1 Algebraic Number Theory Vorlesung 2016 · Reiner, Maximal Orders Stichtenoth, Algebraic Function Fields (Seminar) Chapter 1 Commutative Theory. All rings are associative and have

1

Algebraic Number TheoryVorlesung 2016

Prof. Dr. G. Nebe, Lehrstuhl D fur Mathematik, RWTH Aachen

Contents

1 Commutative Theory. 41.1 The ring of integers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.1.1 The integral closure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.1.2 Norm, Trace and Discriminant. . . . . . . . . . . . . . . . . . . . . . . 6

An algorithm to determine an integral basis of a number field. . . . . . 91.1.3 Dedekind domains. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.2 Geometry of numbers. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.3 Finiteness of the ideal class group. . . . . . . . . . . . . . . . . . . . . . . . . . 171.4 Dirichlet’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191.5 Quadratic number fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

1.5.1 Imaginary quadratic number fields. . . . . . . . . . . . . . . . . . . . . 251.6 Ramification. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

1.6.1 How to compute inertia degree and ramification index ? . . . . . . . . . 291.6.2 Hilbert’s theory of ramification for Galois extensions. . . . . . . . . . . 29

1.7 Cyclotomic fields. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311.7.1 Quadratic Reciprocity. . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

1.8 Discrete valuation rings. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351.8.1 Completion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 371.8.2 Hensel’s Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 381.8.3 Extension of valuations. . . . . . . . . . . . . . . . . . . . . . . . . . . 40

1.9 p-adic number fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 421.9.1 Unramified extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

1.10 Different and discriminant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 471.10.1 Cyclotomic p-adic fields . . . . . . . . . . . . . . . . . . . . . . . . . . 491.10.2 An application to algebraic number fields. . . . . . . . . . . . . . . . . 50

2 Non-commutative theory. 522.1 Central simple algebras. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

2.1.1 Simple algebras. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 522.1.2 The theorem by Skolem and Noether. . . . . . . . . . . . . . . . . . . . 542.1.3 The Brauer group of K . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

2.2 Orders in separable algebras. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 562.2.1 Being a maximal order is a local property . . . . . . . . . . . . . . . . 57

2.3 Division algebras over complete discrete valuated fields. . . . . . . . . . . . . . 592.3.1 General properties. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

2

CONTENTS 3

2.3.2 Finite residue class fields. . . . . . . . . . . . . . . . . . . . . . . . . . 612.3.3 The central simple case: analysis . . . . . . . . . . . . . . . . . . . . . 622.3.4 The central simple case: synthesis . . . . . . . . . . . . . . . . . . . . . 642.3.5 The inverse different. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 652.3.6 Matrix rings. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

2.4 Crossed product algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 662.4.1 Factor systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 662.4.2 Crossed product algebras . . . . . . . . . . . . . . . . . . . . . . . . . . 672.4.3 Splitting fields. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 692.4.4 Field extensions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

Ground field extensions. . . . . . . . . . . . . . . . . . . . . . . . . . . 71Field extensions of L. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

2.4.5 A group isomorphism Br(K) ∼= Q/Z. . . . . . . . . . . . . . . . . . . . 732.5 Division algebras over global fields. . . . . . . . . . . . . . . . . . . . . . . . . 74

2.5.1 Surjectivity of the reduced norm . . . . . . . . . . . . . . . . . . . . . . 772.6 Maximal orders in separable algebras. . . . . . . . . . . . . . . . . . . . . . . . 78

2.6.1 The group of two-sided ideals. . . . . . . . . . . . . . . . . . . . . . . . 792.6.2 The Brandt groupoid. . . . . . . . . . . . . . . . . . . . . . . . . . . . 812.6.3 The finiteness of the class number. . . . . . . . . . . . . . . . . . . . . 822.6.4 The Eichler condition. . . . . . . . . . . . . . . . . . . . . . . . . . . . 842.6.5 Stable equivalence of ideals. . . . . . . . . . . . . . . . . . . . . . . . . 852.6.6 Algorithmic determination of classes and types . . . . . . . . . . . . . . 88

2.7 Automorphisms of algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 972.7.1 Skew Laurent series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 972.7.2 Automorphism groups of algebras . . . . . . . . . . . . . . . . . . . . . 982.7.3 The algebra σA . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 992.7.4 The finite dimensional and central simple case. . . . . . . . . . . . . . . 992.7.5 Generalized cyclic algebras. . . . . . . . . . . . . . . . . . . . . . . . . 992.7.6 Restriction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

2.8 The Brauer group of Q((t)). . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1012.8.1 Discrete valuated skew fields. . . . . . . . . . . . . . . . . . . . . . . . 1012.8.2 Skew Laurent series II . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

Subfields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1022.8.3 Non-crossed products over Q((t)) . . . . . . . . . . . . . . . . . . . . . 1032.8.4 An example where exponent 6= index . . . . . . . . . . . . . . . . . . . 104

3 Exercises. 107Literatur:

Neukirch, Algebraische ZahlentheorieReiner, Maximal OrdersStichtenoth, Algebraic Function Fields (Seminar)

Chapter 1

Commutative Theory.

All rings are associative and have a unit.

1.1 The ring of integers

1.1.1 The integral closure

Definition 1.1.1. An algebraic number field K is a finite extension of Q.

Example. K = Q[√

5] ∼= Q[x]/(x2 − 5).

Remark 1.1.2. Let L/K be a finite extension of fields and let a ∈ L. Then εa : K[x] →L, p(x) 7→ p(a) defines a K-algebra homomorphism with image K[a] (the minimal K-subalgebraof L that contains a). Since K[x] is a principal ideal domain, the kernel of εa is generatedby a monic polynomial Kern(εa) = (µa(x)). The image of εa is an integral domain, soµa(x) ∈ K[x] irreducible. This uniquely determined monic irreducible polynomial µa is calledthe minimal polynomial of a over K.

Example. a = 1+√

52∈ Q[√

5] ⇒ µa = x2 − x− 1 is the minimal polynomial of a over Q.

Definition 1.1.3. If B is a ring and A a subring of the center Z(B) := b ∈ B | bx =xb for all x ∈ B, then B is called an A-algebra.If B is an A-algebra then b ∈ B is called integral over A, if there is n ∈ N and a1, . . . , an ∈ Asuch that

(?) bn + a1bn−1 + . . .+ an−1b+ an = 0.

B is called integral over A, if any element of B is integral over A.

Theorem 1.1.4. Let B be an A-algebra and b ∈ B. The following are equivalent

(a) b is integral over A.

(b) The smallest A-subalgebra a A[b] of B, that contains b is a finitely generated A-module.

(c) b is contained in some A-subalgebra of B, that is a finitely generated A-module.

4

1.1. THE RING OF INTEGERS 5

Proof. (a) ⇒ (b): If b is integral, then (?) implies that A[b] = 〈1, b, . . . , bn−1〉A.(b) ⇒ (c): Clear.(c) ⇒ (a): Let R = 〈b1, . . . , bn〉A ≤ B be some A-subalgebra of B that contains b. Assumewlog that 1 ∈ R. Then there are (not necessarily unique) aij ∈ A such that

bbi =n∑j=1

aijbj for all 1 ≤ i, j ≤ n.

Let f = det(xIn − (aij)) ∈ A[x] be the characteristic polynomial of (aij) ∈ An×n. Thenf ∈ A[X] is monic and f((aij)) = 0 ∈ An×n. Therefore f(b)bi = 0 for all 1 ≤ i ≤ n, sof(b)1 = f(b) = 0, and hence b is integral over A.

Example.(a) α := 1+

√5

2∈ Q[√

5] is integral over Z.(b) 1

2∈ Q is not integral over Z.

Theorem 1.1.5. Let B be a commutative A-algebra and

IntA(B) := b ∈ B | b integral over A.

Then IntA(B) is a subring of B called the integral closure of A in B.

Proof. We need to show that IntA(B) is a ring, so closed under multiplication and addition.Let b1, b2 ∈ IntA(B) and

A[b1] = 〈c1, . . . , cn〉A, A[b2] = 〈d1, . . . , dm〉A.

Since cidj = djci for all i, j and 1 ∈ A[b1] ∩ A[b2] we get

A[b1, b2] ⊂ 〈cidj | 1 ≤ i ≤ n, 1 ≤ j ≤ m〉A.

This is a subring of B that is a finitely generated A-module and contains b1+b2, b1−b2, b1b2.

Theorem 1.1.6. Let C be a commutative ring, A ≤ B ≤ C. If C is integral over B and Bis integral over A, then C is integral over A.

Proof. Let c ∈ C. Since C is integral over B there are n ∈ N and b1, . . . , bn ∈ B such that

cn + b1cn−1 + . . .+ bn−1c+ bn = 0.

Put R := A[b1, . . . , bn]. Since B is integral over A this ring R is a finitely generated A-module.Moreover c ∈ R[c] and R[c] is a finitely generated R-module. So also R[c] is a finitely gener-ated A-module. and hence c is integral over A.

Definition 1.1.7. Let A be an integral domain with field of fraction K := Quot(A).

IntA(K) := x ∈ K | x is integral over A

is called the integral closure of A in K.If A = IntA(K), then A is called integrally closed.

6 CHAPTER 1. COMMUTATIVE THEORY.

Example. Z is integrally closed.Z[√

2] is integrally closed.Z[√

5] is not integrally closed.

Theorem 1.1.8. Let L ⊇ K be a finite field extension and A ⊂ K integrally closed withK = Quot(A). The for any b ∈ L:b is integral over A, if and only if µb,K ∈ A[x].

Proof. ⇐ clear.⇒: Let b ∈ L be integral over A. Then there are n ∈ N and a1, . . . , an ∈ A such that

bn + a1bn−1 + . . .+ an−1b+ an = 0.

Put p(x) = xn + a1xn−1 + . . .+ an−1x+ an ∈ A[x] and L := ZerfK(p) be the spitting field of

p, Then all zeros b ∈ L of p are integral over A. The minimal polynomial µb,K of b over Kdivides p, so also the zeros of µb,K are integral over A. The coefficients of µb,K are polynomialsin the zeros, so also integral over A. Since these lie in K, they indeed lie in IntA(K) = A. Soµb,K ∈ A[x].

Corollary 1.1.9. Let K be an algebraic number field. Then the ring of integers

ZK = IntK(Z) = a ∈ K | µa,Q ∈ Z[x].

Any Z-basis of ZK is called an integral basis of K.

Example. For K = Q[√

2] we obtain ZK = Z[√

2] and (1,√

2) is a Z-basis of K.If K = Q[

√5], then ZK = Z[(1 +

√5)/2] and (1, (1 +

√5)/2) is a Z-basis of K.

In the exercise you prove the more general statement: Let 1 6= d ∈ Z be square free and

K := Q[√d], then α := 1+

√d

2is integral over Z if and only if d ≡4 1. In this case (1, α) is an

integral basis of K, in all other cases (1,√d) is an integral basis.

1.1.2 Norm, Trace and Discriminant.

Remark 1.1.10. Let L/K be a extension of fields of finite degree [L : K] := dimK(L) = n <∞.

(a) Any α ∈ L induces a K-linear map

multα ∈ EndK(L);x 7→ αx.

In particular this endomorphism has a trace, determinant, characteristic polynomialχα,K := χmultα and minimal polynomial µα,K := µmultα.

(b) The map mult: L→ EndK(L) is an injective homomorphism of K-algebras.

(c) The map SL/K : L → K,α 7→ trace (multα) is a K-linear map, called the trace of Lover K.


(d) The map NL/K : L→ K,α 7→ det(multα) is multiplicative, i.e. NL/K(αβ) = NL/K(α)NL/K(β)for all α, β ∈ L. In particular it defines a group homomorphism NL/K : L∗ → K∗ be-tween the multiplicative groups L∗ and K∗ = (K \ 0, ·) of the fields.

(e) Let α ∈ L. Then µα,K ∈ K[X] is an irreducible polynomial of degree d := [K(α) :

K] := dimK(K(α)) dividing n and χα,K = µn/dα,K.

(f) If χα,K = Xn − a1Xn−1 + . . .+ (−1)n−1an−1X + (−1)nan ∈ K[X], then NL/K(α) = an

and SL/K(α) = a1.

Proof. Exercise.

Theorem 1.1.11. Assume that L/K is a finite separable extension and let σ1, . . . , σn : L→K be the distinct K-algebra homomorphisms of L into the algebraic closure K of K (son = [L : K]). Then for all α ∈ L

(a) χα,K =∏n

i=1(X − σi(α)).

(b) µα,K =∏d

i=1(X − αi) where σ1(α), . . . , σn(α) = α1, . . . , αd has order d = [K(α) :K].

(c) SL/K(α) =∑n

i=1 σi(α).

(d) NL/K(α) =∏n

i=1 σi(α).

Proof. (c) and (d) follow from (a) using Remark 1.1.10 (f) above.To see (b) let d := [K(α) : K]. Since L/K is separable, also the subfieldK(α) is separable overK, so µα,K =

∏di=1(X −αi) for d distinct αi ∈ K. The d distinct K-algebra homomorphisms

ϕ1, . . . , ϕd from K(α) into K correspond to the d possible images ϕi(α) = αi ∈ K of α.In particular this proves (a) and (b) if L = K(α).For the more general statement we use the following:Fact.1 Any K-algebra homomorphism τ : E → K of some algebraic extension E of K intothe algebraic closure K extends to an automorphism τ ∈ AutK(K).Let ϕj be such an extension of ϕj for all j = 1, . . . , d and let τ1, . . . , τn/d = HomK(α)(L,K).Then

σ1, . . . , σn = ϕj τi | 1 ≤ j ≤ d, 1 ≤ i ≤ n/d

In particular each ϕj can be extended in exactly n/d ways to a K-homomorphism ϕj τi :L→ K, 1 ≤ i ≤ n/d.

This implies that χα,K = µn/dα,K and also (a) and (b) follow.

Corollary 1.1.12. Let K ⊆ L ⊆M be a tower of separable field extensions of finite degree.Then

SM/K = SL/K SM/L and NM/K = NL/K NM/L

1(1.33) of the script of the Algebra lecture


Proof. Let m := [M : K], ` := [L : K] and n := [M : L]. Then m = `n. Define an equivalencerelation on σ1, . . . , σm = HomK(M,K) by

σj ∼ σi ⇔ (σj)L = (σi)L.

As we have seen in the last proof each equivalence class Aj contains exactly n elements.Therefore for any α ∈M

SM/K(α) =m∑i=1

σi(α) =∑j=1

∑σ∈Aj

σ(α).

Wlog we assume that Aj = [σj]. Then∑σ∈Aj

σ(α) = Sσj(M)/σj(L)(σj(α)) = σj(SM/L(α)).

Therefore SM/K(α) =∑`

j=1 σj(SM/L(α)) = (SL/K SM/L)(α). Similarly for the norm.

Definition 1.1.13. Let L/K be a separable extension and let B := (α1, . . . , αn) be a K-basisof L.

(a) The Trace-Bilinear-Form S : L × L → K, S(α, β) := SL/K(αβ) is a symmetricK-bilinear form.

(b) The discriminant of B is the determinant of the Gram matrix of B, d(B) := det(S(αi, αj)i,j).

Remark 1.1.14. If σ1, . . . , σn = HomK(L,K) then d(B) = det((σi(αj))i,j)2.

Proof. SL/K(αiαj) =∑n

k=1 σk(αi)σk(αj) = [(σk(αi)i,k)tr(σk(αi)i,k)]i,j so (SL/K(αiαj)) = AtrA

with A = (σk(αi)i,k).

Example. If K = Q and L = Q[√d] then B := (1,

√d) is a K-basis of L and d(B) =

2 · (2d) = det

(1√d

1 −√d

)2

Theorem 1.1.15. Let L/K be a separable extension an let B := (α1, . . . , αn) be a K-basis ofL. Then the trace bilinear form is a non-degenerate symmetric K-bilinear form. In particulard(B) 6= 0.

Proof. Choose a primitive element α ∈ L, so L = K(α) and B1 := (1, α, . . . , αn−1) is anotherK-basis of L. By the transformation rule for Gram matrices, d(B) = d(B1)a2 where a ∈ K∗is the determinant of the base change matrix between B and B1. So it is enough to showthat d(B1) 6= 0. By the remark above d(B1) = d(A)2 where

A = ((σi(αj))j=0,..,n−1,i=1,..,n =

1 σ1(α) σ1(α)2 . . . σ1(α)n−1

1 σ2(α) σ2(α)2 . . . σ2(α)n−1

...... . . . . . .

...1 σn(α) σn(α)2 . . . σn(α)n−1


and σ1, . . . , σn = HomK(L,K). By Vandermonde det(A) =∏

i<j(σj(α) − σi(α)), so

d(B1) = (∏

i<j(σj(α)−σi(α)))2 6= 0, since the different embeddings of L into K have differentvalues on the primitive element α.

Definition 1.1.16. Let K be an algebraic number field and B := (α1, . . . , αn) be an integralbasis of K (i.e. a Z-basis of the ring of integers ZK). Then the discriminant of K isdK := d(B).More general let A = 〈β1, . . . , βn〉Z be a free Z-module of full rank in K. Then

dA := d((β1, . . . , βn))

is called the discriminant of A.

Remark 1.1.17. dK and dA are well defined, which means that they do not dependent onthe choice of the integral basis B.If A′ ⊆ A ⊆ K are two finitely generated Z-modules of full rank in K, then by the maintheorem on finitely generated Z-modules (elementary divisor theorem) the index

a := [A : A′] := |A/A′| <∞

and dA′ = a2dA.

Example. K = Q[√d], 0, 1 6= d ∈ Z square-free. Integral basis, Gram matrix, discrimi-

nant.

An algorithm to determine an integral basis of a number field.

Definition 1.1.18. Let V ∼= Rn be an n-dimensional real vector space and Φ : V × V → Ra non-degenerate symmetric bilinear form.

(a) A lattice in V is the set of all integral linear combinations of an R-basis of V .

L = 〈B〉Z = n∑i=1

aibi | ai ∈ Z

for some basis B = (b1, . . . , bn) of V . Any such Z-basis B of L is called a basis of L andthe determinant of the Gram matrix of B with respect to Φ is called the determinantof L.

(b) For a lattice L := 〈B〉Z the set L# := x ∈ V | Φ(x, L) ⊆ Z is called the dual latticeof L (wrt Φ).

(c) L is called integral (wrt Φ), if L ⊆ L#.

Remark. L# is a lattice in V , the dual basis B∗ of any lattice basis B of L is a lattice basisof L#. The base change matrix between B and B∗ is the Gram matrix MB(Φ) = (Φ(bi, bj))of B. In particular det(MB(Φ)) = [L# : L] = |L#/L| for any integral lattice L.


Theorem 1.1.19. Let K be an algebraic number field, O ⊆ ZK a full Z-lattice in K. Then(O, SK/Q) is an integral lattice and

O ⊆︸︷︷︸f

ZK ⊆︸︷︷︸dK

Z#K ⊆︸︷︷︸

f

O#

which yields an algorithm to compute ZK.

Corollary. The ring of integers ZK in an algebraic number field is finitely generated, soany algebraic number field has an integral basis.

1.1.3 Dedekind domains.

Example. Let K = Q[√−5]. Then ZK = Z[

√−5] and

21 = 3 · 7 = (1 + 2√−5) · (1− 2

√−5)

has no unique factorization.Note that the factors above are irreducible but not prime.Reason: The ideals 3ZK = ℘3℘

′3, 7ZK = ℘7℘

′7, (1 + 2

√−5)ZK = ℘3℘7, and (1− 2

√−5)ZK =

℘′3℘′7 are not prime ideals, where

℘3 = (3, 1 + 2√−5), ℘′3 = (3, 1− 2

√−5), ℘7 = (7, 1 + 2

√−5), ℘′7 = (7, 1− 2

√−5)

and so 21ZK = ℘3℘′3℘7℘

′7 is a unique product of prime ideals.

A ring with a unique prime ideal factorisation is called a Dedekind ring:

Definition 1.1.20. A Noetherian, integrally closed, integral domain in which all non-zeroprime ideals are maximal ideals is called a Dedekind domain.

Example. Z[x] is not a Dedekind domain, because (x) is a prime ideal (the quotient isisomorphic to Z) but not maximal, since Z is not a field.

Theorem 1.1.21. Let K be a number field. Then ZK is a Dedekind domain.

Proof. Clearly ZK is integrally closed and an integral domain.We first show that ZK is Noetherian, i.e. any ideal of ZK is finitely generated. Let 0 6= AEZKbe an ideal and choose 0 6= a ∈ A. If B := (b1, . . . , bn) is an integral basis of K, thenaB := (ab1, . . . , abn) ∈ An is also a Q-basis of K. The lattice 〈aB〉Z ⊆ A ⊆ 〈B〉Z = ZKhas finite index in ZK . Therefore also A has finite index in ZK and, by the main theoremon finitely generated Z-modules, A is finitely generated as a Z-module and hence also as aZK-module.The above consideration also applies to non-zero prime ideals 0 6= ℘EZK of ZK , in particularany such prime ideal has finite index in ZK . Therefore ZK/℘ is a finite integral domain, soa field, which means that ℘ is a maximal ideal.

Lemma 1.1.22. Any finite integral domain R is a field.


Proof. Let 0 6= a ∈ R, then multa : R→ R is injective (the kernel is 0, since R is an integraldomain) and hence surjective (since R is finite). In particular there is some x ∈ R such thatmulta(x) = 1.

Definition 1.1.23. Let R be a commutative ring and A,B ER. Then

A+B := a+ b | a ∈ A, b ∈ BER, AB := n∑i=1

aibi | n ∈ N, ai ∈ A, bi ∈ BER.

If A ⊆ B we say that B divides A. The greatest common divisor

ggT(A,B) := (A,B) = A+B

is the ideal generated by A and B.

From now on let R be a Dedekind domain and K = Quot(R).

Main theorem 1.1.24. Any ideal 0 6= I E R in R has a unique factorization into primeideals,

I = ℘1 . . . ℘s, s ∈ N0, ℘i ER prime ideals .

For the proof we need two lemmata:

Lemma 1.1.25. If 0 6= I ER then there are non-zero prime ideals ℘1, . . . , ℘s ER such that℘1 . . . ℘s ⊆ I.

Proof. Let M := I ER | I 6= 0, and for all prime ideals ℘1, . . . , ℘s the product℘1 . . . ℘s is not contained in I

. We

need to show that M = ∅. Assume that M 6= ∅. Since any ascending chain of idealsin R is finite, the set M contains some maximal element A ∈ M. Then A is not a primeideal, hence there are b1, b2 ∈ R such that

b1b2 ∈ A, b1 6∈ A, b2 6∈ A.

Let Ai := (bi)+A. Then Ai ) A but A1A2 ⊂ A. Since A is maximal inM, both Ai containa product of prime ideals, hence also A1A2 and therefore A, a contradiction.

Lemma 1.1.26. Let 0 6= ℘ER be a prime ideal and put

℘−1 := x ∈ K | x℘ ⊆ R.

Then for any non zero ideal 0 6= AER the ideal A℘−1 properly contains A.

Proof. We first show that ℘−1 6= R: Choose some 0 6= a ∈ ℘ and let s ∈ N be minimal with theproperty that there are non-zero prime ideals ℘1, . . . , ℘s in R such that ℘1 . . . ℘s ⊆ (a) ⊆ ℘.(These exist since R is Noetherian.)Claim. There is some i such that ℘i ⊆ ℘.Otherwise there are ai ∈ ℘i \ ℘ for all i = 1, . . . , s, but a1 . . . as ∈ ℘1 . . . ℘s ⊆ ℘ which


contradicts the fact that ℘ is a prime ideal.Assume wlog that ℘1 ⊆ ℘. Since R is a Dedekind domain, the non-zero prime ideal ℘1 ismaximal. Therefore ℘ = ℘1.By the minimality of s we have that ℘2 . . . ℘s 6⊆ (a) so there is some b ∈ ℘2 . . . ℘s such thata−1b 6∈ R. On the other hand

a−1b℘ = a−1b℘1 ⊆ a−1℘1 . . . ℘s ⊆ a−1(a) = R

so a−1b ∈ ℘−1 \R.Now choose some nonzero ideal A E R and assume that A℘−1 = A. Let A = 〈α1, . . . , αn〉R(observe that A is finitely generated, since R is Noetherian). Then for any x ∈ ℘−1 andany i we have xαi =

∑nj=1 xijαj for some matrix (xij) =: X ∈ Rn×n. Therefore the vector

(α1, . . . , αn)tr is in the kernel of (xIn − X) ∈ Kn×n, so the determinant of this matrix is 0.But then x is a zero of some monic polynomial with coefficients in R, so x ∈ IntK(R) = R,since R is integrally closed. This holds for any x ∈ ℘−1 contradicting the fact that ℘−1 6⊆ R.

Corollary 1.1.27. For any non-zero prime ideal 0 6= ℘ER the product ℘℘−1 = R.

Proof. ℘ ( ℘℘−1 ⊆ R. Since R is a Dedekind domain, ℘ is a maximal ideal, so ℘℘−1 = R.

Proof of the main Theorem 1.1.24Existence. LetM := AER | 0 6= A 6= R,A 6= ℘1 . . . ℘s for all prime ideals ℘1, . . . , ℘s and all s ∈N. We need to show that M = ∅. If M 6= ∅, then M contains some maximal element, sayA. Since maximal ideals are prime ideals, the ideal A is not a maximal ideal. There is somemaximal ideal ℘ E R that contains A, so A ⊆ ℘ ⊆ R and hence A ( A℘−1 ⊆ ℘℘−1 = R.Now A 6= ℘ was maximal in M, so there are prime-ideals ℘1, . . . , ℘s such that

A℘−1 = ℘1 . . . ℘s ⇒ A = ℘1 . . . ℘s℘

a contradiction.Uniqueness. (this is analogues to the proof of uniqueness of prime factorization in Z) Wehave seen in the proof of Lemma 1.1.26 that if a prime ideal ℘ divides the product of twoideals, then it divides one of the factors

I1I2 ⊆ ℘⇒ I1 ⊆ ℘ or I2 ⊆ ℘.

So assume thatA = ℘1 . . . ℘s = Q1 . . .Qt

then ℘1 divides Q1 . . .Qt so it divides one of the factors, say Q1. Since Q1 is maximal, thisimplies Q1 = ℘1, so

℘−1A = ℘2 . . . ℘s = Q2 . . .Qt

Definition 1.1.28. A fractional ideal of R is a finitely generated R-submodule 6= 0 of K.

Remark 1.1.29. Let J be a fractional ideal of R. Then there is c ∈ K, A E R, such thatcA = J .

1.2. GEOMETRY OF NUMBERS. 13

Proof. Let J = 〈α1, . . . , αn〉R, αi = βiγi∈ K wit βi, γi ∈ R. Let γ := γ1 . . . γn. Then

A := γJ ER and J = γ−1A.

Theorem 1.1.30. The set of fractional ideal of R is an abelian group, the ideal group ofR.

Proof. The group law is of course ideal multiplication, this is associative, commutative, theunit is (1) = R and the inverse is A−1 = x ∈ K | xI ⊆ R.

Corollary 1.1.31. Any fractional ideal A of R has a unique factorization

A = ℘n11 . . . ℘nss

with non-zero prime ideals ℘1, . . . , ℘s and ni ∈ Z.

Definition 1.1.32. The ideal group of R is denoted by JR. It contains the subgroup(c) | c ∈ K∗ = PR of principal fractional ideals. The quotient ClK := JR/PR is calledthe class group of K.

There is an exact sequence

1→ R∗ϕ1→ K∗

ϕ2→ JRϕ3→ ClK → 1

where ϕ1 is just the inclusion, ϕ2(c) = (c), and ϕ3 is the natural epimorphism. This meansthat ϕ1 is injective, im(ϕ1) = ker(ϕ2), im(ϕ2) = PR = ker(ϕ3), and ϕ3 is surjective.

If R = ZK is the ring of integers in an algebraic number field K, then

• Z∗K is a finitely generated abelian group

• ClK is a finite group, hK := |ClK | is called the class number of K

1.2 Geometry of numbers.

Definition 1.2.1. Let (Rn, (, )) be a Euclidean space. Any Z-module generated by a basisof Rn is called a full lattice in (Rn, (, )). Let Γ := 〈b1, . . . , bn〉Z be a full lattice. ThenB = (b1, . . . , bn) is called a basis of Γ and

E(B) := n∑i=1

λibi | 0 ≤ λi ≤ 1

the fundamental parallelotope of B. The determinant of Γ is det(Γ) := det((bi, bj))and the covolume of Γ is

covol(Γ) := vol(Rn/Γ) := vol(E(B)) =√

det(Γ).

Example. Z2: Different bases yield different E(B) but these have the same covolume.


Remark 1.2.2. E(B) is a fundamental domain for the action of Γ on Rn by translation.this means that

Rn =⋃γ∈Γ

γ + E(B)

and this union is almost disjoint, Γ-translates of E(B) are either equal or intersect only inthe boundary.

Definition 1.2.3. Let ∅ 6= X ⊂ Rn.(a) X is called centrally symmetric, if for any x ∈ X also its negative −x ∈ X.(b) X is called convex, if for any two x, y ∈ X and any t ∈ [0, 1] also x+ t(y − x) ∈ X.

Clear: ∅ 6= X convex and centrally symmetric, then 0 ∈ X.

Theorem 1.2.4. (Minkowski) Let Γ ⊂ (Rn, (, )) be a full lattice in Euclidean space and letX ⊆ Rn be convex and centrally symmetric. If vol(X) > 2n vol(Rn/Γ) then Γ ∩X 6= 0.

Proof. We show that there are γ1 6= γ2 ∈ Γ such that

(1

2X + γ1) ∩ (

1

2X + γ2) 6= ∅

Then there are x1, x2 ∈ X such that 12x1 + γ1 = 1

2x2 + γ2 and hence

1

2(x1 − x2) = γ2 − γ1 ∈ Γ ∩X

is a nonzero vector. Note that 12(x1−x2) is the midpoint of the line between x1 and −x2 and

therefore in X.So assume that the Γ -translates of the set 1

2X = 1

2x | x ∈ X are disjoint,

(1

2X + γ1) ∩ (

1

2X + γ2) = ∅ for all γ1 6= γ2 ∈ Γ

But then also the intersection with the fundamental parallelotope

(E(B) ∩ (1

2X + γ1)) ∩ (E(B) ∩ (

1

2X + γ2)) = ∅ for all γ1 6= γ2 ∈ Γ so

vol(Rn/Γ) = vol(E(B)) ≥∑

γ∈Γ vol(E(B) ∩ (12X + γ)) =∑

γ∈Γ vol((E(B)− γ) ∩ 12X) = vol(1

2X) = 1

2nvol(X)

which contradicts the assumption.

Example. The bound is tight: Take Γ = Z2 and

X := (x1

x2

)∈ R2 | |x1| < 1 and |x2| < 1.

Then vol(X) = vol(X) = 22, covol(Γ) = 1 and X ∩ Γ = 0.We now apply this to number fields K. For this aim we need to embed K into some

euclidean space.

1.2. GEOMETRY OF NUMBERS. 15

Remark 1.2.5. Let K be an algebraic number field of degree [K : Q] =: n. Let

σ1, . . . , σn : K → Q ⊂ C

be the n distinct embeddings of K into the algebraic closure Q of Q which we embed into thefield of complex numbers. This yields an embedding

j : K → KC :=n∏k=1

C = Cσ1,...,σn, x 7→ (σ1(x), . . . , σn(x)) = (xσ1 , . . . , xσn).

The Galois group of C over R Gal(C/R) = 〈〉 ∼= C2 acts on KC via

(xσ1 , . . . , xσn) = (yσ1 , . . . , yσn) with yσj = xσj .

Here σj : K → C, σj(x) := σj(x). We call σ : K → C real, if σ = σ and complex if σ 6= σ.Let

KR := Fix〈〉(KC) := (xσ) ∈ KC | xσ = xσ.Then j(K) ⊂ KR.

Example. K ∼= Q[X]/(X3− 2) = Q[ 3√

2]. Let α ∈ K with α3 = 2. Then α is a primitiveelement of K and the embeddings of K into C are given by

σ1 : α 7→ 3√

2(∈ R), σ2 : α 7→ ζ33√

2, σ3 = σ2 : α 7→ ζ23

3√

2.

Then σ1 is real, σ2 and σ3 are complex and the action of the complex conjugation on KC is

(x, y, z) = (x, z, y).

Therefore we obtain KR = (a, b+ ic, b− ic) | a, b, c ∈ R.

Remark 1.2.6. The mappings

N : KC → C, N(x1, . . . , xn) =∏n

i=1 xiS : KC → C, S(x1, . . . , xn) =

∑ni=1 xi

extend norm and trace, in the sense that for any α ∈ K

NK/Q(α) =n∏i=1

σi(α) = N(j(α)), SK/Q(α) =n∑i=1

σi(α) = S(j(α)).

Remark 1.2.7. Let ρ1, . . . , ρr : K → R ⊂ C be the real places of K and σ1, σ1, . . . , σs, σs :K → C the complex places of K, so n = [K : Q] = r + 2s. Then

m : KR → Rr+2s, (xρ1 , . . . , xρr , xσ1 , xσ1 , . . . , xσs , xσs) 7→ (xρ1 , . . . , xρr ,<(xσ1),=(xσ1), . . . ,<(xσs),=(xσs))

is a R-linear isomorphism that maps the restriction of the standard inner product 〈x, y〉 :=∑ni=1 xiyi on KC to the canonical metric (Minkowski metric)

(x, y) :=r∑i=1

xiyi + 2r+2s∑j=r+1

xjyj.


Proof. Wlog r = 0, s = 1, so KR = (x, x) | x ∈ C. Then

〈(x, x), (y, y)〉 = xy + xy = 2(<(x)<(y) + =(x)=(y)).

In the following we will treat all lattices in KR as lattices in (Rr+2s, (, )) with respect tothe positive definite Minkowski metric.

Theorem 1.2.8. If 0 6= AEZK is an ideal in ZK then Γ := j(A) is a full lattice in KR withcovolume

covol(Γ) =√|dK ||ZK/A|.

In particular det(j(ZK)) = |dK | is the absolute value of the discriminant of K.

Proof. Let B = (α1, . . . , αn) be an integral basis of A. and let A := (σi(αj))ni,j=1 ∈ Cn×n.

Then the Gram matrix of B with respect to the trace bilinear form S is

MB(S) = AtrA.

So dA = det(MB(S)) = det(A)2 = [ZK : A]2dK . On the other hand

(〈j(αi), j(αk)〉)ni,k=1 = (n∑`=1

σ`(αi)σ`(αk))ni,k=1 = A

trA

and therefore vol(KR/Γ) =

√det(A

trA) = | det(A)| =

√|dK |[ZK : A].

Definition 1.2.9. For any nonzero integral ideal 0 6= A E ZK we define the norm of A tobe N(A) := [ZK : A].

Clearly for a ∈ ZK this is the usual norm NK/Q(a) = N((a)).

Remark 1.2.10. For any two nonzero integral ideals A,B we have

N(AB) = N(A)N(B)

so N defines a group homomorphism

N : JK → R>0, N(℘n11 · · ·℘nss ) := N(℘1)n1 · · ·N(℘s)

ns .

Proof. Since A,B have a factorisation into prime ideals it is enough to show the multiplica-tivity in the following two cases(a) gcd(A,B) = 1: But then AB = A ∩ B and by Chinese Remainder Theorem ZK/AB ∼=ZK/A× ZK/B has order

N(AB) = |ZK/AB| = |ZK/A||ZK/B| = N(A)N(B).

(b) powers of prime ideals N(℘n) = N(℘)n. For any prime ideal 0 6= ℘ E ZK , the ideals ofZK/℘n are precisely ℘i/℘n with 0 ≤ i ≤ n. This yields a composition series

ZK ⊇ ℘ ⊇ ℘2 ⊇ . . . ⊇ ℘n−1 ⊇ ℘n

where all composition factors ℘i/℘i+1 are isomorphic to ZK/℘. More precisely for anyp ∈ ℘ \ ℘2 multiplication by p yields an isomorphism between ZK/℘ and ℘/℘2, etc. So|ZK/℘| = |℘/℘2| = . . . = |℘n−1/℘n| = N(℘) and |ZK/℘n| =

∏ni=1 |℘i−1/℘i| = N(℘)n.

1.3. FINITENESS OF THE IDEAL CLASS GROUP. 17

1.3 Finiteness of the ideal class group.

Remark 1.3.1. For any n ∈ N there are only finitely many integral ZK-ideals I E ZK withnorm N(I) ≤ n. Here a fractional ZK-ideal is called integral, if it is contained in ZK, henceif it is an ideal in the usual sense.

Proof. Let I E ZK be an ideal with norm N(I) = |ZK/I| = n. Then nZK ⊆ I ⊆ ZK andI/nZK is one of the finitely many subgroups of the finite abelian group ZK/nZK ∼= Z/nZ[K:Q].

General assumption:K is a number field of degree [K : Q] = r + 2s = n,

σ1, . . . , σr : K → R ⊂ C, σr+1, . . . σr+s, σr+s+1 = σr+1, . . . , σr+2s = σr+s : K → C

the real resp. complex embeddings of K into C. These are also called the places of K.

Theorem 1.3.2. Let 0 6= AE ZK be an ideal. For any i ∈ 1, . . . , r + s let ci = cσi ∈ R>0

such that cr+i = cr+s+i for all 1 ≤ i ≤ s (cσi = cσi) and

r+2s∏i=1

ci >

(2

π

)s√|dK |N(A).

Then there is some 0 6= a ∈ A such that |σi(a)| < cσi for all 1 ≤ i ≤ n. In particular anyintegral ideal contains an element 0 6= a ∈ A, such that |NK/Q(a)| ≤

(2π

)s√|dK |N(A).

Proof. Let X := (x1, . . . , xn) ∈ KR | |xi| ≤ ci for all 1 ≤ i ≤ n. Then X and its imagem(X) is convex and centrally symmetric, where m : KR → Rr+2s,

(x1, . . . , xr, xr+1, . . . , xr+s, xr+s+1, . . . , xr+2s︸︷︷︸=xr+1,...,xr+s

) 7→ (x1, . . . , xr,<(xr+1),=(xr+1), . . . ,<(xr+s),=(xr+s))

and Rr+2s is endowed with the positive definite bilinear form (x, y) :=∑r

i=1 xiyi+2∑2s

j=1 xr+jyr+j.With respect to this metric, the volume of m(X) is

vol(m(X)) = vol(x1, . . . , xn) ∈ Rr+2s | |xi| ≤ ci, x2r+2j−1+x2

r+2j ≤ c2r+j for all 1 ≤ i ≤ r, 1 ≤ j ≤ s =

(∏r

i=1 2ci)∏s

j=1 2πc2r+j = 2r+sπs

∏ni=1 ci > 2r+sπs

(2π

)s√|dK |N(A) = 2r+2s vol(Rn/Γ) whereΓ = j(A). By Minkowski’s lattice point theorem there is some non-zero element in m(X) ∩m(j(A)) = m(X ∩ j(A)).

Theorem 1.3.3. Recall that the class group of K is ClK := JK/PK is the group of equivalenceclasses of fractional ZK-ideals in K, where two ideals A and B are called equivalent, if theydiffer by a principal ideal, so if there is 0 6= x ∈ K such that (x)A = B.

(a) Any ideal class [A] ∈ ClK contains an integral ideal A1 ∈ [A], A1 E ZK such that

N(A1) ≤MK :=

(2

π

)s√|dK |.


(b) The class number of K, hK := |ClK | is finite.

Proof. (a) implies (b) since there are only finitely many integral ideals of norm ≤MK .To see (a), let AE ZK an integral representative of the ideal class. By Theorem 1.3.2 thereis some 0 6= a ∈ A, such that |N(a)| ≤MKN(A). Let A1 := (a)A−1. Then A1 is integral, inthe class of A and N(A1) = |N(a)|N(A)−1 ≤MK .

Example: K = Q[√

5], dK = 5, r = 2, s = 0, so MK =√

5 < 3 and any ideal classcontains some integral ideal of norm 1 or 2.

Norm 1 Then the ideal is (1) = ZK and therefore principal.

Norm 2 If N(I) = 2, IEZK , then 2ZK ⊆ I ⊆ ZK . The ring ZK/2ZK ∼= F2[x]/(x2 +x−1) ∼= F4

has no nontrivial ideals, so there are no ideals of norm 2 (note that N(2ZK) = 4).

So we have seen that ZK = Z[1+√

52

] is a principal ideal domain.

Example: K = Q[√

15], dK = 60, r = 2, s = 0, so MK = 2√

15 < 8 and we have toconsider all integral ideals of norm 2,3,4,5,7.

Norm 2 ℘2 = (2, 1 +√

15) is the unique ideal of norm 2. (ZK/2ZK ∼= F2[X]/(X2 − 15) ∼=F2[X]/(X+1)2 has a unique non-trivial ideal). ℘2 is not a principal ideal since otherwiseZ[√

15] contains an element a = x + y√

5 of norm N(a) = x2 − 15y2 = ±2. Thenx2 ≡5 ±2 which is a contradiction.

Norm 3 ℘3 = (3,√

15) but ℘2℘3 = (3 +√

15) is a principal ideal.

Norm 4 2ZK = ℘22.

Norm 5 ℘5 = (5,√

15) but ℘3℘5 = (√

15) is a principal ideal.

Norm 7 ℘7 = (7, 1 +√

15), ℘′7 = (7, 1 −√

15). These ideals satisfy ℘7℘2 = (1 +√

15) and℘′7℘2 = (1−

√15).

So in total ClK = 〈[℘2]〉 ∼= C2.

Remark 1.3.4. Since any ideal is a product of prime ideals, the class group is generated bythe classes of prime ideals ℘iEZK such that N(℘i) ≤MK. Note that the norm of the primeideal ℘ is a power of the prime p with pZ = ℘ ∩ Z.

Remark 1.3.5. What is known about class numbers? Not much.If K = Q[

√d] (d < 0, d ∈ Z square free) is an imaginary quadratic number field then hK = 1

if and only if d ∈ −1,−2,−3,−7,−11,−19,−43,−67,−163.One conjectures that there are infinitely many real quadratic number fields K (so r = 2, s = 0)for which hK = 1, but one cannot even prove that there are infinitely many number fields(without restricting the degree) with class number 1.

1.4. DIRICHLET’S THEOREM 19

1.4 Dirichlet’s theorem

We start with some preliminary technical remarks on lattices. Let V = (Rn, (, )) alwaysdenote the Euclidean space of dimension n.

Lemma 1.4.1. A subgroup Γ ≤ V is a lattice (i.e. there are R-linear independent elements(v1, . . . , vm) ∈ V m such that Γ = 〈v1, . . . , vm〉Z) if and only if Γ is discrete, which means thatfor all γ ∈ Γ there is some ε > 0 such that Bε(γ) ∩ Γ = γ.

Proof. Let V0 := 〈Γ〉R and B := (γ1, . . . , γm) ∈ Γm a basis of V0. Put Γ0 := 〈γ1, . . . , γm〉Z.Then Γ0 is a lattice. We prove that Γ/Γ0 is finite, because then Γ is finitely generated andby the main theorem on f.g. abelian groups it is free of the same rank as Γ0.Let E(B) be the fundamental parallelotope defined by B, then vol(E(B)) is finite andV0 = ∪γ∈Γ0E(B) + γ. Since E(B) is compact and Γ is discrete, there are only finitelymany points in E(B)∩Γ = x1, . . . , xa. But then Γ = ∪ai=1xi + Γ0 and hence |Γ/Γ0| ≤ a.

Lemma 1.4.2. Let Γ ≤ V be a lattice. Then Γ is a full lattice (i.e. contains a basis of V ),if and only if Γ has finite covolume in V , if and only if there is some bounded set M ⊂ Vsuch that V = ∪γ∈ΓM + γ.

Proof. If Γ is a full lattice, and B a lattice basis of Γ, then M := E(B) is such a boundedset.On the other hand assume that Γ has not full rank in V and choose some v ∈ V \ 〈Γ〉R. IfV = ∪γ∈ΓM + γ for some bounded set M , then for any n ∈ N there is some an ∈ M suchthat nv = an + γn for some γn ∈ Γ. Since M is bounded, limn→∞

1nan = 0, so

v =1

n(an + γn) = lim

n→∞

1

nan + lim

n→∞

1

nγn = lim

n→∞

1

nγn ∈ 〈Γ〉R

because subspaces are closed.

We now want to apply these basic facts on lattices to study the unit group Z∗K of the ringof integers in some algebraic number field.

Recall that the places σ1, . . . , σr+2s of K define an embedding

j : K → KR = (x1, . . . , xr, y1, . . . , ys, y1, . . . , ys) | xi ∈ R, yi ∈ C

and that we identified KR via the mapping m with Rr+2s where

m : KR → Rr+2s, (x1, . . . , xr, y1, . . . , ys, y1, . . . , ys) 7→ (x1, . . . , xr,<(y1),=(y1), . . . ,<(ys),=(ys)).

Note that j is a ring homomorphism so it defines a group homomorphism j : K∗ → K∗R.Define the logarithm

` : K∗R → Rr+s, `(x1, . . . , xr, y1, . . . , ys, y1, . . . , ys) := (log(|x1|), . . . , log(|xr|), log(|y1|2), . . . , log(|ys|2)).

Then ` is again a group homomorphism from the multiplicative group K∗R to the additivegroup of the vector space Rr+s.


Theorem 1.4.3. Let λ := ` j : Z∗K → Rr+s. Then λ is a group homomorphism with

ker(λ) = µK = z ∈ K | za = 1 for some a ∈ N

the group of roots of unity in K. Let Γ := λ(Z∗K) ≤ Rr+s.

Proof. It is clear that λ is a group homomorphism. The image of λ is a subgroup of theadditive group of a vector space, hence torsion free, so all elements of Z∗K that have finiteorder lie in the kernel of λ and therefore µK ⊆ ker(λ). To see equality let x ∈ Z∗K be suchthat λ(x) = 0. Then

j(x) ∈ X := (x1, . . . , xr, y1, . . . , ys) ∈ KR | |xi| = 1, |yi|2 = 1.

So j(ker(λ)) is contained in a bounded subset of KR. On the other hand j(x) ∈ j(ZK) =: Λis contained in the lattice j(ZK) = 〈j(b1), . . . , j(bn)〉Z for any integral basis (b1, . . . , bn) of K.But Λ∩X is always finite, so ker(λ) is finite and hence a torsion group, so contained in µK .

Remark 1.4.4. Since the norm is multiplicative Z∗K = x ∈ ZK | NK/Q(x) = ±1. Notethat if x ∈ ZK satisfies NK/Q(x) = 1 then x−1 ∈ Z[x] can be obtained from the minimalpolynomial of x.Let UK := x ∈ K | N(x) = ±1.Then λ(Z∗K) ⊆ λ(UK) = H := (a1, . . . , ar+s) ∈ Rr+s |

∑r+si=1 ai = 0 ∼= Rr+s−1.

Theorem 1.4.5. Let Γ := λ(Z∗K) ≤ Rr+s. Then Γ ≤ H := (a1, . . . , ar+s) ∈ Rr+s |∑r+si=1 ai = 0 ∼= Rr+s−1 is a full lattice in H.

Proof. We have to show that Γ is a full lattice in H. It is clear that Γ ≤ H is a subgroup.We first show that Γ is discrete. To this aim we show that for any c > 0 the set

Xc := (am) ∈ Rr+s | |am| < c for all m

meets Γ in only finitely many points. But

`−1(Xc) = (x1, . . . , xr, y1, . . . , ys, y1, . . . , ys) ∈ KR | e−c ≤ |xi| ≤ ec, e−c ≤ |yi|2 ≤ ec

is bounded and therefore contains only finitely many points of the lattice Λ = j(ZK) ⊂ j(Z∗K).Therefore also |Γ ∩Xc| <∞.We now show that Γ has finite covolume in H: Choose c1, . . . , cr, d1, . . . , ds ∈ R>0 such that

r∏i=1

ci

s∏j=1

d2j =: C > MK .

Let X := (x1, . . . , xr, y1, . . . , ys, y1, . . . , ys) ∈ KR | |xi| < ci, |yj|2 < dj. Then X ⊂ KR is abounded set.

Since there are only finitely many ideals of a given norm in ZK there are α1, . . . , αN ∈ZK \ 0 such that for any element α ∈ ZK with |N(α)| ≤ C there is some unit u ∈ Z∗K andsome 1 ≤ i ≤ N such that α = uαi.

1.4. DIRICHLET’S THEOREM 21

Let U := y ∈ K∗R | N(y) = ±1 ≤ K∗R. Then `(U) = H and U is the full preimage of Hunder `. Put

T := U ∩N⋃i=1

Xj(α−1i ).

We then claim that U = ∪ε∈Z∗KTj(ε).Let y ∈ U . Then Xy−1 = x ∈ KR | |xi| ≤ c′i where c′i = ci|yi|−1. Since

∏i |yi| = N(y) = 1

also∏

i c′i =

∏i ci = C. By Minkowski’s theorem there is some 0 6= a ∈ ZK such that

j(a) ∈ Xy−1, so j(a) = xy−1 for some x ∈ X. This means that |NK/Q(a)| < C so there issome u ∈ Z∗K and some i ∈ 1, . . . , N such that a = uαi. Then

y = xj(a)−1 = xj(αi)−1j(u)−1 ∈ Tj(u−1).

Corollary 1.4.6. Let t := r + s− 1. Then there are ε1, . . . , εt ∈ Z∗K and µ ∈ µK such that

Z∗K = 〈µ〉 × 〈ε1, . . . , εt〉 ∼= C|µK | × Zr+s−1.

The εi are called fundamental units of K.

Example. K = Z[√

5], ZK = Z[1+√

52

] then Z∗K = 〈−1〉 × 〈1+√

52〉.

Definition 1.4.7. A subset Γ ⊂ K is called an lattice in K, if there is some Q-basis B ofK such that Γ = 〈B〉Z.A subset O ⊂ K is called an order in K, if O is a subring of K that is a lattice.

Example. If Γ ⊂ K is a lattice then

O(Γ) := x ∈ K | xΓ ⊆ Γ

is an order in K.

Clearly any order O is consists of integral elements and hence is contained in the uniquemaximal order ZK of K. Since O also contains a basis of K, the index |ZK/O| is finite.Moreover O∗ = x ∈ O | N(x) = ±1.Theorem 1.4.8. If O is an order in K, then O∗ ≤ Z∗K is a subgroup of finite index.

Proof. The same proof as above proves that also O∗ has t = r + s− 1 fundamental units.

The Regulator

Definition 1.4.9. Let K be a number field and σi (1 ≤ i ≤ r+ s) a complete set of pairwisenon- conjugate embeddings of K in C. Then the regulator of a set ε1, ε2, ..., εr+s−1 ofr + s− 1 elements in K∗ of norm ±1 is defined as

Reg(ε1, ε2, ..., εr+s−1) = det(ni log |σi(εj)|)r+s−1i,j=1 .

Here the integer ni ∈ 1, 2 equals 1 if σi is a real embedding and 2 otherwise. The regulatorReg(R) of an order R in K is the regulator of a system of fundamental units for R∗. Weput Reg(R) = 1 if R∗ is finite, i.e., if R is either Z or an imaginary quadratic order. Theregulator of K is R(K) =Reg(ZK).


By the Dirichlet unit theorem, regulators of orders do not vanish. Unlike the discriminantof the order, which is an integer, the regulator of an order is a positive real number that isusually transcendental as it is an expression in terms of logarithms of algebraic numbers.

A few formulas relating regulator R(K), class number h(K), number of roots of unity|µK | =: ω(K) and discriminant |d(K)|. We keep the notation [K : Q] = n = r + 2s.

Theorem 1.4.10. Let ζK(z) :=∑

A1

N(A)zdenote the Dedekind zeta function of K, where

the sum is over all non-zero integral ideals of ZK. Then ζK has an analytic continuation toC with a simple pole at z = 1.

(a) limz→0 z−(r+s−1)ζK(z) == h(K)R(K)ω(K)−1.

(b) limz→1(z − 1)ζK(z) = 2r(2π)s h(K)R(K)

ω(K)√|d(K)|

.

(c) limx→∞NK(x)x

= 2r(2π)s h(K)R(K)

ω(K)√|d(K)|

where NK(x) denotes the number of integral ideals

of ZK of norm ≤ x.

1.5 Quadratic number fields

Let K = Q[√d], d ∈ Z, d 6= 1, 0 square-free be a quadratic number-field (i.e. an extension of

Q of degree 2). Then ZK = Z[ω] with ω :=

√d d ≡4 2, 3

1+√d

2d ≡4 1

. Note that dK = d if d ≡4 1

and dK = 4d otherwise, in particular dK is either 0 or 1 modulo 4.

Theorem 1.5.1. Let Γ be a full lattice in K.(a) There is some m ∈ Q and γ ∈ K such that Γ = 〈m,mγ〉Z.(b) Let a, b, c ∈ Z, gcd(a, b, c) = 1, a > 0, such that aγ2 +bγ+c = 0. Then aγ = h+kω ∈ ZKand

O(Γ) := x ∈ K | xΓ ⊆ Γ = 〈1, aγ〉Z = 〈1, kω〉Z.

Proof. (a) Is just the Hermite normal form for integral matrices: If Γ = 〈α, β〉Z, then thereare x, y ∈ Q such that 1 = xα+ yβ. Choose m ∈ Q such that u := mx and v := my both liein Z and gcd(u, v) = 1. Then there are r, s ∈ Z such that 1 = us− rv. Put

γ :=rα + sβ

m, then Γ = 〈m,mγ〉Z.

(b) Clearly O(〈m,mγ〉Z) = O(〈1, 1γ〉Z), so wlog assume that m = 1. Then O(Γ) contains aγ,since both, aγ and aγ2 = −bγ − c lie in Γ. On the other hand let x+ yγ =: δ ∈ O(Γ). Thenx+ yγ ∈ Γ, so x, y ∈ Z and yγ ∈ O(Γ), so yγ2 ∈ Γ implying that y is divisible by a.

Corollary 1.5.2. Let O be an order in K. Then O = Of := 〈1, fω〉 for some f ∈ N. Thisnumber f is called the conductor (Fuhrer) of O.We have fZK ⊂ Of ⊂ ZK and d(Of ) = f 2dK.

1.5. QUADRATIC NUMBER FIELDS 23

Remark 1.5.3. Let 〈σ〉 = Gal(K/Q) (so σ(√d) = −

√d). Then for all a ∈ K we have

σ(a) = SK/Q(a)− a and in particular any order O in K satisfies σ(O) = O.

Definition 1.5.4. Let O ⊂ K be an order. Then

M(O) := Γ ⊆ K | Γ is a lattice , O(Γ) = O

Theorem 1.5.5. M(O) is a group with respect to the usual multiplication of ideals. IfΓ = 〈m,mγ〉Z ∈M(O) where γ ∈ K,m ∈ Q, a, b, c ∈ Z are as in Theorem 1.5.1 (b), then wedefine N(Γ) := m2

aand the inverse of Γ is Γ−1 = N(Γ)−1σ(Γ).

Proof. Clearly ideal multiplication is associative, commutative, etc.The unit element in M(O) is O.We first show that the elements in M(O) have an inverse:Let Γ = 〈m,mγ〉 ∈ M(O). Since O(σ(Γ)) = σ(O(Γ)) = O, also the conjugate σ(Γ) is inM(O). Moreover

Γσ(Γ) = m2〈1, γ, σ(γ), γσ(γ)〉 = N(Γ)〈a, aγ, aσ(γ), aγσ(γ)〉

where a, b, c are as in Theorem 1.5.1 (b). Then aγ2 + bγ + c = 0 so b = aγ + aσ(γ) andc = aγσ(γ). In particular

Γσ(Γ) = N(Γ)〈a, b, c, aγ〉 = N(Γ)O.

We now show that the product of two elements of M(O) is again in M(O):Let Γ1,Γ2 ∈M(O). Then O ⊆ O(Γ1Γ2) by the associativity of ideal multiplication. Moreover

O = (Γ1Γ2)(Γ−11 Γ−1

2 ) = N(Γ1)−1N(Γ2)−1(Γ1Γ2)σ(Γ1)σ(Γ2)

so O(Γ1Γ2) ⊆ O(O) = O.

Definition 1.5.6. Let O be an order in K = Q[√d].

(a) e(O) := [Z∗K : O∗].(b) K+ := a ∈ K∗ | N(a) > 0, n(O) := [K∗ : (K+O

∗)].(c) Cl(O) :=M(O)/aO | a ∈ K∗ is called the class group of O.(d) Cl0(O) :=M(O)/aO | a ∈ K+ is called the ray class group of O.

Remark 1.5.7. (a) If d < 0 then K+ = K∗, n(O) = 1.(b) If d > 0 then O∗ = 〈−1〉 × 〈ε〉 and n(O) = 1 if and only if NK/Q(ε) = −1. Otherwisen(O) = 2.(c) The kernel of the map Cl0(O)→ Cl(O) has order n(O).(d) Every class [Γ]0 ∈ Cl0(O) has a representative of the form Γ = 〈1, γ〉 with γ = x + yω,x, y ∈ Q, y > 0. Such a γ ∈ K is called admissible.

Theorem 1.5.8. Let γ1, γ2 ∈ K be admissible and put Γi := 〈1, γi〉. Assume that O(Γ1) =O(Γ2). Then

[Γ1]0 = [Γ2]0 ∈ Cl0(O)⇔ ∃A =

(k `m n

)∈ SL2(Z), such that γ2 =

kγ1 + `

mγ1 + n.


Proof. ⇒: Let [Γ1]0 = [Γ2]0. Then there is some α ∈ K, N(α) > 0 and A =

(k `m n

)∈

GL2(Z) such that(γ2

1

)= A

(αγ1

α

), so

(γ2 σ(γ2)1 1

)= A

(αγ1 σ(α)σ(γ1)α σ(α)

).

Taking the determinant we obtain

(?) γ2 − σ(γ2) = det(A)N(α)(γ1 − σ(γ1)).

Since γ1 and γ2 are admissible, the coefficient of√d is positive on both sides and hence

det(A) > 0 (note that N(α) > 0 by assumption), so A ∈ SL2(Z). Moreover

γ2 =γ2

1=

kαγ1 + α`

mαγ1 + αn=

kγ1 + `

mγ1 + n.

⇐: Put α := 1mγ1+n

. Then

αΓ1 = 〈α, αγ1〉 = 〈A−1

(αγ1

α

)= 〈γ2, 1〉 = Γ2.

Because of (?) and det(A) = 1 we obtain N(α) > 0.

Definition 1.5.9. Let Γ := 〈1, γ〉 ∈ M(O), γ ∈ K admissible and let a, b, c ∈ Z, a > 0,gcd(a, b, c) = 1 such that aγ2 + bγ + c = 0. Then

Fγ := Fγ(X, Y ) :=1

N(Γ)(X − γY )(X − σ(γ)Y ) = aX2 + bXY + cY 2

is called the binary quadratic form defined by γ.

Then Theorem 1.5.8 immediately implies

Theorem 1.5.10. Let Γi = 〈1, γi〉 ∈ M(O), γi admissible i = 1, 2. Then

[Γ1]0 = [Γ2]0 ∈ Cl0(O)⇔ ∃A =

(k `m n

)∈ SL2(Z) such that Fγ1(kX+`Y,mX+nY ) = Fγ2(X, Y ).

Definition 1.5.11. Let F = Fa,b,c = aX2 + bXY + cY 2 be a binary quadratic form.

(a) disc(F ) := −4ac+ b2 = − det

(2a bb 2c

)is called the discriminant of F .

(b) Two forms Fa,b,c and Fa′b′c′ are called properly equivalent, if there is some A ∈ SL2(Z),such

A

(2a bb 2c

)Atr =

(2a′ b′

b′ 2c′

).

(c) For any D ∈ Z we define

Q(D) := Fa,b,c | a, b, c ∈ Z, gcd(a, b, c) = 1, a > 0,−4ac+ b2 = D/ SL2(Z)

to be the set of proper equivalence classes of binary quadratic forms of discriminant D.

1.5. QUADRATIC NUMBER FIELDS 25

Theorem 1.5.12. Cl0(Of ) is in bijection with Q(f 2dK) by mapping [〈1, γ〉]0 to [Fγ] (whereγ is admissible).

Proof. We first show that the map is well defined: If Γ = 〈1, γ〉 and aγ2 + bγ + c = 0 witha, b, c ∈ Z, a > 0, gcd(a, b, c) = 1 then O(Γ) = 〈1, aγ〉 has discrimimant

d(O(Γ)) = det

(2 −b−b b2 − 2ac

)= −4ac+ b2.

Now the inverse bijection is given by assigning to F := Fa,b,c the admissible root γ of F (X, 1).

Then F (X, Y ) = a(X − γY )(X − σ(γ)Y ) with γ ∈ Q[√disc(F )] = Q[

√f 2dK ] = K.

1.5.1 Imaginary quadratic number fields.

Theorem 1.5.13. Let D = f 2dK < 0. Then

R(D) := Fa,b,c | a > 0,−4ac+b2 = D, a, b, c ∈ Z, gcd(a, b, c) = 1, |b| ≤ a ≤ c, and b > 0 if a = c or |b| = a

is a system of representatives for Q(D).

Proof. Let Fa,b,c ∈ [Fa,b,c] ∈ Q(D) such that a is minimal. Then a ≤ c since(0 1−1 0

)(2a bb 2c

)(0 −11 0

)=

(2c −b−b 2a

)Let k := ba−b

2ac. Then(

1 0k 1

)(2a bb 2c

)(1 k0 1

)=

(2a b+ 2ak

b+ 2ak 2(ak2 + bk + c)

)with b′ = b+ 2ak ∈ [−a, a], c′ = ak2 + bk + c and Fa,b′,c′ ∈ R(D).On the other hand any two forms in Q(D) are inequivalent under the action of SL2(Z) (ex-ercise).

Remark 1.5.14. If Fa,b,c ∈ R(D) then a ≤√|D|/3 because |D| = 4ac−b2 ≥ 4a2−a2 = 3a2.

Example. D = −47, then a ≤√

47/3 < 4, so a = 1, 2, 3. Moreover −47 = −4ac+ b2, sob is odd.

a = 1:

(2 11 24

).

a = 2:

(4 11 12

),

(4 −1−1 12

).

a = 3:

(6 11 8

),

(6 −1−1 8

).

Let ω := 1+√−47

2. Then ω2 − ω + 12 = 0 and the corresponding ideals are

ZK = 〈1, ω〉, ℘2 = 〈2,−σ(ω)〉, ℘′2 = 〈2, ω〉, ℘3 = 〈3,−σ(ω)〉, ℘′3 = 〈3, ω〉.

The class group has order 5, so ClK = 〈℘2〉 ∼= C5.


Remark 1.5.15. The integral ideal 〈a, aγ〉 ∈ [〈1, γ〉]0 has norm N with a | N | a2.

The 2-rank of the class group.

This works similarly also for real quadratic number-fields, but we restrict to imaginaryquadratic fields. So let d ∈ Z be squarefree, d > 0, K = Q[

√−d] with ring of integers

ZK = Z[√−d] and discriminant dK = 4d if −d ≡ 2, 3 (mod 4) resp. ZK = Z[1+

√−d

2] and

discriminant dK = d if −d ≡ 1 (mod 4).

Let α :=√−d resp. α := 1+

√−d

2denote a generator of ZK and f its minimal polynomial.

Let σ denote the non-trivial Galois automorphism of K, so σ(√−d) = −

√−d.

Lemma 1.5.16. A prime p is a divisor of dK, if and only if there is a prime ideal ℘ E ZKsuch that ℘2 = pZK. (We say that p is ramified in K.)

Proof. Let p be a prime. Then the prime ideals dividing p correspond to the maximal ide-als of ZK/pZK ∼= Fp[x]/(f). This is a uniserial ring, iff f has a double zero mod p whichis equivalent to p dividing dK . (Treat 2 separately, for odd primes, one may replace f byX2 + d where this is obvious).

Theorem 1.5.17. Cl(K)/Cl(K)2 ∼= Ω2(Cl(K)) = [I] | [I]2 = 1 ∼= Cg−12 where g is the

number of distinct prime divisors of dK. More precisely for each prime divisor pj of dK let℘j be the prime ideal dividing pjZK. Then

Ω2(Cl(K)) = 〈[℘j] | i = 1, . . . , g〉 and

℘1 · · ·℘g =

√−d if − d ≡ 1, 2 (mod 4)

℘2 · · ·℘g =√−d if − d ≡ 3 (mod 4)

where we assumed in the last case that ℘21 = 2ZK.

It is clear that all ramified prime ideals ℘ have order at most 2 in the class group since℘2 = pZK is principal. We need to show that(a) Any class of order 2 contains an ideal A such that A = σ(A).(b) Any such σ-invariant ideal is equivalent (in the class group) to a product of ramifiedprime ideals.(c) There is no other relation between the classes of the ramified prime ideals.

Lemma 1.5.18. (Hilbert 90) Let a ∈ K such that N(a) = aσ(a) = 1. Then there is some

b ∈ K such that a = σ(b)b

.

Proof. If a = −1 then put b =√−d. Otherwise let b := (1 + a)−1. Then

σ(b)

b=

1 + a

1 + σ(a)=

(1 + a)a

(1 + σ(a))a=

(1 + a)a

a+ 1= a.

1.6. RAMIFICATION. 27

Lemma 1.5.19. Let A be a fractional ideal such that σ(A) = A. Then A = rQ wherer ∈ Q>0 and Q is a (possibly empty) product of distinct ramified prime ideals.

Proof. By the uniqueness of the prime ideal decomposition it is enough to show this for primeideals ℘. The non-trivial Galois automorphism σ acts on the zeros of f mod p. If f has adouble zero mod p then σ fixes the prime ideal ℘ dividing p (these are the ramified primes).If f is irreducible mod p, then pZK is a prime ideal.If f is a product of two distinct linear polynomials then σ interchanges the two zeros of fmodulo p and p = ℘σ(℘) is a product of two distinct prime ideals.

Lemma 1.5.20. Let AE ZK. Then Aσ(A) = N(A)ZK.

Proof. Again it is enough to show this for prime ideals where we did this in the last proof.

The above lemma shows that for any ideal A the inverse [A]−1 = [σ(A)] in the class groupof K. In particular [A] = [σ(A)] if and only if [A] has order 1 or 2 in the class group.

Lemma 1.5.21. If [A] = [σ(A)] then this class contains a σ-invariant ideal.

Proof. In this case there is some r ∈ K∗ such that σ(A) = Ar. ThenN((r)) = N(A)−1N(σ(A)) =1 and therefore |N(r)| = 1. But the norm form is positive definite, so N(r) = 1 and there issome b ∈ K∗ with r = b

σ(b). Put

B := Ab.

Then B ∈ [A] satisfiesσ(B) = σ(A)σ(b) = Arσ(b) = Ab = B

To see the last point (c), we need to show that no other product of distinct ramified primeideals is principal. For simplicity we only deal with the case −d ≡ 1, 2 modulo 4 and showthat in this case for any proper divisor 1 < m < d of d the ring ZK does not contain anelement of norm m. If x, y ∈ Z then the norm of x + y

√−d is x2 + y2d = m then (since

0 < m < d) y2 needs to be 0, so m = x2 is a square which is a contradiction. In the case−d ≡ 1 modulo 4 we also have integral elements (x + y

√d)/2 where x and y are both odd.

The norm of this element is 14(x2 + y2d) so (x2 + y2d) = 4m, which is only possible if y = ±1,

then x2 = 4m − d = m(4 − dm

) and dm

= 3. But this contradicts the fact that d and hencealso m is squarefree, in particular m is not a square.

1.6 Ramification.

Let Q ⊂ K ⊂ L be a tower of algebraic number fields and Z ⊂ ZK ⊂ ZL the correspondingring of integers.

Definition 1.6.1. Let 0 6= ℘E ZK be a prime ideal. Then

℘ZL = ℘e11 · · ·℘err


for prime ideals ℘i E ZK and e1, . . . , er ∈ N. Each ℘i defines a field extension

Fq ∼= ZK/℘ → ZL/℘i ∼= Fqfi

of degree fi, since ℘ = ℘i ∩ZK for all i. Then ei is called the ramification index of ℘i andfi is the inertia degree of ℘i.

Example. K = Q, L = Q[√−7], α := 1+

√−7

2.

ramified prime: (√−7)2 = 7ZL, e = 2, f = 1.

inert prime: (3) = 3ZL, e = 1, f = 2.decomposed prime: 2ZL = (α)(1− α), e1 = e2 = 1, f1 = f2 = 1.

Theorem 1.6.2. Let Q ⊂ K ⊂ L and 0 6= ℘ E ZK be a prime ideal with ℘ZL = ℘e11 · · ·℘errfor prime ideals ℘i E ZL and inertia degrees fi = [(ZL/℘i) : (ZK/℘)]. Then

∑ri=1 eifi = n =

[L : K].

Proof. By the Chinese remainder theorem

ZL/℘ZL =r⊕i=1

ZL/℘eii .

Put k := ZK/℘. Then ZL/℘ZL is a vector space over k and

dimk(ZL/℘ZL) =r∑i=1

dimk(ZL/℘eii ) =r∑i=1

eifi.

So we need to show that dimk(ZL/℘ZL) = n = [L : K].To this aim let ω1, . . . , ωm ∈ ZL such that (ω1, . . . , ωm) is a k-basis of ZL/℘ZL.Claim: (ω1, . . . , ωm) is a K-basis of L.linearly independent: Assume that ai ∈ K not all = 0 are such that

∑mi=1 aiωi = 0. Wlog

we may assume that all ai ∈ ZK . LetA := (a1, . . . , am)EZK and choose some a ∈ A−1\A−1℘.Let bi := aai. Then

∑mi=1 biωi = 0 with bi ∈ ZK not all bi ∈ ℘. Reducing this modulo ℘ we

obtain a linear dependence of the ωi which is a contradiction.generating system: This follows essentially from Nakayama’s Lemma: Let

M := 〈ω1, . . . , ωm〉ZK ≤ ZL and N := ZL/M.

Then ZL = M + ℘ZL so ℘N ∼= (℘ZL + M)/M = ZL/M = N . We claim that N is a torsionmodule. Let N = 〈α1, . . . , αs〉ZK with αi =

∑sj=1 aijαj and aij ∈ ℘. Let d := det(A) where

A = (aij)si,j=1 − Is ∈ Zs×sK . Then d ≡ (−1)s (mod ℘) and A∗A = dIs for A∗ ∈ Zs×sK the

adjoint of A. So

0 = A

α1...αs

= A∗A

α1...αs

=

dα1...dαs

and therefore dN = 0, so |N | is finite. Since M is of finite index in ZL it has the same rankas ZL and generates L as a vector space over K.

1.6. RAMIFICATION. 29

1.6.1 How to compute inertia degree and ramification index ?

Let L = K(α) with α ∈ ZL, f := µα the minimal polynomial of α. Then O := ZK [α] ∼=ZK [X]/(f(X)) is an order in L.

Definition 1.6.3. Let Fα := a ∈ ZL | aZL ⊆ ZK [α] be the largest ZL-ideal contained inZK [α]. Then Fα is called the conductor (Fuhrer) of α.

Theorem 1.6.4. Let ℘ E ZK be a prime ideal such that gcd(℘ZL,Fα) = 1. Assume thatµα(X) = p1(X)e1 · · · pr(X)er ∈ ZK/℘[X]. Then ℘i := (℘, pi(α))) E ZL (1 ≤ i ≤ r) are theprime ideals dividing ℘ZL and

℘ZL = ℘e11 · · ·℘err , fi := [ZL/℘i : ZK/℘] = deg(pi).

Proof. Let O := ZK [α]. Then

ZL = Fα + ℘ZL ⊆ O + ℘ZL ⊆ ZL

and hence O/℘O ∼= ZL/℘ZL ∼= k[X]/(µα(X)) with k = ZK/℘. The ideals in this ring canbe read off from the factorization of µα(X) ∈ k[X].

Corollary 1.6.5. There are only finitely many prime ideals ℘ E ZK for which there is aprime ideal ℘i E ZL such that ℘2

i | ℘ZL. (For short: ZL contains only finitely many ramifiedprimes.)

Proof. Since Fα has only finitely many divisors, we may assume that ℘ is prime to Fα. Thenthe polynomial µα(X) ∈ k[X] has multiple factors, iff

gcd(µα(X), µ′α(X) 6= 1⇔ ℘ divides disc(µα) =∏i<j

(αi − αj) ∈ ZK .

where αi are the roots of µα in the algebraic closure of K. But this ideal has only finitelymany prime divisors.

Example: Let f := X4 + 2X3 − 5X2 − 6X − 1 ∈ Q[X], L = Q[X]/(f(X)), α = X ∈ L,so µα = f . Then Z[α] is of index 3 in ZL. dL = 1600, disc(f) = 14400 = 9dL.

f (mod 2) (X2 +X + 1)2 (2) = ℘22 e = f = 2

f (mod 3) (X + 2)2(X2 +X + 2)f (mod 5) (X2 +X + 2)2 (5) = ℘2

5 e = f = 2f (mod 7) (X2 + 4)(X2 + 2X + 5) (7) = ℘7℘

′7 e1 = e2 = 1, f1 = f2 = 2

1.6.2 Hilbert’s theory of ramification for Galois extensions.

Let L ⊇ K be algebraic number fields and assume that L/K is Galois. Let G := Gal(L/K)denote the Galois group.

Remark 1.6.6. For any σ ∈ G we have σ(ZL) = ZL. If ℘E ZL is a prime ideal, then alsoσ(℘)E ZL is a prime ideal and ℘ ∩ ZK = σ(℘) ∩ ZK.


Theorem 1.6.7. The Galois group acts transitively on the set of prime ideals of ZL thatcontain a given prime ideal ℘ of ZK:

℘ZL = ℘e11 . . . ℘err ⇒ for all 1 ≤ i ≤ r there is σi ∈ G, σi(℘1) = ℘i.

Proof. Assume that ℘2 6= σ(℘1) for all σ ∈ G. By the Chinese remainder theorem there issome x ∈ ZL such that

x ≡ 0 (mod ℘2), x ≡ 1 (mod σ(℘1)) for all σ ∈ G.

Then NL/K(x) =∏

σ∈G σ(x) ∈ ℘2 ∩ ZK = ℘.On the other had σ(x) 6∈ ℘1 for all σ ∈ G, so NL/K(x) 6∈ ℘1∩ZK = ℘ which is a contradiction.

Corollary 1.6.8. e1 = . . . = er =: e, f1 = . . . = fr =: f and [L : K] = n = ref .e is called the ramification index of ℘, e = eL/K(℘) = eL/K(℘i) for all i.f is called the inertia degree of ℘, f = fL/K(℘) = fL/K(℘i) for all i.

Definition 1.6.9. Let ℘E ZL be a prime ideal in ZL. Then

G℘ := σ ∈ G | σ(℘) = ℘

is called the decomposition group of ℘ and Z℘ := FixG℘(L) := x ∈ L | σ(x) =x for all σ ∈ G℘ is called the decomposition field of ℘.

Theorem 1.6.10. Let ℘E ZL be a prime ideal in ZL and let ℘Z := ℘ ∩ Z℘, Z := Z℘.(1) ℘ZZL = ℘e.(2) fL/Z(℘) = fL/K(℘), eL/Z(℘) = eL/K(℘) = e.(3) eZ/K(℘Z) = fZ/K(℘Z) = 1.

Proof. (1) G℘ = Gal(L/Z), so the set of all prime ideals of ZL that contain ℘Z is σ(℘) | σ ∈G℘ = ℘.(2) Let r := [G : G℘] and let ℘ = ℘1, . . . , ℘r be the set of prime ideals of ZL that con-tain P := ℘ ∩ ZK . Then ref = |G| = [L : K] where e = eL/K(℘), f = fL/K(℘). Soef = |G℘| = eL/Z(℘)fL/Z(℘). Clearly eL/Z(℘) ≤ eL/K(℘) and fL/Z(℘) ≤ fL/K(℘) from whichone obtains (2).(3) eL/K(℘) = eL/Z(℘)eZ/K(℘Z) and fL/K(℘) = fL/Z(℘)fZ/K(℘Z).

Theorem 1.6.11. Let k(℘) := ZL/℘ and k := ZK/P with P = ℘ ∩ ZK. Then k(℘)/k(P ) isa normal extension and G℘ → Gal(k(℘)/k(P )) is surjective.

Proof. We first note that k ∼= k(℘Z) = ZZ/℘Z so we may assume that Z℘ = K and G℘ = G.Choose α ∈ ZL such that α := α + ℘ ∈ k(℘) is a primitive element, let f := µα,K ∈ ZK [X]and g := µα,k ∈ k[X]. Then g divides f ∈ k[X]. Since L/K is normal, all roots of f lie inZL, so f ∈ ZL[X] is a product of linear factors, and hence also f and therefore g ∈ k(℘)[X]is a product of linear factors, so k(℘)/k is normal.Now let α1 ∈ k(℘) be a zero of g. Then there is α1 ∈ ZL with f(α1) = 0 such that α1 = α1+℘.This yields the existence of some σ ∈ G = G℘ such that σ(α) = α1. This element σ mapsonto the Galois automorphism of k(℘) that maps α to α1.

1.7. CYCLOTOMIC FIELDS. 31

Definition 1.6.12.1→ I℘ → G℘ → Gal(k(℘)/k)→ 1

is a short exact sequence. In particular the inertia group of ℘ is

I℘ := σ ∈ G℘ | σ(x) ≡ x (mod ℘) for all x ∈ ZLEG℘.

The fixed field T℘ := Fix(I℘) is called the inertia field of ℘.

Corollary 1.6.13. T℘/Z℘ is a Galois extension with Galois group

Gal(T℘/Z℘) ∼= Gal(k(℘)/k) ∼= G℘/I℘ ∼= Cf .

L

e︷︸︸︷⊇ T℘

f︷︸︸︷⊇ Z℘

r︷︸︸︷⊇ K

Gal(L/T℘) = I℘ Cf ∼= G℘/I℘ = Gal(T℘/Z℘) G℘ = Gal(L/Z℘)

Example. L = Q[ 3√

2, ζ3], K = Q, Gal(L/Q) = S3.Prime ideal decompositions:5ZL = ℘5℘

′5℘′′5 with fi = 2. Put Z := Q[ 3

√2]. Then 5ZZ = p5p

′5 with f = 1, f ′ = 2, wlog

℘5 = p5ZL then Z = Z℘5 , G℘5 = Gal(L/Z) ∼= C2 and T℘5 = L.For the prime 2 we obtain 2ZL = ℘3

2 = ( 3√

2)3, T℘2 = Q[ζ3], Z℘2 = Q, G℘2 = G, e = 3, f = 2.

1.7 Cyclotomic fields.

Definition 1.7.1. The cyclotomic polynomials are defined recursively by

Φ1(X) := (X − 1),Φn(X) := (Xn − 1)/∏

d|n,1≤d<n

Φd(X)

The roots of Φn are the primitive n-th root of unity.

Remark 1.7.2. In the Algebra class we have seen the following facts:

(a) Φn(X) ∈ Q[X] is an irreducible polynomial with integral coefficients.

(b) Φn(X) =∏

d∈(Z/nZ)∗(X − ζdn) where ζn is any primitive nth root of unity.

(c) deg(Φn(X)) = ϕ(n) = |Z/nZ∗|.

(d) Q[ζn] := Kn is a Galois extension of Q with Gal(Kn/Q) ∼= (Z/nZ)∗ with explicitisomorphism mapping a ∈ (Z/nZ)∗ to σa : (ζn 7→ ζan). Kn is called the n-th cyclotomicfield.

e If n = pa11 · · · pass is a product of powers of distinct primes then

Kn = Kpa11· · ·Kpass , ζn =

s∏i=1

ζpaii


Remark 1.7.3. (cyclotomic units)(a) Assume that n = pa is a prime power and let i, j ∈ N such that p |6 ij. Then (1− ζjn)/(1−ζ in) ∈ Z[ζn]∗.(b) Assume that n is divisible by at least two distinct primes. Then (1 − ζn) ∈ Z[ζn]∗ and∏

j∈Z/nZ∗(1− ζjn) = 1.

Proof. Exercise.

Theorem 1.7.4. If n = pa is a prime power then ZKn = Z[ζn] and d(Kn) = ±ppa−1(ap−a−1).

Proof. LetO := Z[ζn] = 〈1, ζn, . . . , ζp

a−1(p−1)−1n 〉Z ∼= Z[X]/(Φn(X)).

Then ℘ := (1− ζn)EO is a Galois invariant ideal of norm

NKn/Q(1− ζn) =∏p |6 j

(1− ζjn) = Φn(1) = p.

Note that Φn(X) = (Xpa−1)/(Xpa−1−1) = (Y p−1)/(Y −1) = Y p−1 +Y p−2 +. . .+Y +1 withY = Xpa−1

. By comparing norms we obtain ℘d = pO with d = [Kn : Q] = ϕ(n) = pa−1(p−1).So the unique maximal ideal dividing pO is a principal ideal, hence O = O(Jp(O)) is p-maximal. But the determinant of O is the discriminant of Φn which is not divisible by anyprime ` 6= p, since the n-th roots of unity are pairwise distinct modulo `. Therefore O is also`-maximal for all primes ` 6= p and hence a maximal order (Exercise 4, Sheet 2).

In particular we know that O = ZKn and that the discriminant of Kn is a power of p.Put ζ := ζn. Then

d(O) = d(Φn) =∏

i 6=j∈Z/paZ∗(ζ i − ζj) =

∏i∈Z/paZ∗

Φ′n(ζ i) = NKn/QΦ′n(ζ).

Note that Φ′n(X) = ddX

∏i∈Z/paZ∗(X − ζ i) =

∑i∈Z/paZ∗

∏j 6=i(X − ζj). To compute Φ′n(ζ) we

differentiate the equation (Xpa−1 − 1)Φn(X) = (Xpa − 1) to obtain

pa−1Xpa−1−1Φn(X) + (Xpa−1 − 1)Φ′n(X) = paXpa−1.

Evaluating at ζ we obtain (ζpa−1 − 1)Φ′n(ζ) = paζp

a−1 since Φn(ζ) = 0. Now α := ζpa−1

is aprimitive pth root of unity and hence NQ(α)/Q(α− 1) = ±p, so

NKn/Q(α− 1) = ±ppa−1

NKn/Q(ζ) = ±1

NKn/Q(pa) = ±(pa)pa−1(p−1) ⇒

NKn/Q(Φ′n(ζ)) = ±ps

where s = pa−1(ap− a− 1).

Theorem 1.7.5. Let n = pa11 · · · pass ∈ N. Then ZKn = Z[ζn] and d(Kn) =∏s

i=1 d(Kpaii

)ϕ( n

paii

).

1.7. CYCLOTOMIC FIELDS. 33

This follows from the next more general Lemma.

Lemma 1.7.6. Let K,K ′ be number fields of degree n = [K : Q], n′ := [K ′ : Q] anddiscriminants d := d(K) and d′ := d(K ′). Assume that gcd(d, d′) = 1 and L := KK ′ hasdegree nn′ over Q. If B := (w1, . . . , wn) and B′ = (v1, . . . , vn′) are integral bases of K resp.K ′, then BB := (wivj | 1 ≤ i ≤ n, 1 ≤ j ≤ n′) is an integral basis of ZL and d(L) = dn

′(d′)n.

Proof. (a) BB is a Q-basis of L: It is a generating set by definition of L and these elementsare linearly independent since we assumed that [L : Q] = nn′.(b) To compute d(BB) let σ1, . . . , σn : K → C resp. ϕ1, . . . , ϕn′ : K ′ → C be the distinctembeddings. Then σiϕj : L→ C are the embeddings of L and

d(BB) = det(M)2, where M = (σiϕj(wkvl))(i,j),(k,l) = (σi(wk)ϕj(vl))(i,j),(k,l).

This matrix M is easily seen to be the Kronecker product M = A⊗A′ with A = (σi(wk))i,kand A′ = (ϕj(vl))j,l. Hence d(BB) = dn

′(d′)n as claimed.

(c) BB is an integral basis. Basis is clear, also that the elements of BB are integral. So itremains to show that 〈BB〉Z = ZL. Let α =

∑i,j aijwivj ∈ ZL with aij ∈ Q. We need to

show that all aij ∈ Z. Let A′ be as above and put

a := (ϕ1(α), . . . , ϕn′(α))tr, b := (β1, . . . , βn′)tr, where βj =

n∑i=1

aijwi.

Then a = A′b and d′b = det(A′)b = (A′)∗a. Since all entries are integers, the vector d′b onlyhas integral entries, so d′

∑ni=1 aijwi ∈ ZK which implies that d′aij ∈ Z for all i, j. Similarly

we obtain daij ∈ Z for all i, j and hence aij ∈ Z since d and d′ are co-prime.

We now investigate the ramification indices and inertia degrees of primes in Kn.

Theorem 1.7.7. Let n = pa11 · · · pass ∈ N. For any prime p define vp(n) := a if n = pab withp |6 b. Let fp ∈ N be minimal such that pfp ≡ 1 (mod n/pvp(n)). Then

pZ[ζn] = (℘1 · · ·℘r)ϕ(pvp(n)), fKn/Q(℘i) = fp.

Proof. We need to factorise Φn(X) ∈ Fp[X]. Let n = pam with p |6 m, a = vp(n). If αi | 1 ≤i ≤ ϕ(pa) is the set of primitive pa-th roots of unity and βi | 1 ≤ i ≤ ϕ(m) is the set ofprimitive m-th roots of unity then

αiβj | 1 ≤ i ≤ ϕ(pa), 1 ≤ j ≤ ϕ(m)

is the set of primitive n-th root of unity and

Φn(X) =∏i,j

(X − αiβj) ≡p∏j

(X − βj)ϕ(pa) ≡p Φm(X)ϕ(pa).

The m-th roots of unity are distinct mod p and Fpf contains a primitive m-th root of unity,iff m | pf − 1. So all irreducible factors of Φm(X) ∈ Fp[X] have degree fp.


Example. Let n := 45 = 325. Then Gal(Kn/Q) ∼= C6×C4, Kn = K9K5 and 3Z[ζn] = ℘63

is totally ramified in K9 and inert in K5. So e3 = 6, f3 = 4. So the decomposition field isZ3 = Q, the inertia field is T3 = Q[ζ5].Since 3 |6 5 − 1 the prime 5Z[ζn] = ℘4

5 with e5 = 4, f5 = 6. So the decomposition field isZ5 = Q, the inertia field is T5 = Q[ζ3].To compute the inertia degree of 2, we need to find the minimal f = f2 for which 2f − 1 is amultiple of 45. 24 − 1 = 15, so f = 3 · 4 = 12 and 2Z[ζn] = ℘2℘

′2. The decomposition field of

2 is Q[√−15].

For the prime 11 one finds that 45 | 116 − 1 and hence f11 = 6 and 11Z[ζn] = ℘11℘′11℘′′11℘′′′11.

Since 35 ≡11 1, the prime ideals over 11 are

℘11 = (3− ζ5, 11), ℘′11 = (3− ζ25 , 11), ℘′′11 = (3− ζ3

5 , 11), ℘′′′11 = (3− ζ45 , 11).

The decomposition field of 11 is Z11 = Q[ζ5].

Corollary 1.7.8. Let n be either odd or a multiple of 4. Then p is ramified in Z[ζn] if andonly if p | n.

1.7.1 Quadratic Reciprocity.

Theorem 1.7.9. Let ` and p be odd primes and put `∗ := (−1)(`−1)/2`. Then p is (totally)decomposed in Q[

√`∗], if and only if pZ[ζ`] is a product of an even number of prime ideals.

Proof. Since K` has a subfield L of degree 2 over Q and ` is the only prime that ramifies inK`, this is also the only prime that ramifies in this unique quadratic subfield, so L = Q[

√`∗].

Now assume that pZL = ℘1℘2 is a product of two prime ideals in L and let σ ∈ Gal(K`/Q) =:G be such that σ(℘1) = ℘2. Then σ yields a bijection between the set of prime ideals of Z[ζ`]that contain ℘1 and the ones that contain ℘2, in particular the number of prime ideals ofZ[ζ`] that contain p is even.To see the opposite direction let ℘ be a prime ideal of Z[ζ`] such that ℘ ∩ Z = pZ and letG℘ := StabG(℘) be its decomposition group. Since by assumption |℘G| is even, the index[G : G℘] is even. Now G is cyclic, so the unique quadratic subfield L of K` is contained inthe decomposition field L ⊂ Z℘ = Fix(G℘). Putting PZ := ℘ ∩ Z℘ then fZ℘/Q(PZ) = 1 soalso fL/Q(PZ ∩L) = 1. But p does not divide the discriminant of L, so it is not ramified, andtherefore totally decomposed in L.

Definition 1.7.10. Let 2 6= p be a prime, a ∈ Z such that p |6 a.(a

p

):=

1 if a ≡p x2 for some x ∈ Z−1 otherwise.

is called the Legendre symbol of a at p.

Remark 1.7.11. (a)(ap

)= 1 ⇔ (a+ pZ) ∈ (Z/pZ∗)2 ⇔ a(p−1)/a ≡p 1.

(b)(ap

)(bp

)=(abp

).

(c) Let a ∈ Z be squarefree and K := Q[√a]. Then

(ap

)= 1 ⇔ pZK = ℘1℘2 is totally

decomposed.

1.8. DISCRETE VALUATION RINGS. 35

Theorem 1.7.12. (Gauss reciprocity)(a) Let ` and p be distinct odd primes. Then(

`

p

)(p`

)= (−1)

p−12

`−12 .

(b)(−1p

)= (−1)(p−1)/2.

(c)(

2p

)= (−1)(p2−1)/8.

Proof. (b) is clear.To see (c) we compute in Z[i]. Here (1 + i)2 = 2i. And

1 + ip ≡p (1 + i)p = (1 + i)((1 + i)2)(p−1)/2 ≡p (1 + i)2(p−1)/2i(p−1)/2 ≡p (1 + i)

(2

p

)i(p−1)/2

so (1 + i)(

2p

)i(p−1)/2 ≡p 1 + i(−1)(p−1)/2.

If (p− 1)/2 is even, then this reads as (1 + i)(

2p

)(−1)(p−1)/4 ≡p (1 + i). Dividing both sides

by (1 + i) we obtain(

2p

)≡p (−1)(p−1)/4.

If (p−1)/2 is odd, then we have (1+i)(

2p

)(−i)(−1)(p+1)/4 ≡p 1−i and hence

(2p

)(−i)i(−1)(p+1)/4 ≡p

1 because 1+i1−i = i. So

(2p

)≡p (−1)(p+1)/4.

These two congruences may be summarised as in (c).(a) Let `∗ := (−1)(`−1)/2` be as in Theorem 1.7.9. We show that(

`∗

p

)=(p`

)Then (p

`

)=

(`∗

p

)=

(−1

p

)(`−1)/2(`

p

)= (−1)(p−1)/2(`−1)/2

(`

p

).

We have(`∗

p

)= 1 iff p is decomposed in Q[

√`∗] ⇔ p splits in Q[ζ`] into an even number of

prime ideals. Now pZ[ζ`] = ℘1 · · ·℘s with s = `−1f

and f minimal such that pf ≡` 1. So s iseven ⇔

f | `− 1

2⇔ p

`−12 ≡` 1⇔

(p`

)= 1

1.8 Discrete valuation rings.

Definition 1.8.1. (a) A discrete valuation ring R is a local principal ideal domain (com-mutative, without zero divisors) which is not a field.(b) Let K be a field. A discrete valuation of K is a mapping v : K → Z ∪ ∞ such that


(o) There is some x ∈ K∗ such that v(x) 6= 0.(i) v(x) =∞ ⇔ x = 0.(ii) v(xy) = v(x) + v(y) for all x, y ∈ K∗.(iii) v(x+ y) ≥ minv(x), v(y) for all x, y ∈ K.

Clear: v(1) = 0, v(x−1) = −v(x), v : K∗ → (Z,+) is a group homomorphism.

Remark 1.8.2. v(x+ y) = minv(x), v(y) if v(x) 6= v(y).

Proof. First note that v(ζ) = 0 for any ζ ∈ R such that ζn = 1 for some n. In particularv(−1) = 0 and v(−y) = v(y).Assume that v(x) < v(y). Then

v(x) = v(x+ y − y) ≥ minv(x+ y), v(y) ≥ minv(x), v(y) = v(x).

We therefore have equality everywhere and v(x + y) = v(x) (note that v(y) > v(x) by as-sumption).

Example 1.8.3. Let R be a Dedekind domain K := Quot(R) and 0 6= ℘ER a prime ideal.Then the localisation of R at ℘ is

R(℘) := xy∈ K | x, y ∈ R, y 6∈ ℘.

Then R(℘) is a discrete valuation ring with maximal ideal ℘R(℘) = πR(℘) for any elementπ ∈ ℘ \ ℘2.The prime ideal ℘ also defines a valuation v = v℘ : K∗ → Z by putting v(z) = n ∈ Z≥0 if℘n | zR but ℘n+1 |6 zR and v(x

y) = v(x) − v(y) for all z, x, y ∈ R. Then R(℘) = x ∈ K |

v(x) ≥ 0.

Proposition 1.8.4. (a) Let R be a discrete valuation ring with maximal ideal ℘ = πR 6= 0.Then K := Quot(R) =

.∪i∈Z πiR∗ ∪ 0 and the mapping v : K → Z ∪ ∞, v(πiR∗) :=

i, v(0) :=∞ is a discrete valuation of K.(b) If v : K → Z ∪ ∞ is a discrete valuation, then R := x ∈ K | v(x) ≥ 0 is a discretevaluation ring with maximal ideal x ∈ K | v(x) ≥ 1 =: ℘ = πR for any π ∈ K withv(π) ≥ 1 minimal.

Proof. (a) Since R is a local ring the units are R∗ = R \ ℘. Any element a ∈ R is eithera unit (a ∈ R∗) or a multiple of π and then a1 := π−1a ∈ R. Also a1 is either a unit or amultiple of π. Continuing like this, we may write any non zero element of R in a unique wayas a = πnu with u ∈ R∗ and n ∈ Z≥0. Similarly any element 0 6= x = a

b∈ Quot(R) = K can

be written as πiw with w ∈ R∗ and i ∈ Z in a unique way. Therefore v is well defined. Itclearly satisfies (o), (i) and (ii). So it remains to show the strong triangular inequality. Letx ∈ πiR∗, y ∈ πjR∗, i, j ∈ Z, i ≥ j. Then x+y ∈ πjR and so v(x+y) ≥ j = minv(x), v(y).(b) We prove that R is a ring: 0 ∈ R, 1 ∈ R, a, b ∈ R⇒ ab ∈ R and a+ b ∈ R.The unit group of R is R∗ = x ∈ K | v(x) ≥ 0 and − v(x) ≥ 0 = x ∈ K | v(x) = 0.In particular ℘ is the unique maximal ideal of R. Choose π ∈ ℘ such that v(π) is minimal.Then for any z ∈ ℘ we have v(z) ≥ v(π) and hence zπ−1 ∈ R. So ℘ = πR is a principalideal.


Remark 1.8.5. Let R be a discrete valuation ring and x ∈ K = Quot(R). Then eitherx ∈ R or x−1 ∈ ℘. In particular K = R ∪ x−1 | 0 6= x ∈ ℘.

Theorem 1.8.6. A Noetherian integral domain R is a Dedekind domain if and only if alllocalizations R(℘) of R at non-zero prime ideals are discrete valuation rings.

Proof. (Exercise)

1.8.1 Completion

Remark 1.8.7. Let v : K → Z∪ ∞ be a discrete valuation and s ∈ (0, 1). Then v definesan ultra-metric

d : K ×K → R≥0, d(x, y) := sv(x−y)

where s∞ := 0. This means that d satisfies the following three axioms:(i) d(a, b) = 0 if and only if a = b.(ii) d(a, b) = d(b, a) for all a, b ∈ K.(iii) d(a, c) ≤ maxd(a, b), d(b, c) for all a, b, c ∈ K.

Definition 1.8.8. A metric space (M,d) is called complete, if any Cauchy sequence in Mconverges towards a limit in M .

Theorem 1.8.9. Let v : K → Z ∪ ∞ be a discrete valuation of the field K. Put R thering of all Cauchy sequences in K and N the ideal of all sequences in K that converge to 0.Then N ER is a maximal ideal and hence K := R/N is a field. The valuation v extendsto a valuation v of K and K is complete. The mapping ϕ : K → K, a 7→ (a, a, a, a . . .) +Nis injective and the image is dense in K. The field K is called the completion of K. It isunique up to isomorphism.

Proof. See the lecture Computeralgebra.

Theorem 1.8.10. Let v : K → Z∪∞ be a discrete valuation of the field K with valuationring R and maximal ideal πR. Define

S := lim←R/πiR = (a0, a1, . . .) | ai ∈ R/πi+1R, ai + πiR = ai−1.

Then S is an integral domain and ϕ : R→ S, a 7→ (a+ πR, a+ π2R, . . .) is a ring monomor-phism. The valuation v extends uniquely to a valuation v of S, v(a0, a1, . . . , ) := i if ai 6= 0,ai−1 = 0. S is complete with respect to this valuation and K := Quot(S) is the completionof K.

Proof. S is a ring with componentwise operations since the projections a+ πiR 7→ a+ pi−1Rare ring homomorphisms.ϕ is injective because

⋃∞i=0 π

iR = 0.It is clear that v is a valuation that extends the valuation of R (exercise).To see the completeness of S let (xn)n≥0 be a Cauchy sequence in S, so lim

n,m→∞v(xn−xm) =∞


or more concrete that for all k ≥ 0 there is some N(k) ∈ N such that v(xn − xm) > k for alln,m ≥ N(k). Wlog assume that (N(n))n≥0 is monotone increasing. Put x = (xN(k),k)k≥0.Then x ∈ S since

xN(k),k + πkR = xn,k + πkR = xn,k−1 = xN(k−1),k−1

for all n ≥ N(k). Similarly one shows that v(x − xn) → ∞ for n → ∞ so x is the limit ofthe Cauchy sequence.

For an example see the lecture Computeralgebra, where we introduced the p-adic numbersQp, the completion of Q at the p-adic valuation vp.

Example. The completion of K = Q[ζ3] at prime ideals over 2, 3, 7.

1.8.2 Hensel’s Lemma

Theorem 1.8.11. Let K be a discrete valuated complete field with valuation v, valuationring R. Let f ∈ R[X] be a polynomial and a0 ∈ R such that

v(f(a0)) > 2v(f ′(a0))

Then there is some a ∈ R such that f(a) = 0. More precisely the sequence

an+1 := an −f(an)

f ′(an)∈ R

converges towards some a ∈ R such that f(a) = 0 and v(a− a0) ≥ v(f(a0))− v(f ′(a0)) > 0.

Proof. (see also Computeralgebra) Note that f(t + x) = f(t) + f1(t)x + f2(t)x2 + . . ., for

fi(t) ∈ R[t], f1(t) = f ′(t). Define b0 := − f(a0)f ′(a0)

. Then v

(f(a0)f ′(a0)

)= v(f(a0))− v(f ′(a0)) >

v(f ′(a0)) ≥ 0, so a1 ∈ R.Moreover v(f(a0 + b0)) ≥ minv(fi(a0)bi0) | i ≥ 2, since f(a0) + f1(a0) · b0 = 0. Thereforev(f(a1)) ≥ 2v(b0) > v(f(a0)). Now f ′(t+x) = f ′(t)+2xf2(t)+. . . implies v(f ′(a1)−f ′(a0)) ≥v(b0) ≥ v(f ′(a0)), so v(f ′(a1)) = v(f ′(a0)).This shows that f(ai) converges to 0 v(f(ai))→∞).We now show that (ai) is a Cauchy sequence:

v(an+1 − an) = v(bn) = v

(− f(an)f ′(an)

)= v(f(an)) − v(f ′(an)) → ∞, because that first sum-

mand is strictly monotonously increasing (in Z) and the second summand is constant. So ifm > n: v(am−an) = v((am−am−1) + (am−1−am−2) + . . .+ (an+1−an)) ≥ minv(ai−ai−1) |n < i ≤ m → ∞ which means that (ai) is a Cauchy sequence.

Theorem 1.8.12. (Hensel’s Lemma, more general version) Let (K, v) be a complete discretevaluated field with valuation ring R and maximal ideal πR. Put F := R/πR and : R[X]→F [X] the natural epimorphism. Let f ∈ R[X] be monic such that f = h0g0 with gcd(h0, g0) =1. Then there are h(X), g(X) ∈ R[X] such that h = h0, g = g0 and f = gh.


Proof. We use the fact that v can be extended to a complete valuation on the finite dimen-sional K-algebra A := K[X]/(f) and that also this algebra is complete, so that we may usethe usual Hensel procedure to lift zeros of polynomials in A. (see Skript of Computeralgebra).For a more elementary proof I refer to the exercises (see also Neukirch, Kapitel II, (4.6)).By Chinese remainder theorem F [X]/(f) = F [X]/(h0)⊕F [X]/(g0). Let e, e′ := 1− e be theidempotents in F [x]/(f) corresponding to this decomposition and let e0 ∈ Λ := R[X]/(f) bea preimage of e, so e0 = e.We want to lift e0 to an idempotent in Λ. From this we obtain the required factorisation off in R[X] again by Chinese remainder theorem.We apply the usual Newton-Hensel Iteration to p(X) = X2 −X.We have p(e0) ∈ πΛ and p′(e0) = 2e0 − 1 ∈ Λ \ πΛ.Put en+1 := en − p(en)/p′(en) modulo π2n+1

Λ to achieve that e2n − en ∈ π2nΛ. Modulo π2nΛ

we compute(2en − 1)2 = 4e2

n − 4en + 1 ≡ 1 (mod π2nΛ).

Define the sequence (en) ∈ ΛN0 by

en+1 := en = (e2n − en)(2en − 1) = en + kn = 3e2

n − 2e3n

where kn = (e2n − en)(1− 2en).

Claim: For all n ∈ N0 we have e2n − en ∈ π2nΛ and (2en − 1)2 − 1 ∈ π2nΛ.

Proof: This is true for n = 0. If it holds for n then

e2n+1−en+1 = (en+kn)2−(en+kn) = e2

n−2enkn+k2n−en−kn = (e2

n−en)(1+(2en−1)(1−2en))+k2n ∈ π2n+1

Λ.

From this computation we obtain that (en)n∈N is a Cauchy sequence since also kn ∈ π2nΛ.Now K ⊗ Λ is a finite dimensional vector space over the complete field K and hence again

complete (say with respect to the maximum norm, w(∑aiX

i) := minv(ai), but all norms

are equivalent) and therefore (en) converges to some e∞ ∈ Λ with e2∞ = e∞. For this idem-

potent one gets Λ = e∞Λ⊕ (1− e∞)Λ.To obtain the factorization of the polynomial f , let e∞ = a(x)+(f) ∈ Λ, for some a(x) ∈ R[x],then g := ggT(a, f) and h := f

gare the required factors of f in R[x].

Example. Factorise p(x) = x7 − 1 in Z2[x].In Z[x] we compute p(x) = (x− 1)f(x) with f(x) = x6 + x5 + x4 + x3 + x2 + x+ 1. Since F8

contains a 7th root of unity we obtain

f = h0g0 ∈ F2[x] with h0 = x3 + x2 + 1, g0 = x3 + x+ 1.

With the Euclidean algorithm one computes

1 = gcd(h0, g0) = xg0 + (1 + x)h0 also e = xg0.

Put e1 := x4 + x2 + x ∈ Z2[x] then e21 − e1 ≡f −2(x4 + x2 + x+ 1). Put

e2 := 3e21 − 2e3

1 ≡ −x4 − x2 − x− 10(modulo f)

thene2

2 − e2 ≡f 4(5x4 + 5x2 + 5x+ 27).


Put

e3 := 3e22 − 2e3

2 ≡ 595x4 + 595x2 + 595x+ 2178(modulo f)

Since we only need e3 modulo 16 we may reduce coefficients modulo 16 and work withe3 := 3x4 + 3x2 + 3x+ 2. Then

e23 − e3 ≡f −16.

Put

e4 := 3e23 − 2e3

3 ≡ 99x4 + 99x2 + 99x+ 50(modulo f)

Then e24 − e4 ≡f −17152 = 2867. So by accident we have

e2n − en ≡f an ∈ 22n−1Z2

and obtain

en+1 = 3e2n − 2e3

n = −2(e2n − en)en + e2

n ≡f (e2n − en) + (1− 2an)en ≡f an + (1− 2an)en

from which we obtain the recursion (a := an) an+1 = 4a3n − 3a2

n since e2n+1 − en+1 ≡f

(a+(1−2a)en)2−(a+(1−2a)en) = a2−a+2a(1−2a)en−2a(1−2a)e2n+(1−2a)(e2

n−en) = 4a3−3a2.

1.8.3 Extension of valuations.

Lemma 1.8.13. Let (K, v) be a complete discrete valuated field and f(X) = a0Xn+a1X

n−1+. . .+ an−1X + an ∈ K[X] irreducible. Then v(ai) ≥ minv(a0), v(an) for all 0 ≤ i ≤ n.

Proof. Let t := minv(ai) | 0 ≤ i ≤ n and assume that t < minv(a0), v(an). Letr be maximal such that v(ar) = t. Then r 6= 0 and r 6= n and g(X) := a−1

r f(X) =b0X

n + b1Xn−1 + . . . + bn−1X + bn ∈ R[X], br = 1, br+1, . . . , bn ∈ πR and also g(X) ∈ R[X]

is irreducible.The reduction of g modulo π is

g = Xn−r︸︷︷︸g0

(1 + br−1X + . . .+ b0Xr)︸︷︷︸

h0

∈ R/πR[X]

with gcd(g0, h0) = 1. This contradicts the general version of Hensel’s lemma.

Theorem 1.8.14. Let K be a complete discrete valuated field and L/K finite extension ofdegree n = [L : K] Then there is a unique discrete valuation w : L→ 1

nZ∪ ∞ that extends

the valuation of K. This valuation is given by w(α) := 1nv(NL/K(α)) for all α ∈ L and L is

complete.

Proof. Let R := Rv ⊆ K be the valuation ring of K and O := IntR(L) the integral closure ofR in L. So

O = a ∈ L | ∃f ∈ R[X] monic, such that f(a) = 0 = a ∈ L | µa ∈ R[X].


We claim that O = a ∈ L | NL/K(a) ∈ R = a ∈ L | w(a) ≥ 0.If a ∈ L, then µa ∈ K[X] monic and irreducible, so by Lemma 1.8.13

µa ∈ R[X]⇔ µa(0) ∈ R⇔ NL/K(a) ∈ R.

We now show that the map w above is a discrete valuation of L (it clearly extends thevaluation of K). The conditions (o), (i), (ii) are clearly fulfilled by the multiplicativity of thenorm. To show the strong triangle inequality let we need to show that for all α, β ∈ L

w(α + β) ≥ minw(α), w(β)

This is clear if one of them is 0, so assume that both are nonzero and that w(α) ≥ w(β).Then by (ii) w(α

β) ≥ 0 and hence α

β∈ O. But then also α

β+ 1 ∈ O and therefore w(α

β+ 1) =

w(α + β)− w(β) ≥ 0 which proves (iii).So we have established the existence.For the uniqueness we need the following Lemma

Lemma 1.8.15. Let f(X) = Xn+a1Xn−1+. . .+an ∈ K[X] irreducible. Then v(ak) ≥ k

nv(an)

for all 1 ≤ k ≤ n.

Proof. Let L be the splitting field of f and w : L → R ∪ ∞ be the extension of v to Lconstructed above. If f(X) =

∏ni=1(X − βi) ∈ L[X], then w(βi) = 1

nv(an) for all i. The

coefficient ak is a homogeneous polynomial in the βi of degree k, so

v(ak) = w(ak) ≥ kw(βi) =k

nv(an).

Now assume that there is a second (different) extension w′ of the valuation v and chooseα ∈ L such that w(α) 6= w′(α). Wlog we may assume that w(α) < w′(α) (otherwise replaceα by α−1). Let µα := Xm + a1X

m−1 + . . . + am ∈ K[X], then w(α) = 1mv(am) and all

coefficients satisfy v(ak) ≥ kmv(am) = kw(α). Then

w′(akαm−k) = (m− k)w′(α) + v(ak) > mw(α) = v(am) for all k = 0, . . . ,m− 1.

But am = −am−1α− . . .− a1αm−1 − αm and therefore

w′(am) = v(am) ≥ minw′(akαm−k) | k = 0, . . . ,m− 1 > v(am)

a contradiction.

Definition 1.8.16. Let (K, v) be complete, R = Rv, k = R/πR the residue field. Let L/Kbe a finite extension, w : L∗ → R the extension of v, O = Rw the valuation ring with maximalideal ℘O and residue field ` := O/℘O. Then k ≤ `, v(K∗) ≤ w(L∗).[` : k] =: f = f(w/v) is called the inertia degree and[w(L∗) : v(K∗)] =: e := e(w/v) the ramification index of w over v.

Theorem 1.8.17. In the situation of the definition above we have πO = ℘eO and [L : K] =ef .


Proof. Clearly w(L∗) ≤ 1nZ, so w(L∗) = 1

eZ for some divisor e of n and any element ℘ ∈ L

with w(℘) = 1e

is a prime element of O. So πO = ℘xO with x = w(π)/w(℘) = e.To see that [L : K] = ef we construct a K-basis if L. Let (b1, . . . , bf ) ∈ O such that theirimages form a k-basis of `. We claim that

(℘ibj | 0 ≤ i ≤ e− 1, 1 ≤ j ≤ f) is a K-basis of L.

These elements are linearly independent: Assume that there are aij ∈ K such that∑

i,j aij℘ibj =

0 such that not all aij are zero. Put si :=∑

j aijbj. Then not all si are 0 (choose aij ∈ Rand not all in πR and use the fact that the bj form a basis of O/℘O) and if si 6= 0 thenw(si) ∈ v(K∗).From the fact that

∑e−1i=0 si℘

i = 0 and w(si℘i) 6= w(sj℘

j) for all i 6= j for which sisj 6= 0 weobtain that the nonzero summands have distinct valuations and therefore w(

∑e−1i=0 si℘

i) =minw(si℘

i) | 0 ≤ i ≤ e− 1 <∞ a contradiction.Generating set: Put M := 〈℘ibj | 0 ≤ i ≤ e − 1, 1 ≤ j ≤ f〉R. We claim that M = O andhence (℘ibj | 0 ≤ i ≤ e− 1, 1 ≤ j ≤ f) is an integral basis of L.Clearly M + πO = O so

O = M + πO = M + π(M + πO) = M + π2O = . . . = M + πnO for all n ∈ N.

So M is dense in O, R complete and M finitely generated R-module, so also M is completeand so M = O.

1.9 p-adic number fields

Definition 1.9.1. A p-adic number field is a finite extension of Qp.

Note that any p-adic number field K is a complete discrete valuated field. We assume inthe following that K is a p-adic number field with valuation w extending vp and valuationring R and prime element π. The inertia degree is denoted by f and the ramification indexby e. So

d = ef = [K : Qp], FK := R/πR ∼= Fpf , pR = πeR.

Theorem 1.9.2. Let K be a p-adic number field with valuation ring R and prime elementπ. Then

K∗ = 〈π〉 × 〈µq−1〉 × U (1) = 〈π〉 ×R∗

where q = |R/πR|, µq−1 = z ∈ K | zq−1 = 1 ∼= Cq−1, 〈π〉 = πk | k ∈ Z ∼= Z andU (1) = 1 + πR = ker(R∗ → (R/π)∗).

Proof. It suffices to show that Cq−1∼= µq−1 ⊂ K∗. The polynomial Xq−1 − 1 splits com-

pletely in q−1 distinct linear factors in the residue field FK = R/πR. By Hensels lemma thisimplies that all zeros of Xq−1−1 ∈ R[X] already lie in R, so R∗ contains all q−1 roots of 1.

We now aim to obtain an analogue of Dirichlet’s unit theorem for the structure of theunit group of R.

1.9. P-ADIC NUMBER FIELDS 43

Theorem 1.9.3. There is a unique continous group homomorphism log : K∗ → K such thatlog(p) = 0 and log(1 + x) = x− x2

2+ x3

3− . . . for all 1 + x ∈ U (1).

Proof. Since (K,+) has no torsion, we have log(µq−1) = 0 for any group homomorphismlog. To show that the series for log(1 + x) converges note that for 1 + x ∈ U (1) we havew(x) > 0 and so

w(xn

n) = nw(x)− vp(n)→∞ for n→∞

because nw(x) grows linearly in n but vp(n) only logarithmically. Therefore (xn

n)n∈N tends to

zero which is (because of the nice properties of an ultra metric) equivalent to the convergenceof the series. The homomorphism property follows from the identity of formal power series

log((1 + x)(1 + y)) = log(1 + x) + log(1 + y).

Any α ∈ K∗ can be written uniquely as

α = πew(α) ε(α)︸︷︷︸∈µq−1

α︸︷︷︸∈U(1)

.

To define log(π) we first note that the prime element π is not unique, but we have theequation p = πeε(p)p and then put log(π) := −1

elog(p) and hence

log(α) = ew(α) log(π) + log(α).

This defines a continous group homomorphism with log(p) = 0.To see the uniqueness let λ : K∗ → K be a second logarithm such that λ|U(1) = log|U(1) andλ(p) = 0. Then λ(µq−1) = 0 and

0 = λ(p) = eλ(π) + λ(p) = eλ(π) + log(p) implies λ(π) = log(π).

On U (n) the logarithm has a continous inverse, the exponential:

Theorem 1.9.4. For any n > ep−1

=: m the mappings

exp : πnR→ U (n), x 7→ 1 + x+ x2

2+ x3

6+ . . . =

∑∞i=0

xi

i!

log : U (n) → πnR, 1 + x 7→∑∞

i=1xi

i

are mutually inverse continuous group isomorphisms.

Proof. Let w be the unique contiuation of the p-adic valuation vp to K and v := ew be thecorresponding normed valuation, so v(p) = e, v(π) = 1.

(a) log is well defined: We need to show that for v(x) ≥ n and i ∈ N also v(xi

i) ≥ n.

For i = pai′ we have v(i) = evp(i) = ea. For a > 0 (and hence i > 1) we obtain

vp(i)

i− 1=

a

pai′ − 1≤ a

pa − 1=

1

p− 1

a

pa−1 + pa−2 + . . .+ 1≤ 1

p− 1


hence vp(i) ≤ i−1p−1

and so v(i) = evp(i) ≤ m(i− 1) with m = ep−1

as above. Therefore

v(xi

i≥ in−m(i− 1) = (n−m)i+m ≥ n since i ≥ 1, n > m.

(b) exp is convergent and maps U (n) into πnR.Let i = a0 + pa1 + . . .+ prar with 0 ≤ ai < p, si :=

∑rj=0 ai ≥ 1. Then

vp(i!) =i− sip− 1

⇒ v(i!) =e

p− 1(i− si) = m(i− si)

and so v(xi

i!= iv(x)−m(i− si) = i(v(x)−m) + sim ≥ i and therefore exp(x) is convergent.

Moreover for i ≥ 1

v(xi

i!= iv(x)−m(i−si) = v(x)+(i−1)v(x)−(i−si)m ≥︸︷︷︸

si≥1

v(x)+(i−1)(v(x)−m) ≥︸︷︷︸ v(x) ≥ n > mv(x)

so exp(πnR) ⊆ U (n).Now exp log = id and log exp = id since this is an identity of formal power series andhence correct, whenever the series converge.

Theorem 1.9.5. As a Zp-module the group U (1) = 1 + πR ≤ R∗ is canonically isomorphicto

U (1) ∼= Z/paZ× Zdp where Z/paZ = x ∈ R | xp∗ = 1 torsion of U (1)

as a Zp module.

Proof. We first obtain the (continous) Zp-module structure of U (1):The group U (1) is an abelian group and hence a Z-module. Let U (i) := 1 +πiR ≤ U (1). Then

U (1) > U (2) > . . . and U (i)/U (i+1) ∼= (R/πR,+)

since (1 + πia)(1 + πib) = 1 + πi(a + b) + π2iab. So the mapping (1 + πia)U (i+1) 7→ a + πRdefines a group isomorphism U (i)/U (i+1) ∼= (R/πR,+). Now R/πR is a Fp = Z/pZ-module,so U (1)/U (n+1) is a Z/pnZ-module and therefore

U (1) = lim←U (1)/U (n+1) is a Zp = lim

←Z/pnZ module.

More precisely the Zp-action of z = (zi)i∈N ∈ Zp, zi ∈ Z/piZ on U (1) is given by

z ∗ (1 + x) := (1 + x)z := limi→∞

(1 + x)zi .

For any x ∈ πR the mapping z 7→ (1 + x)z,Zp → U (1) is continous: If z ≡ z′ (mod pn), then(1 + x)z ≡ (1 + x)z

′(mod U (n+1)).

To obtain the rank of the Zp-module U (1) note that for n > m = ep−1

the mapping log :

U (n) → πnR is a continous group homomorphism and also a Zp-module homomorphism,since log((1+x)z) = z log(1+x). So U (n) ∼= πnR ∼= Zdp as Zp-module. Since [U (1) : U (n)] <∞we have U (1) ∼= Zdp⊕T with T finite. Torsion in K∗ are roots of unity, and the roots of unity

in U (1) are those that map to 1 mod π and hence these are the p-power roots of unity.

1.9. P-ADIC NUMBER FIELDS 45

Remark 1.9.6. K∗ ∼= Z ⊕ Z/(q − 1)Z ⊕ Z/paZ ⊕ Z[K:Qp]p as Zp-module. Any Zp-module

generating system of K∗ is called a topological generating system.

Example. (Proofs as exercise !!)(a) Let p > 2 be an odd prime. Then Z∗p = Z/(p − 1)Z ⊕ Zp with Zp ∼= U (1) = 〈1 + p〉Zp .e = 1, p− 1 > 1 so n = 1 > e

p−1works here.

(b) For p = 2 there are 2-power roots of 1 in Z∗2 and

Z∗2 = 〈−1〉 × U (2) = 〈−1〉 × 〈1 + 4〉Z2∼= Z/2Z⊕ Z2

(c) Let K = Q5[√

2], so f = 2, e = 1, R = Z5[√

2]. Then K∗ ∼= 〈5〉 × 〈ζ24〉 × U (1) with

U (1) = 〈log(1 + 5), log(1 + 5√

2)〉 ∼= 5R = 〈5, 5√

2〉

indeed U (1) = 〈1 + 5, 1 + 5√

2〉Z5 .(d) Let K = Q5[

√5], so f = 1, e = 2, e

p−1< 1 and therefore

K∗ = 〈√

5〉 × 〈ζ4〉 × 〈1 +√

5, 1 + 5〉Z5 .

(e) Let K = Q3[√

3], so f = 1, e = 2, ep−1

= 1 and

K∗ = 〈√

3〉 × 〈−1〉 × U (1)

but we only know U (2) = 〈1+3, 1+3√

3〉Z3 from the theory. U (1)/U (2) = 1, 1+√

3, 1−√

3 =

〈1 +√

3〉 ∼= C3 with (1 +√

3)3 = 1 + 3√

3 + 3√

32

+√

33 ≡ 1 + 6

√3 modulo U (3), so

U (1) = 〈1 +√

3, 1 + 3〉

(f) Let K = Q3[√−3], so f = 1, e = 2, e

p−1= 1 and

K∗ = 〈√−3〉 × 〈−1〉 × U (1)

but we only know U (2) = 〈1 + 3, 1 + 3√−3〉Z3 from the theory. U (1)/U (2) = 1, 1 +

√−3, 1−√

−3 = 〈1 +√−3〉 ∼= C3. But now (1 +

√−3)3 = 1 + 3

√−3 + 3

√−3

2+√−3

3= −8 so here

U (1) ∼= C3 × U (2).

Corollary 1.9.7. For n ∈ N we have(a) [K∗ : (K∗)n] = npdvp(n)|µn(K)|.(b) [R∗ : (R∗)n] = pdvp(n)|µn(K)|.

As Exercise: explicit examples with n = 2 and n = 3.

1.9.1 Unramified extensions

Definition 1.9.8. Let K be a p-adic number field with valuation ring OK, prime element πK,residue field OK/πKOK =: FK of characteristic p, discrete valuation vK such that vK(K∗) =Z. Let L/K be a finite extension of K, with valuation ring OL, prime element πL, residuefield OL/πLOL =: FL of characteristic p, discrete valuation vL such that (vL)|K = vK.


(a) e(L/K) := vL(πL)−1 = [vL(L∗) : vL(K∗)] is called the ramification index of L overK.

(b) f(L/K) := [FL : FK ] is called the inertia degree of L over K.

(c) L/K is called unramified, if e(L/K) = 1.

(d) L/K is called purely ramified, if f(L/K) = 1.

(e) L/K is called tamely ramified, if p |6 e(L/K).

(f) L/K is called wildly ramified, if p | e(L/K).

Theorem 1.9.9. Let L/K be a finite extension of p-adic fields, q := |FK |, qf := |FL|. Thenthere is a unique subfield K ≤ T ≤ L such that T/K is unramified and [T : K] = f = [FT :FK ]. T := TL/K is called the inertia field of L/K. The field T = ZerfK(Xqf − X) is aGalois extension of K with Galois group Gal(T/K) ∼= Gal(FT/FK) ∼= Cf . Any unramifiedsubfield K ≤M ≤ L with e(M/K) = 1 is contained in T .

Proof. By Hensel’s Lemma all roots of unity in the residue field FL lift to roots of unity inL and hence T := ZerfK(Xqf −X) ≤ L. This extension has degree f over K and is totallyunramified. Totally unramified subfields of L are generated by certain qf − 1 roots of unity(not necessarily primitive) and hence contained in T .

Theorem 1.9.10. Let K be a p-adic number field, |FK | = q. For any f ∈ N there is aunique unramified extension L = TL/K of K of degree f . This is a galois extension given as

L = ZerfK(Xqf −X) and Galois group ∼= Cf . The restriction map

α : Gal(L/K)→ Gal(FL/FK) = 〈Frobq〉, σ 7→ σ|OL mod πLOL

is a group isomorphism. The preimage ˜Frobq of Frobq is a generator of Gal(L/K) and calledthe Frobeniusautomorphism of L over K.

Proof. Clear. The lifting of the Galois automorphisms is proven similarly as in the numberfield case.

Theorem 1.9.11. If L/K is tamely ramified and T := TL/K denotes the inertia field ofL/K, then there is some prime element πT ∈ T such that L = T [ e

√πT ].

Proof. Assume wlog that K = T and let w be an extension of vK to L. Then [w(L∗) :vK(K∗)] = e = [L : K] and for any prime element πL of L we have w(πL) = 1

e. Note that

any prime element πL generates L. We have πeL = πKε for some unit ε ∈ O∗L. Since FK = FLthere is some unit b ∈ O∗K and u ∈ 1 + πLOL such that ε = bu, so πeL = (bπK)u = π′Ku. Thepolynomial f(X) := Xe − u ∈ OL[X] has a zero modulo πL (take 1). Since e is prime to thecharacteristic of FL, the derivative f ′(X) = eXe−1 satisfies f ′(1) = e ∈ O∗L. By Hensel, wemay hence lift 1 to a zero β ∈ O∗L of f(X), so βe = u. Then π′L := πLβ

−1 satisfies (π′L)e = π′K .It is a zero of the Eisenstein polynomial (Xe − π′K) = µπ′L and hence L = K[π′L].

1.10. DIFFERENT AND DISCRIMINANT 47

Remark 1.9.12. The compositum of tamely ramified extensions is again tamely ramified andhence any extension L/K contains a maximal tamely ramified subfield VL/K.

L ≥︸︷︷︸pa

VL/K ≥︸︷︷︸e′

TL/K ≥︸︷︷︸f

K

with f = f(L/K), e = e(L/K) = pae′.

So the tamely ramified extensions of K with ramification index e and inertia degree fcorrespond to O∗T/(O

∗T )e ∼= 〈µq−1〉/〈µeq−1〉 where T is the unramified extension of degree f of

K and q = pf , p = |FK |.Examples K = Q5:

Extensions of degree 2: Q5[√

2] (f=2,e=1), Q5[√

5], Q5[√

10].Extensions of degree 3 Q5[ζ124] (f=3,e=1), Q5[ 3

√5] since Z∗5/(Z∗5)3 = 1.

Exercise: Classify all extensions of degree 4 of Q5.

1.10 Different and discriminant

Let K be a p-adic number field with valuation ring OK , prime element πK and residue fieldFK = OK/πKOK . Let L/K be a finite extension.

Definition 1.10.1. SL/K : L→ K, x 7→ trace(multx) is called the trace of L over K.S : L× L→ K,S(x, y) := SL/K(xy) is called the trace bilinear form.

O#L := x ∈ L | S(x, α) ∈ OK for all α ∈ OL is called the inverse different of L/K.

O#L is a fractional OL-ideal in L, so O#

L = πdLOL for some d ∈ Z, d ≤ 0.The different of L/K is D(L/K) := π−dL OL and the discriminant of L/K is the norm

D(L/K) := NL/K(D(L/K)) = NL/K(a) | a ∈ D(L/K) = πfdk OK EOK .

Theorem 1.10.2. If L/K is unramified then D(L/K) = OL, D(L/K) = OK.

Proof. Let B := (b1, . . . , bn) ∈ OnL be a lift of some FK-basis of FL. Since the trace bilinear

form of FL over FK is non degenerate, the determinant of the Gram matrix of B with respectto S is not a multiple of πL and hence in O∗L. Therefore OL = O#

L .

Theorem 1.10.3. Let K ⊆ L ⊆M . Then

D(M/K) = D(M/L)D(L/K).

Proof. Let O#L := D(L/K)−1 = πaLOL, O#

M := D(M/K)−1 = πcMOM , and D(M/L)−1 =πbMOM . For z ∈M we compute SM/K(zOM) = SL/K(SM/L(zOM)OL) so

z ∈ D(M/K)−1 ⇔ SM/K(zOM) ⊆ OK ⇔ SM/L(zOM) ⊆ D(L/K)−1 = πaLOL

⇔ SM/L(zπ−aL OM) ⊆ OL ⇔ zπ−aL ∈ D(M/L)−1 = πbMOM ⇔ z ∈ πaLπbMOM

So πcMOM = πaLπbMOM = π

b+e(M/L)M OM .


Corollary 1.10.4. Let T := TL/K. Then D(L/K) = D(L/T ).

Theorem 1.10.5. Assume that OL = OK [α] for some α ∈ L and let f :=∑n

i=0 aiXi ∈

OK [X] denote the minimal polynomial of α over K. Then D(L/K) = f ′(α)OL.

Proof. Writef(X)

X − α= b0 + b1X + . . .+ bn−1X

n−1 ∈ L[X].

Then bn−i = αi−1 +an−1αi−2 + . . .+an−i+1 ∈ OL for all i and (b0, . . . , bn−1) is also an OK-basis

of OL. Then we claim that the dual basis of (1, α, . . . , αn−1) is given by 1f ′(α)

(b0, . . . , bn−1) to

deduce that O#L = 1

f ′(α)OL, from which we obtain the theorem. If α1, . . . , αn are the roots of

f then

SL/K(f(X)

X − ααr

f ′(α)) =

n∑i=1

f(X)

X − αiαri

f ′(αi)= Xr for 0 ≤ r ≤ n− 1

as the difference is a polynomial of degree ≤ n− 1 with zeros α1, . . . , αn. Comparing coeffi-cients we find that

SL/K(bjf ′(α

)αi = δij.

for 0 ≤ i, j ≤ n− 1.

Corollary 1.10.6. Let L/K be a totally ramified extension, [L : K] = e(L/K) =: e and letw denote the normalized valuation of L. Then D(L/K) = πsLOL with s = e− 1 if p does notdivide e and

e ≤ s ≤ e− 1 + w(e), if p divides e.

Proof. We have OL = OK [πL] for any prime element πL of L. Moreover w(πL) = 1 andw(K∗) = eZ. Let f :=

∑ei=0 aiX

i be the minimal polynomial of πL over K Then f is anEisenstein Polynomial, i.e. ae = 1, w(a0) = w(NL/K(πL)) = e and the irreducibility of fallows to apply Lemma 1.8.15 to deduce that w(ai) ≥ e for all 0 ≤ i < e. Theorem 1.10.5says that s = w(f ′(πL)) with

f ′(πL) = a1 + 2a2πL + . . .+ (e− 1)ae−1πe−2L + eπe−1

L .

The w-valuations of the summands lie in different congruence classes modulo eZ and hencew(f ′(πL)) is the minimum of these valuations. If w(e) = 0 (i.e. the tamely ramified case)then this minimum is e− 1. Otherwise this minimum s satisfies e ≤ s ≤ e− 1 + w(e).

Corollary 1.10.7. If L/K is ramified of degree ef = n, e = [L/TL/K ] then D(L/K) =πe−1L OL if L/K is tame. If L/K is wildely ramified, then D(L/K) = πsLOL with e ≤ s ≤e− 1 + w(e) where w : L∗ → Z is the normalized valuation of L.

Proof. Because of Corollary 1.10.4 we may assume that K = TL/K and L/K is totally rami-fied of degree e.


Corollary 1.10.8. L/K is ramified if and only if D(L/K) 6= OK.

(i). Q3(√

3)/Q3 is tamely ramified and v3(D(Z3[√

3]/Z3)) = v3(12Z3) = 1.

(ii). Q2(√

3)/Q2 is wildly ramified and v2(D(Z2[√

3]/Z2)) = v2(12Z2) = 2.

(iii). Q2(√

2)/Q2 is wildly ramified and v2(D(Z2[√

2]/Z2)) = v2(8Z2) = 3.

1.10.1 Cyclotomic p-adic fields

Theorem 1.10.9. Let m ≥ 1 and ζ := ζpm. Then

(a) Qp[ζ]/Qp is totally ramified of degree [Qp[ζ] : Qp] = ϕ(pm) = (p− 1)pm−1.

(b) Gal(Qp[ζ]/Qp) ∼= (Z/pmZ)∗.

(c) π := (ζ − 1) is a prime element of Qp[ζ] with norm N(π) = p.

(d) vp(D(Qp[ζ]/Qp)) = pm−1(mp−m− 1).

Proof. ζ is a zero of

h(X) := X(p−1)pm−1

+X(p−2)pm−1

+ . . .+ 1 ∈ Qp[X].

Put g(X) := h(X + 1). Then g(π) = h(ζ) = 0, g(0) = h(1) = p. As

h(X) = (Xpm − 1)/(Xpm−1 − 1) ≡ (X − 1)(p−1)pm−1

(mod pZp[X])

the polynom g is an Eisenstein polynomial and hence irreducible. We hence conclude (a) and(c). Also (b) follows from the irreducibility of h, as the zeros of h are exactly the powers ζa

with a ∈ 1, . . . , pm not divisible by p. The valuation ring of Qp[ζ] is Zp[π] = Zp[ζ], so we maycompute the discriminant as D(Qp[ζ]/Qp) = h′(ζ)Zp[ζ]. Now h(X) = (Xpm−1)/(Xpm−1−1)so

h′(X) =1

Xpm−1 − 1(pmXpm−1 − pm−1Xpm−1−1h(X)

and h′(ζ) = pmζpm−1

ζpm−1−1. Now η := ζp

m−1is a primitive p-th root of unity, so NQp[η]/Qp(η− 1) = p

and hence NQp[ζ]/Qp(η − 1) = ppm−1

. Therefore vp(N(h′(ζ)) = pm−1(p − 1)m − pm−1 =pm−1(mp−m− 1).

Corollary 1.10.10. Let n = pmk ∈ N such that p does not divide k and ζn be a primitiven-th root of unity..

(a) Zp[ζn] is the valuation ring of Qp[ζn].

(b) T := T (Qp[ζn]/Qp) = Qp[ζpm

n ] is the maximal unramified subfield.

(c) f = [T : Qp] is the order of p in (Z/kZ)∗.

(d) e(Qp[ζn]/Qp) = (p− 1)pm−1.

(e) V (Qp[ζn]/Qp) = Qp[ζpm−1

n ] is the maximal tamely ramified subfield.


1.10.2 An application to algebraic number fields.

Let (K, v) be a discretely valuated field with completion Kv. Then v : Kv → Z ∪ ∞ hasa unique extension to a valuation v of the algebraic closure Kv. Now let L/K be a finteextension. Any embedding τ : L → Kv defines a valuation wτ = v τ of L that extends v.The completion of L at w is then Lw = L⊗KKv as the latter is complete with a dense subsetL.

Theorem 1.10.11. All extensions of v to L are of the form wτ = v τ . We have thatwτ = wτ ′ if and only if τ ′ = σ τ for some σ ∈ AutKv(Kv) (then we say that τ and τ ′ areconjugate over Kv).

Proof. Let w be an extension of v to L with corresponding completion Lw; view w as thevalutaion of Lw extending v : Kv → Z ∪∞. As Lw is an algebraic extension of Kv and theuniqueness of extension of valuations for complete fields, we have w = v τ for all embed-dings τ ∈ HomKv(Lw, Kv). Any other such embedding is of the form στ as in the theorem.

Assume that L/K is separable with primitive element α ∈ L, so L = K(α). Let f :=µα,K ∈ K[X] denote the minimal polynomial of α. Then f = g1 · · · gr ∈ Kv[X].

Corollary 1.10.12. The valuations w1, . . . , wr of L that extend v are in bijection with theirreducible factors of f ∈ Kv[X]. For a ∈ L we have

L⊗K Kv =r⊕i=1

Lwi , NL/K(a) =r∏i=1

NLwi/Kv(a), and SL/K(a) =

r∑i=1

SLwi/Kv(a).

Proof. Any K-linear embedding of L into K is uniquely determined by mapping α to somezero β of f . Two such embeddings are conjugate over Kv if and only if these zeros are zerosof the same irreducible factor gi. Clearly

L⊗K Kv∼= Kv[X]/((f(X)) ∼=

r⊕i=1

Kv[X]/(gi(X)) ∼=r⊕i=1

Lwi .

The characteristic polynomial of any a ∈ L is the product of the characteristic polynomialsof the corresponding elements ai ∈ Lwi where (a1, . . . , ar) denotes the image of a ⊗ 1 underthe above isomorphism. From this we obtain the equations for norm and trace.

Let K be an algebraic number field with ring of integers R. Any prime ideal P ER of Rdefines a valuation

vP : K∗ → Z, vP (α) := maxa ∈ Z | α ∈ P a

with valuation ring R(P ) := ab∈ K = Quot(R) | b 6∈ P. The completion KP of K at vP is

a p-adic number field, where pZ = P ∩ Z. If P (R) denotes the set of all maximal ideals ofR, then

R =⋂

P∈P (R)

R(P ).

We apply the above to the separable extension L/K of algebraic number fields:


Remark 1.10.13. Let P ER be a maximal ideal of R. Then

PZL = ℘e11 . . . ℘err

for pairwise distinct prime ideals ℘i E ZL and the inequivalent valuations of L that extendv := vP are

w1 =1

e1

v℘1 , . . . , wr =1

erv℘r .

Then ei is the ramification index of Lwi over Kv and fi := [ZL/℘i : ZK/P ] the inertia degreeof Lwi over Kv. As [Lwi : Kv] = eifi we re-obtain the formula

[L : K] =r∑i=1

[Lwi : Kv] =r∑i=1

eifi.

Corollary 1.10.14. Let M := ZL, R = ZK PZL = ℘e11 . . . ℘err as before. Then

RP ⊗R D(M/R) =r∏i=1

D(Mwi/RP )

andD(M/R) =

∏℘∈P (M)

M ∩ D(M℘/R℘∩R).

In particular one may read of the ℘-component of D(M/R) from D(M℘/R℘∩R).

Proof. More details will be given later, when we treat local-global properties of lattices inPart II.

Corollary 1.10.15. Let K be an algebraic number field. Then the rational prime p ramifiesin K if and only if p divides the discriminant of K.

Chapter 2

Non-commutative theory.

2.1 Central simple algebras.

This section recalls some of the relevant theory on simple algebras. For more details I referto the Algebra skript, in particular to Section 5.

2.1.1 Simple algebras.

Let K be a field and A a finite dimensional K-algebra. Then A is called a division algebra,if any non-zero element in A has an inverse. A is called simple, if A has no non-trivial 2-sidedideals and central simple, if additionally Z(A) = K. The center of any simple K-algebrais an extension field of K, so any simple K-algebra A is a central simple Z(A)-algebra.

One frequently used fact is the following quite easy remark:

Definition 2.1.1. Let A be a K-algebra. Then A is also a left A-module AA, called theleft regular module. The left regular representation ρA : A → EndK(A) is defined byρA(a) : x 7→ ax. The regular norm is NA/K : A → K, a 7→ det(ρA(a)) and the regulartrace is TA/K : A→ K, a 7→ trace(ρA(a)).

Remark 2.1.2. A ∼= EndA(AA)op by a 7→ (x 7→ xa).

Theorem 2.1.3. Any simple K-algebra A is semi-simple, more precisely A ∼= Dn×n for somedivision algebra D, n ∈ N.

Proof. Let M ≤A A be a simple submodule of the regular left A-module AA. Then for anya ∈ A the image Ma = 0 or simple. Moreover

∑a∈AMa E A is a non-zero two-sided ideal

of A. Since A is simple∑

a∈AMa = A. So AA is the sum of simple modules and hencecompletely decomposable, i.e. a semi-simple A-module. This implies that A is semisimple.By Wedderburns characterisation of semisimple algebras we hence obtain A ∼=

⊕si=1D

ni×nii

for division algebras Di. But A is simple, so s = 1.Many facts follow from the useful “double-centralizer-theorem”:

Theorem 2.1.4. Let B be a semisimple subalgebra of the matrix ring Kn×n. Let

C = CKn×n(B) := x ∈ Kn×n|xb = bx for all b ∈ B

be the centralizer of B in Kn×n. Then

52

2.1. CENTRAL SIMPLE ALGEBRAS. 53

(i) C is semi-simple.

(ii) B ∩ C = Z(B).

(iii) CKn×n(C) = B (double-centralizer-theorem)

(iv) B simple ⇔ C simple.

(v) If B ∼= Dβ×β simple, then C ∼= (Dop)γ×γ for some γ ∈ N and Z := B ∩ C = Z(B) =Z(D). Moreover

B · C = 〈b · c | b ∈ B, c ∈ C〉

is a simple subalgebra of Kn×n with B · C ∼= Zk×k for some k ∈ N (k = βγ`2, where`2 = dimZ(D)).

Proof. The proof is somehow technical, I refer to the algebra skript or the literature.

If A1 and A2 are K-algebras then also the tensor product A1 ⊗A2 is a K-algebra, wherethe multiplication is defined by the bilinear extension of

(a1 ⊗ a2)(a′1 ⊗ a′2) := a1a′1 ⊗ a2a

′2.

Examples

A1 = Kn×n, A2 = Km×m ⇒ A1 ⊗ A2∼= Knm×nm(Kronecker product)

D ⊗K Kn×n ∼= Dn×n

If A/K is a finite Galois extension of degree n, then

A⊗K A ∼= A⊕ · · · ⊕ A

a⊗ b 7→ (aσ1(b), · · · , aσn(b))

where Gal(A/K) = σ1, · · · , σn.e.g. A = K[x]/(p(x))

A⊗ A ∼= A⊗K K[x]/(p(x)) ∼= A[x]/(p(x)) ∼=⊕i

A[x]/(x− xi)

if p(x) =∏

(x− xi) in A[x]. In particular A⊗K A is not simple (if n > 1).

Lemma 2.1.5. Let A,B be simple K-algebras, Z(A) = K · 1 Then A⊗K B is simple.

Proof. For a proof I refer to the algebra skript.

Corollary 2.1.6. The dimension of any central simple K-division algebra D is a square:dimK D = n2. The number n is called the index of D.

54 CHAPTER 2. NON-COMMUTATIVE THEORY.

Proof. Let K be the algebraic closure of K. Then K ⊗K D is a dimK(D)-dimensional simpleK-algebra, so isomorphic to Xn×n for some n and some finite dimensional K-divisionsalgebraX. Since K is algebraically closed, any such divisionalgebra is isomorphic to K (the matrix

describing the left multiplication by a ∈ D on DD has an eigenvalue). So K ⊗K D ∼= Kn×n

and dimK(D) = n2.

In particular the dimension of any central simple K-algebra is a square. A = Dk×k ⇒dimA = k2(Index(D))2.

Theorem 2.1.7. Let D be a division algebra with Z(D) = K and dimK(D) = n2. Then anysubfield of D is contained in some maximal subfield of D. Let L ≤ D be such a maximalsubfield. Then

(i) CD(L) = L

(ii) D ⊗K L ∼=L−Algebra Ln×n (L is a splitting field).

(iii) dimKL = n

Proof. (i) Let L1 be a subfield of D. If L1 is not maximal, then CD(L1)⊃6= L1. Choose

x ∈ CD(L1)− L1, L2 = L1[x] etc. until CD(Li) = Li.(ii) and (iii) follow from the double centraliser theorem. (Exercise)

2.1.2 The theorem by Skolem and Noether.

Theorem 2.1.8. (Skolem-Noether) Let A be a central simple K-algebra, B1, B2 simple sub-algebras of A and ϕ : B1 → B2 an algebra homomorphism. Then there is some a ∈ A∗, suchthat ϕ(b1) = a−1b1a for all b1 ∈ B1.

Proof. Let V be a simple A-module, Dop := EndA(V ). Wlog V = Dk×1, dimK(D) = d,n = dk, so B1 ⊆ A → Kn×n. Let C := CKn×n(A) ∼= Dop. Then V is a simple AC-moduleand AC ∼= Kn×n. Turn V into a B1C-module in two different ways:

(i). (b1c)v := b1cv for all v ∈ V, b1 ∈ B1, c ∈ C

(ii). (b1c)v := ϕ(b1)cv for all v ∈ V, b1 ∈ B1, c ∈ C

Since B1C is simple, there is up to isomorphism a unique module of given dimension. Sothere is some a ∈ Kn×n, such that

a−1b1ca = ϕ(b1)c for all b1 ∈ B1, c ∈ C.

Choosing b1 = 1, one gets a ∈ CKn×n(C) = A and choosing c = 1 yields a−1b1a =ϕ(b1) for all b1 ∈ B1.

One important consequence is the following theorem (also attributed to Wedderburn).

Theorem 2.1.9. (Wedderburn) Any finite dimensional division algebra over a finite field iscommutative.

2.1. CENTRAL SIMPLE ALGEBRAS. 55

Proof. Let D 6= Z(D) = Fq be a finite division algebra. Any element of D is contained insome maximal subfield (of order qn). Any two maximal subfields are isomorphic, so by thetheorem of Skolem and Noether these are conjugate in D∗. Hence the finite group D∗ is aunion of conjugacy classes of subgroups. But for a general finite group G we have

G =⋃g∈G

Hg, H ≤ G, [G : H] <∞⇒ G = H

since there are [G : NG(H)] conjugate subgroups of H and |⋃g∈GH

g| ≤ 1 + (|H| − 1) |G||H| =

1 + |G| − |G||H| < |G| if H 6= G.

2.1.3 The Brauer group of K

To define the Brauer group of K we need the following two properties:

Theorem 2.1.10. Let A,B be central simple K-algebras. Then also the tensor productA⊗K B is a central simple K-algebra.

Lemma 2.1.11. Let D be a central simple K-divisionalgebra. Then

D ⊗K Dop ∼= Kn×n

Proof. DD = V ,D ⊆ EndK(V ) ≡ Kn×n, C := CKn×n(D) ∼= Dop, D⊗KDop ∼= D ·C = Kn×n

We now define the Brauer group of K. The elements are the Morita equivalence classes ofcentral simple K-algebras. The multiplication is given by the tensor product.

Definition 2.1.12. Let A and B be central simple K-algebras. A ∼ B are called Brauer-equivalent) ⇔ there are n,m ∈ N such that An×n ∼= Bm×m.

Br(K) := [A] | A central simple K-algebra

is called the Brauergroup of K. The multiplication on Br(K) is defined as [A][B] :=[A⊗K B].

Theorem 2.1.13. Br(K) is a commutative group with

[A]−1 = [Aop] , [K] = 1

Any element of Br(K) is represented by a unique division algebra.

Some examples:If K s algebraically closed then Br(K) = 1.If K is finite, then Br(K) = 1.Br(R) ∼= C2. (Exercise.)


Remark 2.1.14. Let L/K be an extension of fields. Then there is a group homomorphism

Br(K)→ Br(L), [A] 7→ [L⊗K A].

Its kernel is called the relative Brauer group

Br(L/K) := [A] ∈ Br(K) | L⊗K A ∼= Ln×n for some n

Example. Let D be a central K-division algebra and L a maximal subfield of D. Then[D] ∈ Br(L/K).

Ende am 10.10.11

2.2 Orders in separable algebras.

Definition 2.2.1. A K-algebra A is called separabel over K, if there is some separableextension field L of K such that L ⊗K A ∼=L−Algebra

⊕ki=1 L

ni×ni. Then Sp : A → L : a 7→∑ki=1 Sp(ai) (ai ∈ Lni×ni) is called the reduced trace and N : a → L : a 7→

∏ki=1 det(ai)

the reduced norm.

Remark: Reduced norm and trace take only values in K.Remark: Separable algebras are those semisimple algebras A =

⊕si=1D

ni×nii for which

all Zi = Z(Di) are separable over K.

Let R be a Noetherian integral domain with field of fractions K and A a finite dimensionalK-algebra. We assume that R is integrally closed in K, so

R = IntR(K) = a ∈ K | a is integral over R = a ∈ K | R[a] is finitely generated R−module.

One typical example is the ring of integers in a number field. The other example is a discretevaluation ring.

Definition 2.2.2. Let V be a finite dimensional K-vector space. An R-lattice L in V is afinitely generated R-submodule of V , that spans V as a K-vector space.An R-order Λ in A is a lattice in A that is also a subring (multiplicatively closed).An R-order Λ is called a maximal R-order, if for any order Γ in A with Λ ≤ Γ it holdsthat Λ = Γ.

Remark 2.2.3. Any K-algebra contains R-orders: Let B := (b1, . . . , bn) be some K-basis ofA and put L := 〈b1, . . . , bn〉R the R-lattice generated by this basis. Then the left-order

Ol(L) := a ∈ A | aL ⊆ L

of L is an R-order in A, similarly the right-order Or(L) := a ∈ A | La ⊆ L.

Note that L is an order ⇔ L = Or(L) = Ol(L).

Definition 2.2.4. An element a ∈ A is called integral over R, if R[a] is a finitely generatedR-module. IntR(A) := a ∈ A | a integral over R is called the integral closure of R in A.

2.2. ORDERS IN SEPARABLE ALGEBRAS. 57

In general IntR(A) is not a ring. It is a ring, if A is commutative. It is also a ring if A isa division algebra over a complete field (see below).

Remark 2.2.5. For any R-order Λ and any element λ ∈ Λ the ring R[λ] ⊆ Λ is a submoduleof the f.g. module Λ and hence again finitely generated. In particular any element of Λ isintegral over R.

Theorem 2.2.6. Let λ ∈ A. Then λ is integral over R if and only if the minimal polynomialof λ lies in R[X].

Proof. Let R[λ] = 〈b1, . . . , bn〉R and λbi =∑n

j=1 aijbj with M := (aij) ∈ Rn×n. Since 1 ∈ R[λ]the minimal polynomial of M is equal to the minimal polynomial of λ. As a monic divisor ofthe characteristic polynomial of M , which is a monic polynomial in R[X], also this minimalpolynomial is in R[X].

Theorem 2.2.7. Any separable algebra has some maximal order.

Proof. Let Λ be some R-order in A. Then

Λ ⊆ Λ# := a ∈ A | Sp(ax) ∈ R fur alle x ∈ Λ.

Any overorder ∆ of Λ is also an integral lattice with respect to the trace bilinear form, so

Λ ⊆ ∆ ⊆ ∆# ⊆ Λ#.

Not Λ#/Λ. is a finitely generated R-module and hence does not contain infinite extendingchains of submodules since R is Noetherian. Therefore there is some maximal order thatcontains Λ.

Example. A = 〈1, i, j, k〉Q with i2 = j2 = k2 = −1, ij = −ji = k. Λ = 〈1, i, j, k〉Z,Λ# = 1

2Λ, Γ = 〈1, i, j, 1

2(1 + i+ j + k)〉 is maximal order of A that contains Λ.

In the non-separable algebra A = ∆2(Q) = (a b0 c

) ≤ Q2×2 does not contain maximal

orders: The Z-orders

Λn :=

(Z 1

2nZ

0 Z

)form an infinite ascending chain of orders in A.

2.2.1 Being a maximal order is a local property

In this section we show that for a Dedekind domain R any R-order Λ (in the separableK-algebra A) is maximal if and only if all its localizations

(R \ ℘)−1Λ := rsλ | r, s ∈ R, s 6∈ ℘, λ ∈ Λ

at all prime ideals ℘ E R are maximal orders if and only if all its completions R℘Λ aremaximal orders for all prime ideals ℘ of R.


Recall that any prime ideal ℘ defines a discrete valuation v℘ : K∗ → Z, such that thecorresponding discrete valuation ring is the localisation

R℘ := (R \ ℘)−1R = Rv℘ = x ∈ K | v℘(x) ≥ 0.

The completion of R℘ is denoted by R℘.

Definition 2.2.8. A property is called a local property, if it holds for an R-module Mif and only if it holds for all the localisations (R \ ℘)−1M if and only if it holds for allcompletions R℘M for all prime ideals ℘ of R.

Equality of lattices is a local property:

Theorem 2.2.9. Let R be a Dedekind domain with field of fractions K. Let V be a finitedimensional K-vector space and let L,M be two R-lattices in V . Then the following areequivalent:(1) L = M .(2) L℘ := (R \ ℘)−1L = (R \ ℘)−1M =: M℘ for all maximal ideals ℘ER.

(3) R℘ ⊗ L = R℘ ⊗M for all maximal ideals ℘ER.

Proof. (1) ⇒ (2) ⇒ (3) is clear.To see that (3) implies (1) we use contraposition:So assume that L 6= M , wlog L 6⊆ M and let ` ∈ L \M . Multiply ` with some element ofR to achieve that ` 6∈ M but ℘` ⊆ M for some maximal ideal ℘ of R. Then ` 6∈ R℘ ⊗M so

R℘ ⊗ L 6= R℘ ⊗M for this prime ideal ℘ER.

Theorem 2.2.10. Let R be a Dedekind domain, V a K-vectorspace and L some R-lattice inV .(a) For any R-lattice M in V we have M℘ = L℘ for all but finitely may maximal ideals ℘ ofR.(b) Let X(℘) be an R℘-lattice in V for all maximal ideals ℘ of R such that X(℘) = L℘ forall but finitely may ℘. Then M :=

⋂℘X(℘) is a lattice in V such that M℘ = X(℘) for all ℘.

(c) Let L℘ denote the completion L℘ := R℘ ⊗ L which is a lattice in V℘ := K℘ ⊗ V . Then

(i) L = V ∩ (⋂℘ L℘).

(ii) Let X(℘) be an R℘-lattice in V℘ for all maximal ideals ℘ of R such that X(℘) = L℘ for

all but finitely may ℘. Then M := V ∩⋂℘ X(℘) is a lattice in V such that M℘ = X(℘) for

all ℘.(d) Let ℘ be some maximal ideal in R. Then there are bijections

M ≤ L | L/M is ℘-torsion → M ≤ L℘ |M full lattice → M ≤ L℘ |M full lattice

M 7→M℘ 7→ M℘ with inverse mapping M℘ 7→ L ∩M℘ and similarly M℘ 7→ L ∩M℘.

Proof. (Exercise, see Theorem (4.22) and (5.3) in Reiner’s book.)

2.3. DIVISION ALGEBRAS OVER COMPLETE DISCRETE VALUATED FIELDS. 59

Theorem 2.2.11. Let R be a Dedekind domain with field of fractions K and let A be aseparable K-algebra. Let Λ be an R-order in A. Then the following are equivalent:(1) Λ is a maximal R-order in A.(2) (R \ ℘)−1Λ is a maximal R℘-order in A for all maximal ideals ℘ER.

(3) R℘ ⊗ Λ is a maximal R℘-order in K℘ ⊗ A for all maximal ideals ℘ER.

Proof. Now (3) ⇒ (1) is clear so we need to show (1) ⇒ (2) ⇒ (3).Assume that Λ is a maximal R-order in A and that there is some maximal ideal ℘ER suchthat Λ℘ = (R \ ℘)−1Λ is not a maximal order, so there is some R℘-order Γ that properlycontains Λ℘. Choose α ∈ R such that αΓ ⊆ Λ℘. Then ∆ := Λ∩αΓ is a submodule of Λ suchthat Λ/∆ is ℘-torsion. Since αΓ is a two-sided Λ℘-ideal, also ∆ is a two-sided Λ-ideal in Λ.Since Λ is a maximal order, the right order Or(∆) conincides with Λ. Hence

Λ℘ = Or(∆)℘exerc= Or(∆℘) = Γ.

To see that (2) implies (3) we show that Λ℘ is a maximal R℘-order in A if and only if Λ℘ is

a maximal R℘-order in A⊗ K℘. Again the maximality of Λ℘ implies the one of Λ℘. On the

other hand assume that there is some R℘-order Γ ⊇ Λ℘. Then Γ := Γ ∩ A is an R℘-order in

A that contains Λ℘, hence Γ = Λ℘. Since Γ is just the completion of Γ∩A we obtain Γ = Λ℘.

2.3 Division algebras over complete discrete valuated

fields.

2.3.1 General properties.

Let (K, v) be a complete discrete valuated field, v(K∗) = Z, R = Rv := x ∈ K | v(x) ≥ 0the corresponding valuation ring with prime element π ∈ R, v(π) = 1 and maximal ideal πR.Let D be a K-division algebra with [D : K] = m.

Definition 2.3.1. Define ω : D → 1mZ ∪ ∞ by ω(a) := m−1v(ND/K(a)) where ND/K is

the regular norm.

Remark 2.3.2. Let 0 6= a ∈ D and µa be the minimal polynomial of ρD(a). Then ω(a) =[K(a) : K]−1v(µa(0)) = [K(a) : K]−1v(NK(a)/K(a)).

Proof. Since D is a skew field, the minimal polynomial µa is irreducible and χa = µm/na with

n = [K(a) : K]. Then ND/K(a) = (−1)mµa(0)m/n and

ω(a) =1

mv(ND/K(a)) =

1

nv(µa(0)) = [K(a) : K]−1v(NK(a)/K(a)).

Theorem 2.3.3. An element a ∈ D is integral over R, if and only if ND/K(a) ∈ R, if andonly if ω(a) ≥ 0.


Proof. ⇒ is clear.⇐: Assume that ω(a) ≥ 0, so ND/K(a) ∈ R and hence µa = xn +a1x

n−1 + . . .+an−1x+an ∈K[x] satisfies that an ∈ R. Since D is a division algebra, the minimal polynomial is irre-ducible, so by 1.8.13 all coefficients satisfy v(ai) ≥ minv(a0), v(an) ≥ 0 for all 0 ≤ i ≤ nand hence µa ∈ R[x].

Theorem 2.3.4. ω : D → 1mZ is a discrete valuation of D that extends v.

Proof. Clearly ω is a well defined map that extends v. To show that ω is a discrete valuation(on the not necessarily commutative division algebra D) we need to show that

(i) ω(a) =∞ if and only if a = 0.

(ii) ω(ab) = ω(a) + ω(b) for all a, b ∈ D.

(iii) ω(a+ b) ≥ min(ω(a), ω(b)) for all a, b ∈ D.

The first property follows from the fact that for non zero a ND/K(a)ND/K(a−1) = 1 andhence ND/K(a) = 0 if and only if a = 0. The second porperty follows because the norm ismultiplicative. To see (iii) we may assume that a and b are non-zero, ω(b) ≥ ω(a) and divideby a, so we assume that a = 1 and ω(b) ≥ 0. But then b is integral over R, so also 1 + b isintegral over R, i.e. ω(1 + b) ≥ 0 = min(ω(1), ω(b)).

Theorem 2.3.5. ω is the unique valuation on D that extends v.

Proof. Assume that there is a second (different) extension ω′ of the valuation v and chooseα ∈ D such that ω(α) 6= ω′(α). Wlog we may assume that ω(α) < ω′(α) (otherwise replaceα by α−1). Let µα := Xn + a1X

n−1 + . . . + an ∈ K[X], then ω(α) = 1nv(an) and by Lemma

1.8.15 all coefficients satisfy v(ak) ≥ knv(an) = kω(α). Then

ω′(akαn−k) = (n− k)ω′(α) + v(ak) > nω(α) = v(an) for all k = 0, . . . , n− 1.

But an = −an−1α− . . .− a1αn−1 − αn and therefore

ω′(an) = v(an) ≥ minω′(akαn−k) | k = 0, . . . , n− 1 > v(an)

a contradiction.

Definition 2.3.6. Let ∆ := a ∈ D | ω(a) ≥ 0 be the valuation ring of ω.

Corollary 2.3.7. ∆ = IntR(D) is the unique R-maximal order in D.

Definition 2.3.8. The unique value e ∈ N such that ω(D∗) = 1eZ is called the ramification

index of D over K, e = e(D/K).The valuation vD := eω : D → Z ∪ ∞ is called the normalized valuation of ω.Let πD ∈ D be a prime element, i.e. vD(πD) = 1. Then πD generates the unique maximalideal of ∆ and ∆ := ∆/πD∆ is a skew field. The dimension of ∆ over R := R/πR is calledthe inertia degree of D over K.


Similarly as Theorem 1.8.17 we obtain the following result, the proof of which is left asan exercise:

Theorem 2.3.9. ef = [D : K] = m. More precisely let (a1, . . . , af ) ∈ ∆ such that(a1, . . . , af ) form an R-basis of ∆. Then

(aiπjD | 1 ≤ i ≤ f, 0 ≤ j ≤ e− 1)

is an R-basis of ∆.

Proof: Either blackboard or exercise.Example. D = 〈1, ω, i, iω〉Q3 such that ω2 +ω+1 = 0, i2 = −1, ωi = ω2. Then e = f = 2

and ∆ = 〈1, ω, i, iω〉Z3 with prime element πD = 1− ω.With respect to the Z3-basis (1, πD, i, iπD) the right-regular representation (mapping a ∈

∆ to ρ(a) : x 7→ xa ∈ EndZ3(∆) ∼= Z4×43 ) is given by

ρ(πD) =

0 1 0 0−3 3 0 00 0 0 10 0 −3 3

, ρ(i) =

0 0 1 00 0 3 −1−1 0 0 0−3 1 0 0

, ρ(iπD) = ρ(i)ρ(πD) =

0 0 0 10 0 3 00 −1 0 0−3 0 0 0

and the regular norm

ND/Q3(a0 + a1πD + a2i+ a3iπD) = det(a0 + a1ρ(πD) + a2ρ(i) + a3ρ(iπD) =(a2

0 + 3a0a1 + 3a21 + a2

2 + 3a2a3 + a23)2 = N(a0 + a1πD + a2i+ a3iπD)2

In particular the regular norm lies in Z3 if and only if the reduces norm is in Z3. LetL := (∆, N) denote the Z3-lattice in the quadratic space (D ∼= Q4

3, N). Then the Grammatrixof the associated bilinear form

(x, y) 7→ N(x+ y)−N(x)−N(y) = Sp(xy)

with respect to the basis above is given as2 3 0 03 6 0 00 0 2 30 0 3 6

∼Z3 diag(2, 3, 2, 3).

In particular L#/L ∼= F23 is a quadratic F3-space with induced quadratic form x2 +y2. This is

anisotropic (i.e. there is no non-zero (x, y) ∈ F23 such that x2 + y2 = 0) so for any a ∈ ∆# \∆

the reduced norm N(a) does not lie in Z3 and therefore L = (∆, N) is a maximal integrallattice and in particular ∆ is a maximal order in D.

2.3.2 Finite residue class fields.

In the case where R = R/πR is finite we obtain many more nice properties since then∆/πD∆ is a finite skew field, hence a field of which the isomorphism type is determined byits dimension.


So assume from now on that (K, v) is a complete discrete valuated field, v(K∗) = Z,R = Rv := x ∈ K | v(x) ≥ 0 the corresponding valuation ring with prime element π ∈ R,v(π) = 1 and maximal ideal πR. Assume additionally that R = R/πR is finite, so R ∼= Fq.Let D be a K-division algebra with [D : K] = n and ramification index e and let

ω : D∗ → 1

eZ, ω(a) := n−1v(ND/K(a))

be the unique extension of v to D. Then D is a complete discretely valuated skew field withvaluation ring

∆ := a ∈ D | ω(a) ≥ 0 = IntR(D)

the unique maximal R-order in D. Choose πD ∈ ∆ with ω(πD) = 1e. Then πD∆ is the unique

maximal ideal in ∆ and ∆ := ∆/πD∆ is a finte skew field of dimension f = m/e over Fq, so∆ ∼= Fqf is a field.

From algebraic number theory I we know the following structure theorem in the commu-tative case, which will be also relevant in the non-commutative case (including the proof).

Theorem 2.3.10. Assume that D is commutative, hence an extension field of K. Then(1) There is a unique subfield K ≤ L ≤ D with [L : K] = f(L/K) = f(D/K) = f (themaximal unramified subfield or inertia field of the extension). L ∼= K[ζqf−1] is the splitting

field of the polynomial Xqf−1 − 1 ∈ K[X].(2) In particular if f = m, then D = L ∼= K[ζqf−1] and the regular norm ND/K yields a groupepimorphism ND/K : ∆∗ → R∗.(3) Let L be as in (1). Then D/L is totally ramified, e(D/L) = e(D/K) = [D : L] = e,D = L[πD] and the minimal polynomial of πD ∈ L[X] is an Eisensteinpolynomial

µπD = Xe + a1Xe−1 + . . .+ ae−1X + ae ∈ ZL[X], π | ai for all i, π2 6| ae.

For a proof we refer to the algebraic number theory I course or the exercises.We now treat the case where Z(D) = K, so D is a central simple K-division algebra.

2.3.3 The central simple case: analysis

Theorem 2.3.11. Assume that Z(D) = K.(a) [D : K] = m2 and e(D/K) = f(D/K) = m.(b) The maximal totally unramified subfields (inertia fields) of D are conjugate in D andof degree m over K.

Proof. (a) Let a ∈ ∆ be such that ∆ = R[a]. Then K[a] is a commutative subfield of D, so[K[a] : K] ≤ m. On the other hand

m ≥ [K[a] : K] ≥ [R[a] : R] = f

so f ≤ m. Similarly

m ≥ [K[πD] : K] ≥ [ω(K[πD]∗) : ω(K∗)] = e.


Since ef = m2 we obtain e = f = m.(b) ∆ ∼= Fqf = Fq[a] for some a ∈ ∆ such that a is a primitive qf − 1 root of unity. Since∆ is complete we obtain as in the commutative case (using Hensels Lemma) an elementz ∈ R[a] ⊆ ∆ with z = a and zq

f−1 = 1, so z is a primitive qf − 1 root of 1 in ∆. The fieldL := K[z] is a maximal subfield of D that is totally unramified over K.Now let M be any unramified subfield of D, f ′ := [M : K]. Then Fqf ′ = RM/πRm ≤ ∆/πD∆so f ′ divides f and M is isomorphic to the unique subfield L′ with K ≤ L′ ≤ L and[L′ : K] = f ′. By the theorem of Skolem and Noether there is some a ∈ D∗, such thataMa−1 = L′. Then a−1La ∼= L is a maximal subfield of D that is totally unramified, conju-gate to L and contains M .

Theorem 2.3.12. Assume that Z(D) = K, [D : K] = m2, and let ζ ∈ D be a primitiveqm − 1-st root of unity. Then there is some prime element π′D ∈ ∆ such that

(π′D)m = π and π′Dζ(π′D)−1 = ζqr

for some 1 ≤ r ≤ m, gcd(r,m) = 1

The value of r does not depend on the choice of ζ, π′D, and π and hence only depends on D.rm

is called the Hasse-invariant of D.

Proof. (1) By Skolem-Noether there is some α ∈ D∗ such that αζα−1 = ζq. Wlog

α ∈ ∆, vD(α) = mω(α) =: j ∈ 0, . . . ,m− 1.

Put h := mgcd(j,m)

and h := 1 if j = 0 (which will not happen). Then αh = επb for someb ∈ N0, ε ∈ ∆∗ and h is minimal with this property. We compute

ζqh

= αhζα−h = εζε−1 ∗= ζ

∗ because this equality holds in the residue class field ∆, and ζ is the unique power of ζ thatis congruent to ζ modulo πD∆. Since 1 ≤ h ≤ m and ζ is a primitive qm − 1 root of unity,this implies that h = m and hence gcd(j,m) = 1.(2) Put πD := π−tαr where r, t ∈ Z such that rj − tm = 1. Then vD(πD) = 1 and πDζπ

−1D =

ζqr. We want to achieve that πmD = π.

(3) By (2) the element pim

D commutes with ζ and πD and hence

πmD ∈ Z(〈ζ, πD〉K−algebra) = Z(D) = K.

Comparing valuations we obtain πmD = πε ∈ πR∗ for some ε ∈ R∗. (not the same as in (1)).For any λ ∈ K[ζ] = CD(K[ζ]) the element πDλ also satisfies

πDλζλ−1π−1

D = πDζπ−1D = ζq

r

Moreover(πDλ)m = (πDλ)m−2π2

D(π−1D λπD)λ = . . . = NK[ζ]/K(λ)πmD .

Since N(R[ζ]∗) = R∗ (see Theorem 2.3.10) there is some λ ∈ R[ζ] such that NK[ζ]/K(λ) = ε−1.Put π′D := πDλ, then this element has the desired properties and we have shown the existenceof such a π′D.


(4) Uniqueness:(a) r is independent of the choice of ζ:Let ζ ′ be a second primitive qm − 1 root of unity. Then by the Theorem of Skolem andNoether there is some β ∈ D∗ such that βζβ−1 = ζ ′. Put π′′D := βπ′Dβ

−1.(b) r is independent of the choice of π′D and π:Let π′′D ∈ ∆ be some prime element such that

π′′Dζ(π′′D)−1 = ζqs

(1 ≤ s ≤ m).

Then π′′D = γπ′D for some γ ∈ ∆∗. Then

ζqs

= (γπ′D)ζ(γπ′D)−1 = γζqr

γ−1 ≡ ζqr

(mod πD∆)

and so r = s.

2.3.4 The central simple case: synthesis

Theorem 2.3.13. Let 1 ≤ r ≤ m, gcd(r,m) = 1, K complete discretely valuated field withfinite residue class field. Then there is up to isomorphism a unique central K-division algebraD with [D : K] = m2 and Hasse invariant (D) = r

m.

Proof. Uniqueness. This is clear since D = 〈ζ, πD〉K−algebra is uniquely determined by therelations ζ = ζqm−1, πmD = π and πDζπ

−1D = ζq

r.

Existence. Let ζ ∈ K be a primitive qm − 1 root of unity and L := K[ζ]. Then L isa maximal subfield of D and therefore also a splitting field of D and we construct D as aK-subalgebra of Lm×m as follows:Let θ ∈ Gal(L/K) such that θ(ζ) = ζq

rand π be some prime element of R. Then 〈θ〉 =

Gal(L/K) since gcd(r,m) = 1. Define

∗ : L→ Lm×m, α 7→ α∗ := diag(α, θ(α), θ2(α), . . . , θm−1(α)).

Then ∗ is a K-algebra isomorphism and NL/K(α) = det(α∗). Define

πD :=

0 1 0 . . . 0

0 0 1. . . 0

.... . . . . . . . .

...0 . . . . . . 0 1π 0 . . . . . . 0

.

Then πmD = diag(π, . . . , π) and πDζ∗π−1

D = (ζ∗)qr. Let D be the K-subalgebra of Lm×m

generated by πD and ζ∗. Then

B := ((ζ∗)kπjD | k = 0, . . . ,m− 1, j = 0, . . . ,m− 1)


is a K-linear generating set of D.(a) B is a K-basis of D: Let αj ∈ L such that a =

∑m−1j=0 α∗jπ

jD ∈ D. Then

a =

α0 α1 . . . αm−1

πθ(αm−1) θ(α0). . . θ(αm−2)

. . . . . . . . .

πθm−1(α1) . . . πθm−1(αm−1) θm−1(α0)

.

If a = 0 then the first row of this matrix is 0 and so all αj are 0. So (πjD | j = 0, . . . ,m− 1)is independent over L, therefore as an L-vectorspace (not an L-algebra) D has dimension mand hence

dimK(D) = dimK(L) dimL(D) = m ·m = m2.

(b) D has no zero divisors: Assume that a as above is a zero divisor. Replace a by πla sothat all αj lie in RL and αk 6∈ πRL for some k (RL = R[ζ] ⊆ L). Then det(a) = 0. Butdet(a) ≡ NL/K(α0) (mod πRL) so π | α0. But then det(a) ≡ πNL/K(α1) (mod π2RL) soπ | α1 etc. until we obtain π | αj for all j which contradicts our assumption.(c) Z(D) = K. This is easy: CD(L) = diagonal matrices in D = L, so L is a maximalsubfield of D and Z(D) = CL(πD) = K since θ generates the full Galois group of L over K.(d) The Hasse invariant of D is r

mby construction.

Corollary 2.3.14. Let K be a complete discretely valuated field with finite residue class field.Then the Hasse invariant defines a bijection Br(K) ∼= Q/Z.

2.3.5 The inverse different.

Definition 2.3.15. Let S : D×D → K, S(a, b) := S(ab) be the reduced trace bilinear form.Then ∆# := x ∈ D | S(x, λ) ∈ R for all λ ∈ ∆ is called the inverse different of ∆.

Remark 2.3.16. ∆# is a fractional two-sided ∆-ideal in D, so ∆# = π−dD ∆ for some d ∈ N.

Theorem 2.3.17. ∆# = π1−mD ∆ and d(∆/R) := |∆#/∆| = |R/πm(m−1)R| = qm(m−1).

Proof. Let L = K[ζ] be an inertia field in D and RL = R[ζ]. Then there is some ε ∈ RL suchthat SL/K(ε) = trace(ε∗) = 1. Moreover ∆ =

⊕m−1j=0 RLπ

jD and trace(πjD) = 0 if j is not a

multiple of m. Since trace(ε∗) = 1 we have trace(π−mD ε∗) = π−1 and therefore ∆# ⊆ π1−mD ∆.

So we only need to show that π1−mD ∈ ∆#:

For j = 0, . . . ,m− 1 one computes

trace(RLπjDπ

1−mD ) = trace(RLπ

j+1−mD ) =

0 j 6= m− 1SL/K(RL) j = m− 1

Since this trace is in R, we get ∆# = π1−mD ∆. To compute the order note that ∆/πD∆ ∼=

RL/πRL∼= (R/πR)m.


2.3.6 Matrix rings.

Theorem 2.3.18. Let Γ be some maximal R-order in Dn×n. Then there is a matrix a ∈GLn(D) = (Dn×n)∗ such that aΓa = ∆n×n.

Proof. In the exercises we have shown that ∆n×n is a maximal order in A := Dn×n.Let Γ = 〈γ1, . . . , γs〉R with s = dimK(Dn×n) be an R-basis of Γ. Let 0 6= v ∈ V = Dn×1

be some non zero element in the unique simple A = Dn×n-module. Let ∆op be the maximalR-order in Dop = EndA(V ). Consider the ∆op-submodule

L := 〈γ1v, . . . , γsv〉∆op .

Then L is a full R-lattice in V that is invariant under Γ. It is also a right ∆op-module, so isa Dop-basis B := (b1, . . . , bn) of V such that

L = b1∆op ⊕ . . .⊕ bn∆op

(any ∆op/℘-basis of L lifts to such a basis). Then

Γ ≤ End∆op(L) ∼= ∆n×n

Since Γ is maximal, equality holds and any base change matrix from the standard basis toB is such a conjugating matrix a.

2.4 Crossed product algebras

2.4.1 Factor systems

Let L/K be some finite Galois extension with Galois group G := Gal(L/K). (E.g. L =Qp[ζpf−1], K = Qp, G = 〈F 〉 ∼= Cf .) We want to construct a K-algebra A =

⊕σ∈G uσL such

that (uσ : σ ∈ G) is an L-basis of A and such that

xuσ = uσxσ for all σ ∈ G, x ∈ L and uσuτ = uστ fσ,τ︸︷︷︸

∈L

for all σ, τ ∈ G

The fact that A is associative yields conditions on the mapping f : G×G→ L, (σ, τ) 7→ fσ,τ :

(uσuτ )uρ = uστfσ,τuρ = uστuρfρσ,τ = uστρfστ,ρf

ρσ,τ

uσ(uτuρ) = uσuτρfτ,ρ = uστρfσ,τρfτ,ρ

Definition 2.4.1. A map f : G×G→ L∗ such that fστ,ρfρσ,τ = fσ,τρfτ,ρ for all σ, τ, ρ ∈ G is

called factor system or 2-cocycle of G with values in L∗.Z2(G,L∗) := f | f is factor system .

Remark 2.4.2. Z2(G,L∗) is an abelian group via (fg)σ,τ = fσ,τgσ,τ .

2.4. CROSSED PRODUCT ALGEBRAS 67

Proof. Show that the map (σ, τ) 7→ fσ,τgσ,τ is again a 2-cocycle. This is immediate by thecommutativity of L∗.

Of course the algebra above is not changed if we replace uσ by uσcσ with cσ ∈ L∗. Theproduct is then

(uσcσ)(uτcτ ) = uσuτcτσcτ = uστcστfσ,τc

τσcτc

−1στ

Remark 2.4.3. B2(G,L∗) := f : G×G→ L∗ | ∃cσ ∈ L∗(σ ∈ G) such that fσ,τ = cτσcτc−1στ

is a subgroup of Z2(G,L∗), the group of 2-coboundaries, and the factor group

H2(G,L∗) := Z2(G,L∗)/B2(G,L∗)

is called the second cohomology group of G with values in L∗.Two cocycle f, g are called equivalent, if they represent the same class in H2(G,L∗).

Proof. Need to show that (σ, τ) 7→ cτσcτc−1στ is a 2-cocycle and that Z2(G,L∗) is closed under

product (easy) and inverse (easy).

Remark 2.4.4. Any 2-cocycle f ∈ Z2(G,L∗) is equivalent to a normalized 2-cocycle fsuch that fσ,1 = f1,σ = 1 for all σ ∈ G.

Proof. Choose cσ := f−11,σ .

2.4.2 Crossed product algebras

Theorem 2.4.5. Let G = Gal(L,K), f ∈ Z2(G,L∗), A = (L/K, f) =⊕

σ∈G uσL as above.Then A is called a crossed product algebra. A is a central simple K-algebra and L =CA(L) is a maximal subfield of A. Moreover two such crossed product algebras (L/K, f) and(L/K, g) are isomorphic if and only if the factor systems f and g are equivalent.

Proof. It is clear that equivalent factor systems construct isomorphic algebras. So we mayassume that f and g are normalized.Then u1 = 1A and L ∼= u1L = Lu1 → A. Since conjugation by any element in uσL inducesthe Galois automorphism σ on L, we obtain L = CA(L), so L is a maximal subfield of A.Also Z(A) = ` ∈ L | uσ` = ùσ = K.We now show that A is simple:Let 0 6= X E A be some 2-sided ideal, choose 0 6= x ∈ X

x = uσ1a1 + . . .+ uσrar such that r is minimal.

If r > 1 the we may choose b ∈ L such that bσ1 6= bσ2 and replace x by y = x− bx(bσ1)−1 ∈ Xwhich has a smaller representation

y = x− bx(bσ1)−1 = uσ2b2 + . . .+ uσrbr.

So r = 1 and there is some x = uσ1a1 ∈ X. Since a1 ∈ L∗ is a unit in A and also uσ ∈ A∗this implies that X = A.


Now let g ∈ Z2(G,L∗) and B =⊕

σ∈G vσL = (L/K, g). Assume that A ∼= B and letΦ : A → B be some isomorphism. Then Φ(1) = 1, so Φ(u1) = v1 (since we assumed bothfactor systems to be normalized). Therefore Φ(u1L) = v1L

′ for some subfield L′ ≤ B that isisomorphic to L. By the theorem of Skolem and Noether, there is some b ∈ B∗ such that

b−1Φ(u1x)b = v1x for all x ∈ L.

Put wσ := b−1Φ(uσ)b ∈ B for all σ ∈ G. Then B =⊕

wσL, xwσ = wσxσ and wσwτ =

wστfσ,τ . Hence v−1σ wσ =: cσ ∈ CB(L) = L for all σ defines a cocycle that shows that f ∼ g.

Example. L = K[ζqf−1], K complete field with valuation ring R and residue class fieldFq = R/πR. Gal(L/K) ∼= Cf = 〈F 〉. Define uFn := (uF )n for n = 2, . . . , f − 1. So anynormalized factor system

ϕ : 〈F 〉 × 〈F 〉 → L∗

is determined by a := (uF )f , namely

ϕF i,F j =

1 i+ j < fa i+ j ≥ f

(0 ≤ i, j ≤ f − 1).

(This is a general fact for so called cyclic algebras, crossed product algebras for whichGal(L/K) is cyclic, see exercises.)

Corollary 2.4.6. (L/K, f) ∼= Kn×n if and only if f ∼ 1.

Proof. It is enough to show that (L/K, 1) ∼= Kn×n.(L/K, 1) =

⊕σ∈G uσL with uσuτ = uστ . Define

ψ : (L/K, 1)→ EndK(L), uσx 7→ (` 7→ `σx).

Then ψ is a K-algebra homomorphism.ψ is injective, since (L/K, 1) is simple and surjective by comparing dimensions. Therefore(L/K, 1) ∼= EndK(L) ∼= Kn×n with n = dimK(L).

Example. Let D be a central simple Q-algebra of dimension 4. Then any maximalsubfield L of D is a quadratic extension L = Q[

√a] for some a ∈ Q∗ \ (Q∗)2. The Galois

group Gal(L/Q) = 〈σ〉 ∼= C2 and any normalized factor system is given by

f : G×G→ L∗, f1,1 = f1,σ = fσ,1 = 1, fσ,σ =: b ∈ L∗.

Since u2σ = b commutes with uσ, the value b lies in Q∗ and

D = 〈√a, uσ〉 =

(a, b

Q

)= 〈1, i, j, k〉 with i2 = a, j2 = b, ij = −ji = k.

The factor system is trivial, if and only if there is some x ∈ Q[√a] such that xxσ = b.

Theorem 2.4.7. Let [L : K] = n, G = Gal(L/K). Then (L/K, f) ⊗K (L/K, g) ∼=(L/K, fg)n×n.


Proof. Let A := (L/K, f) and B := (L/K, g) with L-basis (uσ | σ ∈ G) resp. (vσ | σ ∈ G)such that the factor systems f and g are normalized. Then the subalgebra u1L of A isisomorphic to v1L ≤ B. In particular C := A⊗K B contains a subalgebra

(?0) S := u1L⊗ v1L ∼= L⊗K L ∼=⊕σ∈G

L, (u1x⊗ v1y) 7→ (xσy)σ∈G.

We use this to construct an idempotent e ∈ S such that eCe ∼= (L/K, fg). Then the theoremfollows, since we already know that the tensor product of two central simple algebras is againcentral simple and by comparing dimensions.To this aim choose a primitive element a ∈ L, so L = K(a) and put

e :=∏

16=σ∈G

u1a⊗ v1 − u1 ⊗ v1aσ

u1(a− aσ)⊗ v1

.

Then e 7→ (1, 0, . . . , 0) under the map in (?0) (calculation at blackboard) and e2 = e ∈ S isan idempotent. Moreover (u1x⊗ v1)e = e(u1 ⊗ v1x) for all x ∈ L.We now compare eCe and (L/K, fg):

eCe =∑σ,τ∈G

e(uσ ⊗ vτ )(L⊗ L)e =∑σ,τ∈G

e(uσ ⊗ vτ )e︸︷︷︸?

e(1⊗ L)ee(L⊗ 1)e︸︷︷︸L′∼=L

? : e(uσ ⊗ vτ )e = uσ ⊗ vτ∏

ψ∈G\1

aσ ⊗ 1− 1⊗ aψτ

(aσ − aψσ)⊗ 1e = uσ ⊗ vτe

∏ψ∈G\1

aσ − aψτ

aσ − aψσ︸︷︷︸=1,τ=σ,=0,τ 6=σ

So eCe =⊕

σ∈GwσL′ with wσ = e(uσ ⊗ vσ)e and L′ = e(1⊗ L)e ∼= L. Moreover

e(x⊗ 1)ewσ = e(x⊗ 1)e(uσ ⊗ vσ)e = wσe(xσ ⊗ 1)e

and

wσwτ = e(uσ ⊗ vσ)(uτ ⊗ vτ )e = e(uστ ⊗ vστ )ee(fσ,τ ⊗ gσ,τ )e = wστ (fσ,τgσ,τ ).

2.4.3 Splitting fields.

Definition 2.4.8. Let A be a central simple K-algebra. An extension field L/K is called asplitting field for A, if A⊗K L ∼= Ln×n for some n ∈ N.

So L is a splitting field if and only if [A] ∈ Br(L/K) = ker(Br(K) → Br(L), [A] 7→[A⊗K L].

Example. If A = (L/K, f) then L is a splitting field for A.Maximal subfields of division algebras are splitting fields.


Theorem 2.4.9. Let D be a central K-division algebra with m2 = [D : K] and let E/K besome finite field extension.(a) If E is a splitting field for D, then m divides [E : K].(b) Let r be minimal such that E → Dr×r. Then E splits D, if and only if E is a maximalsubfield of Dr×r, if and only if CDr×r(E) = E.

Proof. Let S := E ⊗K Dop. Then Z(S) = E and S is a central simple E-algebra.Note that E splits D if and only if E splits Dop if and only if S ∼= Em×m. In generalS ∼= D′m

′×m′ for some E-division algebra D′ with center Z(D′) = E and some m′ ∈ N.Let V ∼= (D′)m

′be the simple S-module. Then

E ⊆ EndDop(V ) ∼= Dr×r =: B

for some r ∈ N and r is minimal with E → Dr×r. The minimality of r follows from thefact that all matrix rings over D arise as endomorphism rings of some Dop-module W . If Eembeds into EndD(W ), then W is an E ⊗K Dop-module. Let E ′ := CB(E). Then

E ⊆ E ′ = EndS(V ) = (D′)op.

For the K-dimension of V and S we obtain

dim(V ) = dim(D′)m′ = dim(D)r = m2r, dim(S) = dim(E)m2 = dim(D′)m′2 = m2rm′

so [E : K] = m′r and dim(D′) = m2r/m′ in particular [E : K][E ′ : K] = [B : K] = m2r2.Now E splits D if and only if D′ = E if and only if E ′ = E and then E = CB(E) is amaximal subfield of B. If E ′ 6= E, then in any case E ⊆ Z(E ′) and any x ∈ E ′ \E generatesa subfield E[x] of B that properly contains E.

Example. Let E := Q[ζ5], D =(−2,−5

Q

). Then E ⊗Q D ∼= E2×2, so E is a splitting field

of D but the unique subfield Q[√

5] of E with [Q[√

5] : Q] = 2 is not a spitting field for D.

Corollary 2.4.10. If [D : K] = m2 and [E : K] = m, then E is a splitting field for D if andonly if E is a maximal subfield of D.

Theorem 2.4.11. Any central simple K-algebra has some splitting field that is Galois overK, so

Br(K) ∼=⋃

L/K Galois

Br(L/K).

Proof. As we have seen in the algebra class, any central simple K-algebra A has some max-imal subfield L′ that is separable over K. Take L to be the normal closure of L′ over K.Then L/K is Galois and L splits A.

Theorem 2.4.12. Let G = Gal(L/K). Then H2(G,L∗) ∼= Br(L/K), [f ] 7→ [(L/K, f)].

Proof. The map is a well defined injective group homomorphism, as we have seen before.So we need to prove surjectivity. Let [A′] ∈ Br(L/K). Then there is some central simple


K-algebra A representing the same class in the Brauer group as A′ such that L is a maximalsubfield of A.Claim: A = (L/K, f) for some f ∈ Z2(G,L∗):For σ ∈ G the map ` 7→ `σ is a K-algebra homomorphism L → A, so by the Theorem ofSkolem and Noether, there is some uσ ∈ A∗ such that `σ = u−1

σ ùσ for all ` ∈ L. We showthat A =

⊕σ∈G uσL. To see this, it is enough to prove that (uσ | σ ∈ G) is linearly indepen-

dent over L, then equality follows by comparing K-dimensions. Let (aσ1 , . . . , aσk) ∈ (L∗)k

such that s :=∑k

i=1 uσiaσi = 0 and choose such a relation with minimal k. Then k 6= 1, sinceuσ ∈ A∗ for all σ ∈ G. Hence there is some b ∈ L∗ such that bσ1 6= bσ2 and 0 = sbσ1 − bsyields a shorter relation.

2.4.4 Field extensions.

Ground field extensions.

Let L/K be some Galois extension, E/K arbitrary finite field extension. Then LE/E isGalois. Let F := E ∩ L. Then

Gal(LE/E) ∼= Gal(L/F ) =: H ≤ G := Gal(L/K).

Theorem 2.4.13. Let f : G × G → L∗ and f ′ := f|H the restriction of f to H. Thenf ′ : H ×H → L∗ is a factor system and

E ⊗K (L/K, f) ∼= (EL/E, f ′).

Proof. As above let F := L ∩ E. We show that(a) F ⊗K (L/K, f) ∼ (L/F, f ′) and(b) E ⊗F (L/F, f ′) ∼ (EL/E, f ′).(a) Let F ′ := C(L/K,f)(F ).Claim: F ′ = (L/F, f ′).Proof: (L/F, f ′) =

⊕σ∈H uσL. For any a ∈ F, σ ∈ H we have uσa = auσ, so (L/F, f ′) ⊆ F ′.

On the other hand for any x ∈ F :

x∑σ∈G

uσaσ =∑σ∈G

uσaσxσ !

=∑σ∈G

uσaσx

if and only if x = xσ for all σ ∈ G such that aσ 6= 0, because the (uσ : σ ∈ G) form an L-basisof (L/K, f). So equality holds for all x ∈ F if and only if aσ = 0 for all σ ∈ G \H.By the double centralizer theorem F ⊗K (L/K, f) ∼= (F ′)s×s which shows (a).(b) Clearly E ⊗F L ∼= EL, e⊗ ` 7→ e`. So

E ⊗F (L/F, f ′) =⊕σ∈H

uσ(E ⊗F L) =⊕

σ∈Gal(EL/E)

uσEL = (EL/E, f ′).


Field extensions of L.

We now assume that E ⊇ L ⊇ K is a tower of Galois extensions, G := Gal(E/K) andH := Gal(E/L)EG so that G/H = G = Gal(L/K). For any factor system f : G×G→ L∗

let f : G×G→ L∗ be the factor system defined by fg,h := fg,h (“obtained by inflation”).

Theorem 2.4.14. (E/K, f) ∼= (L/K, f)r×r =: B, with r = [E : L].

Proof. Let

B := (L/K, f)r×r = (L/K, f)⊗K Kr×r =⊕σ∈G

uσLr×r.

(E/L, 1) =⊕

h∈H vhE∼→ EndL(E) = Lr×r

vh 7→ (x 7→ xh)e 7→ T (e) : x 7→ xe

For any L-basis (e1, . . . , er) of E and σ ∈ G also (eσ1 , . . . , eσr ) is an L-basis of E so there is

some P (σ) ∈ GLr(L) such that

P (σ)

e1...er

=

eσ1...eσr

.

Moreover for σ, ρ ∈ G we compute P (σρ) = P (σ)ρP (ρ): eσρ1...eσρr

=

P (σ)

e1...er

ρ

= P (σ)ρP (ρ)

e1...er

.

Here a matrix ρ means that one applies ρ to all entries of the matrix. For any e ∈ E letT (e) ∈ GLr(L) denote the matrix of x 7→ xe with respect to the chosen basis. Then

T (eσ) = matrix of x 7→ x(eσ) = P (σ)−1T (e)σP (σ).

Put uσ := uσP (σ) for σ ∈ G. Then B =⊕

σ∈G uσE with

uσuτ = uσP (σ)uτP (τ) = uσuτP (σ)τP (τ)= uστfσ,τP (στ) = uστP (στ)fσ,τ = uστfσ,τ .

For x ∈ E and σ ∈ G we compute

T (x)uσ = T (x)uσP (σ) = uσT (x)σP (σ) = uσP (σ)T (xσ) = uσT (xσ).

Therefore B ∼= (E/K, f).

We hence have the following commutative diagrams for a tower of fields L ⊇ F ⊇ K:

H2(G,L∗)resGH→ H2(H,L∗)

∼=↓ ↓∼=Br(L/K)

⊗KF→ Br(L/F )

andH2(G,F ∗)

inflG→ H2(G,L∗)∼=↓ ↓∼=

Br(F/K) → Br(L/K)

Here G = Gal(L/K) and H = Gal(L/F ). For the second diagram we need to assume thatH EG and put G := G/H.


2.4.5 A group isomorphism Br(K) ∼= Q/Z.

Let K be a complete field with valuation ring R and residue class field Fq = R/πR. LetF ∈ Aut(unr(K)) denote the Frobenius automorphism of the maximal totally unramifiedextension of K, mapping any p′-root of unity ζ to ζq.

Theorem 2.4.15. Let D be some central K-division algebra with Hasse invariant rm

. ThenD ∼= (L/K, σ, π) ∼= (L/K,F, πs) with L = K[ζqm−1], σ = F−r, sr ≡ −1 (mod m).Define inv : Br(K)→ Q/Z, (L/K,F, πs) 7→ s

m.

Proof. Choose some maximal subfield L ≤ D isomorphic to K[ζqm−1] and put u := πD,so that πmD = π and ζ := ζqm−1 so that πDζπ

−1D = ζq

ras in Theorem 2.3.12. Then

D =⊕m−1

j=0 πjDL∼= (L/K,F−r, π) since ζπD = πD(π−1

D ζπD) = πDζq−r = πDζ

σ. There-fore conjugation by πsD induces the Frobenius automorphism on L.

Theorem 2.4.16. inv : Br(K)→ Q/Z, (L/K,F, πs) 7→ sm

is a group isomorphism.

Proof. The bijectivity of this map follows from the bijectivity of the Hasse invariant, whichis not a group homomorphism. We need to show the homomorphism property:So let inv(D) = s

m, inv(D′) = s′

m′we need to show that inv(D ⊗D′) = s

m+ s′

m′.

Let m′′ = lcm(m,m′) and L′′ := K[ζqm′′−1]Then [D], [D′] and hence also [D ⊗D′] are in Br(L′′/K).

D = (L/K,F, πs) ∼ (L′′/K, F ′′, πs1) and D′ = (L′/K, F ′, πs′) ∼ (L′′/K, F ′′, πs2)

with s1 = s[L′′ : L] = sm′′

mand s2 = s′[L′′ : L′] = s′m

′′

m′. So the tensor product

D ⊗D′ ∼ (L′′/K, F ′′, πs1+s2) ands1 + s2

m′′=

s

m+

s′

m′.

Corollary 2.4.17. If A = Dn×n with a central K-division algebra D of index m, then[A] = [D] has order m in Br(K).

Concerning field extensions we obtain from the preceeding sections the following commu-tative diagram

Theorem 2.4.18. Let E/K be a finite extension of degree d := [E : K]. Then

Br(K)invK→ Q/Z

E⊗K ↓ ↓ ·dBr(E)

invE→ Q/Z

is commutative.


Proof. Let W := K[ζqm−1] and put A = (W/K,F, πs), so that invK(A) = sm

. Let L := E∩W .Replacing W by a larger unramified extension if necessary we may assume that L/K is themaximal unramified extension of E/K, so f := f(E/K) = [L : K] is the inertia degree of Eand e = e(E/K) = d/f the ramification index. Let m′ ∈ N be such that m = m′f . Then

E ⊗K A ∼ (EW/E,F f , πs) =: B.

To compute invE(B) we first note that py construction F f is the Frobenius automorphism ofEW/E so we only need to replace π = πK by a suitable prime element πE. As πRE = πeERE

the exponent gets mutliplied by the ramification index e, hence invE(B) = sem′

= sdm

.

Corollary 2.4.19. Let D be a central K-division algebra of index m. Then a finite extensionfield E of K splits D if and only if m divides [E : K]. And E is a maximal subfield of D, ifand only if [E : K] = m.

Corollary 2.4.20. If [E : K] = n, then Br(E/K) ∼= 1nZ/Z ∼= Cn. Moreover Br(E/K) =

[A] ∈ Br(K) | [A]n = 1.

2.5 Division algebras over global fields.

Global fields are either finite extensions of Q (algebraic number fields) or function fields overfinite fields. Though most of the theory is parallel for both sorts of global fields, we willrestrict to the algebraic number fields, which have been introduced in the first part of thislectures.

So let K be an algebraic number field with ring of integers R. A place of K is either afinite place given by a maximal ideal ℘ER or an infinite place, i.e. an embedding σ : K → C.An infinite place is called real, if σ(K) ⊆ R. For a place ℘ let K℘ denote the completion ofK at ℘. If ℘ is real then K℘ = R, for complex places K℘ = C and for finite places this is ap-adic number field.

Theorem 2.5.1. Let A be a central simple K-algebra. Then [A℘] 6= 0 only for finitely manyplaces ℘ of K.

Proof. Let Λ ⊆ A be some R-order in A. Then there are only finitely many prime ideals℘ E R that divide the discriminant of Λ. For any prime ℘ not dividing the discriminant ofΛ the order Λ℘ ≤ A℘ is a maximal R℘-order of discriminant 1. For all these places [A℘] = 0.

Remark 2.5.2. Br(K) →⊕

℘ place Br(K℘), [A] 7→ [A℘] where A℘ =: A ⊗K K℘ is a grouphomomorphism.

For finite places we have seen that Br(K℘) ∼= Q/Z via the variant of the Hasse invariantthat is a group homomorphism. Moreover Br(R) ∼= 1

2Z/Z where inv([H]) = 1

2and Br(C) =

0.

2.5. DIVISION ALGEBRAS OVER GLOBAL FIELDS. 75

Definition 2.5.3. Define a group homomorphism

inv : Br(K)→⊕

℘ finite

Q/Z⊕⊕σ real

1

2Z/Z, [A] 7→ (inv[A℘])℘ finite + (inv[A⊗σ(K) R])σ real

and inv℘ : Br(K) → Q/Z, [A] 7→ inv[A℘], for ℘ E R and invσ : Br(K) → 12Z/Z, [A] 7→

inv[A⊗σ(K) R] for any real place σ and invτ : Br(K)→ 0 for the complex places τ of K.

We want to prove a local-global principle for central simple K-algebras (see Theorem2.5.6 below). To prove this theorem we need one result the proof of which needs class fieldtheory, so goes beyond the scope of this lecture:

Theorem 2.5.4. (Hasse Norm Theorem) Let L be a finite cyclic extension (so Gal(L/K)is cyclic) of the number field K and let a ∈ K. For any place ℘ of K we choose some placeP of L that extends ℘. Then for a ∈ K it holds that

a ∈ NL/K(L)⇔ a ∈ NLP /K℘(LP ) for all ℘.

Note that the condition on the right hand side also includes the infinite places.The theorem is false, without the assumption that L/K be cyclic.The direction ⇒ is almost trivial (see blackboard).

Corollary 2.5.5. Let A = (L/K, σ, a) be a cyclic algebra. Then [A] = [K] ∈ Br(K) if andonly if [A℘] = [K℘] ∈ Br(K℘) for all (finite and infinite) places ℘ of K.

Proof. We have seen in the exercises that [(L/K, σ, a)] = [K] if and only if a ∈ NL/K(L∗), soif and only if a is a global norm. On the other hand

A℘ = K℘ ⊗K (L/K, σ, a) ∼ (LP/K℘, σk, a)

for some k such that 〈σk〉 == Gal(LP/K℘) by Theorem 2.4.13. So [A℘] = [K℘] ∈ Br(K℘) ifand only if a ∈ NLP /K℘(L∗P ) is a local norm. So the Corollary follows from the Hasse NormTheorem.

Theorem 2.5.6. (Hasse-Brauer-Noether-Albert-Theorem) Let A be a central simpleK-algebra. Then [A] = [K] ∈ Br(K) if and only if [A℘] = [K℘] ∈ Br(K℘) for all (finite andinfinite) places ℘ of K.So the group homomorphism inv from Definition 2.5.3 is injective.

Proof. ⇒ is clear. So assume that [A℘] = [K℘] ∈ Br(K℘) for all places ℘ of K but [A] 6=[K] ∈ Br(K). Then the index m of A is m > 1. By Theorem 2.4.11 there is a finite Galoisextension L of K that splits A, so L⊗K A ∼= Lk×k for some k. Let p be a prime divisor of mand H a Sylow p-subgroup of G = Gal(L/K). Choose some subnormal series

1 = Hn EHn−1 E . . .EH0 = H, such that [Hj : Hj−1] = p

and let Ej be the fixed field of Hj for j = 0, . . . , n. Then En = L, Ej/Ej+1 is a cyclic extensionand, since p does not divide [E0 : K] the index of A0 := E0⊗KA is not 1. Let F := En−1 and


B := F ⊗K A. Then L splits B, so B is similar to some cyclic algebra B ∼ C := (L/F, σ, b)for some b ∈ F ∗. Now C℘ ∼ F℘ for all places ℘ of F by assumption. This is equivalent to thefact that b ∈ NLP /F℘(L∗P ) for all places ℘. By the Hasse Norm Theorem this implies that bis a global norm and hence C ∼ F and so F splits A. Similarly one obtains that En−2 splitsA etc until E0 splits A so in particular m | [E0 : K] which contradicts our assumption thatp divides m but does not divide [E0 : K].

Remark 2.5.7. One may show that∑

℘ inv℘ = 0 and that this is the only condition on thelocal Hasse invariants of a global central simple algebra, so the following sequence is exact

1→ Br(K)→⊕℘

Br(K℘) ∼=⊕

℘ finite

Q/Z⊕⊕℘ real

1

2Z/Z

∑→ Q/Z→ 0

Theorem 2.5.8. Let A be a central simple K-algebra and put m℘ := inv℘[A] = inv([A℘]) forany (finite or infinite) place ℘ of K. Let L be a finite extension of K. Then L splits A ifand only if for each place P of L

m℘ divides [LP : K℘]

where ℘ := P ∩K = P|K is the restriction of P to K.

Proof. [L⊗K A] is trivial if and only if [LP ⊗K℘ A℘] is trivial for all ℘. Moreover by Corollary2.4.19 this is equivalent to m℘ | [LP : K℘].

Theorem 2.5.9. Let A be a central simple K-algebra and put m℘ := inv℘[A] = inv([A℘])for any (finite or infinite) place ℘ of K. Then the order of [A] ∈ Br(K) is the least commonmultiple of all the local Schur indices m℘.

Proof. Because the order of [A℘] ∈ Br(K℘) is exactly m℘.

We aim to show that the order of [A] in Br(K) is exactly the index of the underlyingdivision algebra. For this we need a very deep theorem which we cannot prove here:

Theorem 2.5.10. (Grunwald-Wang Theorem) Let ℘1, . . . , ℘s be a finite set of placesof the global field K and let (m℘1 , . . . ,m℘s) ∈ Ns be given so that m℘ = 1 if ℘ is a complexplace and m℘ = 1 or 2 if ℘ is a real place. Let n ∈ N be divisible by all m℘i. Then there is acyclic extension L/K with

[L : K] = n, [LPi : K℘i ] = m℘i for all i = 1, . . . , s

where Pi is some prime of L extending ℘i.

Theorem 2.5.11. Let A be a central simple K-algebra and put m℘ := inv℘[A] = inv([A℘])for any (finite or infinite) place ℘ of K. Then the global Schur index of A is the order of[A] ∈ Br(K), i.e. the least common multiple of all the local Schur indices m℘.

2.5. DIVISION ALGEBRAS OVER GLOBAL FIELDS. 77

Proof. Let L be a cyclic extension of K as in the Grunwald-Wang theorem with n := [L :K] = lcm(m℘). Then by Corollary 2.4.19 LP splits A℘ for all places ℘ of K and hence Lsplits A. By Theorem 2.4.9 this implies that the index of A divides [L : K] = n. By Theorem2.5.9 the number n is equal to the order of [A] in Br(K). It holds in general, that the orderof [A] divides the index of A, so equality follows.

Corollary 2.5.12. Every central simple K-algebra A is a cyclic algebra, so there is a cyclicGalois extension L of K with Gal(L/K) = 〈σ〉 and some a ∈ K∗ such that A = (L/K, σ, a).

Proof. Choose L as in the last proof but with n ∈ N such that n2 = [A : K]. Then n is amultiple of all local Schur indices and we may apply again the Grunwald-Wang theorem.

2.5.1 Surjectivity of the reduced norm

Let K be a number field, ℘ a place of K and K℘ the completion of K at ℘.

Remark 2.5.13. Let D℘ be a central K℘-division algebra. If K℘ 6= R then the reduced normD℘ → K℘ is surjective.The image of the reduced norm of

(−1,−1R

)=: H is R≥0.

Proof. Assume that ℘ is a finite place (otherwise nothing to show). Let R℘ be the valuationring in K℘. Then any element a ∈ K∗℘ has a unique expression as a = uπn with n = v(a) ∈ Zand u ∈ R∗℘. Let L℘ ≤ D℘ be a maximal unramified subfield of D℘. Then the reduced normof L∗℘ contains R∗℘. Moreover the reduced norm of a prime element in D℘ is a prime elementin R℘. So u and πn are reduced norms and the surjectivity follows.

Remark 2.5.14. N(Dn×n) = N(D).

Theorem 2.5.15. (Hasse-Schilling-Maass) Let A be a central simple K-algebra and letα ∈ K∗. Then α is a reduced norm of A if and only if σ(α) > 0 for all real places σ of Kthat ramify in A ([A⊗K,σ R] = [H] ∈ Br(R)).

Proof. Let S denote the set of real places of K that ramify in A and put U(A) := α ∈ K |σ(α) > 0 for all σ ∈ S. We need to show that for any β ∈ U(A) there is some a ∈ A suchthat N(a) = β. Let n2 := [A : K]. Then n is even if S is non empty. Let S ′ be a non emptyfinite set of finite primes of K that contains all finite primes of K that ramify in A.Let β ∈ U(A).For all P ∈ S ′ the element β is a reduced norm of some element in the completion AP , so wemay find an irreducible1 polynomial

fP (X) := Xn + a1,PXn−1 + . . .+ an−1,PX + (−1)nβ ∈ KP [X].

If S is non-empty then n is even and we put

fP (X) := Xn + (−1)nβ ∈ KP [X] for all P ∈ S.1argument for this: pp. 286-290 in Reiner


By the strong approximation theorem (Chinese remainder theorem) for any ε > 0 there issome polynomial

f(X) := Xn + c1Xn−1 + . . .+ cn−1X + (−1)nβ ∈ K[X]

such that||ci − ai,P ||P < ε 1 ≤ i ≤ n− 1 for all P ∈ S ′||ci||P < ε 1 ≤ i ≤ n− 1 for all P ∈ S

If ε is small, then f is irreducible in KP [X], since fP is irreducible for all P ∈ S ′ 6= ∅. If S 6= ∅then n is even and since σ(β) > 0 for all σ ∈ S the polynomial σ(f) has no real roots forsufficiently small ε. Let L := K[X]/(f(X)). Then all local degrees [LP : KP ] are multiplesof the local Schur index since [LP : KP ] = n for P ∈ S ′ and LP = C for P ∈ S. So L is asplitting field of A and we may embed L as a maximal subfield of A. In particular there issome a ∈ A such that f(a) = 0. This element has reduced norm β.

2.6 Maximal orders in separable algebras.

The book by Max Deuring: Algebren (Springer Grundlehren) gives a very nice treatment ofmost of the results of the second part of this lecture.

Let K be a number field, R = ZK the ring of integers in K and A a separable K-algebra.(Many things hold in the more general context that R is a Dedekind domain with field offractions K).

If Λ is a maximal R-order in A, then the completion Λ℘ is a maximal R℘-order in A℘.Moreover

A℘ =⊕

Ai, Z(Ai) = Li, Ai simple , Λ℘ =⊕

Λi

where Λi is a (up to conjugacy unique) maximal ZLi-order in Ai.

Definition 2.6.1. (a) Let Λ be an R-order in A. A prime ideal P of Λ is a proper non-zerotwo-sided ideal P E Λ with KP = A such that for any pair S, T of two-sided Λ-ideals

ST ⊆ P ⇒ S ⊆ P or T ⊆ P .

(b) Let M ≤ A be a full R-lattice in A. Then M is called a normal ideal, if Ol(M) isa maximal order in A. An integral ideal is a normal ideal M such that M ⊆ Ol(M). Amaximal integral ideal is an integral ideal which is a maximal left ideal of its left order.

Remark 2.6.2. (a) Maximal 2-sided ideals are prime ideals.(b) Clearly M ⊆ Ol(M) if and only if M ·M ⊆ M which is equivalent to M ⊆ Or(M). Inparticular the notion of integrality does not depend on the choice of left or right order.

Theorem 2.6.3. Let ℘ be some prime ideal of R and Λ℘ a maximal R℘-order in the separableK-algebra A. Then any left Λ℘-lattice L in A is free, i.e. there is some y ∈ A such thatL = Λ℘y.

2.6. MAXIMAL ORDERS IN SEPARABLE ALGEBRAS. 79

Proof. (a) Wlog we may assume that A is central simple: If A =⊕

iAi, then Λ℘ =⊕

Λi withmaximal Ri-orders Λi in Ai. So L =

⊕Li. If Li = Λiyi then L = Λ℘y with y =

∑yi ∈ A.

(b) For the central simple case we pass to the completion of A at ℘. For the complete ringΛ℘-lattices in the simple A℘-module V = A℘e form a chain, so any Λ℘-lattice is generated

by any element that is not in the unique proper maximal sublattice. Since A℘ =⊕

e A℘e the

sum of the generators of L℘e is a generator for L.

For a full R-lattice L in A we define

L−1 = x ∈ A | LxL ⊆ L.

Then L−1 is again a full R-lattice in A and Or(L−1) ⊇ Ol(L), Ol(L

−1) ⊇ Or(L).

Theorem 2.6.4. Let Λ be a maximal order in the separable K-algebra A. Let L be a full leftΛ-lattice in A. Then

LL−1 = Λ, L−1L = Or(L), (L−1)−1 = L, Ol(L−1) = Or(L).

Proof. Since equality of lattices is a local property, it is enough to show the equalities for alllocalisations L℘ at prime ideals of R. But then L℘ is a principal Λ℘-ideal, so L℘ = Λ℘y℘ forsome y℘ ∈ A, Or(L℘) = y−1

℘ Λ℘y℘, and (L℘)−1 = y−1℘ Λ℘, from which one sees all equalities.

Theorem 2.6.5. Let M be a full R-lattice in A. Then Ol(M) is a maximal R-order if andonly if Or(M) is a maximal R-order. So the notion of normality is independent from left andright.

Proof. Being a maximal order is a local property, so we may pass to all localizations. Butlocally all left ideals of a maximal order are free, so locally the right order and the left orderare conjugate and hence also the right order is a maximal order.

2.6.1 The group of two-sided ideals.

We want to show that maximal R-orders Λ in skewfields are Dedekind domains. We couldproceed as in the commutative case: Λ is Noetherian (since it is finite dimensional over R)and all completions of Λ are discrete valuation rings.

Theorem 2.6.6. Let Λ be an R-order in A. Then the prime ideals of Λ are exactly themaximal ideals of Λ. If P is a prime ideal of Λ then ℘ := R ∩ P is a prime ideal of R andΛ := Λ/P is a finite dimensional simple R/℘-algebra.

Proof. Clearly ℘ := R∩P is a prime ideal of R. Let Λ := Λ/P . Then Λ is a finite dimensionalR/℘-algebra and hence Artinian, so the Jacobson radical J(Λ) is a nilpotent ideal. Since Pis a prime ideal this implies that J(Λ) = 0, so Λ is semi-simple. In fact it is simple sinceany product of non-zero ideals is again non-zero. But this means that P is a maximal ideal.


Theorem 2.6.7. Let Λ be a maximal R-order in the central simple K-algebra A. ThenP 7→ ℘ := P ∩ R defines a bijection between the set of prime ideals P of Λ and those of R.Moreover P is the radical of the localisation Λ℘ and for each prime ideal Q 6= ℘ of R we havePQ = ΛQ.

Proof. Let P be a prime ideal of Λ and ℘ := P ∩ R. The ideals of Λ℘ are all powers of theradical J(Λ℘) (take the completion to obtain a matrix ring over a discrete valuation ring).Being a maximal ideal of Λ we obtain P℘ = J(Λ℘). Moreover PQ = ΛQ for all other primesQ of R, since (℘Λ)Q = ΛQ. So

P = J(Λ℘) ∩ Λ

is uniquely determined by ℘. One the other hand, given some prime ideal ℘ of R, the ideal Pwith P℘ = J(Λ℘) and PQ = ΛQ for all Q 6= ℘ is a maximal 2-sided ideal with Λ/P ∼= Λ℘/P℘.

Theorem 2.6.8. Let Λ be a maximal order in the central simple K-algebra A. Then the setof two-sided Λ-ideals I(Λ) in A is the free abelian group on the prime ideals of Λ.

Proof. Pass to the localizations to see that any two-sided Λ-ideal has a unique factorizationas product of prime ideals.

Corollary 2.6.9. The theorem above holds more general for any maximal order Λ in aseparable K-algebra.

Proof. Maximal orders Λ in⊕

Ai are direct sums Λ =⊕

Λi for maximal IntR(Z(Ai))-ordersΛi in the simple Z(Ai)-algebra Ai. Similarly all Λ-lattices decompose into a direct sum.

Theorem 2.6.10. Let K be a number field with ring of integers R, L/K a finite extension.Let Λ be a maximal order in some central simple L-algebra A, S = IntR(L) the ring of integersin L. Given a prime ideal ℘ E R let ℘S = ℘e11 · · ·℘

edd be the prime ideal factorization in S.

Let Pi be the prime ideal of Λ that contains ℘iΛ. Then ℘iΛ = Pmii where mi is the Schurindex of the ℘i-adic completion of A. Moreover ℘Λ =

∏di=1P

eimii . We have that mi = 1 for

almost all ℘i and the dual

Λ# := a ∈ A | trace(aΛ) ⊆ S =∏℘i

P1−mii Λ

where the product runs through the prime ideals ℘i of S for which mi > 1.

Proof. Exercises.

Remark 2.6.11. Let Λ be a maximal order in the central simple K-algebra A. Let P (Λ)denote the set of two-sided Λ-ideals which are principal as left ideals, i.e.

P (Λ) := Λa | a ∈ A∗,ΛaΛ = Λa.


Then P (Λ) ≤ I(Λ). The factor group is called the ideal class group of Λ and its order iscalled the 2-sided classnumber of Λ, H(Λ) := |I(Λ)/P (Λ)|. The 2-sided classnumber of Λ isfinite. More precisely

H(Λ) ≤∏℘

m℘(A)h(R).

Here the product runs over all prime ideals ℘ of R that ramify in A and m℘ is the local Schurindex at ℘.

2.6.2 The Brandt groupoid.

We now pass to one-sided ideals. If M,N are left ideals of the maximal order Λ, then it doesnot make sense to define the product MN . Let Λ1 be a maximal order in A and M a leftideal of Λ1. Then M is normal and hence also Λ2 := Or(M) is a maximal order in A. Wewill indicate this are 1M2, which means that Λ1 1M2Λ2 = 1M2. Given ideals 1M2, 2N3 wemay multiply them as usual to obtain

1I3 = 1M2 2N3 = 〈mn | m ∈M,n ∈ N〉R.

So the set of normal ideals in A forms a groupoid the so called Brandt groupoid.Note that if M = Λ1y is a principal left ideal, then Λ2 = y−1Λ1y is conjugate to Λ1.

Theorem 2.6.12. Let Λ1 and Λ2 be maximal orders in A. Then there is a left Λ1-ideal 1M2

which has right order Λ2. Any such ideal M defines a group isomorphism

ϕ12 : I(Λ1)→ I(Λ2), I 7→M−1IM

between the group of two-sided ideals of Λ1 and Λ2. The isomorphism ϕ12 is independent ofthe choice of M .

Proof. M := Λ1Λ2 is such an ideal with Ol(M) = Λ1 and Or(M) = Λ2. That ϕ12 is a groupisomorphism is clear, the inverse is ϕ21 : I 7→ MIM−1. Let 1N2 be another Λ1 − Λ2-idealand denote the resulting isomorphism by ψ12. Then

ψ−112 ϕ12(I) = NM−1IMN−1 = (MN−1)−1(MN−1)I = I

for all I ∈ I(Λ1), since I(Λ1) is a commutative group and (MN−1) ∈ I(Λ1).

Definition 2.6.13. (classnumber and typenumber) Let Λ be some maximal order inA. Two left Λ-ideals M and N are called isomorphic (or equivalent) if there is somea ∈ A such that Ma = N . The number of isomorphism classes of left Λ-ideals is called theclassnumber h(Λ).Two left Λ-ideals M and N are called weakly equivalent, if there is some two-sided Λ-idealI and some a ∈ A such that IMa = N .The number of conjugacy classes of maximal orders in A is called the typenumber t(A) ofA.

We will show below that h(Λ) is finite for any R-order Λ in A.


Remark 2.6.14. The group of twosided ideals I(Λ) acts on the set of left Λ-ideals by leftmultiplication and also on the set of isomorphism classes of left Λ-ideals. Two left Λ-idealsM and N are weakly equivalent, if and only if they are in the same I(Λ)-orbit on the set ofisomorphism classes of left Λ-ideals.The right orders of weakly equivalent left Λ-ideals are conjugate.

Theorem 2.6.15. t(A) is the number of weak equivalence classes of left Λ-ideal for anymaximal order Λ.

Proof. We have already seen that for any maximal order Γ in A there is some left Λ-ideal Msuch that Or(M) = Γ. So all types of maximal orders occur as right orders of some fixed setof representatives of the weak equivalence classes of left Λ-ideals. Now let M and N be Λ-Γbimodules, so left Λ-ideals with right order Γ. We need to show that M and N are weaklyequivalent left Λ-ideals. Now Γ = M−1M = N−1N and hence M = MΓ = MN−1N = INfor I = MN−1 ∈ I(Λ).

2.6.3 The finiteness of the class number.

Let M be a left Λ-ideal. Then M is equivalent to some integral ideal i.e. there is some a ∈ Asuch that Ma ≤ Λ. This is because M and Λ are finitely generated R-submodules of A andalso full lattices. So they have compatible bases and we may take the denominators.The theorem proven in this subsection (finiteness of the class number) is valid in a moregeneral case that Λ is an order over some Dedekind domain. This is the Theorem by Jordanand Zassenhaus (see for instance Section 26 of Reiner’s book Maximal orders).Here we want to restrict to the case where K is a number field and R = ZK . In particularany R-order is a Z-order and we may use Geometry of numbers to obtain explicit bounds onthe norm of the integral ideals that represent all isomorphism classes of left Λ-ideals as inthe commutative case. This exposition follows the book by Max Deuring (Algebren).So assume that K is a number field.

Definition 2.6.16. Let Λ be a maximal R-order in the separable K-algebra A. The normof an integral left Λ-ideal M is N(M) := |Λ/M |. For an arbitary left Λ-ideal M we mayalways choose some invertible a ∈ A such that Ma ≤ Λ. Then N(M) = N(Ma)NA/Q(a)−1

where NA/Q is the regular norm over the rationals.

Remark 2.6.17. The norm of a two-sided prime ideal is a prime power.Let ℘ be a prime ideal of R, Γ and Λ be maximal R-orders in A and PΓ and PΛ the two-sidedprime ideals of Γ resp. Λ that contain ℘Γ resp. ℘Λ. Then N(PΓ) = N(PΛ).The norm of M is the product of the norms of all localisations N(M) =

∏℘ER |Λ℘/Λ℘M |.

The norm is multiplicative, i.e. N( 1M2 2L3) = N(M)N(L).The norm of a principal ideal Λa is N(Λa) = |NA/Q(a)| where NA/Q(a) is the regular normof a in the Q-algebra A.

Theorem 2.6.18. Let A be a separable K-algebra over the number field K. Then there issome C = C(A) ∈ R>0 so that for any maximal order Λ in A and any left Λ-ideal M ′, thereis some integral left Λ-ideal M = M ′a equivalent to M ′ such that N(M) < C.


Proof. We view A as a Q-algebra and show that M ′′ =: (M ′)−1 contains an element a ∈M ′′

such that NA/Q(a) < CN(M ′′). Then aΛ ⊆M ′′ and so M ′a =: M ⊆ Λ has norm N(M) < C.(a) We first assume that A is a division algebra. Then we can argue as in the case of numberfields:Let M be a left ideal of the maximal order Λ and choose some Z-basis B := (b1, . . . , bn)of Λ. The regular norm of an element

∑ni=1 xibi is a homogeneous polynomial of degree

n = dimQ(A) in the variables xi. Let

C := max|NA/Q(n∑i=1

xibi)| | |xi| ≤ 1 for all i.

Embed A into Euclidean space V = Rn by letting B be an orthonormal basis. Then Λ isa lattice in V of volumne 1 and M is a lattice of volumne N(M) and determinant N(M)2.The point set

X := (x1, . . . , xn) ∈ Rn | |xi| ≤ N(M)1/n for all iis a cube of volumne 2nN(M). By Minkowski’s lattice point theorem there is some non-zeroa ∈M such that a =

∑ni=1 aibi with (a1, . . . , an) ∈ X. Then NA/Q(a) ≤ N(M)C.

(b) If A is a simple algebra, A = Dn×n for some division algebra D, then not every non-zeroelement in A is a unit, so it is not enough to find a non-zero element in the lattice M thatlies in the cube X. If Γ is a maximal order in D then Λ := Γn×n is a maximal order in A.Let M be a left Λ-ideal and let a = (aij) ∈ M . Since Λ contains the matrix units eij, theideal Γa contains all elements b of which the rows are linear combinations of the rows of aand M is determined by all these rows. Let Mi be the submodule consisting of thoses rowsin M of which the first (i− 1) coefficients are 0 and let

mi := xi | (0, . . . , 0, xi, xi+1, . . . , xn) ∈Mi ≤ Λ.

Then mi is a left ideal in Λ and

N(M) = N(m1)nN(m2)n · · ·N(mn)n.

By part (a) of the proof there are elements ai ∈ mi such that ND/Q(ai) ≤ CDN(mi). Thenai is the first coefficient of some row (0, . . . , 0, ai, ai,i+1, . . . , ai,n) and the matrix

a :=

a1 a11 . . . a1n

0 a2 . . . a2n...

. . . . . ....

0 . . . 0 an

∈Mhas norm

NA/Q(a) =n∏i=1

ND/Q(ai)n ≤ Cn2

D N(m1)n · · ·N(mn)n = (CD)n2

N(M).

(c) In the semisimple case, A =⊕

iAi, with simple components Ai, we can apply the argu-ment above to all components Ai to obtain constants Ci and then put C :=

∏iCi.


Corollary 2.6.19. The classnumber of a finite dimensional separable Q-algebra A is finite.

Proof. It is enough to show that a given maximal order Λ in A has only finitely many isomor-phism classes of left ideals in A. This follows from the theorem and the fact that Λ/dCeΛ isa finite additive group and therefore has only finitely many subgroups.

Example. Let Λ := 〈1, ζ3, i, iζ3〉 ≤(−1,−3

Q

). Then N(a1 + a2ζ3 + a3i + a4iζ3) = a2

1 +

a1a2 +a22 +a2

3 +a3a4 +a24 so C = 62 = 36. For p 6= 3, the maximal left Λ-ideals I

(1)p , . . . , I

(p+1)p

containing pΛ are of index p2, since Λ/pΛ ∼= Z2×2p /pZ2×2

p∼= F2×2

p . For p = 3 the ringΛ/pΛ ∼= F9[x]/(x2) and again, the maximal ideal I3 has index p2 = 9. So we need to show

that the maximal ideals over 2, 3, 5 and the product I(j)(2)I3 (j = 1, 2, 3) are principal ideals.

Since I3 is 2-sided, the product is principal if the two factors are principal. Now

I(1)2 = Λ(1 + i), I

(2)2 = Λ(1 + i)ζ3, I

(3)2 = Λ(1 + i)ζ2

3 , I3 = Λ(1− ζ3).

Also the six maximal ideals containing 5 are principal ideals, hence h(Λ) = H(Λ) = 1 andt(A) = 1.

2.6.4 The Eichler condition.

Definition 2.6.20. Let K be a number field. A central simple K-algebra A is called atotally definite quaternion algebra, if for all infinite places σ of K the completion Aσ isisomorphic to the real division algebra H =

(−1,−1R

).

The algebra A satisfies the Eichler condition if A is not a totally definite quaternion algebra.

So if A is a totally definite quaternion algebra then dimK(A) = 4 and K is a totally realfield and A =

(a,bK

)with a, b ∈ K, σ(a) < 0 and σ(b) < 0 for all infinite places of K.

The main result of this subsection is the following theorem due to Eichler:

Theorem 2.6.21. (Eichler) Let A be a central simple K-algebra that satisfies the Eichlercondition. Let L be a normal ideal in A. Then L is a principal left-ideal, L = Ol(L)a, if andonly if the reduced norm of L

Nr(L) := 〈N(`) | ` ∈ L〉R−ideal ER = ZK

is a principal ideal Nr(L) = Rα for some α ∈ U(A).

Recall that

U(A) = a ∈ K | σ(a) > 0 for all real places σ that ramify in A

So the determination of the ideal classes in A is reduced to a computation in the center,where we need to find a so called ray class group,

ClA(R) := I(R)/aR | a ∈ U(A).

This is a task in commutative algebraic number theory and can be solved with essentiallythe same methods as the computation of the class group for algebraic number fields. The


non-Eichler case will be treated in the next subsection. Note that the reduced norm is alwaysa map

Nr : L(Λ)/ ∼→ ClA(R)

since principal left-ideals are mapped to 1.This theorem has a long proof, see Section 34 in Reiners book. The main idea is based on

the very strong approximation theorem and the fact that there is always one infinite placefor which the norm form is indefinite. Since the proof does not reveal too much insight andwe are too short in time (the proof would require at least one full 90-minutes lecture) I willomit the proof here.

2.6.5 Stable equivalence of ideals.

Let Λ be a maximal R-order in the central simple K-algebra A, (K a number field andR = ZK).

Definition 2.6.22. Two left Λ-lattices (Λ-modules that are R-lattices) X, Y are called stablyisomorphic, if there is some r ∈ N such that X ⊕ Λr ∼= Y ⊕ Λr. Let [X] denote the stableisomorphism class of X. Then

Cl(Λ) := [X] | X is a left Λ− ideal in A

denotes the additive class group of Λ.

It is clear that X ∼= Y ⇒ [X] = [Y ] ∈ Cl(Λ) so the additive class of an ideal consists offull isomorphism classes.

Theorem 2.6.23. For any two left Λ ideals M,M ′ in A, there is some left Λ ideal M ′′ in Asuch that M ⊕M ′ ∼= Λ⊕M ′′. Define [M ] + [M ′] := [M ′′] ∈ Cl(Λ). This makes the set Cl(Λ)into an abelian group with identity element [Λ].

Proof. By the Chinese Remainder Theorem we may replace M and M ′ by isomorphic leftideals so that M,M ′ ≤ Λ and Λ/M and Λ/M ′ are coprime. So AnnR(Λ/M)+AnnR(Λ/M ′) =R. Let

ϕ : M ⊕M ′ → Λ, ϕ(m,m′) := m+m′.

Then ϕ is surjective and the kernel M ′′ := ker(ϕ) is a left Λ-ideal in A (to see this, tensorthe ses

0→M ′′ →M ⊕M ′ → Λ→ 0

with K). Now Λ is a projective left Λ-module and hence the ses is split, so M⊕M ′ ∼= Λ⊕M ′′.It is trivial to show that addition is independent from the choice of representatives M,M ′

in the stable isomorphism classes and is associative and commutative and that [Λ] is theidentity element.Let us show the existence of inverse elements:Any left Λ ideal M is projective and hence a direct summand of some free Λ-module, so thereis some Λ-module Y such that

M ⊕ Y = Λr for some r ∈ N.


Tensoring with K yields Y as a submodule of Ar−1 and hence Y is isomorphic to some sub-module of Λr−1 of full rank. Hence (exercise) there are left ideals M1, . . . ,Mr−1 such thatY = M1 ⊕ . . .⊕Mr−1. Repeating the argument from above (M1 ⊕M2

∼= M ′′ ⊕Λ) we obtainY ∼= Λr−2⊕N for some left ideal N ≤ Λ. So M⊕N⊕Λr−2 ∼= Λr and hence [M ]+[N ] = [Λ].

Remark 2.6.24. By Exercise Number 17, the additive class group Cl(R) is isomorphic tothe usual ideal class group of R.

Corollary 2.6.25. Let Λ be a maximal R-order in the separable K-algebra A and let Xbe a left Λ-lattice such that KX ∼= Ar. Then there is some left Λ-ideal J in A such thatX ∼= Λr−1 ⊕ J .

Proof. Follows by applying Theorem 2.6.23 successively.

Corollary 2.6.26. Let Λ be a maximal R-order in the separable K-algebra A. Then anymaximal order Γ in Ar×r is conjugate to

Γ ∼

Λ . . . Λ J−1

......

......

Λ . . . Λ J−1

J . . . J Λ′

for some left Λ-ideal J and Λ′ := Or(J).

Proof. Let V = Ar. This is an A-Ar×r-bimodule. Since Λ and Γ are R-orders there is a Λ-Γ-lattice X in V . As a Λ-module, X is isomorphic to Λr−1 ⊕ J . So there is some a ∈ (Ar×r)∗

such that Xa = Λr−1 ⊕ J and a−1Γa = EndΛ(Xa) is a desired.

Note that for r ≥ 2 it is enough to choose J from a system of representatives of stableisomorphism classes of left Λ-ideals.

Some notation: Let K be a number field and A a central simple K-algebra. Let

U(A) = a ∈ K | σ(a) > 0 for all real places σ that ramify in A

and ClA(R) := I(R)/aR | a ∈ U(A) denote the associated ray class group. Thenσ : ClA(R) → Cl(R), [I] 7→ [I] is an epimorphism whose kernel is an elementary abelian 2-group (exercise or later). Let Λ be a maximal R-order in A. Then the reduced norm Nr(L)of any left Λ ideal L in A is

Nr(L) := 〈N(`) | ` ∈ L〉R−ideal ER = ZK

The aim of the rest of this subsection is to prove the following theorem by Swan:

Theorem 2.6.27. Let K be an algebraic number field. Then the reduced norm defines agroup isomorphism

ν : Cl(Λ)→ ClA(R), ν([L]) := [Nr(L)]

between the additive class group of Λ and the multiplicative ray class group of R.


For the proof we need to deal with matrix rings over A. So let B := Ar×r = EndA(Ar).Then B and A have the same ramified prime ideals and hence U(A) = U(B) ≤ K∗ andtherefore ClA(R) = ClB(R). If Λ is a maximal order in A, then Γ := Λr×r = EndΛ(Λr) is amaximal order in B.

Lemma 2.6.28. To each Λ-lattice X ≤ Ar there corresponds a Γ-lattice

ϕ(X) := HomΛ(Λr, X) ≤ B.

If X = J1 ⊕ . . .⊕ Jr for left Λ ideals Ji ≤ A, then Nr(ϕ(X)) =∏r

i=1Nr(Ji).

Proof. We first remark that

ϕ : Λ−mod→ Γ−mod, X 7→ HomΛ(Λr, X)

is an equivalence of categories, since Λr is a progenerator for Λ −mod with endomorphismring Γ. Wlog we may assume that X ≤ Λr and all ideals Ji are contained in Λ. Thenϕ(X) ∼= X ⊕ . . .⊕X is a left-ideal contained in Γ and for the regular norm

NB/K(ϕ(X)) = ordR(Γ/ϕ(X)) = (ordR(Λr/X))r = (r∏i=1

ordR(Λ/Ji))r = (

r∏i=1

NA/K(Ji))r

where ordR(M) =∏s

j=1 ℘j, if M = M0 > M1 > . . . > Ms = 0 with R-composition factors

Mj−1/Mj∼= R/℘j. If [A : K] = n2 then [B : K] = (rn)2 and NA/K = Nrn, NB/K = Nrrn

from which we obtain the formula for the reduced norms.

Corollary 2.6.29. Let X =⊕r

i=1 Ji, X′ =⊕r

i=1 J′i. Then

(i) X ∼= X ′ as Λ-modules, if and only if ϕ(X) ∼= ϕ(X ′) as Γ-modules.

(ii) If X ∼= X ′ then∏r

i=1[Nr(Ji)] =∏r

i=1[Nr(J ′i)] ∈ ClA(R).

(iii) If A satisfies the Eichler condition or r ≥ 2, then X ∼= X ′ if and only if∏r

i=1[Nr(Ji)] =∏ri=1[Nr(J ′i)] ∈ ClA(R).

Proof. (i) follows from the fact that ϕ is a Morita equivalence.(ii) is clear.(iii) is Eichler’s theorem 2.6.21 from the previous section, where one has to note that Ar×r

always satisfies the Eichler condition, if r ≥ 2.

In particular if A satisfies the Eichler condition over R, then two left Λ ideals M,M ′ areisomorphic if and only if they are stably isomorphic. In general we always have that A2×2

satisfies the Eichler condition. This allows us to conclude that two left Λ ideals M and M ′

are stably isomorphic if and only if Λ⊕M ∼= Λ⊕M ′.We now conclude the proof of Theorem 2.6.27: Proof. The proof of the Lemma above

shows that ν is well defined. If J ⊕ J ′ ∼= Λ ⊕ J ′′ then ν(J)ν(J ′) = ν(J ′′), so ν is a grouphomomorphism. If ν(J) = ν(Λ), then by the above J ⊕ Λ ∼= Λ ⊕ Λ and hence J is stablyisomorphic to Λ, [J ] = 1. So ν is injective.To show the surjectivity of ν we need to find a preimage for all classes of non-zero primeideals ℘ in R. Given such a prime ideal ℘, there is a maximal left Λ-ideal P , such that℘Λ < P < Λ. But then the reduced norm Nr(P ) = ℘.


2.6.6 Algorithmic determination of classes and types

This section deals with totally definite quaternion algebras over totally real number fields andis taken from my paper “Finite quaternionic matrix groups”, in which I classified all maximalfinite sugroups of GLn(D) for definite quaternion algebras such that [D : Q]n ≤ 40. My maintools to deal with definite quaternion algebras are lattices. I had programs to list short vectorsin an integral lattice and to compute isomteries of lattices and their automorphism group(Plesken/Souvignier-algorithm).

So let D be a totally definite quaternion algebra with center K, a totally real number field,[K : Q] = d. Then we may find one maximal order Λ using the radical idealiser process toobtain a hereditary order and then enlarging this order by adding suitable integral elements.Then we can find enough classes of Λ left ideals, by listing integral ideals of small norm. Tocheck completeness we use the well known mass formulas developed by M. Eichler.

Theorem 2.6.30. (Eichler’s Massformula, without proof) Let h be the class number of K,K = Z(D), [K : Q] = d, R = ZK D a totally definite quaternion algebra. Let D thediscriminant of D over K and Λ any maximal order in D. Let (Ii)1≤i≤s be a system ofrepresentatives of left ideal classes of Λ, Λi := x ∈ D | Iix ⊆ Ii the right order of Ii andωi := [Λ∗i : R∗] the index of the unit group of R in the unit group of Λi. Then one has:

s∑i=1

ω−1i = 21−d ·|ζK(−1)|·h·

∏℘|D

(NK/Q(℘)− 1)

where the product is taken over all primes ℘ of R dividing the discriminant D of D.

A proof is for instance given in Vigneras’ lecture notes.If Λi and Λj are conjugate in D, one may choose a new representative for the class of Ij

to achieve that Λi = Λj. Then I−1i Ij is a 2-sided Λi-ideal. Moreover the Λ-left ideals Ii and

Ij are equivalent, if and only if I−1i Ij is principal.

So if one reorders the Λi such that the first t orders Λ1, . . . ,Λt form a system of repre-sentatives of conjugacy classes of maximal orders in D and Hi the number of isomorphismclasses of 2-sided ideals of Λi (1 ≤ i ≤ t), then

s∑i=1

ω−1i =

t∑i=1

ω−1i Hi.

Definition 2.6.31. Let Λ1, . . . ,Λt form a system of representatives of conjugacy classes ofmaximal orders in D. Let : D → D denote the quaternionic conjugation, x := tracered(x)−x, where tracered is the reduced trace. Then tracered(Λi) ⊆ R and hence Λi = Λi for all i.

(a) ωi := [Λ∗i : R∗] is the index of the unit groups. (we will see below that ωi is finite.

(b) ω1i := 1

2|x ∈ Λi | xx = 1| denotes the index of ±1 in the group of units in Λi of norm

1.

(c) Let N : D → K, N(x) := xx and ωnsi := N(Λ∗i )/(R∗)2. Then ωi = ω1

i ·ωnsi .


The algorithmic problems in evaluating these formulas are:a) determine the ideals Ij.b) decide whether two maximal orders are conjugate in D.c) determine the length of the orbit of Λ under the Galois group Gal(K/Q).d) determine ω−1

i Hi.Problem a) is the major difficulty here. There is of course the well known geometric

approach to this question using the Minkowski bound on the norm of a representative of theideal classes. From the arithmetic point of view one may apply two different strategies tofind the ideals Ij:There is a coarser equivalence relation than conjugacy namely the stable isomorphism cf.Reiner (35.5). The theorem of Eichler (see Reiner (34.9)) says that the reduced norm is anisomorphism of the group of stable isomorphism classes of Λ-left ideals onto the narrow classgroup of the center K. This gives estimates for the norms of the ideals Ij.A second arithmetic strategy is to look for (commutative, non full) suborders O of D. Thenumber of the maximal orders Λi containing O as a pure submodule can be calculated usingthe formula (5.12) from Vigneras.

Theorem 2.6.32. Let Λ we a maximal order in D and I1, . . . , Is represent the isomorphismclasses of left Λ-ideals in D. Let Λi := Or(Ii) denote the right order of Ii. Let B = ZL with[L : K] = 2, L a maximal subfield of D and let

mi(B) := |ϕ : B → Λi/Λ∗i |.

For a prime ideal ℘ E ZK that divides the discriminant D of D let(B℘

)= −1 if ℘ is inert

in ZL/ZK and(B℘

)= 0 if ℘ is ramified in ZL/ZK. Then

s∑i=1

mi(B) = h(B)∏℘|D

(1−(B

℘

)where h(B) is the class number of B.

Examples:

D = Q∞,2 Here K = Q and there is a unique maximal order Λ up to conjugacy in D:

Λ = 〈1, i, j, 1 + i+ j + ij

2| i2 = j2 = −1, ij = −ji = k〉.

Λ is euclidean and the unit group is Λ∗ = SL2(3) hence ω = 242

= 12. Evaluating themass formula gives us ζQ(−1) = ζ(−1) = 1

12.

D = Q∞,11 Now the mass formula is1

12(11− 1) =

5

6=

1

2+

1

3

We claim that there are maximal orders Λ1 and Λ2 with Λ∗1∼= C4 and Λ∗2

∼= C6.Then these orders are not isomorphic and hence not conjugate and provide a systemof representatives of conjugacy classes of maximal orders in D. The Λ1-left ideals are


represented by Λ1 and Λ1Λ2.The first observation is that the imaginary quadratic number fields Q[i] and Q[

√−3]

both are maximal subfields of D (compare local invariants, in both fields 11 is inert).So there is a maximal order Λ1 containing an element i of order 4 and some maximalorder Λ2 containing an element w of order 6. Since i and w both generate maximalsubfields of D, the centralizer

C1 := CΛ∗1(i) = Z[i]∗ = 〈i〉 and C2 := CΛ∗2

(w) = Z[w]∗ = 〈w〉.

If Λ∗1 6= C1 then there would be an additional element a ∈ Λ∗1 such that 〈i, a〉Q−algebra =D. The group G := 〈i, a〉 is a finite subgroup of D∗ ≤ GL4(Q) and its Z-span is someorder Λ in D. But the discriminant of Λ is a divisor of |G|4. Since disc(Λ1) = 112, wehence obtain that 11 | |G| which is a contradiction since ϕ(11) = 10 > 4. SimilarlyΛ∗2 = 〈w〉.

D = Q√3,∞,∞ Here D = Q[√

3]⊗Q∞,2 == Q[√

3]⊗Q∞,3 and one obtains maximal orders Λ1 and Λ2

with

G1 := SL2(3) ≤ Λ∗1 and G2 := C12.C2 ≤ Λ∗2.

One may also show that these groups of order 24 are maximal finite subgroups of Λ∗1and Λ∗2. But the mass formula says that

[Λ∗1 : Z[√

3]∗]−1 + [Λ∗2 : Z[√

3]∗]−1 =1

12.

What is wrong here? The answer is that Λ∗i 6= Z[√

3]∗ × G for some (finite) sugroupG, but both unit groups are non-split extensions. For i = 1 and 2 there are elementsgi ∈ NΛ∗i

(Gi) such that g2i = 2 +

√3 = u, the fundamental unit in Z[

√3]. Note that u

is a totally positive element, so may be a norm of some element in Λi, but u is not asquare in Z[

√3]∗.

D := Q√3+√

5,∞ Then the narrow class group of K = Q[√

3 +√

5] has order 2 and is generated bya prime ideal dividing 11. So there are 2 stable isomorphism classes of Λ-ideals onecontaining the ideal classes of I1, I2, and I3, the other one the one of I4. The secondstrategy applied to O = Z[ζ5,

√3] gives that there are 2 orders Λi containing a fifth

root of unity, because the class number of O is 2 (and again a prime ideal dividing 11generates the class group).

The problems b), c), and d) can be dealt with using the normform of D:

Let D be a definite quaternion algebra over K and N be its reduced norm which is aquadratic form with associated bilinear form 〈x, y〉 = tr(xy) where tr is the reduced traceand the canonical involution of D. The special orthogonal group

SO(D, N) := ϕ : D → D | N(ϕ(x)) = N(x) for all x ∈ D, det(ϕ) = 1

is the group of all proper isometries of D with respect to the quadratic form N .


Theorem 2.6.33. With the notation above one has

SO(D, N) = x 7→ a1xa−12 | ai ∈ D∗, N(a1) = N(a2)

is induced by left multiplication with elements of D of norm 1 and conjugation with elementsof D∗.

Proof. Clearly the mapping x 7→ a1xa−12 with ai ∈ D∗ and N(a1) = N(a2) is a proper

isometry of the K-vector space (D, N).

To see the converse inclusion let D = 〈1, i, j, ij = k = −ji〉K with i2 = a and j2 = band ϕ : D → D be an isometry of determinant 1 with respect to N . Then N(ϕ(1)) = 1 andafter left multiplication by ϕ(1)−1 we may assume that ϕ(1) = 1. Let b2 := ϕ(i), b3 := ϕ(j),and b4 := ϕ(k). Then tr(bi1) = 0 and hence bi = −bi for all i = 2, 3, 4, and b2

2 = a, b23 = b,

b24 = −ab. Moreover tr(bib

∗j) = 0 = −tr(bibj) and hence bibj = −bjbi for all 2 ≤ i 6= j ≤ 4.

Thus (b2b3)b4 = b4(b2b3) and therefore b4 ∈ Kb2b3 is an element of trace 0 in the field gen-erated by b2b3. Since b2

4 = (b2b3)2, this implies that b4 = ±b2b3. If b4 = b2b3, then ϕ is anK-algebra automorphism of D and hence induced by conjugation with an element of D∗ andwe are done. In this case ϕ is of determinant 1. Hence if b4 = −b2b3, the mapping ϕ hasdeterminant −1, which is a contradiction.

Corollary 2.6.34. Let Λi (i = 1, 2) be two orders in D. Then Λ1 is conjugate to Λ2 if andonly if the lattices (Λ1, N) and (Λ2, N) are properly isometric.

Proof. Clearly if the two orders are conjugate the lattices are properly isometric, so we showthe converse: let ϕ : Λ1 → Λ2 be a proper isometry with respect to N . By the Propositionthere are elements a1, a2 ∈ D∗ with N(a1) = N(a2) such that a1Λ1a

−12 = Λ2. Since 1 ∈ Λ1 this

implies that a1a−12 is an element of norm 1 in Λ2 and hence Λ2 = a1a

−12 a2Λ1a

−12 = a2Λ1a

−12 is

conjugate to Λ1.

Since is the identity on the subspace K and the negative identity on the 3-dimensionalsubspace 1⊥ consisting of the elements of D with trace 0, one easily sees that is an improperisometry (of determinant -1) of (D, N). Thus, if one of the orders Λ1 or Λ2 is stable under ,one may omit the word ”properly” in the Corollary above. Note that this holds particularlyfor maximal orders.

Corollary 2.6.35. Let Λ be an order in D. The group of proper isometries of the lattice(Λ, N) is induced by the transformations of the form b 7→ axbx−1, where a ∈ Λ is an elementof norm 1 and x ∈ ND∗(Λ) normalizes Λ.

Normalizer and 2-sided ideals.

Remark 2.6.36. Let a ∈ ND∗(Λ). Then aΛa−1 = Λ and hence Λa is a principal 2-sidedideal. Two elements a, b ∈ ND∗(Λ) induce the same automorphism by conjugation, if andonly if ab−1 = c ∈ K∗. This means that the quotient (Λa)(Λb)−1 = Λc ∈ I(Λ) is a centralprincipal ideal.


Remark 2.6.37. Assume that R is a principal ideal domain and let ℘1, . . . , ℘s be the primeideals of R that ramify in Λ and let κ : ND∗(Λi) → Aut(Λi), a 7→ (x 7→ axa−1). Then|κ(ND∗(Λi))| = ωi2

sH−1i . Therefore the order of the isometry group

|Aut(Λi, N)| = ω1i︸︷︷︸

left mult.

ωi2sH−1

i︸︷︷︸κ(normalizer)

· 2︸︷︷︸−1

· 2︸︷︷︸quat.conj.

.

where s is the number of finite primes of K that ramify in D. Now 2ω1i is simply the

number of shortest vectors of the lattice (Λi, N) and can easily be calculated. Hence ω−1i Hi =

|Aut(Λi, N)|−1·2s+2·ω1i can be obtained using automorphism groups and short vectors of lattices

to evaluate the mass formula.

If the quaternion algebra D has the additional property that if a prime ideal ℘ E Rramifies in D then all prime ideals that contain ℘∩Z also ramify in D, then the Galois groupGal(K/Q) acts on D (note that this is the case for endomorphism rings of modules of finitegroups, they have uniformily distributed invariants):

Choose a K-basis (1 =: b1, b2, b3, b4) of D. An element σ ∈ Gal(K/Q) defines an au-tomorphism σ of the Q-algebra D by σ(

∑aibi) :=

∑σ(ai)bi. By the Theorem of Skolem

and Noether the class σInn(D) of the automorphism σ does not depend on the chosen ba-sis. Therefore one gets a well defined action of Gal(K/Q) on the set of conjugacy classes ofmaximal orders in D. This action preserves ωi and Hi.


If ni denotes the length of the orbits of the class of Mi under Gal(K/Q) one gets thefollowing table:

d D∑ni(ω

1i ·ωnsi )−1 ·Hi

1 Q∞,2 12−1

Q∞,3 6−1

Q∞,5 3−1

Q∞,2,3,5 3−1 + 3−1

Q∞,7 2−1

Q∞,11 2−1 + 3−1

Q∞,13 1Q∞,17 1 + 3−1

Q∞,19 1 + 2−1

2 Q√2,∞ 24−1

Q√2,∞,2,3 1Q√2,∞,2,5 3−1

Q√3,∞ (12·2)−1 + (12·2)−1

Q√5,∞ 60−1

Q√5,∞,2,3 5−1 ·2Q√5,∞,2,5 5−1

Q√5,∞,5,3 5−1 + 3−1

Q√6,∞ (12·2)−1 + (6·2)−1 + (4·2)−1

Q√7,∞ (4·2)−1 + (3·2)−1 + (12·2)−1

Q√10,∞ 3−1 + 2−1 + 12−1 + 4−1

Q√11,∞ 12−1 + 2−1

Q√13,∞ 12−1

Q√15,∞ 3−1 + (1·2)−1 + (2·2)−1 + 6−1 + (3·2)−1 + 2−1 + 12−1 + (2·2)−1

Q√17,∞ 6−1

Q√21,∞ 12−1 + 6−1

Q√33,∞ 6−1 + 3−1

3 Qθ7,∞,7 14−1

Qθ7,∞,2 12−1

Qθ7,∞,3 6−1 + 7−1

Qθ9,∞,3 18−1

Qθ9,∞,2 12−1 + 9−1

Qω13,∞,13 1Qω19,∞,19 2−1 + 1 + 3·1


d D∑ni(ω

1i ·ωnsi )−1 ·Hi

4 Qθ15,∞ (30·2)−1 + 60−1

Qθ16,∞ 16−1 + 24−1

Qθ20,∞ (20·2)−1 + (12·2)−1 + 60−1

Qθ24,∞ (24·2)−1 + (8·2)−1 + 24−1

Qη17,∞ 6−1 + 2·12−1

Q√2+√

5,∞ 24−1 + 60−1

Q√2+√

5,∞,2,5 5−1 + 2·1·2Qη40,∞ (10·2)−1 + 60−1 + 5−1 + (2·2)−1 + (12·2)−1 + (4·2)−1

Q√3+√

5,∞ 60−1 + (12·2)−1 + (12·2)−1 + (5·2)−1

Qη48,∞ (6·2)−1 + (2·2)−1 + 2·3−1 + 24−1

+(8·2)−1 + 2·(1·2)−1 + (4·2)−1 + (1·2)−1 + (8·2)−1 + (2·2)−1

5 Qθ11,∞,11 22−1 + 3−1

Qθ11,∞,2 12−1 + 1−1 + 11−1

Qθ11,∞,3 6−1 + 1−1 ·2 + 5·1−1 + 1−1 ·2Qσ25,∞,5 3−1 + 5·3−1 ·2 + 5·1−1 ·2 + 5·1−1 + 5·1−1

In the first column the degree d := [K : Q] is given, in the second one the name ofthe quaternion algebra D by giving the rational places that ramify in D. The third columncontains the relevant dimensions n and in the last column, the mass formula of D is expanded.Here the sum is taken over a system of representatives of the orbits of Gal(K/Q) on theconjugacy classes of maximal orders in D.

For the algebraic numbers the following notation is used:Notation. As usual ζm denotes a primitive m-th root of unity in C and

√m a square

root of m. Moreover θm := ζm + ζ−1m denotes a generator of the maximal totally real subfield

of the m-th cyclotomic field. ωm (resp. ηm, σm) denote generators of a subfield K of Q[ζm]with Gal(K/Q) ∼= C3 (resp. C4, C5).


Q∞,2 = 〈1, i, j, ω = 1+i+j+ij2〉Q, where i2 = j2 = (ij)2 = −1, and the maximal order is

〈1, , i, j, ω〉Z,

Q∞,3 = 〈1, ω, i, ωi〉Q ∼= Q[ω] ⊕ Q[ω]i, where ω2 + ω + 1 = 0, i2 = −1, ωi = ω−1 and themaximal order is

〈1, ω, i, ωi〉Z,

Q∞,5 = 〈1, i, j, ij〉Q, where i2 = −2, j2 = −5, (ij)2 = −10, and the maximal order is

〈1, ω := 2+i−ij4

, ρ := 1+i+j2

, ij〉ZQ∞,7 = 〈1, ρ, i, ρi〉Q ∼= Q[ρ] ⊕ Q[ρ]i, where ρ2 − ρ + 2 = 0, i2 = −1, (iρ)2 = −2 and the

maximal order is

〈1, ρ, i, ρi〉Z,

Q∞,11 = 〈1, ρ, i, ρi〉Q ∼= Q[ρ] ⊕ Q[ρ]i, where ρ2 + ρ + 3 = 0, i2 = −1, (iρ)2 = −3 andrepresentatives for the two conjugacy classes of maximal orders are

O1 := 〈1, ρ, i, ρi〉Z (with unit group C4) and

O2 := 〈1, ρ+ i, 2i, 1+ρi2〉Z (with unit group C6).

Q√2,∞,∞ = 〈1, ζ8, α, ζ8α〉Q[√

2]∼= Q[ζ8] ⊕ Q[ζ8]α, where ζ2

8 −√

2ζ8 + 1 = 0, α2 = −1, and

α−1ζ8α = ζ−18 and the maximal order is

〈1, ζ8,1+α√

2, ζ8

1+α√2〉Z[√

2].

Q√3,∞,∞ = 〈1, ζ12, α, ζ12α〉Q[√

3]∼= Q[ζ12] ⊕ Q[ζ12]α, where ζ2

12 −√

3ζ12 + 1 = 0, α2 = −1,

and α−1ζ12α = ζ−112 and representatives for the two conjugacy classes of maximal orders are

O1 := 〈1, ζ12, α, ζ12α〉Z[√

3] (where the torsion subgroup of the unit group is C12.C2) and

O2 := 〈1, (1+i)(1+√

3)2

, (1+j)(1+√

3)2

, 1+i+j+ij2〉Z[√

3], where i := ζ312 and j := α (where the tor-

sion subgroup of the unit group is SL2(3)).

Q√5,∞,∞ = 〈1, ζ5, α, ζ5α〉Q[√

5]∼= Q[ζ5] ⊕ Q[ζ5]α, where with b5 := −1+

√5

2it holds that

ζ25 − b5ζ5 + 1 = 0, α2 = −1, and α−1ζ5α = ζ−1

5 and the maximal order is

〈1, (1−ζ5)(1+αb5)√5

, α, ζ5α〉Z[b5].

Q√5,∞,∞,2,√

5 = 〈1, ζ5, α, ζ5α〉Q[√

5]∼= Q[ζ5]⊕Q[ζ5]α, where with b5 := −1+

√5

2it holds that

ζ25 − b5ζ5 + 1 = 0, α2 = −2, and α−1ζ5α = ζ−1

5 and the maximal order is

〈1, ζ5, α, ζ5α〉Z[b5].

Q√5,∞,∞,3,√

5 = 〈1, ζ5, α, ζ5α〉Q[√

5]∼= Q[ζ5]⊕Q[ζ5]α, where with b5 := −1+

√5

2it holds that

ζ25 − b5ζ5 + 1 = 0, α2 = −3, and α−1ζ5α = ζ−1

5 and representatives for the two conjugacyclasses of maximal orders are

O1 := 〈1, ζ5, α, ζ5α〉Z[b5] (where the torsion subgroup of the unit group is ∼= C10) and

O2 := 〈1, 1+α2, 2ζ5, ζ5(1 + α)rangleZ[b5] (where the torsion subgroup of the unit group is

∼= C6).


Q√6,∞,∞ = 〈1, i, j, ω := 1+i+j+ij2〉Q[√

6], where i2 = j2 = (ij)2 = −1 and representatives forthe three conjugacy classes of maximal orders areO1 := 〈1, 2+

√6

2(1 + i), 2+

√6

2(1 + j), 1+i+j+ij

2α〉Z[

√6] (where the torsion subgroup of the unit

group is SL2(3)),

O2 := 〈1, 3+√

63

(1 + ω) =: a, 2+√

62

(j − i) =: b, ab〉Z[√

6], where ωb = ω−1 (where the torsion

subgroup of the unit group is S3). and

O3 := 〈1, 2+√

62

(1 + i) =: a, j+ij+√

62

=: b, ab〉Z[√

6] (where the torsion subgroup of the unitgroup is Q8).

Q√7,∞,∞ = 〈1, i, j, ij〉Q[√

7], where i2 = j2 = (ij)2 = −1 and representatives for the threeconjugacy classes of maximal orders are

O1 := 〈1, i, j+√

7i2

, ij+√

72〉Z[√

7] (where the torsion subgroup of the unit group is Q8),

O2 := 〈1, j+√

7i2

, i + ij+√

72

, 2+√

73

(1 + (1 −√

7)i − j+√

7i2− ij+

√7

2)〉Z[√

7] (where the torsionsubgroup of the unit group is C3),

O3 := 〈1, 3+√

72

(1 + i), 3+√

72

(1 + j), 1+i+j+ij2〉Z[√

7] (where the torsion subgroup of the unitgroup is SL2(3)),Q√13,∞,∞ = 〈1, i, j, ω := 1+i+j+ij

2〉Q[√

13], where i2 = j2 = (ij)2 = −1 and the maximalorder is〈1, i, i+j+z+zi

2, 1+i+j+ij

2〉Z[z], where z = 3+

√13

2.

Q√21,∞,∞ = 〈1, ω, i, ωi〉Q[√

21], where ω2 +ω+1 = 0, i2 = −1, ωi = ω−1 and representativesfor the two conjugacy classes of maximal orders areO1 := 〈1, (1−ω)(1+z)

3, i, (1−ω)(1+z)

3i〉Z[z], where z = 1+

√21

2,

and O2 := 〈1, ω, 1+2ω+i+zω3

=: j, ω+ωj+z+zj2

〉Z[z].

2.7. AUTOMORPHISMS OF ALGEBRAS 97

2.7 Automorphisms of algebras

2.7.1 Skew Laurent series

Let F be a field, D some F -division algebra, σ : D → D some F -algebra automorphism.

Definition 2.7.1. D((x, σ)) := ∑

i≥n xidi | n ∈ Z, di ∈ D is called the skew Laurent

series ring defined by σ. D((x, σ)) is an (associative and distributive) F -algebra with dx =xdσ for all d ∈ D.

Remark 2.7.2. • (∑

i≥n xidi)(

∑j≥m x

jd′j) =∑

k≥n+m xkd′′k where d′′k =

∑i+j=k d

σj

i d′j ∈

D is a finite sum, so multiplication is well defined.

• D ∼= 1 ·D ∼= x0 ·D → D((x, σ)) is some F -subalgebra.

• (1− x)(1 + x+ x2 + . . .) = 1

• v : D((x, σ)) → Z ∪ ∞, v(∑

i≥n xidi) := n if dn 6= 0 is a discrete valuation on

D((x, σ)).

• V := z ∈ D((x, σ)) | v(z) ≥ 0 is an F -subalgebra of D((x, σ)) with maximal idealM := z ∈ D((x, σ)) | v(z) ≥ 1 and residue skew field V/M ∼= D.

• For any w ∈ M the series 1 + w + w2 + . . . ∈ V is well defined and (1 − w)(1 + w +w2 + . . .) = 1, so (1− w) ∈ V ∗.

Theorem 2.7.3. D((x, σ)) is a skew field and V is a valuation ring.

Proof. We need to show that any non-zero element of D is invertible. Let 0 6= z =(∑

i≥n xidi) ∈ D((x, σ)) with dn 6= 0. Then z = xndn(1 − w) where w = −

∑j≥1 x

jd′j ∈ M .

Therefore z−1 = (1− w)−1d−1n x−n ∈ D((x, σ)).

Remark 2.7.4. Let Fixσ(D) = a ∈ D | aσ = a. Then

Z(D((x, σ))) = ∑i≥n

xidi | di ∈ Fixσ(D) and didd−1i = dσ

i

for all d ∈ D.

If the restriction of σ to the center K = Z(D) has infinite order, then Z(D((x, σ))) ∼=Fixσ(K).

Proof. (a) z :=∑

i≥n xidi ∈ Z(D((x, σ))) if and only if all monomials xidi commute with all

monomials in D((x, σ)) (Exercise).

(b) (xidi)(xjd′) = xi+jd

(σj)i d′ and (xjd′)(xidi) = xi+j(d′)(σi)di. So if z is central, then for all

i, j and all d′ ∈ D we obtain

d(σj)i d′ = (d′)(σi)di

Putting d′ = 1 yields that dσi = di for all i and then did′d−1i = (d′)(σi) for all i and all d′ ∈ D.


2.7.2 Automorphism groups of algebras

Let F be a field, A some F -algebra, so F → Z(A) ⊆ A.

Definition 2.7.5. • Aut(A) := σ : A→ A | σ is a ring automorphism .

• AutF (A) := σ : A→ A | σ is an F -algebra automorphism . Then AutF (A) ≤ Aut(A)more precisely AutF (A) = σ ∈ Aut(A) | σ(F ) = F and σ|F = id.

• Inn(A∗) := Inna : A→ A, x 7→ a−1xa | a ∈ A∗E Aut(A).

• Out(A) := Aut(A)/ Inn(A∗).

• Aut(F,A) := σ ∈ Aut(F ) | σ has a extension to some σ ∈ Aut(A).

Remark 2.7.6. • There is an exact sequence of groups

1→ Z(A)∗ → A∗ → Aut(A)→ Out(A)→ 1.

• All ring automorphisms of A fix the center Z(A) as a set, so there is a restriction map

|Z(A) : Aut(A)→ Aut(Z(A)), σ 7→ σ|Z(A).

• Aut(A) acts on A∗/Z(A)∗ ∼= Inn(A∗)E Aut(A) by application (aZ(A)∗, σ) 7→ aσZ(A∗)respectively by the usual conjugation (Inna, σ) 7→ σ−1 Inna σ. The mapping Inn :A∗/Z(A)∗ → Inn(A∗) is an equivalence of Aut(A)-groups.

Example. Let F ≤ K ≤ L, Gal(L/F ) abelian, Gal(L/K) = 〈ϕ〉 cyclic, [L : K] = n. Fora ∈ K∗ we consider the cyclic algebra A = (L/K,ϕ, a) =

⊕uiL with ù = u`ϕ and un = a.

Then a Galois automorphism σ ∈ Gal(K/F ) can be extended to some automorphism of Aif and only if there is some λ ∈ L∗ such that aσ

a= NL/K(λ). In particular if a ∈ F , then

Gal(K/F ) ≤ Aut(K,A).

Proof. We first note that by Galois theory any σ ∈ Gal(K/F ) can be extended to someσ ∈ Gal(L/F ). ⇐: Let σ ∈ Gal(L/F ) and define σ ∈ Aut(A) by σ : u 7→ uλ and ` 7→ `σ. Toshow that this defines a ring automorphism we need to check the relations:

(uσ)n = (uλ)n = unNL/K(λ) = aNL/K(λ) = aσ = (un)σ

(ù)σ = `σuσ = `σuλ = uλ`σϕ = (u`ϕ)σ.

⇒: Now let σ ∈ Aut(A) such that σ|K = σ. Let σ1 ∈ Gal(L/K) be an extension of σ. ThenL = σ1(L) and σ(L) are K-isomorphic subalgebras of A, so by Skolem/Noether, they areconjugate and we may assume wlog that σ(L) = L. Let λ := uσu−1. Then λ ∈ L = CA(L)and (un)σu−n = aσ/a = NL/K(λ).

2.7. AUTOMORPHISMS OF ALGEBRAS 99

2.7.3 The algebra σA

Definition 2.7.7. Let A be some F -algebra and σ ∈ Aut(F ). Then σA is the F -algebrawhose underlying ring is A and the embedding of F into Z(A) is given by f 7→ fσ1A.

Theorem 2.7.8. σ ∈ Aut(F,A) if and only if σA ∼= A as F -algebra.

Proof. First note that the identity map id : σA→ A, a 7→ a restricts to id|F = σ. Note thatthe identity is an F -algebra isomorphism, if and only if σ = id.If ϕ : A→ σA is an F -algebra isomorphism, then ϕid : A→ σA→ A is a ring automorphismof A with (ϕid)|F = id|F = σ.On the other hand assume that there is some σ ∈ Aut(A) such that σ|F = σ. Thenσ : A→ σA is an F -algebra isomorphism.

2.7.4 The finite dimensional and central simple case.

Let K = Z(A) be a field and A a finite dimensional central simple K-algebra. Then by thetheorem of Skolem and Noether AutK(A) = Inn(A∗). Therefore

Out(A) = Aut(A)/ Inn(A∗) = Aut(A)/AutK(A) ∼= Aut(K,A)

Corollary 2.7.9. Out(A) ∼= Aut(K,A) only depends on the class of A in the Brauer groupof K.

Proof. Write A = Mn(D)(= Dn×n) for some centralK-division algebraD. ThenD is uniquelydetermined since D ∼= EndA(V ) for the simple A-module V . Let σ ∈ Aut(K). Then

σ ∈ Aut(K,A)⇔ σMn(D) = Mn( σD) ∼=K Mn(D)⇔ σD ∼=K D ⇔ σ ∈ Aut(K,D).

2.7.5 Generalized cyclic algebras.

Let K = Z(A) be a field and A a finite dimensional central simple K-algebra.

Lemma 2.7.10. Let σ ∈ Aut(A), ord(σ|K) = n < ∞. Then there is some α ∈ A∗ withInnα = σn such that ασ = α.

Proof. Let β ∈ A∗ such that Innβ = σn (such β exists by the Theorem of Skolem andNoether). Then Innβ and σ commute in Aut(A) so βK∗ = βσK∗ (see Remark 2.7.6) andb := β1−σ ∈ K∗. Let F := Fixσ(K). Then K/F is cyclic and NK/F (b) = 1. The Theorem 90by Hilbert implies the existence of some a ∈ K∗ such that b = aσ

a. Put α := βa.

Definition 2.7.11. In the situation of the lemma put

B := (A, σ, α) :=n−1⊕i=0

uiA with un = α, au = uaσ

Then (A, σ, α) is an F -algebra, where F = Fixσ(K).


Theorem 2.7.12. Let B be as in Definition 2.7.11 and F := Fixσ(K). Then B is a centralsimple F -algebra, dimF (B) = n2 dimK(A), CB(K) = A.

Proof. The proof is similar as for crossed product algebras. Of course B is a ring, F ⊆ K ⊆A ∼= u0A ⊆ B, and

dimF (B) = dimK(B) dimF (K) = dimK(A) dimA(B) dimF (K) = m2 · n · n = (mn)2.

To show thatB is a simple algebra we proceed as for crossed products. Assume that 0 6= XEBis a twosided ideal of B and choose 0 6= x =

∑ri=0 u

iai ∈ X such that r is minimal. Thena0 6= 0 (otherwise multiplication by u−1 yields a smaller r).If r = 0, then 0 6= x = a0 ∈ A and hence 0 6= x ∈ A ∩ X E A, so A ∩X = A because A issimple. But then X = B.If r > 0 then choose b ∈ K∗ \F such that b 6= b(σr). Then x− bxb(σr)−1

=∑r

i=0 uia′i ∈ X with

a′0 = a0(1− b/(bσr)) 6= 0, a′r = ar − b(σr)arb(σr)−1

= 0

contradicting the minimality of r.

Corollary 2.7.13. Let σ ∈ Aut(K), F := Fixσ(K), [K : F ] = n <∞. Then σ ∈ Aut(K,A)if and only if there is some central simple F -algebra B containing K as a subfield such thatCB(K) ∼= A.

Proof. ⇐: from the construction in the theorem above.⇒: The theorem of Skolem and Noether gives the existence of some u ∈ B∗ such that(Innu)|K = σ. In particular u−1Ku = K and therefore u−1Au = A so the restriction of Innuto A is an automorphism of A that extends σ.

Examples:1) Let A, σ, α be as in Lemma 2.7.10, then (Mn(A), (σ)ij, diag(α, . . . , α)) ∼=F Mn((A, σ, α)).2) Let A0 be a central simple F -algebra and K/F cyclic, Gal(K/F ) = 〈σ0〉 of order n. LetA := A0 ⊗F K, σ := id ⊗ σ0 ∈ AutF (A). Then any α as in Lemma 2.7.10 satisfies α ∈ F ∗and for any α ∈ F ∗ we obtain (A, σ, α) ∼= A0 ⊗F (K/F, σ0, α).To see this note that both tensor factors naturally imbed into the generalized cyclic algebra.The images commute, so we obtain a homomorphism of the tensor product, which is injective,since the tensor product is simple and hence an isomorphism by comparing dimensions.

2.7.6 Restriction

Remark 2.7.14. Let B be as in Theorem 2.7.12. Then B ⊗F K ∼= Mn(A).

Proof. B is a free right A-module of rank n and the left multiplication

λ : B → EndA(B) = Mn(A), b 7→ λ(b) : x 7→ bx

is an F -algebra monomorphism. Also K = Z(Mn(A)) ≤ Mn(A) so we obtain an embeddingof the central simple K-algebra λ ⊗ diag : B ⊗F K → Mn(A). This is an isomorphism bycomparing dimensions.

2.8. THE BRAUER GROUP OF Q((T )). 101

Theorem 2.7.15. Let σ ∈ Aut(K), ord(σ) = n, F := Fixσ(K), A/K central simple,[A : K] = m2 <∞. Then σ ∈ Aut(K,A) if and only if [A] ∈ Im(res : Br(F )→ Br(K)).

Proof. ⇒: Is the remark above.⇐: Let B be some central simple F -algebra such that B ⊗F K ∼= Mr(A). The id ⊗ σ ∈Aut(Mr(A)), so σ ∈ Aut(K,Mr(A)) = Aut(K,A).

Examples:1) If K is a local field, then the restriction map Br(F ) → Br(K) is surjective, since this isjust multiplication by n := [K : F ] from Q/Z to Q/Z. In particular Aut(K,A) = Aut(K).2) If K is a global field, ℘ a place of F and ℘′ a place of K that lies over ℘, then the degreen℘ := [K℘′ : F℘] of the completions does not depend on the choice of ℘′ since K/F is Galois.Let A0 be some central simple F -algebra. Then the local invariants for A0 ⊗F K are

inv℘′(A0 ⊗F K) = inv℘(A0)n℘

so these are constant on the places ℘′ that lie over some fixed place ℘ of K. Since we mayalways find some further place that is not decomposed in K/F (to insure that the sum of thelocal invariants of A0 is 0) we obtain

Remark 2.7.16. For any central simple K-algebra A:

σ ∈ Aut(K,A)⇔ [A] ∈ Im(res : Br(F )→ Br(K))⇔ invσ(℘)(A) = inv℘(A) for all places ℘ of K.

2.8 The Brauer group of Q((t)).

2.8.1 Discrete valuated skew fields.

Let (D, v) be some complete discrete valuated skew field, F := Z(D), V := x ∈ D | v(x) ≥0 the valuation ring with maximal ideal M := x ∈ D | v(x) > 0. Then D := V/M is askew field with

F = V ∩ F/M ∩ F ≤ Z(D) ≤ D = V/M.

Any d ∈ D∗ induces an automorphism of V that fixes M , because v(dxd−1) = v(x) for allx ∈ D. So Innd : V → V,M → M defines Innd ∈ Aut(D), x 7→ d−1xd. For any u ∈ V ∗ weobtain Innu = Innu ∈ Inn(D). We assume wlog that v : D∗ → Z is surjective.

Remark 2.8.1. There is a group homomorphism θ : Z = v(D∗) → Gal(Z(D)/F ), γ 7→(Innd)|Z(D) where d ∈ D∗ is any element with v(d) = γ.

Lemma 2.8.2. Let L ≤ D be a subfield. Then L ∩ Z(D) ⊂ Fix(θ(v(L∗))).

2.8.2 Skew Laurent series II

We apply this to the following situation: D/K is a central division algebra, [D : K] = m2 <∞, σ ∈ Aut(D), ord(σ|K) = n < ∞, F = Fixσ(K). In this situation we know that there issome α ∈ D∗ such that σ(α) = α and Innα = σn (Lemma 2.7.10).


Moreover

D := D((x, σ)) := ∑i≥n

xidi | n ∈ Z, di ∈ D

is a discrete valuated division algebra with valuation ring V = ∑

i≥0 xidi | di ∈ D and

maximal ideal M = ∑

i≥1 xidi | di ∈ D and V/M ∼= D.

Remark 2.8.3. A monomial z = xka ∈ D is in the center of D if and only if k = ni anda = αic for some i ∈ Z and c ∈ F . So

Z(D) = F ((t)) =: Z with t := xnα−1.

Moreover D =⊕n−1

i=0 xiD((t)) = (D((t)), σt, αt) with xn = tα = αt, (Innx)|D((t)) = σt and

dσt = dσ for all d ∈ D and tσt = t. By Theorem 2.7.12 ind(D) = nm.

The map θ : Z→ Gal(K/F ) maps any γ ∈ Z to θ(γ) = (Innxγ )|K = σγ.

Subfields

F

D

LK

K

X

L

KL

m

n

2

f

l

n/l

n/l

<=m

Lemma 2.8.4. Let L be some maximal subfield of D, so [L : Z] = mn. Then LK ≤ D issome maximal subfield of D.

Proof. Let Z = F ((t)) = Z(D), e := [v(L∗) : v(Z∗)] = [`Z : nZ] = n`, f := [L : F ]. Then

ef = [L : Z] = nm so m = f`. Let X := FixK(σ`). Then L ∩ K ≤ X and [X : F ] = `,

[K : X] = n`. Moreover

f

`= m ≥ [LK : K] = [L : L ∩K] and [L : L ∩K][L ∩K : F ] = f

Since [L ∩K : F ] ≤ [X : F ] = ` we obtain equality everywhere.

2.8. THE BRAUER GROUP OF Q((T )). 103

Theorem 2.8.5. Assume that char(F ) = 0. If D((x, σ)) =: D is a crossed product algebra(i.e. there is some maximal subfield that is Galois over F ((t)) = Z(D)), then there is somemaximal subfield M ≤ D such that M/F is Galois.

Proof. Let L ≤ D be a maximal subfield such that L/Z(D) is Galois. Put M := LK =LZ(D) ≤ D. Then M is a subfield of D. By Lemma 2.8.4 it is a maximal subfield. To showthat M/F is Galois it is enough to show that M/F is normal. But L/Z is normal, so byHensels Lemma also L/Z is normal. So L/F is normal, K/F is cyclic and hence normal andso is the compositum M .

2.8.3 Non-crossed products over Q((t))

We know come to the main result of this section, the construction of a division algebra Dwith center Q((t)) that is not a crossed product algebra. To this aim we use the previoustheorem and try to construct a division algebra D with Z(D) = K a number field such thatD does not contain a maximal subfield M that is Galois over Q.

(0) Choose some prime ` > 2.

(1) Choose some other prime p such that p ≡` 1 but p 6≡`2 1.

(2) Construct some cyclic extension K/Q such that [K : Q] = ` and p is totally ramifiedin K (for instance the subfield of degree ` of the p-th cyclotomic number field). Put〈σ〉 := Gal(K/Q).

(3) Choose some prime ` 6= q 6≡` 1 that is inert in K/Q.

(4) Choose some cyclic division algebra D/K with center K such that

invq(D) 6= 0 and invσ(℘)(D) = inv℘(D) for all places ℘ of K.

Under these assumptions there is some σ ∈ Aut(D) such that σ|K = σ.

(5) Put D := D((x, σ)). Then Z(D) = Q((t)) with t = xnα−1.

Theorem 2.8.6. D is not a crossed product algebra.

Proof. We use Theorem 2.8.5. If D is a crossed product algebra, then the algebra D withZ(D) = K contains some maximal subfield M such that M/Q is a Galois extension. Then[M : Q] = `2 and we have two possible situations:1) Gal(M/Q) is cyclic. Then K is the unique proper subfield of M and since p is totallyramified in K/Q it is also totally (and tamely) ramified in M/Q. So the completion isMp = Qp( `2

√π) for some prime element π in Qp. But Mp is only normal over Qp if Qp con-

tains the `2-th roots of unity. This contradicts our assumption (1) that p 6≡`2 1.2) Gal(M/Q) ∼= C` × C`. Then we consider the completion at the prime q. By assumptionq is inert in K, so Kq is the unique unramified extension of degree ` of Qq. Since M is amaximal subfield (and hence a splitting field) of D and invq(D) 6= 0, the completion Mq is


a field and hence Mq = KqNq is the compositum of Kq with any other subfield Nq of index` over Qq. Since Nq 6= Kq the extension Nq/Qq is again totally (and tamely) ramified andnow of degree `. Being Galois implies that Qq contains the `th roots of unity, so q ≡` 1 acontradiction to assumption (3).

Example Choose ` = 3, p = 7, K = Q(ζ7 + ζ−17 ). Then q = 2 works and we may define

D a division algebra with center K and local invariants

inv2(D) =1

3, inv5(D) =

2

3

2.8.4 An example where exponent 6= index

Let σ ∈ Aut(K) of order n, F := Fix(σ) and D0 a central F -division algebra. Put D :=D0 ⊗F K, σ := id⊗ σ, α := 1. Then by Remark 2.8.3

D((x, σ)) = (D((t)), σt, t) ∼= D0((t))⊗F ((t)) (K((t))/F ((t)), σt, t)

where the last isomorphism is the one from Example 2) in section 2.7.5.The exponent ofD0((t)) divides the ind(D0((t))) = ind(D0) The exponent of (K((t))/F ((t)), σt, t)

divides ind((K((t))/F ((t)), σt, t)) = [K : F ] = n. And therefore the exponent of D((x, σ))divides lcm(n, ind(D0)). But the index

ind(D((x, σ))) = n ind(D0) > lcm(n, ind(D0)) if gcd(n, ind(D0)) 6= 1.

Index

R-lattice, 56R-order, 562-coboundaries, 672-cocycle, 66

inertia degree , 41ramification index , 41

A-algebra, 4additive class group, 85admissible, 23algebraic number field, 4

basis, 9, 13binary quadratic form defined by γ, 24Brauer-equivalent, 55Brauergroup, 55

center, 4central simple, 52centrally symmetric, 14class group, 13class group of O, 23class number, 13class number of K, 18classnumber, 81classnumber and typenumber, 81cohomology group, 67complete, 37completion, 37completions, 57complex, 15conductor, 22, 29convex, 14covolume, 13crossed product algebra, 67cyclic algebras, 68cyclotomic polynomials, 31

decomposition field, 30

decomposition group, 30Dedekind domain, 10determinant, 9, 13different, 47discrete valuation, 35discrete valuation ring, 35discriminant, 8, 9, 24, 47divides, 11division algebra, 52dual lattice, 9

Eichler, 84Eichler condition, 84equivalent, 67, 81

Fuhrer, 22factor system, 66fractional ideal, 12Frobeniusautomorphism, 46full lattice, 13fundamental domain, 14fundamental parallelotope, 13fundamental units, 21

greatest common divisor, 11groupoid, 81Grunwald-Wang Theorem, 76

Hasse Norm Theorem, 75Hasse-Brauer-Noether-Albert-Theorem, 75Hasse-invariant, 63Hasse-Schilling-Maass, 77

ideal class group, 81ideal group, 13index, 53inertia degree, 28, 30, 46, 60inertia field, 31, 46inertia fields, 62

105

106 INDEX

inertia group, 31integers, 6integral, 4, 9, 17integral basis, 6integral closure, 5, 56integral ideal, 78integral over R, 56integrally closed, 5, 56inverse different, 47, 65isomorphic, 81

lattice, 9, 21left-order, 56Legendre symbol, 34local property, 58localisation, 36localizations, 57

maximal R-order, 56maximal integral ideal, 78minimal polynomial, 4Minkowski metric, 15

norm, 16, 82normal ideal, 78normalized 2-cocycle, 67normalized valuation, 60

order, 21

p-adic number field, 42place, 74places, 17prime ideal, 78principal fractional ideals, 13properly equivalent, 24purely ramified, 46

ramification index, 28, 30, 46, 60ramified, 26ray class group, 84, 86ray class group of O, 23real, 15, 74reduced norm, 56, 84reduced trace, 56regular module, 52regular norm, 52

regular representation, 52regular trace, 52regulator, 21relative Brauer group, 56right-order, 56

separabel, 56simple, 52skew Laurent series ring, 97splitting field, 54, 69stably isomorphic, 85

tamely ramified, 46topological generating system, 45totally definite quaternion algebra, 84trace, 6, 47trace bilinear form, 47Trace-Bilinear-Form, 8typenumber, 81

ultra-metric, 37uniformily distributed invariants, 92unramified, 46

weakly equivalent, 81wildly ramified, 46

Chapter 3

Exercises.

Blatt 1

Aufgabe 1

Sei d ∈ Z− 0, 1 quadratfrei und K = Q(√d).

(i). Bestimmen Sie eine Ganzheitsbasis von K.

(ii). Bestimmen Sie die Einheitengruppe Z∗K im Fall d < 0.

Aufgabe 2

Sei p > 2 eine Primzahl. Weiter sei d ∈ Fp[X] quadratfrei mit deg d > 0. Bestimmen Sie

den ganzen Abschluss von Fp[X] in Fp(X,√d) = Fp(X)[T ]/(T 2 − d).

Aufgabe 3

(i). Zeigen Sie, dass jeder (kommutative) faktorielle Ring ganzabgeschlossen ist.

(ii). Begrunden Sie warum Z[√

5] kein Hauptidealbereich ist.

Aufgabe 4

Sei L/K eine endliche Korpererweiterung. Zeigen Sie:

(i). Jedes α ∈ L induziert einen Endomorphismus multα : L → L, x 7→ αx des K-Vektorraums L.

(ii). Die Abbildung mult : L→ EndK(L), α 7→ multα ist ein injektiverK-Algebrenmorphismus.

(iii). Die Abbildung SL/K : L→ K, α 7→ Spur(multα) ist K-linear.

107

108 CHAPTER 3. EXERCISES.

(iv). Die Abbildung NL/K : L→ K, α 7→ det(multα) ist multiplikativ.

(v). Fur jedes α ∈ L ist µα,K(X) ∈ K[X] irreduzibel von Grad d := [K(α) : K]. Ferner ist

d ein Teiler von n := [L : K] und erfullt µn/dα,K = χα,K = χmultα .

Blatt 2

Sei K ein algebraischer Zahlkorper und n = [K : Q].

Definition

• Eine Ordnung in K ist ein Teilring von K der auch ein Gitter in (K,SK/Q) ist.

• Zu einem Gitter I in K sei O(I) := a ∈ K | aI ⊆ I die zugehorige Ordnung.

• Zu einer Ordnung R in K und einer Primzahl p bezeichne

Jp(R) := a ∈ R | am ∈ pR fur ein m ≥ 0

das p-Radikal von R.

Aufgabe 1

Zeigen Sie:

(i). Der Ring der ganzen Zahlen ZK ist eine Ordnung und enthalt jede Ordnung von K.(Man sagt ZK ist die Maximalordnung von K.)

(ii). Jedes von (0) verschiedene Ideal einer Ordnung in K ist ein Gitter.

(iii). Ist I ein Gitter in K so ist O(I) eine Ordnung in K.

Im Folgenden sei R eine Ordnung in K und p eine Primzahl. Zeigen Sie:

Aufgabe 2

(i). Jp(R) ist ein Ideal von R.

(ii). Es existiert ein m ≥ 0 so, dass Jp(R)m ⊆ pR. (Es sei Jp(R)0 = R.)

109

Aufgabe 3

Es gilt:pR ⊆ pO(Jp(R)) ⊆ Jp(R) ( R .

Insbesondere ist |O(Jp(R))/R| ein Teiler von pn−1.

Aufgabe 4 (Zassenhaus Round 2)

(i). Es ist S := a ∈ ZK | pka ∈ R fur ein k ≥ 0 eine Ordnung von K.

(ii). Ist R = O(Jp(R)) so gilt R = S.

(iii). p teilt |ZK/R| genau dann wenn R ( O(Jp(R)).

(Hinweis zu (b): Ware R ( S, so existiert ein k ≥ 0 mit Jp(R)k ·S 6⊆ R und Jp(R)k+1 ·S ⊆R. Wahle x ∈ Jp(R)k · S −R und zeige xJp(R) ⊆ Jp(R).)

Blatt 3

Aufgabe 1

Bestimmen Sie die Einheitengruppen Z∗Q(√d)

fur d = 2, 3, 5.

Aufgabe 2

Sei K = Q(ζ5) und ¯ bezeichne den Automorphismus auf K welcher durch ζ5 7→ ζ−15

definiert wird. Zeigen Sie:

(i). ϕ : Z∗K → µ(Z∗K), u 7→ u/u ist ein wohldefinierter Gruppenmorphismus.

(ii). −1 liegt nicht im Bild von ϕ. Insbesondere existiert zu jedem u ∈ Z∗K ein k ∈ Z so,dass ζk5u ∈ FixK(〈〉).

(iii). Es ist Z∗K =⟨−ζ5,

1+√

52

⟩.

Hinweis: In der Vorlesung Algebra wurde ZK = Z[ζ5] gezeigt.Ad (b): Ist u ∈ Z∗K mit u = −u so liegt u im Z-Gitter

⟨ζ5 − ζ−1

5 , ζ25 − ζ−2

5

⟩Z und wird

daher von ζ5 − ζ−15 in ZK geteilt.

Ad (c): Es ist FixK(〈〉) = Q(ζ5 + ζ−15 ) ∼= Q(

√5).

Aufgabe 3

Es sei K ein Zahlkorper mit genau r reellen und 2s echt komplexen Einbettungen. DieMenge der Einbettungen heisse G. Zeigen Sie:


(i). Fur t ∈ R>0 ist Xt :=

(zτ )τ∈G ∈ KR :∑

τ∈G |zτ | < t⊂ KR eine zentralsymmetrische

konvexe Menge mit Volumen 2rπstn/n!.

(ii). In jeder Idealklasse von K gibt es ein ganzes Ideal a mit NK/Q(a) ≤ n!nn

(4π

)s√|dK |.(iii). Ist K 6= Q so ist |dK | > 1.

Aufgabe 4

Bestimmen Sie den Isomorphietyp von Cl(Q(√−17)) sowie Vertreter aller Idealklassen.

Blatt 4

Aufgabe 1

Sei K = Q(√−d) mit d = 14 bzw. d = 30. Bestimmen Sie jeweils den Isomorphietyp

und ein Vertretersystem von Cl(K), Cl(K)2 und Cl(K)/Cl(K)2.

Aufgabe 2

Sei K ein algebraischer Zahlkorper. Zeigen Sie:

(i). Es sei b =∏k

i=1 pnii EZK ein Produkt von paarweise verschiedenen Primidealen. Weiter

seien bi ∈ pnii − pni+1i . Dann gibt es ein x ∈ b mit x ≡ bi mod pni+1

i fur alle 1 ≤ i ≤ k.

(ii). Sind a ⊆ b zwei gebrochene Ideale in K, so ist b = a + (x) fur ein x ∈ b.

(iii). Jedes gebrochene Ideal von K kann mit (hochstens) zwei Elementen erzeugt werden.

Hinweis: (a) Chinesischer Restsatz.(b) Ohne Einschrankung ist b =

∏ki=1 p

nii ganz. Weiter darf man annehmen, dass a ebenfalls

ein Produkt der Ideale p1, . . . , pk ist.Wahle nun x wie in (a) und zeige dass jedes Primidealvon ZK die Ideale b und a + (x) mit der selben Vielfachheit teilt.

Aufgabe 3

Es seien p und ` zwei Primzahlen und K = Q(ζp). Zeigen Sie:

(i). ZK = Z[ζp].

(ii). Xp − 1 ∈ F`[X] hat genau dann eine mehrfache Nullstelle in F` falls p = `.

(iii). p ist die einzige Primzahl welche in ZK verzweigt.

(iv). p := (1− ζp) ist das einzige Primideal von ZK welches p enthalt und es gilt ep = p− 1sowie fp = 1.

111

(v). Sei ` 6= p und q ein Primideal von ZK mit ` ∈ q. Dann ist eq = 1 und fq ist die Ordnungvon ` in F∗p.

Hinweis: Satz 3.56 der Vorlesung Algebra WS10/11.

Aufgabe 4

Bestimmen Sie die Primidealzerlegung von `Z[ζ7] fur ` = 2, 3, 7, 13, 29 sowie die Tragheits-bzw. Verzweigungsgrade und Erzeuger der auftretenden Primideale.

Hinweis: Zum Faktorisieren von Polynomen uber endlichen Korpern darf ein beliebigesComputeralgebrasystem benutzt werden. Die Faktorisierungsroutinen fur Ideale in Dedekin-dringen durfen nicht verwendet werden.

Blatt 5

Aufgabe 1

Sei n ∈ Z>1 und ζn eine primitive n-te Einheitswurzel. Zeigen Sie:

(i). Sei n = pr fur eine Primzahl p. Fur je zwei zu p teilerfremde Zahlen i, j ∈ Z ist dann(1− ζ in)/(1− ζjn) ∈ Z[ζn]∗.

(ii). Sei n keine Primzahlpotenz. Dann ist (1−ζn) ∈ Z[ζn]∗. Genauer gilt∏

i∈(Z/nZ)∗(1−ζ in) =1.

Hinweis zu (b). Seien T = d ∈ Z>1 : d | n und P = t ∈ T | t ist Primzahlpotenz. Dannist n =

∑n−1i=0 1i =

∏i∈P Φi(1) ·

∏i∈T−P Φi(1). Folgere Φi(1) ∈ Z∗ fur alle i ∈ T − P .

Aufgabe 2

Es seien K ⊆ L ⊆ M algebraische Zahlkorper und p E ZM ein Primideal. Weiter seiP = p ∩ ZL. Zeigen Sie:

(i). eM/K(p) = eM/L(p) · eL/K(P)

(ii). fM/K(p) = fM/L(p) · fL/K(P)

Aufgabe 3

Sei K = Q(ζ5,√

2). Weiter sei p ∈ 2, 3, 5, 11 und p E ZK ein Primideal das p enthalt.Bestimmen Sie die Zerlegungs- und Tragheitsgrade sowie die Zerlegungs- und Tragheitsgruppenvon p.

Aufgabe 4

Es seien a und b gebrochene Ideale in einem algebraischen Zahlkorper K. Zeigen Sie:


(i). Es existieren x, y ∈ K∗ so, dass xa und yb ganze teilerfremde Ideale von ZK sind.

(ii). Es ist a⊕ b ∼= ab⊕ ZK als ZK-Moduln.

(iii). a ist ein projektiver ZK-Modul.

Hinweis: Aufgabe 2 von Blatt 4 sowie Ubungen zur Algebra WS10/11, Aufgabe 1 auf Blatt11.

Blatt 6

Fur einen kommutativen Ring R bezeichne Spec(R) die Menge der von (0) verschiedenenPrimideale von R. Ist p ein Primideal in einem Dedekindring R und M ein R-Modul so seiMp := M ⊗R Rp die Komplettierung von M an p.

Aufgabe 1

Es sei R ein Dedekindring. Weiter seien V ein endlich dimensionaler Quot(R)-Vektorraumund L,L′ zwei volle R-Gitter in V . Zeigen Sie:

(i). Ist M ein endlich erzeugter R-Modul mit Mp = 0 fur alle p ∈ Spec(R), so ist M = 0.

(ii). p ∈ Spec(R) | Lp 6= L′p ist endlich.

(iii). Es ist L = L′ genau dann wenn Lp = L′p fur alle p ∈ Spec(R) gilt.

Aufgabe 2

Sei R ein Noetherscher ganzabgeschlossener Integritatsbereich. Zeigen Sie, dass die fol-genden Aussagen aquivalent sind.

(i). Jedes von (0) verschiedene Primideal von R ist maximal, d.h. R ist ein Dedekindring.

(ii). Fur alle p ∈ Spec(R) ist R(p) ein diskreter Bewertungsring.

Aufgabe 3

Sei p eine Primzahl. Bestimmen Sie die Menge der Quadrate in Z∗p, Zp sowie Q∗p. Bes-timmen Sie ferner Erzeuger sowie den Isomorphietyp von Z∗p/(Z∗p)2 bzw. Q∗p/(Q∗p)2.

Hinweis: Im Fall p > 2 fixiere man ein Nichtquadrat ε ∈ F∗p. Der Fall p = 2 ist gesondertzu betrachten.

Aufgabe 4

Faktorisieren Sie X15 − 1 ∈ Zp[X] fur p ∈ P := 2, 3, 11. Die auftretenden p-adischenZahlen sind modulo p4 zu approximieren. Bestimmen Sie ferner die Primidealzerlegung sowiedie Tragheits- und Zerlegungsgrade von pZ[ζ15] in Z[ζ15] fur alle p ∈ P .

113

Blatt 7

Aufgabe 1

Bestimmen Sie (bis auf Isomorphie) alle Erweiterungen von Q5 von Grad 4.

Aufgabe 2

Sei K ein algebraischer Zahlkorper. Zeigen Sie: Die Menge der in K/Q verzweigtenPrimzahlen ist endlich. Ist ferner K 6= Q so verzweigt mindestens eine Primzahl in K.

Aufgabe 3

(i). Bestimmen Sie Q3(√

3)∗.

(ii). Bestimmen Sie Q3(√−3)∗.

(iii). Sei p eine Primzahl. Bestimmen Sie Erzeuger von Q∗p/(Q∗p)3.

Aufgabe 4

(i). Es seien E/L und L/K endliche separable Erweiterungen. Zeigen Sie

D(E/K) = NL/K(D(E/L))D(L/K)[E:L] .

(ii). Es seien K1/K und K2/K endliche separable Erweiterungen und L := K1K2. ZeigenSie: Sind D(K1/K) und D(K2 : K) teilerfremd und gilt [L : K] = [K1 : K][K2 : K], soist

D(L/K) = D(K1/K)[K2:K]D(K2/K)[K1:K] .

(iii). Es seien K1 = Q(i), K2 = Q(√

5), K3 = Q(√−5) und L = Q(i,

√5). Bestimmen Sie

D(L/Q), D(Ki/Q) sowie D(L/Ki) fur i = 1, 2, 3. Gibt es ein Primideal in ZK3 welchesin ZL verzweigt?

Hinweis zu (a): Ohne Einschrankung ist K vollstandig diskret bewertet. Fur (b) siehe Lemma1.7.6.

”Ubung Algebraische Zahlentheorie II

Prof. Dr. Nebe (WS 11/12)

Aufgabe 1. (Die Brauergruppe der reellen Zahlen.)(a) Zeigen Sie, dass Br(R) ∼= C2.(b) Bestimmen Sie die Signatur der Spurbilinearform von R2×2 und von H (den Hamilton


Quaternionen).(c) Entwerfen Sie einen Algorithmus, der den Isomorphietyp einer in ihrer regularen Darstel-lung gegebenen halbeinfachen R-Algebra A bestimmt.(d)∗ Implementieren Sie (c) in einem Computeralgebra-System.

Aufgabe 2. (Zerfallungskorper) Sei D eine K-Divisionsalgebra und L eine maximalkommutative Teilalgebra von D. Zeigen Sie(a) K ≤ Z(D) ≤ L, L = CD(L) und L ist ein Korper.(b) L⊗Z(D) D ∼= Ln×n wobei n der Index von D als zentral einfache Z(D)-Algebra ist.(c) Sei D = 〈1, i, j, k〉Q mit i2 = j2 = k2 = −1, k = ij. Zeigen Sie dass D eine Divisionsalge-bra ist und bestimmen Sie einen expliziten Isomorphismus wie in (b) fur L = Q[i].(d) Sei D wie in (c). Bestimmen Sie ein a ∈ D∗ mit aia−1 = j.(e) Ist D ⊗Q Qp eine Divisionsalgebra fur p = 2, 3, 5?

Aufgabe 3. (Divisionsalgebren uber p-adischen Zahlen)Sei p eine Primzahl, z ∈ Qp ein primitive (p2 − 1)-te Einheitswurzel und

Z :=

(z 00 zp

), P :=

(0 1p 0

)Sei weiter A = 〈I2, Z, P, ZP 〉Qp ≤ Qp[z]2×2.(a) Zeigen Sie, dass A eine Qp-Divisionsalgebra ist.(b) Bestimmen Sie einen expliziten Isomorphismus Qp[z]⊗Qp A→ Qp[z]2×2.(c) Zeigen Sie, dass jede quadratische Erweiterung von Qp ein maximaler Teilkorper von Aist.

Aufgabe 4. Sei R ein Dedekindbereich mit Quotientenkorper K und V ein endlichdimensionaler K-Vektorraum. Sei L ein R-Gitter in V .(a) Ist M ein R-Gitter in V , so gilt M℘ = L℘ fur fast alle maximalen Ideale ℘ER.(b) Sei fur alle ℘ Emax R ein R℘-Gitter X(℘) in V gegeben so dass X(℘) = L℘ fur fast allemaximalen Ideale ℘ E R. Dann ist M :=

⋂℘X(℘) ein Gitter in V mit M℘ = X(℘) fur alle

℘.(c) Ist ℘ ein maximales Ideal von R, so liefert die Abbildung M 7→M℘ eine Bijektion zwischen

M ≤ L | L/M ist ℘-torsion → M ≤ L℘ |M volles Gitter

(d) Die Mengen der R℘-Gitter in V und die der R℘-Gitter in der Vervollstandigung K℘ ⊗ Vstehen in Bijektion.

Aufgabe 5. Sei R Dedekindbereich, ℘ER ein maximales Ideal und S := (R− ℘)−1R dieLokalisierung von R an ℘. Zeigen Sie:Ist

0→ Aa→ B

b→ C → 0

eine kurze exakte Sequenz endlich erzeugter R-Moduln, so ist

0→ S ⊗R Aid⊗a→ S ⊗R B

id⊗b→ S ⊗R C → 0

exakt. Kurz: Lokalisieren ist ein exakter Funktor.Zeigen Sie auch dass Komplettieren ein exakter Funktor ist.

115

Aufgabe 6. (a) Sei Λ eine R-Maximalordnung in der separablen K-Algebra A. Dannist der Matrixring Λn×n ein R-Maximalordnung in An×n, insbesondere ist Rn×n eine R-Maximalordnung in Kn×n.(b) (unabhangig von (a)) Sei R ein Dedekindbereich und M ein volles R-Gitter in A. Dannist Or(M) := a ∈ A | aM ⊆M eine R-Ordnung in A.Zeigen Sie, dass fur jedes maximale Ideal ℘ER gilt

Or(M℘) = Or(M)℘ und Or(M℘) = Or(M)℘.

Aufgabe 7. (Quaternionenalgebren) Sei K ein Korper der Charakteristik 6= 2 und D einezentral einfache K-Algebra der Dimension 4. Zeigen Sie:(a) Es gibt a, b ∈ K∗, i, j ∈ D mit i2 = a, j2 = b, ij = −ji. Bezeichnung: D =

(a,bK

).

(b) Fur die reduzierte Norm gilt N(x+ yi+ zj + tij) = x2 − ay2 − bz2 + (ab)t2.(c) Die Abbildung x + yi + zj + tij 7→ x − yi − zj − tij ist ein K-Algebren Isomorphismuszwischen D und Dop.(d) Ist D eine Divisionsalgebra, so hat [D] Ordnung 2 in Br(K).(e)(a,bK

) ∼= (α,βK ) genau dann wenn die 3-dimensionalen quadratischenK-Vektorraume (K3, diag(−a,−b, ab))und (K3, diag(−α,−β, αβ)) isometrisch sind.(f)D∗/K∗ ∼= SO(diag(−a,−b, ab)) = A ∈ SL3(K) | A diag(−a,−b, ab)Atr = diag(−a,−b, ab)

Aufgabe 8.(a)(a,bR

)ist eine Divisionsalgebra, genau dann wenn a < 0 und b < 0.

(b) Sind a, b ∈ Z, so ist Λ := 〈1, i, j, ij〉Z eine Z-Ordnung in(a,bQ

). Bestimmen Sie Λ# und

|Λ#/Λ|.(c) Zeigen Sie, dass fur a, b ∈ Z die Hasse Invariante von

(a,bQ

)⊗ Qp trivial ist, falls p kein

Teiler von 2ab ist.(d) Bestimmen Sie die Hasse Invarianten von

(a,bQ

)⊗Qp fur alle p und folgende Paare (a, b):

(−1,−1), (−1,−3), (2, 5), (−2,−5), (−2, 5), (2,−5).

(e) Sei D =(−2,−5

Q

). Zeigen Sie, dass E = Q[ζ5] ein Zerfallungskorper fur D ist, jedoch kein

echter Teilkorper von E die Divisionsalgebra D zerfallt.

Aufgabe 9.Sei A = 〈1, ρ, i, ρi〉Q ∼= Q[ρ]⊕Q[ρ]i, wo ρ2 + ρ+ 3 = 0, i2 = −1, (iρ)2 = −3.(a) Sei Λ := 〈1, ρ, i, ρi〉Z. Zeigen Sie, dass Λ eine Maximalordnung in A ist.(b) Bestimmen Sie alle Hasse Invarianten von A.(c) Sei Γ = 〈1, ρ+ i, 2i, 1+ρi

2〉Z. Zeigen Sie, dass Γ eine Maximalordnung ist.

(d) Bestimmen Sie die Einheitengruppen von Γ und von Λ und folgern Sie, dass Λ und Γnicht isomorph sind.

Zyklische Algebren. Sei Gal(L/K) = 〈σ〉 zyklisch der Ordnung n > 1,

A := (L/K, σ, a) :=n−1⊕j=0

ujL, xu = uxσ, un = a ∈ L∗.


Aufgabe 10. Sei g : G × G → L∗ ein normalisiertes Faktorensystem. Dann ist(L/K, g) ∼= (L/K, σ, a) wobei a =

∏n−1j=0 gσ,σj ∈ K∗.

Folgern Sie, dass H2(G,L∗) ∼= K∗/NL/K(L∗). Der Exponent von H2(G,L∗) teilt also ins-besondere die Ordnung von G.

Aufgabe 11. Seien a, b ∈ K∗.(a) (L/K, σ, a) ∼= (L/K, σs, as) fur alle s ∈ Z, ggT(s, n) = 1.(b) (L/K, σ, 1) ∼= Kn×n.(c) (L/K, σ, a) ∼= (L/K, σ, b) ⇔ b = NL/K(c)a fur ein c ∈ L∗.(d) (L/K, σ, a)⊗K (L/K, σ, b) ∼= (L/K, σ, ab)n×n.(e) Jedes [A] ∈ Br(L/K) hat eine Ordnung, die n teilt.

Aufgabe 12.(a) Sei E/K weitere Korpererweiterung, F := E ∩ L, G = 〈σ〉 = Gal(L/K), H := 〈σk〉 =Gal(L/F ) = Gal(EL/E).Dann ist E ⊗K (L/K, σ, a) ∼= (EL/E, σk, a).(b) Sei E ⊇︸︷︷︸

r

L ⊇︸︷︷︸s

K, G = Gal(E/K) = 〈σ〉, H = Gal(E/L) = 〈σr〉, G = Gal(L/K) =

〈σH = σ〉 = G/H.Fur alle a ∈ K∗ ist (L/K, σ, a) ∼ (E/K, σ, ar).

Aufgabe 13. Sei A eine Q-Algebra, A := 〈α, τ〉Q−Alg. mit α4 + 4α2 + 2 = 0, τ 4 =2, τατ−1 = 3α + α3.Schreiben Sie A als Q-Teilalgebra von Q[a]4×4, mit a4 + 4a2 + 2 = 0.Zeigen Sie A ∼= Q4×4.(Hinweis: Q[a]→ Q[a] : a 7→ 3a+ a3 erzeugt Gal(Q[a]/Q) und N(a) = 2.)

Aufgabe 14. Sei K := Q[√−7]. Bestimmen Sie eine zentrale K-Divisionsalgebra D mit

Hasse Invarianten 13

und 23

an den beiden Primstellen uber 2 und 0 an allen anderen Stellen.Bestimmen Sie Dop.Hinweis: Q[ζ7] ist ein maximaler Teilkorper von D.Ist [D] im Bild von Br(Q)→ Br(K)?

Aufgabe 15. Sei A eine Q-Algebra, A := 〈α, τ〉Q−Alg. mit α4 + α3 + α2 + α+ 1 = 0, τ 4 =2, τατ−1 = α2.

(i) Bestimmen Sie Matrizen in Q[ζ5]4×4 (ζ5 bezeichne eine primitive 5-te Einheitswurzel),welche die Relationen von α und τ erfullen und eine zu A isomorphe Q-Teilalgebra vonQ[ζ5]4×4 erzeugen. Zeigen Sie, daß A zentral einfache Q-Algebra ist.

(ii) Zeigen Sie: Q2 ⊗Q A hat Schurindex 4 und bestimmen Sie die Hasseinvariante vonQ2 ⊗Q A.(Hinweis: Q2[α] ist unverzweigt uber Q2.)

(iii) Sei Λ := Z[α, τ ] ≤ A. Bestimmen Sie die Diskriminante von Λ bzgl. der reduziertenSpur.

(iv) Folgern Sie aus (iii) dass Qp⊗QA ∼= Q4×4p fur alle Primzahlen p 6= 2, 5. und dass Z2⊗ZΛ

die Maximalordnung in Q2 ⊗Q A ist.

117

(v) Geben Sie einen Epimorphismus Z5 ⊗Z Λ → F54 an und schließen Sie mit (iii), daßQ5 ⊗Q A eine Divisionsalgebra ist.

(Hinweis: τ 7→ 214 , α 7→ 1.)

(vi) Bestimmen Sie die Hasseinvariante von Q5 ⊗Q A.(Hinweis: πA := α−1 erzeugt das maximale Ideal J(Z5⊗ZΛ) von Z5⊗ZΛ, und Q5[τ ] istein uber Q5 unverzweigter maximaler Teilkorper von Q5⊗QA. Berechnen Sie die Potenzdes Frobeniusautomorphismus, welche πA auf Z5⊗Z Λ/J(Z5⊗Z Λ) ∼= F5⊗Z5 Z5[τ ] ∼= F54

induziert.)

(vii) Zeigen Sie R⊗Q A ∼= R4×4.

Aufgabe 16. Ein Absolutbetrag eines Schiefkorpers K ist eine Funktion | · | : K → Rmit(i) |x| ≥ 0 und |x| = 0⇔ x = 0. (ii) |xy| = |x||y|. (iii) |x+ y| ≤ |x|+ |y|.Zwei Betrage | · |1 und | · |2 heissen aquivalent, wenn sie dieselbe Topologie definieren.(a) Zeigen Sie: | · |1 und | · |2 sind genau dann aquivalent, wenn es ein s ∈ R>0 gibt mit|x|1 = |x|s2 fur alle x ∈ K.(b) Zeigen Sie: | · |1 und | · |2 sind genau dann aquivalent, wenn fur alle x ∈ K gilt |x|1 < 1⇒ |x|2 < 1.(c) (Der schwache Approximationssatz) Sind | · |i paarweise inaquivalente Betrage von K(i = 1, . . . , n) und a1, . . . , an ∈ K gegeben, so gibt es zu jedem ε > 0 ein x ∈ K mit|x− ai|i < ε fur alle i = 1, . . . , n.(Hinweis: Kap. II Abschnitt 3 in Neukirch, algebraische Zahlentheorie.)(d) Verschiedene Stellen eines algebraischen Zahlkorpers K definieren inaquivalente Betrage.(e) Der (sehr) starke Approximationssatz sagt aus, dass fur globale Korper K unter derVoraussetzung von (c) und der zusatzlichen Annahme, dass | · |0 eine von | · |1, . . . , | · |nverschiedene Stelle von K gegeben ist, ein x ∈ K gefunden werden kann, das |x − ai|i < εerfullt fur alle i = 1, . . . , n und |x|k ≤ 1 fur alle Stellen | · |k 6= | · |i (i = 0, . . . , n), also x ganzist bei allen anderen Stellen (ausser einer vorgegebenen | · |0).

Aufgabe 17. (Steinitzinvariante) R sei ein Dedekindbereich, Ji, Ik gebrochene Ideale.Dann gilt:

J1 ⊕ . . .⊕ Jn ∼= I1 ⊕ . . .⊕ Im ⇔ n = m und J1 · · · Jn ∼= I1 · · · Im

in der Idealklassengruppe von R. Folgern Sie, dass die multiplikative Idealklassengruppe vonR und die additive Idealklassengruppe von R isomorph sind.

Aufgabe 18. Sei L/K eine endliche Erweiterung algebraischer Zahlkorper, R = ZK ,S = ZL. Sei Λ eine Maximalordnung in einer zentral einfachen L-Algebra A. Fur einPrimideal P E R sei PS = P e1

1 · · ·Pedd . Sei ℘i das Primideal von Λ das PiΛ enthalt. Dann

ist PiΛ = ℘mii wobei APi = Dki×kii und [Di : LPi ] = m2

i . Weiter ist PΛ =∏d

i=1 ℘eimii und

Λ# := a ∈ A | trace(aΛ) ⊆ S =∏

PiEmaxS

℘1−mii Λ


Definite Quaternionenalgebren: Mit Qα,p1,...,ps bezeichnen wir die definite Quaternione-nalgebra mit Zentrum K := Q[α] (ein total reeller Zahlkorper) die unter Vervollstandigungan der Stelle ℘ genau dann eine Divisionsalgebra ist, wenn ℘ eine der Stellen pi ist.Aufgabe 19.

a) Sei K ein imaginar quadratischer Zahlkorper. Zeigen Sie dass Z∗K = 〈−1〉 außer furK = Q[

√−1] und K = Q[

√−3], wo Z∗K ∼= C4 resp. C6.

b) Zeigen Sie, dass Q∞,3 Typenzahl und Klassenzahl 1 hat. Die Maximalordnung dieserDivisionsalgebra ist ein (Links) Euklidischer Bereich.

c) Bestimmen Sie die Klassenzahl und Typenzahl der Quaternionenalgebra Q∞,13.

d) Bestimmen Sie die Klassenzahl und Typenzahl der Quaternionenalgebra Q∞,2,5,7.

Aufgabe 20. Bestimmen Sie (mit einem CAS, z.B. Magma) Klassenzahl, Typenzahl,das Brandtsche Gruppoid, Vertreter der Konjugiertenklassen von Maximalordnungen fur dieQuaternionenalgebra Q√7,∞,∞.

Hinweis: ζK(−1) = 23

und die Klassenzahl h(ZK) = 1 fur K = Q[√

7] = Z(Q).

Aufgabe 21. Zeigen Sie, dass fur jede Maximalordnung Λi die Ordnung der Automor-phismengruppe des ZK-Gitters (Λi, N) gegeben ist als

|Aut(Λi, N)| = ω1i︸︷︷︸

left mult.

ωi2sH−1

i︸︷︷︸κ(normalizer)

· 2︸︷︷︸−1

· 2︸︷︷︸quat.conj.

.

1 Algebraic Number Theory Vorlesung 2016 · Reiner, Maximal Orders Stichtenoth, Algebraic Function Fields (Seminar) Chapter 1 Commutative Theory. All rings are associative and have

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