Solution to Commutative Ring Theory Anonymous This is the solution of the homework in the course 《Commutative Ring Theory》 . Problems can be founded on http://www.wwli.url.tw/downloads/CommRing-2019-HW.pdf, and I’m glad to any errata to this document. We suppose A is a commutative ring, p> 0 is a prime number. 1 Ring Theory Revisited 1. Suppose x =0, then (1 + x)(1 - x + x 2 -··· +(-1) n−1 x n−1 )=1 ⇒1+ x is a unit. If a ∈ A × ,b ∈ Nil(A), then a −1 b ∈ Nil(A), ⇒ 1+ a −1 b ∈ A × ⇒ a + b = a(1 + a −1 b) ∈ A × 2. (1) (⇐ =): a 1 ,...,a n ∈ Nil(A) ⇒a 1 ,...,a n ∈ Nil(A[x]) ⇒a 1 x,...,a n x n ∈ Nil(A[x]) a 0 ∈ A × ⊂ (A[x]) × ⇒f ∈ (A[x]) × (= ⇒): If there exists g = b 0 + b 1 x + ··· + b m x m such that fg =1, then (denote a k =0 when k>n, b k =0 when k>m) 1
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Solution to Commutative Ring Theory
Anonymous
This is the solution of the homework in the course《Commutative Ring Theory》. Problemscan be founded on http://www.wwli.url.tw/downloads/CommRing-2019-HW.pdf, and I’mglad to any errata to this document.
We suppose A is a commutative ring, p > 0 is a prime number.
1 Ring Theory Revisited
1. Suppose x = 0, then
(1 + x)(1− x+ x2 − · · ·+ (−1)n−1xn−1) = 1
⇒1 + x is a unit.
If a ∈ A×, b ∈ Nil(A), then a−1b ∈ Nil(A),⇒ 1 + a−1b ∈ A× ⇒ a+ b = a(1 + a−1b) ∈ A×
2. (1) (⇐=): a1, . . . , an ∈ Nil(A)
⇒a1, . . . , an ∈ Nil(A[x])
⇒a1x, . . . , anxn ∈ Nil(A[x]) a0 ∈ A× ⊂ (A[x])×
⇒f ∈ (A[x])×
(=⇒): If there exists g = b0 + b1x+ · · ·+ bmxm such that fg = 1, then
• a0(X) = 0 or a1(X) = 0, easy to know f(X,Y ) = 0.
• a0, a1 ∈ C[X] has no common nontrivial factors. Then
ϕ(f) =(a0(T
2 − 1))+(a1(T
2 − 1))(T 3 − T )⇒ X|a0, X|a1
Contradiction!
So ϕ is injective, thus we can view R := C[X,Y ]/ (Y 2 −X2 −X3) as a subring ofC[T ],and
XT − Y = 0
⇒ T is integral over R
⇒ R[T ] is integral over R
R[T ] = C[T 2 − 1, T 3 − T, T ] = C[T ] is a UFD, so normal.
2. Consider a Noetherian ring R with K := Frac(R). Show that y ∈ K is integral over R ifand only if there exists u ∈ R such that u 6= 0 and uyn ∈ R for all n.
证明. Suppose R is a domain.(=⇒): y ∈ K is integral over R
, then there exists u := wm 6= 0, such that for any n ∈ N+,
uxn ∈ u(R+Rx+ · · ·+Rxm−1) ⊆ R+R · · ·+R = R
10 INTEGRAL DEPENDENCE, NULLSTELLENSATZ 11
(⇐=): R is a Noetherian ring
⇒ u−1R is a Noetherian R-module
⇒ R[x] is a finitely generated R-module
⇒ x is integral over R
3. Let R = Q [X1, X2, . . .] (finite or infinitely many variables). Show that nil(R) = rad(R) ={0}.
证明. Claim: For any f ∈ Q[X1, . . . , Xn], there exists (a1, . . . , an) ∈ Q, such thatf(a1, . . . , an) 6= 0.We can prove it by induction on n. Suppose it holds for k < n.WLOG, suppose f =
∑mi=0 gix
βin where gi ∈ Q[X1, . . . , Xn] and gm 6= 0.
Then by induction, there exists (a1, . . . , ak−1) such that gk(a1, . . . , ak−1) 6= 0,now
f(x) := f(a1, . . . , ak−1, x)
has at most m roots. We can choose ak ∈ Q such that f(ak) 6= 0 For any f ∈ R nonzero,there exists n ∈ N, such that f ∈ Q[X1, . . . , Xn].Let
ϕf : R −→ Q g −→ g(a1, . . . , ak, 0, . . .)
It is a surjective homomorphism.
⇒ R/Kerϕf∼= Q
⇒ Kerϕf is a maximal ideal not containingf
⇒ rad(R) ⊆ {0}
⇒ {0} ⊆ nil(R) ⊆ rad(R) ⊆ {0}
⇒ nil(R) = rad(R) = {0}
4. Let R = Q [X1, X2, . . .] (infinitely many variables). Show that R is not a Jacobson ring.
证明. Denote p = (X) in Q[X], R′ = (Q[X])p and an bijection map Ψ : N+ −→ Qr {0}We constrct
ψ : R −→ R′ Xi 7−→1
X −Ψ(i)
10 INTEGRAL DEPENDENCE, NULLSTELLENSATZ 12
which is surjective.Consider all the irreducible polynomials.If R is a Jacobson ring, then so is R′, but
MaxR′ = {p} 6= {p, (0)} = SpecR′
Contradiction!
5. (1) Let A be a subring of an integral domain B, and let C be the integral closure of Ain B. Let f, g be monic polynomials in B[x] such that fg ∈ C[x]. Then f, g are in C[x].
(2) Prove the same result without assuming that B (or A) is an integral domain.
证明. 偷懒抄书,此题不算![2]Take a field (Frac(B)) containing B in which the polynomials f, g split into linearfactors; say f = Π(x− ξ1) , g = Π(x− ηj) . Each ξi andeach η, is a root of fg, hence isintegral over C. Hence the coefficients of f and g are integral over C.[1]LEMMA (14.7).−Let R ⊂ R′ be a ring extension, X a variable, f ∈ R[X] a monicpolynomial. Suppose f = gh with g, h ∈ R′[X] monic. Then the coefficients of g and h
are integral over R.Proof: Set R1 := R′[X]/〈g〉. Let x1 be the residue of X. Then 1, x1, x
21, . . . form a free
basis of R1 over R′ by (10.25) as g is monic; hence, R′ ⊂ R1 . Now,g (x1) = 0; sog factors as (X − x1) g1 with g1 ∈ R1[X] monic of degree 1 less than g. Repeat thisprocess, extending R1. Continuing, obtain g(X) =
∏(X − xi)and h(X) =
∏(X − yj)
with all xi and yj in an extension of R′. The xi and yi are integral over R as they areroots of f. But the coefficients of q and h are polynomials in the xi and yj ; so they tooare integral over R.
6. Let f : A → B be an injective map, with A Noetherian and B integral over A. Assumethat neither A nor B have zero divisors.(1) Show that if A is a field, then so is B.(2) Deduce that a field k is algebraically closed (i.e., every polynomial has a root) if andonly if for every finite field extension k ⊂ k′ i.e., k′ is f.d. as a k -vector space, we havek = k′.(3) Show that if B is a field, then so is A.
11 FLATNESS 13
证明. (1) For any x ∈ B, there exists a0, . . . , an−1 ∈ A, a0 6= 0, such that
xn + an−1xn−1 + · · ·+ a0 = 0
⇒ x · (− 1
a0)(xn−1 + an−1x
n−1 + · · ·+ a1) = 1
⇒ B is a field.
(2) (=⇒): For any x0 ∈ k′, there exists x1, . . . , xn ∈ k, such that
f(x) = (x− x1) · · · (x− xn) & f(x0) = 0
So x0 ∈ k. (⇐=):If not, there exists an irreducible polynomial f which has no root.Then k[t]/(f(t)) is a finite field extension of degree degf .
(3) For any x ∈ A, there exists y ∈ B such that xy = 1.
⇒ yn + bn−1yn−1 + · · ·+ b0 = 0
⇒ y + bn−1 + · · ·+ b0xn−1 = 0
⇒ y ∈ A
So k is a field.
11 Flatness
Facts.
• k[[t]] is a PID. Its ideal can be written as (tm), while for any f ∈ k[[t]], there existsm ∈ Z>0, g ∈ (k[[t]])∗, such that f = gtm.
• A beautiful diagram:
· · · TorA3 (A,M) TorA3 (K,M) TorA3 (K/A,M)
TorA2 (A,M) TorA2 (K,M) TorA2 (K/A,M)
TorA1 (A,M) TorA1 (K,M) TorA1 (K/A,M)
A⊗A M K ⊗A M K/A⊗A M 0
11 FLATNESS 14
1. For a field k, show that k[[t]][Y, Z]/(Y Z − t) is flat over k[[t]].
证明. We only need to prove k[[t]][Y, Z]/(Y Z − t) has no zero divisors except 0. This iseasy: We claim that every item in k[[t]][Y, Z]/(Y Z − t) can be uniquely written as theform
+∞∑n=0
(fn(Y ) + gn(Z) + an)tn
wherefn ∈ k[Y ], gn ∈ k[Z], an ∈ k, fn(0) = gn(0) = 0
If+∞∑n=0
(fn(Y ) + gn(Z) + an)tn 6= 0
then
tm+∞∑n=0
(fn(Y ) + gn(Z) + an)tn 6= 0
gtm+∞∑n=0
(fn(Y ) + gn(Z) + an)tn 6= 0 g ∈ (k[[t]])∗
2. Let N ′, N,N ′′ be A -modules, and 0→ N ′ → N → N ′′ → 0 be an exact sequence, withN ′′ flat. Prove that N ′ is flat ⇔ N is flat.
证明. If 0 → Mφ−→ M ′ is a A-modular exact sequence, then We have the exact se-
quences:
0 Kerϕ⊗ IdN ′ Kerϕ⊗ IdN 0 · · ·
Tor1(M,N ′′) M ⊗N ′ M ⊗N M ⊗N ′′ 0
Tor1(M′, N ′′) M ′ ⊗N ′ M ′ ⊗N M ′ ⊗N ′′ 0
Because Tor1(M,N ′′) = Tor1(M′, N ′′), the upper line is exact. So N ′ is flat⇔ N is flat.
3. A ring A is absolutely flat if every A -module is flat. Prove that the following areequivalent:
(1) A is absolutely flat.
(2) Every principal ideal is idempotent.
(3) Every finitely generated ideal is a direct summand of A.
证明.
(1)⇒(2): Consider the exact sequence
0 −→ I −→ A −→ A/I −→ 0
Tensoring with I, we obtain the exact sequence (I ⊗A I = I2 because I is principalideal)
0 −→ I2 −→ I −→ I ⊗A A/I −→ 0
So I/I2 = I ⊗A A/I = 0.(I ⊗A A/I = 0 because I ⊗A A/I ↪→ A⊗A A/I ∼= A/I is azero map(using the fact that A/I is flat))Especially, when I = (x), then (x)2 = (x), there exists a ∈ A such that x = ax2,now (x) = (ax) is idempotent.
(2)⇒(3): We just need to prove that every finitely generated ideal is principal ideal.(Then,use the decomposition A = (a)⊕ (1− a))We only need to prove 〈a, b〉 = 〈a+ b − ab〉 when a2 = a, b2 = b.
(3)⇒(1): Suppose any I ▹ A is the direct summand, then A/I is flat, we obtain
Tor1(N,R/I) = 0 for any N, I
. So A is absolutely flat.
4. Prove the following properties of absolutely flat:
(1) Every homomorphic image of an absolutely flat ring is absolutely flat.
(2) If a local ring is absolutely flat, then it is a field.
(3) If a ring A is absolutely flat, then every non-unit in A is a zero-divisor.
12 GOING-UP AND GOING-DOWN 16
证明. (1)A is absolutely flat ⇒〈x〉2 = 〈x〉 in A
⇒〈x〉2 = 〈x〉 in A/I
⇒A/I is absolutely flat
(2) Suppose m is the unique maximal ideal of A, then A/m is a field. If there exists〈x〉 ( A, x 6= 0, then 〈x〉 ⊆ m ( A⇒ x = ax2 ⇒ x(1− ax) = 0.However, 1− ax /∈ m⇒ (ax− 1) ∈ A×, so we obtain x ∈ A×.
(3) If there exists 〈x〉 ( A, x 6= 0, then there exists e ∈ A such that e2 = e, (x) = (e).we get a ∈ A such that x = ae(x 6= a)⇒ x(x− a) = 0.So every non-unit in A is a zero-divisor.
12 Going-up and Going-down
5. Let f : A → B be an integral homomorphism of rings, i.e. B is integra over its subringf(A). Show that f# : Spec(B) → Spec(A) is a closed mapping, i.e. that it maps closedsets to closed sets.
证明. When f(B) 6= B, It may fail. Consider
f : Z =⇒ Z/3Z f#((0)) = (3)
is the counterexample.(Notice that Spec f(A) ∼= V (ker f) is closed set of SpecB)When A ⊆ B, Suppose V (I) ⊆ Spec(B) is the closed set (I / B), then we claim:f#(V (I)) = V (f−1I).
(a) For all p ∈ V (I)⇒ p ⊇ I ⇒ f#(p) = f−1(p) ⊇ f−1(I)⇒ f#(p) ∈ V (f−1(I))
(b) For all q ∈ V (f−1(I)) ⊆ Spec(A), there exists p ∈ Spec(B) such that p ∩ SpecA =
q. Easy to find that p ⊇ I.
6. Let A ⊂ B be an extension of rings, making B integral over A, and let p be a prime idealof A. Suppose there is a unique prime ideal q of B with q ∩A = p. Show that
12 GOING-UP AND GOING-DOWN 17
(a) qBp is the unique maximal ideal of Bp := B [(Ar p)−1]
(b) Bq = Bp
(c) Bq is integral over Ap
证明. (a) We have the following commutative diagram:
A B p q
Ap Bp pAp qBp
Bp is integral over Ap, and pAp is maximal in Ap
⇒ qBp is maximal in Bp
If there exists m ▹ Bp is maximal, then f#p (m) = pAp , thus m = qBp.
(b) p ⊆ q⇒ Bq ⊇ Bp.By the universal property, we only need to show
∀ x ∈ B r q, x is invertible.
If x1= q
b
a1where a1 ∈ A r p, there exists a2 ∈ A r p such that a1a2x = a2bq ⇒
x ∈ q.So x
1/∈ qBp is invertible.
(c) Bq = Bp is integral over Ap.
7. Let the integral extension A ⊂ B and the prime ideal p be as above. Suppose that A isa domain and q, q′ are distinct prime ideals of B, both mapping to p under Spec(B) →Spec(A). Show that Bq is not integral over Ap.
证明. Take y ∈ q′rq. If y−1 ∈ Bq is integral over Ap, then there exists a0, . . . an−1 ∈ Ap
with a0 6= 0 such that a0yn + a1yn−1 + · · · + an−1y = −1. Now there exists s ∈
Ar p, a′0, . . . a′n−1 ∈ A with
a′0yn + a′1y
n−1 + · · ·+ a′n−1y = −s
Now the left is in q′ while −s /∈ q′, contradiction!
Graded rings and modules, Filtrations
12 GOING-UP AND GOING-DOWN 18
1. Many basic operations on ideals, when applied to homogeneous ideals in Z -graded rings,lead to homogeneous ideals. Let I be a homogeneous ideal in a Z-graded ring R. Showthat:
(1) The radical of I is homogeneous, that is, the radical of I is generated by all thehomogeneous elements f such that fn ∈ I for some n.
(2) If I and J are homogeneous ideals of R, then (I : J) := {f ∈ R|fJ ⊂ I} is ahomogeneous ideal.
(3) Suppose that for all f, g homogeneous elements of R such that fg ∈ I one of f andg is in I. Show that I is prime.
证明. We need to show that:
Claim 12.1. For any r =∑
i∈Z ri ∈√I, ri ∈ Ri, we have ri ∈
√I
Consider i0 = min{i | ri 6= 0}, we know that
rni0 = (rn)ni0 ∈ I ⇒ ri0 ∈ Ri0 ∩√I
then consider r − ri0 . By induction, we can show that r ∈ 〈f ∈ Ri | fn ∈ I〉.
We need to show that (the difficult part):
Claim 12.2. For any f =∑
i∈Z fi ∈ (I : J), fi ∈ Ri, we have fi ∈ (I : J)
NowfJ ⊆ I ⇒ fg ⊆ I for any g ∈ Jj
⇒fig ⊆ Ii+j ⊆ I for any g ∈ Jj⇒fiJ ⊆ I
When f =∑
i∈Z fi /∈ I, g =∑
j∈Z gj /∈ I, we need to show that
fg =∑i,j∈Z
figj /∈ I.
Choosei0 = min{i | fi /∈ I} j0 = min{j | fj /∈ I}
then(fg)i0+j0 =
∑i
figi0+j0−i ∈ fi0gj0+I ⊆ A \ I
so fg /∈ I.
12 GOING-UP AND GOING-DOWN 19
⇒
2. Suppose R is a Z -graded ring and 0 6= f ∈ R1
(1) Show that R [f−1] is again a Z-graded ring.
(2) Let S = R [f−1]0 , show that S ∼= R/(f − 1), and R [f−1] ∼= S [x, x−1] where x is anew variable.
证明. (1) R[f−1] is a ring which is graded by
(R[f−1])i =
⟨ai+j
f j
⟩ai+j∈Ri+j
.
(2) We know
(R[f−1])0 =
⟨ajf j
⟩aj∈Rj
.
and we have the surjective ring homomorphism
R −→ (R[f−1])0 ri ∈ Ri 7−→rif i
and the kernel of which is (f − 1).Now the homomorphism
R[f−1
]−→ S
[x, x−1
] ri+j
f j∈ (R[f−1])i 7−→
ri+j
f i+jxi
is an isomorphism.
3. Show that if R is a graded ring with no nonzero homogeneous prime ideals, then R0 is afield and either R = R0 or R = R0 [x, x
−1] .
证明. We first state a lemma which can be proved using Zorn’s lemma:∗
Lemma 12.3. Let I be a homogeneous ideal of a graded ring R, I 6= R, then there existsa homogeneous prime ideal which contains I.
Using the lemma to Ra, we get:∗for details, see: https://math.stackexchange.com/questions/385292/homogeneous-ideals-are-contained-in-homogeneous-
prime-ideals
13 COMPLETIONS 20
Corollary 12.4. If R is a graded ring with no nonzero homogeneous prime ideals, thenany homogeneous item a ∈ R \ {0} is invertible.
Now R0 is a field. if R = R0, then everything was done; Otherwise, Suppose i0 = {i ∈Z>0 | Ri 6= 0} there exists x ∈ Ri0 \ {0}, which is invertible.Now:
• If i0 - r, then Rr = 0 by the euclidean division;
• If i0 | r, a ∈ Rr, then a = ax−r/i0 · x−r/i0 ∈ R0 [x, x−1].
So we’re done.Another point: you can take the homogeneous items from prime ideal p to construct ahomogeneous prime ideal.
4. Taking the associated graded ring can also simplify some features of the structure of R.For example, let k be a field, and let R = k [x1, . . . , xr] ⊂ R1 = k [[x1, . . . , xr]] be therings of polynomials in r variables and formal power series in r variables over k, andwrite I = (x1, . . . , xr) , I
′ for the ideal generated by the variables in either ring. Showthat grI R = grI′ R1
证明. We know thatIk/Ik+1 ∼= I ′k/I ′k+1
sogrI R =
⊕k∈Z
Ik/Ik+1 ∼=⊕k∈Z
I ′k/I ′k+1 = grI′ R1
13 Completions
1. Let A be a local ring, m its maximal ideal. Assume that A is m -adically complete. Forany polynomial f(x) ∈ A[x], let f(x) ∈ (A/m)[x] denote its reduction mod. m. ProveHensel’s lemma: if f(x) is monic of degree n and if there exist coprime monic polynomialsg(x), h(x) ∈ (A/m)[x] of degrees r, n−r with f(x) = g(x)h(x), then we can lift g(x), h(x)back to monic polynomials g(x), h(x) ∈ A[x] such that f(x) = g(x)h(x)
证明. See [Hensel’s Lemma, Theorem 1]† Using induction makes sense.†html:http://therisingsea.org/notes/HenselsLemma.pdf
13 COMPLETIONS 21
2. (a) With the notation of Exercise 1, deduce from Hensel’s lemma that if f(x) has asimple root α ∈ A/M, then f(x) has a simple root a ∈ A such that α = a mod M
(b) Show that 2 is a square in the ring of 7 -adic integers.
(c) Let f(x, y) ∈ k[x, y], where k is a field, and assume that f(0, y) has y = a0 as asimple root. Prove that there exists a formal power series y(x) =
∑∞n=0 anx
n suchthat f(x, y(x)) = 0. This gives the ”analytic branch ” of the curve f = 0 throughthe point (0, a0) .
证明. (a) Suppose f = (x − α)h(x), then by the Hensel’s lemma, there exists g(x) ∈A[x], h(x) ∈ A[x] such that
f(x) = g(x)h(x)
A[x] −→ A/M[x] g 7−→ (x− α) h 7−→ h
As a consequence, there exists a ∈ A, g(x) = x − a, a ≡ α mod M (the root isevidently simple.)
(b) x2 − 2 has a simple root 5 ∈ Z/7Z. By (a), there exists a ∈ A to be a simple rootof x2 − 2 (in Z(7)).
(c) Using (a), let A = k[[x]],M = (x), we get
Corollary 13.1. If f(0, y) ∈ k[y] has a simple root a0, then f(x, y) ∈ k[[x]][y] hasa simple root y(x) =
∞∑n=0
bnxn ∈ k[[x]] such that
f(x, y(x)) = 0 b0 = a0
3. Let A be a Noetherian ring, a an ideal in A, and A the a-adic completion. For anyx ∈ A, let x be the image of x in A. Show that x not a zero-divisor in A implies x not azero-divisor in A. Does this imply that if A is an integral domain then A is an integraldomain?
证明. suppose x is not a zero-divisor in A, then
0 −→ A×x−→ A
⇒0 −→ A×x−→ A
⇒0 is not a zero-divisor in A
14 MITTAG-LEFFLER SYSTEMS AND COMPLETIONS FOR NON-NOETHERIAN RINGS22
To take the example where A is an integral domain but A is not an integral domain, wetake
A = Q[x, y]/(y2 − x2 − x3)
which is a domain because (y2 − x2 − x3) is irreducible in Q[x, y],but
A = Q[[x, y]]/(y2 − x2 − x3)
is not a domain because
[y − x(1 + 1
2x− 1
8x2 · · · )][y + x(1 +
1
2x− 1
8x2 · · · )] = 0 in A
4. Let k be a field and consider the quotient of infinite polynomial ring
R :=k [X,Z, Y1, Y2, Y3, . . .]
(X − ZY1, X − Z2Y2, X − Z3Y3, . . .)
Denote by Z the image of Z in R. Show that the ideal I := (Z) of R satisfies⋂
n≥1 In 6=
{0}. Why is this consistent with Krull’s intersection theorem?
证明. For convinience, we omit the bar.
X = ZnYn ∈ In ⇒ X ∈⋂n≥1
In
R is not Noetherian because
(Y1) ( (Y1, Y2) ( (Y1, Y2, Y3) ( · · ·
is a infinite ascending chain in R.
14 Mittag-Leffler systems and Completions fornon-Noetherian rings
1. Consider an inverse system of sets · · · ←− Anφη+1←− An+1 ←− · · · (where n = 1, 2, . . . .
For each j > i, let ϕj,i : Aj −→ Ai be the composition of ϕj , . . . , ϕi. We say thatMittag-Leffler conditions holds for (An, ϕn)n≥1 if for each i, we have
ϕk,i (Ak) = ϕj,i (Aj) whenever j, k � i
14 MITTAG-LEFFLER SYSTEMS AND COMPLETIONS FOR NON-NOETHERIAN RINGS23
Show that if (An, ϕn)n is Mittag-Leffler and An 6= ∅ for each n, then the limit
lim←−n
An :=
{(an)n ∈
∏n
An : ∀n, ϕn+1 (an+1) = an
}
is nonempty as well.
证明. Let Dn = ∩∞k=n+1ϕk,n(Ak) be a set which is nonempty (because
ϕn+1,n(An+1) ⊇ ϕn+1,n(An+1) ⊇ · · · ⊇ · · ·
is stable and nonempty). We have the inverse system
· · · ←− Dnφη+1←− Dn+1 ←− · · ·
and each ϕn is surjective. So lim←−nAn is nonempty.(Find the PREIMAGE).
Remark 14.1. Moreover,
ϕD :∏
Dn −→∏
Dn (an)n∈Z 7−→ (an − ϕn+1(an+1))n∈Z
is surjective. Thenlim←−
1Dn := coker ϕD = 0
2. Suppose we are given an inverse system of short exact sequences of abelian groups, i.e.a commutative diagram with exact rows, where n = 1, 2, . . . Show that if (An, ϕn)n isMittag- Leffler, then
0 −→ lim←−Anlim fn−→ lim←−Bn
lim gn−→ lim←−Cn −→ 0
is exact. You only have to show the surjectivity of lim gn.
证明. We have the exact sequence:
0∏
nAn
∏nBn
∏nCn 0
0∏
nAn
∏nBn
∏nCn 0
θ′ θ θ′′
whereθ′ :
∏n
An −→∏n
An (an)n∈Z 7−→ (an − ϕn+1(an+1))n∈Z
14 MITTAG-LEFFLER SYSTEMS AND COMPLETIONS FOR NON-NOETHERIAN RINGS24
So by the Snake lemma, we have the exact sequence:
0 −→ lim←−Anlim fn−→ lim←−Bn
lim gn−→ lim←−Cn −→ lim←−1An
We only need to prove that θ′ is surjective. Define Dn as in Problem 1, we get the exactsequence
lim←−Dn −→ lim←−An −→ lim←−An/Dn
from
0∏
nDn
∏nAn
∏nAn/Dn 0
0∏
nDn
∏nAn
∏nAn/Dn 0
τ ′ θ′ τ ′′
By 14.1, lim←−Dn = 0; we will show that lim←−An/Dn = 0, thus lim←−An = 0, thus the proofends.
Remark 14.2. The Mittag-Leffler conditions tell us that
∀ n ∈ Z,∃ m > n s.t. ϕm,n(Am) = Dn
⇒ϕm,n : Am/Dm −→ An/Dn is the zero map.
So lim←−An/Dn = 0 (ONLY ZERO SOLUTION)
3. Let R be a ring (not necessarily Noetherian), I be a proper ideal, and ϕ : M −→ N bea homomorphism of R -modules. Prove the following statements.
(a) If M/IM → N/IN is surjective, then so is ϕ : M → N . Here M = limn≥1M/InM
stands for the I-adic completion.(Hint: First, show M/InM → N/InN is surjective by Nakayama’s lemma. Next,set Kn = Ker [M → N/InN ] to get exact sequences 0
→ Kn/InM →M/InM → N/InN → 0
then try to establish the surjectivity of Kn+1/In+1M → Kn/I
nM in order to applyMittag-Leffler.)
(b) If 0→ K → M → N → 0 is an exact sequence of R -modules and N is flat, then0→ K → M → N → 0 is exact.
(c) If M is finitely generated, then the natural homomorphism M ⊗R R → M given bym⊗ (rn)
∞n=1 7→ (rnm)
∞n=1 is surjective.
14 MITTAG-LEFFLER SYSTEMS AND COMPLETIONS FOR NON-NOETHERIAN RINGS25
证明. (a) We’d like to point out the commutative diagram
0 ∗ Kerfn+1 Kerfn
0 InM/In+1M M/In+1M M/InM 0
0 InN/In+1N N/In+1N N/InN 0
0 coker fn+1 coker fn 0
fn+1 fn
Then by induction we know the two surjectivity in the Hint.
(b) We have
0 K/In+1K M/In+1M N/In+1N 0
0 K/InK M/InM N/InN 0
φ
and ϕ is surjective, so 0→ K → M → N → 0 is exact.
(c) Just see the commutative diagram
R⊕N M ⊗R R
M
∃ surj
4. Suppose I is finitely generated. Let M be an R -module. Prove that
InM = Ker[M →M/InM
]= InM
for all n ∈ Z≥1, and M is I -adically complete as an R -module.
证明. (Hint: Fix n and take generators f1, . . . , fr of In. This yields a surjective homo-morphism of R -modules (f1, . . . , fr) :M
⊕r → InM ⊂M Pass to completions and showthat
(f1, . . . , fr)∧: M⊕r → InM = lim
m≥n
InM
ImM' Ker
[M →M/InM
]⊂ M
15 DIMENSION THEORY 26
which is surjective by the previous exercise. The image of (f1, . . . , fr) : M⊕r → M isboth InM and InM to infer that
M/InM ∼= M/InM ∼=M/InM
then we have M∧∧ ∼=M∧, which shows that M∧ is complete.
15 dimension theory
1. Let k be a field and R = k [X0, . . . , Xn] , graded by total degree. Consider the graded R-module S = R/(f) where f is a homogeneous poly-nomial of total degree d ≥ 1. Showthat when m ≥ d,
χ(S,m) := dimk Sm =
(m+ n
n
)−
(m+ n− d
n
)
证明. We have the SES of the graded k-linear spaces:
0 −→ (f) −→ R −→ S −→ 0
Thus
dimk Sm = dimk Rm − dimk(f)m =
(m+ n
n
)−
(m+ n− d
n
)
2. Let R =⊕
nRn be a Z≥0-graded ring, finitely generated over R0. Assume R0 is Artinian(for example, a field) and let M = ⊕nMn be a finitely generated Z≥0−gradedR -module.Define the Hilbert series in the variable T as
HM (T ) :=∑m≥0
χ(M,m)Tm ∈ Z[[T ]]
where χ(M,m) denotes the length of the R0 -module Mm, as usual. In what follows,graded means graded by Z≥0.
(a) Show that if 0→ M ′ → M → M ′′ → 0 is a short exact sequence of graded R
-modules, then HM (T ) = HM ′(T ) +HM ′′(T ).
(b) Relate HM (T ) and HM(k)(T ) for arbitrary k ∈ Z, where M(k)d := Md+k.
15 DIMENSION THEORY 27
(c) Suppose that R is generated as an R0 -algebra by homogeneous elements x1, . . . , xnwith di := degxi ≥ 1. Show that there exists Q ∈ Q[T ] such that
HM (T ) =Q(T )
(1− T d1) · · · (1− T dn)
as elements of Q[T ].(Hint: you may imitate the arguments for the quasi-polynomiality of χ(M,n).)
(d) What can be said about the Zm≥0 -graded case, for general m?
证明. (a) We just need to prove
l(Mn) = l(M ′n) + l(M ′′
n )
this is true even if R0 is not Artinian. ‡
(b)T kHM(k)(T ) =
∑m≥0
χ(M(k),m)Tm+k
=∑m≥0
χ(M,m+ k)Tm+k
=HM (T )−∑
06n<k
χ(M,n)Tn
(c) This is followed by [2]. We prove by induction on n. When n = 0, R = R0 and M isf.g., so Mn = 0 for n >> 0; now suppose n > 0 and the theorem was true for n− 1,we consider the ES induced by the map ×xn:
0 −→ Km −→Mm −→Mm+dn−→ Lm+dn
We haveHK(T )−HM (T ) = HM(dn)(T )−HL(dn)(T )
then easy to find the required form.
(d) Suppose that R is generated as an R0 -algebra by homogeneous elements x1, . . . , xnwith di := degxi 6= (0, 0, . . . , 0). Then there exists Q ∈ Q[T1, . . . , Tm] such that(T := (T1, . . . , Tm))
1. Let Z3 be the 3 -adic completion of the ring Z, so that Z ↪→ Z3 naturally. Evaluate1 + 3 + 32 + 33 + · · · in Z3.
证明.1 + 3 + 32 + 33 + · · · = 1
1− 3= −1
2
2. Let R be a Noetherian local ring. Suppose that there exists a principal prime ideal p inR such that ht (p) ≥ 1. Prove that R is an integral domain.(Hint: Below is one possible approach. Suppose p = (x) for some x ∈ R. Let q ⊂ p bea minimal prime in R. Argue that (i)x /∈ q, (ii) q = xq, and finally (iii) q = {0}.
证明. (i) Otherwise,x ∈ q ⇒ q = p ⇒ ht(p) = 0
(ii) xq ⊆ q is a prime ideal because
xf /∈ xq, xg /∈ xq =⇒ xfxg /∈ xq
Just verify q ⊆ xq is easier and quicker.
(iii) Nakayama lemma applied to (ii).
3. Let k be a field and R = k[[X]] × k[[X]]. Prove that R is a Noetherian semi-local ring,R contains a principal prime ideal of height 1, but R is not an integral domain.(Hint: It is known that k[[X]] is Noetherian local with maximal ideal (X). Argue thatthe ideals in k[[X]] × k[[X]] take the form I × J where I, J are ideals in k[[X]]. Showthat (X)× k[[X]] and k[[X]]× (X) are the only maximal ideals, and both are of height1. )
证明. (a) R is not an integral domain.
(b) k[[X]] is a Dedekind domain, thus Noetherian local with maximal ideal (X).
(c) We know that I = π1(I) × π2(I), where π1(I), π2(I) ▹ k[[X]] and (π1(a), π2(b)) =
(1, 0)a+ (0, 1)b.Obviously(by contradiction) the maximal ideals are (X)×k[[X]] and k[[X]]× (X);
参考文献 29
thus R is semilocal; moreover,(X)× k[[X]] = ((X, 1)) is principal.the prime ideals are