Algebraic Number Theory: Vorlesung 2011

114 Pages·2016·0.726 MB·English

by  Gabriele Nebe

1
Algebraic Number Theory
Vorlesung 2011
Prof. Dr. G. Nebe, Lehrstuhl D fu¨r Mathematik, RWTH Aachen
Contents
1 Commutative Theory. 4
1.1 The ring of integers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.1.1 The integral closure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.1.2 Norm, Trace and Discriminant. . . . . . . . . . . . . . . . . . . . . . . 6
An algorithm to determine an integral basis of a number field. . . . . . 9
1.1.3 Dedekind domains. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
1.2 Geometry of numbers. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
1.3 Finiteness of the ideal class group. . . . . . . . . . . . . . . . . . . . . . . . . . 17
1.4 Dirichlet’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
1.5 Quadratic number fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
1.5.1 Imaginary quadratic number fields. . . . . . . . . . . . . . . . . . . . . 24
1.6 Ramification. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
1.6.1 How to compute inertia degree and ramification index ? . . . . . . . . . 28
1.6.2 Hilbert’s theory of ramification for Galois extensions. . . . . . . . . . . 29
1.7 Cyclotomic fields. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
1.7.1 Quadratic Reciprocity. . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
1.8 Discrete valuation rings. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
1.8.1 Completion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
1.8.2 Hensel’s Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
1.8.3 Extension of valuations. . . . . . . . . . . . . . . . . . . . . . . . . . . 40
1.9 p-adic number fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
1.9.1 Unramified extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
1.10 Different and discriminant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
2 Non-commutative theory. 49
2.1 Central simple algebras. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
2.1.1 Simple algebras. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
2.1.2 The theorem by Skolem and Noether. . . . . . . . . . . . . . . . . . . . 51
2.1.3 The Brauer group of K . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
2.2 Orders in separable algebras. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
2.2.1 Being a maximal order is a local property . . . . . . . . . . . . . . . . 54
2.3 Division algebras over complete discrete valuated fields. . . . . . . . . . . . . . 56
2.3.1 General properties. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
2.3.2 Finite residue class fields. . . . . . . . . . . . . . . . . . . . . . . . . . 58
2.3.3 The central simple case: analysis . . . . . . . . . . . . . . . . . . . . . 59
2
CONTENTS 3
2.3.4 The central simple case: synthesis . . . . . . . . . . . . . . . . . . . . . 61
2.3.5 The inverse different. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
2.3.6 Matrix rings. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
2.4 Crossed product algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
2.4.1 Factor systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
2.4.2 Crossed product algebras . . . . . . . . . . . . . . . . . . . . . . . . . . 64
2.4.3 Splitting fields. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
2.4.4 Field extensions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
Ground field extensions. . . . . . . . . . . . . . . . . . . . . . . . . . . 68
Field extensions of L. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
2.4.5 A group isomorphism Br(K) ∼= Q/Z. . . . . . . . . . . . . . . . . . . . 70
2.5 Division algebras over global fields. . . . . . . . . . . . . . . . . . . . . . . . . 71
2.5.1 Surjectivity of the reduced norm . . . . . . . . . . . . . . . . . . . . . . 74
2.6 Maximal orders in separable algebras. . . . . . . . . . . . . . . . . . . . . . . . 75
2.6.1 The group of two-sided ideals. . . . . . . . . . . . . . . . . . . . . . . . 76
2.6.2 The Brandt groupoid. . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
2.6.3 The finiteness of the class number. . . . . . . . . . . . . . . . . . . . . 79
2.6.4 The Eichler condition. . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
2.6.5 Stable equivalence of ideals. . . . . . . . . . . . . . . . . . . . . . . . . 81
2.6.6 Algorithmic determination of classes and types . . . . . . . . . . . . . . 84
2.7 Automorphisms of algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93
2.7.1 Skew Laurent series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93
2.7.2 Automorphism groups of algebras . . . . . . . . . . . . . . . . . . . . . 94
2.7.3 The algebra A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95
σ
2.7.4 The finite dimensional and central simple case. . . . . . . . . . . . . . . 95
2.7.5 Generalized cyclic algebras. . . . . . . . . . . . . . . . . . . . . . . . . 95
2.7.6 Restriction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96
2.8 The Brauer group of Q((t)). . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97
2.8.1 Discrete valuated skew fields. . . . . . . . . . . . . . . . . . . . . . . . 97
2.8.2 Skew Laurent series II . . . . . . . . . . . . . . . . . . . . . . . . . . . 97
Subfields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98
2.8.3 Non-crossed products over Q((t)) . . . . . . . . . . . . . . . . . . . . . 99
2.8.4 An example where exponent (cid:54)= index . . . . . . . . . . . . . . . . . . . 100
3 Exercises. 103
Literatur:
Neukirch, Algebraische Zahlentheorie
Reiner, Maximal Orders
Stichtenoth, 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 (cid:15) : K[x] →
a
L,p(x) (cid:55)→ p(a)definesaK-algebrahomomorphismwithimageK[a](theminimalK-subalgebra
of L that contains a). Since K[x] is a principal ideal domain, the kernel of (cid:15) is generated
a
by a monic polynomial Kern((cid:15) ) = (µ (x)). The image of (cid:15) is an integral domain, so
a a a
µ (x) ∈ K[x] irreducible. This uniquely determined monic irreducible polynomial µ is called
a a
the minimal polynomial of a over K.
√ √
Example. a = 1+ 5 ∈ Q[ 5] ⇒ µ = x2−x−1 is the minimal polynomial of a over Q.
2 a
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 a ,...,a ∈ A
1 n
such that
((cid:63)) bn +a bn−1 +...+a b+a = 0.
1 n−1 n
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 ((cid:63)) implies that A[b] = (cid:104)1,b,...,bn−1(cid:105) .
A
(b) ⇒ (c): Clear.
(c) ⇒ (a): Let R = (cid:104)b ,...,b (cid:105) ≤ B be some A-subalgebra of B that contains b. Assume
1 n A
wlog that 1 ∈ R. Then there are (not necessarily unique) a ∈ A such that
ij
n
(cid:88)
bb = a b for all 1 ≤ i,j ≤ n.
i ij j
j=1
Let f = det(xI − (a )) ∈ A[x] be the characteristic polynomial of (a ) ∈ An×n. Then
n ij ij
f ∈ A[X] is monic and f((a )) = 0 ∈ An×n. Therefore f(b)b = 0 for all 1 ≤ i ≤ n, so
ij i
f(b)1 = f(b) = 0, and hence b is integral over A. (cid:3)
Example.
√ √
(a) α := 1+ 5 ∈ Q[ 5] is integral over Z.
2
(b) 1 ∈ Q is not integral over Z.
2
Theorem 1.1.5. Let B be a commutative A-algebra and
Int (B) := {b ∈ B | b integral over A}.
A
Then Int (B) is a subring of B called the integral closure of A in B.
A
Proof. We need to show that Int (B) is a ring, so closed under multiplication and addition.
A
Let b ,b ∈ Int (B) and
1 2 A
A[b ] = (cid:104)c ,...,c (cid:105) , A[b ] = (cid:104)d ,...,d (cid:105) .
1 1 n A 2 1 m A
Since c d = d c for all i,j and 1 ∈ A[b ]∩A[b ] we get
i j j i 1 2
A[b ,b ] ⊂ (cid:104)c d | 1 ≤ i ≤ n,1 ≤ j ≤ m(cid:105) .
1 2 i j A
ThisisasubringofB thatisafinitelygeneratedA-moduleandcontainsb +b ,b −b ,b b . (cid:3)
1 2 1 2 1 2
Theorem 1.1.6. Let C be a commutative ring, A ≤ B ≤ C. If C is integral over B and B
is integral over A, then C is integral over A.
Proof. Let c ∈ C. Since C is integral over B there are n ∈ N and b ,...,b ∈ B such that
1 n
cn +b cn−1 +...+b c+b = 0.
1 n−1 n
PutR := A[b ,...,b ]. SinceB isintegraloverAthisringR isafinitelygeneratedA-module.
1 n
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. (cid:3)
Definition 1.1.7. Let A be an integral domain with field of fraction K := Quot(A).
Int (K) := {x ∈ K | x is integral over A}
A
is called the integral closure of A in K.
If A = Int (K), then A is called integrally closed.
A
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 with
K = Quot(A). The for any b ∈ L:
b is integral over A, if and only if µ ∈ A[x].
b,K
Proof. ⇐ clear.
⇒: Let b ∈ L be integral over A. Then there are n ∈ N and a ,...,a ∈ A such that
1 n
bn +a bn−1 +...+a b+a = 0.
1 n−1 n
Put p(x) = xn+a xn−1+...+a x+a ∈ A[x] and L˜ := Zerf (p) be the spitting field of
1 n−1 n K
˜ ˜
p, Then all zeros b ∈ L of p are integral over A. The minimal polynomial µ of b over K
b,K
divides p, so also the zeros of µ are integral over A. The coefficients of µ are polynomials
b,K b,K
in the zeros, so also integral over A. Since these lie in K, they indeed lie in Int (K) = A. So
A
µ ∈ A[x]. (cid:3)
b,K
Corollary 1.1.9. Let K be an algebraic number field. Then the ring of integers
Z = Int (Z) = {a ∈ K | µ ∈ Z[x]}.
K K a,Q
Any Z-basis of Z is called an integral basis of K.
K
√ √ √
Example. For K = Q[ 2] we obtain Z = Z[ 2] and (1, 2) is a Z-basis of K.
√ √ K √
If K = Q[ 5], then Z = Z[(1+ 5)/2] and (1,(1+ 5)/2) is a Z-basis of K.
K
In the exercise you prove the more general statement: Let 1 (cid:54)= d ∈ Z be square free and
√ √
K := Q[ d], then α := 1+ d is integral over Z if and only if d ≡ 1. In this case (1,α) is an
2 √ 4
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] := dim (L) = n <
K
∞.
(a) Any α ∈ L induces a K-linear map
mult ∈ End (L);x (cid:55)→ αx.
α K
In particular this endomorphism has a trace, determinant, characteristic polynomial
χ := χ and minimal polynomial µ := µ .
α,K multα α,K multα
(b) The map mult: L → End (L) is an injective homomorphism of K-algebras.
K
(c) The map S : L → K,α (cid:55)→ trace (mult ) is a K-linear map, called the trace of L
L/K α
over K.
1.1. THE RING OF INTEGERS 7
(d) ThemapN : L → K,α (cid:55)→ det(mult )ismultiplicative, i.e. N (αβ) = N (α)N (β)
L/K α L/K L/K L/K
for all α,β ∈ L. In particular it defines a group homomorphism N : L∗ → K∗ be-
L/K
tween the multiplicative groups L∗ and K∗ = (K \{0},·) of the fields.
(e) Let α ∈ L. Then µ ∈ K[X] is an irreducible polynomial of degree d := [K(α) :
α,K
K] := dim (K(α)) dividing n and χ = µn/d.
K α,K α,K
(f) If χ = Xn −a Xn−1 +...+(−1)n−1a X +(−1)na ∈ K[X], then N (α) = a
α,K 1 n−1 n L/K n
and S (α) = a .
L/K 1
Proof. Exercise. (cid:3)
Theorem 1.1.11. Assume that L/K is a finite separable extension and let σ ,...,σ : L →
1 n
K be the distinct K-algebra homomorphisms of L into the algebraic closure K of K (so
n = [L : K]). Then for all α ∈ L
(a) χ = (cid:81)n (X −σ (α)).
α,K i=1 i
(b) µ = (cid:81)d (X −α ) where {σ (α),...,σ (α)} = {α ,...,α } has order d = [K(α) :
α,K i=1 i 1 n 1 d
K].
(c) S (α) = (cid:80)n σ (α).
L/K i=1 i
(d) N (α) = (cid:81)n σ (α).
L/K i=1 i
Proof. (c) and (d) follow from (a) using Remark 1.1.10 (f) above.
Tosee(b)letd := [K(α) : K]. SinceL/K isseparable,alsothesubfieldK(α)isseparableover
K, so µ = (cid:81)d (X−α ) for d distinct α ∈ K. The d distinct K-algebra homomorphisms
α,K i=1 i i
ϕ ,...,ϕ from K(α) into K correspond to the d possible images ϕ (α) = α ∈ K of α.
1 d i i
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 of K into the
algebraic closure extends to an automorphism τ˜ ∈ Aut (K).
K
Let ϕ˜ be such an extension of ϕ for all j = 1,...,d and let {τ ,...,τ } = Hom (E,K).
j j 1 n/d K(α)
Then
{σ ,...,σ } = {ϕ˜ ◦τ | 1 ≤ j ≤ d,1 ≤ i ≤ n/d}
1 n j i
In particular each ϕ can be extended in exactly n/d ways to a K-homomorphism ϕ˜ ◦τ :
j j i
E → K, 1 ≤ i ≤ n/d.
This implies that χ = µn/d and also (a) and (b) follow. (cid:3)
α,K α,K
Corollary 1.1.12. Let K ⊆ L ⊆ M be a tower of separable field extensions of finite degree.
Then
S = S ◦S and N = N ◦N
M/K L/K M/L M/K L/K M/L
1(1.33) of the script of the Algebra lecture
8 CHAPTER 1. COMMUTATIVE THEORY.
Proof. Let m := [M : K], (cid:96) := [L : K] and n := [M : L]. Then m = (cid:96)n. Define an equivalence
relation on {σ ,...,σ } = Hom (M,K) by
1 m K
σ ∼ σ ⇔ (σ ) = (σ ) .
j i j L i L
As we have seen in the last proof each equivalence class A contains exactly n elements.
j
Therefore for any α ∈ M
m (cid:96)
(cid:88) (cid:88) (cid:88)
S (α) = σ (α) = σ(α).
M/K i
i=1 j=1 σ∈Aj
Wlog we assume that A = [σ ]. Then
j j
(cid:88)
σ(α) = S (σ (α)) = σ (S (α)).
σj(M)/σj(L) j j M/L
σ∈Aj
Therefore S (α) = (cid:80)(cid:96) σ (S (α)) = (S ◦S )(α). Similarly for the norm. (cid:3)
M/K j=1 j M/L L/K M/L
Definition 1.1.13. Let L/K be a separable extension an let B := (α ,...,α ) be a K-basis
1 n
of L.
(a) The Trace-Bilinear-Form S : L × L → K, S(α,β) := S (αβ) is a symmetric
L/K
K-bilinear form.
(b) ThediscriminantofB isthedeterminantoftheGrammatrixofB, d(B) := det(S(α ,α ) ).
i j i,j
Remark 1.1.14. If {σ ,...,σ } = Hom (L,K) then d(B) = det((σ (α )) )2.
1 n K i j i,j
Proof. S (α α ) = (cid:80)n σ (α )σ (α ) = [(σ (α ) )tr(σ (α ) )] so (S (α α )) = AtrA
L/K i j k=1 k i k j k i i,k k i i,k i,j L/K i j
with A = (σ (α ) ). (cid:3)
k i i,k
√ √
Example. If K = Q and L = Q[ d] then B := (1, d) is a K-basis of L and d(B) =
√
(cid:18) (cid:19)2
1 d
2·(2d) = det √
1 − d
Theorem 1.1.15. Let L/K be a separable extension an let B := (α ,...,α ) be a K-basis of
1 n
L. Then the trace bilinear form is a non-degenerate symmetric K-bilinear form. In particular
d(B) (cid:54)= 0.
Proof. Choose a primitive element α ∈ L, so L = K(α) and B := (1,α,...,αn−1) is another
1
K-basis of L. By the transformation rule for Gram matrices, d(B) = d(B )a2 where a ∈ K∗
1
is the determinant of the base change matrix between B and B . So it is enough to show
1
that d(B ) (cid:54)= 0. By the remark above d(B ) = d(A)2 where
1 1
 
1 σ (α) σ (α)2 ... σ (α)n−1
1 1 1
 1 σ (α) σ (α)2 ... σ (α)n−1 
2 2 2
A = ((σ (αj)) =  . . . 
i j=0,..,n−1,i=1,..,n  . . . 
. . ... ... .
 
1 σ (α) σ (α)2 ... σ (α)n−1
n n n
1.1. THE RING OF INTEGERS 9
(cid:81)
and {σ ,...,σ } = Hom (L,K). By Vandermonde det(A) = (σ (α) − σ (α)), so
1 n K i<j j i
d(B ) = ((cid:81) (σ (α)−σ (α)))2 (cid:54)= 0, since the different embeddings of L into K have different
1 i<j j i
values on the primitive element α. (cid:3)
Definition 1.1.16. Let K be an algebraic number field and B := (α ,...,α ) be an integral
1 n
basis of K (i.e. a Z-basis of the ring of integers Z ). Then the discriminant of K is
K
d := d(B).
K
More general let A = (cid:104)β ,...,β (cid:105) be a free Z-module of full rank in K. Then
1 n Z
d := d((β ,...,β ))
A 1 n
is called the discriminant of A.
Remark 1.1.17. d and d are well defined, which means that they do not dependent on
K A
the choice of the integral basis B.
If A(cid:48) ⊆ A ⊆ K are two finitely generated Z-modules of full rank in K, then by the main
theorem on finitely generated Z-modules (elementary divisor theorem) the index
a := [A : A(cid:48)] := |A/A(cid:48)| < ∞
and d = a2d .
A(cid:48) A
√
Example. K = Q[ d], 0,1 (cid:54)= 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 → R
a non-degenerate symmetric bilinear form.
(a) A lattice in V is the set of all integral linear combinations of an R-basis of V.
n
(cid:88)
L = (cid:104)B(cid:105) = { a b | a ∈ Z}
Z i i i
i=1
for some basis B = (b ,...,b ) of V. Any such Z-basis B of L is called a basis of L and
1 n
the determinant of the Gram matrix of B with respect to Φ is called the determinant
of L.
(b) For a lattice L := (cid:104)B(cid:105) the set L# := {x ∈ V | Φ(x,L) ⊆ Z} is called the dual lattice
Z
of L (wrt Φ).
(c) L is called integral (wrt Φ), if L ⊆ L#.
Remark. L# isalatticeinV,thedualbasisB∗ ofanylatticebasisB ofLisalatticebasis
of L#. The base change matrix between B and B∗ is the Gram matrix M (Φ) = (Φ(b ,b ))
B i j
of B. In particular det(M (Φ)) = [L# : L] = |L#/L| for any integral lattice L.
B
10 CHAPTER 1. COMMUTATIVE THEORY.
Theorem 1.1.19. Let K be an algebraic number field, O ⊆ Z a full Z-lattice in K. Then
K
(O,S ) is an integral lattice and
K/Q
O ⊆ Z ⊆ Z# ⊆ O#
K K
(cid:124)(cid:123)(cid:122)(cid:125) (cid:124)(cid:123)(cid:122)(cid:125) (cid:124)(cid:123)(cid:122)(cid:125)
f dK f
which yields an algorithm to compute Z .
K
Corollary. The ring of integers Z in an algebraic number field is finitely generated, so
K
any algebraic number field has an integral basis.
1.1.3 Dedekind domains.
√ √
Example. Let K = Q[ −5]. Then Z = Z[ −5] and
K
√ √
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 3Z = ℘ ℘(cid:48), 7Z = ℘ ℘(cid:48), (1+2 −5)Z = ℘ ℘ , and (1−2 −5)Z =
K 3 3 K 7 7 K 3 7 K
℘(cid:48)℘(cid:48) are not prime ideals, where
3 7
√ √ √ √
℘ = (3,1+2 −5), ℘(cid:48) = (3,1−2 −5), ℘ = (7,1+2 −5), ℘(cid:48) = (7,1−2 −5)
3 3 7 7
and so 21Z = ℘ ℘(cid:48)℘ ℘(cid:48) is a unique product of prime ideals.
K 3 3 7 7
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-zero
prime 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 is
isomorphic to Z) but not maximal, since Z is not a field.
Theorem 1.1.21. Let K be a number field. Then Z is a Dedekind domain.
K
Proof. Clearly Z is integrally closed and an integral domain.
K
WefirstshowthatZ isNoetherian, i.e. anyidealofZ isfinitelygenerated. Let0 (cid:54)= A(cid:69)Z
K K K
be an ideal and choose 0 (cid:54)= a ∈ A. If B := (b ,...,b ) is an integral basis of K, then
1 n
aB := (ab ,...,ab ) ∈ An is also a Q-basis of K. The lattice (cid:104)aB(cid:105) ⊆ A ⊆ (cid:104)B(cid:105) = Z
1 n Z Z K
has finite index in Z . Therefore also A has finite index in Z and, by the main theorem
K K
on finitely generated Z-modules, A is finitely generated as a Z-module and hence also as a
Z -module.
K
The above consideration also applies to non-zero prime ideals 0 (cid:54)= ℘(cid:69)Z of Z , in particular
K K
any such prime ideal has finite index in Z . Therefore Z /℘ is a finite integral domain, so
K K
a field, which means that ℘ is a maximal ideal. (cid:3)
Lemma 1.1.22. Any finite integral domain R is a field.