xa′b′c. Then z = xb′a = xa′b so that z ∈ I∩J . This gives (a′b′c) ⊂ I∩J .Likewise, suppose that z ∈ I ∩ J . Find x, y ∈ R such that z = xa = xa′cand z = yb = yb′c. Then

z = z(sa′ + tb′) = yb′csa′ + xa′ctb′ = (ys+ tx)a′b′c ∈ (a′b′c).

This proves the claim so that a′b′c is a generator for I ∩ J .

3. Let G be a finite group. Let G ⊂ G′ ⊂ G′′ ⊂ . . . be the sequenceG′ = [G,G], G′′ = [G′, G′], etc. Of course for any subgroup H ⊂ G,[H,H] is the subgroup generated by the commutators aba−1b−1 forall a, b ∈ H. Suppose that G′/G′′ and G′′/G′′′ are cyclic. The objectof this problem is to prove G′′ = G′′′.

a) Show that without loss of generality we can assume G′′′ = 1.

Assume the statement holds for G′′′ = 1. Recall that for any normalsubgroup N C G, we have G(i)/N = (G/N)(i). We will show by inductionthat G(i) is normal for i ≥ 0. The i = 0 case is trivial. Suppose that G(i) isnormal. An element of G(i+1) can be written as [a, b] where a, b ∈ G(i). Thenfor g ∈ G,

g[a, b]g−1 = gaba−1b−1g−1

= gag−1gbg−1ga−1g−1gb−1g−1

= gag−1gbg−1(gag)−1(gbg−1)−1

= [gag−1, gbg−1].

By normality, gag−1, gbg−1 ∈ G(i) which shows that g[a, b]g−1 ∈ Gi+1 andgives the claim.

Hence, we have G′′/G′′′ = (G/G′′′)′′ and (G/G′′′)′′′ = G′′′/G′′′ = 1.Hence we may conclude by the assumption that G′′/G′′′ = (G/G′′′)′′ =(G/G′′′)′′′ = 1 and so G′′ = G′′′.

67

b) Assuming G′′′ = 1, let Z be the centralizer of G′′ in G. Showthat G/Z is abelian (Hint: consider Aut(G′′)). Conclude that G′ ⊂Z.

By assumption, G′′ ∼= G′′/G′′′ is cyclic, so write G′′ = 〈x〉 for some x ∈ G.Consider the map Φ : G→ Aut(G′′) given by g 7→ φg where φg(x) = gxg−1.Since conjugation is an automorphism, this is well-defined. Furthermore, forg, h ∈ G, we have φgh(x) = ghxg−1h−1 = (φg φh)(x). Since x generatesG′′, we may conclude that Φ(gh) = Φ(g)Φ(h). Thus Φ is a homomorphism.Next suppose that g ∈ ker(Φ). Then gxg−1 = x so that g commutes withx. Since x generates G′′, we may conclude that g ∈ Z. Conversely, if g ∈ Z,gxg−1 = x which implies that φg is the identity and so g ∈ ker(Φ). HenceZ = ker(Φ). Thus we have an injection G/Z → Aut(G′′). G′′, however, iscyclic and so we may conclude that Aut(G′′) is abelian. This gives G/Z isabelian and so G′ ⊂ Z.

c) Using the assumptions and b), show that G′ is abelian and hencethat G′′ = 1.

Since G′/G′′ is cyclic, we may write G′/G′′ = 〈yG′′〉 for some y ∈ G′. Nowlet a, b ∈ G′. Then aG′′ = yiG′′ for some i and bG′′ = yjG′′ for some j. SinceG′′ = 〈x〉, there is some k, l such that a = yixk and b = yjxl. But, y ∈ G′ sothat, by b), y commutes with x. This gives ab = yixkyjxl = yjxlyixk = ba.Hence G′ is abelian.

4. Let Mn(C) be the ring of n× n matrices over the complex field.Suppose that A,B ∈Mn(C) are such that AB = BA.

a) Show there exists a nonzero column vector v such that Av ∈ Cvand Bv ∈ Cv.

Let λ ∈ C be an Eigenvalue of A and Eλ the corresponding Eigenspace.Then if v ∈ Eλ, Av = λv. Moreover, A(Bv) = B(Av) = B(λv) = λBv sothat Bv ∈ Eλ. Hence if φ is the linear transformation on Cn given by B, we

68 Spring 2004

can restrict φ to a linear transformation φB : Eλ → Eλ. Now find a nonzeroEigenvalue v of φB. Then we have Bv ∈ Cv and Av ∈ Cv (since v ∈ Eλ).

b) Show that there exists an invertible P ∈Mn(C) such that PAP−1

and PBP−1 are both upper triangular.

We will show this result by induction on n. By a), find a nonzero vsuch that Av,Bv ∈ Cv. Pick a basis for Cn such that v is an element:Cn = Cv ⊕W . Let φA be the transformation given by A and let φB be thatgiven by B. Then the matrices of φA and φB with respect our basis have theform (

λA wA0 MA

)and

(λB wB0 MB

)where λA, λB ∈ C and MA,MB ∈Mn−1(C). Then

AB =

(λAλB λAwB + wTAMB

0 MAMB

)and BA =

(λBλA λBwA + wTBMA

0 MBMA

).

Since these are equal, we have that MAMB = MBMA. Hence, by the induc-tion hypothesis, there is an invertible Q such that QMAQ

−1 and QMBQ−1

are both upper triangular. Letting

P =

(1 00 Q

),

we have

P−1 =

(1 00 Q−1

).

This gives

PAP−1 =

(λA wTAQ

−1

0 QMAQ−1

)which is upper triangular. Similarly, PBP−1 is upper triangular and we haveour result.

Chapter 8

Fall 2003

1. Let G be the group of 2 by 2 matrices of determinant oneover the field, F3, of three elements. Group theorists usually writeG = SL2(3). The group operation is matrix multiplication, butwhen we add elements, we are doing it in the full 2 by 2 matrixring M2(F3).

a) If g ∈ G, show that g satisfies the polynomial equation x2−ax+1 =0 for some a ∈ F3. Factor, or prove irreducible, each of these threepolynomials in F3[x].

The characteristic polynomial of g is det(xI−g). Hence the characteristicpolynomial, evaluated at x = 0 gives the determinant of g. Writing x2−a+bas the characteristic polynomial, we have det(g) = b. Hence b = 1 and thecharacteristic polynomial is x2 − a + 1 for some a ∈ F. Since every matrixsatisfies its own characteristic polynomial (the Cayley-Hamilton Theorem),we have the result.

Since 02 + 1 = 2, 12 + 1 = 2 and 22 + 1 = 2, we have that x2 + 1has no root in F3. Hence it is irreducible (as it has degree 2) over F3.x2 − x+ 1 = x2 + 2x+ 1 = (x− 2)2 and x2 − 2x+ 1 = x2 + x+ 1 = (x− 1)2

are both reducible.

b) If g satisfies x2 + 1 = 0, show that g has order 4. If g satisfiesx2 +x+1 = 0, show that g has order 1 or 4. If g satisfies x2−x+1 = 0show that g has order 2 or 6. Show that there is a unique element

70 Fall 2003

of order 2 in G.

First suppose that g satisfies x2 + 1 = 0. Since this polynomial is irre-ducible, it is the minimal polynomial of g so that g satisfies a polynomial ifand only if it is divisible by x2 + 1. First g2 = −I so that the order of gcannot be 1 or 2. Next, we have (x4−1) = (x2−1)(x2 +1) so that g satisfiesg4 = I. This shows that g has order dividing 4. Hence its order is exactly 4.

Next suppose that g satisfies x2 + x + 1 = (x − 1)2. Then the minimalpolynomial of g is either x2 + x+ 1 or x− 1. In the latter case, g has order1. In the former case, we have that g 6= I (as then the minimal polynomialof g would be x− 1) so that g does not have order 1. Furthermore we havex3 − 1 = (x2 + x + 1)(x − 1) so that g3 = I. this shows that the order of gmust divide 3 and so is 3.

Lastly suppose that g satisfies x2 + 2x + 1 = (x − 2)2. Then the min-imal polynomial of g is either x − 2 or x2 + 2x + 1. In the former case,we have g = −I which implies that g has order 2. In the latter case, theminimal polynomial of g is x2 + 2x + 1. If g2 = I, we have −2g − I = Iso that g = −I and the minimal polynomial of g is not (x − 2)2. We havex6 − 1 = (x2 + 2x + 1)(x4 + x3 + 2x + 2) so that g6 = I. Hence the or-der of g divides 6. It cannot be 1 as then its minimal polynomial wouldbe x − 1 and likewise it cannot be 2. Hence it must be either 3 or 6. Butx3 − 1 = (x2 + 2x + 1)(x − 2) + 1 so that g3 = −I 6= I. Hence the order ofg is 6. Not that the only case where we produced and element of order 2 wehad g = −I so that this is the unique element of order 2.

c) Let V = F3 ⊕ F3 and view M3(F3) as EndF3(V ). Show that G hasorder 24. If g ∈ G has order 3, show that the set of v ∈ V such thatg(v) = v is a one dimensional subspace of V .

First, GL2(F3) has order 48 = 24 ·3. To see this, note that for an invertiblematrix, we have 32 − 1 = 8 choices for the first column (the only column wecan’t have is the zero column). The second column can then be anythingother than a scalar multiple of the first column. This gives 32−3 = 6 choices.Hence the group has order 6 · 8 = 48. Since G is a subgroup, the order of Gdivides 48. Now, we have already seen that G has elements of order 3 and 4(although we did not explicitly show that elements of such order exists, wecan take the companion matrices corresponding to the appropriate minimalpolynomials). Hence both 3 and 4 must divide order of G. This gives us 12,

71

24, or 48 for the order of G. Since the diagonal matrix with diagonal entriesgiven by 2 and 1 (in either order) is an element of GL2(F3), but not G, weknow that |G| 6= 48. Hence G has either order 12 or order 24. However, amatrix which has any of the eight nonzero columns as its first column. Wecan insure its determinant is 1 by making the top right entry zero and thebottom right entry the inverse of the top left or by making the bottom rightentry zero and the top right entry the negative inverse of the lower left entry.Hence G has at least 16 elements and so must have order 24.

Now if g ∈ G has order three, we have seen that the minimal polynomialand characteristic polynomial of g is x2 + x + 1. Since the characteristicpolynomial is the product of the invariant factors, we may conclude thatx2 + x + 1 is the only invariant factor. Hence if x acts on V by g, we haveV ∼= F3[x]/(x

2 + x + 1) = F3[x]/(x − 1)2 as and F3[x]-module. To say thatv ∈ V satisfies g(v) = v is to say that xf(x) = f(x) in F3[x]/(x − 1)3

where f(x) corresponds to v in the above isomorphism. This is equivalent tof(x)(x− 1) = 0 which is equivalent to (x− 1)2 divides f(x)(x− 1) in F3[x].Hence f(x) = x−1 is such a vector and f(x) = 1 is not. This shows that thedimension of the space is at least one, but it is not all of V and so it musthave dimension one.

d) Let v ∈ V be nonzero and H ⊂ G the subgroup of g ∈ G withg(v) = v. Show that H has order 6 and has normal subgroup oforder 3. Use this and c) to show G has 4 subgroups of order 3.

The first statement is actually false, if H had order 6, it would have tocontain an element of order 2. But we showed above that the only element oforder 2 in G is −I. This gives −I ∈ H and so −Iv = v or v = −v. Since v isnonzero and we are not working in characteristic two, this is a contradiction.Hence two cannot even divide the order of H which shows that |H| is either1 or 3.

Now G acts on V and so splits it up into orbits. If |H| were one, the orbitof v would be |G|/|H| > 9. This is a contradiction. Hence we have |H| = 3.Thus every nonzero element of V is stabilized by an element of order 3 andits inverse. Hence we count 16 elements of order 3 in this way. By c) eachelement of order 3 is counted precisely twice and so we may conclude thatwe have 8 elements of order three. This gives exactly 4 subgroups of order 3.

e) Let ι ∈ G be the element of order 2, and G = G/ 〈ι〉. Show that G

72 Fall 2003

also has 4 subgroups of order three, and then that G has a normalsubgroup of order 4. Show that G has a normal subgroup of order 8.

Let g ∈ G be an element of order 3 and consider the element gι. Sinceι is in the center of G (as a scalar multiple of the identity) we have thatthat (gι) = gnιn. Hence the order of gι is 6. Now suppose that h ∈ G is anelement of order 3 with gι = hι. Multiplying both sides by ι we obtain g = h.Hence we have at least 8 distinct elements of order 6 (as there are 8 distinctelements of order 3). Since every element of 6 generates a subgroup of ordersix which then has exactly 2 elements of order 6 (which both generate thegroup) we may conclude that there are at least 4 cyclic subgroups of order 6.Since every such subgroup contains an element of order 2, we may concludethat every such subgroup contains 〈ι〉. We may then conclude from theCorrespondence Theorem that there are at least four subgroups of order 3 inG. But |G| = 12 so that, from the Sylow Theorems, the number of subgroupsof order 3 of G must divide 4. Thus there are exactly four subgroups of order3 G.

This implies that there are 8 elements of order 3 in G and so 12 − 8 =4 elements which do not have order 3. We may hence conclude that theremaining 4 elements of G comprise a unique, and therefore normal Sylow2-subgroup. This implies that G has a normal subgroup of order 4. Bythe Correspondence Theorem, G has a normal subgroup of order 8 (whichcontains 〈ι〉).

2. Let K be a subfield of the complex field C. Let f(x) ∈ K[x] be apolynomial of degree n. Let G be the Galois group of the splittingfield, L ⊂ C, of f(x) over K.

a) Suppose α ∈ C is a root of f(x) with the following property:if β ∈ C is another root, there is a polynomial gβ(x) ∈ K[x] withgβ(α) = β. Assuming f(x) irreducible, what can you say about theorder of the group G? For general f(x), what can you say?

Since β = gβ(α), we may conclude that β ∈ K(α). Since this holds for

73

all roots, we may conclude that L ⊂ K(α) but since α ∈ L, we also haveK(α) ⊂ L. Hence L = K(α). [K(α) : K] is the degree of the minimalpolynomial of α over K, which by assumption is n. Hence we may applythe Fundamental Theorem of Galois Theory to conclude that |G| = n. Iff(x) is not irreducible, we may apply the same reasoning to conclude that|G| = k where k is the degree of irreducible component of which α is a root.In particular, |G| ≤ n.

b) In 1829, Abel considered polynomial with the addition prop-erty that for all roots β, γ of f(x), gβ(gγ(α)) = gγ(gβ(α)). In 1877,Kronecker published a paper calling this polynomials “Abelschen”.Where special property of G does this property imply?

Let σ, τ ∈ G. Then σ(α) = β and τ(α) = γ for some roots β and γ off(x). Furthermore, since σ and τ act as the identity on K, they commutewith gβ and gγ, polynomials in K. This gives

σ τ(α) = σ(γ)

= σ gγ(α)

= gγ σ(α)

= gγ(β)

= gγ gβ(α)

= gβ gγ(α)

= gβ(γ)

= gβ τ(α)

= τ gβ(α)

= τ(β)

= τ σ(α).

But we have already seen that L = K(α). Since σ τ and τ σ agree both onα and on K (where they are the identity), we may conclude that τ σ = στ .Therefore, G is abelian.

c) Show, by example, that for K = Q, there exist polynomials ofarbitrary high degree with the property part b).

74 Fall 2003

Consider the polynomial f(x) = xn−1 over Q. The splitting field is Q(ζ),where ζ is a primitive nth root of unity. The roots of f(x) are given by ζk

for 0 ≤ k < n. Hence we can take gζk(x) = xk (where ζ plays the role of αabove). Then for ζk, ζ l roots of f(x), we have gζk gζl(ζ) = ζkl = gζl gζk(ζ).Therefore, f(x) has the property in b). Since this works for any n, we havethe claim.

d) Assume the condition in a) and suppose G is a p-group. Showthat there is a root β 6= α such that gβ(gγ(α)) = gγ(gβ(α)) for eachroot γ of f(x).

First note that this statement requires the assumption that f(x) is irre-ducible (which is probably the intent). Consider the polynomial f(x) = (x2−2)(x2 +4x+2) ∈ Q[x]. Using the quadratic formula, we find that the roots off(x) are ±

√2 and −2±

√2. Hence the splitting field is L = Q(

√2) and G has

order two (making it a p-group). Now, let α =√

2, β = −α, γ = −2 +√

2,and δ = −2 −

√2. Then we may take gβ(x) = −x, gγ(x) = −2 + x, and

gδ(x) = −2− x, and we have met the criteria in part (a). However, we havegβ gγ(

√2) = 2 −

√2 and gγ gβ(

√2) = −2 −

√2 so that neither β nor γ

satisfy the conclusion of the statement. Likewise, gδ gγ(√

2) = −√

2 andgγ gδ(

√2) = −4−

√2. Hence δ does not satisfy the conclusion. Hence the

statement is false if we do not assume f(x) to be irreducible.

Suppose then that f(x) is irreducible. Since G is a p-group, it has anontrivial element σ which lies in the center. Then β = σ(α) is a root off(x) which is distinct from α. Let γ ∈ L be another root and find a τ ∈ G

75

with τ(α) = γ. We have

gβ gγ(α) = gβ(γ)

= gβ τ(α)

= τ gβ(α)

= τ(β)

= τ σ(α)

= σ τ(α)

= σ(γ)

= σ gγ(α)

= gγ σ(α)

= gγ(β)

= gγ gβ(α)

and we have the statement.

3. Let R be a commutative ring.

a) Define what it means to R to be Noetherian.

R is Noetherian if and only if each of its ideals are finitely generated.

b) Prove that if R is Notherian, then the polynomial ring R[x] isNoetherian.

Let I be an ideal in R[x] and let L be the set of all leading coefficientsof polynomials in L. We will show that L ⊂ R is an ideal. Certainly L isnonempty. Let a, b ∈ L and r ∈ R. Find polynomials f(x), g(x) ∈ I whichhave leading coefficients a and b respectively. Assume that that have degreesda and db respectively. Then rxdbf(x) − xdag(x) ∈ I. If ra − b 6= 0 thenthe lead coefficient of this polynomial is ra− b so that ra− b = 0. If ra− bis zero, we already have ra − b ∈ L as the zero polynomial is in I. This

76 Fall 2003

gives the claim. Since L is an ideal in R it is finitely generated and we mayfind a a1, . . . , an such that L = (a1, . . . , an). Additionally, find polynomialsf1(x), . . . , fn(x) such that fi(x) ∈ I has lead coefficient ai. Let N be themaximum of the degrees of the fi(x).

For each 0 ≤ d ≤ N−1, let Ld be the set of lead coefficients of polynomialsin I with degree d together with 0 ∈ R. We will again show that Ld is anideal of R. Suppose that a, b ∈ Ld and r ∈ R. Then we have polynomialsf(x), g(x) ∈ I of degree d with lead coefficients a and b respectively. Ifra− b 6= 0 we have that the lead coefficient of rf(x)− g(x) is ra− b and thatit has degree d. Hence ra − b ∈ Ld (as we have already assumed 0 ∈ Ld).Thus Ld is an ideal and so it is generated by some a1,d . . . , and,d. Let fi,d bea polynomial in I of degree d with lead coefficient ai,d.

Set

I ′ = (f1(x), . . . , fn(x) ∪ fi,d(x)|0 ≤ d ≤ N − 1, 1 ≤ i ≤ nd) .

We claim that I ′ = I. First, I ′ ⊂ I as it is generated exclusively by poly-nomials which lie in I. Now suppose that I 6⊂ I ′ and find a polynomialf(x) ∈ I\I ′ of minimal degree. Let k be its degree and assume that k ≥ n.Then k is greater than or equal to the degree of fi(x) for each i. Hence wemay find ki such that xkifi(x) has degree k. Now the lead coefficient, c off(x) is in L and so we may find b1, . . . , bn such that c = b1a1 + · · · + bnan.Then we have that f(x) − b1a1x

k1f1(x) − · · · − bnanxknfn(x) is an element

of degree strictly smaller than f(x) which lies in I but not I ′. This is acontradiction.

Hence it suffices to assume that k < n. Then the lead coefficient of f(x),c, lies in Lk. Hence we may find c1, . . . , cnk

such that c = c1a1,k+· · ·+cnkank,k.

Then f(x) − c1a1,kf1,k(x) + · · · + cnkank,kfnk,k is a polynomial whose degree

is strictly less than k which lies in I but not I ′. This is again a contradictionand we may conclude that I = I ′. Hence I is finitely generated and so R[x]is Noetherian.1

4. Let R be the ring of continuous real functions f : R → R suchthat f(x + π) = f(x) for all x ∈ R. Let M be the set of continuousfunctions u : R → R such that u(x + π) = −u(x) for all x ∈ R. Notethat, with the usual addition and multiplication of functions, M isan R module (you need not prove this). Let c, s ∈ M be the usual

1This proof of Hilbert’s Basis Theorem is in Dummit and Foote on p.316

77

cosine and sine functions.

a) Show that u ∈M implies that u(a) = 0 for some a ∈ R. (Yes, youcan use calculus.)

Consider u(0). If u(0) = 0, we are done. Otherwise, suppose thatu(0) 6= 0. Then u(π) = −u(0). Of u(0) and −u(0), one is negative andthe other is positive. From this we may conclude, by the Intermediate ValueTheorem, that U(a) = 0 for some a ∈ (0, π) (since u is continuous).

b) Show that M is not generated by one element.

For a ∈ R, let f : R → R be the function given by f(x) = s(x + a) forall x ∈ R. Then f(x + π) = s(x + a + π) = −s(x + a) = −f(x) for eachx ∈ R (since s ∈ M). This gives f ∈ M (certainly f is continuous). Now,f(π − a) = s(π − a + a) = 1 so that f is nonzero at π − a. Hence for eachelement of a ∈ R, we have a function in M which is nonzero at the number.Suppose for a contradiction that M = (g) for some g ∈ M . By a), thereis a number a ∈ R with g(a) = 0. But by the previous argument, we mayfind an f ∈ M with f(a) 6= 0. Hence if there is an r ∈ R with rg = f , wehave f(a) = r(a)g(a) = 0, a contradiction. Thus M is not generated by oneelement.

c) Show that the map (f, g) 7→ (fc+ gs,−fs+ gc) is an isomorphismR⊗R ∼= M ⊗M . Show that M is projective as an R-module.

Denote the map by φ. Since c, s ∈ M and M is an R-module, φ iswell-defined. Now let (f1, g1), (f2, g2) ∈ R then

φ((f1, g1) + (f2, g2)

)= φ(f1 + f2, g1 + g2)

=((f1 + f2)c+ (g1 + g2)s,−(f1 + f2)s+ (g1 + g2)c

)=(f1c+ g1s+ f2c+ g2s,−f1s+ g1c− f2s+ g2c)

= (f1c+ g1s,−f1s+ g1c) + (f2c+ g2s,−f2s+ d2c)

= φ(f1, g2) + φ(f2, g2),

78 Fall 2003

so that φ preserves addition. Next let f, g, r ∈ R. Then we have

φ(r(f, g)

)= φ

(rf, rg

)= (rfc+ rgs,−rfs+ rgc)

= r(fc+ gs,−fs+ gc),

which shows that φ is R-linear.Now let u, v ∈ M . Note that the product of two elements in M is an

element of R. Hence we have uc− vs, us+ vc ∈ R. Then

φ(uc− vs, us+ vc

)=((uc− vs)c+ (us+ vc)s,−(uc− vs)s+ (us+ vc)c

)=(uc2 − vsc+ us2 + vcs,−ucs+ vs2 + usc+ vc2

)=(u(c2 + s2), v(s2 + c2)

)=(u, v),

so that φ is surjective. Finally suppose that φ(f, g) = (0, 0). Then fc+gs = 0and −fs + gc = 0. Multiplying the first equality by c, the other by s andsubtracting, we have f(c2 + s2) = 0 so that f = 0. Similarly, multiplying thefirst equality by s and the other by c gives g = 0. Hence φ is injective andso an R-module isomorphism. M is thus projective as the direct summandof a free R module.

d) Show the f 7→ fs⊗ s+ fc⊗ c is an isomorphism R ∼= M ⊗RM .

Define a map Ψ : M ×M → R by (u, v) 7→ uv/2. Since the product oftwo elements of M is an element of R (and constant functions are elementsof R), Ψ is well-defined. Now, let u1, u2, v ∈M .

Ψ(u1 + u2, v) =(u1 + u2)v

2

=u1v

2+u2v

2= Ψ(u1, v) + Ψ(u2, v),

which shows that Ψ preserves addition in the first coordinate. Similarly, Ψpreserves addition in the second coordinate. It is trivial to check that Ψ

79

satisfies the remaining requirement for R-bilinearity. Hence we may applythe Universal Property of the Tensor Product to conclude that we have anR-linear map ψ : M ⊗M → R which satisfies u⊗ v = uv/2.

Let φ be the map in the problem statement. Then for u ⊗ v ∈ M ⊗M ,we have

φ ψ(u⊗ v) = φ(uv

2

)=uv

2s⊗ s+

uv

2c⊗ c

=u

2⊗ vs2 +

u

2⊗ vc2

=[u2

+u

2

]⊗ v

= u⊗ v,

where we have used the fact that vs, vc ∈ R. Since M ⊗M is generated bythe simple tensors, we may conclude that φ ψ is the identity on M ⊗M .Likewise, suppose that f ∈ R. Then we have

ψ φ(f) = ψ(fs⊗ s+ fc⊗ c)

= fψ(s⊗ s) + fψ(c⊗ c)

= fs2

2+ f

c2

2= f.

Hence ψ φ is the identity on R which shows that φ and ψ are inverses sothat φ is an R-module isomorphism.

5. Let G be a finite group of order 90 with a non-normal subgroup,H, of order 5.

a) Show that G has a subgroup of order 15. Show this subgroup isabelian.

80 Fall 2003

We have |G| = 90 = 2 ·32 ·5. Let n5 be the number of Sylow 5-subgroups.From the Sylow Theorems, n5 must divide 18 and so must be 1, 3, 9, 2, 6,or 18. But only 1 and 6 are equivalent to 1 modulo 5. But n5 6= 1 as thiswould imply that G has a unique and therefore normal subgroup of order 5.Hence n5 = 6. Hence G acts transitively on a set of order 6. Let K be thestabilizer of H. Then we have n5 = |G|/|K| or |K| = |G|/n5 = 90/6 = 15.Hence G has a subgroup order order 15.

Now let let s3 and s5 be the number of Sylow 3-subgroups and and Sylow5-subgroups of K, respectively. Then s3 must divide 5 and so must be either1 or 5. But 5 is not equivalent to 1 modulo 3. Hence s3 = 1 and K hasa normal subgroup N1

∼= Z3 of order 3. Similarly s5 = 1 so that K has anormal subgroup N2

∼= Z5 of order 5. Then we have |N1||N2| = 15 = |K|and N1 ∩ N2 = 1. We may apply the direct product recognition theoremto conclude that K ∼= N1×N2

∼= Z5×Z3∼= Z15. In particular, K is abelian.

b) Let g be an element of order 3 in the subgroup of order 15 in a).Show that the centralizer of g has order 90 or 45. In the later case,show that G has a normal subgroup of order 5, a contradiction.

Sine 〈g〉 is a subgroup of G of order 3, it must be contained in a Sylow3-subgroup, P , which then has order 9. But the the order of P is the squareof a prime and so it must be abelian (this argument is given in #4 Spring2007). Hence the centralizer of g contains P and so 9 divides its order. Froma), the subgroup of order 5 contained in the abelian subgroup of order 15 (ofwhich g is a member) is contained in the centralizer of g. Hence 5 dividesthe order of the centralizer. Since the order of the centralizer must divide|G| = 90, we may conclude that its order is 90 or 45.

In the later case, G acts on the conjugacy class of g by conjugation. Theorder of this conjugacy class is |G|/45 = 2. Hence we have a homomorphismφ : G → S2 which is nontrivial. Then φ must be surjective which impliesthat its kernel is a normal subgroup, N , of G of order 45 = 32 · 5. From theSylow theorems, N has a normal subgroup Q of order 5. If x ∈ G, we havefrom normality that xQx−1 < K. But xQx−1 is a subgroup of order 5 andQ is the unique subgroup of N of order 5. This gives xQx−1 = Q so that Qis normal. This is a contradiction.

c) Suppose the element g of part b) has centralizer the whole groupG. Form the quotient G = G/ 〈g〉. Show the image of H is normal

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in G.

We have |G| = 30 = 2 · 3 · 5. From the Sylow Theorems, the numberof Sylow 5-subgroups of G is either 1 or 6. Suppose it is 6. Then we haveexactly 24 elements of order 5. Likewise, the number of Sylow 3-subgroupsis 1 or 10. It cannot be 10, however, as this would require 20 elements oforder 3 giving our group at least 24+20 = 44 elements. Hence G has normalsubgroup of order 3. By the Correspondence Theorem, this gives a normalsubgroup, R of G of order 9. Then HR is a subgroup of G of order 45. Sinceit has index 2, it must be normal, resulting in the same contradiction as inthe previous part. Hence we have that the number of Sylow 5-subgroups ofG is 1. So that any subgroup of order 5 is normal. But the image of H issuch a subgroup (certainly hi 6∈ 〈g〉 for any h ∈ H\1, 0 ≤ i ≤ 4). Hencethe image of H is normal.

d) Conclude the G has a normal subgroup of order 15. Show thatG has a normal subgroup of order 5, another contradiction.

By the Correspondence Theorem, and c). G has a normal subgroup, N ,of order 15. From the argument above, N is abelian and so has a uniquesubgroup of order 5. Again by an argument given above, this implies that Ghas a normal subgroup of order 5, a contradiction.

e) Okay, so any group of order 90 has a normal subgroup of order5. Show that such a G has a normal abelian subgroup of order 45.

LetH be the normal subgroup of order 5 and let P be a Sylow 3-subgroup.Then HK is a subgroup of order 45 (since |H ∩K| = 1). But then the indexof this subgroup in G is 2 which implies that it is normal in G. We willconclude our argument by showing that any group of order 45 is abelian.First the Sylow Theorems imply that the number of Sylow 5-subgroups andthe order of Sylow 3-subgroups of such a group are both 1. Hence a subgroupof order 45 is the direct product of a group of order 5 and a group or 9, bothof which must be abelian. This gives the claim.

82 Fall 2003

6. Let F be a field of characteristic zero. Suppose V is a finitedimensional vector space over F , and A ∈ EndF (V ) is such thatA2 = A.

a) Show that V is the internal direct sum of the kernel and imageof A.

Let v ∈ V . Then we have A(v−Av) = Av−A2v = Av−Av = 0. Hencev − Av ∈ ker(A). This implies that V = ker(A) + image (A). Now supposethat v ∈ ker(A)∩ image (A). Then we have w ∈ V such that Aw = v. Thenwe have 0 = Av = A2v = Aw = v. This gives V = ker(A)⊕ image (A).

b) Show that the following three properties are equivalent:

i) A 6= 0;

ii) there is a nonzero v ∈ V with Av = v; and

iii) The trace of A is nonzero.

First note that A satisfies x2 − x so the minimal polynomial of A dividesx(x − 1). This implies that the only invariant factors of A are x, x − 1, orx(x− 1). Hence the only invariant factors are x and x− 1. This shows thatA is diagonalizable with the diagonal entries consisting of ones and zeroes.Chose a basis of V with respect to which the matrix of A is a such a matrix(since F contains all the eigenvalues of A).

Now if A 6= 0, the matrix above must have at least one diagonal entryequal to one. Letting v be the basis vector which corresponds to this column,we have Av = v. Next if we assume there is a v ∈ V with Av = v, we haveA 6= 0. Hence the first two statements are the same. Note that we did notuse the fact that F has characteristic zero. Likewise that fact that the thirdstatement implies the first (which is trivial) also does not require that thecharacteristic be zero.

Now assume the trace of A is nonzero. Again considering the above basis,we have that the trace is k, the number of diagonal entries equal to one. Wemust must have k 6= 0 since the characteristic is zero (and we have at leastone nonzero entry).

c) Suppose that V is the internal direct sum of subspaces W1 and

83

W2. Show there is an A as above with kernel W1 and image W2.

Let A = α1, . . . , αk be a basis forW1 and B = β1, . . . , βn be a basis forW2. Then if v ∈ V , we have a = a+b for some a ∈ W1 and b ∈ W2. But thena is a linear combination of A and b is a linear combination of B so that v isa linear combination of A∪B. Likewise suppose that we have ai, bj ∈ F suchthat

∑aiαi+

∑bjβj = 0. Then

∑aiαi = −

∑bjβj ∈ W1∩W2 = 0. Hence

we may conclude that∑aiαi = 0 and

∑bjβj = 0. Linear independence then

implies that ai = 0 for each i and bj = 0 for each j. Hence A ∪ B is a basisfor V .

Thus we may define a linear transformation A : V → V by A(αi) = 0and A(βj) = βj. Then if a ∈ W1, a is a linear combination of A and soA(a) = 0. Conversely, if A(v) = 0 for v ∈ V , write v =

∑aiαi +

∑bjβj.

Then 0 = A(v) =∑bjβj which implies by linear independence that bj = 0

for each j. Hence a ∈ W1 which shows that W1 = ker(A). Next, letv =

∑bjβj ∈ W2. Then A(v) =

∑bjβk = v which shows that v ∈ image (A)

so that W2 ⊂ image (A). Conversely, let v ∈ image (A). Then find w ∈ Vsuch that A(w) = v. Then writing w =

∑aiαi+

∑bjβj, we have v = A(w) =∑

bjβj ∈ W2. Hence W2 = image (A). Finally A2(αi) = A(0) = 0 = A(αi)and A2(βj) = A(βj). Hence A and A2 agree on a basis for V and so areequal. Thus A is the claimed transformation.

d) Suppose that instead F has nonzero characteristic p. Which ofthe conditions in b) is no longer equivalent to the others? Give anexample.

We have seen that even in characteristic p, statements i) and ii) areequivalent and that iii) implies them. We do not have however that theyimply iii). Consider the case that V is p-dimensional and let A be the identitytransformation. Then certainly A2 = A. The trace of A, however, is p = 0and it is not the zero transformation.

84 Fall 2003

Chapter 9

Spring 2003

1. Assume that G is a finite group, p the smallest prime dividingthe order of G, and H a subgroup of index p in G.

(i) Prove that H is a normal subgroup in G.

G acts transitively on G/H, a set of order p. Thus we have a homomor-phism φ : G→ Sp. If an element g ∈ G lies in the kernel of φ, it must preserveH, that is, gH = H. This shows that ker(φ) ⊂ H. Suppose that ker(φ) 6= H.Then [H : ker(φ)] 6= 1 and so is divisible by a prime q ≥ p (as it must dividethe order of H). But then qp divides [G : ker(φ)] = [G : H][H : ker(φ)] whichin turn divides p!. Since q ≥ p, this is a contradiction. Hence ker(φ) = Hand H is normal.1

(ii) Show that if p + 2 is a prime and G is order p(p + 2), then G iscyclic.

Write q = p + 2. From the Sylow Theorems, we have a subgroup Q oforder p + 2. From the above, this subgroup is normal. Likewise, let np bethe number of Sylow p-subgroups. Then np must divide the order of G andcannot have a factor of p. This gives np is 1 or p + 2. But p + 2 6≡ 1 modpand so we must have np = 1. Hence we have a normal subgroup, P , oforder p. But now P and Q are normal subgroups of G with P ∩ Q = 1(by order considerations) and |G| = |P ||Q|. Hence by the Direct Product

1This proof is exercise 22 in Milne’s notes, the solution is given on page 75

86 Spring 2003

Recognition Theorem, G ∼= P × Q. But both P and Q have prime orderwhich then implies that they are cyclic, that is, P ∼= Zp and Q ∼= Zq. HenceG ∼= Zp × Zq = Zpq (Zp × Zq is generated by (1,1), for example). Thus G iscyclic.

2. Let R be an integral domain and let a be a non-zero non-invertible element of R.

(i) Prove that the ideal (a,X) ⊂ R[X] is not principal.

Suppose for a contradiction that (a,X) = (f(X)) for some f(X) ∈ R[X].Then a = g(X)f(X) for some g(X) ∈ R[X]. This implies that the degreeof f(X) is zero. Hence f(X) = c ∈ R. Furthermore X = ch(X) for someh(X) ∈ R[x]. Again from degree considerations, h(X) = αX + β for someα, β ∈ R and we have cα = 1 so that c is a unit. Thus (a,X) = R[X] so thatwe have 1 = Xs(X) + at(X) for some s(X), t(X) ∈ R[X]. Evaluating thisequation at X = 0 gives 1 = at(0), so that a is invertible. Hence we havecontradicted our assumption.

(ii) Use (i) to show that, if K is a field, then K[X, Y ] is not a PID.

By definition K[X,Y ] = K[Y ][X]. Consider then the ideal (X, Y ). Y isa non-zero non-invertible element in K[Y ] (since, when two polynomials inY multiply, their degrees add). Hence we may apply the above to concludethat (X, Y ) is not principal. This shows that K[X, Y ] is not a PID.

3. Let D be a unique factorization domain and let K be its field offractions.

(i) Prove that D is integrally closed in K.

Let r = a/b ∈ K\D. Since D is a UFD, we may assume that that thereis some prime p ∈ D which divides b, but does not divide a (otherwise wecould cancel all the primes from b and have a/b = c/u = cu−1 ∈ D for

87

some unit u ∈ D). Thus we may write a/b = (a/c) · (1/p) where b = pcand p does not divide a. Suppose now that a/b satisfies a monic polynomialf(x) = xn + αn−1x

n−1 + · · ·+ α0 ∈ D[x]. Then we have

f(r) = rn + αn−1rn−1 + · · ·+ α0

=

(a

c· 1

p

)n+ αn−1

(a

c· 1

p

)n−1

+ · · ·+ α0

=an + αn−1cpa

n−1 + · · ·+ α0cnpn

cnpn.

Since this is zero, we must have that the numerator is zero. This givesan = p(−αn−1ca

n−1 − · · · − α0cnpn−1). Hence p divides an but that implies

that p divides a. Since we have assumed this not to be the case, we have acontradiction. Hence D is integrally closed.

(ii) Prove or disprove: If a is a square free element of D, thatis, a is not divisible in D by the square of a prime element, thenD[X]/(X2 − a) is integrally closed in K[X]/(X2 − a).

This is statement is not true. Let D = Z and a = 1 (certainly a squarefree element). Then we have K = Q. Consider the element f(x) = (1/2)x+(1/2) ∈ K[X]/(x2 − a)\D[X]/(x2 − a). We have

[f(x)]2 − f(x) =

[1

2x+

1

2

]2

− 1

2x− 1

2

=1

4x2 +

1

2x+

1

4− 1

2x− 1

2

=1

4+

1

2x+

1

4− 1

2x− 1

2= 0,

where we have used the fact that x2 = 1. This shows that f(x) satisfiesy2 − y ∈ (D[x]/(x2 − a))[y] so that D[x]/(X2 − a) is not integrally closed inK[X]/(x2 − a).

88 Spring 2003

4. Let p be a prime number, k a field, β a nonzero element of k,and K a splitting field over k for the polynomial f(X) = Xp − β.

(i) Prove that K contains a subfield K1 which is a splitting fieldover k for the polynomial Xp− 1, and that α is a root of f(X) in K,then K = K1(α).

First suppose that k has characteristic not equal to p. Then f(X) isseparable over k as f ′(X) = pXp−1 whose only root is zero. Since zero is nota root of f(X), f(X) and f ′(X) can have no factor in common. Hence f(X)has distinct roots α1, α2, . . . , αp over K. Now for each i, we have(

αiα1

)p=β

β= 1.

So that ζi = αi/α1 is a collection of pth roots of unity which lie in K.Furthermore, it is a collection of p distinct roots as αi is a collection of pdistinct elements. Hence all the pth roots of unity lie in K. Setting K1 to bethe field generated over k by these roots, we have the first statement. Thesecond statement follows from αi = ζiα1 and from the fact that α1 could beany root of f(X).

If k has characteristic p, we have Xp− 1 = (X − 1)p so that the splittingfield of this polynomial over k is simply k (certainly 1 ∈ k). If α is a root off(x), we have f(x) = Xp − αp = (X − α)p so that K = k(α).

(ii) Show that either f(X) is irreducible in k[X], or f(X) as a rootin k.

If the characteristic of k is p, we have f(X) = (X − α)p where α is aroot of f(X). Suppose that α 6∈ k. Then if f(X) = g(X)h(X), for somenonconstant g(x), h(x) ∈ k[X], the constant term of g(X) must be αi forsome k < p (as g(X) must be a product of the linear factors of f(X) overK, but cannot be the product of all of them as h(X) is nonconstant). Henceαi ∈ k. But then i and p are relative prime, so we may find s, t ∈ Z suchthat si+ tp = 1. Then (αi)s = α1−tp = α/βt. Since βt ∈ k, this implies thatα ∈ k, a contradiction. Hence if f(X) has no root, it is irreducible.

Likewise suppose that the characteristic of k is not p. First note a productof pth roots of unity is another pth root of unity, (as is the multiplicative

89

inverse of a pth root of unity) so that the set of pth roots of unity is agroup under multiplication. This the group is of prime order, it is generatedby any pth root of unity other than one. Then we have from above thatf(X) =

∏p−1k=0(X − αζk), where ζ is a pth root of unity and α is a root of

f(X). Suppose that f(X) has no root over k and that f(X) = g(X)h(X)where g(X), h(X) ∈ k[X] are nonconstant. Then as above the constant termof g(X) must be a product of i pth roots of unity and αi. We must then havethat the constant term is of the form αiζj ∈ k for i < p. Again find s, t ∈ Zsuch that sk + tp = 1. Then (αiζj)s = (α/βt)ζjs. Since βt ∈ k, this impliesthat αζjs ∈ k, but this is a root of f(X). This proves the statement.

90 Spring 2003

Chapter 10

Fall 2002

1. Let p and q be distinct primes and G a finite group with p3qelements.

a) Prove that G has either a normal p-Sylow subgroup or a normalq-Sylow subgroup unless p = 2 and q = 3.

Suppose that G has neither a normal Sylow p-subgroup nor a normalSylow q-subgroup. Let np and nq be the number of Sylow p-subgroups andSylow q-subgroups, respectively. From the Sylow Theorems, np is either 1 orq. It cannot be 1, however, as this would imply that G had a normal Sylowp-subgroup. Hence we may conclude that np = q. From the Sylow Theorems,this implies that q ≡ 1 modp, or that we may find k ∈ Z such that q−1 = kp.In particular, q > p. Since q > 1 and p > 0, we may conclude that k > 0.By the same arguments, we must have that nq is p, p2, or p3.

Consider first the case that nq = p. Then as above, we must have thatp ≡ 1 mod(q) so that p > q. This is a contradiction. Next assume thatnq = p2. Then q divides p2− 1 = (p+ 1)(p− 1) so that q divides either p+ 1or p − 1. The latter results in the previous contradiction. Hence we havethat q divides p + 1, so that we may find a m > 0 such that mq = p + 1 orp = mq − 1. Combining this with the above, we have q − 1 = k(mq − 1).This gives q − 1 ≥ mq − 1 so that m = 1. Hence p = q − 1 and p and qare consecutive primes. Since 2 and 3 are the only consecutive primes, wehave p = 2 and q = 3. Lastly consider the case nq = p3. Since every Sylowq-Subgroup has q − 1 elements of order q and these groups only intersect atthe identity, we have (q − 1)p3 = p3q − p3 elements of order q. But then

92 Fall 2002

the remaining p3 elements of G must form a Sylow p-subgroup which impliesthat np = 1, contradicting our assumption. This completes all the cases.

b) Prove that the symmetric group S4 does not have a normal 2-Sylow subgroup and does not have a normal 3-Sylow subgroup.

Consider the subgroup H = 〈(1234), (12)(34)〉. First we have that theorder of (1234) is four and that of (12)(34) is two. Furthermore, we have(1234)(12)(34) = (13) = (12)(34)(1432) = (12)(34)[(1234)−1]. Hence wehave a surjective map φ : D4 → H given by (1234) 7→ r and (12)(34) 7→ m(where D4 is the dihedral group of order eight). This shows that |H| <8. We can exhibit eight elements of H by computing elements of the form[(12)(34)]i[(1234)]j for 0 ≤ i ≤ 1 and 0 ≤ j ≤ 3. This gives the groupshown below. Furthermore we may created another subgroup of order eightsimply by choosing an element of S4 which conjugates (1234) to a 4-cycle notpresent in H and then conjugating H by this element. H and another Sylow2-subgroup are shown below.

H1 (1234) (13)(24)

(1432) (12)(34) (24)(14)(23) (13)

(12)H(12)1 (1342) (14)(23)

(1243) (12)(34) (14)(13)(24) (23)

This shows that S4 does not have a unique Sylow 2-subgroup. Likewisewe have 〈(123)〉 and 〈(124)〉 are distinct Sylow 3-subgroups so that it alsodoesn’t have a unique Sylow 3-subgroup.

2. Prove or give a counterexample to each of the statements be-low, where throughout the problem it is assumed that R and S areintegral domains and f : R→ S is a surjective homomorphism.

a) If R is a PID then S is a PID.

This statement is true. Let I be an ideal in S. Then f−1(I) is an idealof R. Hence we may find an element a ∈ R such that f−1(I) = (a). Now let

93

y ∈ I. Then by surjectivity, we have an element x ∈ R with y = f(x).Then x ∈ (a) so that there is some b ∈ R with x = ba. This givesy = f(x) = f(b)f(a) so that y ∈ (f(a)). Hence I ⊂ (f(a)). The reverseinclusion is clear since a ∈ f−1(I) and so f(a) ∈ I. Thus I is principal whichshows that S is a PID (as it is an integral domain by assumption).

b) If R is a UFD then S is a UFD.

Let R = Z[x], and consider the homomorphism R → C given by eval-uation at

√−5. Then the image of this homomorphism is S = Z[

√−5] =

a+b√−5|a, b ∈ Z ⊂ C. Note also, that in S, we have (1+

√−5)(1−

√−5) =

1− (√−5)2 = 6 = 2 · 3. We will show that 2, 3, 1 +

√−5 and 1−

√−5 are

irreducible from which it will follow that S is not a UFD. Note that we havea multiplicative norm N on S given by N(a+b

√−5) = a2 +5b2 (a restriction

of the norm on C).We have N(1+

√−5) = 6. Hence if 1+

√−5 is a product of two elements

in S, either one must have norm 2 or one must have norm 1 (as 6 = 6 · 1 and6 = 3 · 2 are the only ways we can decompose 6 as positive integers). In theformer case we must have c2 + 5d2 = 2 for some c, d ∈ Z. This gives d = 0so that c2 = 2, a contradiction. In the latter case, we have c2 + 5d2 = 1 sothat c = ±1 and d = 0. Either chose for c then results in a unit. Hence wehave shown that 1 +

√−5 is an irreducible. For the same reason, 1−

√−5 is

an irreducible. Now N(3) = 9. Hence if 3 is a product of two elements in S,either they both have norm 3, or one has norm one, thus implying it is a unitas above. In the former case we have integers c and d with c2 +5d2 = 3 whichimplies that c2 = 3, a contradiction. Hence 3 is irreducible. 2 is irreducibleby a similar argument as we have already shown there are no elements ofnorm 2 in S. This gives the result.

c) If R is a PID and f is not an isomorphism then S is a field.

This statement is true. First suppose that P ⊂ R is a prime ideal. Thenwe have P = (p) for prime element p ∈ R. Suppose that M ⊃ P is anotherideal with M = (m). Then we have p = am for some a ∈ R. Now, p is irre-ducible so that either a or m is a unit. In the first case, we have m = a−1pso that P = M . In the second, we have M = R. Hence P is maximal. Nowf : R → S is surjective so S ∼= R/I for some ideal I in R. But S is anintegral domain so that I is prime. From the above, I is then maximal which

94 Fall 2002

implies that S is a field.

d) If R is a UFD and f is not an isomorphism then S is a field.

The statement is false. Set R = Z[x], S = Z, and consider the mapf : R → S given by evaluation zero. This map is certainly onto. Moreoverit is not an isomorphism as x ∈ ker(f). S, however, is not a field.

3. One defines a division ring to be a (not necessarily commutative)ring in which every non-zero element has a multiplicative inverse.Let D be a finite division ring. Let Z = x ∈ D|xy = yx,∀y ∈ D bethe center of Z.

a) Prove that Z is a finite field. If Z = Fq, show that D has qn

element for n.

Suppose that a, b ∈ Z and let x ∈ D. Then (a+b)x = ax+bx = xa+xb =x(a + b) so that a + b ∈ Z. Likewise abx = axb = xab implies that ab ∈ Z.Furthermore, (−a)x = −ax = −xa = x(−a) implies that −a ∈ Z. Lastlya−1xa = a−1ax = x = xaa−1 = axa−1. Multiplying on the right by a−1 givesa−1x = axa−2 = xaa−2 = xa−1. Hence we have shown that a−1 ∈ Z. ThusZ is a subdivision ring of D. Since it is commutative, by definition, we havethat Z is a field. It’s certainly finite.

Now D is an abelian group under addition and we have, for a, b ∈ Z andx, y ∈ D that (abx) = (ab)x, a(x + y) = ax + ay, (a + b)x = ax + bx, and1x = x. This shows that D is a vector space over D. If n is the dimension,we have |D| = qn.

b) Show that, given x ∈ D, the set Z(x) = y ∈ D|xy = yx is a di-vision ring containing Z and conclude that Z(x) has qnx for some nx.

Let a, b ∈ Z(x). Then (a + b)x = ax + bx = xa + xb = x(a + b) sothat a + b ∈ Z(x). Likewise abx = axb = xab implies that ab ∈ Z(x). Fur-thermore, (−a)x = −ax = −xa = x(−a) implies that −a ∈ Z(x). Lastly

95

a−1xa = a−1ax = x = xaa−1 = axa−1. Multiplying on the right by a−1 givesa−1x = axa−2 = xaa−2 = xa−1. Hence we have shown that a−1 ∈ Z(x).Thus Z(x) is a subdivision ring of D, which certainly contains Z. By thesame argument as in a), we have that the order of Z(x) is qnx for some nx.

c) Show that D∗ = D\0 is a group under multiplication. Showthat the centralizer of x ∈ D∗ is Z(x)\0. Apply the class equa-tion to D∗ to conclude that qn−1 = q−1+

∑ qn−1qni−1

for some integers ni.

Multiplication is associative in D∗ as part of the definition of a ring, 1acts as the identity, and every element has an inverse by assumption. Thecentralizer of x is Z(x) ∩D∗ = Z(x)\0. Hence let xi be a collection ofrepresentatives for each conjugacy class in D∗\0 consisting of more thanone element and, by c) find ni so that qni is the order of Z(xi). This gives,from the class equation, that qn − 1 = q − 1 +

∑ qn−1qni−1

, since the center hasorder q − 1.

4. Let F be a field and K/F a finite extension of [K : F ] = n. Letf(x) ∈ F [x] be an irreducible polynomial of degree m. Show thatif n and m are relatively prime then f(x) remains irreducible in K[x].

Let α be root of f(x) in some algebraic closure of F . Then the irreduciblepolynomial of α over F is f(x) and so has degree m. Let g(x) (a factor off(x)) be the irreducible polynomial of α over K and suppose that g(x) hasdegree k. Then

[K(α) : F ] = [K(α) : K][K : F ] = [K(α) : F (α)][F (α) : F ],

so that kn = [K(α) : F (α)]m. Since m and n are relatively prime, we musthave that m|k. But m ≥ k so we must have m = k. This implies thatf(x) = g(x) (as both are monic) so that f(x) is irreducible over K.

96 Fall 2002

Chapter 11

Spring 2002

1. Let K be the splitting field for x4 − 2 over Q.

(a) How many fields F are there with Q ⊂ F ⊂ K? For each suchfield describe one or more generators of F over Q.

Note that over C, the roots of f(x) are 4√

2, − 4√

2, i 4√

2, and −i 4√

2. Takingthe ratio of the first and the third, we see that i ∈ K. Hence Q( 4

√2, i) ⊂ K.

But also by the above we can see that 4√

2 and i generate K so that K =Q( 4√

2, i). Note that i is degree 2 over Q (as a root of x2 + 1) and i 6∈ Q( 4√

2)(as the latter consists only of real numbers). Hence i remains degree 2 overQ( 4√

2). Since 4√

2 is degree 4 over Q (as the above polynomial is irreducibleas it is Eisenstein at 2), we may conclude that [K : Q] = 8.

Let G be the Galois group of K over Q. Then any σ ∈ G is determinedby its image on 4

√2 and i. Moreover σ(i) = ±i and σ( 4

√2) must be one of

the above four roots of x4 − 2. This gives 8 possible combinations and soby order considerations, all the possibilities must be realized. Let σ ∈ G begiven by σ( 4

√2) = i 4

√2 and σ(i) = i. Then we can see that σ4 is the identity

and σ2( 4√

2) = − 4√

2. Hence σ has order 4. Likewise, let τ ∈ G be definedby τ( 4

√2) = 4

√2 and τ(i) = −i. Then τ has order two. We also have that

στ( 4√

2) = i 4√

2 = τσ−1( 4√

2) and στ(i) = −i = τσ−1(i). This shows thatστ = τσ−1 so that τ lies in the normalizer of 〈σ〉. Since 〈σ〉 ∩ 〈τ〉 = 1, wemay conclude that 〈σ〉 〈τ〉 is a subgroup of order 8. This shows that σ andτ generate G. Moreover, we see that G = 〈σ, τ〉 satisfies the relations of D8,the dihedral group of order 8. Hence we have a surjective homomorphismD8 → G. By order considerations, we may conclude that this this is an

98 Spring 2002

isomorphism.We now find all the subgroups of G. A subgroup of order 2 must be gen-

erated by an element of order 2 and so this gives 〈τ〉, 〈τσ〉, 〈τσ2〉, 〈τσ3〉, 〈σ2〉as the five subgroups of order 2. A subgroup of order 4 must be isomorphicto Z4, in which case it is 〈σ〉, or it must be isomorphic to K4 (the Klein4-group) in which case it has 3 elements of order two. Since there is only onepower of σ which has order 2, we must have two elements of the form τσi inour group. Picking two such elements, if their product in either order is σ orσ3, we do not get a subgroup of order four. Eliminating such pairs, we arriveat 〈σ2, τ〉 = 1, τ, σ2, τσ2 and 〈τσ3, σ2〉 = 1, σ2, τσ, τσ3, which are indeedsubgroups. This gives the following lattice of subgroups (with the inclusionrelationships shown).

G

vvvvvvvvv

KKKKKKKKKK

〈σ2, τ〉

xxxx

xxxx

x

GGGG

GGGG

G〈σ〉 〈τσ3, σ2〉

uuuuuuuuu

JJJJJJJJJ

〈τ〉

SSSSSSSSSSSSSSSSSSSSS 〈τσ2〉

IIIIIIIIII〈σ2〉 〈τσ3〉

sssssssssss〈τσ〉

iiiiiiiiiiiiiiiiiiiiiii

1

From the Galois Correspondence, this gives 8 nontrivial intermediatefields. First consider the subgroups of order 2 which correspond to subfield ofdegree 4 over Q. 〈τ〉 fixes 4

√2 and so its fixed field contains Q( 4

√2). Since this

is already degree 4, it must be the fixed field. Likewise 〈τσ〉 gives Q(i 4√

2).The fixed field of 〈σ2〉 contains Q(i,

√2) which also has degree four over Q

(we can see this by consider the tower Q ⊂ Q(√

2) ⊂ Q(i,√

2), the latterextension is degree 2 as i 6∈ Q(

√2), a field consisting of all real numbers).

Lastly 〈τσ3〉 and 〈τσ〉 give

Q([− 1√

2+

1√2i

]4√

2

)and Q

([1√2

+1√2i

]4√

2

)respectively, which have degree four since their generating numbers are rootsof x4 + 2 (which is irreducible by Eisenstein’s Criterion).

Next we can consider the subgroups of order 4 which gives extensionsof degree 2 over Q. First we note that both σ2 and τ preserve

√2 so that

99

the fixed field of 〈σ2, τ〉 contains Q(√

2) and must then equal it by degreeconsiderations. Likewise, the fixed field of 〈σ〉 is Q(i) and that of 〈τσ3, σ2〉is Q(i

√2). Writing

√i =

[1√2

+1√2i

]4√

2,

we have the following lattice of intermediate fields (which corresponds to thelattice of subgroups so that the larger fields re on bottom).

Q

rrrrrrrrrrr

NNNNNNNNNNNN

Q(√

2)

ttttttttt

KKKKKKKKKKQ(i) Q(i

√2)

qqqqqqqqqq

NNNNNNNNNN

Q( 4√

2)

UUUUUUUUUUUUUUUUUUUU Q(i 4√

2)

KKKKKKKKKKQ(i,

√2) Q(i

√i 4√

2)

qqqqqqqqqqQ(√i 4√

2)

hhhhhhhhhhhhhhhhhhhhhhh

Q(i, 4√

2)

(b) Describe the structure of the Galois group of K/Q and describethe structure of the Galois group K/F for each field F in part (a)of this problem.

We have seen in part (a) that the structure of the Galois group of K/Qis D8. From the Galois Correspondence, the structure of the Galois group ofK/F is given by the subgroup H which corresponds to K. We have alreadyseen these structures above. Hence the structure corresponding to Q(

√2)

and Q(i√

2) is K4. That of Q(i) is Z4. Lastly that of all the degree 4 exten-sions over Q (so that K has degree 2 over them) is Z2.

(c) Which of the fields in part (a) of this problem are Galois overQ? For each such field describe the structure of the Galois groupof F/Q.

From the lattice of subfields we see that neither Q( 4√

2) nor Q(i 4√

2) con-tains the other. But these two generators are roots of the same minimal

100 Spring 2002

polynomial over Q (x4−2) and so we may conclude that neither extension isGalois over Q. Likewise neither Q(

√i 4√

2) nor Q(i√i 4√

2) contains the otherand both generators are roots of x4 + 2. Hence neither of these extensionsare Galois over Q.

Next Q(i,√

2) is Galois over Q as the splitting field of (x2 + 1)(x2 − 2)(separability is not an issue since Q has characteristic zero). By the GaloisCorrespondence, the structure of the Galois group is H = G/ 〈σ2〉 (where〈σ2〉 is necessarily normal). Now any element of order 4 in H (rather thantwo or one) necessarily has to come from an element of order 4 in G. Butσ2 ∈ 〈σ2〉 and (σ3)2 ∈ 〈σ2〉 so that H has no element of order four. We mayconclude that H ∼= K4.

Lastly, Q(i), Q(√

2), and Q(i√

2) are Galois over Q as the splitting fieldsof x2 + 1, x2 − 2, and x2 + 2, respectively. Since they have degree 2 over Q,they must then have Galois group Z2.

2. (a) Let U and V be two subsets of a group G, and L a subgroupof G containing U . Show that

UV ∩ L = U(V ∩ L).

Let g ∈ UV ∩ L. Then g ∈ L and we can write g = uv with u ∈ U andv ∈ V . But u ∈ L so that u−1 ∈ L. This gives v = u−1uv ∈ L. Hencev ∈ V ∩ L and g = uv ∈ U(V ∩ L). This implies that UV ∩ L ⊂ U(V ∩ L).Likewise, suppose that g ∈ U(V ∩L) then g = uv with u ∈ U and v ∈ V ∩L.Since v ∈ V , g = uv ∈ UV . Likewise v ∈ L and u ∈ U ⊂ L so thatg = uv ∈ L. This gives the claim.

(b) Let G be a group and H,K be subgroups of G satisfying

H ∩K = 1, H ⊂ NG(K), and K ⊂ NG(H).

Show that every element of H commutes with every element of K.

101

Let h ∈ H and k ∈ K and consider the element hkh−1k−1. SinceH ⊂ NG(K), we have hkh−1 ∈ K so that hkh−1k−1 ∈ K. Likewise,K ⊂ NG(H), so that khk−1 ∈ H and g ∈ H. Thus we have shown thathkh−1k−1 ∈ H ∩K = 1 so that hkh−1k−1 = 1. This implies hk = kh andso we have the claim.

(c) Let G be the semidirect product of subgroups H and K suchthat H is normal in G. Recall that

NG(K) = g ∈ G : gKg−1 = K

andCH(K) = h ∈ h : hk = hk for all k ∈ K.

Show that NG(K) ∩ H commutes element-wise with K and thatNG(K) ∩H = CH(K). Deduce that NG(K) = KCH(K).

We have NG(K) ∩H ⊂ NG(K). Next we will show that

K ⊂ NG(NG(K) ∩H).

Let k, k′ ∈ K and h ∈ NG(K)∩H. Consider khk−1k′kh−1k−1. kk′k−1 ∈ K sothat hkk′k−1h−1 ∈ K. This shows that khk−1k′(khk−1)−1 = khk−1k′kh−1k−1 ∈K and so we may conclude that khk−1 ∈ NG(K). By normality, we havekhk−1 ∈ H. This gives the claim. Now (NG(K) ∩H) ∩K ⊂ H ∩K = 1.Hence we may conclude from part (b) that every element of NG(K) ∩ Hcommutes with every element of K.

Hence we have NG(K) ∩H ⊂ CH(K). Now, by definition, CH(K) ⊂ H.Certainly if an element commutes with each element of K, it preserves K,hence CH(K) ⊂ NG(K). This gives CH(K) = NG(K) ∩H. Lastly, we haveKCH(K) = K(NG(K) ∩H). But K ⊂ NG(K) so that we may apply (a) toconclude that K(NG(K) ∩H) = NG(K) ∩KH = NG(K) since KH = G.

3. Let I be an ideal in the commutative ring R. Define

rad I = r ∈ R : rn ∈ I for some positive integer n

102 Spring 2002

and Jac I to be the intersection of all maximal ideals that contain I.

(a) Prove that rad I and Jac I are ideals in R.

Jac I is an ideal as an intersection of ideals. Suppose then that x, y ∈rad I. Then we may find positive integers nx, ny such that xnx , yny ∈ I.Taking n to be the maximum of nx, ny, we have xn, yn ∈ I. Now in theexpansion of (x+y)2n, each term has either x or y to a power of at least n sothat we have (x+y)2n ∈ I. This gives x+y ∈ rad I. Next suppose that r ∈ R.Then (rx)n = rnxn ∈ I so that rx ∈ rad I. This shows that rad I is an ideal.

(b) Prove that I ⊂ rad I ⊂ Jac I.

If x ∈ I, x1 ∈ I so that x ∈ rad I. This shows the first containment.Let J be a maximal ideal which contains I. Suppose that y ∈ rad I. Thenyn ∈ I ⊂ J for some n. But J is maximal and therefore prime. We mayhence conclude that y ∈ J . This shows rad I ⊂ J . This gives rad I ⊂ Jac I.

(c) Let k be a field, R = k[x1, . . . , xn], and I = 〈f〉 the principal ideal ofR generated by f . If f = fa1

1 · · · famm is the factorization into a prod-

uct of distinct irreducible polynomials, show that rad I = 〈f1 · · · fm〉.

Let a = maxa1, . . . , am. Then we see that f divides (f1 · · · fm)a sothat f1 · · · fm ∈ rad I. Likewise, suppose that g ∈ rad I. Then we mayfind a positive integer n such that gn is divisible by f . Hence gn is divisibleby fi. But fi is irreducible and therefore prime. We may hence concludethat fi divides g. That is, fi appears in the prime factorization of g. Sincethis holds for each i, we may conclude that f1 · · · fm divides g. This givesrad I = 〈f1 · · · fm〉.

4. (a) Let R be a commutative ring with identity element 1R. LetM be an R-module and suppose that N1, . . . , NJ are R-submodulesof M . Prove that the following assertions are equivalent:

103

(i) The map

π : N1 ×N2 · · · ×NJ → N1 +N2 + · · ·+NJ

defined by π(a1, . . . , aJ) = a1 + · · · + aJ is an R-module isomor-phism.

(ii) Every element x ∈ N1 + · · ·+NJ can be written uniquely in theform x = a1 + · · ·+ aJ with ai ∈ Ni for each i = 1, . . . , J.

First we will show that the map in assertion (i) is always a surjectiveR-module homomorphism. Write N = N1×· · ·×Nj and L = N1 + · · ·+NJ .If (a1, . . . , aJ), (b1, . . . , bJ) ∈ N , we have

π((a1, . . . , aJ) + (b1, . . . , bJ)

)= π

(a1 + b1, . . . , aJ + bJ

)= a1 + b1 + · · ·+ aJ + bJ

= a1 + · · ·+ aJ + b1 + · · ·+ bJ

= π(a1, . . . , aJ) + π(b1, . . . , bJ).

Likewise, if r ∈ R, we have

π(r(a1, . . . , aJ)

)= π

(ra1, . . . , raJ

)= ra1 + · · ·+ raJ

= r(a1 + · · ·+ aJ)

= rπ(a1, . . . , aJ).

Hence π is an R-module homomorphism. Next suppose that a1+· · ·+aJ ∈ L.Then π(a1, . . . , aJ) = a1 + · · ·+ aJ . Thus π is surjective.

Now suppose that assertion (i) holds. Then π is injective. Since everyx ∈ L can be written in the form a1 + · · · + aJ with ai ∈ Ni, it suffices toshow that this expression is unique. Suppose than that x = b1 + · · · + bJfor bi ∈ Ni. Then π(a1, . . . , aJ) = x = π(b1, . . . , bJ) which shows that(a1, . . . , aJ) = (b1, . . . , bJ). That is, ai = bi and the expression is unique.Conversely suppose that (ii) holds and that (a1, . . . , aJ) ∈ kerπ. Thena1 + · · ·+ aJ = 0. But 0 = 0 + · · ·+ 0 with 0 ∈ Ni is another such expressionfor 0. Hence by uniqueness, we have ai = 0. This gives (a1, . . . , aJ) = 0 and

104 Spring 2002

so π is injective. By the above we have that π is an R-module isomorphism.

(b) Let T ∈M(3, Q) be the matrix 6 −2 −26 −1 −34 −2 0

.

Compute the characteristic polynomial, the minimal polynomial,the invariant factors, and the rational canonical form of T .

We have that det(xI−T ) = x3−5x2+8x−4 so that this is the character-istic polynomial of T . This polynomial factors to (x−2)2(x−1). We find that(T−I)(T−2I) is zero but T 6= 2I and T 6= I. Hence the minimal polynomialof T is (x− 1)(x− 2). Since this is the largest invariant factor of T and thecharacteristic polynomial is the product of all the invariant factors, we mayconclude that the invariant factors are (x−2) and (x−1)(x−2) = x2−3x+2.From this we may conclude that the rational canonical form of T is 2 0 0

0 0 −20 1 3

.

Chapter 12

Fall 2001

1. Suppose that k is a field of characteristic p > 0.

(i) If a is an element of k, show that the polynomial f(x) = xp − aeither splits in k[x] or is irreducible in f [x].

First suppose that f(x) has a root in k, α. Then αp = a so that(x − α)p = xp − αp = xp − a = f(x). Hence f(x) splits in k. Conversely,suppose that f(x) has no root in k. Let L be the slitting field of f(x) overK and let α be a root. Then as above f(x) = (x−α)p over L. Suppose thatwe can write f(x) = g(x)h(x) where g(x) and h(x) are nonconstant (monic)polynomials over k. Then g(x) = (x − α)i for some 1 ≤ i < p, (i < p sinceh(x) is nonconstant). In particular the constant term of g(x) is αi, whichmust then lie in k. But i and p are relatively prime, so we may find s, t ∈ Zsuch that si + tp = 1. This gives (αi)s = α/αtp = α/at. Since at ∈ k, thisimplies that α ∈ k, a contradiction. Hence f(x) is irreducible.

(ii) Let σ : k → k be the Frobenius endomorphism defined byσ(α) = αp. Assume that σ is an automorphism, that is, assumeσ is surjective. Prove that every irreducible polynomial in k[x] isseparable.

First suppose that f(x) is a polynomial over k and that f(x) has a re-peated root α in some extension, K, of k. Then over K, we have f(x) =(x − α)2f0(x) for some polynomial f0(x). Then f ′(x) = 2(x − α)f0(x) +(x − α)2f0(x) so that f ′(α) = 0. Hence α is a root both of f(x) and of

106 Fall 2001

f ′(x). Therefore, over k, the minimal polynomial of α divides both f(x) andf ′(x). In particular, they are not relatively prime. Hence to show that f(x)is separable, it suffices to show that it is relatively prime to its derivative.

Now let f(x) ∈ k[x] be irreducible. Then the degree of f ′(x) is strictly lessthan f(x). Hence f ′(x) is either 0, a constant polynomial, or a nonconstantpolynomial of degree less than f(x). Since f(x) is irreducible, the last twoconditions imply that the greatest common divisor of f(x) and f ′(x) is 1.Suppose then that f ′(x) = 0. Since the derivative of axk is (ak)xk−1 (andthe derivative preserves sums), this implies that each monomial in f(x) hasorder divisible by p. Hence we may write f(x) = anx

pn + an−1x(n−1)p + · · ·+

a0. Furthermore, we may find αi ∈ k such that αpi = ai, as the Frobeniusendomorphism is surjective. This gives

f(x) = anxpn + an−1x

(n−1)p + · · ·+ a0

=(αnx

n)p

+(αn−1x

(n−1))p

+ · · ·+(α0

)p=(αnx

n + αn−1x(n−1) + · · ·+ α0

)p,

which shows that f(x) is not irreducible, a contradiction. Hence f(x) isseparable.

2. Let G be a finite group.

(i) Suppose that G has order n and p is the smallest prime divisorof n. Show that any subgroup of index p is normal.

See Spring 2003 #1(i).

(ii) Suppose that G has order 255. Prove that G has a normal sub-group of order 17, and prove that G has a cyclic normal subgroupof order 85. From these facts deduce that any group of order 255is cyclic.

107

We have that |G| = 3 · 5 · 17. From the Sylow Theorems, we have thatthe number of Sylow 17 subgroups must divide 15 and be equivalent modulo17 to 1. Hence the number of Sylow 17-subgroups is 1 and there is a unique,and therefore normal, subgroup, N , of order 17. Next let H be a Sylow5-subgroup. Since N is normal, HN is a subgroup of order 85 (as the inter-section is trivial from order considerations). But since 17 is not equivalentto 1 modulo 5, H must be a normal subgroup of HN . Hence by the DirectProduct Recognition Theorem, we have that HN ∼= H×N ∼= Z85. Moreoverthe index of HN in G is 3, the smallest prime dividing its order. We mayconclude from (a) that HN is normal in G.

Now we haveG ∼= HNoK whereK is a Sylow 3-subgroup. Hence we havea map ψ : K → Aut(HN). The order of Aut(HN) is φ(85) = 16 · 4 = 256.But since 3 does not divide 256, any element of order 3 in K would have tobe mapped to the identity by ψ. From this, we may conclude that ψ is trivialand so G ∼= HN ×K ∼= Z255.

3. Let R be a principal ideal domain, M a finitely generated freeR-module and S ⊂ M a submodule. Show that the following areequivalent. (State clearly any theorems you use about modulesover a PID).

(a) The submodule S is complemented, that is, there exists asubmodule T ⊂M with

M ∼= S ⊕ T.

(b) The quotient module M/S is a free R-module.

(c) If x ∈ S and x = ay for some y ∈M , a ∈ R (a 6= 0), then y ∈ S.

First we will show the equivalence of the first two. If S has a complement,T , we have M/S ∼= (S ⊕ T )/S ∼= T (via the map induced by S ⊕ T →T , (s, t) 7→ t). Now T is a submodule of a free module and so free itself(from the Structure Theorem for Modules over a PID or by Fall 2005 #2(i)).

108 Fall 2001

Conversely, suppose that M/S is free and let α1 + I, . . . , αk + I be a basis.We will show that T = (α1, . . . , αk) is a complement to S in M . First supposethat x ∈M . Then we may find ri ∈ R such that r1α1+ . . .+rkαk+I = x+I.Then there is an s ∈ S such that x = r1α1 + . . .+ rkαk + s. This shows thatM = S+T . Next suppose that x ∈ S∩T . Then x = r1α1+. . .+rkαk for somer1, . . . , rk ∈ T . But then x+S = 0+S so that r1α1 + . . .+ rkαk +S = 0+S.By linear independence, we have ri = 0. This gives x = 0 and so we haveS ∩ T = 0 and so M = S ⊕ T .

Thus the first two conditions are equivalent. Suppose the second conditionholds. Suppose that x ∈ S and x = ay for some y ∈ M and a ∈ R. SinceM/S is free, it is torsion free. Thus ay = 0 in M/S implies y = 0 in M/S. Inother words, y ∈ S. Conversely, if the third condition holds, M/S is torsionfree. But it is finitely generated (by generators for M , for example) so that,by the Structure Theorem for Modules over a PID, it is free.

4. Let Q denote the field of rational numbers.

(i) Find a Galois extension of Q with Galois group isomorphic toZ/3Z (the cyclic group of order three).

Let ζ be a primitive 9th root of unity. Consider K = Q(ζ + ζ8). We cansee that ζ + ζ8 is a root of f(x) = x3 − 3x+ 1. Hence K has at most degree3 over Q. Moreover, the rational roots test gives only ±1 as possible rootsof f(x). Since neither is a root and f(x) is cubic, we may conclude that it isirreducible. Hence K has degree 3 over Q. Lastly, K is an intermediate fieldbetween Q and Q(ζ), which has abelian Galois group (Z/9Z)∗ and so K isGalois over Q. Thus the Galois group of K over Q is Z/Z3.

(ii) Find a Galois extension of Q with Galois group isomorphic toZ/3Z× Z/3Z.1

1The outline of how to do this for any abelian group is given in Dummit and Foote, p.600

109

Let ξ be a primitive 7th root of unity. We will show that Q(ξ)∩Q(ζ) = Q.ξζ is a primitive 63rd degree of unity so that Q(ξζ) has degree φ(63) over Qand Q(ξζ) ⊂ Q(ξ)Q(ζ). Since this latter composite automatically has degreeless than or equal to the product of the two degree of the fields which compriseit, we have [Q(ζ)Q(ξ) : Q] = φ(63). But the Galois group of Q(ζ)Q(ξ) overQ is isomorphic to a subgroup of Gal(Q(ξ)/Q) × Gal(Q(ζ)/Q), a group oforder 6 · 6 = 36 = φ(63). Hence we may conclude that it is entire group.This implies the claim.

Now consider the extension F = Q(ξ+ ξ6). As above, we find that ξ+ ξ6

is a root of g(x) = x3 + x2 − 2x − 1, which by the rational roots test isirreducible. Hence F has degree three over Q and is an intermediate field inthe abelian extension Q(ξ) over Q. We may conclude that F is Galois overQ with Galois group Z/3Z. Now F ∩ K is contained in Q(ξ) ∩ Q(ζ) andso must equal Q. Thus Gal(FK/Q) is the direct product of Gal(F/Q) andGal(K/Q) and so it isomorphic to Z/3Z×Z/3Z. FK is the splitting field ofthe polynomial

(x3 + x2 − 2x− 1)(x3 − 3x+ 1) = x6 + x5 − 5x4 − 3x3 + 7x2 + x− 1.

110 Fall 2001

Chapter 13

Spring 2001

1. Let |S| denote the cardinality of a finite set S. Let G be a finitegroup, and let C a be subgroup of G.

(a) Suppose K is a normal subgroup of G. Suppose p is a primenumber with DOES divide |C| but does NOT divide |C ∩K|. Showp DOES divide |G/K|.

Since K is normal, we may apply the Second Isomorphism Theorem toconclude that CK is a subgroup and CK/K ∼= C/(C ∩ K). In particular,we have [CK : K] = |C|/|C ∩ K|. Since p divides the top, but not thebottom of this fraction, we may conclude that p divides [CK : K]. Then[G : K] = [G : CK][CK : K] is divisible by p.

(b) Let G′ be the commutator subgroup of G. Suppose p is a primenumber which divides |C| but does not divide |C ∩ G′|. Show thatG has a subgroup of index p.

By (i), p divides |G/G′|. Writing |G/G′| = np, and noting that G/G′ isabelian we may find a subgroup K of G/G′ of order n (the fact the everyfinite abelian group has a subgroup of any order dividing the order of thegroup is proved in Fall 2005 #1(ii)). By the Correspondence Theorem, Ghas a subgroup H (which contains G′) such that K = H/G′. This gives[G : G′] = [G : H][H : G′] or np = [G : H]n. We may conclude that H hasindex p.

112 Spring 2001

(c) Now let K be a normal subgroup of G and let p be a primenumber which divides |K|. Let Sp(G) (respectively Sp(K)) denotethe set of Sylow p-subgroups of G (respectively, K). Show

Sp(K) = P ∩K|P ∈ Sp(G).

See Fall 2005 #1(iii).

(d) Show |Sp(K)| divides |Sp(G)|.

See Fall 2005 #1(iv).

2. Let Z[x] denote the ring of polynomials in one variable withinteger coefficients.

(a) Describe the maximal ideals in Z[x] by giving generators for themaximal ideals, and explain why your answer is correct.

Let I be a maximal ideal in Z[x] and consider I ∩ Z. If ab ∈ I ∩ Z, wehave ab ∈ I so that either a ∈ I ∩ Z or b ∈ I ∩ Z. Likewise, a, b ∈ I ∩ Z andn ∈ Z implies that a + b ∈ I ∩ Z and na ∈ I ∩ Z. Hence I ∩ Z is a primeideal in Z and so most be equal to (p) for some number p which is zero or aprime (I ∩ Z 6= (1) as this would imply I = Z[x]).

First suppose that p = 0 let f(x) be a polynomial in I. Since Z[x] isa UFD, we may factor f(x) into irreducibles. But since I is prime, one ofthe irreducibles must lie in I. Since I ∩ Z = 0, this irreducible must benonconstant. Hence we may find an irreducible nonconstant f(x) ∈ I. Nowsuppose that g(x) ∈ I and f(x) 6 |g(x). By factoring out the content of g(x),we may assume that g(x) has content 1. Now suppose g(x) = f(x)q(x) inQ[x]. Clearing denominators, we may find n ∈ Z such that nf(x) = g(x)q1(x)for some q1(x) ∈ Z[x]. But then the content of q1(x) must be n. Thisgives q1(x)/n ∈ Z[x] so that f(x) = g(x)q1(x)/n in Z[x], contradicting ourassumption. Thus f(x) does not divide g(x) in Q[x] which implies that their

113

greatest common divisor over Q[x] is 1 (as f(x) is irreducible over Z[x] andso is irreducible over Q[x]). Thus we may find s(x), t(x) ∈ Q[x] such thats(x)f(x) + t(x)g(x) = 1. Now clearing denominators, we may find N ∈ Z(nonzero) such that Ns(x), Nt(x) ∈ Z[x]. This gives N = Ns(x)f(x) +Nt(x)g(x) ∈ I, a contradiction. Hence we have shown I = (f(x)).

Now, viewing f(x) as an element of Q[x], there are only finitely manyk ∈ Z such that f(k) = 0. Likewise there are only finitely many k ∈ Z suchthat f(k) = ±1 (as these are zeros of f(x)∓1). Hence we may find k ∈ Z andprime p such that p divides f(k). Consider the ideal (I, p). By assumptionthis ideal strictly contains I. Now suppose for a contradiction that (I, p) =Z[x]. Then we may find g(x), h(x) ∈ Z[x] such that pg(x) + f(x)h(x) = 1.This gives pg(k) + f(k)h(k) = 1. But now p divides the left hand side andnot the right hand side, a contradiction. Hence (p, I) is proper which showsthat I is not a maximal ideal.

Hence we must have I ∩ Z = (p) where p is a prime. Then we may con-sider Z[x]/(p) ∼= Zp[x]. Zp[x] is a PID and so its maximal ideals are exactlyideals of the form (f(x)) where f(x) is irreducible in Zp(x). Hence applyingthe Correspondence Theorem we see that the maximal ideals in I must beof the form (p, f(x)) where f(x) is irreducible modulo p (certainly maximalideals correspond to maximal ideals under the correspondence in the Cor-respondence Theorem by considering their characterization as ideals whosequotients are fields).

(b) If R is a commutative ring with identity element 1, recall thatan ideal P of R with P 6= R is called a PRIME IDEAL if for everya, b ∈ R,

ab ∈ P ⇒ a ∈ P or b ∈ P.

Show that the ideal (3, x2 + x + 1) in Z[x] is not a prime ideal byexhibiting f(x), g(x) ∈ Z[x] with f(x)g(x) ∈ (3, x2 + x + 1) but withf(x) 6∈ (3, x2 + x+ 1) and g(x) 6∈ (2, x2 + x+ 1).

We have (x− 1)2 = x2 − 2x + 1 = (x2 + x + 1)− 3(x) ∈ (3, x2 + x + 1).Suppose that (x − 1) ∈ (3, x2 + x + 1). Then we have g(x), h(x) ∈ Z[x]with (x − 1) = 3g(x) + (x2 + x + 1)h(x). Reducing this equation modulo3, we see that (x − 1) = (x2 + x + 1)h(x) so that x2 + x + 1 divides x − 1in Z3[x]. But degree considerations render this impossible. Hence we that(x− 1) 6∈ (3, x2 + x+ 1) which shows that (3, x2 + x+ 1) is not prime.

114 Spring 2001

(c) Is every nonzero prime ideal in Z[x] a maximal ideal ideal?If yes prove it. If not give an example (with proof) of a nonzeroprime ideal which is not maximal.

Consider the ideal I = (x−1) in Z[x]. Since I∩Z = 0, we have alreadyseen above that I is not maximal. However, (x − 1) is irreducible in Z[x](as a polynomial of degree one) so that, since Z[x] is UFD, it is prime. Thisimplies that I is a prime ideal and so the answer is no.

3. Let f(x) be a separable irreducible monic polynomial over afield F , and let K be the splitting field of f(x) over F . Supposedeg f(x) = p, where p is a prime number. Let G be the Galois groupof K over F . Let P be a Sylow p-subgroup of G. Let N be thenormalizer of P in G. SUPPOSE P = N . (The goal of this problemis to show P = G.)

(a) Show that |G| = ps with s an integer not divisible by p.

See Fall 2005 #4(i).

(b) With s as in part (a), show there are exactly s(p− 1) elementsof order p in G.

See Fall 2005 #4(ii) and (iii).

(c) Now let α and β be roots of f(x) in K. Let Hα be the Ga-lois group of K over F [α], and let Hβ be the the Galois groupof K over K[β]. Show that G is the disjoint union of Hα andσ ∈ G|the order of σ is p.

See Fall 2005 #4(iv) and (v).

(d) Show Hβ = Hα and F [α] = F [β].

115

See Fall 2005 #4(vi) and (vii).

(e) Show K = F [α], and conclude P = G.

See Fall 2005 #4(vii).

4. (a) Let Cn×n denote the set of n× n matrices with entries in C,and let A ∈ Cn×n and B ∈ Cn×n. Prove there exists a nonsingularT ∈ Cn×n such that TAT−1 and TBT−1 are diagonal (i.e. A and Bcan be diagonalized simultaneously) if and only if AB = BA.

Note that we most first assume that A and B are diagonalizable. Withoutthis assumption, we could take A to be any matrix which is not diagonalizableand let B be the n × n identity. Under these circumstances, A and B willcertainly commute (AB = A = BA), but to find such a T , we would haveTAT−1 diagonal, a contradiction.

First we will prove a small lemma, suppose that A and B commute. thenB preserves the eigenspaces of A. To see this suppose that λ is an eigenvalueof A with eigenspace E. Then we have A(Bv) = B(Av) = B(λv) = λBv,which proves the claim.

Now suppose that A and B commute. We will proceed by induction onn. Certainly if n = 1, every matrix is already diagonal. Hence suppose thatA and B are n× n matrices, the statement having been shown for all k < n.We have two cases. Suppose first that both A and B each have a singleeigenvalues (which of course most lie in C as it is algebraically closed). Thenthe diagonal form of A is λAI for some λA ∈ C. But then for any invertiblematrix T we have TAT−1 = λATIT

−1 = λAI. In particular A is diagonal.Likewise B is diagonal. Hence A and B are simultaneously diagonalizable(by the identity, or by any invertible matrix).

Hence we may assume that one of A orB has at least two distinct eigenval-ues. Without loss of generality, assume that A has two distinct eigenvalues.By picking a basis for Cn, view A and B as linear transformations. Let E bethe eigenspace of one of the eigenvalues, λ of A. Let W be the complement,

116 Spring 2001

that is, let W be a subspace of V with V = E⊕W . Then W is a direct sumof the other eigenspaces of A (since A has a basis of eigenvectors). Since Apreserves E, we may restrict A to AE : E → E and still have a well-definedlinear transformation. Likewise, we may restrict A to AW : W → W . Like-wise we have BE : E → E and BW : W → W . Now BW and AW commuteon W and since dimW < dimV , we may apply the induction hypothesisto find a basis of W with respect to which BW and AW are both diagonal.Likewise, we may a basis of E with respect to which both AE and BE arediagonal. The union of these two bases is a basis for V (this is proved inFall 2003 #6 c)). A and B will be diagonal with respect to this basis. Thisis equivalent to saying there is an invertible matric T such that TAT−1 andTBT−1 are both diagonal (viewing A and B as matrices).

Conversely, suppose that such a T exists. Then we have TAT−1 is diag-onal and TBT−1 is diagonal. The product of two diagonal matrices, how-ever, is a diagonal matrix whose entries are the product of the two orig-inal entries. Hence, diagonal matrices commute and we have TABT−1 =TAT−1TBT−1 = TBT−1TAT−1 = TBAT−1. Multiplying on the right by Tand the left by T−1 gives AB = BA.

(b) Let k be a field, and let A ∈ kn×n be nonsingular. Prove thatthere exists a polynomial q(x) ∈ k[x] such that A−1 = q(A).

Suppose for a contradiction that x divides the minimal polynomial of A.Then x divides the characteristic polynomial of A. But the characteristicpolynomial of A is det(xI − A) and the assumption that x divides meansthat x = 0 is a zero. This gives det(−A) = 0. So that (−1)n det(A) = 0 anddet(A) = 0, a contradiction to the assumption that A is nonsingular. Hencethe minimal polynomial of A has the form xn + an−1x

n−1 + · · · + a0 wherea0 6= 0. This implies that An+ · · ·+a0 = 0 so that (−An−· · ·−a1A)/a0 = I.Then A(−An−1 − · · · − a1)/a0 = I and so

q(x) = − 1

a0

xn−1 − · · · − a1

a0

is such a polynomial.

Chapter 14

Fall 2000

1. a) Let G be a group of order 12. Show that either the 2-Sylowsubgroup or the 3-Sylow subgroup is normal.

Let n2 and n3 be the number or 2-Sylow and 3-Sylow subgroups, respec-tively. From the Sylow Theorems, n2 divides 3 and so equals 1 or 3. Likewise,n3 divides 4 and is equivalent to 1 modulo 3. Hence it must equal 1 or 4.Assume for a contradiction that G has no normal Sylow subgroup. Thenn2 = 3. Pick a Sylow 2-subgroup, H. Then H has 3 elements of order 2 or4. Each of the remaining 2 Sylow 2-subgroups must be distinct from H andso we have two more elements of order 2 or 4. Hence G contains at least 5elements of order 2 or 4. Likewise all the Sylow 3-subgroups are isomorphicto Z3 so that they are generated by any of their elements other than the iden-tity. Since they are distinct, they then intersect only at the identity. Thisgives 2 · 4 = 8 elements of order 3. Hence we have shown |G| ≥ 8 + 5 = 13,a contradiction.

b) For the rest of this problem, suppose G is a group of order 24with neither the 2-Sylow subgroup nor the 3-Sylow subgroup nor-mal. Show that G has a non-normal subgroup, H, of order 6.

Suppose first for a contradiction that G has a subgroupH of order 6 whichis normal. From they Sylow Theorems, H has a normal and therefore uniquesubgroup, K, of order 3. Let g ∈ G. Then from normality, gKg−1 < H andsince gKg−1 has order 3, we may conclude that gKg−1 = K. Hence K isa normal subgroup of G of order 3, a contradiction. Thus G cannot have a

118 Fall 2000

normal subgroup of order 6.Now we will show that G has a subgroup of order 6. This combined with

the argument above will prove the statement. Let n3 be the number of 3-Sylow subgroups of G. From the Sylow Theorems, n3 must equal 1 or 4. Byassumption, n3 6= 1 so that n3 = 4. Now G acts transitively on the collectionof Sylow 3-subgroups by conjugation. Since the order of this set is 4, theorder of a stabilizer of a Sylow 3-subgroup must be 24/4 = 6. Hence G hasa subgroup, H, of order 6.

c) Show that the subgroup H of part b) contains no subgroup nor-mal in G except the identity.

Suppose for a contradiction that K < G is a normal subgroup of order 2.Then G/K is a subgroup of order 12. By a), it must have either a normalsubgroup of order 4 or one of order 3. By the Correspondence Theorem,G then has a normal subgroup of order either 8 or 6. The former casecontradicts our assumption and the latter contradicts the argument given inb).

A proper nontrivial subgroup of H must be of order 2 or 3. We haveassumed that G has no normal subgroup of order 3. The above shows that Ghas no normal subgroup of order 2. Hence H contains no normal subgroup(including itself) other than the 1.

d) Show that G is isomorphic to S4.

Again G acts on the collection of 3-Sylow subgroups which has order 4.This gives a homomorphism φ : G → S4. Now, H is the stabilizer of aparticular 3-Sylow subgroup, hence if g ∈ G acts trivially on each 3-Sylowsubgroup, we have g ∈ H. This implies that ker(φ) ⊂ H. But ker(φ) isnormal in G and so we may conclude from c) that ker(φ) = 1. By orderconsiderations, we may conclude that φ is an isomorphism so that G ∼= S4.

2. Let R be a commutative integral domain with the two propertiesi) and ii) below.

119

i) Every nonzero ideal I can be written uniquely in the formI = P e1

1 · · ·P enn , where all the Pi are prime ideals and the ei

are positive integers. Of course, “unique” means only up toorder.

ii) If I ⊂ J are two ideals then there is an ideal K such thatJK = I.

In the discussion, for convenience, we will also use the factoriza-tions where ei = 0. Of course P 0 = R and IR = I for all I and P .

a) If I = P e11 · · ·P en

n and J = P f11 · · ·P fn

n , state a necessary and suffi-cient condition on the ei, fi for I ⊂ J to hold.

The condition is ei ≥ fi. Suppose first that the condition holds. Thenevery element x of I can be written in the form

x =n∏i=1

ei∏j=1

a(i)j

for a(i)j ∈ Pi. But then since fi ≤ ei, we have that x is divisible

y =n∏i=1

fi∏j=1

a(i)j ,

which is an element of J by definition. Hence we have x = ry for r ∈ R.This gives x ∈ J so that I ⊂ J .

Conversely, suppose that I ⊂ J . Then from property ii), we may find anideal K ⊂ R such that JK = I. If P is a prime ideal which appears (non-trivially) in the factorization of K, it must then appear in the factorizationof I. Hence the factorization of K can be written in the form P d1

1 · · ·P dnn

(for di ≥ 0). Then I = JK (and the fact that R is commutative) impliesei = fi + di so that ei ≥ fi.

b) Show that R is Noetherian.

Let I1 ⊂ I2 ⊂ · · · be an increasing sequence of ideals. As above, anyprime ideal which appears nontrivially in the factorization of Ij for any j

120 Fall 2000

must appear in that of I1. Let

Pe(1)i

1 · · ·P e(1)n

n

be the factorization of I1. Then the factorization of Ij can be written as

Pe(j)i

1 · · ·P e(j)n

n .

From a), e(j)i j∈N is a a decreasing sequence of positive integers and so itmust stabilize at some mi. Letting M = maxm1, . . . ,mn, we see thatIM = Ij for j ≥ N .

c) If P ⊂ Q are prime ideals, show that P = Q.

Note that any PID is an example of a ring which satisfies properties i)and ii). In particular Z is such a ring. Note that (0) ⊂ (2) are two primeideals of Z but that (0) 6= (2). Hence we must add the assumption thatP 6= 0. If Q = 0, we have trivially that P = 0. Hence we may assumethat both P and Q are nonzero so that they have factorizations.

Now the prime factorization of Q is Q so that Q must appear nontriviallyin the prime factorization of P . But the prime factorization of P is P andso we have P = Q.

d) If x, y ∈ R, let (x), (y), and (x, y) be the ideals generated byx, y, and the set x, y respectively. Suppose (x) = P e1

1 · · ·P enn and

(y) = P f11 · · ·P fn

n . Show that (x, y) = P g11 · · ·P gn

n where gi is the mini-mum of ei and fi.

Since (x) ⊂ (x, y) we have that any prime which appears nontrivially inthe factorization of (x, y) must appear in that for (x). Hence the factorizationof (x, y) has the claimed form. Now, (x) ⊂ (x, y) implies that gi ≤ ei by a).Likewise, gi ≤ fi. Hence gi is less than or equal to the minimum of ei andfi. Let

z ∈ (x, y) = (x) + (y) = P e11 · · ·P en

n + P f11 · · ·P fn

n .

Then z can be written as

n∏i=1

ei∏j=1

a(i)j +

n∏i=1

fi∏j=1

b(i)j ,

121

where a(i)j , b

(i)j ∈ Pi. Since R is commutative, both of the above products lie

inP

mine1,f11 · · ·Pminen,fn

n ,

which shows that z does as well. Thus (x, y) is contained in the ideal above.This shows that gi ≥ minei, fi and gives the result.

e) Suppose x 6∈ (y). Show that x2 6∈ y(x, y).

Let(x) = P e1

1 · · ·P enn and (y) = P f1

1 · · ·P fnn .

Since x 6∈ (y), we have (x) 6⊂ (y) so that ei < fi for some i by a). By reorder-ing, we may assume that e1 < f1. By commutativity, y(x, y) = (y)(x, y) and,by d), the factorization of the latter is

P g1+f11 · · ·P gn+fn

n ,

where gi = minei, fi. Since e1 < f1, g1 = e1 and so the power of P1 in thefactorization of (y)(x, y) is e1 + f1. The factorization of (x2) = (x)2 is

P 2e11 · · ·P 2en

n .

Since 2e2 < e1 + f1, we may conclude from a) that (x2) 6⊂ y(x, y). Thisimplies that x2 6∈ y(x, y), as claimed.

3. Let R be a not necessarily commutative ring which is a finite di-mensional algebra over the complex field C. Let J be the Jacobsonradical of R and assume that R/J is a commutative ring. Recallthat an irreducible (or simple) module M is a nonzero module suchthat all submodules are (0) or M .

a) Show that R/J has the form C⊕ · · · ⊕ C as a ring.

Since R is a C-algebra, we have a ring homomorphism C → R given byz 7→ z ·1R. Since C is a field and this is not the zero map (1·1R = 1R), we may

122 Fall 2000

conclude that it is an injection. Hence C is a subring of R. Since J is a twosided ideal, rx ∈ J for all r ∈ R, x ∈ J . In particular, zx ∈ J for all x ∈ Jand z ∈ C. Hence J is a C-subspace of V . Then R/J is a finite dimensionalC vector space as the quotient of a finite dimensional vector space. By thesame argument as above C is a subring of R/J . Suppose that I1 ⊃ I2 ⊃ . . .is a sequence of left ideals in R/J . Then, it is in particular, a sequence ofC-subspaces of R/J so that it must stabilize by dimension considerations.This implies that R/J is left Artinian. Moreover the Jacobson Radical ofR/J is zero. Hence we have that R/J is left semi-simple.

Hence by the Wedderburn Theorem, R/J , as a C-algebra, is isomorphicto a (finite) direct sum of matrix rings over division rings. But R/J is com-mutative and so each of the matrix rings must be 1× 1 (as the collection ofn × n matrices over any division ring is noncommutative for n ≥ 2) and allthe division rings must be fields. Hence we may write R/J ∼= L1⊕· · ·⊕Ln forsome n and fields Li. But now C is (isomorphic to) a subring of L1 ⊕ · · ·Lnso that each Li, as an ideal, is also a finite dimensional vector space over Cand so (as above) we have a map C → L which is injective. This impliesthat Li is a finite extension of C. In particular Li is an algebraic extensionof C. Hence if α ∈ L, we may find a polynomial over C to which α is a root.But C is algebraically closed and so this implies that α ∈ C so that Li = C.This gives the result.

b) Suppose that V is an irreducible left R module. Show thatJV = 0.

Since V is irreducible, V 6= 0 and V has no proper nonzero R-modules.Let a ∈ V be nonzero. Then Ra is a nonzero R-submodule of V so that wemust have Ra = V . This shows that V is finitely generated. Moreover, JV isa submodule of V and so it must equal V or 0. In the former cause, however,we may apply Nakayama’s Lemma to conclude that V = 0, a contradiction.Hence we have JV = 0.

c) Suppose V is as above. Show that V is one dimensional over C.

We have seen that R/J ∼= C⊕· · ·⊕C in (a). Hence, every simple left idealin R/J is isomorphic to C. From the characterization of simple modules overa semi-simple ring, this shows that every simple (that is, irreducible) module,L, over R/J is isomorphic to C as an R/J-module, where the multiplication

123

is given by (z1, . . . , zn) · z = z1z (in general it is a fixed factor that acts, butsince they are all the same we may as well assume that it is the first). SinceC viewed as a subring of R/J as above is (z, . . . , z), we see that the C-module structure on L is essentially the same as the R/J-module structure.Hence every simple R/J-module is isomorphic as a C-module to C. That is,it is one-dimensional over C.

Now V is a simple R-module. We can give V an R/J-module structureby (r + J) · v = rv since JV = 0, this is well-defined. Moreover, let Mbe an R/J-submodule of V . Then (r + J)M ⊂ M for all r ∈ R. But bydefinition, this gives rM ⊂M so that, since M is automatically closed underaddition, we have that M is an R-submodule of V . Hence V is simple as anR/J-module. From this we may conclude that V is one-dimensional over C.

d) Suppose M is any left R module finite dimensional over C. Showthat there is a sequence of submodules 0 = M0 ⊂M1 ⊂ . . . ⊂Mr = Msuch that Mi+1/Mi is one dimensional over C.

Let n the dimension of M over C. Suppose that N and L are R-moduleswith N ⊂ L and N 6= L. Then N and L are also C-vector spaces and so wemust have dimC L > dimCN . Hence if N1 ⊂ N2 ⊂ . . . is a strictly increasingsequence of R-submodules of M , dimCNi is a strictly increasing sequence.Hence such a sequence can have at most n+ 1 terms.

We will describe a process for finding such a sequence. Consider first thesequence 0 ⊂ M . Find an R-module M1 such that 0 ⊂ M1 ⊂ M (and thecontainments are strict). If no such M1 can be found, terminate the processand let 0 ⊂M be the final sequence. Suppose now that we have the sequence0 = M0 ⊂ M1 ⊂ . . . ⊂ Mk−1 ⊂ Mk = M . Find a submodule N (strictly)between Mi and Mi+1 for the smallest value of i for which this is possible.If it is not possible for any i, terminate the process with the above sequenceas the final sequence. If it is possible for some i, add this submodule to thesequence, renumber and repeat step 2. We are guaranteed that the processwill terminate since such a sequence can have at most n+ 1 modules by theabove.

Hence we have a sequence 0 = M0 ⊂M1 ⊂ . . . ⊂Mr = M such that thereare no R-submodules strictly between Mi and Mi+1. Hence, by the Corre-spondence Theorem, Mi+1/Mi is simple and so by c), it is one-dimensionalover C.

124 Fall 2000

e) Let M be a finitely generated faithful left R module. Show thatthere is an injection φ : R → EndC(M). Show that M has a basissuch that, with respect to this basis, the matrices correspondingto all φ(r) are in upper triangular form.

First we will show that M is finite dimensional over C. Since M is finitegenerated over R, we may find m1, . . .mk ∈ M such that each m ∈ M isR an R-linear combination of the mi. Likewise, R is finite dimensional overC and so we can find r1, . . . , rl ∈ R such that every r ∈ R is a C linearcombination of the ri. Combining these two, we see that each m ∈ M is aC-linear combination of rimj so that M is finitely generated over C. This,of course, implies that it is finite dimensional.

Now, let φ : R → EndC(M) be given by r 7→ fr where fr(m) = rm. Forz ∈ C, m,n ∈M , we have fr(zm) = rzm = zrm = zfr(m) and fr(m+ n) =r(m+n) = rm+rn = fr(m)+fr(n) so that φ is well-defined. Next if r, s ∈ Rand m ∈M , we have fr+s(m) = (r+s)m = rm+sm = fr(m)+fs(m) whichimplies that φ(r + s) = φ(r) + φ(s). Likewise, frs(m) = rsm = fr fs(m)so that φ(rs) = φ(r)φ(s). Finally, for z ∈ C, we have fz(m) = zm so thatφ(z) = z, where we view C as a subring of EndC(M) in the usual way. Henceφ is a C-algebra homomorphism. Suppose now that r ∈ R is mapped to zero.Then rm = 0 for all m ∈ M . Since M is a faithful R-module, this impliesthat r = 0 and we see that φ is injective.

By d), we have a sequence 0 = M0 ⊂ M1 ⊂ . . . ⊂ Mr = M such thatMi+1/Mi is one dimensional over C. Pick a elements b1, . . . , br ∈ M suchthat bi ∈ Mi −Mi−1. We will show by induction that b1, . . . , bi generatesMi over C. Since M1 is dimensional over C and b1 6= 0, we have that b1generates M1 over C. Suppose then that the statement holds for some i.Then bi+1 + Mi is not zero in Mi+1/Mi. Hence bi+1 + Mi must generateMi+1/Mi over C. That is, if mi+1 ∈ Mi, we can write mi+1 = ai+1bi+1 +Mi

for some ai+1 ∈ C. But then mi+1 = ai+1bi+1 +mi for some mi ∈Mi, which,by the induction hypothesis, is a linear combination over C of b1, . . . , bi.This gives the claim. In particular b1, . . . , br generates M .

Suppose now that a1b1+· · ·+arbr = 0 for some ai ∈ C. Suppose that thereis some index i such that ai is nonzero and let j be the largest such index.Then we have bj = (1/aj)(a1b1 + · · · + aj−1bj−1) ∈ Mj−1, a contradiction.Hence we have shown that b1, . . . , br is a basis for M . This also showsthat b1, . . . , bi is linearly independent for each i so that it is a basis forMi. Now let r ∈ R. Then, since Mi is an R-submodule of M , we have

125

φ(r)(bi) = rbi ∈Mi and so φ(r)(bi) is a linear combination of b1, . . . , bi. Thisshows that with respect to our basis, the matrix of φ(r) is upper triangular.

4. Let F be the field of q elements where q is prime. Set E = F (x)to be the field of fractions of the polynomial ring F [x]. If 0 6= a ∈ F ,and b ∈ F , define sa,b : F [x] → F [x] by setting sa,b(f(x)) = f(ax+ b).

a) Show that the sa,b extend to automorphisms of E and the set ofall sa,b form a subgroup, G, of the automorphism group of E. Showthat s1,b|b ∈ F forms a normal cyclic subgroup of order q.

For, a, b ∈ F , a 6= 0, write (abusing notation) sa,b : E → E as f(x)/g(x) 7→f(ax+b)/g(ax+b). We see that sa,b is well-defined since f(x)k(x) = h(x)g(x)implies f(ax+b)k(ax+b) = h(ax+b)g(ax+b) for f(x), g(x), h(x), k(x) ∈ F [x](as evaluation of polynomials is a homomorphism). Viewing F [x] as a subsetof E in the usual way, we can see that this is an extension of the above.Again if f(x), g(x), h(x), k(x) ∈ F [x], we have

sa,b

(f(x)

g(x)+h(x)

k(x)

)= sa,b

(f(x)k(x) + h(x)g(x)

g(x)k(x)

)=f(ax+ b)k(ax+ b) + h(ax+ b)g(ax+ b)

g(ax+ b)k(ax+ b)

=f(ax+ b)

g(ax+ b)+h(ax+ b)

k(ax+ b)

= sa,b

(f(x)

g(x)

)+ sa,b

(h(x)

k(x)

)and

126 Fall 2000

sa,b

(f(x)

g(x)· h(x)k(x)

)= sa,b

(f(x)h(x)

g(x)k(x)

)=f(ax+ b)h(ax+ b)

g(ax+ b)k(ax+ b)

=f(ax+ b)

g(ax+ b)· h(ax+ b)

k(ax+ b)

= sa,b

(f(x)

g(x)

)· sa,b

(h(x)

k(x)

),

since evaluation of polynomial is a homomorphism. Hence sa,b is a homo-morphism. Now consider sa−1,−ba−1 . For f(x), g(x) ∈ F [x], we have

sa−1,−ba−1 sa,b(f(x)

g(x)

)= sa−1,−ba−1

(f(ax+ b)

g(ax+ b)

)

=f(a(a−1x− ba−1) + b

)g(a(a−1x− ba−1) + b

)=f(x)

g(x)

and

sa,b sa−1,−ba−1

(f(x)

g(x)

)= sa,b

(f(a−1x− ba−1)

g(a−1x− ba−1)

)

=f(a−1(ax+ b)− ba−1

)g(a−1(ax+ b)− ba−1

)=f(x)

g(x),

which shows that sa,b and sa−1,−ba−1 are inverses so that sa,b is an automor-phism of E.

Let G be the collection of such automorphisms. We have already shown

127

that s ∈ G implies s−1 ∈ G. Now let sa,b, sc,d ∈ G. Then we have

sa,b sc,d(f(x)

g(x)

)= sa,b

(f(cx+ d)

g(cx+ d)

)

=f(c(ax+ b) + d

)g(c(ax+ b) + d

)=f(acx+ bc+ d)

g(acx+ bc+ d)

= sac,bc+d

(f(x)

g(x)

),

which shows that sa,b sc,d = sac,bc+d ∈ G. Thus G is a subgroup of theautomorphism group of E. Next suppose that sa,b = sc,d. Then in particular,sa,b(x) = sc,d(x) so that ax + b = cx + d. This shows that a = c and b = dand so we may conclude that the order of G is q(q − 1).

Finally let N = s1,b|b ∈ F. Then if sa,c ∈ G, we have

sa,c s1,b s−1a,c = sa,c+b sa−1,−ca−1

= s1,ba−1

∈ N.

This shows that N CG. Since s1,b 6= s1,c for b 6= c, we have that N has orderq and so is cyclic.

b) Set K = f ∈ E|sa,b(f) = f for all 0 6= a, b ∈ F . Show that E/K isa Galois extension of degree q(q − 1).

Consider the polynomial over E given by

m(y) =∏g∈G

(y − gx).

For h ∈ G, we have

hm(y) =∏g∈G

(y − (hg)x) =∏g∈G

(y − gx),

so that m(y) ∈ F [y]. Now, the roots of m(y) are ax + b for a, b ∈ F , a 6= 0.This shows that m(y) is separable and that E contains all the roots of m(y).

128 Fall 2000

If E ′ is the splitting field of m(y) over K, we have E ′ ⊂ E. Note also thatF ⊂ K ⊂ E ′ and x ∈ E ′. Hence E ′ contains F (x) = E. Thus E ′ = E andE is the splitting field of m(y), a separable polynomial, over K. This showsthat E is Galois over K.

Furthermore, the degree of m(y) is q(q − 1) which shows that the degreeof x is at most q(q − 1) over K. Since E is generated by x over K, thisimplies [E : K] ≤ q(q − 1). But the G is contained in the Galois group of Eover K so that [E : K] ≥ q(q − 1). Hence we have the claim. Note that wealso have shown that G is the Galois group of E over K.

c) Show that (xq − x)q−1 ∈ K. Show that K = F ((xq − x)q−1).

Let a, b ∈ F with a 6= 0. Then

sa,b

((xq − x)q−1

)=((ax+ b)q − (ax+ b)

)q−1

=(aqxq + bq − ax− b

)q−1

=(ax(aq−1xq−1 − 1)q−1 + (bq − b)

)q−1

=(ax(xq−1 − 1)

)q−1

= aq−1(x(xq−1 − 1)

)q−1

= (xq − x)q−1,

where we have used the facts that α 7→ αq is a homomorphism in character-istic q, α ∈ F× implies αq−1 = 1, and α ∈ F implies αq = α. This showsthat (xq − x)q−1 ∈ K.

Hence, writing β = (xq − x)q−1, we have a tower F (β) ⊂ K ⊂ E and somay conclude that

[E : F (β)] = [E : K][K : F (β)].

From the above, [E : K] = q(q − 1). Moreover, E is generated over F , andtherefore over F (β), by x. But x satisfies the polynomial (yq − 1)q−1 − β inF (β)[y]. Since the degree of this polynomial is q(q − 1), we may concludethat [E : F (β)] ≤ q(q − 1). Hence q(q − 1)[K : F (β)] ≤ q(q − 1). We mayhence conclude that [K : F (β)] = 1 and we have the statement.

129

d) What is the Galois group of E/L where L = F (xq − x)? Showthat L is Galois.

We have K ⊂ L ⊂ E from c). Moreover, suppose that b ∈ F . Then wehave that

s1,b(xq − x) =

((x+ b)q − (x+ b)

)= (xq − x+ bq − b)

= xq − x,

which shows that L is contained in the fixed field of N . By the GaloisCorrespondence the fixed field has degree q − 1 over K.

Now L is generated over K by xq − x, which satisfies the polynomialm(y) = yq−1 − (xq − x)q−1 over K. Hence to show that L is the fixed fieldof N , it suffices to show that this polynomial is irreducible. If q = 1, thisis trivial as it is a degree 1 polynomial. Hence we may assume that q > 2.We have already shown the roots of m(y) over E are a(xq − x) for a ∈ F×.Hence if m(y) = f(y)g(y) where f(y), g(y) ∈ K[y] are nonconstant, we havea(xq−x)l ∈ K for some 1 ≤ l < q−1. Since a ∈ K, this implies (xq−x)l ∈ K.Now since q 6= 2, F× is generated by a nontrivial element c. We have sc,0 ∈ Gand

sc,0

((xq − x)l

)=((cx)q − (cx)

)l= clxl(cq−1xq−1 − 1)l

= cl(xq − 1)l.

But then cl = 1, a contradiction to the fact that c has order q − 1. Hencem(y) is irreducible and we have shown that L is the fixed field of N . Thisshows that E/L has Galois group N (the cyclic group of order q). Since Nis normal, it also shows that L/K is Galois (with Galois group G/N).

e) Suppose L′ is a field, K ⊂ L′ ⊂ E, L′ 6= E, and L′/K is Galois.Show that L′ ⊂ L.

Suppose that H is a normal subgroup of G which does not contain N .Note that |G| = q(q − 1). Since q does not divide q − 1, N is a Sylow q-subgroup of G. Thus N is the unique subgroup of G of order q. If q divided

130 Fall 2000

the order of H, Cauchy’s Theorem would imply that H has a subgroup oforder q so that N < H, which we have assumed not to be the case. Hence |H|and |N | are relatively prime so that H ∩N = 1. The Second IsomorphismTheorem implies that NH is a group and the Direct Product RecognitionTheorem implies that NH ∼= N×H. Hence H is contained in the centralizerof N . From above, sa,c s1,b s−1

a,c = s1,ba−1 . If this equals s1,b, we have a = 1so that sa,c ∈ N . Hence the centralizer of N is N and we have H = 1.From the Galois Correspondence, we may conclude that only the fixed fieldof 1, E, is not contained in L.

Chapter 15

Spring 2000

1. Let f(x) = x3 − 7 in each part below.

a) Let K be the splitting field of f(x) over Q, the field of rationalnumbers. Describe the Galois group Gal(K/Q), and describe theintermediate fields between K and Q. Which intermediate fieldsare not Galois over Q?

First note that f(x) is irreducible over Q as it is Eisenstein at 7. Letα1, α2, and α3 be the roots of f(x). We have that (αi/α1)

3 = 7/7 = 1 sothat αi/α1 is a primitive 3rd root of unity. Since there are three distinctroots (as every irreducible polynomial over Q, a field of characteristic zero isseparable), we have that K contains all three roots of unity. Furthermore, ifwe let 3

√7 be any root of f(x) (the real root if we want to view K as a subset

of C), we find that ζ 3√

7 and ζ2 3√

7, where ζ is a primitive 3rd root of unity(exp(2πi/3) if K ⊂ C) are also roots of f(x). Hence K = Q( 3

√7, ζ). We

also see that K is the composition of the fields Q( 3√

7) and Q(ζ) which haverelatively prime degrees 3 and 2, respectively, over Q. Hence K has degree 6over Q.

Applying the Fundamental Theorem of Galois Theory, we see that G hasorder six. Now any element σ ∈ G must map ζ to ζ or ζ2 and must map3√

7 to 3√

7, ζ 3√

7, or ζ2 3√

7. Since the images of these two numbers determineσ and we have six total choices, all the choices must lie in the Galois groupand comprise the Galois group. Let σ ∈ G be the map which preserves ζand maps 3

√7 to ζ 3

√7. The σ2 preserves ζ and maps 3

√7 to ζ2 3

√7. σ3 is the

identity. Likewise, let τ ∈ G be the map which preserve 3√

7 and maps ζ

132 Spring 2000

to ζ2. The τ 2 is the identity. By the Sylow Theorems, any group of order6 has a unique subgroup of order 3 which is normal. Hence by the SecondIsomorphism Theorem, 〈τ〉 〈σ〉 = G so that σ and τ generate G. Lastly wehave

στ(ζ) = ζ2 = τσ2(ζ)

and

στ(3√

7) = ζ3√

7 = τσ2(3√

7).

Hence we see that G ∼= D3∼= S3, where the isomorphism is τ 7→ (12) and

σ 7→ (123).

Now any nontrivial proper subgroup of S3 must have order 2 or 3. Fromthe Sylow Theorems, there are either 1 or 3 subgroups of order 3. But 〈(12)〉,〈(13)〉, and 〈(23)〉 are all distinct and of order 2. Hence these are all sub-groups of order 2 (and they are not normal). Likewise, 〈(123)〉 is the unique(and so normal) subgroup of S2 or order 3. Now (12) corresponds to τ . Thefixed field of 〈τ〉must have degree 3 a contain Q( 3

√7). Hence it equals Q( 3

√7).

Likewise (13) = (12)(123)2 corresponds to τσ2 and the subgroup generatedhas fixed field Q(ζ2 3

√7) (which has degree three as it is degenerated by a root

of f(x)). (23) = (12)(123) corresponds to τσ and gives fixed field Q(ζ 3√

7).〈(123)〉 corresponds to 〈σ〉 which must have a fixed field of degree two andcontain Q(ζ) and so equals Q(ζ). Hence the intermediate fields are Q, Q(ζ),Q( 3√

7), Q(ζ 3√

7), Q(ζ2 3√

7), and K = Q( 3√

7, ζ). The ones that are Galois areQ (trivially), Q(ζ), and K (we have applied the Fundamental Theorem ofGalois Theory extensively).

b) Let L be the splitting field of f(x) of over R, the field of realnumbers. Describe the Galois group Gal(L/R).

Over R, we have f(x) = (x− 3√

7)(x2 + 3√

7x+[ 3√

7]2). The discriminant ofthe latter polynomial is ( 3

√7)2−4( 3

√7)2 = −3( 3

√7)2 which is negative. Hence

applying the quadratic formula we see that its roots do not lie in R (or fromthe fact that ζ 6∈ R and our work above). Hence the splitting field of f(x)has degree 2 over R. By the Fundamental Theorem of Galois Theory, wehave G ∼= Z2 the cyclic group of order 2 (where the unique nontrivial mapswitches the two roots of the irreducible quadratic). They are no nontrivialproper intermediate fields.

133

c) Let M be the splitting field of f(x) over F13, the finite field withthirteen elements. Describe the Galois group Gal(M/F13).

(Sarah Hanusch provided this solution.) We can check by hand that f(x)has no root in F13 (that is 7 is not a cube modulo 13). Hence, as f(x) hasdegree three, we may conclude that f(x) is irreducible over F13. Moreover,notice that 1, 3, and 9 are cube roots of unity modulo 13. Hence, if α is aroot of f(x), the roots of f(x) will be α, 3α, and 9α. This gives M = F13(α).But f(x) is the minimal polynomial of α and so F13(α) is a degree 3 extensionof F13. We may conclude that the Galois group is Z3 as this is only group oforder 3.2. What is the order of GL3(Z/4Z), the group of 3 × 3 invertiblematrices over Z/4Z.

Note that I = 0, 2 is an ideal in Z4 with Z4/I ∼= Z2 (as rings). Henceany matrix over Z4, we may consider as a matrix over Z and the determinantwill be a unit in Z4 exactly when it is 1 in Z2 (as the both of the units inZ4 map to 1 in the above isomorphism). Since there exactly two ways torepresent each equivalence class in Z4/I, there will be exactly 29 matricesover Z4 which map to each matrix in Z2. Hence it suffice to find the numberof invertible matrices over Z2 and multiply by 29.

Recall that a matrix over a field (such as Z2) is invertible if and only ifthere is no (linear) dependence relation among its columns. Consider thenan invertible matrix A over Z/2Z. For the first column, we can pick anythingother than the zero column. Since there are three slots in the column and2 choices this gives 23 − 1 = 7 choices. The second column can then beanything by a scalar multiple of the second column. This gives 23 − 2 = 6choices. Finally, the third column can be anything but a linear combinationof the first two columns (of which there are four). This gives 4 choices. Hencethe number of invertible matrices over Z2 is 7 · 6 · 4 = 7 · 3 · 24. Hence thenumber of invertible matrices over Z4 is 7 · 3 · 213 = 172, 032.

134 Spring 2000

3. Let G be a non-abelian group of order p3, p prime. Prove thatthe center Z(G) of G is of order p and that Z(G) = [G,G], where[G,G] is the commutator subgroup of G, that is the group gener-ated by xyy−1x−1, x, y ∈ G.

We have seen previously (Spring 2007 #2) that if G/Z(G) is cyclic, thenG is abelian. Hence we cannot have |Z(G)| = p2 (or |Z(G) = p3, obviously).But a p-group always has a nontrivial center. Hence we must have |Z(G)| =p. This gives |G/Z(G)| = p2, and so G/Z(G) is abelian (Spring 2007 #2).Hence Z(G) contains the commutator subgroup. Hence the commutatorsubgroup has order p or 1. Order 1, however, implies that G is abelian.Hence we have Z(G) = [G,G].

4. Let R be a PID and F its field of fractions. Suppose S is a ringwith R ⊂ S ⊂ F .

a) Show that all elements α ∈ S can be written as a/b where a, b ∈ Rand 1/b ∈ S.

Suppose a/b ∈ S. Then since R is a PID, we may assume (by cancelingfinitely-many irreducibles) that the greatest common divisor of a and b is 1.Hence we may find s, t ∈ R with sa+ tb = 1. This gives (1/b)− t = sa/b ∈ S(since s ∈ R ⊂ S). Hence 1/b = sa/b+ t ∈ S as required.

b) Show that S is a PID.

Let J be an ideal in S and consider the collection I ⊂ R of elementsa ∈ R such that a/1 ∈ J . It is trivial to check that I is an ideal of R. Letr ∈ R be a generator for I. We will show that r (that is, r/1) generates J .First suppose that a/b ∈ I. Let d be the greatest common divisor of a andb. Then a = da′ and b = db′ so that a/b = a′/b′ is in lowest terms. By a),we have 1/b′ ∈ S. Furthermore, a′ = b′(a′/b′) ∈ J so that a′ ∈ I. Hencewe must have some x ∈ R such that a′ = xr. Then a/b = (x/b′)r so thatJ ⊂ Sr′. Conversely, we have r′ ∈ I so r′ ∈ J which implies that Sr′ ⊂ J .

135

Hence J is principal and so R is a PID.c) Show that if S is finitely generated as an R-module then S = R.

Suppose that S is finitely generated as an R-module and a/b ∈ S is inleast terms with b not a unit, that is, with a/b 6∈ R. Then 1/b ∈ S. Let pbe an irreducible dividing b. Then among the generators of S, let k be themaximum power of p which appears among any denominator (which must befinite since each denominator is divisible by p a finite number of times andthere are finitely-many denominators). Than any R-linear combination ofthese generators cannot have a power of p higher than k in its denominator.But (1/b)k+1 has a power of p higher than k. Hence we have reached acontradiction. This shows that no such element can be an element of R, thatis, R = S.

136 Spring 2000

Chapter 16

Fall 1999

1. Assume that V is a commutative integral domain with quotientfield F and that V has the property that ∀α ∈ F , with α 6= 0, eitherα ∈ V or α−1 ∈ V . [Such a V is called a valuation domain.]

a) Let I, J be ideals of V . Show that either I ⊂ J or J ⊂ I. [For fu-ture reference, note that this implies that V has a unique maximalideal.]

Suppose that J 6⊂ I. That is, we may find α ∈ J\I. In particular, α 6= 0.Let x ∈ I. We may assume that x 6= 0 as then x ∈ J trivially. Now we haveeither α/x ∈ V or x/α ∈ V . In the former case, we have α = (α/x)(x) ∈ I,which is a contradiction. Hence we must have the latter which implies thatx = (x/α)(α) ∈ J . This gives I ⊂ J .

b) Suppose that I is a finitely generated ideal of V . Show that I isa principal ideal.

First suppose that I = (a, b) for some a, b ∈ V . If either a or b is zero, wehave that I is principal trivially. Hence we have either a/b ∈ V or b/a ∈ V .Assume without loss of generality that we are in the former case. Thena = (a/b)b so that (a, b) = (b). Now we will prove the claim by induction onthe minimal number of generators for I. The base case is trivial. Assumethen that I is generated by n + 1 elements, x1, . . . , xn+1. Then (xn+1, xn)is generated by either xn or xn+1. After reordering, assume it is generatedby xn. Hence I is generated by x1, . . . , xn so that by the (strong) induction

138 Fall 1999

hypothesis, I is principal.

c) Let 0 6= α ∈ F and suppose there exists a0, . . . , an−1 ∈ V such thatαn + an−1α

n−1 + · · ·+ a0 ∈ V . Show that α ∈ V .

Suppose that α 6∈ V . Then α−1 ∈ V is not a unit. Hence it is containedin the unique maximal ideal M of V . By subtracting the value of αn+· · ·+a0

from a0, we may assume that αn + · · ·+ a0 = 0. Let β = α−1. We have(1

β

)n+ an−1

(1

β

)n−1

+ · · ·+ a0 =1 + an−1β + · · ·+ a0β

n

βn

If this is zero, we must have 1 = −an−1β−cdots−a0βn ∈M , a contradiction.

Hence we must have α ∈ V .

d) Let W be an integral domain with V ⊂ W ⊂ F . Show W is alsoa valuation domain.

Let Q be the field of fractions of W . Each time we have an injective mapof W into a field, this map must factor through Q. But the inclusion mapW → F is such a map and so we must have Q ⊂ F . Likewise, V ⊂ W sothat F ⊂ Q. Hence Q = F . Hence if α ∈ Q is nonzero we have either α ∈ Vor α−1 ∈ V . This implies that either α ∈ W or α−1 ∈ W .

e) Let V ⊂ W ⊂ F as in (d) and let I be a proper ideal of W . Showthat I ⊂ V .

Suppose that α ∈ I and α 6∈ V . Then α 6= 0 so we must have α−1 ∈ V .But then α−1 ∈ W which implies that 1 = α−1α ∈ I so that I is not a properideal. This is a contradiction.

f) Let M be the unique maximal ideal of W . Show that

W = as|a ∈ V, s ∈ V −M.

Let a/b ∈ W be nonzero (0 = 0/1 trivially satisfies the conclusion).Either a/b ∈ V or b/a ∈ V . Suppose that we are in the latter case. Since

139

b/a is a unit in W , b/a 6∈ M . Hence we have a/b = 1/(b/a) where 1 ∈ Vand b/a ∈ V −M . In the former case, a/b = (a/b)/1, where a/b ∈ V and1 ∈ V −M .

2. Let G be a group of order 5075(= 52 · 7 · 29). Let P be a Sylow5-subgroup of G, let Q be a Sylow 29-subgroup of G, and let R bea Sylow 7-subgroup of G.

a) Show that P is a normal subgroup of G.

From the Sylow Theorems, the number of Sylow 5-subgroups must divide7 · 29 = 203 and be equivalent to 1 modulo 5. There first requirement gives1, 7, 29, and 203 as possibilities. Of these only 1 satisfies the second require-ment. Hence there is a unique and therefore normal Sylow 5-subgroup whichmust then be P .

b) Show that G has a normal subgroup H of order 52 · 29.

G/P is a group of order 7 · 29. From the Sylow Theorems, it has a sub-group of order 29, but then this subgroup has index given by the smallestprime dividing the order of G/P and so is normal (this argument is givenin Spring 2003#1(i)). By The Correspondence Theorem, G has a normalsubgroup of order 52 · 29.

c) Show that Q is a normal subgroup of G.

Now since |H| = 52 · 29, the Sylow Theorems imply that H has a uniqueand normal subgroup Q′ of order 29. Consider gQ′g−1 for g ∈ G. Since H isnormal, gQG−1 < H and has order 29 so that gQ′g−1 = Q′. Hence Q′ is anormal subgroup of order 29 in G. This also gives Q = Q′.

d) Show that G has a subgroup, K, of order 52 · 7.

Again |G/P | = 7 ·29 and so the Sylow Theorems give a subgroup of order

140 Fall 1999

7. The Correspondence Theorem then gives a subgroup K of G of order 52 ·7(which contains P ).

e) Show that K is normal if and only if G is abelian.

If G is abelian, K is certainly normal. Hence suppose that K is normal.Then Q and K intersect trivially (by order considerations) and we have|Q||K| = |G|. From the Direct Product Recognition Theorem, we haveG ∼= Q × K. Q is abelian as it has prime order. P is a normal subgroupof G and so it is also normal in K. The Sylow Theorems imply that K hasa normal subgroup L of order 7, which is normal. Hence K ∼= P × L. L isabelian as it has prime order. P is abelian as it has prime squared order (seeSpring 2007 #2). This implies that K is abelian and so G is abelian.

3. Let F ⊂ F (α) ⊂ E with E/F a finite Galois extension. Let G bethe Galois group of E/F and let H be the Galois group of E/F (α).

a) Suppose σ and τ are both in G and let σH and τH be the cor-responding left cosets of H. Show that σH = τH if and only ifσ(α) = τ(α).

σH = τH if and only if there is an µ ∈ H with σ = τµ. If there is sucha µ, µ preserves α so that σ(α) = τµ(α) = τ(α). Conversely, if σ(α) = τ(α)then τ−1σ preserves α and so writing τ−1σ = µ, we have the statement.

b) Now let σ1, . . . , σn be a complete set of left coset representa-tives for H in G. Let τ be an arbitrary element of G. Show thatτσ1, . . . , τσn is a complete set of left coset representatives for H in G.

We have that the index of H in G is n (if any of the σi lie in the samecoset, we may remove the repeats and renumber without affecting anything;we can replace the repeats later if we really want to). Now suppose thatτσiH = τσjH for some i 6= j. Then we have µ ∈ H such that τσiµ = τσj,applying τ−1 to the left, we have σiµ = σj, a contradiction (since we have

141

removed repeats). Hence we have representatives for n distinct cosets whichproves the statement.

c) Let g(X) = (X − σ1(α))(X − σ2(α)) · (X − σn(α)). Show that g(X) ∈F [X].

We have seen that σH = τH if and only if σ(α) = τ(α). Hence any setof coset representative for H in G gives a collection of the (distinct) Galoisconjugates of α (here we must assume that σiH 6= σjH for i 6= j). Letτ ∈ G. Then σ1(α), . . . , σn(α) and τσ1(α), . . . , τσn(α) are equal as sets.This shows that τg(X) = g(X) so that g(X) is in the fixed field of G. Thestatement follows.

d) Show that g(X), as defined in part (c), is the minimal polyno-mial for α over F .

We have that [G : H] = n. This shows that the degree of the fixed fieldof H, F (α), is degree n over F . Hence, the irreducible polynomial of α overF has degree n. But g(X) is a monic polynomial over F of degree n whichhas α as a root. The claim follows.

4. Let F be a field and R = Mn(F ) the ring of n × n matrices overF . Let V = F n be the space of column vectors of length n, viewedas a left R-module in the usual way.

a) Show that the only submodules of V are 0 and V .

Suppose M is a submodule and x ∈ M is nonzero. Then certainly xis independent and so we may find a basis B which contains x as a member.Let y ∈ V be any vector. Then from the universal property of free modules,we may find an R-linear map φ : V → V which maps x to y. Let A thematrix of this transformation with respect to the standard basis on V . Thenwe have Ax = y so y ∈ V . This implies M = Y .

142 Fall 1999

b) Show that R is a direct sum of left ideals Li, each of which isisomorphic to V as an R-module.

Let Li be the set of matrices which have nonzero entries only on the ithcolumn. Certainly the sum of two matrices in Li lies in Li. Now, supposea1, . . . , an are column vectors, as is b. Then [a1|a2| . . . |an]T ∈ R is arbitraryand [0| · · · |b| · · · 0] ∈ Li (assuming of course that b is in the ith column).Then we have

[a1|a2| . . . |an

]T·[0| · · · |b| · · · 0

]=

0 · · · a1· b · · · 00 · · · a2 · b · · · 0...

......

0 · · · an · b · · · 0

,where · represents the usual dot product of two vectors. Hence we can seethat Li is an ideal of R. Certainly any matrix is a sum of these matrices andLi ∩ Lj = 0. Hence R = L1 ⊕ · · · ⊕ Ln. Lastly the map φ : Li → V givenby [0| · · · |b| · · · 0] 7→ b is trivially an R-module isomorphism given that theabove calculation (along with the similar calculations for v ∈ V ) shows thatit preserves R-multiplication.

c) Let W be a left R-module and let x ∈ W . Let L be one the idealsLi in (b). Show that if Lx 6= 0, then Lx is a submodule of Wwhich is isomorphic to V .

Let v ∈ V be the column vector consisting of all ones and define a mapφ : Lx → V by cx 7→ cv. This map is R-linear as we have φ(rcx) = rcv =rφ(cv) for r ∈ R and φ(c1x+ c2x) = (c1 + c2)x = c1x+ c2x = φ(c1x)+φ(c2x)for c1, c2 ∈ L. Next suppose that φ(c1x) = 0. Then cv = 0. But cv isthe (possibly) nonzero column of c. Hence we have c = 0 and so cx = 0.This shows that φ is injective. Lastly the image of φ is a nonzero (since φ isinjective from a nonzero module) submodule of V . By a), we may concludethat the image of φ is V and so φ is an R-module isomorphism.

d) Let W be as in (c) and further assume that the dimension ofW as a vector space over F is finite (viewing F as the subring ofR consisting of scalar matrices). Prove that W is isomorphic as anR-module to a finite direct sum of copies of V .

143

Let x ∈ W be nonzero. Then Ix 6= 0, where I is the identity matrix.Hence if Ei ∈ Li is the matrix which has a 1 in the i, i entry and zeroselsewhere, we have E1x + . . . + Enx 6= 0. Therefore at least one Eix isnonzero so that at least one Lix is nonzero. From the above Lix is thenisomorphic to V .

Let S be the collection of submodules of M which are isomorphic todirect sums of copies of V . Partially order S with respect to the propertythat M ≤ N if N is the (internal) direct sum of M and and another elementof S (or of course if M = N). Note that the dimension (over F ) of a directsum of k copies of V is at least k. Hence if n is the dimension of W overF , n is an upper bound on how many copies of V we may have. We havejust shown that S is nonempty. Let T = Mi be a chain of submodules inS. Since each Mi contains at most n copies of V , there is a submodule ofT which contains the maximal amount of copies of V among elements of T .Thus submodule is trivially the maximal submodule of T . Hence by Zorn’sLemma S has a maximal element, N .

Suppose for a contradiction that N 6= W . Then W/N is nonzero. Lety ∈ W be an element which does not map to zero in W 7→ W/N . By theabove, we can find Li such that Liy is nonzero. Now Liy is isomorphic to Vand so is a simple R-module. Hence, if there is a nonzero v ∈ Liy which liesin N , we must have Liy ⊂ N . But this contradictions our assumption on y.Hence Liy ∩ N = 0. This shows that Liy + N = Liy ⊕ N ∈ S is strictlygreater than N , contradicting maximality. Hence N = W and we have theclaim.

144 Fall 1999

Chapter 17

Fall 1998

1. a) Let G be a finite group and let H be a subgroup of G of indexn. Let Sn be the symmetric group on n elements. Show there is ahomomorphism ψ from G to Sn such that if K is the kernel of ψ,then K ⊂ H, and if N is a normal subgroup of G with N ⊂ H ⊂ G,then N ⊂ K.

Let S = g1H, g2H, . . . , gnH be the n cosets of H in G (with g1 = e).Then we can define an action of G on S by g · (gi)H = ggiH. This is welldefined: if xH = yH then there is some h ∈ H such that y = xh and so,for g ∈ G, gyH = gxhH = gxH. Moreover egiH = giH and for a, b ∈ G,(ab) · giH = abgiH = a · (b · giH). Hence identifying giH with i, we have ahomomorphism ψ : G → Sn where g maps to its permutation on 1, . . . , nunder this action.

Now g ∈ ker(ψ) = K if and only if g acts trivially on all the cosets ofH. In particular, g ∈ K implies gH = H so that g ∈ H. This shows thatK ⊂ H. Next let N be a normal subgroup of G contained in H. Then forx ∈ N , g ∈ G, we have g−1xg ∈ N ⊂ H so that g−1xgH = H which impliesxgH = gH. This shows that x acts trivially on each coset and so x ∈ K.Hence we have N ⊂ K.

b) Suppose G is a group of order 210 = 2 · 3 · 5 · 7, and let B be asubgroup of G of order 5. Suppose B is not normal in G. Let N bethe normalizer of B.

i) Show that the order of N is either 10 or 15.

146 Fall 1998

From the Sylow Theorems, we have that the number of Sylow 5-subgroupsof G, that is the number of subgroups of order 5, divides 2 · 3 · 7. Hence itmust equal 1, 2, 3, 7, 6, 14, 21, or 42. But it also must be equivalent to 1modulo 5 so that is must equal 1, 6, or 21. By assumption, it is not equal to 1(since B is not normal). Thus it is 6 or 21. In either case, G acts transitivelyon the collection of subgroups of order 5 by conjugation so that the numberof such subgroups is |G|/|N | (since N is the stabilizer under this actions ofone of the subgroups). Thus |N | is either 35 or 10.

ii) Show that if G contains a subgroup of order 10, then the orderof N is 10.

Suppose K < G has order 10. Then from the Sylow Theorems applied toK, K has a normal subgroup, P , of order 5. Hence K normalizes P . But Pis a subgroup of order 5 in G and so by the previous argument its normalizerhas either order 10 or order 35. Since its normalizer contains K and 10 doesnot divide 35, we may conclude that the normalizer of P has order 10. Thisimplies that orbit of P under the conjugation action of G has order 21. Butthe orbit of P is that same as that of B (from the Sylow Theorems) and wemay conclude that |N | = 10.

iii) Show that G contains either a subgroup of order 10, or a sub-group of order 35, but not both.

If G contains a subgroup, K, of order 35, we may apply the Sylow The-orems to K to conclude that K has a normal subgroup of order 5. We maythen apply the same argument as in ii) to conclude that |N | = 35. From i),G contains a subgroup of order 10 or one of order 35. If however, G containsa subgroup of order 10, ii) says that |N | = 10 and so the previous argumentshows that G has no subgroup of order 35. Likewise the presence of a sub-group of order 35 forbids the presence of a subgroup of order 10.

iv) Show that G contains a subgroup of order 35 if and only if Gcontains a normal subgroup of order 7.

Suppose that G contains a normal subgroup P or order 7. Then, trivially,B is contained in the normalizer of 7. This implies, from the Second Isomor-

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phism Theorem, that BP is a subgroup of G and that, since B ∩ P = 1by order considerations, |BP | = 35.

Conversely, suppose that G contains a subgroup H of order 35. Thenfrom a), we have a homomorphism G → S6 such that the kernel, K, iscontained in H. Then K has order 1, 5, 7, or 35. If |K| = 1, we have thatG is isomorphic to a subgroup of S6. But |S6| = 6! is not divisible by 210(since it is not divisible by 7). This is a contradiction. Likewise if |K| = 5,we have a normal (and so unique) subgroup of G of order 5, contradictingthe assumption that B is not normal. If |K| = 35, K = H and so H isnormal. From the Sylow Theorems, H has a unique subgroup, P , of order5. Then for g ∈ G, gPg−1 < H by normality and so by order considerations,gPg−1 = P . Hence G has a normal subgroup of order 5 and we arrive at thesame contradiction as before. We may hence conclude that |K| = 7 and soG has a normal subgroup of order 7.

2. Let G be a group and let G′ be the commutator subgroup ofG; (recall that G′ is the subgroup of G generated by all elementsof the form aba−1b−1 with a and b in G). Let H be any subgroupof G. A well known theorem says that G′ ⊂ H if and only if His a normal subgroup of G and G/H is abelian. You may use thistheorem without proving it. In what follows, Galois extensions areautomatically assumed to be finite.

a) Let F ⊂ E be a Galois extension of fields. Show there is a fieldA, with F ⊂ A ⊂ E, such that

i) A/F is Galois,

ii) the Galois group of A/F is abelian, and

iii) if K is any field with F ⊂ K ⊂ E, with K/F Galois, and withthe Galois group of K/F abelian, then K ⊂ A; (this A is theunique largest field between F and E satisfying i) and ii)).

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We will use the Galois Correspondence extensively throughout this prob-lem. Let G be the Galois group of E/F . Let A be the fixed field of G′.Then, since G′ is a normal subgroup of G (by the above), A/F is Galois.Moreover, the Galois group of A/F is isomorphic to G/G′ and so is abelian(again by the above). Lastly, suppose that F ⊂ K ⊂ E and K/F is Galoiswith abelian Galois group. Let N be the subgroup of G corresponding to K.Then G/N is abelian and so (by the above) we have G′ ⊂ N . This impliesthat K ⊂ A. Hence A satisfies the properties given.

b) With F ⊂ A ⊂ E as in part a), suppose that K is a field withF ⊂ K ⊂ E. Show that the following are equivalent:

i) K/F is Galois with abelian Galois group and

ii) K ⊂ A.

That the first property implies the second, we have already shown. Sup-pose then that K ⊂ A and let H be the subgroup of G corresponding to H.Then G′ ⊂ H, which implies (by the above) that H is normal and G/H isabelian. This implies that K/F is Galois and has abelian Galois group.

c) Let F be a field having algebraic closure Ω. Let F ⊂ E ⊂ Ω andF ⊂ L ⊂ Ω. Suppose both E and L are Galois extensions of F . Showthat E ∩ L is a Galois extension of F .

First E ∩ L is a finite extension of F as it is contained in E, a finiteextension of F . Next suppose that α ∈ E ∩ L. Then α ∈ E and E is aseparable extension of F . Hence the minimal polynomial of α over F has norepeated roots in Ω. This implies that E ∩ L is separable. Finally, supposethat f(x) is an irreducible polynomial over F which has a root α in E ∩ L.Then α ∈ E and, since E is a normal extension of F , all the roots of f(x) liein E. Likewise all the roots of f(x) lie in L. Hence all the roots of f(x) liein E ∩ L which shows that E ∩ L is a normal extension of F . Thus we haveshown that E ∩ L is Galois.

d) Let F,L,E and Ω be as in part c); (in particular, we assume thatE/F and L/F are both Galois). Considering the Galois extensionF ⊂ E, let A(E) be the field A discussed in part a). Analogously

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define A(L) and A(E ∩ L); (so, for instance, A(E ∩ L) is the largestfield between F and E ∩ L such that A(E ∩ L)/F is Galois and hasan abelian Galois group). Show that A(E ∩ L) = A(E) ∩ A(L).

For simplicity, call an extension ‘abelian’ if it is Galois with abelian Galoisgroup. We have that A(E)/F and A(L)/F are abelian. Now F ⊂ A(E) ∩A(L) ⊂ A(E) so that, by b), the Galois group of [A(E)∩A(L)]/F is abelian.But A(E) ⊂ E and A(L) ⊂ L so that A(E) ∩ A(L) ⊂ E ∩ L. By a), thesetwo facts imply A(E) ∩ A(L) ⊂ A(E ∩ L).

Likewise, A(E ∩ L) ⊂ E ∩ L ⊂ E and A(E ∩ L)/F is abelian so thatA(E∩L) ⊂ A(E). The same argument gives A(E∩L) ⊂ A(L). This impliesthat A(E ∩ L) ⊂ A(E) ∩ A(L) and so A(E ∩ L) = A(E) ∩ A(L).

3. Let R ⊂ T be commutative integral domains both containing(the same) multiplicative identity, 1. Let

I = r ∈ R|rt ∈ R ∀t ∈ T;

(incidentally I is called the conductor of T to R).

a) Show that I is the largest subset of R which is an ideal in T .That is, I is an ideal in T and I contains any subset of R which isan ideal of T .

Let x, y ∈ I then if t ∈ T , we have (x+ y)t = xt+ yt ∈ R since xt, yt ∈ Rand R is a subring of T . Likewise if x ∈ I and s ∈ T then for each t ∈ T ,(xs)t = x(st) ∈ R so that xs ∈ I (since xs ∈ R as x ∈ I). This shows thatI is an ideal of T . Let J ⊂ R be an ideal of T . Let x ∈ J . Then if t ∈ T ,we have xt ∈ J ⊂ R. This shows that x ∈ I and we we have J ⊂ I. Thestatement follows.

b) For the remainder of this problem, P will be a prime idealof R, and Q1 and Q2 will both be prime ideals of T such thatQ1 ∩ R = P = Q2 ∩ R (possibly Q1 = Q2). Show that if I 6⊂ P ,

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then Q1 = Q2.

If I 6⊂ P , find an element x ∈ I − P . Let a ∈ Q1. Then ax ∈ Q1 sinceQ1 is an ideal of T and ax ∈ R since x ∈ I. Hence ax ∈ Q1 ∩ R = Q2 ∩ R.Hence ax ∈ Q2. Since Q2 is prime, we have a ∈ Q2 or x ∈ Q2. In thelatter case, x ∈ Q2 ∩ R = P , which we have assumed not to be the case.Hence a ∈ Q2 and so Q1 ⊂ Q2. By symmetry, Q2 ⊂ Q1 and the claim follows.

c) Let Q be any prime ideal of T with Q ∩ R = P , and let u ∈ Q.Consider an arbitrary element in the ring R[u]. Of course, such anelement has the form

anun + an−1u

n−1 + · · ·+ a1u+ a0,

where the coefficients ai come from R. Show that this element isin Q ∩R[u] if and only if a0 ∈ P .

Since the element lies in R[u] by definition, saying that it is contained inQ ∩ R[u] is equivalent to saying that it lies in Q. Note that, since u ∈ Qand Q is an ideal, anu

n + · · · + a1u ∈ Q. Hence the element in questionhas the form x + a0 where x ∈ Q. Suppose then that it lies in Q. Thena0 = (x+ a0)− x ∈ Q. Since a0 ∈ R, this gives a0 ∈ P . Conversely supposethat a0 ∈ P . Then a0 ∈ Q so that x+ a0 ∈ Q.

d) Returning to Q1 and Q2 (which here may or may not be equal),pick u ∈ Q1 ∩Q2, and show that Q1 ∩R[u] = Q2 ∩R[u].

Now any element in R[u] and so any element in either of the sets in ques-tion has the form x = anu

n+ · · ·+a0 with ai ∈ R. By c), both the statementx ∈ Q1 ∩ R[u] and the statement x ∈ Q2 ∩ R[u] are equivalent to the state-ment a0 ∈ P . Hence the two statements are equivalent and the two sets areequal.

e) Now suppose that Q1 6= Q2. Also suppose that for any ring Swith R ⊂ S ⊂ T , that Q1∩S 6= Q2∩S except when S = R; (obviouslywhen S = R the two intersections are equal, as they both equal P ).Show under these hypotheses, that P = Q1 ∩Q2 = I (the conductorof T in R).

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First, since Q1 ∩ T 6= Q2 ∩ T , we have Q1 6= Q2. By b), we may concludethat I ⊂ P . We also have P ⊂ Q1 ∩ Q2 trivially. Next suppose thatu ∈ Q1 ∩ Q2. By d), we have Q1 ∩ R[u] = Q2 ∩ R[u]. Since R ⊂ R[u] ⊂ Tand R[u] is a ring, we may conclude that R[u] = R. But then u ∈ R[u] ⊂ R.This shows that Q1 ∩Q2 ⊂ R. But Q1 ∩Q2 is an intersection of ideals in Tand so is an ideal in T which is contained in R. By a), we may conclude thatQ1 ∩ Q2 ⊂ I. Thus we have shown I ⊂ P ⊂ Q1 ∩ Q2 ⊂ I. We may henceconclude that all containments are equalities and the proposition follows.

4. Let R be a ring which is left Artinian, meaning it has the de-scending chain condition on left ideals. Let M be a left R-modulewhich is faithful. That is, if r ∈ R and rM = (0), then r = 0.

a) Suppose that m1,m2, . . . ,mk ∈M . Show that

r ∈ R|rmi = 0, 1 ≤ i ≤ k

is a left ideal of R.

Let I be the set in question. Suppose x, y ∈ I. Then (x + y)mi =xmi + ymi = 0 for each i. Likewise if r ∈ R, then (rx)mi = r(xmi) = 0.This shows that I is a left ideal in R.

b) Show there are finitely many elements m1,m2, . . . ,ms ∈ M suchthat

r ∈ R|rmi = 0, 1 ≤ i ≤ s = 0.

Let S be the collection of left ideals in R of the form

r ∈ R|rmi = 0, 1 ≤ i ≤ k

for m1, . . . ,mk ∈ M Since R is left Artinian, this collection has a minimalelement (with respect to conclusion). Let

I = r ∈ R|rmi = 0, 1 ≤ i ≤ s

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(with m1, . . . ,ms ∈M) be the minimal ideal and suppose for a contradictionthat it is nonzero. Pick then a nonzero x ∈ I. SinceM is a faithful R-module,we may find ms+1 ∈M such that xms+1 is nonzero. Then set

J = r ∈ R|rmi = 0, 1 ≤ i ≤ k.

By definition J ∈ S. Furthermore J ⊂ I and x ∈ I − J . This contradictsthe choice of I and so we may conclude that I = 0.

c) Show R is isomorphic as an R-module to a submodule of M ⊕M ⊕ · · · ⊕M for some finite number of copies of M .

Consider the map φ : R → M ⊕ · · · ⊕ M (s copies of M) given byr 7→ (rm1, . . . , rms) where m1, . . . ,ms ∈ M are as in b). Then for r, t ∈ R,we have

φ(r + t) =((r + t)m1, . . . , (r + t)ms

)= (rm1 + sm1, . . . , rms + tms)

= (rm1, . . . , rms) + (tm1, . . . , tms)

= φ(r) + φ(t)

and

φ(rt) =((rt)m1, . . . , (rt)ms

)= r(tm1, . . . , tms)

= rφ(t).

Hence φ is an R-module homomorphism. Now if r ∈ ker(φ), we have rmi = 0for each i. This implies that r = 0. Hence φ is injective and we have thestatement.

d) Suppose 0 = N0 ⊂ N1 ⊂ N2 ⊂ . . . ⊂ Nt = M is a composition seriesfor M . Suppose also that V is a simple left R-module. Show thatV is module isomorphic to Ni/Ni−1 for some i with 0 < i ≤ t.

We will need a lemma first. Suppose that M1, . . . ,Mk is a collection ofR-modules with Ni ⊂Mi. Then, viewing N1 ⊕ · · · ⊕Nk as a a submodule of

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M1 ⊕ · · · ⊕Mk in the obvious way, we have(M1 ⊕ · · · ⊕Mk

)/(N1 ⊕ · · · ⊕Nk

)∼= (M1/N1)⊕ · · · ⊕ (Mk/Nk).

To see this, consider the map φ : M1⊕· · ·⊕Mk → (M1/N1)⊕· · ·⊕ (Mk/Nk)given by (m1, . . . ,mk) 7→ (m1 +N1, . . . ,mk +Nk). φ is a well-defined surjec-tive R-module homomorphism, essentially as a direct sum of quotient maps.Furthermore, (m1, . . . ,mk) lies in the kernel if and only if mi ∈ Ni for each i.That is, if and only if it lies in N1⊕· · ·⊕Nk. Applying the First IsomorphismTheorem, we have the result.

Now let a ∈ V be nonzero. Then since Ra is a nonzero submodule ofV , we must have V = Ra. Hence R → V , r 7→ ra is a surjective R-modulehomomorphism. Hence, setting I to be the kernel of this map, a left R-module, we have R/I ∼= V by the First Isomorphism Theorem. By c), let Kbe a submodule of M ⊕· · ·⊕M to which R is isomorphic. Then we may finda submodule N ⊂ K such that K/N ∼= V .

Now consider the series

N0 ⊂ N1 . . . ⊂ Nt ⊂ Nt ⊕N1 ⊂ . . . ⊂ Nt ⊕Nt ⊂ . . . Nt ⊕ · · · ⊕Nt,

where for everything is viewed as a submodule of M⊕· · ·⊕M in the obviousway. Applying the above lemma, we have that each factor is isomorphic toNi+1/Ni for some i. Hence since each of these modules are simple, we findthat the series above is a composition series and so cannot be refined.

Finally, consider the series 0 ⊂ N ⊂ K ⊂ M ⊕ · · ·M . By the SchreierRefinement Theorem, we may conclude that this series and the series abovehave equivalent refinements. since the series above has no refinements, 0 ⊂N ⊂ K ⊂ M ⊕ · · ·M has a refinement which is equivalent to it. Note alsothat any refinement of 0 ⊂ N ⊂ K ⊂ M ⊕ · · ·M cannot have a submodulebetween N and K since N/K ∼= V is simple. Hence (up to isomorphism) Vis a factor of any refinement of 0 ⊂ N ⊂ K ⊂ M ⊕ · · ·M . In particular, itis a factor of a refinement which is equivalent to the above series. Since allthe factors above are isomorphic to Ni+i/Ni for some i, we may conclude theclaim.

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Chapter 18

Spring 1998

1. Let G be a group of order 36.

(a) Show that if G does not contain a normal subgroup of order 9,then G contains a normal subgroup K of order 3 and G/K ∼= A4;(you may use with proof the fact that A4 is the only subgroup ofS4 of order 12).

From the Sylow Theorems, the number of Sylow 3-subgroups of G is 1or 4. Since we have assumed that G has no normal Sylow 3-subgroup, wemay conclude that G has 4 Sylow 3-subgroups. Letting S be the collectionof Sylow 3-subgroups, we have a transitive action of G on S. Hence we havethe corresponding homomorphism G → S4. Let K be the kernel of this ho-momorphism. Then K is the intersection of the normalizers of the Sylow3-subgroups. In particular K is contained in such a normalizer, which hasorder 36/4 = 9. Thus the order of K is 1, 3, or 9. |K| = 1 is impossible since|S4| = 24 < |G|. |K| = 9 is impossible since G has no normal subgroup oforder 9. Thus |K| = 3 so that G has a normal subgroup of order 3. Further-more, by the First Isomorphism Theorem, G/K is isomorphic to a subgroupof S4 of order 12. Hence we must have G/K ∼= A4.

(b) Show that A4 has a normal subgroup of order 4.

Consider the set H = 1, (12)(34), (13)(24), (14)(23) in S4. Since all theelements (aside from the identity) are a product of disjoint transpositions,we may conclude that H is contained in A4. Moreover all the elements (aside

156 Spring 1998

from the identity) have order 2 and the product of any pair of the nontrivialelements gives the other nontrivial element. Hence H < A4 is a subgroup.But H contains all the products of disjoint transpositions in S4 and so is aunion of conjugacy classes in S4 (since conjugatation of an element of S4 beanother element simply applies the second permutation to the cycle decom-position of the first; so that the cycle type itself is preserved). This showsthat H is a normal subgroup of S4 and so in particular of A4.

(c) Show G has a normal subgroup N of order 12 containing K.

This follows from the Correspondence Theorem applied to the normalsubgroup in G/K of order 4.

(d) Show that either N is abelian or Z(N) = 2; (here Z(N) denotesthe center).

Since K < N , K is a normal subgroup of N . Let H be a Sylow 2-subgroup. Then G is the semidirect product of HK. Hence, hence we havea map φ : H → Aut(N) ∼= Z2. Pick two elements a, b ∈ H which are notinverses and are not the identity. Then if φ(a) and φ(b) are both not trivial,φ(ab) is. Hence we have a nontrivial element a ∈ H which commutes withevery element of K. Since every element of G can be written as hk withh ∈ H and k ∈ K (and H itself as abelian), we have that a ∈ Z(N). Hencethe center of N is nontrivial and contains an element of order 2 or 4.

Now |Z(N)| cannot be 4 or 6 as either implies that |N/Z(N)| is prime sothat N/Z(N) is cyclic and |Z(N)| = 12. |Z(N)| also cannot be 3 as Z(N)contains an element of order 2 or 4. Hence |Z(N)| = 2 or |Z(N)| = 12.

(e) Show that Z(N) CG (since this is incompatible with |Z(N)| = 2,we have proved that every group of order 36 is the semidirect prod-uct of Sylow subgroups).

Let a ∈ Z(N). Let g ∈ G. By the normality of N , gag−1 ∈ N . Letx ∈ N . Then from normality, g−1xg ∈ N so that it commutes with a. This

157

gives

(gag−1)x(gag−1)−1 = gag−1xga−1g−1

= gg−1xgaa−1g−1

= x.

This shows that gag−1 commutes with x so that gag−1 ∈ Z(N). Hence Z(N)is normal in G.

2. Let A ⊂ B be an extension of integral domains. Let L ⊂M be thecorresponding extension of quotient fields. Assume M/L is a finiteGalois extension, with group G. Assume B is the integral closureof A in M , i.e., B is the collection of elements in M which satisfya monic polynomial with coefficients in A. Also assume B ∩ L = A(or equivalently that A is integrally closed). Recall that a primeP ⊂ B is said to lie over the prime ideal P ∩ A of A.

(a) Let I ⊂ B be an ideal. Suppose I is contained in the union of afinite collection of prime ideals. Show that I is contained in one ofthe prime ideals.

Write I ⊂ P1∪ . . .∪Pn where P1, . . . , Pn are prime ideals. For each i 6= j,we have one of two cases: either there is an xij ∈ I such that xij ∈ Pi − Pjor all the elements of I which lie in Pi also lie in Pj. In the latter case, wemay remove Pi from the union and still have the containment. Hence afterpossible removal of prime ideals we have I ⊂ P1 ∪ . . . ∪ Pk, with xij ∈ Isatisfying xij ∈ Pi − Pj. If k = 1, we are done, so assume that k ≥ 2.

Now, consider xj =∏

i6=j xij ∈ I (since k ≥ 2, this is meaningful). SincePi is an ideal, xj ∈ Pi for i 6= j. But xj is a product of element which donot like in Pj so that, since Pj is prime, xj 6∈ Pj. Now x =

∑j xj ∈ I (again

since k ≥ 2, this is well-defined) so that x ∈ P1 ∪ · · · ∪ Pk. Hence x ∈ Pj forsome Pj. Then

xj = x−∑i6=j

xi ∈ Pj.

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This is a contradiction. Hence we have k = 1.

(b) Let I be an ideal of B. Show that I ∩A = 0 if and only if I = 0.

Certainly one implication is trivial. Assume then that I ∩ A = 0. Sup-pose α ∈ I is nonzero. Since α is integral over A, we may find a monicpolynomial f(x) = xn + . . . + a0 over A which is satisfied by α. This givesa0 = −αn − · · · − a1α ∈ I. But a0 ∈ A. Hence a0 = 0 and we haveα(αn−1 + · · · + a1) = 0. Since B is an integral domain and α 6= 0, we musthave αn−1 + · · ·+ a1 = 0 so that α satisfies a monic polynomial over A of de-gree n− 1 < n. This contradicts the choice of f(x). Hence we may concludethat no such α exists, so that I = 0. Note that to show this result, we onlyneeded the facts that B is integral over A and that B is an integral domain.

(c) Show there is no inclusion between primes ideals of B lying overa fixed prime ideal of A. That is, if P1∩A = P2∩A and P1 ⊂ P2 thenP1 = P2.

Consider the integral domain B = B/P1. By the Correspondence The-orem, P2 = P2/P1 is an ideal. Moreover, by the Second Isomorphism The-orem (and again the Correspondence Theorem), A = (A + P1)/P1 is asubring of B. Now B is integral over A (for each α + P1 ∈ B, find amonic polynomial over A of which α is a root; passing to the quotient, weget a polynomial over A of which α + P1 is a root). Next suppose thatz ∈ P2 ∩ (A+P1). Then z = x+ y for some x ∈ A and y ∈ P1. Furthermore,x = (x+ y)− y ∈ P2 so that x ∈ P2∩A. Thus z ∈ (P2∩A)+P1. This showsthat A∩ P2 ⊂ [(P2 ∩A)+P1]/P1. By the Second Isomorphism Theorem, thelatter is isomorphic to (P2 ∩A)/(A∩P1) = 0. Hence A∩ P2 = 0 and we mayapply (b) to conclude that P2 = 0 which gives P2 = P1.

(d) Let P be a prime ideal of B. Let x ∈ P . Show that the productof elements g(x), as g runs over elements of G is in P ∩ A.

Write a =∏

g∈G g(x). Then h(a) =∏

g∈G hg(x) =∏

g∈G g(x) = a. Hencea lies in the fixed field of G which is to say it lies in L. But L ∩ B = A soa ∈ A.

Since x ∈ B, x is integral over A and so we may find a monic polynomial,f(y), over A of minimal degree of which x is a root. Let m(y) be the irre-

159

ducible polynomial of x over L. Then m(y) divides f(y) so that all the rootsof m(y) are also roots of f(y). But for g ∈ G, β = g(x) is a root of m(y),as β is a Galois conjugate of x. Hence β is a root of m(y) and so a root off(y). In particular, β is integral over A which shows that β ∈ B. Then a isthe product of elements in B, one of which lies in P (namely x = e(x) wheree ∈ G is the identity). Hence a ∈ P and we have the claim.

(e) For a fixed prime ideal Q of A, G permutes the primes of P lyingover Q (you don’t need to prove this). Show G acts transitively onthis set of primes.

Let P1, P2, . . . , Pn be an orbit of G and let P be a prime of B lyingover Q. Then if x ∈ B, we may apply (d) to conclude that a =

∏g∈G g(x)

lies in Q. Hence a ∈ P1 ∩A so that a ∈ P1. But P1 is a prime ideal so thereis some g0 ∈ G such that g0(x) ∈ P1. Now for some i, Pi = g−1

0 P1. Hencex = g−1

0 g0(x) ∈ Pi. Thus we may conclude that P ⊂ P1 ∪ . . . ∪ Pn. By (a),P ⊂ Pj for some j. By (c), we may conclude that P = Pj. Hence G has onlyone orbit.

3. Let K/k be a finite field extension.

(a) Suppose k is finite. What kind of a group is the Galois groupAutkK (no proof needed)? Show that K/k is a primitive extension,i.e., K = k(x) for some x ∈ K.

Let Fp be the base field of k. Then K/Fp is Galois with cyclic Galoisgroup (generated by the Frobenius automorphism). Since Fp ⊂ k ⊂ K, theGalois group of K over k is isomorphic to a subgroup of a cyclic group andso is cyclic.

To prove the second statement, we will prove a lemma: If G is a finitesubgroup of the multiplicative group of a field F , G is cyclic. Let G be sucha group. Then G is finite and abelian and so, by the Structure Theorem for

160 Spring 1998

abelian groups, we have

G ∼= Zn1 × Zn2 × · · · × Znk

for some n1|n2| · · · |nk (with ni ≥ 2). We see that the order of G is n1n2 · · ·nkand that the order of each element in G divides nk. Hence every element inG is a solution of the polynomial ynk − 1 over F . But this can have at mostnk roots and so we see that n1 · · ·nk ≤ nk. This gives k = 1 and so G iscyclic as claimed.1

Thus we have that the multiplicative group K× is cyclic. Let θ be agenerator. Then since 0 ∈ k, we have that θ generates K over k.

(b) Now suppose k is infinite and K has only finitely many subfieldscontaining k. Show that K is a primitive extension.

Choose an x ∈ K such that [k(x) : k] is maximal, (this is possible be-cause such a degree is always less than [K : k]). Now suppose that y ∈ Kand consider the collection of fields of the form k(x + ty) for t ∈ k. Sincethere are finitely many such fields (and k is infinite), we may find a s, t ∈ ksuch that s 6= t with k(x + ty) = k(x + sy). Now we have (t − s)x =t(x+sy)−s(x+ty) ∈ k(x+ty) which shows that x ∈ k(x+ty) (since s−t ∈ k isnonzero) and so k(x) ⊂ k(x+ty). Since [k(x) : k] is maximal, this implies thatk(x) = k(x+ty). We also have (t−s)y = (x+ty)−(x+sy) ∈ k(x+ty) = k(x)which implies that y ∈ k(x). Thus k(x) = K and we have the result.2

(c) Assume for the rest of the problem that K = k(x) is a primitiveextension. Let k ⊂ E ⊂ K be a subfield. Let g(X) ∈ E(X) be theminimal polynomial of x over E. Show that E is generated over kby the coefficients of g(X).

Let E ′ be field generated over k by the coefficients of g(X). Thensince g(x) ∈ E[x], we have E ′ ⊂ E. Now, K is generated over bothE and E ′ by x (since it is generated by x over k). This implies that[K : E] = deg(g(X)) = [K : E ′] which gives E ′ = E(as g(x) is irreducibleover E ′ since it is irreducible over E).

1The proof of this claim is given in Dummit and Foote on p. 3142This proof is given in Dummite and Foot on p. 594

161

(d) Show that K has only finitely many subfields containing k.

We see from (c) that every distinct subfield gives a different divisor of theminimal polynomial of x over k. Since there are only finitely many divisors,we must have only finitely many subfields.

4. Let R be a Wedderburn ring (an Artinian ring with J(R) = 0)and M a left ideal. Suppose M is faithful as a module and nosubideal is faithful. [Note: I made M a left R-module instead of aright R-module and flipped everything around because there is nochance I can multiply on the right]

(a) Explain why M can be decomposed into a direct sum of simplemodules, i.e., M ∼= I1⊕ · · · ⊕ Ik, with each Ii being a simple module.

That every R-module is completely reducible is one of the equivalenceformulations of a Wedderburn ring. M is a left R-module as a left ideal.

(b) Let Ji = Ann (Ii), the annihilator of Ii− so Ii is a simple faithfulR/Ji-module. Describe R/Ji.

Another equivalent formulation of a Wedderburn Ring is that it can bedecomposed as a ring into a direct sum of matrix rings over division rings:

R ∼= Mn1(∆1)⊕Mn2(∆2)⊕ · · · ⊕Mns(∆s),

where the minimal (and so simple as R-modules) left ideals are collections ofelements whose nonzero entries lie only in a fixed column of one of the matrixrings. Moreover, every simple R-module is isomorphic to one of these leftideals. Hence, up to isomorphism and after renumbering, we may considerIi to be equal to collection of elements whose nonzero entries lie in the firstcolumn of Mn1(∆1). Then since Ji is a two-sided ideal of R, and matrix ringsover division rings are simple (as rings), we may conclude that the annihila-tor is a direct sum of the above matrix rings (since an ideal in a direct sum is

162 Spring 1998

a sum of ideals in the individual factors). But for j 6= 1, Mnj(∆j) certainly

annihilates any element of Mni(∆1) and so annihilates Ii. Hence Ji contains

all the factors but the first one. Since Ii 6= 0, 1 ∈ R does not lie in Ji andso we may conclude that Ji 6= R. Hence Mn1(∆1) 6⊂ Ji as otherwise wewould have the whole ring. This gives (again under the above isomorphism)Ji = Mn2(∆2) ⊕ · · · ⊕ Mns(∆s). Hence we have that R/Ji ∼= Mn1(∆1), asimple (Artinian) ring.

(c) Show that the obvious ring homomorphism

φ : R→ R/J1 × · · · ×R/Jk

is an isomorphism.

The ring homomorphism is given by r 7→ (r+J1, . . . , r+Jk). This is triv-ially a ring homomorphism and we can also see that it is surjective. Henceit suffices to show that it is injective. Suppose then that r ∈ R satisfiesr ∈ Ji for all i. Then r annihilates every element of Ii for each i. Now form ∈ M , we can, since M is the direct sum of the Ii’s, find xi ∈ Ii such thatm = x1 + · · ·+ xk. But then rm = rx1 + . . .+ rxk = 0. Since M is faithful,we may conclude that r = 0 and so φ is an injection.

(d) Show MR = R.

Again, it suffices to consider the case where

R ∼= Mn1(∆1)⊕Mn2(∆2)⊕ · · · ⊕Mns(∆s)

is a direct sum of matrix rings over division rings. From (c),

R ∼= R/J1 × · · · ×R/Jk

where, by (b), R/Ji is isomorphic to a matrix ring over a division ring foreach i. Hence by the uniqueness statement in Wedderburn’s Theorem, wemay conclude that k = s. That is, the number of simple modules whichdecompose M is the same as the number of matrix factors which decomposeR. Suppose now that we have matrix factor D with no Ii lying in it (i.e., noIi is the collection of elements which have all their nonzero entries in a fixedcolumn of D). Then taking any nonzero element of D, we find an element

163

which annihilates all the elements of M . This is a contradiction (also if afixed matrix ring contained more than one ideal, we could remove of themand have an subideal which was still faithful). Hence every matrix factormust contain exactly one Ii. After reordering, assume Ii ⊂Mni

(∆i).Now consider one of the matrix rings Mni

(∆i) over a division ring. Sup-pose that the ideal Ii is the collection of elements which lie in the jth column(with the obvious meaning here). Then if Ek,l is the matrix with a 1 inthe kth row and the lth column, and zeros elsewhere, we have Ek,j ∈ Iifor each 1 ≤ k ≤ ni. Now, for 1 ≤ k, l ≤ ni and δ ∈ ∆i, we haveEk,j(δEj,l) = δEk,l so that δEk,l ∈ IiMni

(∆i). Since Mni(∆i) is certainly

comprised of sums of elements of this form, we have IiMni(∆i) = Mni

(∆i).This gives Mni

(∆i) ⊂ MR. Since every element of R is a finite sum ofelements of Mni

(∆i), we may conclude that MR = M .

164 Spring 1998

Chapter 19

Fall 1997

1. a) Let S4 denote the group of permutations of 4 things. Describethe conjugacy classes in S4, determine the normal subgroups of S4,and show that S4 is solvable.

Since conjugation of an element σ ∈ S4 by an element τ ∈ S4 appliesthe permutation associated with τ to the representation of σ given by itscycle decomposition, we can conclude that conjugacy classes in S4 are thecollections of elements with the same cycle decomposition. Hence they arethe identity, the 2-cycles, the 3-cycles, the 4-cycles and the products of twodisjoint 2-cycles. These classes are shown below.

Type No. Order Elements of S4 .(·) 1 1 1(··) 6 2 (12), (13), (14), (23), (24), (34)

(· · ·) 4 3 (123), (124), (134), (234)(· · ··) 6 4 (1234), (1243), (1324), (1342), (1423), (1432)(··)(··) 4 2 (12)(34), (13)(24), (14)(23)

Now any normal subgroup must be both a subgroup and a union of con-jugacy classes. Any normal subgroup which contains the 4-cycles must alsocontain the products of disjoint 2-cycles as an element of this type is thesquare of a 4-cycle. This gives a subgroup of order at least 11 (adding theidentity). Hence it must be either 12 or 24. Since there is no conjugacy classother than the identity which has only one element it must then be 24. Henceno proper normal subgroup contains the 4-cycles. Likewise, no proper nor-mal subgroup can contain the 2-cycles as (12)(13)(14) = (1234) and so this

166 Fall 1997

subgroup would contain a 4-cycle. Next a normal subgroup cannot consistof the identity and the 3-cycles alone as this would have order 5, which doesnot divide 24. This leaves the normal subgroup A4 which is the collectionconsisting of the identity, the 3-cycles, and the products of disjoint 2-cyclesand the normal subgroup V which is the collection of the identity and theproducts of disjoint 2-cycles (and of course 1 and S4 itself).

Lastly, we have the composition series 1 CH C V CA4 C S4, where H isthe subgroup 1, (12). Since V is order four it is abelian which implies thatH is a normal subgroup of V . Now, from order considerations, the factors ofthese series are isomorphic to Z2, Z2, Z3, and Z2, in that order. Since theseare all simple and abelian, S4 is solvable.

b) Suppose G is a subgroup of order 24, which is not isomorphicto S4, and in which a Sylow 3-subgroup is not normal. Show therethere is a homomorphism φ : G → S4 whose kernel is of order 2,and that the Sylow 2-subgroup of G is normal.

From the Sylow Theorems, the number of Sylow 3-subgroups must divide8 and be equivalent to 1 modulo 3. This gives 1 and 4 as possibilities. Sincewe have assumed that the number is not 1, we may conclude that it is 4.Hence G acts on the set of Sylow 4-subgroups by conjugation and this givesa homomorphism G → S4. Let K be the kernel. Since G is not isomorphicto S4, the kernel is not trivial. Moreover, the kernel is the intersection ofthe normalizers of the Sylow 3-subgroups. In particular, it is contained in anormalizer, which has order 24/4 = 6. Hence |K| divides 6.

If |K| = 3, K is a normal Sylow 3-subgroup, which contradicts our as-sumption. Suppose then that |K| = 6. Then from the Sylow Theorems, Khas a unique subgroup, H, of order 3. Then for g ∈ G, gHg−1 is in K bynormality and so equals H by order considerations. Hence H is a normalSylow 3-subgroup, giving the same contradiction as before. Thus we mayconclude that |K| = 2 so that K is a normal subgroup of G of order 2.

Consider then the group G/K, of order 12. By Fall 2000 #1 a), G/K haseither a normal subgroup of order 3 or one of order 2. In the former case, wemay conclude that G has a normal subgroup of order 6 from the Correspon-dence Theorem and arrive at the same contradiction as we did above. HenceG has a normal subgroup of order 4. Applying the Correspondence Theoremagain, we see that G has normal subgroup of order 8, which is then a Sylow2-subgroup.

167

c) Give an explicit example of a group of the type considered in b).

Consider the group G = Z2 × A4. Since G has a nontrivial center (Z2 iscertainly contained in the center), we see that is not isomorphic to S4 (since,by the above, the only conjugacy class of S4 of order 1 is the identity).Moreover, A4 has more than one subgroup of order 3: 1, (123), (132) and1, (124), (142) are both such subgroups. Hence, viewing A4 as a subgroupof G in the usual way, we see that G has more than one subgroup of order 3and so it has no normal Sylow 3-subgroup.

2. Let Fq denote the finite field of q elements. Consider the fieldFq(x) of rational functions with coefficients in Fq.

a) Show that, for any a, b ∈ Fq, a 6= 0, there exists a unique auto-morphism σa,b of Fq(x) with σa,b = ax+ b and σa,b(c) = c for all c ∈ Fq.

Existence is shown in Fall 2000 #4 a). Uniqueness follows from the factthat Fq(x) is generated as a field by Fq and x.

b) Show that the set G = σa,b|a, b ∈ Fq, a 6= 0 is a group of orderq(q − 1).

This is shown also in Fall 2000 #4 a).

c) Show that the fixed field of G in Fq(x) is Fq((xq − x)q−1).

This is shown in Fall 2000 #4 c).

d) Let K be a subfield of Fq(x) which properly contains Fq. Showthat Fq(x) is finite over K.

Since K properly contains Fq, we can find relatively prime polynomialsf(x) and g(x) such that f(x)/g(x) ∈ K and f(x) and g(x) are not both

168 Fall 1997

constant (and of course g(x) 6= 0). Now, the polynomial g(y) ∈ K[y] givenby g(y)(f(x)/g(x)) − f(y) is nonconstant in y and has x has a root. Hencethe degree of k(x) over K is at most the degree of g(y). This gives the result.

3. Let f(x) = x6 + 3.

a) Show that f(x) is irreducible in Q[x].

f(x) is irreducible as it is Eisenstein at 3.

b) Let α ∈ C be a root of f(x). Show that f(x) splits over Q(α).

Let 6√

3 be the real, positive 6th root of unity. Then α has the formα = 6

√3β where β6 = −1 (this is seen easily by dividing α by 6

√3 and ras-

ing it to the 6th). Then α3 =√

3β3. But (β3)2 = −1 so that β = ±i.Hence α3 = ±i

√3. Either way, we have i

√3 ∈ Q(α). But now a primi-

tive 6th root of unity is ζ = (1/2) + i(√

3/2). Hence ζ ∈ Q(α). But thenαζ i for 0 ≤ i ≤ 5 are six distinct roots of f(x) in Q(α) and we have the result.

c) Compute the Galois group of Q(α)/Q.

With notation as above, we have α2 = 3√

3β2. Now (β2)6 = (β6)2 =(−1)2 = 1 so that β2 is a 6th root of unity. This implies that β2 ∈ Q(α) (bythe above) so that 3

√3 ∈ Q(α). Hence we have shown that Q(ζ, 3

√3) ⊂ Q(α).

But Q(ζ) has degree 2 over Q (as φ(6) = 2) and Q( 3√

3) has degree 3 overQ (as it is generated by a root of x3 − 3, which is irreducible since it isEisenstein). Hence, their composite Q(ζ, 3

√3) has degree 6 over Q (as 2 and

3 are relatively prime). This implies that Q(α) = Q(ζ, 3√

3).Now let G be the Galois group. Then an element of G is determined by

its image on ζ and 3√

3. 3√

3 must be mapped to one of 3√

3, ζ2 3√

3, and ζ4 3√

3as these are the roots of its minimal polynomial over Q. Likewise ζ must bemapped to one of ζ and ζ5. This gives a total of 6 possibilities. Since theGalois group has order 6, all these possibilities must be realized and give theentire Galois group.

169

Let σ ∈ G be the map which satisfies σ( 3√

3) = ζ2( 3√

3) and σ(ζ) = ζ.Likewise, let τ be given by τ( 3

√3) = 3

√3 and τ(ζ) = ζ5. Then σ3( 3

√3) =

σ2(ζ2 3√

3) = σ(ζ4 3√

3) = 3√

3 and σ3(ζ) = ζ so that σ has order 3. Likewise,τ 2( 3√

3) = 3√

3 and τ 2(ζ) = τ(ζ5) = (ζ5)5 = ζ. Hence τ has order 2. We mayconclude that 〈σ, τ〉 has order divisible by both 3 and 2 so that it must be6 and we have G = 〈σ, τ〉. Finally we have στ( 3

√3) = ζ2 3

√3 = τσ2( 3

√3) and

στ(ζ) = ζ5 = τσ2(ζ) so that στ = τσ2. Hence we see that G =< σ, τ > satis-fies the relations of the dihedral group of order 6 and so we have a surjectivehomomorphismD6 → G. SinceG has order 6, we may conclude thatG ∼= D6.

d) Find all subfields of Q(α) and carefully show that the list is com-plete.

First any subfield must contain 1 and so contain the field, Q, generatedby 1. Thus it suffices to find all intermediate fields Q ⊂ F ⊂ Q(α). Fromthe Galois Correspondence, it suffices to find all the subgroups of G and thefixed field of each subgroup. Now the G has order 6 and so any (nontrivialproper) subgroup must have order 2 or 3. Hence, to find all the subgroups, weneed only find the elements of order 2 and 3 and consider the subgroups thatthey generate. From above, G = 1, σ, σ2, τ, τσ, τσ2. σ and σ2 have order 3and both generate the subgroup H1 = 1, σ, σ2. The other three nontrivialelements have order two and so generate H2 = 1, τ, H3 = 1, τσ andH4 = 1, τσ2. Hence we have the following lattice of subgroups.

G

3333

3333

3333

3333

FFFFFFFFFFFFFFFFFFFFFF

〈σ〉

〈τ〉

MMMMMMMMMMMMM 〈τσ〉

zzzz

zzzz

〈τσ2〉

lllllllllllllllll

1Now both σ and σ2 fix ζ and so Q(ζ) lies in the fixed field of H1. But

since |H1| = 3, we may apply the Galois Correspondence to conclude that thedegree of the fixed field over Q is 2. Hence Q(ζ) is the fixed field. Likewisethe degree of the remaining fixed fields over Q is 3 and so it suffices to find a

170 Fall 1997

field of degree 3 over Q which is contained in them. Now τ fixes 3√

3 we mayconclude that the field field of H2 is Q( 3

√3). Next, τσ(ζ2 3

√3) = τ(ζ4 3

√3) =

ζ2 3√

3 so that the fixed field of H3 is Q(ζ2 3√

3). Finally, τσ2 fixes ζ4 3√

3 andthe fixed field of H4 is Q(ζ4 3

√3). Hence we have the corresponding lattice of

subfields which comprises all the fields contained in Q(α).

Q

CCCC

CCCC

CCCC

CCCC

CCCC

C

OOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO

uuuuuuuuuuuuuuuuuuuuuuuuu

Q(ζ)

Q( 3√

3)

SSSSSSSSSSSSSSSS Q(ζ2 3√

3)

nnnnnnnnnnnnQ(ζ4 3

√3)

ggggggggggggggggggggggg

Q(α) = Q(ζ, 3√

3)

4. Let C denote the field of complex numbers. Let R be a finitedimensional algebra over C and V a simple R-module which is faith-ful; (recall that V is faithful if whenever r ∈ R and rv = 0 for allv ∈ V , then r = 0).

a) Prove that dimC(V ) is finite.

Find a ∈ V which is nonzero. Then since V is simple, we must haveV = Ra. Now since R is finite dimensional over C, we may find r1, . . . , rkwhich generate R as a C vector space. Hence every element of R is a C-linearcombination of r1, . . . , rk. But then if v ∈ V , we may find r ∈ R such thatv = ra. Writing r = c1r1+· · ·+ckrk for ci ∈ C, we have v = c1r1a+· · ·+ckrka.Hence we may conclude that r1a, . . . , rka generate V as C-module. Thisshows that V is finite dimensional over C.

b) Prove that the ring EndR(V ) of endomorphisms of the R-moduleV is a division algebra which is finite dimensional over C.

Let φ ∈ EndR(V ) be nonzero. Since ker(φ) is a submodule of V , it must

171

be zero (since by assumption it is not V ). Likewise the image of φ must beV . Hence we may conclude that φ is an isomorphism. This implies that φhas an inverse and so EndR(V ) is a division ring.

Now we have a map C → EndR(V ) given by z 7→ φz where φz(v) = zv.This is trivially a ring homomorphism. Furthermore if z ∈ C and zv = 0for all v ∈ V , we have z = 0 since V is faithful. Thus the map above isinjective and we may view C as a subring of EndR(V ). Furthermore, forφ ∈ EndR(V ), z ∈ C, and v ∈ V , we have φzφ(v) = zφ(v) = φ(zv) = φφz(v)which shows that C lies in the center of EndR(V ).

Next consider the map Φ : EndR(V ) → V given by φ 7→ φ(a). We see thatfor φ, ψ ∈ EndR(V ), Φ(ψ + φ) = ψ(a) + φ(a) = Φ(ψ) + Φ(φ) and for c ∈ Cthat Φ(cφ) = cφ(a) = cΦ(φ) so that Φ is C-linear. Moreover, if φ(a) = 0then φ(ra) = rφ(a) = 0 for each r ∈ R. Since V = Ra, this shows that φ iszero. Hence the map is injective and EndR(V ) is a isomorphic to a subspaceof V , a finite dimensional C-vector space. This implies that EndR(V ) is finitedimensional over C (essentially this works since EndR(V ) ⊂ EndC(V )).

c) Prove that EndR(V ) = C.

Let α ∈ EndR(V ). The division subring of EndR(V ) generated by α andC is commutative so that it is a field. Hence C(α) is an extension of C. But itis contained in a finite dimensional vector space and so is a finite dimensionalover C. Hence it is algebraic and so we may find some f(x) ∈ C[x] such thatα is a root. But f(x) splits in C which shows that α ∈ C. This gives theresult.

d) Describe, without proofs, the structure of R and of the mostgeneral R-module.

The representation R → EndC(V ) given by r 7→ φr where φr(v) = rv isfaithful and simple. Hence it is an isomorphism which gives R ∼= Mn(C),the ring of n × n matrices over C. From Wedderburn’s Theorem, the mostgeneral R-module is a direct sum of the R-module L, the set of n×1 columnvectors.

172 Fall 1997

Chapter 20

Spring 1997

1. a) Define what it means for a group G to be (i) solvable and (ii)nilpotent.

The upper central series of a group G is defined inductively: Z0(G) = 1,Z1(G) = Z(G), and Zi+1(G) is the subgroup Zi(G) < Zi+1(G) < G such thatZi+1(G)/Zi(G) = Z(G/Zi(G)). G is nilpotent if Zc(G) = G for some c. G issolvable if it has a series 1 = H0 CHi C · · ·CHs = G such that each factorHi+1/Hi is abelian.

b) Give examples (with explanations) of (i) a group which is notsolvable, and (ii) a group which is solvable but not nilpotent.

S5 is not solvable. A composition series of S5 is 1 C A5 C S5. SupposeS5 has a series of the above form, it would imply, by the Schrier RefinementTheorem, that the two series had a mutual refinement. Every refinement,however, of a series where all the factors are abelian is another series whereall the factors are abelian (applying the Isomorphism Theorems). Hence1CA5 CS5 has a refinement with all abelian factors. But it is a compositionseries and so cannot be refined. Since A5 is not abelian, this is a contradiction.

S4 is solvable, but not nilpotent. To see this, we have Z1 = Z(S4) = 1(the conjugacy classes of S4 are given in Fall 1997 #1 a), from which it fol-lows that the center is trivial). Now if i ≥ 1 and we have Zi(S4) = 1,then Zi+1(S4) is the group which satisfies Zi+1(S4)/1 = Z(S4/1) = 1.Hence Zi+1(S4) = 1 and so by induction, Zi(S4) = 1 for all i which showsthat S4 is not nilpotent. That S4 is solvable is shown in Fall 1997 #1 a).

174 Spring 1997

c) Prove that a finite group whose order is a prime power pn > 1has a non-trivial center.

Let G act on itself by conjugation. Then for every orbit which containsmore then one element (that is, consisting of elements which do not lie inthe center), we have that the order of the orbit is |G|/|CG(x)| where x is anelement of the orbit. But since CG(x) < G is not equal to G, p must dividethe order of the orbit. Hence if S is the collection of elements which lie insuch an orbit, we have |G| = |Z(G)|+ |S| (we are essentially using the classequation). Hence |Z(G)| = |G| − |S| is divisible by p. But |Z(G)| ≥ 1 (since1 ∈ Z(G)) and so we may conclude that |Z(G)| ≥ p and in particular is nottrivial.

d) Prove that a finite group of prime power order is nilpotent.

Let Zi(G) be the upper central series of G, where G is a group of orderpk. Then for each i ≥ 1, if G/Zi(G) is not trivial, it also has order which isa power of p. Hence it has a nontrivial center. Thus |Zi+1(G)/Zi(G)| ≥ p ifZi(G) 6= G so that |Zi(G)| is a strictly increasing sequence until it reaches|G|. Since |G| is finite, we must have some j such that |Zj(G)| = |G| whichimplies that G is nilpotent.

2. In this problem R is a commutative ring with 1. The follow-ing notation and terminology will be used. Spec (R) is the setof prime ideals of R. For R to be 0-dimensional means that ifP,Q ∈ Spec (R) and P ⊂ Q then P = Q. For R to be reduced meansthat the intersection of all prime ideals of R is 0. Finally, e ∈ Ris an idempotent if e2 = e.

a) Explain why if R is 0-dimensional, R/P is a field for everyP ∈ Spec (R).

If P is a prime ideal, it is contained in some maximal ideal M . This maxi-

175

mal ideal will then be a prime ideal and so by the definition of 0-dimensional,we may conclude that P = M is maximal. Thus R/P is a field.

b) Suppose for every element a ∈ R, there is an idempotent e ∈ Rsuch that aR = eR. Show that R is 0-dimensional and reduced.

Suppose that P is a prime ideal. We will show that P is a maximal ideal.Let a ∈ R with a + P 6= 0 + P , that is, a 6∈ P . Then we may find anidempotent e ∈ R such that Ra = Re. Then there is some u ∈ R such thata = eu and v ∈ R such that e = av. This gives e = euv. Now passing tothe quotient R/P , we have e+P = euv+P , but R/P is an integral domainand so has the cancelation property. Hence we have (u+P )(v+P ) = 1 +Pso that v + P and u + P are units in R/P . Next, e2 + P = e + P so thate(e− 1) +P = 0 +P . Hence we have e+P = 0 +P or e+P = 1 +P . Now,a+P 6= 0+P and v+P 6= 0+P (as a unit) so that e+P = av+P 6= 0+P .Hence e + P = 1 + P and so a + P is a unit in R/P (with inverse v + P ).Since a + P nonzero was arbitrary, we may conclude that R/P is a fieldand so P is maximal. Thus if P is contained in any proper ideal, and so inparticular another prime ideal, they ideals are equal. This shows that R is0-dimensional.

Suppose that a ∈ R is contained in the intersection of all prime ideals inR. First we will show that a is nilpotent. Suppose the contrary and let S bethe collection of (proper) ideals which contain no positive power of a. Thenif I1 ⊂ I2 ⊂ . . . is a chain of elements in I, their union is an ideal whichdoes not contain any positive power of a. Hence, by Zorn’s Lemma, S has amaximal element (with respect to conclusion) P which we will show is prime.Suppose then that x, y ∈ R − P . Then by maximality, (x) + P 6∈ S so thatam ∈ (x) + P for some positive m. Likewise, an ∈ (y) + P for some positiven. Then aman ∈ (xy) + P so that we cannot have xy ∈ P . Thus we havethat P is prime. Since P contains no positive power of a, it cannot containa. This gives the claim.

Now find u ∈ R such that e = ua. Since a is nilpotent, we may find somen such that an = 0. Then e = en = (ua)n = unan = 0. Now we also havesome v ∈ R such that a = ev. This shows that a = 0. Hence R is reduced.

c) Let (Fi)i∈I be a family of fields. Show that the direct productring

∏i∈I Fi is 0-dimensional and reduced.

176 Spring 1997

Suppose that P ⊂∏

i∈I Fi is a prime ideal. Then P ∩ Fi (where we viewFi as an ideal of the product in the obvious way) is an ideal so that it equals0 or Fi. Hence P is a direct product of factors which consist either of theentire field or 0. Let ei be the element in

∏i∈I Fi which has a 1 in the

i factor a zero elsewhere. Similarly defined ej. Then eiej = 0 ∈ P so thatei ∈ P or ej ∈ P . Hence P cannot have two factors of zero. Since P cannotbe the whole ring, it must then be the direct product

∏i∈I Ai where for some

fixed j, Aj = 0 and Ai = Fi for i 6= j. Certainly there is no inclusionrelation among sets of this form. This shows that

∏i∈I Fi is 0-dimensional.

Likewise all sets in the above form are in fact prime ideals. Let P havethe form above. Then it is an ideal as the direct product of ideals. If(∏

i∈I αi)(∏

i∈I β) ∈ P , we have αjβj = 0 so that one of αj and βj is zero.Assume without loss of generality that αj = 0. Then

∏i∈I αi ∈ P . Thus

P is a prime ideal. Now the intersection of all these sets is zero as for eachfactor, this is a prime ideal whose intersection with that factor is zero. Thus∏

i∈I Fi is reduced.

d) Suppose R is 0-dimensional and reduced. For P ∈ Spec (R),let FP = R/P . Describe the natural ring homomorphism φ : R →∏

P∈Spec (R) FP , and prove that it is injective.

We have the natural quotient map R→ R/P , which is given by r 7→ r+P .Hence, from the Universal Property of Direct Products, we have a ring ho-momorphism R→

∏P∈Spec (R) FP which is given by r 7→

∏P∈Spec (R)(r+ P ).

Suppose then that r ∈ R is mapped to zero. Then r ∈ P for all P ∈ Spec (R).Since R is reduced, we may conclude that r = 0 and the map is injective.

e) Show that if Spec (R) is finite, then the map φ in part d) isbijective.

Write Spec (R) = P1, . . . , Pk. Since R is 0-dimensional, we may findxij ∈ Pi − Pj for each i 6= j. Now, let xj =

∏i xij Then xi is a product of

elements which do not lie in Pj and so xj 6∈ Pj, but xj lies in every otherideal. Since xj 6∈ Pj and R/Pj is a field, we may find cj ∈ R such thatcjxj ∈ 1 + Pj. Write yj = cjxj. Then φ(yj) is the element with a 1 in jth

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factor and a 0 elsewhere. Now for∏

i ri + Pi ∈∏

iR/Pi, we have that

φ

(∑i

riyi

)=∑i

φ(ri)φ(yi)

=∑i

[∏j

(rj + Pj)φ(yj)

]=∏j

rj + Pj,

which shows that φ is surjective.

3. Let P (X) be an irreducible polynomial in Q[x]. Let α ∈ C be aroot of P (X). Let n be the degree of P (X) and let r be the numberof roots of P (X) in the field Q(α). Let G be the group of P (X), i.e.,the Galois group of a splitting field of P (X) over Q.

a) Show that r divides n.

Let S = α1, . . . , αn be the collection of distinct roots of F (X) overC (it has n distinct roots since it is irreducible over Q, a field of charac-teristic zero). Since each αi has the same degree over Q, we can concludethat αj ∈ Q(αi) implies that Q(αi) = Q(αj). Hence the relation definedby αi ∼ αj if and only if αj ∈ Q(αi) is an equivalence relation as we maypartition S in this way. Furthermore, since each αi is a root of the same irre-ducible polynomial over Q, all the fields of the form Q(αi) are isomorphic asQ-algebras. In particular, f(X) (a polynomial over Q) has the same numberof roots in each field. Hence each set in our partition has the same numberof elements. Since r is the number of elements in one of the sets, we canconclude that r|n (as n is the order of S).

b) Show that if G is abelian, then r = n.

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Let H be the subgroup of G which corresponds to Q(α) in the GaloisCorrespondence. Since G is abelian, we may conclude that H is normal sothat Q(α) is Galois over Q. f(x), however is an irreducible polynomial overQ which has a root in Q(α). Hence it must have all its roots in Q(α) and wemay conclude that r = n.

c) Determine r if P (X) = X6 − 2, giving the reason behind you an-swer.

By the argument given in a), it suffices to choose a particular root in Cand determine r for it. Let α = 6

√2 be real positive 6th root of 2 and consider

the extension Q( 6√

2). Now, we have − 6√

2 ∈ Q( 6√

2) so that r ≥ 2. Next, the6 roots of P (X) over C are ζ i 6

√2, where 0 ≤ i ≤ 5 and ζ = exp(πi/3) is the

canonical primitive 6th root of unity. Hence if we have a third root, it mustbe ζ 6

√2, ζ2 6

√2, ζ4 6

√2, or ζ5 6

√2 (since we already have ζ0 6

√2 and ζ3 6

√2). Now,

if one of the above lies in Q( 6√

2), we can divide it by 6√

2 and conclude thatζ, ζ2, ζ4, or ζ5 lies in Q( 6

√2). But each of these numbers are complex and

Q( 6√

2) consists only of real numbers. Hence we may conclude that r = 2.

4. Let F be an algebraically closed field, V a finite dimensional vec-tor space over F , and T : V → V a linear map of V into itself (i.e.,T ∈ EndF (V )). Recall that T is called nilpotent if T k = 0 for someinteger k, and T is called unipotent if it is of the form I +N whereI is the identity and N is nilpotent. Also T is called semisimple ifits minimal polynomial had distinct roots (i.e., T is diagonalizable).

a) Prove that there exist S,N ∈ EndF (V ) such that S is semisimple,N is nilpotent, SN = NS and T = N +S. (Note: there is more thanone way to do this. It follows from a result on the syllabus for thisexam. It also can be proved using the ideas in part c) below.)

Since F is algebraically closed, we can find a basis b1, . . . , bn of V withrespect to which the matrix of T is in Jordan Canonical Form. Let A bethe diagonal matrix which has the same diagonal entries, λ1, . . . , λn, as the

179

matrix of T . Let B be the matrix which has a 1 in each place above thediagonal that the matrix of T has a 1 and zeros elsewhere. Now let S be thetransformation given by A and let T be that for B.

Trivially, then, T = S + N (as the corresponding matrices satisfy this).Furthermore S is obviously diagonalizable as it is already diagonal. The de-terminant of xI−S isXn, as the determinant of an upper triangular matrix isthe product of the diagonal entries (just consider again the matrix xI−B andexpand about the leftmost column repeat for each sub-matrix). Hence this isthe characteristic polynomial of S. This shows Sn = 0 so that S is nilpotent.Finally, for each bi, S(bi) = 0 or S(bi) = bi−1. In the former case, we haveTS(bi) = T (0) = 0 and ST (bi) = S(λibi) = λiS(bi) = 0. In the latter, we canconclude that the (i− 1)st column and the ith column are part of the sameJordan block so that λi = λi−1. Hence we have ST (bi) = λiS(bi) = λibi−1

and TS(bi) = T (bi−1) = λi−1bi−1 = λibi−1. Hence we have TS = ST and theresult follows.

b) Prove that S and T have the same characteristic polynomial.

Of course it suffices to show this result for any matrix of S and T (as thecharacteristic polynomial does not depend on basis). Now, the matrices for Sand T above have the same diagonal entries. That for S is upper triangularand that for T is diagonal. Hence xI−S and xI−T are also upper triangularand have the same diagonal entries. From this we may conclude that theyhave the same determinant so that S and T have the same characteristicpolynomial.

c) The purpose of this part is to show that the S and the N thatyou found in part a) are unique. Thus suppose that S and N areany elements of EndF (V ) satisfying the conditions: S is semisimple,N is nilpotent, NS = SN and T = S +N . For each λ ∈ F , let

Vλ = v ∈ V |Sv = λv

and let

Wλ = v ∈ V |(T − λ)kv = 0, for some integer k > 0.

Prove:

(i) Vλ ⊂ Wλ and

180 Spring 1997

(ii) Vλ = Wλ.

Conclude that S and N are uniquely determined by T .

Let λ ∈ F . First note that N commutes with T − λ: N(T − λ) =N(S + N − λ) = NS + N2 − Nλ = (T − λ)N . Now let v ∈ Vλ. ThenSv = λv. Furthermore, (T − λ)v = (S + N − λ)v = Nv + (S − λ)v = Nv.Suppose that for i ≥ 0, (T−λ)iv = N iv. Then (T−λ)i+1 = (T−λ)(T−λ)i =(T − λ)N iv = N i(T − λ)v = N i+1(v). Thus, by induction, (T − λ)kv = Nkvfor each k. Now, the latter is zero for some k since N is nilpotent. Thisshows that the former is zero for some k and so v ∈ Wλ. This gives (i).

Next we will show that V is the direct sum both of the Vλ’s and of theWλ’s. First make V into an F [x]-module by having x act as T in the usualway. Then, as an F [x] module, V is isomorphic to M which is a directproduct of factors of the form F [x]/(x − λ)i. Collect together the factorswhich have the same λ ∈ C (but not necessarily the same power) and let Uλbe the sum of these factors. Then, by unique factorization of polynomials, anelement in M is killed by some power of x− λ only if if lies in Uλ. Certainlythe converse is also true. Hence we see that under the isomorphism, Wλ

corresponds to Uλ so that V is the direct sum of the Wλ’s.We can use a similar argument to show that V is the direct sum of the

Vλ’s. We can, however, show this directly. Since S is diagonalizable, we mayconclude that V has a basis of eigenvectors of S. Hence each such vector liesin Vλ for some λ so that V is the sum of the Vλ’s. Furthermore, fix λ ∈ Cand find the basis vectors b1, . . . , bk which lie in it. Let bk+1, . . . , bn be theremaining eigenvectors. For w ∈ V , write w = c1b1 + . . .+ cnbn and supposethat w ∈ Vλ. Then Sw = λc1b1 + . . . λckbk + λk+1bk+1 + . . . λncnbn. For someλi 6= λ. Then by linear independence, we have ci = ci for each 1 ≤ i ≤ kand λci = λici for each k+ 1 ≤ i ≤ n. We may conclude that ci = 0 for eachk + 1 ≤ i ≤ n so that w is a linear combination of the elements of the basiswhich lie in λ. Certainly no eigenvector (and therefore basis vector) can havetwo different eigenvalues. Hence any vector which lies in some Vλ and also inthe direct sum of Vλ’s distinct from the first is a linear combination of twodisjoint sets of basis vectors. This can be true only of the zero vector and sothe sum is direct.

Suppose then that w ∈ Wλ−Vλ. Then we may write w = u+v1 + . . .+vkwhere u ∈ Vλ and vi ∈ Vλi

for some λi 6= λ. Then vi ∈ Wλi. Since w ∈ Wλ,

we may conclude that vi = 0 and so w = u ∈ Vλ, a contradiction. Hence we

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have shown (ii). Now the Wλ’s are completely determined by T . Hence so arethe V ′

λs. The latter, however, determine the eigenvalues and eigenvectors ofS. Since S has a basis of eigenvectors, we may conclude that it is determinedby T . Hence so is N = T − S.

d) Suppose T ∈ GL(V ) (i.e., T is invertible). Prove that there ex-ist unique S, U ∈ GL(V ) such that S is semisimple, U is unipotent,SU = US, and T = SU .

To see existence, let S and N be as above. Now the characteristic poly-nomial for T is the same as that for S. But to say that T is invertible isto say that x is not a factor of its characteristic polynomial (as the latterhappens if and only if T has determinant zero). Hence S is also invert-ible. Set U = I + S−1N . We have SS−1N = N = NSS−1. This givesSS−1N = SNS−1 so that canceling S from both sides, we see that N andS−1 commute. Hence if Nk = 0, we have (S−1N)k = (S−1)kNk = 0. Thisshows that U is unipotent. Now we have SU = S + SS−1N = S + N = T .Likewise US = (1 +S−1N)S = S +S−1NS = S +N = T so that SU = US.Finally, U = TS−1 is invertible. Hence U and S satisfy the criteria.

Next suppose that V and R also satisfy the criteria (with R being thesemisimple one). Then V = I + M where M is nilpotent. Furthermore,RM = R(I +M)− R = (I +M)R − R = MR so that R and M commute.This shows that RM is nilpotent. Then T = RV = R + RM , where Ris semisimple, RM is nilpotent, and RM commutes with R (certainly if Rcommutes with M it commutes with RM). Hence we may apply the aboveuniqueness to conclude R = S and SM = RM = N . Then V = I + M =I + S−1N = U . Hence we have shown uniqueness.

182 Spring 1997

Chapter 21

Fall 1996

1. Let V be a finite dimensional vector space over the rational fieldQ. Suppose A ⊂ V is a finitely generated subgroup under addition.

a) Show that A is generated (as a group) by some subset of a basisof V .

Since A is a finite abelian group, we may apply the Structure Theorem ofAbelian Groups to conclude that A is a direct product of cyclic groups. More-over, if one of these cyclic groups were finite, we would have some nonzeroa ∈ A and nonzero n ∈ N such that na = 0. But V is a module over Q andso is torsion free (as it is free). Hence we must have A ∼= Zn for some n. Foreach i, let ai ∈ A be the element which maps to the element in Zn with a 1in the ith coordinate and zeroes elsewhere. Then since the latter elementsgenerate Zn, ai generates A (as an abelian group). Moreover, suppose thatr1a1 + · · ·+ rnan = 0 for some ri ∈ Q. Then, clearing denominators, we havec1a1 + . . . + cnan = 0 for some ci ∈ Z. This gives (c1, . . . , cn) = 0 by theabove isomorphism and so we may conclude that ci = 0. Hence ai is alinear independent set in V and so is a subset of some basis.

b) Suppose G ⊂ GL(V ) is a finite subgroup of invertible linear en-domorphisms of V over Q. Suppose B ⊂ V is a finitely generatedabelian group. Let B′ be the subgroup generated by all g(B) forall g ∈ G. You may assume the (easy) facts that B′ is finitely gen-erated, B ⊂ B′ ⊂ V and g(B′) ⊂ B′ for all g ∈ G. Let Z be the ringof integers and let R be a commutative ring. GLn(R) is the group

184 Fall 1996

n× n matrices over R with inverses also matrices over R. SupposeGLn(Q) is the group of invertible n × n matrices and G ⊂ GLn(Q)is a finite subgroup. Show that there is P ∈ GLn(Q) such thatPGP−1 = PgP−1|g ∈ G ⊂ GLn(Z).

Let B the subgroup of V = Qn generated by the elements of a basis overQ. The since V is finite dimensional, B is a finitely generated subgroup.Hence B′ is finitely generated and g(B′) ⊂ B′ for all g ∈ G. By a), wemay find a basis of V of which a subset generates B′ as an abelian group.Furthermore, B′ contains a basis for V so that the subset which generatedB′ as an abelian group also generates V as a Q-vector space.

Hence we have a set b1, . . . , bn which generates B′ as an abelian groupand is a basis of V . Now consider g(bi) for g ∈ G. By the above g(bi) ∈ B′ sothat is is a integral linear combination of b1, . . . , bn. This shows that withrespect to this basis, every matrix for the transformation given by a g ∈ Ghas integer entries. This is equivalent to saying that we have a P ∈ GLn(Q)such that PgP−1 has integer entries for any g ∈ G. But then g−1 ∈ G alsohas integer entries. Hence we have shown that PGP−1 ⊂ GLn(Z).

c) If f : R → S is a ring homomorphism, there is a natural grouphomomorphism GLn(f) : GLn(R) → GLn(S) defined by letting f acton each entry. Let m ≥ 2. Show that the kernel of GLn(Z/pmZ) →GLn(Z/pm−1Z) is abelian.

Suppose that A is the in the kernel of this map. Then all the diagonalentries are mapped to one so that they are of the form I + kpm−1. Like-wise all the off diagonal entries are of the form kpm−1. Hence we can finda matrix A′ over Z/pmZ such that A = I + pm−1A′. Let B = I + pm−1B′.Then AB = (I + pm−1A′)(I + pm−1B′) = I + pm−1A′ + pm−1B′ + p2m−2AB.Since m ≥ 2, 2m − 2 ≥ m. Hence we may conclude that the above isI + pm−1A′ + pm−1B′. Similarly, BA = I + pm−1B′ + pm−1A′ and so we havethe kernel is abelian.

d) Suppose G ⊂ GLn(Q) is a finite simple group. Show that thereis an injection G→ GLn(Z/pZ) for any prime p.

Since conjugation is a homomorphism, we can apply b) to find a G′ ⊂GLn(Z) such that G ∼= G′. Then Z → Z/pZ is a homomorphism so we

185

can apply the fact in the problem statement of c) to find a homomorphismG′ → GLn(Z/pZ). Composing these maps, we get a homomorphism of theabove form. Since G is simple, this map must be an injection.

2. Let G be a group.

a) Let G have order 30. Show that G has a normal subgroup oforder 15. Show that every subgroup of G of order 3, 5, or 15 isnormal.

From the Sylow Theorems, the number of Sylow 3-subgroups divides 10and is equivalent modulo 3 to 1. Hence it must be 1 or 10. Likewise, thenumber of Sylow 5-subgroups must be 1 or 6. Suppose that neither is 1.Then, since distinct subgroups of prime order cannot intersect nontrivially,we have 20 elements of order 3 and 24 elements of order 5. Thus |G| ≥ 44,a contradiction. Thus we have either a normal subgroup of order 3 or oneof order 5. Let P be a Sylow 3-subgroup and Q a Sylow 5-subgroup. Thenone is contained in the normalizer of the other. Hence PQ is a subgroup ofG of order 15 (by the Second Isomorphism Theorem and the fact that theyintersect trivially from order considerations).

Now if a subgroup of G has order 15, its index is the smallest prime whichdivides |G|. We may apply Spring 2003 #1(i) to conclude that it is normal.In particular, PQ is a normal of G of order 15. Now applying the SylowTheorems to PQ, we find that it has a unique and so normal subgroup, H,of order 3. Then for g ∈ G, gHg−1 < PQ is contained in PQ by normalityand so must equal H. Hence H is a normal subgroup of order 3. Since itis a Sylow 3-subgroup, this shows that it is the only subgroup of order 3.Likewise, a Sylow 5-subgroup of PQ is a normal subgroup of G of order 5,which shows that it is the only subgroup of G of order 5.

b) Suppose H ⊂ G is a subgroup of the group G. Define coreG(H) ⊂G to be the normal subgroup of G such that if N ⊂ H is a normalsubgroup of G, then N ⊂ coreG(H). You may assume coreG(H) ex-ists. Suppose L/F is a Galois extension of fields with Galois group

186 Fall 1996

G. Let α ∈ L have minimal polynomial f(x) ∈ F [x] over f . LetK = F (α) and H ⊂ G be the subgroup of G associated to K. Showthat L is the splitting field of f(x) if and only if coreG(H) = 1.

First note that coreG(H) = 1 is equivalent to saying that H contains nonontrivial normal subgroup (including itself). Suppose then that H containsno nontrivial normal subgroup. Since L contains all the roots of f(x) (as itis a normal extension which contains one of the roots and f(x) is irreducible)and since K is generated by a root of f(x) over F , the splitting field off(x), E, must satisfy K ⊂ E ⊂ L. However, from our assumption and anapplication of the Galois Correspondence, no such field is normal over Fother than E = L. Since a splitting field must be normal, we can concludethat E = L.

Conversely suppose that L is the splitting field of f(x) over F . Supposethere is a field, E, which is a normal extension of F and satisfies K ⊂ E ⊂ Land E 6= L. Then E contains all the roots of f(x), since it contains one ofthem. Hence the field generated over F by the roots of f(x) is contained inE. But this shows that that this field is the splitting field and does not equalL, a contradiction. Hence there is no such field. By the Galois Correspon-dence, we may conclude that H contains no normal subgroup.

c) Suppose G has order 30, and L is the splitting field of f(x). Showthat f(x) has degree 15 or 30.

Since L contains a root of f(x), the degree of f(x) must divide 30. Henceit must be 1, 2, 3, 5, 6, 10, 15, or 30. The first case implies that F , not L, isthe splitting field of F . The next 4 cases imply that K has the correspondingdegree over F so that, from the Galois Correspondence, |H| is 15, 10, 6, 5.In all these cases H contains a subgroup of order either 3 or 5, which bya), must be normal. This shows that coreG(H) 6= 1 and so L is not thesplitting field of f(x). Thus the degree of f(x) is 15 or 30.

d) Now suppose that G has order 18. Show that if L is the split-ting field of a polynomial of degree 6, it is the splitting field of apolynomial of degree 9.

Let f(x) be the polynomial of degree 6. First suppose f(x) is reducible.If f(x) splits into linear factors, than L cannot be its splitting field (as K is).

187

Likewise, if f(x) splits into a polynomial of degree k and 1 or more linearfactors, the splitting field has degree dividing k! over F for k < 6 and socannot be 18. Next, if f(x) splits into two factors of degrees k and n andotherwise linear factors, the degree of the splitting field divides k!n! for somen+ k ≤ 6. The only way that 18 can divide this if 3 divides both k and n sothat k = n = 3. Let f(x) = g(x)h(x) where both g(x) and h(x) have degree3. Then L is also the splitting of g(x)2h(x), a polynomial of degree 9. Lastlyif the factorization of f(x) has three or more nonlinear factors, they must allhave degree 2 so that the degree of splitting field divides 8 and so cannot be18.

Hence it suffices to assume that f(x) is irreducible. Let α ∈ L be aroot and write K = F (α). Now [K : F ] = 6 so that, if H is the subgroupassociated to α, we have |H| = 3. By the above coreG(H) = 1 so that H isnot itself normal. Now suppose for a contradiction that G contains a normalsubgroup N of order 2. Then G/N has order 9 = 32 and so is abelian. Henceevery subgroup of G/N is normal so that, by the Correspondence Theorem,every subgroup of G which contains N is normal. Now since HN is sucha subgroup and has order 6 (applying the Second Isomorphism Theorem).Hence, from the Sylow Theorems, HN has a unique and so normal subgroupof order 3. Since H is a subgroup of HN of order 3, this subgroup must beH. Now for g ∈ G, gHg−1 is contained in HN by normality and, since ithas order 3, must equal H. Thus H is normal in G, a contradiction. HenceG cannot have a normal subgroup of order 2.

By the Sylow Theorems, G has a subgroup P of order 2. Hence P isa nonnormal subgroup of order 2. Now the only proper subgroup of P is1 and so coreG(P ) = 1. Let K ′ be the fixed field of K. Then from thePrimitive Element Theorem, K ′ = F (θ) for some θ ∈ K ′ (as an intermediatefield of a finite Galois extension). Let g(x) be the minimal polynomial ofθ over F . From the Galois Correspondence, [F (θ) : F ] = 9 so that g(x)has degree 9. Moreover, since coreG(P ) = 1, we may apply the above toconclude that L is the splitting field of g(x). This gives the claim.

3. Let F be a field and G a finite group. Let F ∗ denote the groupof nonzero elements in F under multiplication. G will be called F -

188 Fall 1996

pure if for all subgroups H ⊂ G and all homomorphisms λ : H → F ∗,λ(h) = 1 for all h ∈ H.

a) If |G| is odd, show that G is Q-pure for the rational field Q.

Let H ⊂ G and suppose that λ : H → Q∗ is a homomorphism. Leth ∈ H. Since h ∈ G, the order of h is some odd number n. This gives[λ(h)]n = λ(hn) = λ(1) = 1. Hence λ(h) is an nth root of unity for n odd.But the only nth root of unity (over C) which lies in R is 1. Since Q ⊂ R,we must have λ(h) = 1.

b) From now on assume G is a finite subgroup of the automor-phism group AutF (F [x1, . . . , xn]), where AutF indicates the rings ofhomomorphisms that is the identity on F . Show that for g ∈ G,g(α) ∈ F [x1, . . . , xn] is a prime element if and only if α is.

For simplicity, write R = F [x1, . . . , xn]. Suppose that α is reducible sothat α = βγ for some nonconstant β, γ ∈ R. Then g(α) = g(β)g(γ) asg is an automorphism. Since g−1 is the identity on F , g cannot map anonconstant to a constant (as otherwise g−1 would have to map an elementof F to a nonconstant and in particular not to itself). Hence g(α) and g(β)are nonconstant so that g(α) is reducible.

Conversely suppose that g(α) is reducible so that g(α) = βγ for somenonconstant β, γ ∈ R. Then α = g−1(β)g−1(γ), where, by as a similar ar-gument as above, g−1(β) and g−1(γ) are nonconstant. Thus α is reducible.Hence we have shown that α is reducible if and only if g(α) is reducible. Thisimplies the claim.

c) We say that two prime elements π ∈ F [x1, . . . , xn] are distinct ifand only if there is no unit u such that uπ1 = π2. Also, from now weassume G is F pure. Show that if π is prime, then either g(π) = πor g(π) and π are distinct primes.

Suppose that there is some unit u such that π = ug(π). Then u isnonzero and constant so that u ∈ F ∗. Now note that if gn = 1, we haveπ = gn(π) = unπ (since g(u) = u). Since R is an integral domain, wemay conclude that un = 1. Hence we have a well defined homomorphismλ : 〈g〉 → F ∗ given by g 7→ u. Since G is F -pure, we may conclude that

189

u = 1. Hence if g(π) and π are associates, π = g(π).

d) Assume π ∈ F [x1, . . . , xn] is a prime element. Let G(π) be theproduct of distinct primes of the form g(π). Show that G(π) isfixed by G. If α ∈ F [x1, . . . , xn] is fixed by G, and π divides α showthat G(π) divides α.

Write G(π) = g1(π) · · · gn(π) where gi(π) is the collection of distinctimages of π under G. Let h ∈ G. Then for each i, hgi(π) = gj(π) forsome j as it is an image of π under an element of g (and applying c)). Nowsuppose that hgi(π) = hgj(π). Then gi(π) = gj(π) and so gi = gj. Hence thecollection gi(π) is that same as hgi(π) ignoring order so that, since R iscommutative, their product is the same. Hence

hG(π) = h(g1(π) · · · gn(π)

)= hg1(π) · · ·hgn(π)

= g1(π) · · · gn(π)

= G(π).

Thus G(π) is fixed by G.Suppose that π divides α ∈ R and α is fixed by G. Then α = βπ for

some β ∈ R and α = g(α) = g(β)g(π). Hence g(π) divides α for any g ∈ G.Thus G(π) is a product of distinct primes, each of which divide α. We mayconclude that G(π) divides α.

e) Assume π is a prime element of F [x1, . . . , xn]. Show that G(π) isa prime element of the fixed ring

F [x1, . . . , xn]G = f ∈ F [x1, . . . , xn]|g(f) = f for all g ∈ G.

Assume that G(π) divides αβ for some in α, β ∈ RG. Then π, a factorof G(π), divides αβ in R so that, since π is prime, π divides α or β in R.Assume without loss of generality that π divides α. Then since α is fixed byG, we may apply d) to conclude that G(π) divides α in R. Thus we mayfind γ ∈ R such that α = G(π)γ. Let g ∈ G. By d), G(π) is fixed by g so we

190 Fall 1996

have G(π)γ = α = g(α) = G(π)g(γ). Hence we may conclude that γ = g(γ)so that γ ∈ RG. This shows that G(π) divides α in RG and so is prime.

4. Let R ⊂ K be such that R is a commutative domain and K is itsfield of fractions.

a) Suppose y ∈ R and dy divides y. Show that d is a unit.

Notice that the definition of divides requires that dy 6= 0 so that y 6= 0.Find c ∈ R such that y = cdy. Then from the cancelation property of adomain, we have cd = 1. Since R is commutative, we may conclude that dis a unit (with inverse d).

b) Suppose F is a free module over R. Show that if 0 6= m ∈ F andrm = 0 then r = 0. If 0 6= d ∈ R is a nonunit, show there there is anonzero x ∈ R such that the equation (dx)y = m has no solution fory ∈ F .

Find a basis for F and let m ∈ F . Then we may find ri ∈ R and basisvectors bi such that m = r1b1 + · · ·+rkbk. Since m 6= 0, we must have at leastone ri not equal to zero and after reordering, we may assume r1 6= 0. Thenif rm = rr1b1 + ·rrkbk = 0, we must have, by linear independence, rr1 = 0.This implies that r = 0 (as R is a domain).

Again write m = r1b1 + · · ·+ rkbk with r1 6= 0. Suppose that (dr1)y = mhas a solution y ∈ F . Then writing y as a linear combination y = s1b1 +· · · skbk + sk+1bj+1 + · · · + bn, we may use linear independence to concludethat dr1s1 = r1. Hence dr1 divides r1 so that, by a), d is a unit. This is acontradiction. Hence x = r1 is such an element.

c) Consider K to be an R module. Show that if K is the submoduleof a free R module, then R = K. Show that if K is projective as anR module, then R = K.

Let K ⊂ F where F is free. Then 1 ∈ F . Suppose for a contradiction

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that 0 6= d ∈ R has no inverse in R. Then we may find nonzero x ∈ R suchthat dxy = 1 has no solution for y ∈ F . But 1/(dx) ∈ K ⊂ F is a solutionso that we have a contradiction. Hence we have shown that every elementof R has an inverse in R so that R is a field. This shows that R = K. Fur-thermore one of the equivalent formulations of a projective module is that isa direct summand of a free module. Hence if K is a projective module, it isa submodule of a free module so that F = K.

d) Suppose I ⊂ R is a proper ideal and R/I is projective as an Rmodule. Show that R/I = 0.

Suppose x ∈ I ⊂ R is nonzero. Since R/I is a submodule of a free mod-ule, it is torsion free. But x(1+ I) = x+ I = 0+ I in R/I. Since I is proper,1+I 6= 0+I and so this is a contradiction. Hence no such x exists and I = 0.

e) Suppose R ⊂ M ⊂ K is such that M is finitely generated as anR module. Show that there is an 0 6= r ∈ R such that rM ⊂ R.Suppose K is finitely generated as an R module. Show that K = R.

Suppose thatM is generated by a1/b1, . . . , ak/bk ⊂ K. Let b = b1 · · · bk.Since bi 6= 0 for each i, b 6= 0. Furthermore, ci = b/bi ∈ R for each i. Ifm ∈M , we can find ri ∈ R such that m = r1a1/b1 + · · ·+rkak/bk. This givesbm = c1r1a1 + · · ·+ ckrkak ∈ R. Hence bM ⊂ R.

Suppose that K is a finitely generated R module. Then by the previousargument, we may find r ∈ R such that rK ⊂ R. Let a ∈ R be nonzero.Then 1/a = r · (1/(ar)) ∈ R so that a has an inverse in R. This shows thatR is a field so that R = K.

192 Fall 1996

Chapter 22

Spring 1996

1. Let G be a subgroup of S5 which contains a 5-cycle and a 2-cycle.Show that G = S5.

Let the two cycle be given by σ = (ij) where of course 1 ≤ i, j ≤ 5 andi 6= j. Likewise, let τ = (acde) be the 5-cycle. Then we see that τ(a) = b,τ 2(a) = c, etc., so that 〈τ〉 acts transitively on the set 1, 2, 3, 4, 5. Inparticular, there is some power of τ which maps i to j. Since this power τwill also be contained in G, we may as well assume that τ = (ijabc) (sinceevery power of 5-cycle is either the identity or another 5-cycle). Next, itsuffices to show that our subgroup G has order 120. Hence if we conjugateby an element of S5 and show that the resulting subgroup is S5, the resultwill follow. Conjugate G then by the permutation which maps i to 1, j to 2,a to 3, b to 4, and c to 5. Then our new subgroup is generated by τ = (12345)and σ = (12). But that these two elements generate S5 is a well known fact.

2. Let F2 be the field of two elements and consider the ringR = F2[x]/(x

2). Describe all isomorphism classes of R-modules with16 elements.

Suppose M is an R-module, then since F2 is a subring of R, M is inparticular a vector space over F2. Moreover we have an F2-linear map T :

194 Spring 1996

M → M given by m 7→ xm. Thus we can make M into an F2[x]-moduleby having x act as T in the usual way. Since we must have T 2 = 0, this iswell-defined.

Now applying the Structure Theorem for Modules over a PID, we musthave M isomorphic as an F2[x]-module to a direct product of factors of theform F2[x]/(f(x)). But since x2 acts as zero on M , it must act as zero oneach factor of M which implies that f(x) divides x2. Hence each factor musthave the form F2[x]/(x

2) or F2[x]/(x) (ignoring trivial factors). Now thefirst factor has 4 elements and and the second has 2. Certainly two moduleswhich are isomorphic as F2[x]-modules are isomorphic as R-modules. Hencewe have the possibilities

R⊕R, R⊕R/(x)⊕R/(x), and R/(x)⊕R/(x)⊕R/(x)⊕R/(x).

The first module above has exactly 4 elements which are killed by x (namely(x, x), (x, 0), (0, x), and (0, 0)). The second has 8 (those with a 0 or an xin the first coordinate). Finally x kills every element of the last module.Hence these three are not isomorphic as R-modules and so represent the 3equivalence classes.

3. a) Define nilpotent and solvable groups.This is done in Spring 1997 #1 a).

b) Either prove or give a counterexample to the following state-ments.

(i) A nilpotent group is solvable.

(ii) A solvable group is nilpotent.

Suppose G is nilpotent. Then Zn(G) = G for some n ∈ N. SinceZi+1(G)/Zi(G) = Z(G/Zi(G)), we see that Zi(G) CZi+1 and Zi+1(G)/Zi(G)is abelian. Hence

1 = Z0(G) C Z1(G) C · · ·C Zn(G) = G

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is a series in which every factor is abelian and so shows that G is solvable.Hence statement (i) is true. Statement (ii) is false and a counterexample isgiven in Spring 1997 #1 b) (S4 is the counterexample).

c) Let G be the group with identity e generated by elements a, b, csatisfying a2 = b3 = c7 = e, ab = ba, and ac = ca. Determined whetheror not G is solvable and whether G is nilpotent.

Before show this, we will prove the following lemma: A7 is generated bythe elements σ = (1234567) and τ = (123). Let H be the subgroup of A7

that they generate. We will first show that 8 divides |H|. First we see that(234) = στσ−1 ∈ H, (345) = σ2τσ−2 ∈ H, and (456) = σ2τσ−3 ∈ H. Hencewe have (12345) = (123)(345) ∈ H so that (12345)(456) = (1234)(56) ∈ H.Likewise, (24)(56) = (245)(456) ∈ H. Consider now the subgroup K =〈(24)(56), (1234)(56)〉. We see that that the order of (1234)(56) is 4 and thatof (24)(56) is 2. Since [(1234)(45)]2 = (13)(24) 6= (24)(56), we can concludethat |K| ≥ 5. Since 4 divides K, we may conclude that |K| ≥ 8. Moreover,(1234)(56)(24)(56) = (12)(34) = (24)(56)[(1234)(56)]3. Hence K satisfies therelations of D8, the dihedral group of order 8 and so we have a surjectionhomomorphism φ : D8 → K given by r 7→ (1234)(45) and m 7→ (24)(56).Since |K| ≥ 8, we may conclude that this map is injective. We may henceconclude that K is a subgroup of H of order 8 and so gives the first claim.

Now, |H| is divisible by 3, 5, and 7 because it contains (123), (12345),and (1234567), respectively. Thus |H| is divisible by |A7|/3. Hence the indexof H in A7 is either 1 or 3. Suppose for a contradiction that is the latter.Then A7 acts nontrivially on the cosets of H in A7 and we have a nontrivialhomomorphism A7 → S3. By order considerations, however, we must have anontrivial kernel and so a nontrivial proper normal subgroup of A7. This isa contradiction to the simplicity of A7. Hence we have shown that the indexof H in A7 is 1 so H = A7.

Let F be the free group on the elements a, b, c. By the Universal Prop-erty of Free Groups, we have a homomorphism F → A7 given by a 7→ 1,b 7→ (123) and c 7→ (123467). Now these elements satisfy (trivially) allthe defining relations on G. Hence this homomorphism factors through Gand we have a homomorphism G → A7 given by a 7→ 1, b 7→ (123) andc 7→ (1234567). By the above, this map is surjective and we may apply theFirst Isomorphism Theorem to conclude that G/N ∼= A7 for some normalsubgroup N in G. But now A7 is not solvable (as it is simple and not abelian)

196 Spring 1996

and so a quotient of G is not solvable. Since the quotient of a solvable groupis solvable, we may conclude that G is not solvable. By b), it is not nilpotenteither.

4. Let f(x) = x8 + 1.a) Let K be the splitting field of Q, the field of rational numbers.Determine the Galois group of K/Q.

Suppose ζ is a root of x8 + 1. Then ζ8 = −1. Now, (ζ8)2 = 1. Henceζ is 16th root of unity. Since ζ8 6= 1, ζ is a primitive 16th root of unity.Now the order of ζ is φ(16) = 8. Furthermore, the extension Q(ζ) is degree8 over Q and is Galois with Galois group (Z16)

×, where the isomorphism(Z16)

× → Gal(F/Q) is given by a 7→ σa where σa(ζ) = ζa. Hence Q(ζ) is thesplitting field of f(x) and the Galois group is as above.

Working modulo 16, we see that 32 = 9, 33 = 11, and 34 = 1 so that3 and 11 have order 4 in (Z16)

∗ and 9 has order 2. We also have 52 = 9,53 = 13, and 54 = 1 so that 5 and 13 have order 4. Lastly, 7 and 15 haveorder 2. Hence (Z16)

× is an abelian group of order 8 which is not cyclic buthas a factor of Z4. This gives that the Galois group is Z4 × Z2.

b) How many subfields of K are of degree 4 over Q? How many ofthese are Galois over Q? Explain.

From the Galois Correspondence, a degree 4 subfield over Q correspondsto a subgroup of (Z16)

× of order 2. Such a subgroup must be generated byan element of order 2. Hence, by thr above, the subgroups of order 2 are1, 7, 1, 9, and 1, 15. Thus we have 3 such subfields. Since (Z16)

× isabelian, every subgroup is normal and so every subfield of K, in particularthose of degree 4 over Q, is Galois.

c) Let L be the splitting field of f(x) over F41, the field of 41 ele-ments. Determine the Galois group of L/F41.

As above, a solution to f(x) is a primitive 16th root of unity. We have

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|F×41| = 40. Since 16 does not divide this number, there is no element F×41 oforder 16. Hence, F41 does not contain a primitive 16 root of unity. Considernext F412 . We have |F×412| = 1681− 1 = 1680 = 16(105) so that F×412 , a cyclicgroup, has an element of order 16. This is a primitive 16th root of unity.This element raised to the 8th is a square root of unity which doesn’t equal1. Hence it must equal −1, so that powers of this element, relatively primeto 16 give the 8 roots of the polynomial over F412 . Hence the Galois grouphas order 2 over F41 and the Galois group is Z2.

198 Spring 1996

Chapter 23

Fall 1995

1. a) Let K ⊂ L be a Galois extension of fields of degree 60. Showthat there is an irreducible polynomial of degree 12 in K[x] with aroot in L. Show that this polynomial has all its roots in L.

By the Galois Correspondence, there are finitely many subfields betweenK and L (as they correspond to the finitely many subgroups of the Galoisgroup, a group of order 60). Hence by the Primitive Element Theorem, everysubfield between K and L is a primitive extension of K. By Cauchy’s The-orem, there is a subgroup of the Galois group of order 5. This correspondsto a subfield K ⊂ F ⊂ L with [F : K] = 12. By the above F = K(θ). Thisimplies that the minimal polynomial of θ over F is an irreducible polynomialin K[x] of degree 12 with a root in L. Since L is a normal extension of K,we may conclude that it contains all the roots of f(x).

b) Suppose L/K is Galois with group G of order 60 and G is solv-able. Show that there is an irreducible polynomial of degree 4 inK[x] with a root in L.

By the above argument, it suffices to show that G has a subgroup oforder 15. Now since G is solvable it must either have a nontrivial propernormal subgroup or be abelian, in which case it has a nontrivial propernormal subgroup. Suppose first that G has a normal subgroup of order 3or 5. The from Cauchy’s Theorem, it has a subgroup of the other order.From the Second Isomorphism Theorem, the product of these two groups isa subgroup of order 15. Hence it remains to check the cases when G has a

200 Fall 1995

normal subgroup of order 2, 4, 6, 10, 12, 20, and 30.

Suppose that G has a normal subgroup N of order 6. Then from theSylow Theorems applied to N , we see that N has unique subgroup, H, oforder 3. For g ∈ G, we have that gHg−1 < N by normality and so thenmust equal H by order considerations. Thus H is a normal subgroup of G oforder 3 and so we may apply the argument above. Likewise if N is a normalsubgroup of G order 10 or 20, we may apply the Sylow Theorems to concludethat N has a unique subgroup of order 5 which will then be normal in G andreduce to the case above.

If G has a subgroup, N , of order 30, we may apply the argument givenin Fall 1996 #2 a), to conclude that N has a unique subgroup of order 3.We may then apply the above to conclude that this subgroup is a normalsubgroup of G of order 3. If G has a normal subgroup, N , or order 2, we havethat G/N is a subgroup of order 30 and so has a normal subgroup of order 3.By the Correspondence Theorem, we have a normal subgroup of G of order6 and so may apply the previous argument. Next suppose that G has a sub-group N of order 4. Then G/N has order 15 and so, by the Sylow Theorems,has a normal subgroup of order 5. Thus gives a normal subgroup of G oforder 20 and we may apply the argument above. Lastly suppose that N hasa group of order 12. Then by Fall 2000 #1 a), N has a unique subgroup of ei-ther 3 or 4. We may hence reduce to the previous cases. This covers all cases.

c) Show that there is a group G of order 60, such that if L/K abovehas group G, then there is no irreducible polynomial in K[x] of de-gree 4 with a root in L.

If such a polynomial exists, the field generated over K by a root hasdegree 4 so that by the Galois Correspondence, G has a subgroup of order15. Hence it suffices to find a group of order 60 with no subgroup of order 15.Consider the group A5 and suppose for a contradiction that it has a subgroupH of order 15. Then H has index 4 in A5 and so we have a nontrivial actionof A5 on a set of order 4. This gives a nontrivial homomorphism A5 → S4.By order considerations this homomorphism has a nontrivial kernel whichshows that A4 is not simple. This is a contradiction and so gives the claim.

201

2. Let G be a finite group acting on a finite set X. For u ∈ X, set[u] = v ∈ X|gv = v for some g ∈ G and set Gu = g ∈ G|gu = u.

a) Show that the cardinality of [u] is the index [G : Gu] of Gu.

Let S be the collection of cosets of Gu in G. Consider the map φ : [u] → Sgiven by gu 7→ gGu. To see that this is well-defined, let, suppose thatgu = hu. Then g−1hu = u so that g−1h ∈ Gu which shows gGu = hGu.Likewise, if gGu = hGu, we have g−1h ∈ Gu so that g−1hu = u and hu = gu.This shows that φ is injective. Lastly, φ is trivially surjection as every cosetin S of the form gGu for some g ∈ G. Hence we may conclude that the twosets have the same cardinality which gives the result.

b) If v ∈ [u] then there is some g ∈ G such that gGug−1 = Gv.

First note that if v ∈ [u], we have [u] = [v]. Find g ∈ G with v = gu.Then for hu ∈ [u], we have hu = hg−1v ∈ [v] and for hv ∈ [v], we havehv = hgu ∈ [u]. Now if h ∈ Gu, we have ghg−1v = ghu = gu = v so thatghg−1 ∈ Gv. Hence, gGug

−1 ⊂ Gv. By the above and a), we have |Gu| = |Gv|which implies gGug

−1 = Gv.

c) We say u ∈ X is a fixed point if gu = u for all g ∈ G. Let p be aprime and G a p-group. If X has order prime to p, show that thereis a fixed point. If X has order divisible by p, and X has a fixedpoint, show that it has more than one.

Consider the relation on X given by u ∼ v if and only if [v] = [u]. Thisis clearly an equivalence relations so that we may conclude that sets of thisform (orbits) for a partition of G.

Suppose first that p does not divide |X|and suppose for a contradictionthat there is no fixed point. Let ui be a collection of elements in X suchthatX = [u1]∪. . .∪[un] and the union is disjoint. The we have |X| =

∑|[ui]|.

But |[ui]| = |G|/|Gui| from a). Since Gui

is clearly a subgroup of G and byassumption it is not all of G, we see that |[ui]| is divisible by p. Hence |X| isdivisible by p. This is a contradiction and so we may conclude that there isa fixed point.

Next assume |X| is divisible by p. As above write X as a disjoint unionX = [u1]∪. . .∪[un]∪[vi]∪. . .∪[vk] where each ui is not a fixed point and each

202 Fall 1995

vj is (so that |[vj]| = 1). Now, as above we have |X| =∑

j[vj]+∑

[ui] = k+lpfor some l ∈ Z. Since |X| is divisible by p, we must have k divisible by p.Since it is not zero, this implies that it is at least 2 and the statement secondfollows.

d) Let G be a group of order divisible by p, and X the subset ofG×· · ·×G (p times) of all tuples (g1, . . . , gp) such that g1 · · · gp = 1 ∈ G.Let Cp be the cyclic group of order p generated by σ. Show thatσ((g1, . . . , gp)) = (g2, . . . , gp, g1) defines an action of Cp on X. Concludefrom c) above that X has at least two fixed points. Show that thismeans G has an element of order p.

For gi ∈ G, we have

σi(σj((g1, . . . , gp))) = σi((gj, . . . , gp, g1, . . . gj−1))

= (gi+j, . . . , gp, gi, . . . , gi+j−1)

= σi+j((g1, . . . , gp))

(where we consider our subscripts to modulo p). Next if g1 · · · gp = 1, wehave g2 · · · gpg1 = g−1

1 g1g2 · · · gpg1 = g−11 g1 = 1. Hence applying σ to such a

tuple produces another one. Combining these two, we see that every powerof sigma produces another tuple when acting on a tuple of this type. Addingthe fact that e(g1, . . . , gp) = σp(g1, . . . , gp) = (g1, . . . , gp), we see that we havea well-defined action.

Now the tuple (e, . . . , e) is a fixed point of this action. Hence we mayapply c) to conclude that there is another fixed point. Let this fixed pointbe given by (g1, . . . , gp). Then (g2, . . . , gp, g1) = σ(g1, . . . , gp) = (g1, . . . , gp).Hence we may conclude that gi = gi+1 for each i. So that the tuple is(g, . . . , g) for some g ∈ G, g 6= e (as this would give the previous fixed point).This element then satisfies gp = 1. Its order must divide p and so must beeither 1 or p. Since it is not 1, we may conclude that g is an element of orderp in G.

3. Let F = GF (q) be the finite field of q elements. For f(x) ∈ F [x],

203

define the function φF,f : F → F by setting φF,f (a) = f(a). (i.e. thevalue of f(x) with a substituted for x). The point here is that f(x)is a formal polynomial and φF,f is the associated function on thefield F .

a) Find f(x) and g(x) such that f(x) 6= g(x) but φF,g = φF,f .

Since F× has q−1 elements, we see that aq−1 = 1 for every a ∈ F×. Henceaq − a = a(aq−1 − 1) = 0 for every a ∈ F . Thus the function associated tothe polynomial xq − x gives the zero function as does the zero polynomial.These polynomials are certainly not equal.

b) Show that φF,f = 0 if and only if f(x) is divisible by xq − x.

Let I be the collection of polynomials which gives zero such that the cor-responding function is the zero function. Then it is trivial to check that I isan ideal. Hence, I is generated by the unique monic polynomial of minimaldegree which it contains. Suppose then that f(x) is this polynomial. Thenfor each a ∈ F , we must have (x−a) that divides f(x). By unique factoriza-tion, we may conclude that the product of all such polynomials divides f(x).Hence f(x) has a degree of at least q. By a), however, xq − x ∈ I. Hence wemay conclude that I = (xq − x) and the statement follows.

c) Suppose that for any finite field K ⊃ F , φK,f = φK,g. Show thatf(x) = g(x).

First we have that φK,f−g = φK,f − φK,g = 0 (again since evaluation is ahomomorphism). Now if K = Fqn , we may apply a similar argument as ina) and b) to conclude that a polynomial gives the zero function on K if andonly if it is divisible by xq

n − x in K[x]. Hence, f(x) − g(x) is divisibly byxq

n − x for any n and so, if it isn’t zero, it has degree greater than qn foreach n, which is a contradiction. Hence f(x)− g(x) = 0 and so f(x) = g(x).

d) Suppose φ : F → F is any function. Show that there is an f(x)with φ = φF,f .

Let for a ∈ F , let fa(x) ∈ F [x] be given by the product of all linearpolynomials of the form x − b with b ∈ F and a 6= b. Then we see that

204 Fall 1995

fa(a) 6= 0 and fa(b) = 0 for b 6= 0. Define f(x) ∈ F [x] by

f(x) =∑a∈F

φ(a)

fa(a)fa(x).

We see then that φF,f = φ.

4. Suppose R is a ring and A,P,Q are left R modules. Assume weare given R module surjective morphisms f : P → A and g : Q→ A.Let K,L be the kernels of f, g respectively. Assume P is projective,which by definition implies that there is an R module homomor-phism h : P → Q with g h = f .

a) Set I = (p, q)|f(p) = g(q) ⊂ P ⊕Q. Show that L⊕ P ∼= I.

Consider the map φ : L ⊕ P → I given by (l, p) 7→ (p, l + h(p)). First,that for l ∈ L and p ∈ P , we have g(l + h(p)) = g(l) + f(p) = f(p) so thatthe map is well-defined. Next, let m,n ∈ L, a, b ∈ P and r ∈ R. Then wehave that

φ((m, a) + (n, b)

)= φ(m+ n, a+ bh)

=(a+ b,m+ n+ h(a+ b)

)=(a+ b,m+ h(a) + n+ h(b)

)=(a,m+ h(a)

)+(b, n+ h(b)

)= φ(m, a) + φ(n, b)

205

and

φ(r(m, a)

)= φ(rm, ra)

=(ra, rm+ h(ra)

)= r(a,m+ h(a)

)= rφ(m, a).

Hence φ is an R-module homomorphism. Next, suppose that φ(m, a) = 0.Then (a,m + h(a)) = 0 and so a = 0. This gives h(a) = 0 which givesm = 0. Hence φ is injective. Lastly suppose that (p, q) ∈ I the f(p) = g(q).We also have g(h(p)) = f(p) so that there must be some l ∈ L such thatq = l + h(p). Then φ(l, p) = (p, q) and so φ is surjective. Hence φ is an Rmodule isomorphism and we have the claim.

b) From now on assume that Q is projective. Show that L ⊕ P ∼=K ⊕Q.

If Q is projective, we may apply the same argument above to concludethat K ⊕Q ∼= I. Applying the result of part a) gives the claim.

c) If K is projective show that L is also.

One of the equivalent formulations of a module M being projective isthat it is the direct summand of a free module. Hence we may find a freemodule FK such that FK ∼= BK for some module BK . Likewise we have afree module FQ with FQ ∼= Q⊕ BQ. Then F ∼= FQ ⊕ FK is certainly free (adirect product of direct products of copies of R is a direct product of copiesof R) and we have F ∼= K ⊕ Q ⊕ BK ⊕ BQ

∼= L ⊕ P ⊕ BK ⊕ BQ. HenceL is a direct summand of the free module F which shows that it is projective.

d) Assume R is a Dedekind domain and A is a cyclic module overR. If f : P → A is as above, show that K is projective.

The Cartan-Eilenberg Theorem states that every submodule of a projec-tive module over a Dedekind domain is projective. This implies that that Kis projective (I don’t see why the assumption that A is cyclic is given).

206 Fall 1995

5. Let k be a field, and let R be a k algebra that is finite dimen-sional as a vector space over k. If M is a left module over R, letgen(M) be the minimal number of generators of M as a moduleover R.

a) Show that gen(M) is finite if and only if M is finite dimensionalover k. Give an upper bound for dimk(M) in terms of dimk(R) andgen(M).

Let n be the dimension ofR over k and find a basis b1, . . . , bn ofR over k.Suppose first that gen(M) = s is finite. Then we may find m1, . . . ,ms ∈ Mwhich generate M as an R-module. Let m ∈ M . Then we may find anr1, . . . , rs ∈ R such that m = r1m1 + · · · + rsms. Moreover we may findci,j ⊂ k such that ri = ci,1b1 + · · ·+ ci,kbk for each i. This gives

m =∑i,j

ci,jbjmi

so that bjmi generates M as a k-vector space. Since this is a finite set, wemay conclude that M is finite dimensional over k. Also note that we havedimk(M) ≤ dimk(R)gen(M). Conversely, suppose that M is finite dimen-sional over k. Then we can find a finite set which generates M over k. Inparticular, this set will generate M over R and so gen(M) will be boundedby the cardinality of this set and so be finite.

b) Find a relationship between dimk(R), gen(M), and dimk(M) thatis equivalent to M being free over R.

First let mii∈I be a set of generators for M (over R) and b1, . . . , bn abasis for R over k. Suppose that we have some nontrivial dependence relation∑rimi = 0 among the generating set. Then we have some ci,j ⊂ k such

that ri = ci,jbj so that∑ci,jbjmi = 0. Since we have at least one ri nonzero,

we must have at least one ci,j nonzero which implies that bjmi is nota linearly independent set over k. Conversely suppose that we have somenontrivial dependence relation

∑ci,jbjmi among the set bjmi. Then the R

coefficient on mi is∑ci,jbj. Since we have at least one ci,j nonzero, we may

apply linear independence of the basis bj over R to conclude that at leastone of the coefficients on and mi is nonzero. This shows that mi is not

207

linearly independent over R. Hence we see that mi is linearly independentover R if and only if bjmi is linearly independent over k.

Now suppose that M is free. Then we have some generating set miwhich is independent over R. In particular this is a minimal generating set(if we were able to write one of the elements in terms of the others, we wouldhave a nontrivial dependence relation). Hence from a), bjmi is a generatingset forM over k which is independent over k by the above. If gen(M) happensto be finite we have then have the relationship dimk(M) = dimk(R)gen(M).Conversely suppose that we have this relationship and gen(M) is finite. Letmi be a minimal generating set. Then bjmi generates M over k andhas the same cardinality of a basis and so must be a basis. In particular,it is linearly independent, which by the above, guarantees that mi is lin-early independent over R and so a basis for M over R. Hence M is free.Thus we have shown that, in the finite case, M is free over R if and onlyif dimk(M) = dimk(R)gen(M). For obvious reasons, this cannot general-ize to the infinite case. In that case, we have M is free if and only if it hasa generating set mi over R such that mibj is linearly independent over k.

c) Assume from now on that R is a simple algebra (has no non-trivial ideals). Recall that x ∈ R|xr = rx for all r ∈ R is called thecenter of R. Show that the center of R is a field.

Let Z be the center of R. First, if x, y ∈ Z, and r ∈ R, we have r(x+y) =rx+ ry = xr+yr = (x+y)r so that the center of R is closed under addition.Likewise, rxy = xry = xyr so that the center is closed under multiplication.Certainly 1 ∈ Z which shows that Z is a commutative subring of R. It thensuffices to show that each element of the center has a multiplicative inversein the center.

Suppose then that x ∈ Z. Consider the set Rx. If r1, r2 ∈ R thenr1x + r2x = (r1 + r2)x so that the set is closed under addition. Moreover,if rx ∈ Rx and s ∈ R, we have srx ∈ Rx and rxs = rsx ∈ R (sincex ∈ Z). Hence Rx is a two sided ideal in R. Since R is simple, Rx mustequal R and so we have some u ∈ R such that ux = 1. Since x is inthe center of R, we also have ux = 1. Finally, suppose that r ∈ R. Thenrux = r = uxr = urx. Hence we have (ru−ur)x = 0 so that (ru−ur)ux = 0which gives (ru− ur) = 0. From this we may conclude that ur = ru and sou ∈ Z. Hence Z is a field.

208 Fall 1995

d) Suppose further that k is algebraically closed. If R is a divisionalgebra, show that R = k.

Again let Z be the center of R. By the definition of algebra, we have aring homomorphism k → Z (in particular 1 7→ 1). Since k is a field, thismap must be injective and so we may conclude that k is (isomorphic to) asubring of R which is contained in the center. Now suppose that α ∈ R. LetK = k(α) be the subdivision ring generated by α and k. Since k lies in thecenter of R, k commutes with α so that K is commutative. Hence k ⊂ K isan extension of fields. Moreover, K is contained in a finite vector space overk (namely R) and so must be finite dimensional itself. In particular, K is analgebraic extension of k, an algebraically closed field. This gives K = k sothat α ∈ k. We may hence conclude that R = k.

e) We continue to assume R is finite dimensional, simple over k andk is algebraically closed. Show that R is isomorphic to to the n× nmatrix ring Mn(k) for some n.

If I1 ⊃ I2 ⊃ . . . is a decreasing sequence of left ideals in R, then it is, inparticular, a decreasing sequence of k-subspaces of R. By dimension consid-erations (since R is finite dimensional over k), this sequence must stabilize.Hence R is Artinian. Now R is simple and Artinian so that in particular it issemisimple. Hence, by the Wedderburn Theorem, it is isomorphic to a directsum of matrix rings over division rings. Now the first factor of this directsum is an ideal, so that, since R is simple, it must equal this first factor.Hence R ∼= Mn(∆) for some n and division ring ∆.

Now the center of a matrix ring is contained in the collection of diagonalmatrices. Hence since Mn(∆) is a k-algebra (since R is a k-algebra), we havea ring homomorphism k → ∆ where we identify ∆ with the collection ofdiagonal matrices. But then by c), we have ∆ = k. Hence we may concludethat R ∼= Mn(k), a claimed.

Chapter 24

Spring 1995

1. Let G be a group and let H be a group of finite index.

(a) If g is an element of G show that there exists a smallest positiveinteger k such that gk ∈ H and show that k divides every integer mfor which gm ∈ H.

Consider the collection of cosets giH. Since H has finitely many cosetsin G, we may conclude that at least two of the above is equal. Hence we havesome i < j such that giH = gjH. This gives gj−iH = H so that gj−i ∈ H.Hence the set of positive integers m such that gm ∈ H is nonempty. By theWell-Ordered Property of the positive integers, we may conclude that thereis least such integer, k.

Suppose that gm ∈ H. Let d be the (positive) greatest common divisorof m and k. Then there is some s, t ∈ Z such that d = sm + tk. This givesgd = (gm)s(gk)t ∈ H so that k ≤ d. But we already have d ≤ k. Hence wemay conclude that the greatest common divisor of k and m is k which showsthat k divides m.

(b) Show that if H is normal in G, the number k in (a) must dividethe index of H in G.

If H is normal, G/H is a group and k is the order of gH in G/H and so,by Lagrange’s Theorem, must divide the order of the group. The order ofthe group is the index of H in G.

210 Spring 1995

(c) Show by a counter-example that the statement in (b) does notalways remain true, if H is not normal in G.

Consider the group S3 with H = 1, (12). We see that |H| = 2 so thatthe index of H in S3 is 3. Now (23) 6∈ H, but (23)2 = 1 ∈ H. Hence we seethat the number k from (a) which corresponds to (23) is 2 which does notdivide 3, the index of H in S3.

2. Quoting any general theorem you wish about modules over prin-cipal ideal domains, or otherwise, prove the following two theorems.

(a) Let G be a finite abelian group. Then G is cyclic if and only ifits order is the smallest positive integer m such that gm = 1 for allg ∈ G.

First suppose that G is cyclic. Then every group raised to the orderof G is 1. Likewise there is an element of order |G| in G (a generator ofG) so that no smaller positive integer will give the result. This gives oneimplication. Conversely suppose that G is finite and abelian and its ordersatisfies the statement above. Then from the Structure Theorem for FiniteAbelian Groups (which is a special case of the Structure Theorem for Modulesover a PID), we have

G ∼= Zn1 × Zn2 × Znk

for some ni ≥ 2 with n1|n2| · · · |nk. Now we see that gnk = 1 for each elementof G and that the order of G is n1n2 · · ·nk. Hence we must have k = 1 sothat G ∼= Zn1 for some n1. Thus G is cyclic.

(b) Let V be a finite dimensional vector space over a field F , and letT : V → V be a linear transformation. We say that T acts cycliclyon V if and only if there exists and element v0 ∈ V such that theelements T iv0, i = 0, 1, 2, . . . span V . Show T acts cyclically on V ifand only if dimV is the degree of the monic polynomial f(x) ∈ F [x]of least degree such that f(T )v = 0 for all v ∈ V .

211

Suppose first that the statement about the degree of the polynomial holds.Now in the usual way, T makes V into an F [x]-module and, applying theStructure Theorem for Modules over a PID, we have

V ∼= F [x]/(a1(x))⊕ · · · ⊕ F [x]/(an(x))

as an F [x]-module for some nonconstant a1(x)| · · · |an(x) (there are no freefactors as this would imply that V is infinite dimensional over F ). We seethat the dimension of V over F is deg(a1(x)) + · · · + deg(an(x)) and thatan(x) acts trivially on all elements of the module. Hence an(T )v = 0 forall v ∈ V . This gives deg(a1(x)) + · · · + deg(an(x)) ≤ deg an(x) from whichme may conclude that n = 1 so that V ∼= F [x]/(a1(x)). Now, the latter isspanned by xi · 1 which shows that V is spanned by T iv0 where v0 is thevector which corresponds to 1.

Conversely, suppose that T acts cyclically on V . Then again finding anisomorphism

V ∼= F [x]/(a1(x))⊕ · · · ⊕ F [x]/(an(x))

of the above form, we see that xiv0 spans the latter for some v0. Now, an(x)acts as zero each vector. an(x)v0 = 0 shows that xkv0, for k ≥ deg(an(x))can be written in terms of those with powers less than deg(an(x)). This givesdimF V ≤ deg(an(x)). But the degree of F [x]/(a1(x))⊕· · ·⊕F [x]/(an(x)) is∑

deg(ai(x)). Hence we may conclude that n = 1 so that the dimension ofV is deg(an(x)), which is the degree of the monic polynomial of least degreedescribed above.

3. F ⊂ E be a field extension.

(a) Show that these is a unique F -linear transformation

φ : E ⊗F E → HomF (E,E)

such that for a, b, c ∈ E,

(φ(a⊗ b)) (c) = abc.

212 Spring 1995

Consider the map Φ : E × E → HomF (E,E) given by Φ(a, b)(c) = abc.First we will show that this is well defined. Let a, b, c, d ∈ E and r ∈ F .Then

Φ(a, b)(c+ d) = ab(c+ d)

= abc+ abd

= Φ(a, b)(c) + Φ(a, b)(d)

and

Φ(a, b)(rc) = abcr

= rabc

= rΦ(a, b)(c),

which shows that Φ is well-defined. Now, suppose that a1, a2, b1, b2, c ∈ E.Then we have

Φ(a1 + a2, b1)(c) = (a1 + a2)b1c

= a1b1c+ a2b1c

= Φ(a1, b1)(c) + Φ(a2, b1)(c),

which shows that Φ(a1 + a2, b1) = Φ(a1, b1) + Φ(a2, b1). Likewise, we haveΦ(a1, b1+b2) = Φ(a1, b1)+Φ(a1, b2). Next suppose that a, b, c ∈ E and r ∈ F .Then we have

Φ(ra, b)(c) = rabc

= arbc

= Φ(a, rb)(c)

= arbc

= rabc

= rΦ(a, b)(c),

which shows that Φ(ra, b) = Φ(a, rb) = rΦ(a, b). Hence Φ is F -bilinear.Now, we may apply the Universal Property of the the Tensor Product toconclude that there is a unique F -linear transformation as above.

213

(b) Suppose the extension is finite; let n = [E : F ] be its degree.What is the dimension of the image of φ?

Suppose∑ai ⊗ bi ∈ E ⊗F E. Then for c ∈ E, we have

φ(∑

ai ⊗ bi

)(c) =

∑φ(ai ⊗ bi)(c)

=∑

(aibic)

=(∑

aibi

)c.

Hence every element in the image of φ is of the form fa for some a ∈ E wherefa(c) = ac for c ∈ E. Certainly every such map is in the image (it is theimage of (1, c) for example). Let V be this subset of HomF (E,E). Then issuffices to find the dimension of V over F .

Consider the map ψ : V → E given by fa 7→ fa(1) = a. Then forfa, fb ∈ V we have ψ(fa + fb) = a + b = ψ(fa) + ψ(fb). For r ∈ F , we haveψ(rfa) = rfa(1) = ra = rψ(fa). Hence ψ is an F -linear map. Now, supposefa maps to 0. Then a = 0 so that fa = 0. Certainly the map is surjective.Hence it is an isomorphism, we may conclude that dimF V = dimF E = n.Thus the dimension of the image is n.

(c) What is the dimension of the kernel of φ?

From the First Isomorphism Theorem, we have E⊗FE/ ker(φ) ∼= image (φ).Hence, by (b), we may conclude that E ⊗F E/ dimF ker(φ) has dimension n.Now the dimension of E ⊗F E is n2. Hence we have n2 − dimF ker(φ) = n.This gives dimF ker(φ) = n2 − n.

(d) Give a basis for the kernel of φ in the special case F = R (reals)and E = C (complexes).

From the above the dimension of the kernel is 22−2 = 2. Now, a basis forC over R is 1, i. Hence a basis for C⊗R C over R is 1⊗1, 1⊗ i, i⊗1, i⊗ i.Now for c ∈ C, we have φ(1⊗1+ i⊗ i)(c) = c+ i2c = 0. Hence 1⊗1+ i⊗ i ∈ker(φ). Next, φ(1⊗ i− i⊗ 1)(c) = ic− ic = 0. Hence 1⊗ i− i⊗ 1 ∈ ker(φ).Now if 1⊗ 1 + i⊗ i, 1⊗ i− i⊗ 1 were linearly dependent, we would havea, b ∈ R such that a(1 ⊗ 1) + a(i ⊗ i) + b(1 ⊗ i) − b(i ⊗ 1) = 0. By linearindependent of the basis for C ⊗R C, we have a = 0 and b = 0. Hence

214 Spring 1995

1⊗ 1 + i⊗ i, 1⊗ i− i⊗ 1 is linear independent so that it forms a basis forker(φ).

4. Let p be a prime. Let F be a field such that Xp − 1 has a rootz 6= 1 in F .

(a) Prove that the characteristic of F is not p.

Suppose that F has characteristic p, then α 7→ αp is a ring homomor-phism. This gives Xp − 1 = Xp − 1p = (X − 1)p. Hence the only solutionto Xp − 1 in characteristic p is 1. Since we have assumed this not to be thecase, F does not have characteristic p.

(b) Prove that Xp − 1 =∏p

i=1(X − zi).

Since zp = 1, we see that the order of z as an element of the group F×

divides p. By assumption, z 6= 1. Hence the order of z is p. This shows thatzipi=1 is a distinct set. Moreover (zi)p = (zp)i = 1. Hence the above setgives the p distinct solutions to Xp − 1. This gives the claim.

(c) Let α be an element of an extension of F such that α 6∈ F butαp = a ∈ F . Show F (α) is a Galois extension of p.

From above, we have αzipi=1 is a distinct set. We have also (αzi)p =αp = a. Hence the collection gives the p roots of Xp − a. This gives thatthe splitting field of Xp− a over F is contained in F (α). Since α is a root ofXp−a, we also have the reverse inclusion. This also shows that α is separableso that F (α)/F is a separable extension. Hence F (α)/F is Galois.

(d) Let G = Gal(F (α)/F ) be the Galois group. Show that for eachσ ∈ G there is a unique integer i(σ) in the interval 1 ≤ i(σ) ≤ p suchthat σ(α) = zi(σ)α.

We will show that Xp − a is irreducible over F . Note that by the above,

215

Xp−a =∏p

i=1(X−ziα). Hence ifXp−a is to split into two nonconstant poly-nomials, the constant term of one of them must be αkzj for some 1 ≤ k < pand some j ∈ F . Now, we have now k and p are relative prime, so we havesome s, t ∈ Z such that sk + tp = 1. This gives (αkzj)s = α/(αtp)zjs ∈ F .Since zjs ∈ F and αtp = at ∈ F , we may conclude that α ∈ F , a contradic-tion. Hence Xp−a is irreducible. This shows that α must be mapped to oneof αzipi=1. This shows existence of the above integer. Uniqueness followsfrom the fact that αzipi=1 is a distinct set (by the above).

(e) Show that the map σ 7→ i(σ) (mod p) is an isomorphism of Gonto Z/pZ.

Let φ be the map. Then φ is well-defined by (e). Suppose then thatσ, τ ∈ G. Then

στ(α) = σ(zi(τ)α)

= zi(τ)σ(α)

= zi(τ)zi(σ)α

= zi(τ)+i(σ)α,

which shows that φ(στ) = i(σ) + i(τ) = φ(σ) + φ(τ) (of course we are view-ing all the exponents module p). Next if i(σ) = 0, σ(α) = α so that σ isthe identity (as F (α) is of course generated over F by α). Then the map isinjective. Lastly we have |G| = p since, by (d), the minimal polynomial ofα over F has degree p. This shows that the groups have equal order and wemay hence conclude that the map is surjective and so an isomorphism.

(f) Conversely, suppose now that E/F is Galois with group G cyclicof order p. Let σ ∈ G, σ 6= 1. Show that there is an element α ∈ E,α 6= 0 such that σ(α) = zα.

Let τ = σ−1 and r : E → E be given by

r(β) = β + zτ(β) + z2τ 2(β) + · · ·+ zp−1τ p−1(β).

Since 1, τ, . . . , τ p−1 are distinct automorphisms on E, there restrictionsto E× are distinct characters E× → E×. Hence by the fact that distinctcharacters are linearly independent, we may conclude that these functions

216 Spring 1995

are linearly independent. In particular, r : E → E cannot be the zerofunction. Hence there is some β ∈ E such that r(β) 6= 0. Now we have

τ(r(β)

)= τ(β + zτ(β) + z2τ 2(β) + · · ·+ zp−1τ p−1(β)

)= τ(β) + zτ 2(β) + z2τ 3(β) + · · ·+ zp−2τ p−1(β) + z−1β

= z−1(β + zτ(β) + z2τ 2(β) + · · ·+ zp−1τ p−1(β)

)= z−1r(β),

since τ preserves z. Hence σ(r(β)) = τ−1(r(β)) = zr(β). Hence α = r(β) issuch an element.1

(g) Show α 6∈ F , but αp ∈ F and E = F (α).

α 6∈ F since σ preserves every element of F and z 6= 1. Likewise, σ(αp) =σ(α)p = zpαp = αp shows that αp ∈ F (since then every power of σ will alsopreserve αp). Finally, E/F is Galois with group of prime order. Hence thereare no proper nontrivial subgroups of the Galois group, which shows, bythe Galois Correspondence, that there are no nontrivial intermediate fieldsbetween E and F . Hence F (α) = E or F (α) = F . The latter case isimpossible since α 6∈ F . Hence we have E = F (α).

1This proof is given in Dummit and Foote on p. 626

Part II

Stanford University

Chapter 25

Stanford Fall 2006

25.1 Morning Session

1. In parts (a) and (c), let G be a nonabelian group of order 56.

(a) Prove that G has a normal 2-Sylow subgroup or a normal 7-Sylow subgroup.

We have |G| = 56 = 7 · 23. From the Sylow Theorems, the number 2-Sylow subgroups is 1 or 7. Likewise the number of 7-Sylow subgroups iseither 1 or 8 (as 2 and 4 are not equivalent to 1 modulo 7). Suppose wedo not have a normal 7-Sylow subgroup. Then we have 8 · 6 = 48 elementsof order 7 as the Sylow 7-subgroups have prime order and so must intersecttrivially. Hence we have only 56− 48 = 8 elements which do not have order7. These elements thus comprise the unique subgroup of order 8 and so wehave unique 2-Sylow subgroup.

(b) Let Zn denote a cyclic group of order n. Compute the order ofAut(Q) when Q = Z8, Z4 × Z2, and Z2 × Z2 × Z2.

By definition Z8∼= 〈x|x8 = 1〉. Let F be the free group generated by x.

Then from the Universal Property of Free Groups, we have a homomorphismF → Z8 given by x 7→ xk where 0 ≤ k ≤ 7. Now, (xk)8 = 1 so thatthis map factors through the normal subgroup generated by x8. Hence wehave a homomorphism Z8 → Z8 given by x 7→ xk. Next assume that k is

220 Stanford Fall 2006

relatively prime to 8. Then we have s, t ∈ Z such that sk + 8t = 1. Thisgives (xk)s = x1−8t = x. Hence x is in the image of this map so that themap is surjective. By order considerations, the map is also injective and soan automorphism. Now suppose that Z8 → Z8 is an automorphism then xmust map to xk for some k. Suppose that k is not relatively prime to 8, thatis, it is divisible by 2. Then 4k is divisible by 8 and so we have (xk)4 = 1.This implies that xk does not generate Z8, so that our map is not surjective,a contradiction. We may hence conclude that |Aut(Z8)| = 4.

Now Z4 × Z2 is generated by (1, 0) and (0, 1). Now the image of (1, 0) inan automorphism must be another element of order 4. This gives the choices(1, 0), (1, 1), (3, 0), and (3, 1). Likewise, (0, 1) must map to (0, 1), (2, 0), and(2, 1), but it cannot map to the square of the element to which (1, 0) maps(as then the image of our automorphism will only be a subgroup of order 4and hence not the entire group). This gives 8 possible automorphisms. Nowwe have

Z4 × Z2 =⟨x, y|x4 = y2 = xyx−1y−1 = 1

⟩.

Hence, by a similar argument as above, all of the maps above are well-definedhomomorphisms. Furthermore, they are surjective since the image has ordergreater then 4. Thus we have |Aut(Z4 × Z2)| = 8.

Finally Z2×Z2×Z2 is a 3-dimensional vector space over Z2 = F2. Henceits automorphism group is isomorphic to GL3(F2). Now to determine theorder of this group, suppose we are choosing such a matrix. Then we maypick any column other than the zero column for the first column, 7 choices.We may then pick an column for the second column that is not an F2-linearcombination of the first, 6 choices. Finally, the third column may be anythingbut a linear combination of the first two columns, giving 4 choices. Hencethe order of the group is 7 · 6 · 4 = 168.

(c) How many isomorphism classes are there of nonabelian groupsof order 56 with a normal abelian 2-Sylow subgroup? Explain.

Before we give our result, we will need to prove a lemma. Suppose thatK is a cyclic group of prime order p and H is any group. Suppose thatφ1, φ2 : K → Aut(H) are homomorphisms such that φ1(K) and φ2(K) areconjugate subgroups in Aut(H). Then H oφ1 K

∼= H oφ2 K. To see this,find σ ∈ Aut(H) such that σφ1(K)σ−1 = φ2(K). Since K is of prime order,φ1(K), and φ2(K) are either trivial or isomorphic to K. By the fact that

25.1 Morning Session 221

they are conjugate, they have the same order. Hence if φ1 is trivial, φ2 istrivial so that we have H oφ1 K

∼= H ×K ∼= H oφ2 K. Suppose then thatφ1 is not trivial. Then φ2 is not trivial. Hence we have φ1(K) ∼= φ2(K) ∼=K. Let x be a generator of K. Then φ2(K) is generated by φ2(x) so thatσφ1(x)σ

−1 = φ2(x)a for some 0 < a < p (if a were 0, we have would have

φ1(x) = 1 so that φi(K) would be trivial).Now consider the map ψ : H oφ1 K → H oφ2 K given by ψ(h, k) =

(σ(h), ka). To see that this is a homomorphism, let h1, h2 ∈ H and xa1 , xa2 ∈K. Then

ψ((h1, x

a1)(h2, xa2))

= ψ(h1φ1(x

a1)(h2), xa1xa2

)=

(σ(h1φ1(x

a1)(h2)), (xa1xa2)a

)=

(σ(h1)σφ1(x

a1)(h2), xaa1xaa2

)=

(σ(h1)φ2(x)

aa1σ(h2), xaa1xaa2

)=

(σ(h1)φ2(x

a1a)σ(h2), xaa1xaa2

)=(σ(h1), x

aa1

)(σ(h2), x

aa2

)= ψ(h1, x

a1)ψ(h2, xa2),

which shows that ψ is a homomorphism. Now since σ is an automorphism,its surjective. Likewise since a is relative prime to p, k 7→ ka is surjectiveon K (as above). This shows that ψ is surjective. Likewise, if ψ(h, k) = 1,we have σ(h) = 1 and ka = 1 which implies that h and k are the identity,since both individual maps are injective. Hence ψ is injective and so ψ is anisomorphism. The claim follows.1

Now suppose that G is a nonabelian group of order 56 with a normalabelian 2-Sylow subgroup, H. Let K be a 7-Sylow subgroup. From orderconsiderations, H ∩ K = 1 and HK = G. Hence G is the semidirectproduct of H and K with H normal. This gives a homomorphism θ : K →Aut(H), which cannot be trivial (if it is trivial, G ∼= H ×K which is abelianas both H and K are abelian). But if H is isomorphic to either Z4 × Z2 orZ8, 7 does not divide the order of Aut(H), from part (b). Hence θ cannot be

1This proof is outline in an exercise in Dummit and Foot, p.184 #6

222 Stanford Fall 2006

nontrivial in this case. Hence we have H ∼= Z2 × Z2 × Z2 and Aut(H) hasorder 168. Now any choice for a nontrivial θ gives |θ(K)| = 7. Hence θ(K) isa 7-Sylow subgroup of Aut(H). By the above lemma, all such choices resultin isomorphic groups. Hence there is only one such group up to isomorphism.It is constructed by finding an element of Aut(H) of order 7 and mapping x tothis element, where x is a generator of K and then taking the correspondingsemidirect product.

2. Let G be the following group of order 42:

G =⟨x, y|x7 = y6 = 1, yxy−1 = x2

⟩.

Determine the conjugacy classes of G and the degrees of its irre-ducible characters. Compute the values of at least one irreduciblecharacter of degree greater than 1.

First note that the relation yx = x2y can be used to put any word in xand y in the form xiyj for some i, j. Since x7 = y6 = 1, there are at most 42distinct expressions of this form. Since we are given that the group has order42, they must all be distinct and so the group consists of this collection of 42elements. In particular the order of x is 7 and that of y is 6. Furthermore, therelation yx = x2y also allows us to conclude that conjugating xiyj by eitherx or y will result in an expression of the form xkyj. Hence the conjugacyclass of xiyj is contained in xkyj6

k=0.Now let 0 ≤ j ≤ 5 be such that 2j 6= 1 mod7. Then for 0 ≤ k ≤ 6, we

have

xkyj(xk)−1 = xkyjx7−k

= xkyj−1(x2)7−ky

= xkyj−1x2(7−k)y

= . . .

= xkx2j(7−k)yj

= xk+2j(7−k)yj.

25.1 Morning Session 223

The Conjugacy Classes of G.

1y3

x, x2, x4

x3, x5, x6

xy3, x2y3, x4y3

x3y3, x5y3, x6y3

y, xy, x2y, x3y, x4y, x5y, x6yy2, xy2, x2y2, x3y2, x4y2, x5y2, x6y2

y4, xy4, x2y4, x3y4, x4y4, x5y4, x6y4

y5, xy5, x2y5, x3y5, x4y5, x5y5, x6y5

Suppose that for 0 ≤ k, l ≤ 6, two show such expressions are equal. Thenk + 2j(7 − k) and l + 2j(7 − l) are equivalent modulo 7 which implies that(k− l) + 2j(l− k) = (k− 1)(1− 2j) is zero modulo 7. Since Z7 is an integraldomain, and we have assumed that 2j is not 1 modulo 7, we may concludethat k = l modulo 7. Since 0 ≤ k, l < 6, we may conclude that k = l.Hence we have 7 distinct expressions which shows that for 2j 6= 1 mod7, theconjugacy class of yj contains xiyj for each i. By the above, then, this is theconjugacy class.

Next note that yxy−1 = x2, yx2y−1 = x4, and yx4y−1 = x. Hence thecollection x, x2, x4, of conjugate elements, is preserved when conjugated byy. It is certainly preserved (as it is preserved pointwise) when conjugated byy. Hence it is a conjugacy classes in G. Likewise, yx3y−1 = x6, yx6y−1 =x5, and yx5y−1 = x3 so that x3, x5, x6 is a conjugacy class. Moreover,xy3x−1 = xy3x6 = xx48y3 = y3 so that y3 lies in the center of G and so y3is a conjugacy class. Of course 1 is also a conjugacy class.

Finally, we have y(xy3)y−1 = x2y3, y(x2y3)y−1 = x4y3, and y(x4y3)y−1 =xy3 so that the conjugacy class of xy3 contains xy3, x2y3, x4y3. Moreoverthis collection is preserved by y. Since y3 commutes with x all of the aboveelements commute with x. Hence x preserves the collection and so it is a con-jugacy class. Since all the other elements are accounted for, we may concludethat the remaining conjugacy classes must be contained in x3y3, x5y3, x6y3.But we have y(x3y3)y−1 = x6y3 and y(x5y3)y−1 = x5y3. Hence this collectionis a conjugacy class. A table of the conjugacy classes is given below.

Now let ζ be a 6th root of unity (not necessarily primitive) and let F be

224 Stanford Fall 2006

the free group generated by x and y. Then from the Universal Property ofFree Groups, we have a homomorphism F → C× given by x 7→ 1 and y 7→ ζ.Moreover, we have 17 = ζ6 = 1 and ζ · 1 · ζ−1 = 1 = 12 so that the normalsubgroup generated by x7, y6, and yxy−1x−2 lies in the kernel of this map.Hence the map factors through this subgroup and we have a homomorphismG→ C× given by x 7→ 1 and y 7→ ζ. Since there are 6 6th roots of unity, thisgives 6 distinct one-dimensional (and so irreducible) representations. Sincethere are 10 conjugacy classes in G, we have 4 remaining representations. Ifa, b, c, d are the degrees of the representations, we must then have a2+b2+c2+d2 = 42− 6 = 36. Now, if one of these degrees is 6 or more, we already havethat its square is at least 36, which clearly can’t happen. Hence each of thesedegree must be less than 3 (as the degree must divide the order of the group).Now 4 · 9 = 36 so no degree can be less than 3. Hence we have 6 degree 1irreducible representations and 4 degree 3 irreducible representations.

Now let ξ be 7th root of unity and consider the matrices

X =

ξ 0 00 ξ2 00 0 ξ4

and Y =

0 1 00 0 11 0 0

.

We can check that X7 = 1, Y 6 = 1, and Y XY −1 = X2. Hence, as above, wehave a homomorphism G → GL3(C) given by x 7→ X and y 7→ Y . We canthen compute the character of this representation which is as follows, whereθ = ξ + ξ2 + ξ4 and ψ = ξ3 + ξ5 + ξ6. We can check that the norm of thisrepresentation is 1 so that is irreducible.

g 1 y3 x x3 xy3 x3y3 y y2 y4 y5

χ(g) 3 3 θ ψ θ ψ 0 0 0 0

3. Find an extension E of Q with Gal(E/Q) ∼= (Z/3Z)× (Z/3Z).

See Fall 2001 #4(ii).

25.1 Morning Session 225

4. Suppose J is an n × n matrix over an algebraically closed fieldof characteristic not equal to 3 and minimal polynomial (T − λ)n

where λ 6= 0. Find the Jordan canonical form of J3.

Suppose first that (J3 − λ3I)k = 0 for some k. Then we have (J −λI)k(J − λζI)k(J − λζ2)k = 0, where ζ is a primitive 3rd root of unity.Hence f(x) = (x − λI)k(x − λζI)k(x − λζ2)k is divisible by the minimalpolynomial (x−λ)n for J . We but since we are not working in characteristic3, ζ, ζ2 6= 1 and so, since λ 6= 0, we have ζλ, ζ2λ 6= λ. Hence we mayapply unique factorization to conclude that k ≥ n. Likewise, by the samefactorization, we have (J3 − λ3)n = 0. Hence the minimal polynomial for J3

is (x− λ3)n. From this we may conclude that the Jordan Canonical form ofJ3 consists of 1’s above the diagonal and has diagonal entries equal to λ3.

5. In this exercise, “commutative ring” means “commutative ringwith unit,” and it is assumed that a ring homomorphism f : A→ Bsatisfies f(1A) = 1B. If A, B, and C are commutative rings, we saythat C is a coproduct of A and B if there exist ring homomorphismsα : A → C and β : B → C such that if D is any commutative ring,and f : A→ D and g : B → D are ring homomorphisms, there existsa unique ring homomorphism φ : C → D such that f = φ α andg = φ β.

(a) If A and B are commutative rings, regard them as Z-modules,and let A ⊗ B = A ⊗Z B. Explain briefly why A ⊗Z B naturally hasthe structure of the commutative ring.

First any tensor product of modules naturally has the structure of anabelian group (in fact it has an R-module structure if both rings are R-modules and the tensoring is over R). We can also define a multiplication onA⊗B by (a⊗ b)(c⊗ d) = (ac)⊗ (bd), extending in the obvious way to sumsof simple tensors by using the ring axioms. This gives A⊗B the structure ofa Z-algebra. Since multiplication in A and B is commutative, the multiplica-tion in A⊗B is clearly commutative. Moreover, the element 1⊗1 acts as a 1.

226 Stanford Fall 2006

(b) Prove that C is a coproduct of A and B if and only if C ∼= A⊗B.

First we will shot that A⊗B is a coproduct of A and B from which oneof the implications will follow. Define α : A → A ⊗ B as a 7→ a ⊗ 1. Thenfor a1, a2 ∈ A, we have that α(a1 + a2) = (a1 + a2)⊗ 1 = a1 ⊗ 1 + a2 ⊗ 1 =α(a1)+α(a2) and that α(a1a2) = (a1a2)⊗1 = (a1⊗1)(a2⊗1) = α(a1)α(a2) sothat α is a ring homomorphism (certainly 1A is mapped to 1A⊗B). Likewiseif β : B → A⊗B is b 7→ 1⊗ b, β is a ring homomorphism.

Now suppose that D is a commutative ring and that f : A → D andg : B → D are ring homomorphisms. Define a map Φ : A × B → C by(a, b) 7→ f(a)g(b). Suppose that a1, a2 ∈ A and b ∈ B. Then we see that

Φ(a1 + a2, b) = f(a1 + a2)g(b)

= f(a1)g(b) + f(a2)g(b)

= Φ(a1, b) + Φ(a2, b),

which shows that Φ preserves addition in the first coordinate. Similarly,Φ preserved addition in the second coordinate and so we may Apply theUniversal Property of Tensor Products to conclude that there is a grouphomomorphism φ : A ⊗ B → D such that φ(a ⊗ b) = f(a)g(b). Now,A⊗B is generated as an abelian group by the simple tensors. Hence to showthat φ preserves multiplication, it suffices to show it preserves multiplicationbetween simple tensors. Let then a, c ∈ A and b, d ∈ B. Then φ(a⊗b·c⊗d) =f(ac)g(bd) = f(a)g(b)f(c)g(d) = φ(a ⊗ b)φ(c ⊗ d), where we have used thefact that D is commutative. Furthermore, φ(1 ⊗ 1) = f(1)g(1) = 1. Henceφ is a ring homomorphism. Now, for a ∈ A, we have φ α(a) = φ(a⊗ 1) =f(a)g(1) = f(a) so that f = φ α. Likewise g = φ β and so such a φ exists.

Finally suppose that φ : A⊗B → D is another ring homomorphism suchthat f = φα and g = φ(β). Then for a ∈ A, f(a) = φ(a⊗1) and for b ∈ B,g(b) = φ(1 ⊗ b). Hence φ(a ⊗ b) = φ(a ⊗ 1)φ(1 ⊗ b) = f(a)g(b) = φ(a ⊗ b).Since again the tensor product is generated by the simple tensors, we mayconclude that φ = φ so that φ is unique. We may conclude that A⊗ B is acoproduct of A and B.

Next we will show that any two coproducts of A and B are isomorphicfrom which the other implication will follow (below whenever I say map, Imean ring homomorphism). Suppose then that C1 is a coproduct of A andB with corresponding maps α1 : A→ C1 and β1 : B → C1. Likewise suppose

25.1 Morning Session 227

that C2 is a coproduct of A and B under α2 : A→ C1 and β2 : B → C2. Thenhaving C2 play the role of D in the definition, we have a map φ : C1 → C2

such that α2 = φ α1 and β2 = φ β2. Reversing the roles of C1 and C2, weget a map ψ : C2 → C1 such that α1 = ψ α2 and β1 = ψ β2. Considerthen ψ φ : C1 → C1, we have α1 = (ψ φ) α1 and β1 = (ψ φ) β1 so thatψ φ is the map which comes from α1 and β1 playing the roles of f and g.But the identity on C1 will also accomplish this and so, by uniqueness, wehave that ψ φ is the identity. The symmetric argument gives that φ ψ isthe identity and we have shown that C1

∼= C2, as claimed.

228 Stanford Fall 2006

25.2 Afternoon Session

1. Let G be a p-group and H a nontrivial normal subgroup. Showthat H ∩ Z(G) has at least p elements.

Since H is normal, it is a union of conjugacy classes (of G). Pick repre-sentatives x1, . . . , xn for the conjugacy classes in H which consist of morethen one element. The remaining singlet conjugacy classes consists of exactlythose elements in H ∩ Z(G). Now G acts on H via conjugation, the orbitsbeing conjugacy classes, and the stabilizers being commutators. Hence wemay apply the Class Equation to conclude that

|H| = |H ∩ Z(G)|+n∑i=1

|G|/|CG(xi)|.

Now, for each xi, |G|/|CG(xi)| is a power of p which by assumption is not 1.Hence the sum on the right of the above is divisible by p. |H| 6= 1 is also apower of p which shows that it is divisible by p. Hence |H∩Z(G)| is divisibleby p. Since 1 ∈ H ∩ Z(G), we may conclude that |H ∩ Z(G)| ≥ p.2

2. Let A ⊂ B be finite abelian groups, and let χ : A → C× be ahomomorphism (linear character). Show that χ can be extendedto B, and the number of extensions equals [B : A].

Consider first the case when B = 〈x〉 is cyclic of order n. Then A = 〈xa〉for some 0 ≤ a < n. Moreover, if d is the greatest common divisor of a andn, we have that xa has order n/d so that the index of A is d. Now since allthe irreducible representations of an abelian group are 1-dimensional (anda 1-dimensional representation is always irreducible), we must have exactlyn distinct homomorphisms φ : B → C× and n/d exactly homomorphismsψ : A → C×. Certainly each map φ determines (not necessarily uniquely)a map ψ, by restriction. Suppose for a contradiction that we have d + 1distinct maps φ1, . . . φd+1 → C× such that each φi restricted to A gives a

2This proof is in Dummite and Foote, p.188

25.2 Afternoon Session 229

fixed map ψ : A → C×. Now φi(x) must be an nth root of unity and byassumption, [φi(x)]

a = ψ(xa). But d is the greatest common divisor of nand a so that we may find s, t ∈ Z such that d = ns + at. This gives[φi(x)]

d = [φi(x)]ns + [φi(x)]

at = 1 + [ψ(xa)]t. Hence each φi(x) is a root ofthe polynomial yd−1− [ψ(xa)]t ∈ C[y]. But this polynomial has only d rootsso we have a contradiction. Hence each ψ is given by at most d φ’s. Thisimplies by the above that each ψ is given by exactly d φ’s. In other words,every homomorphism A → C can be extended to a homomorphism B → Cin exactly [B : A] ways.

I screwed up the general case (the argument I had in mind was ridiculouslywrong) so I haven’t really done this one I’m not really sure if building upfrom cyclic groups is even the way to go, it might be better to attack it withcharacter theory, but I don’t feel like it.

3. Let q = pn where p is a prime, and let Fq denote the finitefield with q elements. Show that the Frobenius automorphismσ : Fq → Fq defined by φ(x) = xp is diagonalizable as a linear trans-formation over Fp if and only if n divides p− 1.

First, the Galois group of Fq over Fp is cyclic of order n, generated by σ.In particular, σn = 1. Hence the minimal polynomial over Fp must dividef(x) = xn − 1. Suppose for a contradiction that there is some polynomialg(x) ∈ Fp[x] such that g(σ) = 0 and deg(g) < n. Then g(σ)(α) = 0 for allα ∈ Fq. This gives g(αp) = 0 for all α ∈ Fp so that g(xp) has pn roots. Butthe degree of g(xp) is pdeg(g) < pn. This is a a contradiction. Hence f(x)must be the minimal polynomial of σ over Fp. Since it has degree n, it isalso the characteristic polynomial of σ.

Now suppose that σ is diagonalizable over Fp. Then in particular, Fpmust contain all the eigenvalues of Fp. Hence xn− 1 must have all of it rootsover Fp. Moreover xn − 1, the minimal polynomial of σ, must have distinctroots. This implies that there are exactly n elements of order dividing nin F×p , a cyclic group. Since all the elements of order diving n the form asubgroup we may conclude that F×p , a group of order p − 1 has a subgroupof order n. Hence n divides p− 1.

230 Stanford Fall 2006

Conversely, suppose that n divides p−1. Then F×p , a cyclic group of orderp−1 contains an element of order n. This element then generates a subgroupof n elements of order dividing n. Thus f(x), the minimal polynomial overFp has all its distinct roots in Fp. From this we may conclude that σ isdiagonalizable over Fp.

4. Determine the splitting field K and the Galois group Gal(K/Q)of the polynomial x4 − 2 over the field Q. Find all quadratic exten-sions of Q contained in K.

This is done in Spring 2002 #1.

5. Let F ⊂ K be fields. Let R be the polynomial ring F [X], whereX is an indeterminate, and similarly let S = K[X].

(a) Show that if f and g are monic polynomials in S, and S/fS ∼=S/gS as S-modules, then f = g.

Let φ be isomorphism S/fS → S/gS. Then we have φ(g + fS) =φ(g[1 + fS]) = gφ(1 + fS) = 0. Since φ is injective, we may concludethat g+ fS = 0 + fS. That is, g ∈ fS. But applying the same reasoning toφ−1(f + gS), we may conclude that f ∈ gS. Hence f and g are associates,so that they differ by a constant. They are both monic however, so we mayconclude that f = g (note that this argument works even if either polyno-mial is zero, in which case the monic assumption is irrelevant). This fact isalso a consequence of the uniqueness statement in the Structure Theorem forModules over a PID.

(b) Show that if x ∈ R, then S ⊗R (R/xR) ∼= S/xS.

This isomorphism is an isomorphism of S-algebras and is shown in Spring2005 #1 a).

(c) Suppose that M and N are finitely generated R-modules. Show

25.2 Afternoon Session 231

that if S ⊗RM ∼= S ⊗R N as S-modules then M ∼= N as R-modules.

Applying the Structure Theorem for Modules over a PID, we may con-clude that M ∼= M1 ⊗ · · · ⊗Mk (as R-modules), where Mi = R/fiR, fi ∈ Ris a nonconstant monic polynomial, and fi|fi+1. Likewise N ∼= N1⊗· · ·⊗Nr

for Ni = R/giR, gi ∈ R nonconstant and monic, and gi|gi+1. Now by the factthat the tensor product is distributive over direct sums and the S-modulehomomorphism in (b), we have

S/f1S ⊕ · · · ⊕ S/fkS ∼= (S ⊗R R/f1R)⊕ · · · ⊕ (S ⊗R R/fkR)∼= S ⊗RM∼= S ⊗R N∼= (S ⊗R R/g1R)⊕ · · · ⊕ (S ⊗R R/grR)∼= S/f1S ⊗ · · · ⊗ S/frS,

as S-modules. Hence by the uniqueness statement in the Structure Theorem(and the fact that divisibility of polynomial does not depend on which fieldin an extension in which we work), we may conclude that r = k and S/fiS ∼=S/giS as S-modules. By (a), we may conclude that fi = gi. Hence the aboveexpression for M and N are R-modules are the same and we may concludethat M ∼= N as R-modules.

232 Stanford Fall 2006

Chapter 26

Stanford Spring 2006

26.1 Morning Session

1. Suppose that H is a subgroup of G of index n. Show that G hasa normal subgroup of index less than or equal to n!. Use this toprove that there is no finite simple group of order 2430 = 2 · 35 · 5.

First we should note that every group has normal subgroup of index 1,namely the group itself. Hence the question should probably read that Ghas a proper subgroup of the kind mentioned; in this case we must alsoassume that H 6= G. Now G acts on the collection of cosets of H by leftmultiplication. Hence we have a nontrivial action of G on a set of n elementswhich gives a homomorphism G → Sn. If K is the kernel of this map, wehave, from the First Isomorphism Theorem, G/K ∼= S, where S is a subgroupof Sn. Now K is a proper normal subgroup and its index, the order of S,divides n!.

Now let |G| = 2 · 35 · 5. Then if H is a 3-Sylow subgroup, H has index 10and so we have a nontrivial map φ : G → S10. Suppose for a contradictionthat this is injective. Then we have that G is isomorphic to a subgroup ofS10. Hence |G| divides |S10| = 10!. 3, however, only divides 10! 4 times andit divides |G| 5 times. This is a contradiction. Hence the kernel of φ is anontrivial proper subgroup of G so that G is not simple.

234 Stanford Spring 2006

2. Let G be the following group of order 28:⟨x, y|x7 = y4 = 1, yxy−1 = x−1

⟩.

Determine the conjugacy classes of G an computer its charactertable.

First note that the relation yxy−1 = x−1 allow us to put each word inx and y in the form xiyj for some i, j. Since x7 = y4 = 1, we may take0 ≤ i < 7 and 0 ≤ j < 4. Since the order of the group is 28, all such wordsmust be distinct. In particular, y has order 4 and x has order 7. Now wemay again apply yxy−1 = x−1 to conclude that conjugation of xiyj by eitherx or y results in a word of the form xkyj. Hence the conjugacy class of xiyj

is contained in xkyj6k=0.

Now suppose that 6j 6≡ 1 mod7. Let 0 ≤ k < 7. Then we have

xkyj(xk)−1 = xkyjx7−k

= xkx(7−k)6j

yj

= xk−6jkyj.

Suppose that two such expressions, for 0 ≤ k, l < 7 are equal. Then wehave k − 6jk ≡ l − 6jl mod7, so that (k − l)(1 − 6j) ≡ 0 mod7. Since Z7

is an integral domain and since 1 − 67 6≡ 0 mod7, we may conclude thatk ≡ l mod7. By the restrictions on k and l, we have k = l. Hence we have 7distinct conjugates and so, by the above, we may conclude that the conjugacyclass for yj is equal to xkyj6

k=0 for j = 1, 3.Now we have yxy−1 = x6 and yx6y−1 = x so that the set x, x6 of

conjugate elements is preserved under conjugation by y. Since it is triviallypreserved under conjugation by x, we have that it is a conjugation class.Likewise, we have yx2y−1 = x5 and yx5y−1 = x2 so that x2, x5 is a conju-gacy class. yx3y−1 = x4 and yx4y−1 = x3 implies that x3, x4 is a conjugacyclass. Certainly 1 is a conjugacy class.

Next we have xy2x−1 = y2 so that y2 is a conjugacy class (as y2 lies inthe center of Z(G)). Hence we have xk(xiy2)x−k = xiy2. The relationshipsy(xy2)y−1 = x6y2, y(x6y2)y−1 = xy2, y(x2y2)y−1 = x5y2, y(x5y2)y−1 = x2y2,y(x3y2)y−1 = x4, and y(x4y2)y−1 = x3y3 then allow us to conclude (as above)that the remaining conjugacy classes of G as xy2, x6y2, x2y2, x5y2, andx3y2, x4y2. The conjugacy classes of G are summarized in the table below.

26.1 Morning Session 235

The Conjugacy Classes of G

1y2

x, x6

x2, x5

x3, x4

xy2, x6y2

x2y2, x5y2

x3y2, x4y2

y, xy, x2y, x3y, x4y, x5y, x6yy3, xy3, x2y3, x3y3, x4y3, x5y3, x6y3

Next let ω be a 4th root of unity in C. Let F be the free group gen-erated by x and y. Then, from the Universal Property of Free Groups,we have a homomorphism F → C× given by x 7→ 1 and y 7→ ω. Since17 = ω4 = 1 and ω1ω−1 = 1 = 1−1, this map factors through G and sowe have homomorphism G → C× given by x 7→ 1 and y 7→ ω. Con-versely suppose that φ : G → C× is a homomorphism. Then we haveφ(x) = φ(y)φ(x)φ(y−1) = φ(yxy−1) = φ(x)−1. Hence φ(x)2 = 1. Thusφ(x) = 1 or φ(x) = −1. But the latter case contradicts 1 = φ(x7) = [φ(x)]7.Hence we must have φ(x) = 1. Likewise [φ(y)]4 = 1 so that φ(y) is a 4throot of unity. Hence we have exactly four 1-dimensional representations ofG. Let χi be the character of the representation of this kind which maps yto ii−1.

Next let ζ be primitive 7th root of unity and consider the matrices

Xk =

(ζk 00 ζ−k

)and Y =

(0 11 0

).

We can see that X7k = Y 4 = 1 and Y XkY

−1 = X−1k . Hence we have a

2-dimensional representation G → GL2(C) given by x 7→ Xk and y 7→ Y .Computing the character corresponding to this representation (see the chartbelow), we can see that its norm is 1 for each k. Hence it is irreducible.Taking the trace of Xk for k = 1, 2, 3 give three distinct values, so we have3 irreducible 2-dimensional representations of this kind. Let χ5, χ6, andχ7 be the character of that which maps x to Xi, X2, and X3, respectively.Now we also have that the product of characters is a character. Hence we

236 Stanford Spring 2006

Character Table of G

g 1 y2 x x2 x3 xy2 x2y2 x3y2 y y3

χ1(g) 1 1 1 1 1 1 1 1 1 1χ2(g) 1 −1 1 1 1 −1 −1 −1 i −iχ3(g) 1 1 1 1 1 1 1 1 −1 −1χ4(g) 1 −1 1 1 1 −1 −1 −1 −i iχ5(g) 2 2 θ1 θ2 θ3 θ1 θ2 θ3 0 0χ6(g) 2 2 θ2 θ3 θ1 θ2 θ3 θ1 0 0χ7(g) 2 2 θ3 θ1 θ2 θ3 θ1 θ2 0 0χ8(g) 2 −2 θ1 θ2 θ3 −θ1 −θ2 −θ3 0 0χ9(g) 2 −2 θ2 θ3 θ1 −θ2 −θ3 −θ1 0 0χ10(g) 2 −2 θ3 θ1 θ2 −θ3 −θ1 −θ2 0 0

may conclude that χ8 = χ2χ5, χ9 = χ2χ6, and χ10 = χ2χ7 are all characterof G. Calculating the norm of each, we see that they give the remainingirreducible characters of G. Hence we have the character table shown be-low, where θ1 = ζ + ζ6 = 2 cos(2π/7), θ2 = ζ2 + ζ5 = −2 cos(3π/7), andθ3 = ζ3 + ζ4 = −2 cos(π/7).

3. (i) Suppose that d > 1 is a square-free integer and d ≡ 1 mod4.Determine (with proof) the ring of algebraic integers in Q(

√d).

First we will prove a lemma. Suppose that E is an extension of Q andα ∈ E is an algebraic integer. Then if m(x) ∈ Q[x] is the minimal polynomialfor α over Q, we have m(x) ∈ Z[x]. Find a monic polynomial f(x) ∈ Z[x]of minimal degree of which α is a root. Then suppose that f(x) is reducibleover Q with f(x) = a(x)b(x) for a(x), b(x) ∈ Q[x] nonconstant. Then fromGauss lemma, we may find nonconstant polynomials A(x), B(x) ∈ Z[x] suchthat f(x) = A(x)B(x). This implies that α is a root of either A(x) orB(x), contradicting the choose of f(x). Hence f(x) is irreducible which gives

26.1 Morning Session 237

m(x) = f(x) ∈ Z[x], as claimed. In particular, since the minimal polynomialof a/b ∈ Q is x− a/b, we have a/b algebraic over Z if and only if a/b ∈ Z.

Now let R be the ring of integers and set ω = 1+√d

2. Then we see that

ω2 − ω + (1 − d)/4 = 0 which shows that Z[ω] ⊂ R (since by assumption(1 − d)/4 ∈ Z). Conversely suppose that r = a + b

√d ∈ R for a, b ∈ Q.

If b = 0, we have r ∈ Q so that r ∈ Z ⊂ Z[ω]. If b 6= 0, r 6∈ Q and theminimal polynomial of r is x2 − 2ax+ (a2 − b2d). By the above, 2a ∈ Z anda2 − b2d ∈ Z. This gives (2a)2 − (2b)2d = 4(a2 − b2d) ∈ Z so that 4b2d ∈ Z.Since d is square-free, we must have 2b ∈ Z. Now we may write a = x/2 andb = y/2 for some x, y ∈ Z. This gives (x2 − y2d)/4 = a2 + b2d ∈ Z so thatx2 − y2d ≡ 0 mod4. Now since d ≡ 1 mod4 and since 0 and 1 are the onlysquares modulo 4, we must have x2 and y2 are both 0 or both 1 modulo 4.In the former case, x and y are both even. In the latter, they are both odd.Hence we may write r = a + b

√d = t + sω where t = (x − y)/2 and s = y

are both integers. This gives r ∈ Z[ω] and completes the proof.1

(ii) Explain how the principal ideals (2), (3), and (13) factor intoprime ideals in the ring of algebraic integers in Q(

√13).

Note the 1, ω are linearly independent over Z. Suppose that k ∈ Z isnot a unit in Z. Then if a + bω ∈ Z[ω], we have k(a + bω) = ka + kbω. Ifthis were to equal one, we would have ka = 1, a contradiction since k is nota unit in Z. In particular, we may conclude that 3, 3, and 13 are not units inZ[ω] so that (2), (3), and (13) are proper ideals. Note also that ω2 = 3 + ω.

First we will show that (2) is maximal. Let a + bω ∈ Z[ω]\(2). LetI = (a+ bω, 2). Since a+ bω 6∈ (2), a and b are not both even. Suppose thena is even and write a = 2k. Then bω = a + bω − 2k ∈ I. Now b is odd, sowe can find s, t ∈ Z such that 2s + bt = 1. This gives ω = 2sω + btω ∈ I.Hence, 3 = ω2 − ω ∈ I. Thus, 1 = 3 − 2 ∈ I so that I = Z[ω]. Likewise,suppose that b is even. Then a is odd so that we can find s, t ∈ Z such that1 = 2s + at. This gives 1 + btω = 2s + t(a + bω) ∈ Z[ω]. Since b is even,we may write bt = 2k. This gives 1 = 1 + btω − 2kω ∈ I. Thus we haveshown (2) is maximal. In particular, it is prime which shows that it is itsown factorization into prime ideals.

Next note that 3 = (ω − 1)ω. Hence we have (3) = (ω − 1)(ω) as ideals.Now, we will show that both (ω−1) and (ω) are maximal and so prime. First

1This proof is from Dummit and Foote, p. 698

238 Stanford Spring 2006

suppose that a+bω ∈ Z[ω]\(ω−1). Then since 3 ∈ (ω), 3 cannot divide botha and b. Let I = (1−ω, a+ bω). Next, we have a+ b = a+ bω+ b(1−ω) ∈ I.Hence is 3 does not divide a + b, we we can write 1 as an integral linearcombination of 3 and a + b and conclude that 1 ∈ I. Hence it suffices toassume that 3 divides a + b. in this case, 3 cannot divide neither a nor b asthen it would divide the other which is not the case. Then a is equivalent toeither 1 or 2 modulo 3. First suppose that it is equivalent to 1. Then since3 divide a + b, b is equivalent to 2 modulo 3. Hence can find s, t such that2 = b + 3s and 1 = a + 3t. Now a + bω = 1 + 2ω − 3t − 3sω. However,1 + 2ω = 1− ω + 3ω ∈ I, a+ bω ∈ Z[ω], a contradiction (to our assumptionthat 3 divides a + b). Thus we must have that a is equivalent to 2 modulo3. Hence b is equivalent to 1 modulo 3. As above, we may find s, t ∈ Z suchthat a+ bω = 2 + ω − 3t− 3sω. Now, 2 + ω = −(1− ω) + 3 ∈ I. Hence wehave reached a similar contradiction as before and so we may conclude that(1− ω) is maximal.

Likewise suppose that a+ bω ∈ Z[ω]\(ω). Then a = a+ bω− bω ∈ I. If ais divisible by 3, we have a + bω ∈ I. Likewise, if a is not divisible by 3, wemay write 1 as an integral linear combination of a and 3 and conclude that1 ∈ I. Hence (1 − ω) is maximal and so prime and we may conclude that(3) = (ω − 1)(ω) is the factorization of (3) into prime ideals.

Lastly, we have (2ω − 1)2 = 13 so that (13) = (2ω − 1)2. Again we willshow that (2ω − 1) is maximal. Suppose then that a + bω ∈ Z[ω]\(2ω − 1).Now −6−ω = (1−2ω)ω ∈ I. Furthermore, (2a+b)ω = a(2ω−1)+a+bω ∈ I.Hence −12a− 6b = (2a+ b)ω+(2a+ b)(−6−ω) ∈ I. Now as above, we mayassume that 13 divides −6(2a+ b) = −12a− 6b. Hence 13 divides 2a+ b sothat 2a+ b ∈ (2ω − 1). This gives 2(a+ bω) = 2a+ b− b(1− ω) ∈ (2ω − 1).Hence we may find k ∈ Z[ω] such that 2(a + bω) = (2ω − 1)k. Hence 2divides (2ω − 1)k. 2, however is prime and so we may conclude that itdivides either k or 2ω − 1. Since it does not divide the latter, it must dividek. Hence we have n ∈ Z[ω] such that 2(a + bω) = (2ω − 1)2n. Thusa + bω = n(2ω − 1) ∈ (2ω − 1), a contradiction. Thus (2ω − 1) is maximalwhich shows that the above factorization is prime.

4. Find the Galois group of x4 + 1 over Q, Q(√

2), F2, F3, F5, and F7.

26.1 Morning Session 239

In all cases except characteristic 2, let ζ be a root of x4+1. Then ζ4 = −1so that ζ8 = 1. Hence the order of ζ as an element of the multiplicative groupof the algebraic closure of the field in question divides 8. Since ζ4 6= 1 (as wenot in characteristic 2), we may conclude that the order is 8. Hence ζ i7

i=0

is a collection of 8 8th roots of unity. Thus ζ is a primitive 8th root of unity.Moreover ζ i for i ∈ Z×

8 is a root of x4 + 1. Thus the Galois group in all casesis the field adjoin ζ, a primitive 8th root of unity.

The Galois group of Q(ζ) over Q is σa, where, for a ∈ Z×8 , σa(ζ) = ζa.

The Galois group is isomorphic to Z×8 via the map σa 7→ a. Z×

8 , in turn, isisomorphic to Z2 × Z2 since it is a group of order 4 and every element hasorder 2. Now i is a 4th root unity so that i ∈ Q(ζ). Furthermore,

(1√2

+1√2i

)8

= 1,

so that this element of C lies in Q(ζ). Hence

√2 =

[1

2

(1√2

+1√2i

)(1− i)

]−1

∈ Q(ζ).

Thus we have Q ⊂ Q(√

2) ⊂ Q(ζ). Since Q(√

2) has degree 2 over Q, wemay conclude that Q(ζ) has degree 2 over Q(

√2). Hence the Galois group

in this case is Z2.

Over F2, we have (x + 1)4 = x4 + 4x3 + 6x2 + 4x + 1 = x4 + 1. Hencex4 + 1 splits over F2 and so the Galois group is the trivial group.

Suppose that F is a finite field which does not have characteristic 2. Then,from above, F contains all the roots of x4 + 1 if and only if its multiplicativesubgroup F× contains an element of order 8. Since F× is cyclic, this occursif and only if 8 divides F×. Now, we have |F×3 | = 2 so that x4 + 1 does nothave a root in F3. Next, |F×9 | = 8 so that x4 + 1 splits over F×9 and we havethat the Galois group has order 2 and so must be Z2. Finally, 8 divides 624,but not 124, 24, or 4. Hence x4 + 1 does not split over F5, F25, or F125, butit does split over F625. This shows that the splitting field of x4 + 1 over F5 isF625. This is a extension (of a finite field) of order 4 and so has Galois groupZ4.

240 Stanford Spring 2006

5. How many similarity classes are there of rational matrices withcharacteristic polynomial (x3+1)3(x2+1)3 and a minimal polynomialof degree 10.

Note first that we have the factorization x3 + 1 = (x + 1)(x2 − x + 1).Furthermore, if f(x) = x2 − x + 1, we have f(x + 2) = x2 + 3x+, whichis Eisenstein and p = 3. We may conclude that f(x) is irreducible. Hencemonic polynomial which divides (x3 + 1)3(x2 + 1)3 must be a product ofpowers of (x+ 1), (x2 − x+ 1), and (x2 + 1).

In particular, the invariant factors and the minimal polynomials are pow-ers of the three polynomials above. The minimal polynomial, which has thesame roots as the characteristic polynomial, must have a power of at least 1of each factor. Hence (x + 1)(x2 − x + 1)(x2 + 1) divides the minimal poly-nomial. Since, by assumption, the minimal polynomial has degree 10, theremaining factors must have degrees adding to 5. Now if (x2+1) is one of theremaining factors, we can have the remaining factors being (x2 + 1)2(x+ 1),(x2 + 1)(x2 − x + 1)(x + 1), or (x2 + 1)(x + 1)3. The last choice, however,is impossible and this would imply (x+ 1) dividing the minimal polynomialand therefore the characteristic polynomial 4 times. Likewise, if x2−x+1 isone of the remaining factors and (x2 +1) is not, we have (x2−x+1)2(x+1)or (x2−x+1)(x+1)3. Again the last choice is impossible for the same reasonas above. If neither of the second degree polynomials divide the remainingfactors, we have that their product must be (x + 1)5, which is impossible.Hence the choices for the minimal polynomial are (x+1)2(x2+1)3(x2−x+1),(x+ 1)2(x2 + 1)2(x2 − x+ 1)2, (x+ 1)2(x2 + 1)(x2 − x+ 1)3.

Now the characteristic polynomial is the product of the invariant factors.In the first case above, then, the remaining invariant factors must multiplyto (x+ 1)(x2− x+ 1)2. The second largest invariant factor can then be (x+1)(x2−x+1)2 or (x+1)(x2−x+1), since, from the divisibility requirements,it must have all the irreducible factors which (x+ 1)(x2 − x+ 1) has. In thelatter case, the final invariant factor must be (x2 − x + 1). Hence we have2 choices for the invariant factors in this case Assume next that we are inthe second case above, then the remaining invariant factors must multiplyto (x+ 1)(x2 + 1)(x2 − x+ 1). By the divisibility requirement, this must bean invariant factor. Hence we have only 1 choice in this case. Lastly in thethird case, the remaining invariant factors multiply to (x + 1)(x2 + 1)2. Asin the first case, we have 2 choices. Thus there are 5 choices total. Sincethe similarity class of a matrix is determined by its invariant factors, we may

26.2 Afternoon Session 241

conclude that there are 5 equivalence classes.

26.2 Afternoon Session

1. Let V be a finite-dimensional vector space over a field F of char-acteristic p and let T : V → V be a linear transformation such thatT p = I is the identity map.

(i) Show that T has an eigenvector in V .

We have that T satisfies xp−1 = (x−1)p. Hence the minimal polynomialand so all the invariant factors of T are factors of (x− 1). Thus, as an F [x]-module via T , V contains a factor of F [x]/(x − i)i for some i. Now in thismodule, we have (T−1)·(x−1)i−1 = (x−1)i = 0. Thus T ·(x−1)i−1 = (x−1)i.Hence the vector in V which corresponds to (x− 1)i−1 under the above iso-morphism (which is nonzero) is an eigenvector of T (there is also probablyan argument which is more pure linear algebra).

(ii) Show that T is upper triangular with respect to a suitable basisof V .

Choose a basis for V and let A be the matrix given by T . From above,the only eigenvalue of A is 1. Hence the Jordan Canonical Form, B, of Aover some algebraic closure of F is upper triangular consisting of only 1’sand 0’s. Thus A and B are similar over F . This implies that they havethe same Rational Canonical Form over F . But both matrices consist onlyof elements of F and so this Rational Canonical Form is also over F . Thisshows (by uniqueness) that A and B have the same Rational Canonical Formover F and so a similar over to F . This implies that there is a basis of Vwith respect to which T is given by B, an upper triangular matrix (I thinkthis can also be proved by induction on the dimension of V , through the useof (i) so that we do not need to invoke the Jordan Canonical Form).

242 Stanford Spring 2006

2. Let G be a finite group, and let H be a subgroup of index two.Let x ∈ H and let CG(x) be the centralizer of x in G.

(i) Prove that if CG(x) 6⊂ H then the conjugacy class of x in G equalsthe H conjugacy class of x; on the other hand, if CG(x) ⊂ H thenthe conjugacy class of x in G is contained in H but splits into twoH-conjugacy classes.

First note that H is normal since its index is the smallest prime dividing|G| (see Spring 2003 #1 (i) for this result). Now suppose that CG(x) 6⊂ H.Then CG(x)H is a subgroup of G from the Second Isomorphism Theorem.Furthermore, from the Correspondence Theorem, the only subgroup of Gwhich contains H is G, as G/H ∼= Z2 has no nontrivial proper subgroups.Since CG(x)H contains but does not equal H, we may conclude that it isG. Hence we may apply the Second Isomorphism Theorem to conclude thatCG(x)/[H ∩ CG(x)] ∼= G/H. Since H ∩ CG(x) = CH(x), may concludethat |CG(x)| = 2|CH(x)|. Now the order of the G conjugacy class of x is|G|/|CG(x)| = 2|H|/(2|CH(x)|) = |H|/|CH(x)|, the order of the H conjugacyclass of x. Since the H conjugacy class is contained in the G conjugacy class,the result follows.

Next suppose that CG(x) ⊂ H. Then CG(x) = CH(x) and the order of theG-conjugacy class of x is |G|/|CG(x)| = 2|H|/|CH(x)|, twice the order of theH-conjugacy class. Let g ∈ G\H and consider gxg−1, an element of H. Sup-pose c ∈ CG(gxg−1). Then cgxg−1c−1 = gxg−1 so that g−1cgxg−1c−1g = x.We may hence conclude that g−1cg ⊂ CG(x) ⊂ H. Since H is normal,this gives c ∈ H so that CG(gxg−1). We may conclude from the previousargument that the H-conjugacy class of gxg−1 has order half that of theG-conjugacy class. Lastly suppose for a contradiction that gxg−1 and x areH-conjugate. Then we have h0 ∈ H such that h0gxg

−1h−10 = x. Since H has

index two in G, every element in G can be written as either hg or h for someh ∈ H. We have hgxg−1h−1 = hh−1

0 xh0h−1 so that every G-conjugate of x is

also an H-conjugate. This is a contradiction. Thus the H conjugacy classesof x and gxg−1 are disjoint, are contained in the G-conjugacy class of x andhave orders which add to the order of the G-conjugacy class of x. We mayconclude that the G-conjugacy class of x splits into these two H-conjugacyclasses.

(ii) Let G be the symmetric group S9 and G be A9. Determine the

26.2 Afternoon Session 243

G-conjugacy classes of even permutations that split into two con-jugacy classes.

Of course the S9-conjugacy classes are given by distinct cycle types. Thefirst conjugacy class is the identity which cannot split into two conjugacyclasses. Next suppose that our cycle type has as an even cycle. Then takinga representative of this conjugacy class, we have that the even cycle givencommutes with the representative so that the centralizer is not contained inA9 and so the class does not split. Hence our cycle type must have only oddcycles.

First suppose that it is is a single cycle. Then we have the cycle types of(123), (12345), (1234567), and (123456789). The cycle (89) commutes witheach of the first three cycles and so we may conclude that these conjugacyclasses do not split. On the other hand, note that there are 8! 9-cycles in S9.To see this, note that there are 9! ways to pick an ordered list of the numbers1-9 of length 9 which does not repeat . Each 9-cycle is given by 9 of theselists as we get such a list by starting at any of the 9 numbers contained in thethe 9-cycle. Thus we have 9!/9 = 8! 9-cycles as claimed. This implies thatthe centralizer of a fixed cycle has order 9!/8! = 9 and so is the subgroupgenerated by the cycle. Since this subgroup is then contained in A9, we mayconclude that this conjugacy class splits.

The final case to consider is a product of multiple odd cycles. Suppose firstwe have only 3-cycles. Then our conjugacy class is represented (123)(345)or (123)(456)(789), both of which commute with (14)(25)(36). Hence theseconjugacy classes do not split. A conjugacy class with multiple odd cyclescannot contain a 7-cycle or a 9-cycle as then we would not have enoughelements. Hence the final class is (12345)(678) (any more 3-cycles or 5-cyclescauses too many elements). Now, as above the conjugacy class of this elementcontains 9!/(15) elements so that its centralizer is the subgroup generated byit. Hence, as before, we have that this conjugacy class splits.

3. Let A be a Noetherian commutative ring containing a field kand ideal I such that if J =

√I is the radical of I then A/J is a

finite dimensional k-vector space. Prove that A/I is also a finite

244 Stanford Spring 2006

dimensional vector space.

I never figured this one out before I took the prelim, let me know if yousolve it.

4. Let G be a finite p-group, and λ : G → C× a homomorphism.Assume that the order of λ is a prime p so that H = ker(λ) is asubgroup of index p. Let θ be an irreducible character of G suchthat λθ = θ. Show that 〈θ, θ〉H = p and deduce that θ is inducedfrom a character of H.

Let g ∈ G\H, then λ(g) 6= 1. But, λ(g)θ(g) = θ(g) so we must haveθ(g) = 0. Hence

〈θ, θ〉H =1

|H|∑h∈H

|θ(h)|2

=1

|H|∑g∈G

|θ(g)|2

=|G||H|

〈θ, θ〉G

= p,

since 〈θ, θ〉G = 1.The rest of this problem I will do when we cover it in Representation

Theory.5. Let ζ = e2πi/7. Find an element α of Q(ζ) such that [Q(α) : Q] = 3.Show that there does not exist β ∈ Q(α), β 6∈ Q such that β3 ∈ Q.

The Galois group of Q(ζ) over Q is isomorphic to Z×7 , where the map from

Z×7 is given by a 7→ σa, where σa(ζ) = ζa. Consider the element α = ζ + ζ6.

Now we can check that α is a root of the polynomial f(x) = x3 +x2− 2x− 1over Q. From the rational roots test, this cubic polynomial has no rationalroots and so is irreducible. We may hence conclude that [Q(α) : Q] = 3.

26.2 Afternoon Session 245

Suppose for a contradiction that such a β exists. Consider the polynomialf(x) = x3 − β3 ∈ Q[x]. The roots of f(x) over C are β, βω, and βω2, whereω is a primitive cube root of unity. Since β 6∈ Q, either f(x) is irreducibleor it split into an irreducible quadratic of of the form (x − β)(x − βωk) fork = 1, 2 and another linear term. In either case, we may conclude that theirreducible polynomial of β over Q has a root of the form βωk. Now, Q(α) isan intermediate field in a Galois extension with abelian Galois group. Hence,by the Galois Correspondence, it is Galois over Q. Since β ∈ Q(α), we mayconclude by normality that βωk ∈ Q(α). This gives ωk ∈ Q(α). Since kis 1 or 2, ωk is a primitive cube root of unity and we may conclude thatω ∈ Q(α). This gives Q ⊂ Q(ω) ⊂ Q(α). So that 2 = [Q(ω) : Q] divides3 = [Q(α) : Q], a contradiction. Thus no such β exists.

246 Stanford Spring 2006

Chapter 27

Stanford Fall 2005

27.1 Morning Session

1. Let p and q be primes with p, q 6= 2 and suppose that p dividesq + 1.

(a) Show that there exists a nonabelian group G of order pq2 whoseSylow q-subgroup is not cyclic.

First note that p and q are distinct as p cannot divide p + 1 (p + 1 isequivalent to 1 and so not zero modulo p). Consider then group Zq × Zq.Since every element in this group has order dividing q, we can make Zq ×Zq

into a vector space over Fq in the usual way. Certainly it is then a twodimensional vector space. In this way Aut(Zq×Zq) is isomorphic to GL2(Fq)(a group homomorphism is automatically Fq-linear by the definition of themultiplication of Fq). Now, suppose we are choosing a matrix in GL2(Fq).Then we may choose any column for the first column aside from the zerocolumn, giving q2 − 1 choices. Likewise, we may choose any column for thesecond except an Fq-linear combination of the first, q2− q choices. Hence theorder of this group is (q2 − 1)(q2 − q) = q(q + 1)(q − 1)2.

By assumption then, p divides the order of Aut(Zq×Zq) and so the lattercontains, by Cauchy’s Theorem, and element φ of order p. Hence we have anontrivial homomorphism ψ : Zp → Aut(Zq × Zq) where 1 maps to φ. Thuswe may form the semidirect product G = Zp nψ (Zq×Zq). This group is notabelian since ψ is not trivial. It also has order pq2 and its Sylow q-subgroup,

248 Stanford Fall 2005

Zq × Zq, is not cyclic.

(b) Show that if G is a nonabelian group of order pq2 then it has anormal q-Sylow subgroup Q, and if Q is not cyclic then a p-Sylowsubgroup of Aut(Q) is cyclic.

From the Sylow Theorems, the number of q-Sylow subgroups must dividep and so be 1 or p. However, p ≤ q + 1. Since q is odd, q + 1 is even sothat we cannot have p = q + 1. Thus p ≤ q. Thus p cannot be equivalentto 1 modulo q and we may conclude that p is not the number of q-Sylowsubgroups. Thus a q-Sylow subgroup, Q, of G must be normal.

Suppose then that Q is noncyclic. Since Q has order q2, it must be abelianand so isomorphic to Zq × Zq (this argument is given in Spring 2007 #2).Hence, by the above, the order of Aut(Q) is q(q+ 1)(q− 1)2. Since p dividesq − 1, it is equivalent modulo q to −1 and so cannot by equivalent moduloq to 1, as q 6= 2. Thus p only divides the order of Aut(Q) once, so that ap-Sylow subgroup of Aut(Q) has order p and so is cyclic.

(c) Show that any two nonabelian groups of order pq2 with non-cyclic Sylow q-subgroups are isomorphic.

From the above we have that any such group has a normal subgroup,Q ∼= Zq × Zq of order q2 and a subgroup, P ∼= Zp of order p (a Sylow p-subgroup). Hence, G is the semidirect product of P and Q. Let G1 = Pnφ1Qand G2 = P nφ2 Q be two such groups, with φ1, φ2 : P → Aut(Q). Since φ1

and φ2 are both not trivial, we may conclude by the above that their imagesare both Sylow p-subgroups of Aut(Q) and so are conjugate. By a lemmaproved in Stanford Fall 2006 Morning Session #1 (c), we may conclude thatG1

∼= G2 are isomorphic and the claim follows.

2. Suppose that f(t) ∈ Q[t] is an irreducible polynomial with ex-actly 3 real roots. Let K be the splitting field of f(t) over Q. Showthat Gal(K/Q) ∼= S5. Prove any nonobvious facts you use about S5.

27.1 Morning Session 249

We may may view Gal(K/Q) as a s subgroup of S5 by considering itspermutation on the five roots of f(t). Let α be any root of f(t). Then[Q(α) : Q] = 5 since f(t) is irreducible. But α ∈ K so the tower Q ⊂ Q(α) ⊂K implies that 5 divides [K : Q] and so divides |Gal(K/Q)|. By Cauchy’sTheorem, Gal(K/Q) must contain an element of order 5. Since the onlyelements of order 5 in S5 are 5-cycles, we may conclude that Gal(K/Q) hasa 5-cycle.

Likewise let τ : C → C be conjugation. Then τ must interchange thetwo complex roots of f(t) as f(t) has real coefficients. Since τ preserves theremaining three roots and Q, we see that τ restricts to an automorphismof K. As an element of S5, this is a 2-cycle. Thus the Galois group is asubgroup of S5 which contains a 2-cycle and 5-cycle. By Spring 1996 #1, wemay conclude that the Galois group is all of S5.

1

3. Let A be a commutative ring. The ring A is called Artinian ifit satisfies the decreasing chain condition : if I1 ⊃ I2 ⊃ I3 ⊃ . . . isa sequence of ideals then for some N we have IN = IN+1 = IN+2 = · · · .

(a) If A is an Artinian integral domain, show that A is a field.

Suppose that a ∈ A is nonzero. We have a descending sequence of ideals(a) ⊃ (a2) ⊃ (a3) ⊃ . . ., so that for some N we have (aN) = (aN+1). Inparticular, we may find b ∈ A such that aN = baN+1. Now since A is an inte-gral domain, and a is nonzero, aN is nonzero as we may use the cancelationproperty of a domain so conclude that 1 = ba. Then b is the inverse of a andwe may conclude that A is a field.

(b) If A is an Artinian commutative ring, show that every primeideal in A is maximal.

First suppose that I is an ideal in A. Then, from the Correspondence

1This proof is from Dummit and Foote, p. 629-630

250 Stanford Fall 2005

Theorem, every sequence of decreasing ideals in A/I is of the form

J1/I ⊃ J2/I ⊃ J3/I ⊃ . . . ,

where Ji are ideals in A and J1 ⊃ J2 ⊃ J3 . . . is a descending sequence. SinceA is Artinian, the latter sequence must stabilize so that the former one doesas well. Hence we may conclude that A/I is Artinian. Now, if P is a primeideal, A/P is an Artinian integral domain so that it is a field by (a). Thisimplies that P is maximal.

4. Let G be a a group of odd order. Prove that if χ is a complexirreducible character of G and χ(g) is real for all g ∈ G then χ(g) = 1.

Let σ : G → GL(V ) be the representation whose character is χ. Letg ∈ G. Then we have [φ(g)]n = I for some odd n. This implies that theminimal polynomial of φ(g) divides xn − 1. Hence φ(g) is diagonalizablewhose eigenvalues are nth roots of unity. Thus with respect to some basis,the matrix of φ(g) is diag(ζk1 , . . . , ζkm), where ζ is a primitive nth root ofunity. The inverse of this matrix is diag(ζ−k1 , . . . , ζ−km) = diag(ζk1 , . . . , ζkm).Hence we may conclude that χ(g−1) = χ(g) = χ(g).

Now, assume for a contradiction that φ is not the trivial representationand let χ1 be the trivial character. Then we have 〈χ, χ1〉 = 0 since χ isirreducible. This gives

∑g∈G χ(g) = 0 so that

∑g 6=1 χ(g) = χ(1). Since no

element of G has order 2, we may partition G\1 into two sets S1 and S2

where the inverse of each element in S1 lies in S2 and vice-versa. By theabove, we have

∑g∈S1

χ(g) =∑

g∈S2χ(g2). This gives 2

∑g∈S1

χ(g) = χ(1).Now χ(1) is the degree of φ which must divide |G| and so be odd. Thus∑

g∈S1χ(g) = χ(1)/2 ∈ Q\Z.

∑g∈S1

χ(g), however, is a sum of roots of unityand so is integral over Z (as the sum of two integral elements is an integralelement). But Z is integrally closed (in Q) and so we have a contradiction.Thus φ is the trivial representation and χ = 1.

27.1 Morning Session 251

5. Let T, U ∈ Matn(F ) where F is any field. Prove that if T andU are nilpotent matrices and rank(T k) = rank(Uk) for all k, thenT = AUA−1 for some A ∈ Matn(F ).

By an argument given in Spring 2007 #3, the Rational Canonical formof T is determined by dim ker((T − µI)i) for i ∈ N and µ ∈ F . Since Tis nilpotent, it it satisfies xn for some n. Hence its minimal polynomialdivides xn and has only the root zero. Thus the characteristic polynomialof T has only the root zero. Hence T − µ and so (T − µ)k is nonsingularfor µ 6= 0. This gives dim ker((T − µ)i) = 0 for µ 6= 0. Likewise for U sothat dim ker((T − µ)i) = dim ker((U − µ)i) for µ 6= 0. By assumption, wehave dim ker(T i) = dim ker(U i). Thus we may conclude that the RationalCanonical Form of T is the same as that for U . We may conclude thatT = AUA−1 for some A ∈ Matm(F ).

252 Stanford Fall 2005

27.2 Afternoon Session

1. Let G be the group of order 18 with generators x, y, and zsubject to the relations

x3 = y3 = y2 = 1, xy = yx, zxz−1 = y, and zyz−1 = x.

Determine the conjugacy classes of G and compute its charactertable.

Consider a word in x, y, and z. We can use the relations zy = xz andzx = yz to put all the z’s to the right. Then since x and y commute, we canput the word in the form xiyjzk for 0 ≤ i, j ≤ 2 and 0 ≤ k ≤ 1. Hence everyelement of G can be put in this form. Since G has order 18, we may concludethat each of these forms give distinct elements. In particular, x−1 = x2,y−1 = y2, and z−1 = z.

Now zxz−1 = y and zyz−1 = x also shows that 〈x, y〉 is normal and so aunion of conjugacy classes. Of course 1 is a conjugacy class. Next x, yis preserved under conjugation by z as this switches the two elements. Sincex and y commute, it is preserved point-wise under conjugation by x and y.Hence it is a conjugacy class. Likewise x2, y2 is a conjugacy class. Nextzxiyiz−1 = (zxz−1)i(zyz−1)i = yixi = xizi. This shows that z commuteswith xy and x2y2. Since x and y certainly commute with the latter twoelements, we may conclude that they lie in the center of G. Hence xy, andx2y2 are conjugacy classes. This also gives zx2yz−1 = xy(zxz−1) = xy2

and zxy2z−1 = x2y so that x2y, xy2 is a conjugacy class (as again it ispreserved point-wise under conjugation by x and y).

Now, the commutator of xiz contains 1, xy, x2y2, and xiz. Hence itsorder is at least 4 and divides 18. This implies that its order is at least6. Hence the order of the conjugacy class of xiz is at most 3. We havex(xiz)x−1 = xi+1y2z and y(xiz)y−1 = xi+2yz so that the conjugacy class ofz contains xiz, xi+2yz, xi+1y2z and so this must be the conjugacy class ofz. Hence we have the conjugacy classes of G are given in the table below.

Let ω be a cuberoot of unity and let ζ be a square root of unity (i.e.,ζ = ±1). Define a map φ : G → C× by x, y 7→ ω and z 7→ ζ. Then sinceω3 = ζ2 = 1, ζωζ−1 = ω, and ω commutes with itself, we may conclude thatφ is well-defined. This gives 6 distinct one-dimensional (and so irreducible)representations of G. Label the corresponding characters χi for 1 ≤ i ≤ 6.

Now let ω = exp(2π/3) be a primitive cube root of unity and consider

27.2 Afternoon Session 253

The Conjugacy Classes of G

1xyx2y2

x, yx2, y2

x2y, xy2

z, x2yz, xy2zxz, yz, x2y2zx2z, y2z, xyz

the matrices

X :=

(ω 00 ω2

)and Z :=

(0 11 0

).

We can check that they satisfy the relations to give a well-defined homomor-phism G → GL2(C) which satisfies x 7→ X, y 7→ X2 = X−1, and z 7→ Z.Denote the corresponding character χ7. Finally we get two addition char-acters by taking χ8 = χ7χ2 and χ9 = χ7χ3. By taking the norm, we cansee that these characters are irreducible. Thus we may conclude that thecharacter table for G is given below.

254 Stanford Fall 2005

Character Table for G

g 1 xy x2y2 x x2 x2y z xz x2zχ1(g) 1 1 1 1 1 1 1 1 1χ2(g) 1 ω2 ω ω ω2 1 1 ω ω2

χ3(g) 1 ω ω2 ω2 ω 1 1 ω2 ωχ4(g) 1 1 1 1 1 1 −1 −1 −1χ5(g) 1 ω2 ω ω ω2 1 −1 −ω −ω2

χ6(g) 1 ω ω2 ω2 ω 1 −1 −ω2 −ωχ7(g) 2 2 2 −1 −1 −1 0 0 0χ8(g) 2 2ω2 2ω −ω −ω2 −1 0 0 0χ9(g) 2 2ω 2ω2 −ω2 −ω −1 0 0 0

2. Let p be an odd prime and ζ = e2πi/p. Show that there exists aunique subfield K of Q(ζ) such that [K : Q] = 2. Let χ : (Z/pZ)× →±1 be the unique nontrivial homomorphism and let

α =

p−1∑a=1

χ(a)ζa.

Show that α2 = (−1)(p−1)/2p and conclude that K = Q(α).

The Galois group of Q(ζ) is isomorphic to (Zp)× via the map a 7→ σa

where σa(ζ) = ζa. By a lemma proved in Spring 1998 #3 (a), (Zp)× is

cyclic as a finite subgroup of the multiplicative group the field Zp. Hence(Zp)

× ∼= Zp−1. Now, since p is odd, (p − 1)/2 is an integer which dividesp − 1. Hence there is a unique subgroup of Zp−1 of order (p − 1)/2. Hencethere is a unique subgroup of the Galois group of index 2. From the GaloisCorrespondence, there is a unique subfield of Q(ζ) containing Q of degree 2over Q.

Let σ be a generator of the Galois group of Q(ζ) over Q. By identifyingthe Galois group with (Z/pZ)×, we have that χ is the unique homomorphismfrom the Galois group to ±1. Then χ(σ) = −1 as otherwise χ would betrivial. Also write σ(ζ) = ζk. Then k is a generator for Z/pZ so that everya ∈ (Z/pZ)× can be written as kn for 0 ≤ n < p− 2. Now, χ(kn) = 1 if and

27.2 Afternoon Session 255

only if n is even. Thus, we have

α =

p−2∑n=0

(−1)nζkn

=

p−2∑n=0

(−1)nσn(ζ).

Next we have that

ασ−1(α) =

[p−2∑i=0

(−1)iσi(ζ)

][p−2∑j=0

(−1)jσ−j(ζ)

]

=

p−2∑i,j=0

(−1)i(−1)jσi(ζ)

σj(ζ)

=

p−2∑i,j=0

(−1)i−jσj(σi−j(ζ)

ζ

)

=

p−2∑k=0

(−1)kp−2∑j=0

σj(σi−j(ζ)

ζ

),

since the fact that σkp−2k=0 is a group under multiplication allows us to con-

clude that σi−j runs through σk as i runs through 0 to p− 1 for each fixedj. Now for 0 < k ≤ p − 2, σk(ζ) 6= ζ, we may conclude that σk(ζ)/ζ is aprimitive pth root of unity (as it is a pth root of unity but is not 1). Hence

σp−2j=0σ

j

(σk(ζ)

ζ

)is a sum of all the pth roots of unity aside from 1. This sum is −1 (as the sumof all the pth roots of unity is 0). If, on the other hand, k = 0, σk(ζ)/ζ = 1.Hence, we have

σp−2j=0σ

j

(σk(ζ)

ζ

)= p− 1.

Now we have of the p − 1 choices for k, half are odd and half are even(including 0). Thus we man conclude that the the total double sum above isασ−1(α) = p− 1 + [(p− 1)/2](−1) + [(p− 1)/2− 1](−1) = p.

Now letH be the unique subgroup of the Galois group of index 2. Supposethat σ−1 ∈ H. Then σj−1 = σ−1σj ∈ H if and only if σj ∈ H. Now in theexpansion for α, σn(ζ) has a plus sign if and only if n is even, that is, if and

256 Stanford Fall 2005

only if σn ∈ H. This implies that σ−1(α) = α as the same elements whichhad a plus sign before still do. Likewise, if σ−1 6∈ H, we may conclude thatσ−1(α) = −α.

Now suppose that c ∈ H. Then c ≡ kn modp for some even n so that cis a square modulo p. Conversely suppose that c is a square modulo p. Thenwrite c ≡ a2 modp. Find l such that a ≡ kl modp, we have c ≡ k2l modp.Hence c ∈ H. In particular, −1 ∈ H if and only if −1 is a square modulop. Now suppose that a2 = −1 modulo p, then a4 = 1 so that a has order4 as an element of (Z/pZ)×. Conversely if a ∈ (Z/pZ)× has order 4, a2 hasorder 2. Hence a is a square root of unity over Z/pZ which is not 1 so thatit must be −1. Hence −1 is a square if and only if Z/pZ, a cyclic group,has an element of order 4. This occurs if and only if p ≡ 1 mod4. Thusσ−1(α) = α if and only if p ≡ 1 mod4 and σ−1(α) = −α otherwise. Thatis, σ−1(α) = (−1)(p−1)/2α. Now from above, (−1)(p−1)/2α2 = ασ−1(α) = p sothat α2 = (−1)(p−1)/2p as claimed. Furthermore, α is a root of x2± p over Qwhich is irreducible by Eisenstein’s Criterion which gives [Q(α) : Q] = 2 sothat K = Q(α).2

3. Let A be a Noetherian integral domain with field of fractions F .If f ∈ A is not a unit, prove that the ring A[f−1] generated by f−1

and A is not a finitely-generated A-module.

Suppose that A[f−1] is a finitely generated A-module. Then we can findb1, . . . , bm ∈ A[f−1] which generate A[f−1] as an A-module. Now each bihas an expression (not necessarily unique) as a polynomial over A in f−1.Choose such an expression for each bi and let m be the highest degree off−1 which occurs in any of these expressions. Then each bi is an A-linearcombination of 1, . . . , f−k. This implies that 1, . . . , f−k generates A[f−1]as an A-module. Now, f−k−1 ∈ A[f−1] so we may find a0, . . . , ak ∈ A suchthat f−k−1 = a0 + a1f

−1 + · · · + akf−k. Multiplying this by fk, we have

f−1 = a0fk + a1f

k−1 + · · ·+ ak ∈ A. Thus f is a unit in A.

2Some hints for this problem are in Dummite and Foote, p. 637, Exercise #11

27.2 Afternoon Session 257

4. Let G be a finite group of odd order. Prove that if g ∈ G isconjugate to g−1 then g = 1.

Suppose that we have some x ∈ G such that xgx−1 = g−1. Then(xg−1x−1)(xgx−1) = 1 so that xg−1x−1 = g. Thus x2gx−2 = xg−1x−1 = g.Now since G is of odd order, x must have odd order. Find n such thatx2n+1 = 1. This gives g = x2n+1gx−2n−1 = xx2ngx−2nx−1 = xgx−1 = g−1.We may conclude that g2 = 1 so that g has either order 1 or 2. Since itcannot have order 2 (as a member of a group of odd order), we have g = 1.

5. Let n > 1 be odd. Let A and B be matrices in GL2(C) such thatAn = 1, BAB−1 = A−1 and A 6= I. Suppose that X commutes withboth A and B. Prove that X is a scalar matrix.

First note that A and B cannot commute as this would give A = A−1

so that A2 = I. Since A 6= I and A does not have order 2 (as the order ofA must divide n), this is a contradiction. Now, since An = 1, the minimalpolynomial of A divides xn − 1 and so has no repeated roots. Hence, wemay apply Spring 2005 # 2 a) to conclude that A is diagonalizable. Thuswe may find a basis of eigenvalues v1, v2 of A for C2. Write A(vi) = aivi.If a1 = a2, A is a scalar and so commutes with B, a contradiction. Hencea1 6= a2 and C2 = Ea1 ⊕ Ea2 where Eai

= Cvi is the Eigenspace of aiwith respect to A. Now, A(Xvi) = X(Avi) = ai(Xvi). Hence Xvi is anEigenvector of A with Eigenvalue ai. This gives Xvi ∈ Eai

= Cvi. WriteXvi = xivi and assume for a contradiction that x1 6= x2. Then as above, wehave C2 ∼= Sx1 ⊕ Sx2 where Sxi

= Cvi is the Eigenspace of xi with respectto X. Then we have X(Bvi) = BXvi = xiBvi so that Bvi is an Eigenvectorof X with Eigenvalue xi. As above, this gives B(vi) = bivi for some bi ∈ C.But then AB(vi) = aibivi = biaivi = BA(vi) so that A and B commute on abasis. This implies that they commute, a contradiction. Hence x1 = x2 andwe may conclude that X is a scalar matrix.

258 Stanford Fall 2005

Chapter 28

Stanford Spring 2005

28.1 Morning Session

1. Suppose p and q are odd primes and p < q. Let G be a finitegroup of order p3q.

(a) Prove that G has a normal Sylow subgroup.

This is done in Fall 2002#1 a).

(b) Let np and nq denote the number of p-Sylow and q-Sylow sub-groups of G. Determine, with proof, all ordered pairs (np, nq) thatare possible for groups of order p3q.

By the argument in Fall 2002#1 a), nq cannot equal neither p nor p2.Hence, since one of np and nq is 1, we may conclude that the three possibilitiesare (q, 1), (1, p3), and (1, 1). The third possibility is realized in the groupZq ⊗ Zp3 .

Next, define a map φ : Z3 × Z3 × Z3 → Z6 by φ(a, b, c) = 2a. It is simpleto check that φ is a nontrivial group homomorphism. Now the automorphismgroup of Z7 is Z×

7∼= Z6. Hence we have a nontrivial homomorphism Z3 ×

Z3 × Z3 → Aut(Z7) and we may use this map to form a semidirect product(Z3 × Z3 × Z3) n Z7 which is not direct. Since Z7 is normal, Z3 × Z3 × Z3

is not, as this would imply a direct product. Hence we may conclude thatn7 = 1 and n3 6= 1. Hence this must be a realization of the possibility (q, 1).

260 Stanford Spring 2005

Likewise, consider the group Aut(Z3 × Z3 × Z3). Z3 × Z3 × Z3 is a 3-dimensional vector space over F3 with the obvious multiplication. Moreover,a map from Z3 × Z3 × Z3 to itself if a group homomorphism if and only if itis F3-linear. Hence to find the order of Aut(Z3 × Z3 × Z3), it suffices to findthe order of GL3(F3). Consider the later as a group of matrices and supposewe are choosing such a matrix. Then the first column can be anything otherthan a the zero column, giving 33 − 1 possibilities. The second column, inturn, can be anything other than a F3 scalar multiple of the second column33 − 3 possibilities. Likewise, the third column can be anything other thana F3 linear combination of the first two columns, 33 − 32 possibilities. Hencethe order of the group is (33− 1)(33− 3)(33− 32) = 25 · 33 · 13. In particular,we may apply Cauchy’s Theorem to find and element of Aut(Z3 × Z3 × Z3)of order 13 and so we have a nontrivial map Z13 → Aut(Z3×Z3×Z3) whichmaps 1 to this element. As above we may construct the a semi-direct productZ7 n (Z3 × Z3 × Z3) and conclude that the possibility (1, p3) is realized.

2. Let f(X) ∈ Q[X] be a monic irreducible polynomial of degree 4with roots α, β, γ, and δ. Recall that the discriminant of a poly-nomial with roots r1, . . . , rn is

∏i<j(ri − rj)

2.

(a) Prove that λ = αβ + γδ is the root of a monic cubic polynomialg(X) ∈ Q[X] whose discriminant is the same as the discriminant off(X).

Identifying the roots of f(X) with the numbers 1 to 4 (α is identifiedwith 1, β with 2, etc.), we may consider the Galois group of f(X) to be asubgroup of S4. Now, consider the elements

λ1 = αβ + γδ,

λ2 = αγ + βδ, and

λ3 = αδ + βγ.

We have (1234)(λ1) = λ3, (1234)(λ2) = λ2, and (1234)(λ3) = λ1, so that(1234) permutes the λi. Likewise, (12)(λ1) = λ1, (12)(λ2) = λ3 and (12)(λ3) =

28.1 Morning Session 261

λ2, so that (12) also permutes the λi. S4, however, is generated by (12) and(1234) so the may conclude that all of S4, and so the Galois group, permutesthe λi. Hence the elementary symmetric functions in λ1, λ2, λ3 are pre-served by the Galois group and we may conclude that they lie in Q. Thisgives g(X) = (X − λ1)(X − λ2)(X − λ3) ∈ Q[x].

Finally we have that

λ1 − λ2 = (α− δ)(β − γ),

λ1 − λ3 = (α− γ)(β − δ), and

λ2 − λ3 = (α− β)(γ − δ),

so that the discriminant of g(X) is

(α− β)2(α− γ)2(α− δ)2(β − γ)2(β − δ)2(γ − δ)2,

the same as the discriminant for f(X).

(b) If f(X) ∈ Z[X], prove that g(X) ∈ Z[x].

If f(X) ∈ Z[X], α, β, γ, and δ are all integral over Z. Since the integralelements form a subring, we may conclude that λ1, λ2, and λ3 are integralover Z. By a lemma prove in Stanford Spring 2006 Morning Session #3 (i),we may conclude that the minimal polynomial of λi has integral coefficientsfor each i. Now, factoring g(X) into irreducible factors gives g(X) has aproduct of minimal polynomial of these λi. Hence g(X) is a product ofpolynomials with integer coefficients and so has integer coefficients itself.

3. Let M be a finitely generated module over the Noetherian com-mutative ring R. Prove that if f : M → M is an R-module homo-morphism, and if f is surjective, then f is also injective.

Since M is finitely generated and R is Noetherian, M is Noetherian asan R-module.1 Now since f , is an R-module homomorphism, fn is an R-

1If this is not considered a well known fact, it is proved in Rotman, p. 438

262 Stanford Spring 2005

module homomorphism for each n. Hence, we may conclude that ker(fn) is asubmodule. Certainly we have the containment ker(f) ⊂ ker(f 2) ⊂ . . .. SinceM is Noetherian, this sequence must stabilize. This implies that ker(fn) =ker(fn+1) for some n. Let m ∈ M and suppose that f(m) = 0. Since fis surjective, fn is surjective for each n (if a ∈ M , find a1 ∈ M such thatf(a1) = a and then find a2 ∈ M such that f(a2) = a1; this gives f 2(a2) = aand we may proceed likewise). Hence find x ∈ M such that fn(x) = m.This gives fn+1(x) = f(m) = 0. So that x ∈ ker(fn+1) = ker(fn). Hencem = fn(x) = 0. This shows that f is injective.

4. Let G be the nonabelian group of order 16 with generators xand y subject to the relations

x8 = y2 = 1 and yxy−1 = x3.

Determine the conjugacy classes of G and compute is character ta-ble.

First note that the relation yx = x3y allows us to put any word in x andy in the form xiyj for 0 ≤ i ≤ 7 and 0 ≤ j ≤ 1. Since G has order 16 all ofthese expressions must be distinct. Furthermore y2 = 1 implies y = y−1 andx8 = 1 implies x−1 = x7. Now xi commutes with every power of x so thatis centralizer has order at least 8. Since the order of its conjugation class isthe order of G divided by the order of its centralizer, we may conclude thatits conjugacy class has order at most 2. Hence the relations yxy−1 = x3,yx2y−1 = x6y, and yx5y−1 = x7 allow us to conclude that x, x3, x2, x6,and x5, x7 are all conjugacy classes in G. Next we have yx4y−1 = x4 sothat x4 lies in the center of G and x4 is a conjugacy class. Certainly 1is a conjugacy class.

Now y commutes with 1, y, x4, and x4y, so that, as above its conjugacyclass has order at most 4. Furthermore, xyx−1 = x6y, x(x6y)x−1 = x4y,and x(x4)yx−1x2y. This implies that y, x2y, x4y, x6y is a conjugacy class.Finally we have x(xy)x−1 = x7y, x(x7y)x−1 = x5y, and x(x5y)x−1 = x3yshow that the remaining four elements of G are conjugate and so form the

28.1 Morning Session 263

conjugacy class xy, x3y, x5y, x7y. The conjugacy classes of G are given inthe table below.

The Conjugacy Classes of G

1x4

x, x3

x2, x6

x5, x7

y, x2y, x4y, x6yxy, x3y, x5y, x7y

Now let w, z ∈ ±1. Then we have w8 = z2 = 1 and zwz−1 = w = w3.Hence we have a well defined group homomorphism G→ Z× given by x 7→ wand y 7→ z. There are four such maps and we will write χi for i ≤ i ≤ 4as the corresponding (necessarily irreducible) characters. Likewise, considerthe matrices

X1 =

(0 1-1 0

)and Y1 =

(-1 00 1

).

We can check that they also satisfy the necessary relations to allow a grouphomomorphism G→ GL2(C) given by x 7→ X1 and y 7→ Y1. Write χ5 as thecorresponding character. It is simple to check that the norm of χ5 is 1 sothat it is irreducible.

Next let

ζ =1

2

√2 +

1

2i√

3

be the standard 8th root of unity. Then we can check that

X1 =

(ζ 00 ζ3

)and Y1 =

(0 11 0

)satisfy the necessary relations to give a map G→ GL2(C) defined by x 7→ X2

and y 7→ Y2. We define the correspond character as χ6 and can check that it isirreducible. Lastly, we define χ7 = χ2χ6 and can check that it is irreducible.Hence the character table of G is as below.

264 Stanford Spring 2005

The Character Table of G

g 1 x4 x x2 x5 y xyχ1(g) 1 1 1 1 1 1 1χ2(g) 1 1 −1 1 −1 1 −1χ3(g) 1 1 1 1 1 −1 −1χ4(g) 1 1 −1 1 −1 −1 1χ5(g) 2 2 0 −2 0 0 0

χ6(g) 2 −2 i√

2 0 −i√

2 0 0

χ7(g) 2 −2 −i√

2 0 i√

2 0 0

5. If B is a positive-definite symmetric real matrix, show that thereexists a unique positive-definite symmetric real matric C such thatC2 = B.

First we will show that an n × n matrix B over R is symmetric if andonly if 〈x,By〉 = 〈Bx, y〉 for all x, y ∈ Cn (viewing B as a matrix over C),where 〈, 〉 is the standard inner on C. First suppose that B is symmetric sothat we can write it as (bij) with bij = bij. Let x = (xi), y = (yi) ∈ Cn,considered as column vectors. Then Bx = (

∑j bijxj) and Ay = (

∑j bijyj).

Hence 〈Bx, y〉 =∑

i

∑j bijxjyi and 〈x,By〉 =

∑i

∑j bijxiyj =

∑j

∑i bjixjyi

as bij = bij. Since bij = bji, we may conclude that 〈x,By〉 = 〈Bx, y〉.Conversely if the latter condition holds, it hold for x = ei and y = ej whereek is the standard orthonormal basis on Cn. By the above calculation, thisgives bij = bji. Note that this lemma implies that symmetry of a matrix is aproperty of the correspond linear transformation and so does not depend onbasis.

From now on we assume B is positive-definite and symmetric. Nextsuppose that λ ∈ C is an Eigenvalue of B with x a nonzero correspondingEigenvector. Then Ax = λx so that λ 〈x, x〉 = 〈Ax, x〉 = 〈x,Ax〉 = λ 〈x, x〉.Since x is nonzero, 〈x, x〉 is also nonzero and we may conclude that λ = λ sothat every Eigenvalue of B is real. Hence every root of det(xI − B) is real

28.1 Morning Session 265

so that det(xI − B) splits over R. Now view B as matrix over R. Taking aroot of det(xI −B) allows us to conclude that B has a nonzero Eigenvectorin Rn. Dividing this vector by its norm, we may conclude that B has anEigenvector of norm 1.

We will show by induction on n that Rn has an orthonormal basis ofEigenvectors of B. The base case is trivial by the above. Now, fix an nand, by the above, find an Eigenvector e of norm 1. Let W be the subspaceof Rn of vectors which are orthogonal to e. If w ∈ W , we have 〈Bw, e〉 =〈w,Be〉 = 〈w, λe〉 = λ 〈w, e〉 = 0. Hence B can be restricted to a lineartransformation B : W → W . Since the standard inner product on W willbe then be a restriction of that on Cn (as e has norm 1 and it orthogonal toevery vector in K), we may conclude that B satisfies the symmetric conditionabove. Hence we may apply the induction hypothesis to conclude that W hasan orthonormal basis of Eigenvalue of B. Hence combining this basis with e,we have an orthonormal basis, B, of Rn of Eigenvalues of B. In particular,B is diagonalizable.

Let D = diag(λ1, . . . , λn) be the diagonalization of B. Hence if write x ∈Rn as x = (x1, . . . , xn) and with resect to B, we have Bx = λ1x1 + · · ·+λnxn.Furthermore, since B is orthonormal, 〈Bx, x〉 = λ1x

21+· · ·+λnx2

n. Choosing xto be the jth basis vector, we have 〈Bx, x〉 = λj. Since B is positive-definite,we must have λj > 0. Hence, letting T = diag(

√λ1, . . . ,

√λn) ∈ Mn(R), we

have T 2 = D. Thus if C is the matrix, with respect to the standard basis,of the linear transformation given by T , we have C2 = B. Furthermore,〈x, Tx〉 =

√λ1x

2i + · · · +

√λnx

2n > 0 for each x ∈ Rn. This implies that T ,

and so C, is positive-definite. Finally if y ∈ Rn is written (y1, . . . , yn) withrespect to B, we have 〈x, Ty〉 = λ1x1y1 + · · · + λnxnyn = 〈Tx, y〉 so that Tis a symmetric transformation. This implies that C is as well. This is theuniqueness side of the proof.

Likewise, suppose that A is another positive-definite symmetric matrixwith A2 = C. Then as above we may find an orthonormal basis A withrespect to which A has the form S = diag(µ1, . . . , µn), with µi > 0. ThenS2 = (diag)(µ2

1, . . . , µ2n) so that (diag)(µ2

1, . . . , µ2n) is the diagonalization of

B. This gives µi =√λi. Hence, changing back to the standard basis, we

must have B = A and so have uniqueness.

266 Stanford Spring 2005

28.2 Afternoon Session

1. Suppose that A ⊂ B is an integral extension of commutativerings with unit.

(a) If q is a maximal ideal of B, prove that p = q ∩ A is a maximalideal of A.

Suppose first that we have an integral extension R ⊂ S of commutativerings with unit such that S is a field. Suppose that r ∈ R. Then r−1 ∈ S isby assumption integral over R so that we may find a0, . . . , am ∈ R such thatf−m+am−1f

−m+1+· · ·+a0 = 0. This gives r−1 = −(am−1+· · ·+a0rm−1) ∈ R

and we may hence conclude that R is a field.

Now, from the Second Isomorphism Theorem, A/p = A/(A ∩ q) ∼=(A + q)/q can be viewed as a subring of B/q. Taking an element b ∈ Bwe have a monic polynomial over A which it satisfies. Hence b+q ∈ B/q willsatisfy the polynomial in (A + q)/q that we get when we reduce to originalpolynomial modulo q. Hence A/p ⊂ B/q is an integral extension. q, how-ever, is maximal so B/q is a field and we may hence conclude that A/p is afield, by the above. This gives that p is maximal.2

(b) Outline the proof that for any prime ideal p ⊂ A there exists aprime ideal q of B with p = q ∩ A.

Let D = A\p. Then D is a multiplicatively closed set in both A and Bso the may define the localizations D−1A and D−1B and get the natural ringhomomorphisms πA : A→ D−1A and πB : B → D−1B. Moreover D−1A is asubring of D−1B in the obvious way. Hence it is easy to see that the diagram

AπA //

ι

D−1A

D−1ι

BπB // D−1B

commutes, where ι and D−1ι are the inclusion maps. It also simple to checkthat D−1B is integral over D−1A (for b/d ∈ D−1B, just take the polynomial

2This proof is given in Dummit and Foot on p. 694

28.2 Afternoon Session 267

overA which b satisfies and consider its image inD−1A[x] to find a polynomialit satisfies).

Now let m be any maximal ideal in D−1B. Then from (a), m ∩D−1A isa maximal ideal in D−1A. But since D = A\p, D−1A is a local ring withunique maximal ideal given by the set of elements whose numerator lies in p

(an element not in p is trivially a unit in D−1A). Moreover, the contractionof such and ideal is p. Hence we have that π−1

A (D−1ι)−1(m) = p. Now q =π−1B (m) is a prime ideal in S (which does not intersect D). By commutativity,

we may conclude that ι−1(q) = p. This is to say that q ∩ A = p.3

2. Let F = Z/2Z, and let F [X,Y ] be the polynomial ring in twovariables. Let I be the ideal generated by X5 + X3 + X and Y 3 +(X3 + 1)Y + 1, and let R be the quotient ring F [X,Y ]/I. Determinethe number of maximal ideals in the ring R.

By the Correspondence Theorem, it suffices to find the number of maxi-mal ideals in F [X, Y ] which contain I. Suppose thatM is such an ideal. Notethat we have X5+X3+X = X(X2+X+1)2. Hence since M is maximal, it isprime and so must contain either X or X2+X+1. Suppose first that X ∈M .Then Y 3+Y +1 = [Y 3+(X3+1)Y +1]+X3Y ∈M and so (X, Y 3+Y +1) ⊂MSimilarly, I ⊂ (X, Y 3 + Y + 1). Now F [X, Y ]/(X, Y 3 + Y + 1) is triviallyisomorphic to F [Y ]/(Y 3 +Y +1) as F -algebras. But Y 3 +Y +1 is irreducibleover F as it is cubic and has not root in F . Hence, F [Y ]/(Y 3 + Y + 1) isa field which shows that (X, Y 3 + Y + 1) is a maximal ideal. This givesM = (X, Y 3 + Y + 1).

Suppose next that X2+X+1 ∈M . Then X3+1 = (X+1)(X2+X+1) ∈M . Hence (Y+1)(Y 2+Y+1) = Y 3+1 = [Y 3+(X3+1)Y+1]+(X3+1)Y ∈M .Thus we must have either Y + 1 ∈ M or Y 2 + Y + 1 ∈ M . Suppose firstthat Y + 1 ∈ M . Then we have I ⊂ (X2 + X + 1, Y + 1) ⊂ M . Again,F [X,Y ]/(Y + 1, X2 + X + 1) is isomorphic to F [X]/(X2 + X + 1) (as F -algebras) via the map X 7→ X and Y 7→ 1. The latter is a field sinceX2 + X + 1 is irreducible as it has no root. Hence (Y + 1, X2 + X + 1) is

3This proof is in Dummit and Foot on p. 720

268 Stanford Spring 2005

maximal and we have M = (Y + 1, X2 +X + 1).Likewise suppose that Y 2 + Y + 1 ∈ M . Then over the field K =

F [X,Y ]/M , the images of X and Y are both solutions to the the poly-nomial Z2 + Z + 1 ∈ K[Z]. But if α is a solution, we have (α + 1)2 =α2 + 1 = α = (α + 1) + 1 so that α + 1 is also a root. Hence we must haveeither Y = X + 1 or Y = X in K. Suppose we are in the first case. ThenY +X+1 ∈M . This gives I ⊂ (Y +X+1, X2+X+1, Y 2+Y +1) ⊂M . NowF [X,Y ]/(Y +X+1, X2+X+1, Y 2+Y +1) is isomorphic to F [X]/(X2+X+1)via the map X 7→ X and Y 7→ X + 1. Hence, since the latter is a field, wemay conclude that M = (Y +X + 1, X2 +X + 1, Y 2 + Y + 1).

Finally suppose that X + Y ∈M . Then we have I ⊂ (X + Y,X2 +X +1, Y 2 + Y + 1) ⊂ M . Again F [X, Y ]/(X + Y,X2 + X + 2, Y 2 + Y + 1) isisomorphic as an F -algebra to F [X]/(X2 +X + 1) via the map X, Y 7→ X.Since the latter is a field, we have M = (X + Y,X2 + X + 1, Y 2 + Y + 1).Thus we have shown that there are exactly 4 maximal ideals in R.

3. If G is a permutation group acting on a set S, we say G isn-transitive if |S| ≥ n and whenever x1, . . . , xn are distinct elementsof S and y1, . . . , yn are distinct elements of S, there exists g ∈ G suchthat g(xi) = yi. We will denote by χ(g) the number of fixed pointsof g. Prove that a necessary and sufficient condition for G to be3-transitive is that

1

|G|∑g∈G

χ(g)3 = 5.

It is routine to check that we can define an action of G onto S3 by g ·(s1, s2, s3) = (g · s1, g · s2, g · s3) for g ∈ G and (s1, s2, s3) ∈ S3. Hence, forg ∈ G, a point in S3 is fixed if an only if each coordinate is fixed by g. Thisgives χ(g)3 such points. Hence, from Burnside’s Lemma, the above sum is 5if and only if G has 5 orbits on S3.

Suppose than that G has 5 orbits on S3. Note that any element in Gsends a point of the form (s, s, s) ∈ S3 to another point of the same form.Hence (s, s, s)|s ∈ S is a union of orbits. Likewise, (s, s, t)|s, t ∈ S, s 6= t,

28.2 Afternoon Session 269

(s, t, s)|s, t ∈ S, s 6= t, (t, s, s)|s, t ∈ S, s 6= t are all unions of orbits. Theremaining points in S3, that is, those with distinct coordinates must thenalso be a union of orbits. But since there are only 5 orbits, all of these setsmust be orbits. In particular, G acts transitively on the collection of pointsin S3 which distinct points, which is to say that G is 3-transitive.

Conversely, ifG is 3-transitive, it is certainly also transitive and 2-transitive.Hence all the sets above must again be orbits under the action of G onto S3.This allow us to conclude that G has 5 orbits. Hence the sum being 5 isequivalent to G being 3-transitive.

4. Suppose that A is an n× n matrix over C with minimal polyno-mial (X − λ)n where λ 6= 0. Find the Jordan form of A2. What ifλ = 0?

Let λ = 0 and T be the transformation given by A. Then we can canfind an ordered basis b1, . . . , bn of Cn which respect to which the matrixof T has all of entries zero and a 1 on the diagonal just above the maindiagonal (as this is the Jordan Form of T ). That is, we have T (b1) = 0 andT (bi) = bi−1 for i ≥ 2. Hence, T 2(b1) = T 2(b2) = 0 and T 2(bi) = bi−2 fori ≥ 3.

Suppose first that n is even. Construct a new ordered basis c1, . . . , cnfor Cn by writing ci = b2i−1 for 1 ≤ i ≤ n/2 and ci = b2(i−n/2) for n/2 < i ≤ n.Now as i runs from i to n/2, 2i−1 runs through all the odd numbers from 1 ton−1. Likewise, as i runs from n/2+1 to n, 2(i−n/2) runs through all the evennumbers from 2 to n. Hence c1, . . . , cn is just a reordering of b1, . . . , bnand so is a basis. Now T 2(c1) = T 2(b1) = 0 and T 2(n/2 + 1) = T 2(b2) = 0.Furthermore, for 1 < i ≤ n/2, T 2(ci) = T 2(b2i−1) = b2i−3 = ci−1 as 1 ≤ i−1 ≤n/2. Likewise, for n/2 + 1 < i ≤ n, T 2(ci) = T 2(b2(i−n/2)) = b2(i−1−n/2) = ci.Thus the matrix of T 2 with respect to this basis consists of 2 Jordan blocks,each corresponding to the polynomial n/2. This matrix is then the Jordanform of A2.

Now suppose that n is odd. Again construct a new order basis c1, . . . , cnof Cn by setting ci = b2i−1 for 1 ≤ i ≤ (n + 1)/2 and ci = b2i−(n+1) for(n + 1)/2 < i ≤ n. Again the first collection of basis elements hits all the

270 Stanford Spring 2005

odd numbers from 1 to n and the second hits all the even numbers from2 to n − 1 so that this basis is just a reordering of the original basis. Wehave T 2(c1) = T 2(b1) = 0 and T 2(c(n+1)/2+1) = T 2(b2) = 0. Likewise, for1 < i ≤ (n + 1)/2, we have T 2(ci) = T 2(b2i−1) = b2i−3 = ci−1 and, for(n+ 1)/2 + 1 < i ≤ n, T 2(ci) = T 2(b2i−(n+1)) = b2i−(n+1)−2 = ci−1. Thus thematrix of T with respect to this basis consist of 2 Jordan blocks, one whichcorresponds to x(n+1)/2 and the other which corresponds to x(n+1)/2. Thismatrix is the Jordan form of A.

Now assume that λ 6= 0. Then, we see that (A2 − λ2)n = (A − λ)n(A +λ)n = 0, as (A − λ)n = 0. Hence A2 satisfies (x − λ2)n. Suppose thatA2 satisfies (x − λ2)k for some k < n. Then we have (A − λ)k(A + λ)k =(A2 − λ2)k = 0 so that A satisfies (x − λ)k(x + λ)k. Since λ 6= 0, thispolynomial is not divisible by (x−λ)n, the minimal polynomial of A. Hencewe have a contradiction and may conclude that (x − λ2)n is the minimalpolynomial of A2. Hence its Jordan form is one Jordan block consisting ofthis polynomial.

5. Find the Galois group of the polynomial X5 + 99X − 1 over thefields Z/2Z, Z/3Z, Z/5Z, Z/11Z, and Q.

Let f(X) be the polynomial in question. Then, over Z2, we have thatf(X) = X5 +X+1 = (X2 +X+1)(X3 +X2 +1). Now both of these factorsare irreducible over Z2 as they have degree two and three and no roots overZ2. Thus if α is a root of the former and β is a root of the latter, we have[Z2(α) : Z2] = 2 and [Z2(β) : Z2] = 3. Furthermore, both of these extensionsare Galois (as a finite extension of a finite field). We must have that theGalois group of Z2(α) is Z2 and that of Z2(β) is Z3. Now Z2(α)∩Z2(β) = Z2

as the degrees of these two extensions over Z2 have relatively prime degree.This gives that the Galois group of Z2(α)Z2(β) over Z2 is Z3 × Z3. Thiscomposite, however, is the splitting field of f(x) so we have that the Galoisgroup is Z6 (generated by the Frobenius automorphism).

Next, over Z3, we have f(X) = X5− 1 so that a splitting field of f(X) isgenerated by ζ, a 5th root of unity. Now ζ will have order 5 as an elementof Z3(ζ)

× so that we may conclude that 5 divides the order of the latter.

28.2 Afternoon Session 271

Hence Z3(ζ) cannot have degrees 1, 2, or 3 over Z3 as these fields have order3, 9, and 27, respectively. Hence their multiplicative groups have order 2, 8,and 26, none of which are divisible by 5. Consider, then the field F34 . Wesee that the multiplicative group has order 80, which by Cauchy’s Theoremmust contain an element of order 5. Thus Z4(ζ) = F34 is the splitting fieldfor f(X). We may conclude that the Galois group is Z4 (generated by theFrobenius automorphism).

Over Z5, we have f(x) = X5−X−1. Suppose that α is a root of f(x) anda ∈ Z5, then we have f(α+a) = (α+a)5−(α+a)−1 = (α5−α−1)+a5−a = 0.In particular, f(X) has not root in Z5 because then 0 would be a root, whichit is not. Hence if f(X) is reducible over Z5, it must split as f(x) = g(x)h(x)where g(x) is an irreducible quadratic and h(x) is an irreducible cubic. Nowif α is a root of g(x), we have [Z5(α) : Z5] = 2 so that Z5(α) does notcontain any of the roots of h(x) (as each root of later generates a degree 3extension). But we have already seen that the roots of f(x) and so the rootsof h(x) must have the form α + a for a ∈ Z5. Since an element of this formlies in Z5(α), we have a contradiction. Hence f(x) is irreducible so that thesplitting field has degree 5 (adjoining one root gives a degree 5 extension andthis extension contains all the roots by the above, or by the fact that everyfinite extension of Z5 is Galois). Thus we must have that the Galois groupis Z5 (again generated by the Frobenius automorphism).

Over Z11, we have f(x) = x5 − 1. We also have f(1) = 0, f(3) = 0,f(4) = 0, f(5) = 0, and f(9) = 0. Hence f(x) splits over Z11. Hence theGalois group is the trivial group.

Lastly consider f(x) over Q. Since f(x) is irreducible modulo 5 (and sinceit is separable modulo 5), we see that the Galois group, viewed as a subgroupof S5 contains a 5-cycle. Likewise, the factorization over Z2 implies that theGalois group contains an element which is a disjoint product of a 2-cycleand a 3-cycle. Cubing this element, we see that the Galois group contains a2-cycle. Hence by Spring 1996 #1, we may conclude that the Galois groupis S5.

Index

Fields and Galois Theorypth roots, 214

in characteristic p, 88, 105Abelian Galois groups over the ra-

tionals, 108Abelschen polynomials, 72adjoining square roots to the ra-

tionals, 51algebraic closure

of a finite field, 15algebraically closed, 8commutator subgroup of a Galois

group, 147core of a group, 185cosets in a Galois group, 140extensions of degree 60, 199extensions of the rationals adjoined

with√

2, 63field of fractions for a group, 47finite fields

algebraic closure, 15extension of degree 2, 29

formal polynomial, 202Frobenius automorphism, 105minimal polynomial, 140, 159number of roots, 177orderp2q Galois group, 2318, 18530, 185

polynomials over the field of order3, 69

Primitive Extension Theorem, 159rational functions over a finite field,

125, 167relatively prime extensions, 95splitting field

of x3 − 7, 131of x4 − 2, 97of x6 + 3, 168of x8 + 1, 196over the rationals, 7

Sylow subgroups of a Galois group,38

tensor product, 41, 211

GroupsA4, 155F -pure, 187S4, 91, 155, 165S5, 193p-group, 173abelian subgroups, 79alternating group, 54Burnside’s Lemma, 3center, 33centralizer, 79conjugates of a subgroup, 3cyclic, 85, 210derived series, 66fixed points of an action, 201

INDEX 273

nilpotent, 173, 194normal subgroup, 85, 145, 155,

185normal subgroups, 106orderp3, 13412, 11724, 9130, 18536, 15590, 79210, 145255, 1065075, 139

outer automorphisms, 54presentation, 23, 194pure, 187semi-direct product, 100solvable, 173, 194subgroup of finite index, 209Sylow subgroup, 185Sylow subgroups, 33, 54, 79, 91,

106, 117, 139, 145, 165of a normal subgroup, 111

symmetric group, 54with four subgroups, 4

Linear AlgebraMn(F ), 141SL2(F3), 69semisimple, 178characteristic polynomial, 58, 178commuting matrices, 67, 115cyclic transformation, 210Eigenvalue, 36endomorphisms, 36, 69, 82idempotent, 36idempotents, 82

linear maps as matrices, 9matrices with complex entries, 67minimal polynomial, 58nilpotent, 178over F2, 51simulaneously diagonalizable, 115subgroups of rational vector spaces,

183trace, 82transpose, 6unipotent, 178universal maps of the tensor prod-

uct, 31upper triangular matrices, 3

Modulesalgebras

complex, 170finite-dimensional, 206

annihilator, 161characteristic polynomial, 13characteristic polynomials, 102determinant, 25exact sequence, 35faithful, 151, 161finitely generated submodules, 35free, 35invariant factors, 102minimal polynomial, 13, 102overMn(F ), 141a field of fractions, 190a PID, 107a simple ring, 121a Wedderburn Ring, 161an Artinian Ring, 151the field of order 2, 193the integers, 25

274 INDEX

projective, 190, 204rank, 35tensor product, 59, 76

and ideals, 19of a field extension, 211of the rationals, 59

Ringsconductor, 149discrete valuation ring, 21division ring

finite, 94example of R ∼= R⊗R R, 76factorization of prime ideals, 118Hilbert Basis Theorem, 75ideals, 65integers adjoined the square root

of 2, 26integral closure, 157integrally closed, 86Jacobsen radical, 101Jacobson radical, 121Noetherian, 75PID

field of fractions, 134maps onto, 92quotients of, 92

polynomialsin infinitely-many variables, 10in two variables, 86over the integers, 112

prime elements, 187prime spectrum, 174radical of an ideal, 101simple, 121simple subalgebras of the ring of

matrices with comples entries,45

UFDbut not a PID, 86non-Noetherian, 10PID criterion, 12quotient of, 92

valuation domain, 137