Download An Introduction to Algebraic Number Theory, and the Class Number

An Introduction to Algebraic Number Theory, and the Class
Number Formula
Sam Chow
Department of Mathematics and Statistics
University of Melbourne
Parkville VIC 3010 Australia
[email protected]
SUPERVISOR: DR CRAIG WESTERLAND
Honours thesis, October 2011.
Abstract
Algebraic number theory studies algebraic properties of the ring of algebraic integers in a number field.
We describe various algebraic invariants of number fields, as well as their applications. These applications
relate to prime ramification, the finiteness of the class number, cyclotomic extensions, and the unit theorem.
Finally, we present an exposition of the class number formula, which generalizes a result about the Riemann
zeta function. The class number formula is a deep connection between algebraic number theory and analytic
number theory, and its proof employs many of the techniques developed in the main body of the thesis.
i
Acknowledgments
Thanks to my supervisor, Craig Westerland, for a very interesting project. The year has presented me with
many challenges, and Craig has always been there to provide emotional support and sage advice. Thanks to
John Groves for teaching me the relevant commutative algebra over the summer. I am extremely grateful
to Trithang Tran, who generously gave up his time to read through most of this document, find mistakes,
and suggest ways to improve it. Thanks to Kim Ramchen for drawing my attention to a few typographical
errors. Thanks to the fellow students who listened to me talk about number theory every week; they really
motivated me to make steady progress. Finally, I would like to thank my partner, Julia, as well as my
parents and friends, for being so supportive of my career and life choices, as well as for making me happy in
general.
ii
Contents
1 Background (algebra)
1
1.1
Rings and modules of fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1.2
Tensor products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
1.3
Chinese remainder theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
1.4
Field theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8
1.5
Noetherian rings and modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12
2 Algebraic number theory
12
2.1
Rings of integers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13
2.2
Norms, traces and discriminants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18
2.3
Dedekind domains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
26
2.4
The ideal class group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
38
2.5
Discrete valuations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
41
2.6
Factorization in extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
43
2.7
Norms of ideals, and finiteness of the class number . . . . . . . . . . . . . . . . . . . . . . . .
52
2.8
Cyclotomic extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
60
2.9
Lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
68
2.10 Units in the ring of integers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
74
2.11 Fundamental units and regulators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
80
3 The class number formula
81
iii
1
Background (algebra)
This chapter defines notations and presents basic results from abstract algebra. See Atiyah-Macdonald [1],
chapters 1-3 in particular.
1.1
Rings and modules of fractions
Let a and b be ideals in a ring A (denoted a, b C A). Their product is
X
n
ab :=
ai bi : ai ∈ a, bi ∈ b, n ∈ Z≥0 ,
i=1
which is an ideal in A.
Let A be a commutative unital ring, and let X ⊆ A. The ideal generated by X is
X
n
ai xi : ai ∈ A, xi ∈ X, n ∈ Z≥0 .
(X) :=
i=1
Let f : A → B be a ring homomorphism. Let a C A and b C B. The extension ae of a is
X
n
bi f (ai ) : bi ∈ B, ai ∈ a, n ∈ Z≥0 ,
ae := Bf (a) :=
(1.1)
i=1
which is an ideal in B. The contraction bc of b is
bc := f −1 (b),
(1.2)
which is an ideal in A.
Lemma 1.1. Let f : A → B be a homomorphism between commutative unital rings, and let b be a prime
ideal in B. Then bc is a prime ideal in A.
Proof. (Proper ideal:) Let x, y ∈ bc and a ∈ a. Then f (x), f (y) ∈ b, so f (x + y) = f (x) + f (y) ∈ b and
f (−x) = −f (x) ∈ b, so x + y, −x ∈ bc . Moreover, f (ax) = f (a)f (x) ∈ b (as b C B), so ax ∈ bc . Hence
bc C A. Further, b 6= B ⇒ 1 ∈
/b⇒1∈
/ bc . Hence bc is a proper ideal in A.
(Prime:) Let x, y ∈ bc . Then f (x)f (y) = f (xy) ∈ b. As b is a prime ideal, f (x) ∈ b or f (y) ∈ b, so x ∈ bc
or y ∈ bc . Hence bc is a prime ideal in A.
However, if a is a prime ideal, then ae need not be a prime ideal, for instance
f : Z → Q,
x 7→ x
has (2Z)e = Q, which is not a prime ideal in Q.
Let A be a commutative unital ring. A multiplicatively closed subset of A is a subset S ⊆ A such that:
1
• 1 ∈ S.
• If a, b ∈ S, then ab ∈ S.
The ring of fractions is
S
−1
A :=
a
: a ∈ A, s ∈ S
s
=,
where as = bt if there exists u ∈ S such that u(at − bs) = 0. If 0 ∈ S then S −1 A is trivial, so henceforth we
assume that 0 ∈
/ S.
Let A be a commutative unital ring, let S be a multiplicatively closed subset of A, and let M be an A-module.
The module of fractions is
m
S −1 M :=
: m ∈ M, s ∈ S
=,
s
where
m
s
=
n
t
if there exists u ∈ S such that u(mt − ns) = 0.
We can define extended and contracted ideals in rings of fractions by considering
f : A ,→ S −1 A
f (x) =
x
.
1
If a C A, then
ae := a(S −1 A) = S −1 a.
Proof. (⊆:) Let n ∈ Z≥0 . For i = 1, 2, . . . , n, let a0i ∈ a and asii ∈ S −1 A. Then
element of a(S −1 A).
n
n
X
X
ai
a0i ai ti
a0i =
∈ S −1 A,
s
s
i
i=1
i=1
Qn
Q
where s = j=1 sj and ti = j6=i sj .
(⊇:) Let a ∈ a, s ∈ S. Then
a
s
Pn
i=1
a0i asii is an arbitrary
(1.3)
is an arbitrary element of S −1 A, and
a
1
= a ∈ a(S −1 A).
s
s
(1.4)
Lemma 1.2. Let A be a commutative unital ring, and let S be a multiplicatively closed subset of A. Then
every ideal in S −1 A is an extended ideal.
Proof. Let b C S −1 A. Let a := bc = {x ∈ A : x1 ∈ b}. Then b = S −1 a:
(⊆:) Let xs ∈ b. Then x1 = s · xs ∈ b, so x ∈ a. Hence xs ∈ S −1 a.
(⊇:) Let as ∈ S −1 a. Then a ∈ a, so a1 ∈ b, so as = a1 1s ∈ b.
Thus b = ae is an extended ideal, so every ideal in S −1 A is an extended ideal.
2
Proposition 1.3. Let A be a commutative unital ring, and let S be a multiplicatively closed subset of A.
Then p 7→ S −1 p is a bijection from the set of prime ideals in A which do not meet S to the set of prime
ideals in S −1 A.
Proof. (Well defined:) Let p be a prime ideal in A which does not meet S. As p C A and S is multiplicatively
/ S −1 p, so S −1 p 6= S −1 A. Let xs yt ∈ S −1 p, where xs , yt ∈ S −1 A.
closed, S −1 p C S −1 A. As p 6= A, 1 ∈
/ p, so 11 ∈
Then xy ∈ p, so x ∈ p or y ∈ p. Thus xs ∈ S −1 p or yt ∈ S −1 p. Hence S −1 p is a prime ideal in S −1 A.
(Injective:) Let I and J be prime ideals in A which do not meet S, and assume that S −1 I = S −1 J. We
show that I ⊆ J, and the reverse inclusion will follow by symmetry. Let p ∈ I. Then p1 ∈ S −1 I = S −1 J, so
there exist q ∈ J and s ∈ S such that p1 = qs . Now there exists u ∈ S such that u(ps − q) = 0. ups = uq ∈ J,
but us ∈ S, so us ∈
/ J. As J is a prime ideal, us ∈
/ J, and (us)p ∈ J, it follows that p ∈ J. Hence I ⊆ J,
and by symmetry I = J.
(Surjective:) Let J be a prime ideal in S −1 A.
• (J c is a prime ideal in A that does not meet S:) As 1 ∈
/ J, 1 ∈
/ J c , so J c is a proper ideal in A. Suppose
x, y ∈ A and xy ∈ J c . Then x1 y1 ∈ J. As J is a prime ideal in S −1 A, x1 ∈ J or y1 ∈ J. Thus x ∈ J c or
y ∈ J c . Hence J c is a prime ideal in A. It remains to prove that J c ∩ S = ∅. Proof by contradiction:
assume that there exists s ∈ S ∩ J c . Then 1s ∈ J and 1s = 1 ∈ S −1 A, so J = S −1 A, contradicting the
fact that J is a prime ideal in S −1 A. Hence J c is a prime ideal in A that does not meet S.
• (J ce = J:) By definition, J ce ⊇ J. The ideal J ce C S −1 A is generated by { x1 : x ∈ J c } ⊆ J, and
J C S −1 A, therefore J ce = J.
Lemma 1.4. Let S be a multiplicatively closed subset of a commutative unital ring A. Let I, J C A. Then
S −1 (IJ) = (S −1 I)(S −1 J).
Proof. (⊆:) Let n ∈ Z≥0 , a1 , . . . , an ∈ I, b1 , . . . , bn ∈ J, s ∈ S. Then
Pn
i=1
ai bi
s
=
n
X
ai bi
i=1
s 1
Pn
i=1
ai bi
s
∈ (S −1 I)(S −1 J).
∈ S −1 (IJ) is arbitrary and
(1.5)
Pn
(⊇:) Let n ∈ Z≥0 , a1 , . . . , an ∈ I, b1 , . . . , bn ∈ J, s1 , . . . , sn , t1 , . . . , tn ∈ S. Then i=1 asii btii ∈ (S −1 I)(S −1 J)
is arbitrary and
Pn
n
X
ai bi xi
ai bi
= i=1
,
(1.6)
s
t
X
i
i
i=1
Qn
Q
where X = j=1 sj tj and xi = j6=i sj tj (for i = 1, 2, . . . , n). As S is multiplicatively closed, X ∈ S. As
IJ C A, ai bi xi ∈ IJ (for i = 1, 2, . . . , n). Hence
n
X
ai bi
i=1
si ti
Pn
=
a i bi x i
∈ S −1 (IJ).
X
i=1
3
(1.7)
A ring is local if it has precisely one maximal ideal.
Lemma 1.5. Let p be a prime ideal in a commutative unital ring A, and let S = A \ p. Then S is a
multiplicatively closed subset of A, and (the localization of A at p)
Ap := S −1 A
is a local integral domain whose maximal ideal is pAp .
Proof. (S is multiplicatively closed:) 1 ∈
/ p therefore 1 ∈ S. Let a, b ∈ S = A \ p. We need to show that
ab ∈ S. Proof by contradiction: assume that ab ∈
/ S. Then ab ∈ p, so a ∈ p or b ∈ p, p is a prime ideal.
Thus a ∈
/ S or b ∈
/ S, contradiction. Hence ab ∈ S, so S is multiplicatively closed.
(Ap is a local ring whose maximal ideal is pAp :) We need to show that
o
np
: p ∈ p, s ∈
/p
pAp :=
s
/ pAp , pAp 6= Ap .
is the only maximal ideal in Ap . As p C A and S is multiplicatively closed, pAp C Ap . As 11 ∈
u
It now suffices to prove that pAp contains all non-units in Ap . If s ∈ Ap \ pAp , then u ∈ A \ p, so us is a unit
(with inverse us ) in Ap . Hence pAp contains all non-units in Ap , and therefore contains any proper ideal in
Ap .
(Ap is an integral domain:) Ap is a subring of the field of fractions of A, and
integral domain.
1.2
1
1
∈ Ap , therefore Ap is an
Tensor products
Some of the proofs are omitted; Atiyah-Macdonald [1] is a good reference.
Let A be a commutative unital ring, and let M and N be A-modules. The tensor product of M and N is an
A-module T with an A-bilinear mapping g : M × N → T such that: if P is an A-module and f : M × N → P
is an A-bilinear mapping, then there exists a unique A-linear mapping f 0 : T → P such that f = f 0 ◦ g.
Remark
(i) The tensor product M ⊗ N := M ⊗A N := T exists, and is unique up to isomorphism.
(ii) If {xi }i∈I generate M and {yj }j∈J generate N , then {xi ⊗ yj }i∈I,j∈J generate M ⊗ N . In particular,
the tensor product of two finitely generated modules is finitely generated. Moreover, if M and N are
A-modules with respective bases {ai } and {bj }, then {ai ⊗ bj } is a basis for M ⊗ N .
(iii) Tensor products commute with direct sums via the canonical isomorphism
(⊕i Mi ) ⊗A (⊕j Nj ) → ⊕i,j (Mi ⊗A Nj )
X
X
X
(
mi ) ⊗ (
nj ) 7→
(mi ⊗ nj ).
i
j
i,j
4
Lemma 1.6. Let A be a commutative unital ring, and let P be an A-module. Let
M 0 → M → M 00 → 0
be an exact sequence of A-modules. Then
M 0 ⊗A P → M ⊗A P → M 00 ⊗A P → 0
is an exact sequence of A-modules.
Lemma 1.7. Let A be a commutative unital ring, and let M, N, K be A-modules such that M ∼
= N . Then
M ⊗A K ∼
= N ⊗A K.
Proof. Let f : M → N and g : N → M be inverse module homomorphisms. Then id ⊗ f and id ⊗ g are
inverse module homomorphisms between M ⊗A K and N ⊗A K.
Corollary 1.8. Let A be a commutative unital ring. Let M be an A-module, and let a C A. Then
M ∼A
= ⊗A M.
aM
a
(1.8)
Proof. Take the exact sequence
0→a→A→
A
→ 0,
a
and tensor with M over A, to give the exact sequence
a ⊗A M → A ⊗A M →
A
⊗A M → 0.
a
(1.9)
By the first isomorphism theorem,
A ⊗A M
A
⊗A M ∼
.
=
a
a ⊗A M
It remains to show that
A ⊗A M ∼ M
=
a ⊗A M
aM
[a ⊗ m] 7→ [am]
is an isomorphism.
P
P
(Well defined:) Assume that ai ∈ A, xi ∈ M and [ ai ⊗ xi ] = [ bi ⊗ yi ]. Then
X
X
X
1⊗
(ai xi − bi yi ) =
ai ⊗ xi −
bi ⊗ yi ∈ a ⊗A M,
P
so (ai xi − bi yi ) ∈ aM , so [ai xi ] = [bi yi ].
(Injective:) If ai ∈ A, xi ∈ M and [
P
P
ai xi ] = [0], then
ai xi ∈ aM , so
X
X
ai ⊗ xi = 1 ⊗
ai xi ∈ a ⊗ M.
(Surjective:) Let m ∈ M . Then [1 ⊗ m] 7→ [m].
Hence,
M ∼ A ⊗A M ∼ A
=
= ⊗A M.
aM
a ⊗A M
a
5
(1.10)
1.3
Chinese remainder theorem
Let A be a commutative unital ring. Ideals I, J C A are relatively prime if I + J = A. If I C A, then elements
p, q ∈ A are congruent modulo I (denoted p ≡ q mod I) if p − q ∈ I.
Theorem 1.9 (Chinese remainder theorem). Let A be a commutative unital ring. Let n ∈ Z>0 and let
a1 , . . . , an C A be relatively prime in pairs.1 If x1 , . . . , xn ∈ A, then the congruences
x ≡ xi
mod ai ,
i = 1, . . . , n
(1.11)
have a simultaneous solution x ∈ A. Moreover, if x is one solution, then y ∈ A is a solution if and only if
x−y ∈
∩ni=1 ai
=
n
Y
ai .
i=1
Moreover,
A
Qn
i=1
ai
∼
=
n
Y
A
.
a
i=1 i
(1.12)
Proof. (Existence:) First consider the situation with two ideals. Let I, J be relatively prime ideals in A, and
let x, y ∈ A. We need to find an element k ∈ A such that k ≡ x mod I and k ≡ y mod J. As I + J = A,
there exist a ∈ I and b ∈ J such that a + b = 1. If k := ay + bx then
k = ay + bx ≡ bx ≡ (1 − a)x ≡ x
mod I
k = ay + bx ≡ ay ≡ (1 − b)y ≡ y
mod J.
Now consider existence in the general case. Let n ∈ Z>2 and let a1 , . . . , an C A be relatively prime in pairs.
Let x1 , . . . , xn ∈ A. We need to show that there exists k ∈ A such that k ≡ xi mod ai for i = 1, 2, . . . , n.
For i = 1, 2, . . . , n, we will find yi such that
yi ≡ 1
mod ai ,
yi ≡ 0 mod aj ,
Then k =
Pn
i=1
j 6= i.
xi yi will satisfy the necessary condition.
By symmetry, it suffices to find y ∈ A such that y ≡ 1 mod a1 and y ≡ 0 mod aj for j = 2, 3, . . . , n. For
Qn
i = 2, 3, . . . , n, there exist ai ∈ a1 and bi ∈ ai such that ai + bi = 1. Then i=2 (ai + bi ) = 1 and
n
Y
(ai + bi ) ≡
i=2
n
Y
bi
mod a1 =⇒
i=2
n
Y
(ai + bi ) ∈ a1 +
i=2
n
Y
ai ,
(1.13)
i=2
Qn
so a1 + i=2 ai = A. We showed existence for two relatively prime ideals, therefore there exists y ∈ A such
Qn
Qn
that y ≡ 1 mod a1 and y ≡ 0 mod i=2 ai . For j = 2, 3, . . . , n, y ∈ i=2 ai ⊆ aj , i.e. y ≡ 0 mod aj .
Hence there exists k ∈ A such that k ≡ xi mod ai for i = 1, 2, . . . , n.
1 That
is, if i 6= j then ai + aj = A.
6
(Uniqueness modulo ∩ni=1 ai :) Let n ∈ Z>0 and let x, y ∈ A satisfy x ≡ y mod ai for i = 1, 2, . . . , n. Then
x − y ∈ ai for i = 1, 2, . . . , n, therefore
x − y ∈ ∩ni=1 ai .
Thus, the solutions to the congruences (1.11) are congruent modulo ∩ni=1 ai .
We also want to show that
∩ni=1 ai =
n
Y
ai .
i=1
(⊇:) Let x ∈
Qn
i=1
ai . Then
x∈
n
Y
ai ⊆ aj ,
j = 1, 2, . . . , n
i=1
∴ x ∈ ∩ni=1 ai .
Hence
∩ni=1 ai ⊇
n
Y
ai .
i=1
(⊆:) Proof by induction.
• Assume that I, J C A and I + J = A. Then there exist a ∈ I and b ∈ J such that a + b = 1. If x ∈ I ∩ J
then x = (a + b)x = ax + bx ∈ IJ (as ax ∈ IJ and bx ∈ IJ). Hence I ∩ J ⊆ IJ.
Qn
• Now let n ∈ Z>2 and let a1 , . . . , an C A be relatively prime in pairs, and assume that ∩ni=2 ai ⊆ i=2 ai .
Qn
We showed in the proof of existence that a1 + i=2 ai = A, therefore
∩ni=1 ai = a1 ∩ (∩ni=2 ai ) ⊆ a1 ∩
n
Y
ai ⊆ a1 ·
i=2
n
Y
i=2
ai =
n
Y
ai .
i=1
By induction, we conclude that if n ∈ Z>0 and a1 , . . . , an C A are relatively prime in pairs, then
∩ni=1 ai ⊆
n
Y
ai .
i=1
Hence
∩ni=1 ai =
n
Y
ai .
(1.14)
i=1
Finally, the following is an isomorphism:
n
Y
A
→
a
a
i=1 i
i=1 i
A
Φ : Qn
[a] 7→ ([a], . . . , [a]).
(Well defined:) If x, y ∈ A and [x] = [y] ∈
QnA
i=1
x−y ∈
n
Y
ai
then
ai = ∩ni=1 ai =⇒ ([x], . . . , [x]) = ([y], . . . , [y]) ∈
i=1
7
n
Y
A
.
a
i=1 i
Hence Φ is well defined.
(Injective:) Assume that a ∈ A and ([a], . . . , [a]) = 0 ∈
a ∈ ∩ni=1 ai =
n
Y
Qn
A
i=1 ai .
Then
A
ai =⇒ [a] = 0 ∈ Qn
i=1
i=1
ai
.
Hence Φ is injective.
Qn
(Surjective:) Let x1 , . . . , xn ∈ A. Then ([x1 ], . . . , [xn ]) is an arbitrary element of i=1 aAi , and we want to
show that [x1 , . . . , xn ] is in the image of Φ. By the first part of the theorem, there exists x ∈ A such that
x ≡ xi mod ai for i = 1, 2, . . . , n. Then
n
Y
A
[x] 7→ ([x1 ], . . . , [xn ]) ∈
,
a
i=1 i
so Φ is surjective.
Thus, Φ is an isomorphism, so
A
Qn
i=1
1.4
ai
∼
=
n
Y
A
.
a
i=1 i
Field theory
We begin with some basic results. See Milne [7], or any introductory text on abstract algebra.
E/F is a field extension (also denoted E ≥ F ) if E and F are fields and E ⊇ F . The degree (or dimension)
of the extension, [E : F ], is the degree of the vector space E over F . The extension E/F is finite if [E : F ]
is finite.
Let E/F be a field extension, and let α ∈ E. α is algebraic if there exists g ∈ F [X] \ {0} such that g(α) = 0.
The extension E/F is algebraic (or E is algebraic over F ) if every element of E is algebraic over F .
Let E/F be a field extension, and let α ∈ E be algebraic over F . The minimal polynomial of α over F ,
denoted irr(α, F ), is the unique monic irreducible polynomial f ∈ F [X] such that f (α) = 0. The set of
polynomials g ∈ F [X] such that g(α) = 0 forms a nonzero principal ideal in F [X], which is generated by
irr(α, F ).
Let E/F be a field extension, and let α ∈ E. Let f ∈ F [X] be a monic, irreducible polynomial. F [α] is a
stem field for f if f (α) = 0.
Lemma 1.10. Let E/F be a field extension, and let α ∈ E be algebraic over F . Let f ∈ F [X] be a monic
polynomial such that f (α) = 0. Then the following are equivalent:
(a) f is irreducible.
(b) If g ∈ F [X] and g(α) = 0 then f divides g.
8
(c) If g ∈ F [X] and g(α) = 0 then deg(f ) ≤ deg(g).
Moreover,
F [X]
→ F [α]
(f )
[X] 7→ α
is an F -isomorphism.2
Corollary 1.11. Let F be a field, and let F [α] and F [β] be stem fields for F . The following is an F isomorphism
F [α] → F [β]
α 7→ β.
Lemma 1.12. A field extension E/F is finite if and only if E is algebraic and finitely generated (as a field)
over F .
Lemma 1.13. Let F be a field, and let f ∈ F [X] be a monic irreducible polynomial of degree n. Then
F [X]
(f )
n−1
is a field extension of F of degree n. A basis for F(f[X]
]}. Consequently (from
) over F is {[1], [X], . . . , [X
n−1
1.10), a stem field F [α] for F is of degree n over F , with basis {1, α, . . . , α
}.
Lemma 1.14. Let F be a field, and let α be algebraic over F , and let f be the minimal polynomial of α
over F . Let Ω be a field containing F . Then the following is a bijection:
{F-homomorphisms φ : F (α) → Ω} → {roots of f in Ω}
φ 7→ φ(α).
Proof. (Well defined:) Let φ : F (α) → Ω be an F -homomorphism. Then
f (φ(α)) = φ(f (α)) = φ(0) = 0,
(1.15)
so φ(α) is a root of f . Hence the function is well defined.
(Injective:) Let φ and ψ be F -homomorphisms F (α) → Ω such that φ(α) = ψ(α). As φ and ψ agree on F
and α, they agree on F (α). Hence φ = ψ, so the function is injective.
(Surjective:) Let β ∈ Ω be a root of f . The F -homomorphism α 7→ β : F (α) → Ω is mapped to β. Hence
the function is surjective.
2 That
is, an isomorphism which fixes elements of F . F embeds naturally into
9
F [X]
(f )
and F [α] via c 7→ [c] and c 7→ c.
Let E/F be a field extension, and let f ∈ F [X]. E splits f , or f splits in E[X], if there exist m ∈ Z≥0 and
α1 , . . . , αm ∈ E such that
m
Y
f (X) =
(X − αi ),
i=1
and E is a splitting field for f if, further,
E = F [α1 , . . . , αm ].
Let F be a field. A polynomial f ∈ F [X] is separable over F if none of its irreducible factors has a multiple
root (in a splitting field). An algebraic extension E/F is separable if the minimal polynomial of every element
of E is separable (otherwise it is inseparable). F is perfect if every finite extension of F is separable.
We shall also use some higher-level results without proof, as they will make later proofs more concise. See
Lang [4] for details.
Let E/F be an algebraic extension, and let σ : F ,→ L be an embedding of F into an algebraically closed
field L. Let Sσ be the set of extensions of σ to an embedding of E into L. The separable degree of E over F
is
[E : F ]s = |Sσ |.
Then
[E : F ]s ≤ [E : F ],
with equality if and only if E/F is separable. Now assume that E/F is finite. The inseparable degree of E
over F is
[E : F ]
.
[E : F ]i =
[E : F ]s
A finite extension [E : F ] is separable if and only if [E : F ]i = 1.
Lemma 1.15. Let E/k be a finite extension, and let E ≥ F ≥ k be a tower of fields. Then
(a) [E : k]s = [E : F ]s · [F : k]s .
(b) [E : k]i = [E : F ]i · [F : k]i .
(c) E/k is separable if and only if E/F and F/k are separable.3
An algebraic extension E/F is normal if the minimal polynomial (over F ) of every element of E splits in
E[X].
Lemma 1.16.
(a) Let E ≥ F ≥ k be a tower of fields. If E/k is a normal extension, then E/F is a normal extension.
(b) Let E and F be a normal extensions of a field k. Then EF and E ∩ F are normal extensions of k.
Theorem 1.17 (Primitive element theorem). Let E be a finite, separable extension of a field k. Then there
exists a primitive element, i.e. an element α ∈ E such that E = k(α).
3 Separable
extensions form a distinguished class of extensions, since E/k is separable if and only if each step of the tower
is separable.
10
√ √
√
√
Example Q[ 2, 3] = Q[ 2 + 3]:
(⊇:)
√
(⊆:)
2+
√
√ √
√
√
√ √
3 ∈ Q[ 2, 3] ∴ Q[ 2 + 3] ⊆ Q[ 2, 3].
√
√
√
1
√ ∈ Q[ 2 + 3]
2+ 3
√
√
√
√
√
√
√
( 2 + 3) − ( 3 − 2)
2=
∈ Q[ 2 + 3]
2
√
√
√
√
√
√
√
( 2 + 3) + ( 3 − 2)
3=
∈ Q[ 2 + 3]
2
Hence
3−
√
2= √
√ √
√
√
Q[ 2, 3] ⊆ Q[ 2 + 3].
√ √
√
√
We conclude that Q[ 2, 3] = Q[ 2 + 3]. It is not obvious how to find a primitive element in general.
Lemma 1.18. An algebraic extension E/F is normal and separable if and only if, for each α ∈ E, the
minimal polynomial of α has [F [α : F ]] distinct roots in E.
Theorem 1.19. Let E/F be a field extension, and let Aut(E/F ) be the group of F -automorphisms of E.
The following are equivalent:
(a) E is the splitting field of a separable polynomial f ∈ F [X].
(b) F = E G for some finite group G of automorphisms of E.
(c) E is normal and separable, and of finite degree, over F .
(d) F = E Aut(E/F ) .
If these statements hold, the extension E/F is Galois,4 and we define the Galois group
Gal(E/F ) := Aut(E/F ).
If E/F is a Galois extension, then
|Gal(E/F )| = [E : F ],
so the Galois group is finite.
Let L/K be an algebraic extension. Let α ∈ L, and let f := irr(α, K). The conjugates of α in L are the
roots of f in L.
Lemma 1.20. Let L/K be a finite separable extension. Then there exists a field E ⊇ L such that E/K is
Galois.
4 Some
other definitions differ.
11
Proof. By the primitive element theorem (1.17), there exists α ∈ L such that L = K(α). Let f ∈ K[X] be
the minimal polynomial of α, and let α = α1 , . . . , αm be the conjugates of α. Then E = L[α1 , . . . , αm ] is
the splitting field of f ∈ K[X], and f is separable (being the minimal polynomial of α ∈ L, where L/K is
separable), so E/K is Galois.
Let L/K be a finite separable extension. The Galois closure of L/K is the minimal (by inclusion) field
E ⊇ L such that E/K is Galois.
1.5
Noetherian rings and modules
See Atiyah-Macdonald [1] for the proofs.
A commutative unital ring A is Noetherian if it satisfies the following equivalent conditions:
P
• Every non-empty set of ideals in A has a maximal element, i.e. if
is a non-empty set of ideals in A,
P
P
then there exists I ∈
such that if J ∈
and J ⊇ I then J = I.
• If (Ik ) is a sequence of ideals in A such that I1 ⊆ I2 ⊆ . . ., then there exists N ∈ Z>0 such that if
k ≥ N then Ik = IN .5
• Every ideal in A is finitely generated.
A module M is Noetherian if it satisfies the following equivalent conditions:
P
• Every non-empty set of submodules of M has a maximal element, i.e. if
is a non-empty set of
P
P
submodules of M , then there exists I ∈
such that if J ∈
and J ⊇ I then J = I.
• If (Mk ) is a sequence of submodules of M such that M1 ⊆ M2 ⊆ . . ., then there exists N ∈ Z>0 such
that if k ≥ N then Mk = MN .6
• Every submodule of M is finitely generated.
Lemma 1.21. Let A be a Noetherian ring, and let M be a finitely generated A-module. Then M is a
Noetherian A-module.
2
Algebraic number theory
Most of the ideas in this chapter can be attributed to Dedekind, Kummer and Noether. The approach
presented follows Milne [8] fairly closely.
5 In
6 In
words, every ascending chain of ideals is stationary.
words, every ascending chain of submodules of M is stationary.
12
2.1
Rings of integers
Algebraic integers are a generalization of integers. Specifically, the algebraic integers in the number field Q
are the elements of Z, but larger number fields contain more algebraic integers. We will see that the algebraic
integers in a number field form a ring with some nice properties.
Let A be an integral domain, and let L ⊇ A be a field. An element α ∈ L is integral over A if it is the root
of a monic polynomial with coefficients in A. We want to prove that the elements of L that are integral over
A form a ring. The proof we will give, which is due to Dedekind, relies on the following lemma.
Lemma 2.1. Let A be an integral domain, and let L ⊇ A be a field. An element α ∈ L is integral over A if
and only if there exists a nonzero finitely generated A-submodule M of L such that αM ⊆ M .
Proof. (⇒:) Let α ∈ L be integral over A. Then there exist n ∈ Z>0 and a1 , . . . , an ∈ A such that
αn + a1 αn−1 + . . . + an−1 α + an = 0.
Let M be the A-module generated by 1, α, . . . , αn−1 . Then M is a finitely generated A-submodule of L such
that αM ⊆ M .
(⇐:) Assume that there exists a nonzero finitely generated A-submodule M of L such that αM ⊆ M . Let
n ∈ Z>0 and let M be the A-module generated by (nonzero) v1 , . . . , vn . For i = 1, 2, . . . , n, there exist
Pn
ai1 , . . . , ain ∈ A such that αvi = j=1 aij vj . Thus
Cv = 0,
(2.1)
where

α − a11

 −a21
C=
 ..
 .
−an1
−a12
α − a22
..
.
...
...
...
..
.
−an,n−1

−a1n

−a2n 
,
..


.
α − ann


v1
 
 v2 

v=
 ..  .
.
vn
In order to show that α is integral over A, it suffices to prove that det(C) = 0. As Cv = 0, multiplying both
sides by the adjugate matrix of C yields
0 = adj(C)Cv = det(C)In v = det(C)v.
As v1 6= 0, we conclude that det(C) = 0. Hence α is integral over A.
Theorem 2.2. Let A be an integral domain, and let L ⊇ A be a field. Then the elements of L that are
integral over A form a ring.
Proof. Let α, β ∈ L be integral over A. Then 1 and −α are integral over A, so it remains to show that α + β
and αβ are integral over A.
By 2.1, there exist nonzero finitely generated A-submodules M and N of B such that αM ⊆ M and βN ⊆ N .
Consider
X
n
M N :=
ai bi : n ∈ Z≥0 , a1 , . . . , an ∈ M, b1 , . . . , bn ∈ N .
i=1
13
• M N 6= 0, since M 6= 0, N 6= 0.
• (M N is an A-submodule of L:) Let
n
X
a i bi +
i=1
m
X
Pn
i=1
ai bi ,
Pm
j=1 ci di
ci di ∈ M N,
a
j=1
n
X
∈ M N and a ∈ A. Then
ai bi =
i=1
n
X
(a · ai )bi ∈ M N.
i=1
• (M N is a finitely generated A-module:) Let m, n ∈ Z>0 . Let M be the A-module generated by
u1 , . . . , um , and let N be the A-module generated by v1 , . . . , vn . Then M N is generated by
ui vj : i ∈ {1, 2, . . . , m}, j ∈ {1, 2, . . . , n} .
• ((α + β)M N ⊆ M N and (αβ)M N ⊆ M N :) Let n ∈ Z>0 and let a1 , . . . , an ∈ M, b1 , . . . , bn ∈ N . Then
Pn
i=1 ai bi is an arbitrary element of M N .
n
n n
n
X
X
X
X
(α + β)
ai bi =
(αai )bi + ai (βbi ) ,
αβ
ai bi =
(αai )(βbi ).
i=1
i=1
i=1
i=1
As αM ⊆ M and βN ⊆ N , we have αai ∈ M and βbi ∈ N for i = 1, 2, . . . , n. Thus
(α + β)
n
X
ai bi ∈ M N,
αβ
i=1
n
X
ai bi ∈ M N.
i=1
Hence, by 2.1, α + β and αβ are integral over A. Thus, the set of elements of L that are integral over A
form a ring.
Let A be an integral domain, and let L ⊇ A be a field. The integral closure of A in L is the set B of elements
of L that are integral over A. B is a ring, by 2.2. B is an integral domain, since 1 ∈ B and B is a subring
of a field (namely L). A ring C ⊇ A is integral over A if each element of C is integral over A. Let K ⊇ Q
be a field. An element α ∈ K is an algebraic integer if α is integral over Z.
An algebraic number field (or number field ) is a finite (and hence algebraic) extension of Q. If L is a number
field, the ring of integers in L, denoted OL , is the integral closure of Z in L.
Any field of characteristic 0 is perfect. In particular, Q is perfect, so any number field is a finite, separable
extension of Q. By 1.17, this implies that any number field L has a primitive element over Q, i.e. L = Q(α)
for some α ∈ L.
Proposition 2.3. Let A be an integral domain, and let K be the field of fractions of A. Let L be a field
containing K, and let α ∈ L be algebraic over K. Then there exists d ∈ A such that dα is integral over A.
Proof. Let m ∈ Z>0 and let k1 , . . . , km ∈ K be such that
αm + k1 αm−1 + . . . + km = 0.
(2.2)
Let d ∈ A be the product of the denominators of k1 , . . . , km . Then
(dα)m + k1 d(dα)m−1 + . . . + km dm = 0,
Hence dα is integral over A.
14
k1 d, . . . , km dm ∈ A.
(2.3)
An integral domain is integrally closed if it is its own integral closure in its field of fractions.
Proposition 2.4. Let A be a unique factorization domain. Then A is integrally closed.
Proof. Since any element in A is integral over A, it remains to show that if α is in the field of fractions of A
and α is integral over A then α ∈ A. Let a, b ∈ A, b 6= 0. We can write ab in ‘simplest terms’ by writing a
and b as products of primes, and then cancelling. Thus we may assume that no prime element of A divides
both a and b. Suppose that ab is integral over A. To show: b is a unit. Proof by contradiction: assume that
b is not a unit. Then there exists a prime π ∈ A such that π | b but π 6 | a. Let
a n
a n−1
+ an−1
+ . . . + a0 = 0,
(2.4)
b
b
where n ∈ Z≥0 and a0 , . . . , an−1 ∈ A. Then
an + an−1 an−1 b + . . . + a0 bn = 0.
(2.5)
Now π|an , so π|a, since π is prime. This contradicts the assumption that a and b are relatively prime, so b
is a unit and ab ∈ A. Hence A is integrally closed.
Consequently, the ring of integers in Q is Z.
Proposition 2.5. Let A be an integrally closed integral domain with field of fractions K, and let L/K be a
finite extension. Then an element α ∈ L is integral over A if an only if its minimal polynomial over K has
coefficients in A.
Proof. Let α ∈ L.
(⇐:) Assume that the minimal polynomial f ∈ K[X] of α over K has coefficients in A. As f is monic, α is
integral over A.
(⇒:) Conversely, assume that α is integral over A. Let f ∈ K[X] be the minimal polynomial of α over K.
We need to show that the coefficients of f lie in A. As the coefficients of f lie in K, and A is integrally closed,
it suffices to prove that the coefficients of f are integral over A. By the Vieta formulae, the coefficients of f
are sums and products of the roots of f . As the integral closure of A in L is a ring, it suffices to prove that
the roots of f are integral over A. Let β ∈ L be any root of f .
Let n ∈ Z>0 and a1 , . . . , an ∈ A satisfy
g(α) = αn + a1 αn−1 + . . . + an−1 α + an = 0,
(2.6)
where g ∈ A[X] is defined by g(X) = X n + a1 X n−1 + . . . + an−1 X + an . The map
σ : K[α] → K[β]
α 7→ β
(2.7)
g(β) = g(σ(α)) = σ(g(α)) = σ(0) = 0,
(2.8)
is a K- isomorphism by 1.11. Now
so β is integral over A. Thus all roots of f are integral over A, and we conclude from that f ∈ A[X].
15
Example This example demonstrates that rings of integers are not necessarily unique factorization domains.
√
√
We shall see that the ring of integers of Q[ −5] is Z[ −5]. Consider
√
√
2 × 3 = (1 + −5)(1 − −5).
√
√
√
We will show that the elements 2, 3, 1 + −5 and 1 − −5 are irreducible in Z[ −5]. Consider the map7
√
Nm : Z[ −5] → Z≥0
√
a + b −5 7→ a2 + 5b2 .
We prove some properties of this map.
(The map Nm is multiplicative:) Let a, b, c, d ∈ Z. Then
√
√
√
Nm((a + b −5)(c + d −5)) = Nm((ac − 5bd) + (ad + bc) −5) = (ac − 5bd)2 + 5(ad + bc)2
√
√
= a2 c2 + 5a2 d2 + 5b2 c2 + 25b2 d2 = (a2 + 5b2 )(c2 + 5d2 ) = Nm(a + b −5) · Nm(c + d −5).
√
Hence, the map Nm : Z[ −5] → Z≥0 is multiplicative.
√
Let α ∈ Z[ −5]. The following are equivalent:
(a) Nm(α) = 1.
(b) αᾱ = 1.
√
(c) α is a unit in Z[ −5].
√
Proof. Let α = a + b −5, a, b ∈ Z.
• ((a) =⇒ (b):)
√
√
Nm(α) = (a + b −5)(a − b −5) = αᾱ.
Thus, if Nm(α) = 1, then αᾱ = 1.
√
√
• ((b) =⇒ (c):) Assume that αᾱ = 1. Then α is a unit in Z[ −5] with inverse ᾱ ∈ Z[ −5].
√
√
• ((c) =⇒ (a):) Assume that α is a unit in Z[ −5] with inverse β ∈ Z[ −5]. Then
Nm(α) · Nm(β) = Nm(αβ) = Nm(1) = 1,
therefore Nm(α) = ±1. As Nm(α) ≥ 0, we must have Nm(α) = 1.
√
If 2 = xy with x, y ∈ Z[ −5], then Nm(x)Nm(y) = Nm(2) = 4. Without loss of generality Nm(x) ≤
Nm(y). As 2 ∈
/ Image(Nm), it follows that Nm(x) = 1, so x is a unit. Hence 2 is irreducible in
√
√
Z[ −5]. Similarly, 3 ∈
/ Image(Nm), so 3 is irreducible in Z[ −5]. Similarly, as 2 ∈
/ Image(Nm) and
√
√
√
√
√
Nm(1 + −5) = Nm(1 − −5) = 6, it follows that 1 + −5 and 1 − −5 are irreducible in Z[ −5]. Thus
√
√
2 × 3 = (1 + −5)(1 − −5)
7 This
is the norm map that we will soon encounter, which is the product of the (not necessarily distinct) Galois conjugates.
16
√
constitutes two distinct factorizations of 6 ∈ Z[ −5] into irreducible factors.8
Though rings of integers are not necessarily unique factorization domains, they do have many nice properties.
We now aim to establish that the ring of integer in a number field is integrally closed, which essentially
involves proving that integral closures are integrally closed.
Lemma 2.6. Let A, B, C be rings such that A ⊆ B ⊆ C. If B is finitely generated as an A-module and C
is finitely generated as a B-module, then C is finitely generated as an A-module.
Proof. Assume that B is finitely generated as an A-module, and that C is finitely generated as a B-module.
Let B be the A-module generated by {β1 , . . . , βm }, and let C be the B-module generated by {γ1 , . . . , γn },
where m, n ∈ Z>0 .
Let D be the A-module generated by βi γj : i ∈ {1, . . . , m}, j ∈ {1, . . . , n} . In order to show that C is
finitely-generated as an A-module, it suffices to prove that C = D.
• (C ⊇ D :) As C is a B-module, C contains βi γj for i ∈ {1, 2, . . . , m} and j ∈ {1, 2, . . . , n}. As C is an
A-module (being a B-module), C ⊇ D.
Pn
• (C ⊆ D :) Let γ ∈ C. Then there exist b1 , . . . , bn ∈ B such that γ = j=1 bj γj . For j = 1, 2, . . . , n,
Pm
there exist a1j , . . . , amj ∈ A such that bj = i=1 aij βi . Now
X
γ=
aij (βi γj ),
aij ∈ A,
i,j
so C ⊆ D.
Thus C = D, so C is finitely generated as an A-module.
Lemma 2.7. Let A be an integral domain, and let B ⊇ A be a ring which is integral over A and finitely
generated as an A-algebra. Then B is finitely generated as an A-module.
Proof. Proof by induction on r, for rings of the form A[γ1 , . . . , γr ].
• Let C = A[γ], and assume that C is integral over A. Then there exist n ∈ Z>0 and a1 , . . . , an ∈ A
such that
γ n + a1 γ n−1 + . . . + an−1 γ + an = 0.
Let x ∈ C = A[γ]. Then there exist N ∈ Z>0 and c0 , c1 , . . . , cN ∈ A such that
x = c0 + c1 γ + . . . + cN γ N .
(2.9)
Substituting γ n = −(a1 γ n−1 + . . . + an−1 γ + an ) into equation (2.9) yields an expression of the form
x = d0 + d1 γ + . . . + dn−1 γ n−1 ,
d0 , d1 , . . . , dn−1 ∈ A.
Hence C is the A-module generated by {1, γ, . . . , γ n−1 }.
8 These
factors do not differ by multiplication by a unit (consider their norms).
17
• Assume that D ⊇ A is a ring that is finitely generated as an A-module, and let E = D[α] be integral
over A. To show: E is finitely generated as an A-module.
E is integral over A, and D ⊇ A, therefore E is integral over D. From the base step of our induction,
E is therefore finitely generated as a D-module. Additionally, D is finitely generated as an A-module,
so, by 2.6, E is finitely generated as an A-module.
Hence, by induction, B is finitely generated as an A-module.
Lemma 2.8. Let A, B and C be integral domains such that B is integral over A and C is integral over B.
Then C is integral over A.
Proof. Let γ ∈ C. To show: γ is integral over A. By 2.1, it suffices to find a nonzero, finitely generated
A-submodule E of C such that γE ⊆ E. As γ is integral over B, there exist n ∈ Z>0 and b1 , . . . , bn ∈ B
such that
γ n + b1 γ n−1 + . . . + bn−1 γ + bn = 0.
Let D = A[b1 , . . . , bn ], and let E = D[γ]. Then E is a nonzero A-submodule of C (as γ, b1 , . . . , bn ∈ C), and
γE ⊆ E (as γ ∈ E). It remains to show that E is finitely generated as an A-module.
D is integral over A and finitely generated as an A-algebra. Thus, by 2.7, D is finitely generated as an
A-module. By 2.6, it therefore suffices to prove that E = D[γ] is finitely generated as a D-module. As
γ n + b1 γ n−1 + . . . + bn−1 γ + bn = 0,
b1 , . . . , bn ∈ D,
E is the D-module generated by {1, γ, . . . , γ n−1 }. Hence E is finitely generated as a D-module. We conclude
that γ is integral over A. Thus, C is integral over A.
Theorem 2.9. Let A be an integral domain with field of fractions K. Let L/K be an algebraic extension.
The integral closure of A in L is integrally closed.
Proof. Let B be the integral closure of A in L, and let C be the integral closure of B in L. To show: B
is integrally closed. As L is a field containing B, it suffices to prove that C = B. We know that C ⊇ B,
therefore it suffices to prove that C ⊆ B. Thus, it suffices to prove that C is integral over A. As B is integral
over A and C is integral over B, C is integral over A by 2.8. We conclude that B is integrally closed.
By 2.9 (with A = Z and L as any number field), the ring of integers in any number field is integrally closed.
2.2
Norms, traces and discriminants
Invariants are at the heart of algebraic number theory. We will encounter many powerful theorems involving
algebraic invariants, as well as a few surprising relationships between them. The main goals of this section
are to provide definitions and to introduce formulae for later use. These formulae have been taken from
Milne [8], which is where any of the omitted proofs in this section can be found. We will also show that
rings of integers are finitely generated as Z-modules, allowing us to define an integral basis. Finally, using
18
√
the discriminant, we will completely describe the rings of integers for number fields of the form Q[ d], where
d is a square-free integer.
Let A ⊆ B be rings such that B is a free A-module of rank n. Then any β ∈ B defines an A-linear map
B→B
x 7→ βx,
which, if we choose a basis for B as an A-module, can be represented by a matrix. The trace (TrB/A β) and
norm (NmB/A β) of β in the extension B/A are respectively the trace and determinant of this matrix. The
trace and norm are well defined, as choosing a different basis would give rise to a similar matrix.
If a ∈ A and β, β 0 ∈ B then
• Tr(β + β 0 ) = Tr(β) + Tr(β 0 )
• Tr(aβ) = aTr(β)
• Tr(a) = na
• Nm(ββ 0 ) = Nm(β) · Nm(β 0 )
• Nm(a) = an .
Lemma 2.10. Let E/M and M/F be finite field extensions. Then
(a) TrE/F = TrM/F ◦ TrE/M
(b) NmE/F = NmM/F ◦ NmE/M .
Proof. See Lang [4], Theorem 5.1. This is also discussed less formally in Milne [7], Proposition 5.44.
Proposition 2.11. Let L/K be a field extension of degree n, and let β ∈ L. Let f ∈ K[X] be the minimal
polynomial of β over K and let β1 = β, β2 , . . . , βm be the roots of f in the algebraic closure of K. Then
TrL/K (β) = r(β1 + . . . + βm ),
where r := [L : K[β]] =
NmL/K (β) = (β1 . . . βm )r ,
n
m.
Proof. The matrix of x 7→ βx : K[β] → K[β] with respect to the basis {1, β, . . . , β m−1 } for K[β] over K is


Qm
0 0 0 . . . 0 (−1)m i=1 βi


 1 0 0 ... 0

?


..


 0 1 0 ... 0

.

.
 .

.
..
 ..

?






Pm
0 0 ... 0 1
β
i
i=1
19
The trace of this matrix is
Pm
i=1
βi and the determinant is
TrK[β]/K (β) =
m
X
βi ,
Qm
i=1
βi . Hence
NmK[β]/K (β) =
i=1
m
Y
βi .
i=1
By 2.10,
TrL/K (β) = TrK[β]/K (TrL/K[β] (β)) = TrK[β]/K (rβ),
as β ∈ K[β]
= rTrK[β]/K (β) = r(β1 + . . . + βm )
and
NmL/K (β) = NmK[β]/K (NmL/K[β] (β)) = NmK[β]/K (β r ),
as β ∈ K[β]
= (NmK[β]/K (β))r = (β1 . . . βm )r
Corollary 2.12. Let L/K be a finite field extension, and let β ∈ L. Then TrL/K (β), NmL/K (β) ∈ K.
Proof. Let f ∈ K[X] be the minimal polynomial of β over K. Up to sign, the sum and product of the
conjugates of β are coefficients of f , and therefore lie in K. By 2.11, the trace and norm lie in K.
Corollary 2.13. Let A be an integrally closed integral domain with field of fractions K, and let L/K be a
finite extension. If β ∈ L is integral over A, then TrL/K (β), NmL/K (β) ∈ A.
Proof. Let β ∈ L be integral over A. By 2.12, TrL/K (β), NmL/K (β) ∈ K. As A is integrally closed, it suffices
to prove that TrL/K (β) and NmL/K (β) are integral over A.
Let f ∈ K[X] be the minimal polynomial of β over K. By 2.5, f ∈ A[X]. As, in addition, f is monic, the
roots of f are integral over A. By 2.11, TrL/K (β) and NmL/K (β) are sums and products of these roots, and
are therefore integral over A (since the integral closure of A in L is a ring).
Thus TrL/K (β), NmL/K (β) ∈ A.
Corollary 2.14. Let L/K be a separable extension of degree n ∈ Z>0 , and let β ∈ L. Let {σ1 , . . . , σn } be
the set of distinct K-homomorphisms L ,→ Ω, where Ω/K is a Galois extension and Ω ⊇ L. Then
TrL/K (β) = σ1 β + . . . + σn β,
NmL/K (β) = σ1 β · · · σn β.
(2.10)
Proof. Let f ∈ K[X] be the minimal polynomial of β over K, and let β1 = β, β2 , . . . , βm be the roots of f
(distinct because L/K is a separable extension). By 2.11, we need to show that each βi occurs r times in
n
the multiset {σ1 (β), . . . , σn (β)}, where r = [L : K[β]] = m
.
By 1.14, each βi occurs once in the set {τ1 β, . . . , τm β}, where τ1 , . . . , τm are the distinct K-homomorphisms
K[β] → Ω. These are defined by9
τj : K[β] → Ω
β 7→ βj .
9 See
1.14.
20
Being injective, each τj is an embedding of K[β] into Ω. The extension L/K[β] is separable, by 1.15, so
[L : K[β]]s = [L : K[β]] = r.
Hence each τj has r distinct extensions to a K-homomorphism L → Ω. As rm = n, every K-homomorphism
L → Ω can be obtained in this way (as there must be [L : K]s = [L : K] = n of these K-homomorphisms
n
L → Ω). We therefore conclude that each βi occurs r times in the multiset {σi β}, where r = [L : K[β]] = m
,
and therefore 2.11 yields
TrL/K (β) = σ1 β + . . . + σn β,
NmL/K (β) = σ1 β · · · σn β.
Lemma 2.15. Let K be a number field. An element α ∈ OK is a unit if and only if NmK/Q (α) = ±1.
Proof. (⇒:) Assume that α ∈ OK is a unit. Then there exists β ∈ OK such that αβ = 1. Thus
Nm(α)Nm(β) = 1. The minimal polynomials of α and β over Q have coefficients in Z, by 2.5. These
have constant coefficients equal to ±Nm(α) and ±Nm(β) respectively, so Nm(α), Nm(β) ∈ Z. Since
Nm(α)Nm(β) = 1, it now follows that Nm(α) = ±1.
(⇐:) Assume that α ∈ OK and Nm(α) = ±1. Define i : K ,→ C, x 7→ x. Now
Y
Y
± 1 = Nm(α) =
σα = α ·
σα.
σ:K,→C
Then α has inverse equal to ±
Q
σ6=i
(2.11)
σ6=i
σα ∈ OK , so α ∈ OK is a unit.
Let L/K be a finite field extension, and let {e1 , . . . , em } be a basis for L over K. The discriminant of L/K
is the discriminant of the symmetric bilinear form
L×L→K
(α, β) 7→ TrL/K (αβ),
i.e. the determinant of the matrix (TrL/K (ei ej )).
Let A and B be rings such that B ⊇ A and B is a free A-module of rank m. The discriminant of β1 , . . . , βm ∈
B is
D(β1 , . . . , βm ) = det(TrB/A (βi βj )).
(2.12)
Lemma 2.16. Let A, B be rings such that B ⊇ A and B is a free A-module of rank m. Let β1 , . . . , βm ∈ B.
Pm
For j = 1, 2, . . . , m, let γj = i=1 aji βi where aij ∈ A for i, j ∈ {1, 2, . . . , m}. Then
D(γ1 , . . . , γm ) = det(aij )2 · D(β1 , . . . , βm ).
Proof. For k, l ∈ {1, 2, . . . , m},
TrB/A (γk γl ) =
X
TrB/A (aki βi alj βj ) =
i,j
X
i,j
21
aki TrB/A (βi βj )alj .
(2.13)
Hence
(TrB/A (γk γl )) = M · TrB/A (βi βj ) · M T ,
M = (aij ).
Thus
det(TrB/A (γk γl )) = det(M )2 · det(TrB/A (βi βj ))
D(γ1 , . . . , γm ) = det(aij )2 · D(β1 , . . . , βm ).
If {β1 , . . . , βm } and {γ1 , . . . , γm } are bases for B over A, then det(aij ) is a unit. Thus the discriminant of a
basis is well defined up to a multiplication by the square of a unit in A. In particular, the ideal in A that it
generates is independent of the choice of basis. The discriminant of B over A is
disc(B/A) = [D(β1 , . . . , βm )] ∈
A
,
(A× )2
where {β1 , . . . , βm } is a basis for B over A. We often regard disc(B/A) as a representative D(β1 , . . . , βm ) ∈ A
of disc(B/A) in (AA× )2 .
Corollary 2.17. Let A and B be integral domains such that B ⊇ A. Assume that B is a free A-module of
rank m, and that disc(B/A) 6= 0. Then γ1 , . . . , γn ∈ B form a basis for B as an A-module if and only if
(D(γ1 , . . . , γm )) = (disc(B/A)) C A.
Proposition 2.18. Let B be an integral domain such that B ⊇ Z, and assume that B is a free Z-module of
rank m. Then
(a) Elements γ1 , . . . , γm ∈ B generate a Z-submodule
10
N of finite index in B if and only if
D(γ1 , . . . , γm ) 6= 0.
(b) If γ1 , . . . , γm ∈ B and D(γ1 , . . . , γm ) 6= 0 then
11
D(γ1 , . . . , γm ) = (B : N )2 · disc(B/Z).
(2.14)
Proposition 2.19. Let L/K be a finite separable extension of degree m, and let σ1 , . . . , σm be the distinct
K-homomorphisms L ,→ Ω, where Ω is a Galois extension of K containing L.12 If {β1 , . . . , βm } is a basis
for L over K, then
D(β1 , . . . , βm ) = det(σi βj )2 6= 0.
(2.15)
Proposition 2.20. Let K be a field of characteristic 0, let β be algebraic over K, and let L = K[β]. Let
Qm
f (X) be the minimal polynomial of β over K, and let f (X) = i=1 (X − βi ) in the Galois closure of L/K.
Then
Y
D(1, β, . . . , β m−1 ) =
(βi − βj )2 = (−1)m(m−1)/2 · NmL/K (f 0 (β)).
(2.16)
1≤i<j≤m
The discriminant of f (X) is D(1, β, . . . , β m−1 ).
10 That
is, an additive subgroup.
{β1 , . . . , βm } be a basis for B as a Z-module. The motivation for this formula is that (B : N ) is the ratio of the volumes
of the corresponding fundamental domains, which is the ratio of the determinants of the matrices with rows being βi and γi
respectively, which is the determinant of the change of basis matrix.
12 [L : K] = [L : K] = m.
s
11 Let
22
This coincides with our standard notion of the discriminant of a quadratic polynomial:
√
√
2
−a + a2 − 4b −a − a2 − 4b
= a2 − 4b.
disc(X 2 + aX + b) =
−
2
2
For a cubic with no quadratic term,
disc(X 3 + aX + b) = −4a3 − 27b2 .
Lemma 2.21. Let m ∈ Z>0 , and let V be an m-dimensional vector space over a field K. Let {e1 , . . . , em }
be a basis for V , and let
ψ :V ×V →K
be a nondegenerate13 bilinear form. Then there exists a basis {e01 , . . . , e0m } for V , such that
ψ(ei , e0j ) = δij ,
i, j ∈ {1, 2, . . . , m},
where δ is the Kronecker delta.
Proposition 2.22. Let A be an integrally closed integral domain with field of fractions K. Let L/K be a
separable extension of degree m ∈ Z>0 , and let B be the integral closure of A in L. Then
(a) There exist free A-submodules M and M 0 of L, of rank m, such that M ⊆ B ⊆ M 0 .
(b) If A is a Noetherian ring, then B is a finitely generated A-module.
(c) If A is a principal ideal domain, then B is a free A-module of rank m.
Proof.
(a) Let {α1 , . . . , αm } be a basis for L over K. By 2.3, there exist d1 , . . . , dm ∈ A such that di αi ∈ B for
Qm
i = 1, 2, . . . , m. Let d = i=1 di . Then d ∈ A, βi := dαi ∈ B for i = 1, 2, . . . , m, and {β1 , . . . , βm } is a
basis for L over K. The bilinear form
ψ :L×L→K
(x, y) 7→ TrL/K (xy)
0
is nondegenerate, since if x ∈ L× then ψ(x, x1 ) = 1 6= 0. Thus, by 2.21, there exists a basis {β10 , . . . , βm
}
for L over K such that
TrL/K (βi βj0 ) = δij ,
i, j ∈ {1, 2, . . . , m},
where δ is the Kronecker delta.
0
It remains to show that the free A-submodules M := Aβ1 + . . . + Aβm and M 0 := Aβ10 + . . . + Aβm
of L
0
satisfy M ⊆ B ⊆ M . As β1 , . . . , βm ∈ B, it follows immediately that M ⊆ B. Now it remains to show
that B ⊆ M 0 . Let β ∈ B, and write
β=
m
X
bj βj0 ,
b1 , . . . , bj ∈ K.
j=1
13 That
is, there does not exist v ∈ V \ {0} such that if x ∈ V then ψ(v, x) = 0.
23
It remains to prove that bi ∈ A for i = 1, 2, . . . , m. Let i ∈ {1, 2, . . . , m}. By 2.13, TrL/K (β · βi ) ∈ A.
Moreover,
m
m
m
X
X
X
TrL/K (β · βi ) = TrL/K (
bj βj0 βi ) =
bj TrL/K (βj0 βi ) =
bj δij = bi ,
j=1
so bi ∈ A. Hence β =
Pm
0
j=1 bj βj
j=1
j=1
∈ M 0 , so we conclude that B ⊆ M 0 .
(b) Assume that A is a Noetherian ring. From (a), there exists a free A-module M 0 of rank m such that
B ⊆ M 0 . As M 0 is finitely generated A-module, 1.21 implies that M 0 is a Noetherian A-module. Hence
B is finitely generated as an A-module, being a submodule of the Noetherian A-module M 0 .
(c) Assume that A is a principal ideal domain. Then A is a Noetherian ring, since every ideal in A is finitely
generated. Hence, there exist free A-submodules M and M 0 of L, of rank m, such that M ⊆ B ⊆ M 0 .
Then B is a free A-module, being a finitely generated submodule of the free module M 0 over the principal
ideal domain A. Moreover, m = rank(M ) ≤ rank(B) ≤ rank(M 0 ) = m, so rank(B) = m.
In particular, applying 2.22(c) with A = Z reveals that rings of integers are finitely generated free Z-modules.
Let K be a number field. An integral basis for K is a basis for OK as a Z-module.
Let K be a number field, [K : Q] = m. Then OK is free of rank m over Z, so disc(OK /Z) is a well defined
integer (choosing two integral bases for K, their discriminant differs by multiplication by a square of a unit
in Z, i.e. multiplication by 1). This integer is called the discriminant of K, and is sometimes denoted DK
or ∆K .
Lemma 2.23. Any Z-independent set of size m := [K : Q] (such as an integral basis for K) is a basis for
K/Q.
Proof. Let α1 , . . . , αm be Z-independent elements in K. It suffices to prove that they are Q-independent
(since there are m = [K : Q] of them).
Let
p1
pm
q1 , . . . , qm
∈ Q (so p1 , . . . , pm ∈ Z and q1 , . . . , qm ∈ Z \ {0}) satisfy
X pi
qi
αi = 0.
Q
P
Q
Multiplying both sides by
qi yields
(pi Si )αi = 0, where Si := j6=i qj . Each pi Si ∈ Z, therefore
pi Si = 0 for i = 1, . . . , m (since α1 , . . . , αm are Z-independent). For i = 1, . . . , m, Si 6= 0 ∴ pi = 0. Hence
{α1 , . . . , αm } is an independent set of vectors in K (as a vector space over Q). As |{α1 , . . . , αm }| = [K : Q],
{α1 , . . . , αm } is a basis for K as a vector space over Q.
Consequently,
TrOK /Z = TrK/Q NmOK /Z = NmK/Q ,
OK
.
(2.17)
OK
Lemma 2.24. Let K be a number field. Then the sign of disc(K/Q) is (−1)s , where 2s is the number of
homomorphisms K ,→ C with image not contained in R.
24
Theorem 2.25 (Stickelberger’s theorem). Let K be an algebraic number field. Then disc(OK /Z) is congruent to 0 or 1 modulo 4.
We can use the discriminant to determine some rings of integers:
Theorem 2.26. Let m be a square-free integer. Then the ring of integers of the algebraic number field
√
Q[ m] is
√
(a) Z[ m] if m ≡ 2, 3 mod 4,
√
(b) Z[ 1+2 m ] if m ≡ 1 mod 4.
√
Proof. Let K = Q[ m].
√
√
(a) Assume that m is congruent to 2 or 3 modulo 4. D(1, m) = disc(X 2 − m) = 4m. As m ∈ OK , 2.18
yields
√
√
(2.18)
4m = D(1, m) = disc(OK /Z) · (OK : Z[ m])2 .
√
• Case: (OK : Z[ m) is even. Then
m = disc(OK /Z) ·
As
√
(OK :Z[ m])
2
∈ Z, we know that
√
(OK :Z[ m])
2
2
√
(OK : Z[ m])
.
2
(2.19)
2
is congruent to 0 or 1 modulo 4. By Stickelberger’s
theorem, 2.25, disc(OK /Z) is congruent to 0 or 1 modulo 4. Hence m is congruent to 0 or 1 modulo
4, contradicting the assumption that m is congruent to 2 or 3 modulo 4.
√
√
√
• Case: (OK : Z[ m]) is odd. Then gcd(4, (OK : Z[ m])2 ) = 1, so, by Euclid’s lemma, (OK : Z[ m])2
√
√
divides m. As m is squarefree, (OK : Z[ m])2 = 1, so OK = Z[ m].
√
We conclude that OK = Z[ m].
(b) Assume that m is congruent to 1 modulo 4. Note that
is X 2 − X + 1−m
∈ Z[X]. Now 2.18 yields
4
√
1+ m
2
∈ OK , as the minimal polynomial of
√ 2
√
1 − m
1+ m
= D(1, m) = disc(OK /Z) · OK : Z
.
m = disc X 2 − X +
4
2
As m is squarefree and
OK
√
1+ m
2
(2.20)
√ 2
OK : Z 1+2 m
divides m,
so OK = Z
√
1+ m
2
√ 2
1+ m
:Z
= 1,
2
.
25
(2.21)
2.3
Dedekind domains
In this section, we prove that rings of integers in number fields are Dedekind domains. We study Dedekind
domains more generally and, in particular, prove that nonzero proper ideals in Dedekind domains factor
uniquely into products of nonzero prime ideals.
√
√
√
Example By 2.26, the ring of integers in Q[ −5] is Z[ −5]. In Z[ −5],
(2, 1 +
In fact (2, 1 +
√
√
−5)(2, 1 −
−5) and (2, 1 −
√
√
√
√
−5) = (4, 2 − 2 −5, 2 + 2 −5, 6) = (2).
√
−5) are prime ideals in Z[ −5].
√
For instance, to show that (2, 1 + −5) is prime, we need to show that
fact it is a field, since the following map is a ring isomorphism:
√
Z[ −5]
Z
√
→
Φ:
2Z
(2, 1 + −5)
√
Z[ −5]
√
(2,1+ −5)
is an integral domain. In
[x] 7→ [x].
• (Well defined:)√Assume that x, y ∈ Z and [x] = [y] ∈
Z[ −5]
√
. Hence Φ is well defined.
[x] = [y] ∈ (2,1+
−5)
• (Injective:) Assume that x ∈ Z and [x] 7→ [0] ∈
there exist p, q, r, s ∈ Z such that
(2.22)
Z
2Z .
Then x − y ∈ 2Z ⊆ (2, 1 +
√
Z[ −5]
√
.
(2,1+ −5)
Then x ∈ (2, 1 +
√
√
√
−5)Z[ −5], so
√
−5) C Z[ −5]. Thus
√
√
√
√
x = 2(p + q −5) + (1 + −5)(r + s −5) = (2p + r − 5s) + (2q + r + s) −5.
(2.23)
x ∈ Z, therefore 2q + r + s = 0. Thus,
x = 2p + r − 5s = 2p + r − 5(−2q − r) = 2p + 10q + 6r ∈ 2Z.
Hence [x] = 0 ∈
(2.24)
Z
2Z ,
and we conclude that Φ is injective.
√
√
• (Surjective:) Let a, b ∈ Z[ −5]. We need to show that [a + b −5] is in the image of the map.
[a − b] 7→ [a − b] = [a − b + b(1 +
√
√
−5)] = [a + b −5].
Hence Φ is surjective.
√
Z[ −5]
√
(2,1+ −5)
√
which is a field. Hence (2, 1 + −5) is a maximal (and therefore prime)
√
√
√
ideal in Z[ −5]. Similarly, (2, 1 − −5) is a prime ideal in Z[ −5].
Thus
is isomorphic to
Z
2Z ,
√
√
This example shows how to factorize (2) C Z[ −5] as the product of nonzero prime ideals in Z[ −5]. This
factorization is unique, though we have not shown it.
Proposition 2.27. Let A be a principal ideal domain. Then the following are equivalent.
(a) A has exactly one nonzero prime ideal.
26
(b) Up to associates, A has exactly one prime element.
(c) A is local and not a field.
Proof.
• ((a) ⇒ (b):) Assume that A has exactly one nonzero prime ideal, (y), where y ∈ A. Then y ∈ A is
prime. Thus, we need to show that if x ∈ A is prime then x ∼ y. Let x ∈ A be prime in A. Then (x)
is a nonzero prime ideal in A, so (x) = (y). Hence x ∼ y. Thus A has exactly one prime element, up
to associates.
• ((b) ⇒ (c):) Assume that up to associates, A has exactly one prime element. Fields have no nonzero
proper ideals, and therefore no nonzero prime ideals, and therefore no prime elements, so A is not a
field. It remains to show that A is local.
Let (x) and (y) be maximal ideals in A. Then x, y ∈ A are prime, so x ∼ y, i.e. (x) = (y). Hence A
has at most one maximal ideal. It remains to show that A contains a maximal ideal.
As A is not a field, there exists z ∈ A such that z 6= 0 and z is not a unit in A. Now (z) is a nonzero
proper ideal in A. Hence (z) is contained within a (nonzero) maximal ideal in A.14
• ((c) ⇒ (a):) Since A is local, A has precisely one maximal ideal (m). m 6= 0, since A is not a field15 .
Hence (m) is a nonzero prime ideal in A. It remains to show that (m) is unique in this sense.
Let (x) be a prime ideal in A. Then (x) is contained within a maximal ideal, and (m) is the only
maximal ideal in A, so (x) ⊆ (m). Thus m|x. Let x = am, where a ∈ A. Since x is prime, it is
irreducible, so a must be a unit (note that m is not a unit, since (m) is a proper ideal in A). Hence
x ∼ m, so (x) = (m).
A discrete valuation ring is a principal ideal domain satisfying the three equivalent conditions in 2.27. If A
is a discrete valuation ring with prime element π, then any nonzero element can be uniquely expressed as
uπ m , where m ∈ Z≥0 and u ∈ A is a unit. The only nonzero prime ideal in A is p = (π), so any nonzero
ideal in A has the form pm for a well defined integer m ≥ 0.
Proposition 2.28. An integral domain A is a discrete valuation ring if and only if the following three
conditions hold:
(a) A is a Noetherian ring,
(b) A is integrally closed, and
(c) A has exactly one nonzero prime ideal.
Proof.
(⇒:) Assume that A is a discrete valuation ring.
14
This is a direct application of Zorn’s lemma, ordering by inclusion the (non-empty) set of ideals in A containing (z). See
corollary 1.4 of Atiyah-Macdonald [1].
15 Let y ∈ A \ {0} be a non-unit. Then (0) ⊆ (y) 6= A, so (0) is not a maximal ideal in A.
27
• (A is a Noetherian ring:) Let p be the unique nonzero prime ideal, and let I1 ⊆ I2 ⊆ . . . be a sequence
of ideals in A. If I1 = I2 = . . . = 0, then the sequence (Ik ) is stationary.
Otherwise, let n = min{k ∈ Z>0 : Ik 6= 0}, and let In = pm , where m ∈ Z>0 . Then
{Ik : k ∈ Z≥n } ⊆ {p, p2 , . . . , pm } =: S,
which is a finite set. Therefore some element of S appears infinitely often in (Ik ), so the ascending
chain (Ik ) is eventually stationary. Hence A is a Noetherian ring.
• (A is integrally closed:) A is a principal ideal domain, therefore A is a unique factorization domain.
By 2.4, A is integrally closed.
• A is a discrete valuation ring, therefore A has exactly one nonzero prime ideal.
(⇐:) Let A be an integral domain which is Noetherian and integrally closed, and assume that A has exactly
one nonzero prime ideal. To show: A is a discrete valuation ring. As A is an integral domain with exactly
one nonzero prime ideal, it remains to show that every ideal in A is principal. As A = (1) and {0} = (0), it
remains to show that every nonzero ideal in A is principal.
First we’ll determine the nonzero prime ideal in A. Let c ∈ A \ {0} be a non-unit. For each m ∈ A \ {0},
Im := {a ∈ A : c | am}
P
P
is a proper ideal in A. Let
:= {Im : m ∈ A \ (c)}. As A is a Noetherian ring,
has a maximal element
p := Ib = {a ∈ A : c | ab}, where b ∈ A \ (c). Then c ∈ p, so p 6= (0).
P
(p is a prime ideal in A:) p ∈ , therefore p is a proper ideal in A. Assume, for the sake of contradiction,
that p is not a prime ideal in A. Then there exist x, y ∈ A \ p such that xy ∈ p. Then yb 6= 0 (as y 6= 0, b 6= 0
and A is an integral domain). Now
A 6= Iyb ⊇ p + (x) ) p,
as x ∈ Iyb ⇐ c|xyb ⇐ xy ∈ Ib = p and if a ∈ p then c|ab ⇒ c|ayb ⇒ a ∈ Iyb . We have contradicted the
P
maximality of p in , therefore p is a prime ideal in A. Thus, p is the unique nonzero prime ideal in A, as
A has precisely one nonzero prime ideal.
Let K be the field of fractions of A. Note that b ∈ A \ (c) ⇒ c 6 |b ⇒
b
c
∈
/ A.
(p = A cb , and therefore cb ∈ A:) In order to show that p = A cb , it suffices to show that p · cb = A (then
p = A cb ). By definition, pb ⊆ (c), so p · cb ⊆ A. Moreover, p · cb is an ideal in A (begin an A-submodule of
A). Assume, for the sake of contradiction, that p · cb 6= A. By Zorn’s lemma, there exists a maximal ideal
containing p · cb . However, A has exactly one nonzero prime ideal (namely p), and p · cb 6= (0), therefore
p·
b
⊆ p.
c
(2.25)
However, p is a nonzero A-submodule of K (as pCA), and is finitely generated (as pCA and A is a Noetherian
ring), and cb ∈
/ A, so (2.25) contradicts 2.1. Hence p · cb = A, so p = A cb . Finally, cb ∈ p ⊆ A.
28
Let a be a nonzero ideal in A, and we want to show that a is principal. Let π = cb , so that p = (π), π ∈ A,
and π −1 ∈ K \ A. Consider the sequence
a ⊆ aπ −1 ⊆ aπ −2 ⊆ . . .
(2.26)
of A-submodules of K.
(The sequence (2.26) is strictly increasing:) Proof by contradiction: assume that the sequence is not strictly
increasing. Then there exists r ∈ Z≥0 such that aπ −r = aπ −r−1 . Now π −1 a = a. However, a is a nonzero,
finitely generated (as a C A and A is a Noetherian ring) A-submodule of K, so π −1 is integral over A (by
2.1). As π ∈ K and A is integrally closed, π −1 ∈ A, contradiction. Hence the sequence (2.26) is strictly
increasing.
As A is a Noetherian ring, this implies that the sequence (2.26) is not a sequence of ideals in A. Let
m := min{n ∈ Z≥0 : aπ −n ⊆ A}.
Now aπ −m C A. Moreover, aπ −m−1 ( A, so aπ −m ( (π) = p. As aπ −m 6= (0) is not contained in p (which
is the only maximal ideal in A), Zorn’s lemma implies that aπ −m = A, so a = Aπ m = (π m ) is principal.
Hence A is a principal ideal domain with exactly one nonzero prime ideal, and is therefore a discrete valuation
ring.
A Dedekind domain is an integral domain A which is not a field, and which satisfies the following conditions:
(a) A is Noetherian.
(b) A is integrally closed.
(c) Every nonzero prime ideal in A is maximal.
Lemma 2.29. Let A be a local integral domain. Then A is a Dedekind domain if and only if A is a discrete
valuation ring.
Proof. (⇒:) Assume that A is a Dedekind domain. By 2.28, it remains to show that A has exactly one
nonzero prime ideal. A is local, so let M be the unique maximal ideal in A. Let I be a nonzero prime ideal
in A. Then I is a maximal ideal, since A is a Dedekind domain. Since I 6= 0, this implies that I = M .
Hence M is the only nonzero prime ideal in A, so A is a discrete valuation ring.
(⇐:) Assume that A be a discrete valuation ring. By 2.28, it remains to show that every nonzero prime
ideal in A is maximal. By 2.28, A has exactly one nonzero prime ideal, I. To show: I is maximal. Note that
there exists a maximal ideal J which contains I (see footnote 14). Being a maximal ideal, J 6= (0) must also
be a prime ideal, so J = I. Hence I is a maximal ideal, so A is a Dedekind domain.
Thus, a local integral domain is a Dedekind domain if and only if it is a discrete valuation ring.
We now aim to show that localizing a Dedekind domain at a nonzero prime ideal yields a discrete valuation
ring. The motivation for this is as follows. Later in this section we will study the factorization of prime
29
ideals in extensions, which is of great importance in algebraic number theory. A useful tool will be to take
an equation involving ideals, and to localize both sides of the equation at a prime ideal.16 The idea is that
discrete valuation rings have stronger properties than general Dedekind domains.
Lemma 2.30. Let A be an integral domain, and let S be a multiplicatively closed subset of A.
(a) If A is a Noetherian ring, then S −1 A is a Noetherian ring.
(b) If A is integrally closed, then S −1 A is integrally closed.
Proof.
(a) Assume that A is a Noetherian ring. Let S −1 I (where I C A) be an arbitrary ideal in S −1 A (by 1.2).
Since A is Noetherian, I is finitely generated in A, say I = (a1 , . . . , an ), where a1 , . . . , an ∈ A. Then
S −1 I = ( a11 , . . . , a1n ) is finitely generated in S −1 A. Thus every ideal in S −1 A is finitely generated, so
S −1 A is a Noetherian ring.
(b) Assume that A is integrally closed. A and S −1 A have the same field of fractions, say K. Let α ∈ K be
integral over S −1 A. We want to show that α ∈ S −1 A.
αm + am−1 αm−1 + . . . + a1 α + a0 = 0,
where m ∈ Z>0 and a0 , . . . , am−1 ∈ S −1 A. Thus, there exist s1 , . . . , sm ∈ S such that si ai ∈ A for
i = 1, 2, . . . , m. Multiplying both sides by sm , where s = s0 s1 . . . sm−1 , yields
(sα)m + sam−1 (sα)m−1 + . . . + sm−1 a1 (sα) + sm a0 = 0.
(2.27)
In (2.27), all of the coefficients lie in A. Hence sα is integral over A, so sα ∈ A, i.e. α ∈ S −1 A.
Proposition 2.31. Let A be a Dedekind domain, and let p be a nonzero prime ideal in A. Then Ap is a
discrete valuation ring.
Proof. Ap is a local integral domain (by 1.5), so by 2.29 it suffices to show that Ap is a Dedekind domain.
By 2.30, Ap is Noetherian and integrally closed. Ap is not a field because it has a nonzero maximal ideal
pAp . It remains to show that every nonzero prime ideal in Ap is maximal.
Let IAp be an arbitrary nonzero prime ideal in Ap , where I is a nonzero prime ideal in A such that I ⊆ p
(using 1.3 with S = A \ p). Then I must be a maximal ideal in A, since A is a Dedekind domain. However,
I ⊆ p, and p is a proper ideal in A, therefore I = p. Now IAp = pAp , which is the maximal ideal in Ap .
Hence every nonzero prime ideal in Ap is a maximal ideal. We conclude that Ap is a discrete valuation
ring.
We now move towards proving the following important theorem:
16 Localization is an important concept in abstract algebra generally, because the result is a local ring. In our case we get a
discrete valuation ring, which is even ‘nicer’ !
30
Theorem 2.32. Let A be a Dedekind domain, and let a be a nonzero proper ideal in A. Then a can be
written in the form
a = pr11 . . . prmm ,
(2.28)
where m ∈ Z>0 , p1 , · · · , pm are pairwise distinct nonzero prime ideals and r1 , . . . , rm ∈ Z>0 . Moreover, the
pi and ri are uniquely determined, up to rearrangement.
Lemma 2.33. Let A be a Noetherian ring. Then every nonzero ideal in A contains a product of nonzero
prime ideals.
Proof. Proof by contradiction: assume that there exists a nonzero ideal in A which does not contain a
P
product of nonzero prime ideals. Let
be the set of nonzero ideals in A which do not contain a product
P
of nonzero prime ideals. As
is non-empty and A is a Noetherian ring, there exists a maximal element
P
I ∈ .17 Then I is not a prime ideal, and I 6= A, therefore there exist x, y ∈ A \ I such that xy ∈ I. Since
I + (x) ) I and I + (y) ) I, the maximality of I implies that I + (x) and I + (y) contain a product of prime
ideals. Now I = (I + (x))(I + (y)) contains a product of prime ideals, contradiction. Hence every nonzero
ideal in A contains a product of nonzero prime ideals.
Lemma 2.34. Let A be a commutative unital ring, and let a and b be relatively prime ideals in A. Let
m, n ∈ Z>0 . Then am and bn are relatively prime.
Proof. As a + b = A, there exist a ∈ a and b ∈ b such that a + b = 1. Now
1 = (a + b)m+n =
m+n
X
i=1
m+n
m X m + n
X
m + n i m+n−i
m + n i m+n−i
ab
=
ai bm+n−i +
ab
∈ am + bn
i
i
i
i=m+1
i=1
∴ am + bn = A,
so am and bn are relatively prime.
Lemma 2.35. Let A be a commutative unital ring, and let p be a maximal ideal in A. Let m ∈ Z>0 , and
let q = pAp C Ap . Then the map
A
Ap
Φ: m → m
p
q
a
[a] 7→
1
is an isomorphism.
Proof. Let S := A \ p. By 1.4,
qm := (pAp )m = (S −1 p)m = S −1 (pm ).
(Well defined:) Let a, b ∈ A, and assume that [a] = [b] ∈
A
pm .
Then a − b ∈ pm , so
a b
a−b
− =
∈ S −1 pm = qm ,
1 1
1
17 That
is, if J ∈
P
and J ⊇ I then J = I.
31
as 1 ∈ S. Thus
a
1
=
b
1
∈
Ap
qm ,
so Φ is well defined.
A
(Injective:) Assume that x ∈ A and [ x1 ] = [0] ∈ qmp . Then x1 ∈ qm = S −1 (pm ), so there exist b ∈ pm and
t ∈ S such that x1 = bt ∈ S −1 A. Thus there exists u ∈ S such that u(xt − b) = 0. With s := ut ∈ S (since S
is multiplicative),
sx = utx = ub ∈ pm .
We want to show that [x] = [0] ∈
A
pm ,
i.e. that x ∈ pm .
As s ∈
/ p, the ideal (s) + p properly contains p. As p is a maximal ideal in A, it follows that (s) + p = A, so (s)
and p are relatively prime ideals in A. By 2.34, (s) and pm are relatively prime ideals in A, so (s) + pm = A.
Thus, there exists a ∈ A such that
as ≡ 1 mod pm .
Now
sx ∈ pm =⇒ asx ∈ pm =⇒ x ∈ pm ,
as 0 ≡ asx ≡ x mod pm . Thus [x] = [0] ∈
A
pm ,
so Φ is injective.
(Surjective:) Let a ∈ A and s ∈ S. Now as is an arbitrary element of Ap , and we want to show that [ as ] ∈
is in the image of Φ. As (s) + pm = A,18 there exists x ∈ A such that xs ≡ 1 mod pm . Now
asx ≡ a
Thus
Ap
qm
mod pm ∴ asx − a ∈ pm .
asx − a
ax a
− =
∈ S −1 pm ,
1
s
s
a
ax
Ap
=
= Φ([ax]) ∈ m .
s
1
q
so
Hence Φ is surjective.
We conclude that Φ is an isomorphism.
Now we prove 2.32. We use the standard motif of localizing and working with discrete valuation rings.
Proof of 2.32.
• (Existence:) By 2.33, there exist m ∈ Z>0 , r1 , . . . , rm ∈ Z, and distinct nonzero prime ideals
p1 , . . . , pm C A such that
a ⊇ b := pr11 . . . prmm .
By 2.34, the ideals pr11 , . . . , prmm are relatively prime, so by the Chinese remainder theorem (1.9), the map
m
Y
A
A
→
ri
b
p
i=1 i
[x] 7→ ([x], . . . , [x])
18 As
s∈
/ p, the ideal (s) + p properly contains p. As p is a maximal ideal in A, it follows that (s) + p = A, so (s) and p are
relatively prime ideals in A. By 2.34, (s) and pm are relatively prime ideals in A, so (s) + pm = A.
32
is an isomorphism. For i = 1, 2, . . . , m, define qi := pi Api , which is the unique maximal ideal in the
discrete valuation ring Api (by 1.5). By 2.35, the map
m
m
Y
Y
Api
A
ri →
p
q ri
i=1 i
i=1 i
([x1 ], . . . , [xm ]) 7→
h x i
h x i
1
m
,...,
1
1
(2.29)
is an isomorphism. Thus, the map
m
Y
A
Api
→
b
q ri
i=1 i
h x i
h x i
[x] 7→
,...,
1
1
is an isomorphism, being a composition of isomorphisms.
Φ:
(2.30)
For i = 1, 2, . . . , m, the rings Api are discrete valuation rings with corresponding maximal ideal qi .
Qm
Qm
Therefore the image of a in the map A → i=1 Api is i=1 qsi i , for some s1 , . . . , sm ∈ Z with si ≤ ri
for i = 1, 2, . . . , m. Thus,
m
ps1 · · · psm a Y
qsi i
m
1
=
Φ
Φ
=
,
(2.31)
ri
b
q
b
i=1 i
where
a
:= {x + b : x ∈ a},
b
qsi i
:= {x + qri i : x ∈ qsi i },
qri i
i = 1, 2, . . . , m,
and
ps11 · · · psmm
:= {x + b : x ∈ ps11 · · · psmm }.
b
As Φ is an isomorphism, equation (2.31) yields
a
ps1 · · · psmm
= 1
.
b
b
(2.32)
a = ps11 · · · psmm ,
(2.33)
As a ⊇ b and ps11 · · · psmm ⊇ b,
so the factorization exists.
• (Uniqueness:) Assume that
ps11 · · · psmm = pt11 · · · ptmm ,
where m ∈ Z>0 , s1 , . . . , sm , t1 , . . . , tm ∈ Z≥0 , and p1 , . . . , pm are distinct nonzero prime ideals in A.19
For i = 1, 2, . . . , m, localizing at pi yields
m
Y
m
Y
s
t
pj j Api =
pjj Api
j=1
m
Y
j=1
(pj Api )sj =
j=1
19 We
m
Y
j=1
add factors p0i if necessary.
33
(pj Api )tj
(2.34)
(If i, j ∈ {1, 2, . . . , m} and i 6= j then pi + pj = A:) Let i, j ∈ {1, 2, . . . , m}, with i 6= j. To show:
pi + pj = A. Proof by contradiction: assume that pi + pj 6= A. As A is a Dedekind domain, the
nonzero prime ideals pi and pj are maximal. Hence pi = pi + pj = pj , so i = j, contradiction. Thus,
pi + pj = A.
For i, j ∈ {1, 2, . . . , m} with i 6= j, there therefore exists x ∈ pj such that x − 1 ∈ pi . Thus,
1 ∈ pj Api C Api , so pj Api = Api . Now equation (2.34) becomes
qsi i = qtii ,
where qi := pi Api is the unique maximal ideal in the discrete valuation ring Api (by 1.5). Hence si = ti
for i = 1, 2, . . . , m, so the factorization is unique.
Let A be a Dedekind domain. Let I be a nonzero proper ideal in A, and let p be a nonzero prime ideal in A.
Then p divides I if p occurs in the prime factorization of I. The following result provides useful equivalent
definitions.
Lemma 2.36. Let A be a Dedekind domain, let I be a nonzero proper ideal in A, and let p be a nonzero
prime ideal in A. Then the following are equivalent:
(a) p divides I.
(b) IAp 6= Ap .
(c) I ⊆ p.
Proof. Let
I = pe11 . . . pegg ,
(2.35)
where g, e1 , . . . , eg ∈ Z>0 , and where p1 , . . . , pg are prime ideals in A. Now
g
Y
IAp =
(pi Ap )ei .
(2.36)
i=1
• ((a) ⇒ (c):) Assume that p occurs in the prime factorization of I. Then there exists i ∈ {1, 2, . . . , g}
e
such that p = pi . Thus, p ⊇ pe11 · · · pgg = I.
• ((c) ⇒ (b):) Assume that I ⊆ p. Then IAp ⊆ pAp ( Ap , so IAp 6= Ap .
• ((b) ⇒ (a):) Assume that IAp 6= Ap . Then
g
Y
(pi Ap )ei 6= Ap ,
i=1
so there exists i ∈ {1, 2, . . . , n} such that pi Ap 6= Ap .
Claim: pi ⊆ p. Proof by contradiction: assume that there exists x ∈ pi \ p. Then
Ap , so pi Ap = Ap , contradiction. Thus pi ⊆ p.
x
1
∈ pi Ap is a unit in
However pi is a prime ideal in A, and is therefore a maximal ideal in A, since A is a Dedekind domain.
As p 6= A, this implies that p = pi , so p divides I.
34
Many more technical results follow from 2.32. We discuss them here, but the reader may wish to skip them
for now, and return to them when they are needed.
Lemma 2.37. Let A be a Dedekind domain, and let a, b C A. Then a ⊆ b if and only if aAp ⊆ bAp for all
nonzero prime ideals p in A. Thus, a = b if and only if aAp = bAp for all nonzero ideals p in A.
Proof. The cases a = (0), a = A, b = (0) and b = A are easily checked, so we exclude them henceforth.
• (⇒:) Assume that a ⊆ b, and let p be a nonzero prime ideal in A. Then aAp ⊆ bAp .
• (⇐:) Assume that aAp ⊆ bAp for all nonzero prime ideals p in A. Let
a = pr11 . . . prmm ,
b = ps11 . . . psmm ,
(2.37)
where m ∈ Z>0 , r1 , . . . , rm , s1 , . . . , sm ∈ Z≥0 , and where p1 , . . . , pm are nonzero prime ideals in A (add
factors p0i if necessary). For i = 1, 2, . . . , n, localize both sides of the equations in (2.37) at pi , and the
assumption aAp ⊆ bAp becomes
(pi Api )ri ⊆ (pi Api )si ,
so ri ≤ si for i = 1, 2, . . . , n. Hence a ⊆ b.
Therefore, a ⊆ b if and only if aAp ⊆ bAp for all nonzero prime ideals p in A. Consequently, a = b if and
only if aAp = bAp for all nonzero ideals p in A.
Corollary 2.38. Let A be a Dedekind domain, and let a, b be nonzero ideals in A such that a ⊇ b. Then
there exists a ∈ A such that a = b + (a).
Proof. Let
b=
m
Y
pri i ,
a=
i=1
m
Y
psi i ,
(2.38)
i=1
where m ∈ Z>0 , r1 , . . . , rm , s1 , . . . , sm ∈ Z≥0 , and where p1 , . . . , pm are nonzero prime ideals in A (add
factors p0i if necessary). Then ri ≥ si for i = 1, 2, . . . , m (as a ⊇ b).
For i = 1, 2, . . . , m, choose xi ∈ A such that xi ∈ psi i \ psi 1 +1 . By the Chinese remainder theorem, there exists
a ∈ A such that a ≡ xi mod pri i (for i = 1, 2, . . . , m). To show: a = b + (a).
By 2.37, it suffices to prove that
aAp + bAp = (a)Ap
(2.39)
for all nonzero prime ideals p in A. By 2.36 (b), both sides of equation (2.39) are equal to Ap if p does not
occur in the prime factorization of a. Hence it suffices to show that if i ∈ {1, . . . , m}, r = ri , s = si and
p = pi then ps Ap = pr Ap + (a)Ap .
Let i ∈ {1, . . . , m}, r = ri , s = si and p = pi . It remains to show that
ps Ap = pr Ap + (a)Ap .
35
(2.40)
• (⊇:) Note that a ∈ ps , so (a)Ap ⊆ ps Ap . Also, r ≥ s, therefore pr Ap ⊆ ps Ap . Hence,
pr Ap + (a)Ap ⊆ ps Ap .
• (⊆:) Note that r ≥ s.
– Case: r = s. Then ps Ap = pr Ap ⊆ pr Ap + (a)Ap .
– Case: r ≥ s + 1. Then
a ≡ xi
mod pr =⇒ a − xi ∈ pr ⊆ ps+1 =⇒ a ≡ xi
mod ps+1
,
i
so a ∈ ps \ ps+1 . Hence (a)Ap ⊆ ps Ap and (a)Ap ( ps+1 Ap .
As Ap is a discrete valuation ring with unique prime ideal pAp , we know that any nonzero ideal
in Ap has the form pn Ap , for some n ∈ Z>0 . Moreover, (a)Ap ⊆ ps Ap and (a)Ap ( ps+1 Ap ,
therefore (a)Ap = ps Ap . Now
pr Ap + (a)Ap = pr Ap + ps Ap ⊇ ps Ap .
Hence ps Ap = pr Ap + (a)Ap , and we conclude that a = b + (a).
Corollary 2.39. Let A be a Dedekind domain, let a be a nonzero ideal in A, and let a ∈ a \ {0}. Then there
exists b ∈ a such that a = (a, b).
Proof. Apply 2.38 with a ⊇ (a) 6= 0.
Lemma 2.40. Let A be a Dedekind domain, and let a be a nonzero ideal in A.
(a) Let a ∈ a. Then there exists an ideal a∗ in A such that aa∗ = (a).
(b) Let c be a nonzero ideal in A. Then there exists an ideal a∗ in A such that aa∗ is a principal ideal in A
and c + a∗ = A.
Proof.
(a) If a = 0, then a∗ = (0) satisfies aa∗ = (a). If a = A, then a∗ = (a) sayisfies aa∗ = (a). Henceforth,
assume that a = 0 and that a 6= A. Now
(a) = pr11 . . . prmm ,
a = ps11 . . . psmm ,
(2.41)
where m ∈ Z>0 , r1 , . . . , rm , s1 , . . . , sm ∈ Z≥0 , and where p1 , . . . , pm are nonzero prime ideals in A (add
factors p0i if necessary). Now (a) ⊆ a implies that ri ≥ si for i = 1, 2, . . . , m.20 Now
a∗ =
n
Y
i=1
satisfies aa∗ = (a).
20 See
the proof of 2.37 if this is not clear.
36
pi ri −si
(2.42)
(b) a ⊇ ac 6= (0), so, by 2.38, there exists a ∈ A such that a = ac + (a). Moreover, from part (a), there exists
an ideal a∗ in A such that aa∗ = (a). Thus,
a = ac + aa∗ = a(c + a∗ ),
(2.43)
therefore c + a∗ = A.21
Importantly, we are able to reconcile our definition of relatively prime ideals with the intuitive notion:
Lemma 2.41. Let A be a Dedekind domain, and let a and b be nonzero proper ideals in A. Then the
following are equivalent:
(a) a + b = A.
(b) There does not exist a nonzero prime ideal p C A such that p is in the prime factorization of a and b.
Proof. By 2.36, a prime ideal p C A divides a and b if and only if it contains a and b.
• If a + b = A and p contains a and b, then p = p + b ⊇ a + b = A, contradiction.
• If a + b 6= A, then there exists a maximal ideal p ⊇ a + b. Then p is prime and p contains a and b.
We will prove that rings of integers in number fields are Dedekind domains, but first we need the following
lemma:
Lemma 2.42. Let k be a field, and let B ⊇ k be an integral domain which is algebraic over k. Then B is a
field.
Proof. It remains to show that every nonzero element of B has an inverse in B. Let 0 6= β ∈ B. As β is
algebraic over k, the ring k[β] is a finite-dimensional vector space over k. The map
k[β] → k[β]
x 7→ βx
is an injective linear transformation between vector spaces of the same dimension, and therefore surjective.
Hence there exists x ∈ k[β] ⊆ B such that βx = 1.
Theorem 2.43. Let A be a Dedekind domain with field of fractions K, and let L be a finite separable
extension of K. Let B be the integral closure of A in L. Then B is a Dedekind domain.
Proof.
• By 2.9, B is an integrally closed integral domain.
21 If
c + a∗ 6= A, then factorizing a and c + a∗ and substituting into equation (2.43) will yield a contradiction.
37
• (B is a Noetherian ring:) As A is a Noetherian ring, and B is a finitely generated A-module, B is a
Noetherian A-module (by 1.21). Any I C B is therefore finitely generated as an A-module, and the
same finite set of generators generates I as a B-module, i.e. as an ideal in B. Hence B is a Noetherian
ring.
• (Every nonzero prime ideal in B is maximal:) Let q be a nonzero prime ideal in B. In order to show
that q is maximal, we need to show that Bq is a field.
First we establish that q ∩ A is a maximal ideal in A. As A is a Dedekind domain, it suffices to prove
that q ∩ A is a nonzero prime ideal in A.
– (q ∩ A 6= 0:) Let β ∈ q, β 6= 0. Then there exist n ∈ Z>0 and a1 , . . . , an ∈ A such that
β n + a1 β n−1 + . . . + an = 0,
(2.44)
and an 6= 0 (otherwise the polynomial β n + a1 β n−1 + . . . + an is reducible). Then
0 6= an = −(β n + a1 β n−1 + . . . + an−1 β) ∈ q ∩ A.
(2.45)
Hence q ∩ A 6= (0).
– (q ∩ A 6= A:) 1 ∈
/ q ⊇ q ∩ A, therefore q ∩ A 6= A.
– (q ∩ A is a prime ideal in A:) Assume that x, y ∈ A and xy ∈ q ∩ A. As q is a prime ideal in B,
we know that x ∈ q or y ∈ q. Thus x ∈ q ∩ A or y ∈ q ∩ A. Therefore q ∩ A is a prime ideal in A.
Hence q ∩ A is a nonzero prime ideal in A. As A is a Dedekind domain, q ∩ A is a maximal ideal in A.
Thus,
A
p
is a field, where p := q ∩ A. The image of the injective map
B
A
→
p
q
[a] 7→ [a]
B
is a isomorphic to A
p , by the first isomorphism theorem, and is therefore a field contained in q .
Moreover, Bq is an integral domain, since q is a prime ideal in B. By 2.42, we conclude that Bq is a
field. Thus, q is a maximal ideal in B.
Hence B is a Dedekind domain.
Corollary 2.44. Let K be a number field. Then OK is a Dedekind domain.
Proof. In 2.43, let A = Z, and let L be any number field.
2.4
The ideal class group
Let A be a Dedekind domain with field of fractions K. A fractional ideal of A is a nonzero A-submodule a
of K such that
da := {da : a ∈ a} ⊆ A
38
for some d ∈ A \ {0}. In this case, da ⊆ A is an A-module, so da C A.
Fractional ideals of A are not bona fide ideals unless they are contained in A, so we shall refer to ideals as
integral ideals if there is any confusion. A fractional ideal is principal if it takes the form
(b) := bA := {ba : a ∈ A},
for some b ∈ K.
We shall see that the product of two fractional ideals is a fractional ideal:
a·b=
n
nX
o
ai bi : ai ∈ a, bi ∈ b, n ∈ Z≥0 .
(2.46)
i=1
For principal fractional ideals, (a)(b) = (ab).
The following theorem results in unique prime factorization of fractional ideals, allowing negative powers,
and neatly describes what all of the fractional ideals are. The fact that fractional ideals have inverses is
extremely useful, and nontrivial.22
Theorem 2.45. Let A a be a Dedekind domain with field of fractions K. Then, with the group operation as
in equation (2.46), the set Id(A) of fractional ideals of A is the free abelian group on the set of prime ideals
in A.
Proof. First we show that Id(A) is a group. Let a, b, c ∈ Id(A) be such that da, eb, f c ⊆ A, where d, e, f ∈ A.
• (ab ∈ Id(A):) Note that (de)ab ⊆ A. Moreover, ab is an A-submodule of K, since it is closed under
addition and scalar multiplication, and that it contains additive inverses. Hence ab is a fractional ideal
of A.
• The operation is commutative because K is commutative:
ab =
n
nX
n
o nX
o
ai bi : ai ∈ a, bi ∈ b, n ∈ Z≥0 =
bi ai : ai ∈ a, bi ∈ b, n ∈ Z≥0 = ba.
i=1
i=1
• A is the identity.
• (Inverses exist:) We first prove this for integral ideals. Let (0) 6= a C A and 0 6= a ∈ a. By 2.40, there
exists a∗ C A such that aa∗ = (a). Now a · (a−1 a∗ ) = A, so a−1 a∗ ∈ Id(A) is an inverse for a.
Now let a ∈ Id(A). Then there exists d ∈ A \ {0} such that da C A. Now d · (da)−1 is an inverse for a.
Hence Id(A) is a group. To complete the proof, we need to show that any fractional ideal of A can be
expressed uniquely in the form
pt11 · · · ptmm ,
where m ∈ Z>0 , t1 , . . . , tm ∈ Z, and p1 , . . . , pm are prime ideals in A.
Let a ∈ Id(A), and let d ∈ A be such that da C A.
22 That
is, without the powerful technical framework that we have developed.
39
(Existence:)
da = pr11 . . . prmm ,
(d) = ps11 . . . psmm ,
Now
da = (d)a =⇒ a = (da) · (d)−1 =
ri , si ≥ 0.
n
Y
pri i −si .
i=1
(Uniqueness:) Assume that n ∈ Z>0 and
n
Y
pri i =
i=1
Then
n
Y
psi i ,
ri , si ∈ Z.
i=1
n
Y
i
=
pc+r
i
i=1
n
Y
i
,
pc+s
i
i=1
where c = |ri | + |si |. Then c + ri , c + si ∈ Z≥0 , so 2.32 yields ri = si for all i.
Let A be a Dedekind domain. The ideal class group of A is the quotient
Cl(A) =
Id(A)
,
P(A)
where P(A) is the subgroup of principal fractional ideals in Id(A). If Cl(A) is finite, the class number of A
is the order of Cl(A).
Let K be a number field. The ideal class group of K is Cl(OK ), and the class number of K is the order of
the ideal class group of K. We shall see that this is always finite.
The ideal class group and the class number of a number field are important invariants in algebraic number
theory. Intuitively, they measure the failure of prime factorization of elements in the ring of integers.23 This
is motivated by the following result:
Proposition 2.46. Let A be a Dedekind domain. Then A is a principal ideal domain if and only if A is a
unique factorization domain.
Proof. (⇒:) Assume that A is a principal ideal domain. Then A is a unique factorization domain, as any
principal ideal domain is a unique factorization domain.
(⇐:) Assume that A is a unique factorization domain. {0} = (0) and A = (1), so it remains to show that
every nonzero proper ideal in A is principal. By prime factorization of ideals (2.32), it suffices to prove that
every nonzero prime ideal in A is principal.
Let p be a nonzero prime ideal in A. Let 0 6= a ∈ p. Then a is not a unit in A. Since A is a unique
factorization domain, a factors as a product of prime elements in A:
a = π1 · · · πn ,
23 The
n ∈ Z>0 ,
π1 , . . . , πn prime.
task of finding precise ways to describe this is the subject of current research. See Martin [6], 2011.
40
As p is prime, there exists i ∈ {1, . . . , n} such that πi ∈ p. Now p ⊇ (πi ). Moreoever, (πi ) is a maximal ideal
in A (since it is a nonzero prime ideal in A and A is a Dedekind domain), therefore p = (πi ). Hence p is a
principal ideal in A.
We conclude that A is a principal ideal domain.
2.5
Discrete valuations
Previously, we defined discrete valuation ring without defining discrete valuations. We shall see that this
terminology is no coincidence, but more importantly we give examples of discrete valuations which we use
in later sections, and which are useful more broadly in number theory.
Let K be a field. A discrete valuation on K is a nonzero group homomorphism v : K × → (Z, +) such that
if a, b ∈ K × then
v(a + b) ≥ min{v(a), v(b)}.
A discrete valuation is normalized if it is surjective. Note that Image(v) is a nonzero subgroup of Z, so there
exists m ∈ Z>0 such that Image(v) = mZ. If v is a discrete valuation, then x 7→ m−1 v(x) is a normalized
discrete valuation.
Let K be a field, and let v : K × → Z be a discrete valuation. By induction, if m ∈ Z≥2 and a1 , . . . , am ∈ K
then
v(a1 + . . . + am ) ≥ min{v(ai ) : i ∈ {1, . . . , m}}.
Note that v(1) = 0, as v(1) = v(1) + v(1). Also, v(−1) = 0, as v(−1) + v(−1) = v(1) = 0. Thus, if x ∈ K ×
then v(−x) = v(−1) + v(x) = v(x). It is sometimes convenient to define v(0) := ∞.
Example Let A be a Dedekind domain with field of fractions K, and let p be a nonzero prime ideal in A.
For x ∈ K × , let ordp (x) be the exponent of p in the factorization of (x) ∈ Id(A). Then ordp : K × → Z is a
normalized discrete valuation.
Proof. We first show that ordp : K × → Z is a discrete valuation:
• The function ordp : K × → Z is well defined, since Id(A) is the free abelian group on the set of prime
ideals in A (from 2.45).
• Let x, y ∈ K × . Factorizing (x) and (y) as in 2.45 yields
ordp (xy) = ordp (x) + ordp (y),
(2.47)
so ordp : K × → Z is a group homomorphism.
• We show that ordp (x + y) ≥ min{ordp (x), ordp (y)}. Proof by contradiction: assume that
ordp (x + y) < ordp (x),
41
ordp (x + y) < ordp (y).
(2.48)
Let m := ordp (x + y). Then ordp (x) ≥ m + 1 and ordp (y) ≥ m + 1. Now
x, y ∈ pm+1 =⇒ x + y ∈ pm+1 =⇒ ordp (x + y) ≥ m + 1,
(2.49)
contradiction. Hence ordp (x + y) ≥ min{ordp (x), ordp (y)}.
Therefore ordp : K × → Z is a discrete valuation.
It remains to show that ordp is normalized, i.e. that 1 ∈ Image(ordp ). By 2.40, there exists p∗ C A such that
pp∗ is a principal ideal in A and p + p∗ = A. The p∗ 6= (0), so let x ∈ A \ {0} ⊆ K × be such that
pp∗ = (x).
(2.50)
Now ordp (x) = 1, since p does not occur in the prime factorization of p∗ (by 2.41). Hence ordp : K × → Z is
a normalized discrete valuation.
Proposition 2.47. Let K be a field, and let v : K × → Z be a discrete valuation on K. Then
A := {0} ∪ {a ∈ K × : v(a) ≥ 0}
(2.51)
is a discrete valuation ring with unique nonzero prime ideal
m := {0} ∪ {a ∈ K × : v(a) > 0}.
(2.52)
Moreover, if v(K × ) = mZ, m > 0, and v(π) = m, then m = (π).
Proof. We want to establish that A is a discrete valuation ring. By 2.27, it suffices to prove that A is a local
principal ideal domain which is not a field.
(A is a subring of K, and is therefore an integral domain:) Let x, y ∈ A \ {0}.24
• 0 ∈ A, by definition.
• 1 ∈ A, as v(1) = 0.
• v(x + y) ≥ min{v(x), v(y)} ≥ 0 ⇒ x + y ∈ A.
• v(−x) = v(x) ≥ 0 ⇒ −x ∈ A.
• v(xy) = v(x) + v(y) ≥ 0 ⇒ xy ∈ A.
Hence A is a subring of the field K, and is therefore an integral domain.
(m C A:) Let x, y ∈ m \ {0} and a ∈ A \ {0}.25
• 0 ∈ m by definition.
• v(−x) = v(x) > 0 ⇒ −x ∈ m.
24 The
case x = 0 and the case y = 0 are trivial.
the remaining cases are trivial.
25 Again,
42
• v(x + y) ≥ min{v(x), v(y)} > 0 ⇒ x + y ∈ m.
• v(ax) = v(a) + v(x) ≥ v(x) > 0 ⇒ ax ∈ m.
Hence m C A.
(A is a local ring with unique maximal ideal m:) 1 ∈ A \ m, so m 6= A. It suffices to prove that m contains
all non-units in A. Let x be a non-unit in A. Then x1 ∈ K × \ A, so v( x1 ) < 0. Now
1
= v(1) − v(x) = −v(x)
∴ v(x) > 0,
0>v
x
so x ∈ m. Hence m contains all non-units in A. We conclude that A is a local ring with maximal ideal m.
(A is a principal ideal domain:) Let I be a nonzero proper ideal in A. There exists x ∈ I such that if y ∈ I
then v(y) ≥ v(x). Then I = (x): if y ∈ I then v( xy ) = v(y) − v(x) ≥ 0 ⇒ xy ∈ A ⇒ y ∈ (x). Hence A is a
principal ideal domain.
From the above paragraph, we conclude that if v(K × ) = mZ, m > 0, then m = (π) for any π ∈ K × such
that v(π) = m. Finally, since A is a local principal ideal domain which is not a field,26 we conclude that A
is a discrete valuation ring.
2.6
Factorization in extensions
Let A be a Dedekind domain with field of fractions K. Let L/K be a finite separable extension, and let B
be the integral closure of A in L. Let p be a nonzero prime ideal in A. B is a Dedekind domain, by 2.43, so
pB factorizes uniquely into a product of nonzero prime ideals in B (by 2.32):
pB = Pe11 · · · Pegg ,
ei ≥ 1.
(2.53)
The nonzero prime ideal p C A ramifies in B (or L) if there exists i ∈ {1, 2, . . . , g} such that ei > 1, and
ramifies totally if g = 1 and e1 > 1. Let P be a prime ideal in B. The prime ideal P divides p if P divides
pB. The ramification index, e(P/p),
h
iis the exponent of P in the prime factorization of pB. The residue class
B
27
degree is given by f (P/p) := P
: A
The prime ideal p splits in B (or L) if ei = fi = 1 for all i, and is
p .
inert in B (or L) if pB is a prime ideal in B.
Lemma 2.48. Let A be a Dedekind domain with field of fractions K. Let L/K be a finite separable extension,
and let B be the integral closure of A in L. Let p be a nonzero prime ideal in A, and let P be a nonzero
prime ideal in B. Then P divides p if and only if p = P ∩ K.
Proof. By 2.36, P divides p if and only if pB ⊆ P.
(⇒:) Assume that P divides p.
• p ⊆ pB ⊆ P and p ⊆ A ⊆ K. Hence p ⊆ P ∩ K.
26 As
v is nonzero, there exists x ∈ K × such that v(x) > 0 (if y ∈ K × and v(y) < 0 then v( y1 ) > 0. Then x ∈ A \ {0} but
v( x1 ) < 0 ⇒ x1 ∈
/ A.
27 p and P are (nonzero)
prime and hence maximal ideals in the respective Dedekind domains A and B, therefore
are fields. The map [x] 7→ [x] :
A
p
→
B
P
is an inclusion, so we can regard
43
B
P
as a vector space over
A
.
p
B
P
and
A
p
• As A is integrally closed, P ∩ K ⊆ B ∩ K = A.
• K is an A-module (being a K-module) and P is an A-module (being a B-module), therefore P ∩ K is
an A-module. Since P ∩ K ⊆ A, this means that P ∩ K C A.
• 1∈
/ P ⊇ P ∩ K, so P ∩ K 6= A.
Thus P ∩ K is a proper ideal in A which contains p. Since p is a maximal ideal (as p is a nonzero prime
ideal in the Dedekind domain A), we conclude that p = P ∩ K.
(⇐:) Assume that p = P ∩ K. Then
pB = (P ∩ K)B ⊆ PB = P,
so P divides p.
Lemma 2.49. Let
φ
ψ
→B−
→C→0
0→A−
be an exact sequence of vector spaces over a field. Then dim B = dim A + dim C.
Proof. Apply the rank-nullity theorem to φ and ψ, noting exactness:
dim B = dim(Imageψ) + dim(kerψ) = dim C + dim(Imageφ)
= dim C + dim A − dim(kerφ) = dim C + dim A.
The following formula links the ramification indices and residue class degrees to the degree of the extension:
Theorem 2.50. Let A be a Dedekind domain with field of fractions K. Let L/K be a finite separable
extension of degree m, and let B be the integral closure of A in L. Let p be a prime ideal in A, and let pB
prime factorize in B as follows:
pB = Pe11 · · · Pegg ,
ei ≥ 1.
Then
(a)
g
X
B A
ei fi =
:
= m.
pB p
i=1
(2.54)
(b) If L/K is a Galois extension, then all of the ramification indices are equal, and all of the residue class
degress are equal, so
ef g = m.
(2.55)
We now aim to prove 2.50.
44
Lemma 2.51. Let A be a Dedekind domain with field of fractions K. Let L/K be a finite separable extension
of degree m, and let B be the integral closure of A in L. Let p be a nonzero prime ideal in A, and let P be
a nonzero prime ideal in B which divides p. Let f = f (P/p) and let r ∈ Z>0 . Then PBr is a vector space of
dimesion rf over A
p.
Proof. A
p is a field because p is a (nonzero) prime (and therefore maximal) ideal in the Dedekind domain
B
B
A
A. The inclusion [x] 7→ [x] : A
p → Pr makes Pr a vector space over p . We prove by induction that its
dimension is rf .
B A
: p = f.
• By definition, P
B
28
• Let n ∈ Z>0 , and assume that PBn : A
To show: Pn+1
:A
p = nf .
p = (n + 1)f .
Applying 2.49 to the short exact sequence (of vector spaces over
0→
A
p)
B
B
Pn
→ n+1 → n → 0
n+1
P
P
P
(2.56)
yields
B Pn B Pn
A
dim
= dim
+ dim
=
:
+ nf.
Pn+1
Pn+1
Pn
Pn+1 p
n
n
Thus, it suffices to prove that PPn+1 : A
p = f . By 2.38, there exists y ∈ P such that
Pn = Pn+1 + (y) C B.
(2.57)
(2.58)
which implies that ordP (y) = n, as y ∈ Pn \ Pn+1 . The map
Φ:
B
Pn
→ n+1
P
P
[x] 7→ [xy]
is an isomorphism:
– (Well defined:) If a, b ∈ B and [a] = [b] ∈
B
P,
then
a − b ∈ P =⇒ ay − by = (a − b)y ∈ Pn+1 =⇒ [ay] = [by] ∈
Pn
.
Pn+1
Hence Φ is well defined.
– (Injective:) If x ∈ B and xy ∈ Pn+1 , then
ordP (x) = ordP (xy) − ordP (y) ≥ (n + 1) − n = 1 ⇒ x ∈ P.
Hence Φ is injective.
– (Surjective:) Let t ∈ Pn . From equation (2.58), there exists x ∈ B such that
t − xy ∈ Pn+1 .
Now [x] 7→ [xy] = [t] ∈
28 This
the field
Pn
Pn+1 ,
so Φ is surjective.
notation does not mean to imply that
B
Pn
is a field; it merely describes the dimension of
A
.
p
45
B
Pn
as a vector space over
Thus we can regard
Pn
Pn+1
as a vector space of dimension 1 over
We conclude that
By induction,
B
Pr
:
A
p
B
Pn+1
:
A
p
B
P.
Now
n
A
B
Pn
P
B A
:
:
=
·
= f.
:
Pn+1 p
Pn+1 P
P p
(2.59)
= (n + 1)f .
= rf .
Lemma 2.52. Let A be a Dedekind domain with field of fractions K, and let p be a nonzero prime ideal in
A. Let L/K be a Galois extension with G := Gal(L/K), and let B be the integral closure of A in L. Then
any element of G restricts to a bijection B → B, and G acts transitively on the set of prime ideals in B
which divide p.
Proof. We first show that the action is well defined. Let σ ∈ G.
(σB = B:)
• (⊆:) Let b ∈ B. Then f (b) = 0 for some monic polynomial f ∈ A[x], so f (σ(b)) = σ(f (b)) = 0, so
σ(b) ∈ B.
• (⊇:) Let b ∈ B. Then b = σ(σ −1 (b)) ∈ σB.
Hence σB = B. As σ|B : B → B is a bijective homomorphism, σ|B : B → B is an isomorphism, so it takes
prime ideals to prime ideals: if P is a prime ideal in B, then σP is a prime ideal in B.
(If P divides p, then σP divides p:) Assume that P divides p. Then
σP ∩ K = σP ∩ σK = σ(P ∩ K),
= P ∩ K,
as σ is injective
as σ ∈ Gal(L/K)
= p,
by 2.48,
(2.60)
so σP divides p (by 2.48). Hence the action of G on the set of prime ideals in B which divide p is well
defined.
(The action is transitive:) Proof by contradiction: assume that β, τ are prime ideals in B dividing p, and
assume that β and τ are not conjugate. For i = 1, 2, . . . , m, there exists ci ∈ τ \ σi β.29 By the Chinese
remainder theorem (by 2.41, distinct nonzero prime ideals are relatively prime), there exists x ∈ B such that
x≡0
mod τ
x ≡ c1
mod σ1 β
..
.
x ≡ cm
mod σm β.
29 By
2.43, B is a Dedekind domain, so every nonzero prime ideal in B is maximal. In particular, τ is a maximal ideal in B,
so it cannot be strictly contained in σi β.
46
Now x ∈ τ , but x ∈
/ σi β for i = 1, 2, . . . , n. Define
Y
b :=
σx = NmL/K (x) ∈ K,
(2.61)
σ∈G
by 2.14 and 2.12. As x ∈ τ ,
Y
b=x
σx ∈ τ.
(2.62)
σ∈G\{id}
Thus, b ∈ τ ∩ K = p (by 2.48). Let σ ∈ G. Then x ∈
/ σ −1 β, so σx ∈
/ β. However,
Y
σx = b ∈ p ⊆ pB ⊆ β,
σ∈G
contradicting the fact that β is a prime ideal in B. Hence the action is transitive.
Proof of 2.50. (a)
h
Pg
B
•
i=1 ei fi = pB :
A
p
i :
By the Chinese remainder theorem,
Y B
B
B
= Q ei ∼
.
=
pB
Pei i
Pi
Using 2.51,
•
h
B
pB
:
A
p
i
= m:
X
X
g g
B A
B
A
:
=
=
ei fi .
ei :
pB p
Pi
p
i=1
i=1
h
i
B
– We first establish that if A is a principal ideal domain, then pB
:A
p = m:
Assume that A is a principal ideal domain. By 2.22, B is a free A-module of rank m. Then
there is an isomorphism Am ∼
= B of A-modules, so (using 1.7)
By 1.8, this becomes
Thus, if A is a PID, then
h
B
pB
:
A
p
i
A
A
⊗ A Am ∼
= ⊗A B.
p
p
(2.63)
B ∼ Am ∼ A m
.
=
=
pB
pAm
p
(2.64)
= m.
– Now relax the assumption that A is a PID. Let S := A \ p. Then S is a multiplicatively closed
subset of A and of B, and 0 ∈
/ S. Let A0 := S −1 A = Ap and B 0 := S −1 B. Then K is the field
of fractions of A0 and L is the field of fractions of B 0 .
(B 0 is the integral closure of A0 in L:)
∗ (⊆:) If α ∈ B 0 , then there exists s ∈ S such that sα ∈ B. As B is integral over A, there
exist n ∈ Z>0 and a1 , . . . , an ∈ A such that
(sα)n + a1 (sα)n−1 + . . . + an = 0
a1
an
∴ αn + αn−1 + . . . + n = 0,
s
s
0
so α is integral over A .
47
(2.65)
(2.66)
∗ (⊇:) Assume that α ∈ L is integral over A0 . Then there exist n ∈ Z>0 , a1 , . . . , an ∈ A and
s1 , . . . , sn ∈ S such that
a1
an
αn + αn−1 + . . . +
=0
(2.67)
s1
sn
n
Y
sn
s
s :=
∴ (sα)n + a1 (sα)n−1 + . . . + an = 0,
si .
(2.68)
s1
sn
i=1
The coefficients of sα lie in A, so sα ∈ B, so α ∈ B 0 .
Hence B 0 is the integral closure of A0 in L.
Moreover, A0 is a discrete valuation ring (by 2.31). Now
S −1 (pB) = S −1
g
Y
Pei i
(2.69)
i=1
pB 0 =
g
Y
(Pi B 0 )ei .
(2.70)
i=1
Hence
g
X
"
#
B0
A0
ei fi =
:
,
pB 0 pA0
i=1
(2.71)
i
h
h 0
i
A0
B
= P
: A
as PBi B 0 : pA
= fi .30 Moreover, A0 is a principal ideal domain with field of
0
p
fractions K, and L is the field of fractions of B 0 , so
"
#
B0
A0
:
= m.
(2.72)
pB 0 pA0
From equations (2.71) and (2.72),
g
X
"
#
B0
A0
ei fi =
:
= m,
pB 0 pA0
i=1
h
i
B
and we had already shown that pB
:A
p = m. Hence,
g
X
(2.73)
B A
ei fi =
:
= m.
pB p
i=1
(b) Assume that L/K is a Galois extension, and let G = Gal(L/K) = {σ1 , . . . , σm }. Let σ ∈ G, and let P
be a prime ideal in B which divides p. By 2.52, σ restricts to an isomorphism B → B.
• (e(σP/p) = e(P/p) :)
pB =
g
Y
Pei i .
(2.74)
(σPi )ei .
(2.75)
i=1
Applying σ to both sides yields31
pB =
g
Y
i=1
0
B
A0
n = fi . If {[x1 ], . . . , [xn ]} is a basis for P
over A
, then { x11 , . . . , x1n } is a basis for PBB 0 over pA
0 ).
p
i
i
is an ideal in B which contains σ(p) = p (note that σ fixes K ⊇ p), so σ(pB) ⊇ pB. Similarly, pB ⊆ σ −1 (pB) ⇒
σ(pB) ⊆ pB. Hence σ(pB) = pB.
30 Let
31 σ(pB)
48
Hence e(σP/p) = e(P/p).
• (f (σP/p) = f (P/p) :) A basis
for
for
B
P
{β1 , . . . , βf }
(2.76)
{σβ1 , . . . , σβf }
(2.77)
corresponds to a basis
B
σβ .
Hence f (σP/p) = f (P/p).
Let i, j ∈ {1, . . . , g}. By 2.52, there exists σ ∈ G such that Pj = σPi . Now
ej = e(Pj /p) = e(σPi /p) = e(Pi /p) = ei
(2.78)
fj = f (Pj /p) = f (σPi /p) = f (Pi /p) = fi ,
(2.79)
and
so all of the ramification indices and residue class degrees are equal. Hence,
ef g = m.
The following is an important criterion to decide whether or not a prime ramifies:
Theorem 2.53. Let K be a number field, and let L be a finite separable extension of K. Let A be a Dedekind
domain with field of fractions K,32 and let B be the integral closure of A in L. Assume that B is a free
A-module. Let p be a nonzero prime ideal in A. Then p ramifies in L if and only if p|disc(B/A).33 In
particular, only finitely prime ideals ramify.
We need some preliminary results to prove 2.53.
Lemma 2.54. Let A be a commutative unital ring and let B be a commutative ring containing A. Assume
that B admits a finite basis {e1 , . . . , em } as an A-module, and let a ∈ A. Then {[e1 ], . . . , [em ]} is a basis for
B
the A
a -module aB , and
A
D([e1 ], . . . , [em ]) = [D(e1 , . . . , em )] ∈ .
(2.80)
a
Proof. The isomorphism
Am → B
m
X
(a1 , . . . , am ) 7→
ai ei
i=1
gives, when tensored (over A) with
A
a,
an isomorphism
A m
a
32 e.g.
→ Am ⊗
A
A
B
→B⊗ →
a
a
aB
A = OK
means that p contains any representative of disc(B/A), or that p divides the ideal generated by disc(B/A).
33 This
49
([a1 ], . . . , [am ]) 7→ (a1 , . . . , am ) ⊗ [1] 7→
m
X
ai ei ⊗ [1] 7→
i=1
Hence {[e1 ], . . . , [em ]} is a basis for the
m
X
ai [ei ].
(2.81)
i=1
B
A
a -module aB .
D([e1 ], . . . , [em ]) = det Tr
B
A
aB / a
[ei ej ]
i,j∈{1,2,...,m}
= [D(e1 , . . . , em )] ∈
h
i
= det TrB/A (ei ej )
A
.
a
(2.82)
Lemma 2.55. Let A be a commutative unital ring, and let B1 , . . . , Bg be commutative rings containing A
that are finitely generated free A-modules. Then
disc
Y
g
Bi
Y
g
A =
disc(Bi /A).
i=1
i=1
Proof. For i ∈ {1, 2, . . . , g}, let {ei1 , . . . , eini } be a basis for Bi . Then their union, {e11 , . . . , egng }, is a basis
Qg
for the A-module i=1 Bi .
D(e11 , . . . , egng ) = det M,
where M is a block matrix, with a block

Tr(e2i1 )

 Tr(ei1 ei2 )

..


.
Tr(eini ei1 )
Tr(ei1 ei2 ) . . .
Tr(e222 )
...
..
..
.
.
Tr(eini ei2 ) . . .
Tr(ei1 eini )
Tr(ei2 eini )
..
.
Tr(e2ini ).






(2.83)
for each i ∈ {1, 2, . . . , g}.34 Now
D(e11 , . . . , egng ) =
g
Y
D(ei1 , . . . , eini ),
(2.84)
i=1
so
disc
Y
g
Bi
Y
g
A =
disc(Bi /A).
i=1
i=1
Let A be a commutative unital ring. An element α ∈ A is nilpotent if there exists m ∈ Z>0 such that
αm = 0. A commutative unital ring is reduced if it has no nonzero nilpotent elements.
Lemma 2.56. Let k be a perfect field, and let B be a commutative k-algebra which has finite dimension as
a vector space over k. Then B is reduced if and only if disc(B/k) 6= 0.
34 This
follows from the fact that if i 6= j then eia ejb = 0 ∈
Qg
i=1
50
Bi .
Proof. (⇐:) We prove the contrapositive statement. Assume that B is not reduced. Then B has a nilpotent
element β 6= 0. We can extend {β} to a basis {β = e1 , . . . , em } for B (as a vector space over k). Thus, if
j ∈ {1, 2, . . . , m} then βej is nilpotent, so the map
B→B
x 7→ βej x
is nilpotent. In particular, its matrix is nilpotent, and is therefore traceless.35 Hence every entry in the first
row of the matrix (Tr(ei ej )) is zero, so
D(e1 , . . . , em ) = det(Tr(ei ej )) = 0.
(⇒:) Assume that B is reduced. B is a commutative unital ring, so the intersection of the prime ideals in B
(the nilradical) is the set of nilpotent elements in B, which is {0}.36
(Every prime ideal in B is maximal:) Let p be a prime ideal in B. Then Bp is an integral domain containing
k.37 To show: Bp is a field. By 2.42, it remains to show that Bp is algebraic over k. If b ∈ B, then b is the
root of some f ∈ k[X].38 Then [b] ∈ Bp is algebraic. Hence Bp is algebraic over k. By 2.42, Bp is a field. Thus
p is a maximal ideal in B. Hence, every prime ideal in B is maximal.
Let p1 , . . . , pr be disinct prime ideals in B. Since they are maximal, they are pairwise relatively prime.
Hence, by the Chinese remainder theorem,
B
∩ri=1 pi
Now
=
r
Y
B
.
p
i=1 i
X
r B
B
[B : k] ≥
:k =
: k ≥ r.
:k =
∩ri=1 pi
p
pi
i=1 i
i=1
B
Y
r
(2.85)
(2.86)
Therefore B only has finitely many prime ideals, say p1 , . . . , pg , where ∩gi=1 pi = {0}. Now
B=
B
∩gi=1 pi
=
g
Y
B
.
p
i=1 i
(2.87)
Each pBi is a finite (since B is a finite-dimensional vector space over k) separable (as k is a perfect field)
extension of k. By 2.19,
B
disc
k 6= 0.
(2.88)
pi
Using 2.55,
Y
Y
g
g
B
B
k =
disc
k 6= 0.
disc(B/k) = disc
p
p
i
i=1
i=1 i
(2.89)
35 Let M be the matrix of the map, and let M a = 0. The minimal polynomial of M divides X a , and therefore has the form
X b . The characteristic polynomial of M is divisible by X b , and therefore has the form X c . The trace of M is the sum of the
eigenvalues of M , which is the sum of the roots of the characteristic polynomial of M , which is 0 (by the Vieta formulae).
36 See Atiyah-Macdonald [1], chapter 1.
37 To show that B ⊇ k, we need to show that p ∩ k = {0}. If x ∈ p ∩ k then x ∈ p is not a unit in B, so x is not a unit in k,
p
so x = 0. Hence B
⊇ k.
p
38 Let B have dimension m as a k-vector space. Then 1, b, b2 , . . . , bm+1 are dependent vectors over k.
51
Proof of 2.53. We will regard disc(B/A) as an element in A representing disc(B/A) ∈
A
(A× )2 .
By 2.54,
A
A
B
∈ .
[disc(B/A)] = disc
pB p
p
B
Note that pB
is an algebra over
dimension is [L : K]).
A
p,
and that
B
pB
is finite-dimensional as a vector space over
(2.90)
A
p
(by 2.50, its
(A
p is a perfect field:) The (equivalent) definition of a perfect field given in Lang [4], p252, is: a field k of
characteristic p is perfect if k p = k, i.e. if every element of k is a pth power. In our case, K and A
p both
contain 1, so they have common characteristic t.39 Let x ∈ A ⊆ K. Then there exists y ∈ K such that
A
x = y t . Now y t − x = 0, x ∈ A, y ∈ K, and K is integrally closed, so y ∈ A. Now [x] = [y]t ∈ A
p , so p is a
perfect field.
Now by 2.56, [disc(B/A)] = [0] ∈ A
p if and only if
B
and only if pB is not reduced. Let
B
pB
pB =
is not reduced. In other words, p divides disc(B/A) if
g
Y
Pei i .
i=1
Since A is a Dedekind domain, the Pi are maximal, and therefore relatively prime in pairs. By the Chinese
remainder theorem,
g
B ∼Y B
.
(2.91)
=
pB
Pei i
i=1
• If p ramifies, then there exists i ∈ {1, 2, . . . , g} such that ei > 1. Let x ∈ Pi \ P2i . Then [x] is a nonzero
Qg
nilpotent in PBei , so we can lift [x] to a nonzero nilpotent ([0], . . . , [0], [x], [0], . . . , [0]) in i=1 βBei . Thus
B
pB
i
is not reduced, so p|disc(B/A).
Qg
B
• If p does not ramify, then pB
= i=1
reduced,40 so p 6 | disc(B/A).
i
B
e
βi i
is a direct sum of fields, so it is a field. In particular, it is
Hence p ramifies in L if and only if p | disc(B/A). This implies that only finitely many prime ideals ramify,
since we can factorize (disc(B/A)) as a product of nonzero prime ideals in the Dedekind domain A.
2.7
Norms of ideals, and finiteness of the class number
In this section, we introduce norms of ideals, which appear in the class number formula. We use norms of
ideals to establish that the class number of any number field is finite, as well as some other important results.
Lemma 2.57. Let A be a Dedekind domain with field of fractions K. Let L/K be a finite separable extension,
and let B be the integral closure of A in L. Let P be a nonzero prime ideal in B, and let p := P ∩ A. Then
(a) p is a prime ideal in A.
39 Every
field of characteristic 0 is perfect, so assume that t > 0.
k be a field. If 0 6= x ∈ k and t ∈ Z≥2 and xt = 0, then x = xt (x−1 )t−1 = 0, contradiction. Hence k has no nonzero
nilpotents. We conclude that any field is reduced.
40 Let
52
(b) p 6= (0).
(c) P divides p.
Proof.
(a)
• If x, y ∈ p then −x, x + y ∈ p.
• If x ∈ p, a ∈ A, then ax ∈ P (as P C B) and ax ∈ A, so ax ∈ p. Hence p C A.
• 1∈
/ P ⊆ p ∴ p 6= A. Hence p is a proper ideal in A.
• If x, y ∈ A and xy ∈ p, then x ∈ P or y ∈ P (as P is a prime ideal in B), so x ∈ p or y ∈ p.
Hence p is a prime ideal in A.
(b) Let 0 6= x ∈ P. Let n := [L : K], and let G := {σ1 , . . . , σn } be the set of distinct K-homomorphisms
L ,→ Ω, where Ω ⊇ L is a Galois extension of K. By 2.14,
NmL/K (x) = σ1 (x) · · · σn (x).
(2.92)
(Each Galois conjugate of x lies in B:) Let σ ∈ G, and let f ∈ A[X] be the minimal polynomial of x
over K (by 2.5). Then f (σ(x)) = σ(f (x)) = 0, so σ(x) ∈ B (by 2.5). Thus, each Galois conjugate of x
lies in B.
Define id : L ,→ Ω, x 7→ x. From equation (2.92),
NmL/K (x) = σ1 (x) · · · σn (x) = x
Y
σ(x) ∈ Bx.
(2.93)
σ∈G\{id}
Then Nm(x) ∈ Bx ⊆ P and Nm(x) ∈ A (by 2.13), so Nm(x) ∈ P ∩ A = p. Moreover, Nm(x) 6= 0, from
equation (2.92). Hence p 6= (0).
(c) By 2.36, P ⊇ pB implies that P divides p.
Recall that Id(B) is the free abelian group on the set of nonzero prime ideals in B. Define a homomorphism
N = NL/K : Id(B) → Id(A)
P 7→ pf ,
B
where p = P ∩ A and f = f (P/p) = P
:
for nonzero prime ideals P C B,
A
p .
This generalised definition coincides with a much simpler definition in the case of rings of integers in number
fields. Let K be a number field, and let (0) 6= I C OK . The numerical norm of I is N (I) := (OK : I),41
which we shall prove to be finite.
41 This
is the index of the subgroup I of the additive group OK .
53
Proposition 2.58. Let K be a number field.
(a) Let (0) 6= I C OK . Then NK/Q (I) = N (I) · Z.
(b) Let K ∈ R≥0 . Then there are finitely many nonzero ideals I C OK such that N (I) ≤ K.
(c) Let a, b ∈ Id(OK ), and assume that a ⊇ b. Then (a : b) = N (a−1 b).
Proof. (a) Let
g
Y
I=
pri i
(2.94)
i=1
be the factorization of I into prime ideals in the Dedekind domain OK , and define pi ∈ Z>0 by
Zpi = Z ∩ pi .
(2.95)
By the Chinese remainder theorem,
g
Y OK
OK
=
,
I
pri
i=1 i
so
N (I) =
g
Y
(2.96)
(OK : pri i ).
(2.97)
i=1
By 2.51,
OK
r
pi i
Z
Zpi
is a vector space of dimension ri fi over
(where fi = f (pi /Zpi )), therefore
(OK : pri i ) = pri i fi .
(2.98)
Thus,
N (I) =
g
Y
(2.99)
(pi Z)ri fi = N (I).
(2.100)
=
i=1
N (I) · Z =
g
Y
pri i fi
(OK :
pri i )
i=1
g
Y
i=1
(b) Assume that I C OK is a nonzero ideal such that N (I) ≤ K. Let
I=
g
Y
pri i
(2.101)
i=1
be the factorization of I into prime ideals in the Dedekind domain OK , and define pi ∈ Z>0 by
Zpi = Z ∩ pi .
Then
K ≥ N (I) =
g
Y
pri i fi .
(2.102)
(2.103)
i=1
Thus, there are finitely many possibilities for g and the ri . Moreover, there are finitely many possible
pi , and therefore finitely many possible pi (prime ideals in OK which divide pi OK ). Hence, there are
Qg
finitely many possibilities for I = i=1 pri i .
Therefore, there are finitely many nonzero ideals I C OK such that N (I) ≤ K.
54
(c) There exists d ∈ OK \ {0} such that da, db C A. The map
K→K
x 7→ dx
is an isomorphism of additive groups, therefore (da : db) = (a : b). Define I := da and J := db, so
I, J C A and I ⊇ J 6= (0). Now
(a : b) = (I : J) =
(OK : J)
N (J)
=
= N (I −1 J) = N ((da)−1 (db)) = N (a−1 b).
(OK : I)
N (I)
(2.104)
The map N is transitive:
Lemma 2.59. Let A be a Dedekind domain with field of fractions K. Let E ≥ L ≥ K be a tower of fields
such that each step of the tower is a finite separable extension. Let B be the integral closure of A in L, and
let C be the integral closure of B in E. Then
NL/K ◦ NE/L = NE/K : Id(E) → Id(K).
(2.105)
Proof. Note that B and C are Dedekind domains, by 2.43. Both NL/K ◦NE/L and NE/K are homomorphisms
Id(E) → Id(K), and Id(E) is the free abelian group generated by the set of nonzero prime ideals in C, so it
suffices to prove that if P is a nonzero prime ideal in C then NL/K (NE/L (P )) = NE/K (P ).
Let P is a nonzero prime ideal in C. Let P = P ∩ B, and let p = P ∩ A. Then p = P ∩ A, and
C B
B A
C A
:
:
:
=
·
= f (P/P)f (P/p),
f (P/p) =
P p
P P
P p
(2.106)
so
NL/K (NE/L (P )) = NL/K (Pf (P/P) ) = NL/K (P)f (P/P) = pf (P/p)f (P/P) = pf (P/p) = NE/K (P ).
(2.107)
Thus
NL/K ◦ NE/L = NE/K : Id(E) → Id(K),
i.e. N is transitive.
Proposition 2.60. Let A be a Dedekind domain with field of fractions K. Let L/K be a finite separable
extension, and let B be the integral closure of A in L.
(a) Let (0) 6= a C A and m := [L : K]. Then
N (aB) = am .
(2.108)
(b) Assume that L/K is Galois. Let P be a nonzero prime ideal in B, and let p = P ∩ A. From 2.50, we
know that the ramification indices of p in B are equal, so let
pB = (P1 · · · Pg )e
55
(2.109)
be the prime factorization of pB in B. Moreover, 2.50 also tells us that the residue class degrees are
equal, so let f be the common residue class degree. Then
Y
N (P) · B = (P1 · · · Pg )ef =
σP.
(2.110)
σ∈Gal(L/K)
(c) Let 0 6= β ∈ B. Then
Nm(β) · A = N (β · B).
(2.111)
(d) The following diagram commutes:
b7→(b)
L× −−−−→ Id(B)




Nmy
yN
(2.112)
K × −−−−→ Id(A).
a7→(a)
Proof.
(a) (If p is a nonzero prime ideal in A, then N (pB) = pm :) Let p be a nonzero prime ideal in A, and prime
Qg
factorize pB as pB = i=1 Pei i . Then
N (pB) = N
g
Y
Pg
Pei i = p i=1 ei fi = pm ,
(2.113)
i=1
since p = Pi ∩ A for i = 1, . . . , g,42 and using 2.50.
Now prime factorize a in the Dedekind domain A as a =
N (aB) = N
h
Y
i=1
Qh
i=1
pri i .
h
h
h
Y
Y
m
Y
i
(pi B)ri =
N (pi B)ri =
pmr
=
pri i
= am .
i
i=1
i=1
(2.114)
i=1
(b) As N (P) = pf ,
N (P) · B = (pB)f = (P1 · · · Pg )ef .
(2.115)
Q
Let G := Gal(L/K). In order to show that (P1 · · · Pg )ef = σ∈G σP, it suffices to show that each Pi
occurs ef times in the family {σP : σ ∈ G}. As |G| = m and ef = m
g , it suffices to show that each Pi
occurs the same number of times in the family {σP : σ ∈ G}.
Let i, j ∈ {1, . . . , g}, i 6= j. By 2.52, G acts transitively on the set {P1 , . . . , Pg }. Hence there exists
τ ∈ G such that τ Pi = Pj . If σ ∈ G, then σP = Pi if and only if τ σP = Pj . Hence the number of
times Pi occurs in the family {σP : σ ∈ G} is the same as the number of times Pj occurs in the family
{τ σP : σ ∈ G}. Since G is a group, τ G = G, so {τ σP : σ ∈ G} = {σP : σ ∈ G} as families. Hence Pi
and Pj occur the same number of times in the family {σP : σ ∈ G}. We conclude that
Y
N (P) · B = (P1 · · · Pg )ef =
σP.
σ∈Gal(L/K)
42 p
⊆ pB ⊆ Pi and p ⊆ A, so p ⊆ Pi ∩ A. Moreover, Pi ∩ A ⊆ Pi ∩ K = p (by 2.48).
56
(c) Let E be the Galois closure of L/K,43 let d := [E : L], and let G := Gal(E/K). Let C be the integral
closure of B in E, and let c := β · C. Define
φ : Id(A) → Id(C)
a 7→ a · C.
• (φ is injective:) Assume that I, J ∈ Id(A) and I · C = J · C. Factorize I and J as
I=
g
Y
pri i ,
J=
g
Y
psi i ,
ri , si ∈ Z,
(2.116)
i=1
i=1
where p1 , . . . , pg are pairwise distinct nonzero prime ideals in A. Then
g
Y
(pi C)ri = I · C = J · C =
i=1
g
Y
(pi C)si .
(2.117)
i=1
Factorize each side of equation (2.117) into the product of nonzero prime ideals in C. Any nonzero
prime ideal in C divides at most one nonzero prime ideal in A,44 therefore ri = si for i = 1, . . . , g.
Hence I = J, so φ is injective.
• (NE/K (c) = NmE/K (β) · A:) As φ is injective, it suffices to show that NE/K (c) · C = NmE/K (β) · C.
Qg
Let c = i=1 pri i be the factorization of c in C. Using part (b),
g
g
g Y
g
Y
Y
Y Y
ri Y
NE/K (c) · C =
N (pi )ri · C =
N (pi ) · C
=
σ(pi )ri =
σ
pri i
i=1
=
Y
σ∈G
i=1
Y
σ(c) =
i=1 σ∈G
σ(β · C) =
σ∈G
Y
σ∈G
i=1
σβ
· C = NmE/K (β) · C.
(2.118)
σ∈G
We conclude that
NE/K (c) = NmE/K (β) · A.
(2.119)
Now
NL/K (β · B)d = NL/K (NE/L (β · C)),
= NE/K (β · C),
using part (a)
using the transitivity of N
= NmE/K (β) · A,
from equation (2.119)
= NmL/K (NmE/L (β)) · A,
using the transitivity of Nm
d
= NmL/K (β) · A ∈ Id(A),
using 2.58 and part (a).
(2.120)
Thus,
Id(A) is torsion-free, so
−1 d
NL/K (β · B) NmL/K (β) · A
= A ∈ Id(A).
(2.121)
−1
NL/K (β · B) NmL/K (β) · A
= A ∈ Id(A), i.e.
Nm(β) · A = N (β · B).
43 In
particular, E ⊇ L and E/K is Galois.
p and q be nonzero prime ideals in A. Then pB + qB = (p + q)B = A · B = B, so pB and qB are relatively prime ideals
44 Let
in B.
57
(d) Let b1 , b2 ∈ B \ {0}, so that
NmL/K
b1
b2
is an arbitrary element of L× . Using part (c),
b 1
b2
A=
Nm(b1 )
Nm(b1 )A
N (b1 B)
A=
=
=N
Nm(b2 )
Nm(b2 )A
N (b2 B)
b1
B .
b2
Hence, the following diagram commutes:
b7→(b)
L× −−−−→ Id(B)




Nmy
yN
K × −−−−→ Id(A).
a7→(a)
Corollary 2.61. Let K be a number field, and let α ∈ K × . Then
N (α · OK ) = |NmK/Q (α)|.
(2.122)
N (α · OK ) · Z = NK/Q (α · OK ) = NmK/Q (α) · Z
(2.123)
Proof. Using 2.58 and 2.60,
∴ N (α · OK ) = |NmK/Q (α)|.
The fact that the class number of any number field is finite is a straightforward corollary of an important
theorem regarding the norms of ideals. The proof of this theorem is omitted here (see Milne [8]), but we will
state the theorem and provide some applications.
Theorem 2.62. Let K/Q be an extension of degree n ∈ Z>0 . Let D be the discriminant of K,45 and let 2s
be the number of non-real complex embeddings of K. Then there exists a set of representatives for the ideal
class group of K consisting of nonzero integral ideals I C OK satisfying
N (I) ≤
n! 4 s p
|D|.
nn π
(2.124)
In the terminology of the above theorem, the Minkowski bound is
BK :=
n! 4 s p
|D|,
nn π
(2.125)
n! 4 s
.
nn π
(2.126)
and the Minkowski constant is
CK :=
Theorem 2.63. Let K be a number field. Then the class number of K is finite.
45 Recall
that this is defined to be the discriminant of OK /Z, which is a well defined integer.
58
Proof. By 2.62, it suffices to prove that there exist only finitely many nonzero integral ideals I COK such that
Qg
N (I) ≤ BK . Let (0) 6= I C OK , and assume that N (I) ≤ BK . Let I = i=1 pri i be the prime factorization
of I. Then
g
Y
(2.127)
BK ≥ N (I) =
pri i fi ,
i=1
where pi ∈ Z>0 is defined by Zpi = pi ∩ Z and fi = f (pi /Zpi ) for i = 1, . . . , g. Thus, there are finitely many
possibilities for pi . Each of these yields finitely many possibilities for pi (prime ideals in OK which divide
pi OK ), and finitely many possibilities for ri (as pi > 1). Hence there are finitely many possible I, so the
class number of K is finite.
The above proof gives us an approach to compute the ideal class group of a number field:
√
Example K = Q[ −5]. In the context of the Minkowski bound, n = 2, s = 1, D = disc(X 2 + 5) = −20,
so there exists a set S of representatives
for the ideal class group of K consisting of nonzero integral ideals
√
I C OK satisfying N (I) ≤ BK = 2 π20 < 3.
Let I ∈ S, and let
I=
g
Y
Pri i ,
ri ∈ Z>0
(2.128)
i=1
be the factorization of I into a product of prime ideals in OK . For i = 1, 2, . . . , g, define pi ∈ Z>0 by
pi Z = Pi ∩ Z, and let fi := f (Pi /pi Z). Then
2 ≥ N (I) =
g
Y
pri i fi .
(2.129)
i=1
I = OK is one possibility. Assume that I 6= OK . Then g = 1, p1 = 2, r1 = f1 = 1, so I = P1 divides 2OK .
As
√
(2) = (2, 1 + −5)2
√
is the prime factorization of (2) in OK , this implies that I = (2, 1 + −5). Hence the only elements of
√
S are (1) and (2, 1 + −5). It remains to show that these are distinct in the ideal class group, i.e. that
√
√
(2, 1 + −5) is not a principal ideal in OK = Z[ −5] (see 2.26).
√
√
Assume, for the sake of contradiction, that there exists α ∈ Z[ −5] such that (2, 1 + −5) = (α). Let
√
α = m + n −5, m, n ∈ Z.
√
√
Z[ −5]
√
(2.130)
= {[0], [1]} =⇒ N ((2, 1 + −5)) = 2.
(2, 1 + −5)
Now, using 2.58 and 2.60,
2Z = N ((2, 1 +
√
−5)) = N (α) = Nm(α)Z,
(2.131)
√
which is a contradiction as Nm(α) = m2 + 5n2 6= ±2. Hence (2, 1 + −5) is not a principal ideal in
√
OK = Z[ −5].
√
We conclude that the ideal class group of K is {[(1)], [(2, 1 + 5)]}, and that the class number is 2.
59
Let K be a number field. A finite extension L of K is unramified if no nonzero prime ideal in OK ramifies
in OL .
Theorem 2.64. There does not exist a nontrivial unramified extension of Q.
Proof. Let K be a number field. Let n := [K : Q], and let 2s be the number of non-real complex embeddings
of K. Let D be the discriminant of K. Assume that K is unramified. To show: n = 1.
By 2.53, |D| = 1 (otherwise there exists a prime divisor p of D, in which case pZ ramifies). According to
2.62, there exists a set S of representatives of the ideal class group of K, consisting of nonzero integral ideals
integral ideals I C OK such that
n! 4 p
N (I) ≤ n ( )s |D|.
n π
Now N (I) ≥ 1 yields
p
nn π s
nn π n/2
1 = |D| ≥
( ) ≥
( ) .
(2.132)
n! 4
n! 4
For k ∈ Z>0 define ak :=
kk π k/2
.
k! ( 4 )
Then (an )n=2,3,... is an increasing sequence:
√ π
ak+1
1 k
=
1+
> 1,
ak
2
k
k ≥ 2.
(2.133)
Recall that we need to show that n = 1. Proof by contradiction: assume that n ≥ 2. Then
1 ≥ an ≥ a2 =
π
> 1,
2
(2.134)
contradiction, so n = 1. We conclude that there does not exist a nontrivial unramified extension of Q.
2.8
Cyclotomic extensions
Cyclotomic extensions can be applied to Fermat’s last theorem (see Milne [8], chapter 6), and are ubiquitous
in class field theory (being abelian extensions46 of Q). We follow Milne [8].
Let n ∈ Z>0 . A primitive nth root of unity is an element ζ ∈ C× of order n.
Lemma 2.65. Let G be a group, and let α ∈ G be of order n. Then αm is of order n in G if and only if
gcd(m, n) = 1.
Proof. (⇒:) Let αm be of order n in G. If d ∈ Z>0 is a common factor of m and n, then
n
1 = αnm/d = (αm )n/d =⇒ n =⇒ d = 1.
d
Hence gcd(m, n) = 1.
(⇐:) Let gcd(m, n) = 1.
• (αm )n = (αn )m = 1, therefore the order of αm is at most n.
46 An
abelian extension is a Galois extension whose Galois group is abelian.
60
• If d ∈ {1, 2, . . . , n} and (αm )d = 1 then
αmd = 1 =⇒ n|md =⇒ n|d,
(2.135)
as gcd(m, n) = 1, so d = n.
Hence αm is of order n in G.
Corollary 2.66. Let n ∈ Z>0 , and let ζ be a primitive nth root of unity. Then ζ m is a primitive nth root
of unity if and only if gcd(m, n) = 1.
Let n ∈ Z>0 . The nth cyclotomic polynomial is
Y
Φn (X) =
(X − ζ m ) =
Y
(X − τ ),
(2.136)
m∈(Z/nZ)×
where τ ranges over primitive nth roots of unity.
Note that Φn (ζ) = 0, and that the degree of Φn is φ(n), where φ denotes the Euler totient function. We will
later prove that Φn (X) is the minimal polynomial of ζ over Q. There is also a useful identity
Y
Xn − 1 =
Φd (X),
(2.137)
d|n
as
• Each nth root of unity is a primitive dth root of unity for precisely one positive divisor d of n.
• If d|n, then any primitive dth root of unity is an nth root of unity.
Let K = Q[ζ], where ζ is a primitive nth root of unity. Then K is the splitting field of X n − 1 (a
separable polynomial in Q[X]), so K/Q is Galois. Let G = Gal(K/Q). Then there is a well defined injective
homomorphism
×
Z
G→
nZ
if σζ = ζ m .
σ 7→ [m],
Proof.
• Well defined: Let σ ∈ G. Then σζ is a root of Φn (X), so there exists m ∈ (Z/nZ)× such that σζ = ζ m .
If ζ m = ζ k then ζ m−k = 1, so n|(m − k). Thus m is unique modulo n.
• Injective: As K = Q[ζ], σ ∈ G is determined by σ(ζ).
The following theorem, is the crucial result in this section. Part (a) implies that the above map is an
isomorphism,47 and that the nth cyclotomic polynomial is the minimal polynomial of any nth root of unity,48
for n ∈ Z>0 .
47 An
48 Φ
n
injection between two finite sets of the same size must be bijective.
is monic and of degree equal to deg(ζ, Q), where ζ is a primitive nth root of unity.
61
Theorem 2.67. Let n ∈ Z>0 , and let ζ be a primitive nth root of unity. Let φ denote the Euler totient
function. Then
(a) [Q[ζ] : Q] = φ(n).
Z ×
) .
(b) Gal(Q[ζ]/Q) ∼
= ( nZ
(c) OQ[ζ] = Z[ζ].
(d) Assume that p ∈ Z>0 is a prime such that (p) ramifies in Q[ζ]. Then p|n, and the prime factorization
of (p) C OQ[ζ] is
r
(p) = (P1 · · · Ps )φ(p ) ,
(2.138)
where P1 , . . . , Ps are distinct prime ideals in OQ[ζ] .
We prove 2.67 by induction on the number of distinct prime divisors of n. We begin with the base case:
Proposition 2.68. Let p ∈ Z>0 be prime, let r ∈ Z>0 , let ζ be a primitive pr th root of unity, and let
K = Q[ζ]. Let φ denote the Euler totient function. Then
(a) [K : Q] = φ(pr ) = pr−1 (p − 1).
(b) OK = Z[ζ].
(c) π := 1 − ζ is a prime element of OK , and (p) = (π)e , where e := φ(pr ).
(d) disc(OK /Z) = ±pc , where c = pr−1 (pr − r − 1). Thus p is the only prime to ramify in Q[ζ].
Proof. First we prove (a) and (c). Note that ζ ∈ OK ⇒ Z[ζ] ⊆ OK . If τ is a pr th root of unity, then τ = ζ s
and ζ = τ t for some s, t ∈ Z>0 which are not divisible by p. Then
1−τ
= 1 + ζ + . . . ζ r−1 ∈ Z[ζ].
1−ζ
Similarly,
1−ζ
1−τ
∈ Z[ζ], so
1−τ
1−ζ
is a unit in Z[ζ], and therefore also a unit in OK . Using equation (2.137),
Qr
r
Xp − 1
j=0 Φpj (X)
Φpr (X) = Qr−1
= pr−1 −1
X
j=0 Φpj (X)
=
r−1
tp − 1
,
t = Xp
t−1
= 1 + t + . . . + tp−1 ,
(2.139)
so
p = Φpr (1) =
where e := φ(pr ) and u :=
Y
(1 − τ ),
where τ ranges over primitive pr th roots of unity,
Y 1 − τ
=
(1 − ζ) = u · (1 − ζ)e ,
1−ζ
Q 1−τ
1−ζ
(2.140)
is a unit in Z[ζ] (so u is also a unit in OK ). Thus
(p) = (π)e ,
π := 1 − ζ,
To show: (a) [K : Q] = φ(pr ).
62
e := φ(pr ).
(2.141)
• (≤:) Let G = Gal(K/Q). We constructed an injective homomorphism G →
Z ×
,
pr Z
so
Z × [K : Q] = |G| ≤ r
= φ(pr ).
p Z
• (≥:) We have (p) = (π)e in OK . By 2.50,
[K : Q] ≥ e = φ(pr ).
Hence [K : Q] = φ(pr ), proving (a). As Φpr is monic and has degree equal to [Q[ζ] : Q] = deg(ζ, Q), it
follows that Φpr = irr(ζ, Q).
For (c), we have already shown that (p) = (π)e , where π := 1 − ζ and e := φ(pr ), so it remains to show that
π is a prime element of OK . To show: (π) is a prime ideal in OK .
e
Let (π) = τ1e1 · · · τg g be the prime factorization of (π) in OK . Then
(p) =
g
Y
e ·φ(pr )
τi i
.
(2.142)
i=1
Using 2.50,
φ(pr ) = [K : Q] ≥
g
X
ei · φ(pr ) · f (τi /(p)),
(2.143)
i=1
so g = 1, e1 = 1, and f (τ1 /(p)) = 1. Now (π) = τ1 is a prime ideal in OK , so π is a prime element in OK ,
r
Z
K
proving (c). Now (p) = pφ(p ) in OK , where p = (π). Further, f (p/(p)) = 1, so the map (p)
→ O
(π) is an
isomorphism.
Consider statement (d). We first show that disc(OK /Z) is, up to sign, a power of p, and only compute it
later. By 2.18 (choose an integral basis for K),
disc(OK /Z) · (OK : Z[ζ])2 = disc(Z[ζ]/Z).
(2.144)
Hence it suffices to show that disc(Z[ζ]/Z) is, up to sign, a power of p. By 2.20,
disc(Z[ζ]/Z) = ±NmK/Q (Φ0pr (ζ)).
(2.145)
We first compute Φ0pr (ζ). Differentiating
(X p
r−1
r
− 1) · Φpr (X) = X p − 1
at X = ζ yields
(2.146)
r
Φ0pr (ζ) =
pr ζ p −1
.
ζ pr−1 − 1
(2.147)
Now we compute NmK/Q (Φ0pr (ζ)) up to sign.
• By 2.11, NmK/Q (ζ) is (up to sign) the product of the primitive pr th roots of unity, which is ±1 as they
occur in conjugate pairs (apart from 1, and possibly −1). Hence
NmK/Q (ζ) = ±1.
63
(2.148)
• As p ∈ Q,
r
NmK/Q (p) = p[K:Q] = pφ(p ) .
(2.149)
NmK/Q (−1) = (−1)[K:Q] = ±1.
(2.150)
• As −1 ∈ Q,
s
s
• We aim to show that NmK/Q (1 − ζ p ) = ±pp for 0 ≤ s < r.
– (Base case NmK/Q (1 − ζ) = ±p:) The minimal polynomial of 1 − ζ is Φpr (1 − X). By 2.11,
Nm(1 − ζ) is (up to sign) the product of the roots of Φpr (1 − X), which is (by the Vieta formulae)
Φpr (1) = p. Thus,
NmK/Q (1 − ζ) = ±p.
(2.151)
s
– Let 0 ≤ s < r. Note that ζ p is a primitive pr−s th root of unity. The base case49 then shows that
s
NmQ[ζ ps ]/Q (1 − ζ p ) = ±p.
(2.152)
By 2.10 (the transitivity of Nm),
s
s s
NmK/Q (1 − ζ p ) = NmQ[ζ ps ]/Q NmK/Q[ζ ps ] (1 − ζ p ) = NmQ[ζ ps ]/Q (1 − ζ p )d ,
where
s
d = [K : Q[ζ p ]] =
φ(pr )
pr−1 (p − 1)
[K : Q]
=
= s−1
= ps .
s
p
r−s
[Q[ζ ] : Q]
φ(p )
p (p − 1)
(2.153)
(2.154)
Now
s
s
s d
s
NmK/Q (1 − ζ p ) = NmQ[ζ ps ]/Q ((1 − ζ p )d ) = NmQ[ζ ps ]/Q (1 − ζ p ) = ±pd = ±pp .
(2.155)
In particular,
NmK/Q (1 − ζ p
r−1
) = ±pp
r−1
.
Finally,
r
disc(Z[ζ]/Z) = ±NmK/Q (Φ0pr (ζ)) = ±
Nm(p)r · Nm(ζ)p −1
= ±pc ,
Nm(1 − ζ pr−1 ) · Nm(−1)
where c = r · φ(pr ) − pr−1 = r(p − 1)pr−1 − pr−1 = pr−1 (pr − r − 1).
Equation (2.144), which is
disc(OK /Z) · (OK : Z[ζ])2 = disc(Z[ζ]/Z),
now implies that
• disc(OK /Z) is a power of p, and
• [OK : Z[ζ]] is a power of p.
49 Replace
(2.156)
s
K with Q[ζ p ].
64
Once we prove (b), the proof of (d) will also be completed (by equation (2.144)), so it remains to prove
statement (b), that OK = Z[ζ].
Recall that
Z
OK
→
(p)
(π)
[x] 7→ [x]
is an isomorphism. In particular, it is surjective. Thus,
OK = Z + πOK
∴ OK = Z[ζ] + πOK .
(2.157)
By induction, we’ll show that if m ∈ Z>0 then OK = Z[ζ] + π m OK :
• OK = Z[ζ] + πOK .
• Assume that n ∈ Z>0 and OK = Z[ζ] + π n OK .
π n OK = π n Z[ζ] + π n+1 OK
(2.158)
∴ OK = Z[ζ] + π n Z[ζ] + π n+1 OK = Z[ζ] + π n+1 OK .
(2.159)
Hence, by induction, if m ∈ Z>0 then OK = Z[ζ] + π m OK .
As [OK : Z[ζ]] is a power of p (say pn ), there exists n ∈ Z>0 such that pn OK ⊆ Z[ζ].50 Note that
pn OK = (p)n OK = (π)en OK = π en OK ,
(2.160)
OK = Z[ζ] + π en OK = Z[ζ] + pn OK = Z[ζ].
(2.161)
so
Having shown the base case, we now complete the proof of 2.67.
Proof of 2.67. First we show that if q ∈ Z>0 is a prime such that (q) ramifies in Q[ζ], then q|n.51 Let q ∈ Z>0
be a prime such that (q) ramifies in Q[ζ]. By 2.53, q | disc(O/Z), where O := OQ[ζ] . From 2.18,
disc(O/Z) | disc(Z[ζ]/Z).
(2.162)
Hence q|disc(Z[ζ]/Z). Let f (X) = irr(ζ, Q) ∈ Z[X]. By 2.20,
disc(Z[ζ]/Z) = ±NmQ[ζ]/Q (f 0 (ζ))
∴ q | NmQ[ζ]/Q (f 0 (ζ)) = NmO/Z (f 0 (ζ)).
Moreover,
f (X) | g(X) := X n − 1 ∈ Z[X],
50 The
51 This
K
divides the order of the group, which is pn .
order of any element of the group O
Z[ζ]
part of the proof is inspired by Lang [5], p73.
65
(2.163)
so there exists h(X) ∈ Z[X]52 such that g(X) = h(X) · f (X), so
nζ n−1 = g 0 (ζ) = h(ζ) · f 0 (ζ) + f (ζ) · h0 (ζ) = h(ζ) · f 0 (ζ).
(2.164)
± n = NmO/Z (nζ n−1 ) = NmO/Z (h(ζ)) · NmO/Z (f 0 (ζ)),
(2.165)
Now
since Nm(ζ) = ±1 (by 2.15). As h(ζ) ∈ Z[ζ] ⊆ O ⇒ NmO/Z (h(ζ)) ∈ Z, equation (2.165) yields
NmO/Z (f 0 (ζ)) | n.
(2.166)
From equations (2.163) and (2.166),
q | NmO/Z (f 0 (ζ)) | n.
Thus, if q ∈ Z>0 is a prime such that (q) ramifies in Q[ζ], then q|n.
For the rest of the proof, we continue the induction on the number of distinct prime divisors of n.53 Let
n = pr · m, where p 6 | m. By the inductive hypothesis, we may assume that the theorem holds with m in
r
place of n. Note that ζpr := ζ m is a primitive pr th root of unity, and that ζm := ζ p is a primitive mth root
of unity. Moreover, Q[ζ] = Q[ζpr , ζm ] :
• (⊆:) As gcd(pr , m) = 1, there exist x, y ∈ Z such that xpr + ym = 1. Then
ζ = ζ xp
r
+ym
x
∈ Q[ζpr , ζm ]
= ζpyr · ζm
∴ Q[ζ] ⊆ Q[ζpr , ζm ].
r
• (⊇:) ζpr = ζ m ∈ Q[ζ] and ζm = ζ p ∈ Q[ζ], so Q[ζ] ⊇ Q[ζpr , ζm ].
Further, Q[ζpr , ζm ] = Q[ζpr ] · Q[ζm ], by definition.54 Thus
Q[ζ] = Q[ζpr , ζm ] = Q[ζpr ] · Q[ζm ].
(2.167)
From the base case 2.68,
(p) = pφ(p
r
)
in Z[ζpr ],
(2.168)
where p is some nonzero prime ideal in Z[ζpr ]. Note that (p) does not ramify in Q[ζm ] (as p 6 |m), so let
(p) = p1 · · · ps
in Z[ζm ],
(2.169)
where s ∈ Z>0 , and where p1 , . . . , ps are distinct nonzero prime ideals in Z[ζm ]. Thus, in O,
(pO)φ(p
r
)
=
s
Y
pi O.
(2.170)
i=1
The pi O are relatively prime55 , so pi O must be a power of φ(pr ). As deg(Φpr ) = φ(pr ),
[Q[ζ] : Q[ζm ]] = deg(ζpr , Q[ζm ]) ≤ φ(pr ),
(2.171)
so 2.50 implies that
52 Assume
that h(X) ∈
/ Z[X]. Then there exists t ∈ Z≥2 such that t · h(X) ∈ Z[X] is primitive. But t · h(X) · f (X) = t · g(X)
is not primitive, which contradicts Gauss’ lemma.
53 2.68 is the base case.
54 Let K and L be fields contained in some field M . The compositum K · L of K and L (in M ) is the smallest field F such
that K ∪ L ⊆ F ⊆ M . Reference: Lang [4].
55 p O + p O = (p + p )O = Z[ζ ]O = O.
m
i
j
i
j
66
• [Q[ζ] : Q[ζm ]] = φ(pr ), since p1 O is a power of φ(pr ) in O.
φ(pr )
• For i = 1, 2, . . . , s, there exists a nonzero prime ideal Pi C O such that pi O = Pi
.
Now
r
pO = (P1 · · · Ps )φ(p ) ,
(2.172)
where P1 , . . . , Ps are distinct prime ideals in O, which completes the proof of (d).56
By the inductive hypothesis, [Q[ζm ] : Q] = φ(m). Also, gcd(pr , m) = 1, so
[Q[ζ] : Q] = [Q[ζ] : Q[ζm ]] · [Q[ζm ] : Q] = φ(pr ) · φ(m) = φ(pr m) = φ(n),
(2.173)
proving (a). The injective homomorphism
Gal(Q[ζ]/Q) →
Z ×
nZ
is therefore between finite sets of the same size, and is therefore an isomorphism, proving (b). It remains to
show (c), that O = Z[ζ]. As Z[ζ] ⊆ O (since ζ is an algebraic integer), it remains to show that O ⊆ Z[ζ].
K := Q[ζpr ] and L := Q[ζm ] are vector spaces of respective dimensions s := φ(pr ) and t := φ(m) over Q,
and K · L = Q[ζ]. By 2.68 and the inductive hypothesis, OK = Z[ζpr ] and OL = Z[ζm ]. Thus, it remains to
show that O ⊆ OK · OL (note that O = OK·L ). Let {α1 , . . . , αs } and {β1 , . . . , βt } be integral bases for K
and L respectively. By 2.23, they are also bases for K and L (respectively) over Q, so {αi βj } is a Q-basis
for K · L. Let γ ∈ K · L. Then there exist aij ∈ Z (for i = 1, 2, . . . , s and j = 1, 2, . . . , t) and d ∈ Z>0 such
that
X aij
γ=
αi βj ,
(2.174)
d
i,j
where no prime factor of d divides all of the aij , and it remains to show that d = 1.
(Each of the [K : Q]sep = [K : Q] = s embeddings K ,→ C extends uniquely into an embedding K · L ,→ C
which fixes L:) Let τ : K ,→ C. Then any extension σ : K · L = Q[ζpr , ζm ] ,→ C of τ which fixes each element
of L is defined by
σ : K · L = Q[ζpr , ζm ] ,→ C
ζpr 7→ τ (ζpr )
ζm 7→ ζm .
Thus, each of the s embeddings K ,→ C extends uniquely to an embedding K · L ,→ C which fixes L.
Call these extensions σ1 , . . . , σs . For k = 1, 2, . . . , s,
X aij
σk (αi )βj ,
σk (γ) =
d
i,j
(2.175)
and for i = 1, 2, . . . , s define
xi :=
t
X
aij
j=1
d
56 If
βj ∈ L.
(2.176)
q ∈ Z>0 is a prime factor of m, then we can get a similar factorization of qO by writing n = q a b, with a, b ∈ Z>0 , q 6 |b,
and using the inductive hypothesis for b.
67
Then we obtain a system of s equations
s
X
σk (αi )xi = σk (γ),
k = 1, 2, . . . , s.
(2.177)
i=1
Let i ∈ {1, 2, . . . , s}, and let Mi be the result of replacing the ith column of the s × s matrix (σj (αi )) with


σ1 (γ)
 . 
 . .
 . 
σs (γ)
Let D = det(σj (αi )), and let Di = det(Mi ). By Cramer’s rule,
Dxi = Di .
(2.178)
∆ · xi = D · Di .
(2.179)
By 2.19, D2 = ∆ := disc(OK /Z), so
Note that D, Di ∈ O, since they are determinants of matrices with entries in the ring O, so ∆ · xi ∈ O.
Pt ∆·a
Moreover, ∆ ∈ Z and xi ∈ L, so ∆ · xi ∈ O ∩ L = OL . As ∆ · xi = j=1 d ij βj and {β1 , . . . , βt } is an
∆·a
integral basis for OL , it follows that d ij ∈ Z for i = 1, 2, . . . , s and j = 1, 2, . . . , t. As no prime factor of d
divides all of the aij , it follows that d | ∆.
Thus d | disc(OK /Z), and, similarly d | disc(OL /Z). By 2.53, (d) ramifies in both OK and OL . Now, by
part (d) of this theorem, d | m, and by 2.68, d | pr , so d = 1 (as p 6 |m and d ∈ Z>0 ). We conclude that
O = Z[ζpr ] · Z[ζm ] = Z[ζ], completing the proof of (c).
2.9
Lattices
Let V be an n-dimensional real vector space, n ∈ Z>0 . A lattice Λ in V is a subgroup of the form
Λ = Ze1 + . . . + Zer ,
(2.180)
where e1 , . . . , er are linearly independent elements of (vectors in) V , with r ∈ {0, 1, 2, . . . , n}. Λ is full if
r = n. Note that a basis for V gives an isomorphism V → Rn , which induces a topology on V . A subgroup
Λ of V is discrete if its induced topology is the discrete topology.
Proposition 2.69. Let Λ be a subgroup of an n-dimensional real vector space V , n ∈ Z>0 .57 Then the
following are equivalent:
(a) Λ is a lattice in V .
(b) There exists an open subset U of V such that U ∩ Λ = {0}.
(c) Λ is a discrete subgroup of V .
(d) Each compact subset of V intersects Λ in a finite set.
57 A
vector space is an abelian group with the action of an underlying field.
68
(e) Each bounded subset of V intersects Λ in a finite set.
Proof.
• ((a) ⇒ (b):) Let Λ be a lattice in V . Let r ∈ {1, 2, . . . , n}, and let
Λ = Ze1 + . . . + Zer ,
where e1 , . . . , er are linearly independent elements of (vectors in) V . Then
n
o
1
U := b1 e1 + . . . + bn en : |bi | < for i = 1, 2, . . . , n
2
is an open subset of V such that U ∩ Λ = {0}.
• ((b) ⇒ (c):) Assume that there exists an open subset W of V such that W ∩ Λ = {0}. To show
that Λ is discrete, we need to show that if α ∈ Λ then there exists an open subset U ⊆ V such that
U ∩ Λ = {α}. Let α ∈ Λ, and let U be the translation
U := α + W := {α + w : w ∈ W }
of W . Since a translation is a homeomorphism W → W , U is an open subset of V . Moreover,
β ∈ U ∩ Λ ⇔ β − α ∈ W ∩ Λ = {0} ⇔ β = α,
so U ∩ Λ = {α}. Thus U is an open subset of V and U ∩ Λ = {α}, so Λ is a discrete subgroup of V .
• ((c) ⇒ (d):) Assume that Λ is a discrete subgroup of V . Let C be a compact subset of V . Then C ∩ Λ
is both discrete and compact, so C ∩ Λ is finite.58 Thus, each compact subset of V intersects Λ in a
finite set.
• ((d) ⇒ (e):) Assume that each compact subset of V intersects Λ in a finite set. Let B be a bounded
subset of V . Then the closure B of B in V is compact by the Heine-Borel theorem (as V ∼
= Rn ), so
B̄ ∩ V is finite. Hence B ∩ V ⊆ B ∩ V is finite.
• ((e) ⇒ (a):) This is essentially a construction, though formally we use induction. Assume that each
bounded subset of V intersects Λ in a finite set. Let {e1 , . . . , er } be a maximal R-independent subset
of Λ. By induction on r, we prove that Λ is a lattice in V :
– Suppose r = 0. Then Λ = {0} is a lattice in V .
– Suppose r = 1. Then Λ ⊆ Re1 . As {be1 : |b| < 2} ∩ Λ is finite,
a := min{b ∈ R>0 : be1 ∈ Λ}
exists. It remains to show that Λ = Zae1 . As ae1 ∈ Λ, and Λ is an additive group, it follows that
Zae1 ⊆ Λ, so it remains to show that Λ ⊆ Zae1 . Proof by contradiction: assume that there exists
kae1 ∈ Λ \ Zae1 , where k ∈ R \ Z. Then (k − bkc)ae1 ∈ Λ, and 0 < (k − bkc)a < a, contradicting
the minimality of a. Hence Λ = Zae1 is a lattice in V .
58 We prove the contrapositive. Assume that C ∩ Λ is infinite. Let (a ) is a sequence of distinct elements of C ∩ Λ. The
n
collection of {a1 , . . . , an }, for n ∈ Z>0 , is an open (using the relative topology) cover for C ∩ Λ which has no finite subcover,
so C ∩ Λ is not compact.
69
– Suppose r > 1, and assume the inductive hypothesis: if t ∈ {0, 1, . . . , r − 1} and Γ is a subgroup
of a finite-dimensional real vector space V 0 with at most t R-independent elements such that each
bounded subset of V 0 intersects Γ in a finite set, then Γ is a lattice in V 0 . We need to show that
Λ is a lattice in V .
Let V 0 := Re1 + . . . + Rer−1 , and let Γ := Λ ∩ V 0 . Then Γ is a subgroup of V 0 with at most r − 1
R-independent elements such that each bounded subset of V 0 intersects Γ in a finite set, so, by
the inductive hypothesis, Γ is a lattice in V 0 . Let
Γ = Zf1 + . . . + Zfr−1 ,
(2.181)
where f1 , . . . , fr−1 ∈ V 0 are linearly independent over R. As dim(V 0 ) = r − 1, it follows that
{f1 , . . . , fr−1 } is a basis for V 0 . The idea is now to append some (independent) fr ∈ Λ to
{f1 , . . . , fr−1 } so that Λ = Zf1 + . . . + Zfr .
Define a homomorphism φ : Λ → R by φ(α) = a if
α = a1 f1 + . . . + ar−1 fr−1 + aer ,
a1 , . . . , ar−1 , a ∈ R,
(2.182)
which is well defined since {f1 , . . . , fr−1 , er } is a basis for Re1 + . . . + Rer ⊇ Λ.
We now show that φ(Λ) is a lattice in R. Note that φ(Λ) is a subgroup of R. By the inductive
hypothesis, it therefore suffices to prove that each bounded subset of R intersects φ(Λ) in a finite
set. For M ∈ R>0 , we need to show that {a ∈ φ(Λ) : |a| < M } is a finite set. Let
F = {a1 f1 + . . . + ar−1 fr−1 + aer ∈ Λ :
0 ≤ a1 , a2 , . . . , ar−1 < 1,
|a| < M }.
(2.183)
The set F is the intersection of Λ with a bounded subset of V , so F is a finite set. Thus,
{a ∈ φ(Λ) : |a| < M } = φ(F ) is a finite set. Hence φ(Λ) is a lattice in R, so there exists fr ∈ Λ
such that φ(Λ) = Zφ(fr ).
(f1 , . . . , fr are independent:) Proof by contradiction: assume that f1 , . . . , fr are dependent. Then
φ(fr ) = 0, so Λ ⊆ V 0 . As dim(V 0 ) = n − 1, this contradicts the fact that Λ has r independent
vectors. Hence f1 , . . . , fr are independent.
It remains to show that Λ = Zf1 + . . . + Zfr . As f1 , . . . , fr ∈ Λ, and as Λ is an additive group, it
follows that Λ ⊇ Zf1 + . . . + Zfr . Thus, it remains to show that Λ ⊆ Zf1 + . . . + Zfr . Let α ∈ Λ.
Then φ(α) = aφ(fr ) for some a ∈ Z. Now φ(α − afr ) = 0, so α − afr ∈ ker(φ) = Γ. Thus,
α − afr = a1 f1 + . . . + ar−1 fr−1 ,
a1 , . . . , ar−1 ∈ Z
α = a1 f1 + . . . + ar−1 fr−1 + afr ∈ Zf1 + . . . + Zfr .
(2.184)
(2.185)
Therefore Λ ⊆ Zf1 + . . . + Zfr . Now Λ = Zf1 + . . . + Zfr is a lattice in V , proving the inductive
step.
Hence, by induction, Λ is a lattice in V .
70
Pn
Let V be an n-dimensional real vector space, and let Λ = i=1 Zei be a full lattice in V . A fundamental
domain for Λ is a set
n
n
o
X
D = λ0 +
ai ei : 0 ≤ a1 , . . . , an < 1 ,
(2.186)
i=1
where λ0 ∈ Λ. The fundamental domains partition V ∼
= Rn . For a full lattice
Λ = Ze1 + . . . + Zen ,
the volume of a fundamental domain D for Λ is
µ(D) = | det(e1 , . . . , en )|.
(2.187)
The volume measure µ is Haar measure (so µ is translation-invariant). In particular, the volume measure for
a lattice in Euclidean space is Lebesgue measure, corresponding to our intuitive understanding of volume.
More generally, volumes in V correspond to volumes in Rn , which are defined up to multiplication by a
nonzero constant (depending on the choice of basis). This also implies that volume varies linearly under
directional scaling.
Lemma 2.70. Let V be a finite-dimensional real vector space. Let Λ be a full lattice in V , and let D0 be
a fundamental domain for Λ. Let S be a measurable subset of V such that µ(S) > µ(D0 ). Then there exist
α, β ∈ S such that β − α ∈ Λ \ {0}.
Proof. Summing over all fundamental domains D for Λ,
X
µ(S) =
µ(S ∩ D),
(2.188)
D
since S is the disjoint union of the sets S ∩ D. For each fundamental domain D, there exists a unique λ ∈ Λ
such that λ + D = D0 , in which case λ + (S ∩ D) ⊆ D0 . Any such set λ + (S ∩ D) has measure equal to
µ(S ∩ D), since the measure µ is translation-invariant. The sum of the measures of the sets λ + (S ∩ D) is
µ(S) > µ(D0 ), and the sets λ + (S ∩ D) are subsets of D0 , therefore the sets cannot be pairwise disjoint.
Hence there exist elements α, β ∈ S, and distinct elements λ, λ0 ∈ Λ, such that
α − λ = β − λ0 .
(2.189)
∴ β − α = λ0 − λ ∈ Λ \ {0}.
(2.190)
A subset T of a finite-dimensional real vector space V is symmetric in the origin if the following condition
is satisfied: if α ∈ T then −α ∈ T .
The following theorem, proven by Minkowski in 1889, is the foundation for the ‘geometry of numbers’ —
a branch of number theory. We will later use the geometry of numbers to prove the analytic class number
formula.
71
Theorem 2.71 (Minkowski). Let V be an n-dimensional real vector space, n ∈ Z>0 . Let Λ be a full lattice
in V , and let D be a fundamental domain for Λ. Let T ⊆ V be compact, convex, and symmetric in the
origin.59 If
µ(T ) ≥ 2n µ(D),
(2.191)
then T ∩ Λ contains a point other than the origin.
Proof. Assume that µ(T ) ≥ 2n µ(D). First we establish that (1 + k1 )T contains a point in Λ other than the
origin, for k ∈ Z>0 . Let k ∈ Z>0 , and let U = (1 + k1 )T . Then µ(U ) > µ(T ) ≥ 2n µ(D). Moreover, U is
compact, convex, and symmetric in the origin. We want to show that U ∩ Λ contains a point other than the
origin.
Let S := 21 U := { 12 u : u ∈ U }. Then
1
µ(U ) > µ(D),
2n
so, by 2.70, there exists α, β ∈ S such that β − α ∈ Λ \ {0}. Then 2α, 2β ∈ 2S = U , so −2α ∈ U , since U is
symmetric about the origin. As U is convex,
µ(S) =
β−α=
1
(2β + (−2α)) ∈ U.
2
Now β − α ∈ U ∩ Λ \ {0}.
For each k ∈ Z>0 , we can therefore choose xk ∈ (1 + k1 )T ∩ Λ \ {0}. Note that the conditions on T imply
that if 0 < λ ≤ ρ then λT ⊆ ρT . Thus, (xk ) is a sequence in the set 2T ∩ Λ \ {0}, which is finite since 2T is
compact (by 2.69). Hence (xk ) as a subsequence (xkr ) which converges to a point x ∈ 2T ∩ Λ \ {0}.
(x ∈ T :) Proof by contradiction: assume x ∈
/ T . As T is closed,
1
T = ∩k∈Z>0 1 +
T,
k
(2.192)
so there exists l ∈ Z>0 such that x ∈
/ (1 + k1l )T . Then V \ (1 + k1l )T is a neighbourhood of x. Since
xkr converges to x as r → ∞, the neighbourhood V \ (1 + k1l )T contains infinitely many points xkr . This
contradicts that fact that
xkr ∈ (1 + xkr )T ⊆ (1 + xkl )T
for all r ∈ Z≥l . Hence x ∈ T .
Thus, x ∈ T ∩ Λ \ {0}, so T ∩ Λ contains a point other than the origin.
The following is a useful volume computation, which we shall also need to prove the analytic class number
formula.
Lemma 2.72. Let K be a number field of degree n over Q. Let K have r real embeddings {σ1 , . . . , σr } and
2s non-real complex embeddings {σr+1 , σr+1 , . . . , σr+s , σr+s }. Define an embedding
φ : K ,→ Rr × Cs
59 T
is a measurable subset of V , being compact.
72
α 7→ (σ1 α, . . . , σr+s α).
r
s
n
We can identify V := R × C with R using the basis {1, i} for C. Let 0 6= I C OK . Then
(a) φ(I) is a full lattice in V .
(b) The volume of a fundamental domain for φ(I) is
2−s · N (I) ·
p
|∆K |,
(2.193)
where
• ∆K is the discriminant of K and
• N (I) := (OK : I) is the numerical norm of I.
Proof. Recall that OK is a free group (Z-module) of rank n. I is also free of rank n, being a finite-index
subgroup of OK (the index being N (I)). Let {a1 , . . . , an } be a basis for I as a Z-module.
(a) φ(I) = φ(Za1 + . . . + Zan ) = Zφ(a1 ) + . . . + Zφ(an ). In order to show that φ(I) is a (full) lattice, it
suffices to prove that φ(a1 ), . . . , φ(an ) are independent in V ∼
= Rn . Let A ∈ Mn (R) be the matrix whose
ith row is
(σ1 (ai ), . . . , σr (ai ), Re(σr+1 (ai )), Im(σr+1 (ai )), . . . , Re(σr+s (ai )), Im(σr+s (ai ))).
Now it suffices to prove that det(A) 6= 0. We shall show that the determinant of A is related to the
determinant of the matrix B ∈ Mn (C) whose ith row is
(σ1 (ai ), . . . , σr (ai ), σr+1 (ai ), σr+1 , σr+s (ai ), σr+s ).
By 2.23, {a1 , . . . , an } is a basis for K over Q, so 2.19 yields
det(B)2 = D(a1 , . . . , an ) 6= 0,
(2.194)
so det(B) 6= 0. The matrix A is obtained from the matrix B via the following (ordered) column
operations: for t = 1, . . . , s,
• Add column r + 2t to column r + 2t − 1 (doesn’t change the determinant).
• Multiply column r + 2t − 1 by
1
2
(halves the determinant).
• Subtract column r + 2t − 1 from column r + 2t (doesn’t change the determinant).
• Multiply column r + 2t by i (multiplies the determinant by i).
Hence det(A) = ( 2i )s det(B) 6= 0, so φ(I) is a full lattice in V .
(b) Define the matrices A and B as in (a). From 2.18,
D(a1 , . . . , an ) = (OK : I)2 · disc(OK /Z) = N (I)2 · ∆K
(2.195)
∴ |D(a1 , . . . , an )| = N (I)2 · |∆K |.
(2.196)
| det(B)|2 = | det(B)2 | = |D(a1 , . . . , an )|.
(2.197)
Moreover,
Finally, the volume of the fundamental domain P for φ(I) is
p
p
µ(P ) = | det(A)| = 2−s | det(B)| = 2−s |D(a1 , . . . , an )| = 2−s · N (I) · |∆K |.
73
2.10
Units in the ring of integers
×
Following Milne [8], we study the multiplicative group of units U = OK
in the ring of integers of a number
field K. We focus on a famous theorem, which enables us to understand U as a finitely generated abelian
group.60
Theorem 2.73 (Dirichlet’s Unit Theorem). Let K be a number field. Let r be the number of real embeddings
×
of K and let 2s be the number of non-real complex embeddings of K. Then OK
is finitely generated with
rank r + s − 1.
The proof is in two parts. In the first part, we show that U is finitely generated with rank less than or equal
to r + s − 1. In order to do this we need some preliminary results, the first of which describes torsion.
×
Let K be a number field, and let U = UK = OK
be the group of units in OK . The torsion subgroup of UK
is the group µ(K) of roots of unity in K.
Lemma 2.74. Let K be a number field. Then
(a) µ(K) is finite.
(b) µ(K) is cyclic.
Proof. For k ∈ Z>0 , let ζk denote an arbitrary primitive kth root of unity.
(a) It suffices to show that there are finitely many m ∈ Z>0 such that ζm ∈ K. By 2.67, [Q[ζm ] : Q] = φ(m).
Thus, if ζm ∈ K then φ(m) | [K : Q]. Hence there are finitely many possibilities for φ(m) if ζm ∈ K,
so it suffices to show that φ(m) = d has finitely many solutions m ∈ Z>0 for all d ∈ Z>0 . This is true,
since m ≤ 2φ(m)3 for all m ∈ Z>0 .61 Hence µ(K) is finite.
(b) As µ(K) is finite, we can let n ∈ Z>0 be maximal such that ζn ∈ µ(K). Then e2πi/n ∈ µ(K), and it
remains to prove that µ(K) = he2πi/n i. Let ζm ∈ µ(K) be a primitive mth root of unity. In order to
show that ζm ∈ he2πi/n i, it suffices to show that e2πi/m ∈ he2πi/n i, i.e. that m | n.
Proof by contradiction: assume that m does not divide n. Then there exist a, b ∈ Z such that n = am + b
and 0 < b < m.
n−am
b
(2.198)
e2πi mn = e2πi mn = e2πi/m (e2πi/n )−a ∈ µ(K).
Let d := gcd(b, mn). Then b = xd and mn = yd, where x, y ∈ Z>0 and gcd(x, y) = 1. Now
x
b
e2πi y = e2πi mn =∈ µ(K).
(2.199)
x
As gcd(x, y) = 1, e2πi y is a primitive yth root of unity, so y ≤ n. Now
mn = yd ≤ nd ∴ m ≤ d ≤ xd = b < m,
contradiction. Hence m | n, and we conclude that µ(K) is cyclic.
60 The
61 For
result is attributed to Dirichlet, though he only proved it for rings of the form Z[α].
more details on the solutions to the equation φ(m) = d, see Coleman [3].
74
(2.200)
Lemma 2.75. Let m, M ∈ Z>0 . Then the set of algebraic integers α ∈ C such that
• deg(α, Q) ≤ m and
• if α0 ∈ C is a conjugate of α then |α0 | < M
is finite.
Proof. It suffices to prove that there are finitely many monic polynomials f ∈ Z[X] of degree at most m
such that if α ∈ C and f (α) = 0 then |α| < M . Equivalently, it suffices to prove that if r ∈ {0, 1, . . . , m}
then there are finitely many integer polynomials
f (x) = xr + a1 xr−1 + . . . + ar−1 x + ar ∈ Z[X]
(2.201)
such that if α ∈ C and f (α) = 0 then |α| < M . Let r ∈ {0, 1, . . . , m}, and let f (x) = xr + a1 xr−1 + . . . +
ar−1 x + ar ∈ Z[X] be such that if α ∈ C and f (α) = 0 then |α| < M . For k = 1, 2, . . . , r, Vieta’s formula
yields
X
X
r
|ak | = αs1 · · · αsk ≤
αs1 · · · αsk ≤
M k,
(2.202)
k
1≤s1 <...<sk ≤r
1≤s1 <...<sk ≤r
where α1 , . . . , αr ∈ C are the roots of f . Thus, there are finitely many possibilities for the coefficients of
f , so there are finitely many possible monic polynomials f ∈ Z[X] such that if α ∈ C and f (α) = 0 then
|α| < M . Hence the set of algebraic integers α ∈ C such that
• deg(α, Q) ≤ m and
• if α0 ∈ C is a conjugate of α then |α0 | < M
is finite.
Corollary 2.76. Let α ∈ C be an algebraic integer, all of whose conjugates in C have absolute value equal
to 1. Then α is a root of unity.
Proof. It suffices to prove that {α, α2 , . . .} is finite. Let k ∈ Z>0 . Then αk ∈ Q[α], so
deg(αk , Q) = [Q[αk ] : Q] ≤ [Q[α] : Q].
Thus {α, α2 , . . .} is a subset of the set of algebraic integers β ∈ C such that
• deg(β, Q) ≤ [Q[α] : Q] and
• if β 0 is a conjugate of β then β 0 is absolute value equal to 1,
which is finite by 2.75. Hence {α, α2 , . . .} is finite, so α is a root of unity.
Lemma 2.77. Let K be a number field with r real embeddings and 2s non-real complex embeddings, and let
×
U = OK
. Let
{σ1 , . . . , σr , σr+1 , σr+1 , . . . , σr+s , σr+s }
be the set of embeddings K ,→ C. Define a homomorphism
ψ : K × → (Rr+s , +)
75
α 7→ (log |σ1 α|, . . . , log |σr α|, 2 log |σr+1 α|, . . . , 2 log |σr+s α|)
and a hyperplane
H := (x1 , . . . , xr+s ) ∈ Rr+s : x1 + . . . + xr+s = 0 .
Then
(a) ψ(U ) is a lattice in H.
(b) ker(ψ|U ) = µ(K).
Proof.
(a) We first show that ψ(U ) ⊆ H. Let u ∈ U . The sum of the coordinates of ψ(u) is
X
Y
log |σ(u)| = log
|σ(u)| = log |Nm(u)| = log 1 = 0,
σ:K,→C
(2.203)
σ:K,→C
using 2.15, so ψ(u) ∈ H. Thus ψ(U ) ⊆ H.
It is easy to check that ψ(U ) is an additive group. Thus, ψ(U ) is a subgroup of H. By 2.69, in order to
show that ψ(U ) is a lattice in H, it suffices to prove that each bounded subset of H intersects ψ(U ) in
a finite set. It therefore suffices to show that if M ∈ R>0 then there are finitely many u ∈ U such that
ψ(U ) ⊆ [−M, M ]r+s . Let M ∈ R>0 . If u ∈ U and ψ(U ) ⊆ [−M, M ]r+s , then
• each conjugate of u has absolute value less than or equal to eM , and
• deg(u, Q) ≤ [K : Q].
Hence, by 2.75, there are finitely many u ∈ U such that ψ(U ) ⊆ [−M, M ]r+s . We conclude that ψ(U )
is a lattice in H.
(b) (ker(ψ|U ) ⊆ µ(K):) If α ∈ ker(ψ|U ), then each conjugate of α has absolute value equal to 1, so α is a
root of unity (by 2.76). Additionally, α ∈ U ⊆ K, so α ∈ µ(K). Hence ker(ψ|U ) ⊆ µ(K).
(µ(K) ⊆ ker(ψ|U ):) Let α ∈ µ(K). Any conjugate of α must also be a root of unity, and therefore have
absolute value 1, so α ∈ ker(ψ|U ) (since each σi (α) is a conjugate of α). Hence µ(K) ⊆ ker(ψ|U ).
Thus ker(ψ|U ) = µ(K).
Proposition 2.78. Let K be a number field with r real embeddings and 2s non-real complex embeddings,
×
and let U = OK
. Then the rank of U is less than or equal to r + s − 1.
Proof. Let
{σ1 , . . . , σr , σr+1 , σr+1 , . . . , σr+s , σr+s }
be the set of embeddings K ,→ C. Define a homomorphism
ψ : K × → (Rr+s , +)
α 7→ (log |σ1 α|, . . . , log |σr α|, 2 log |σr+1 α|, . . . , 2 log |σr+s α|)
76
and a hyperplane
H := (x1 , . . . , xr+s ) ∈ Rr+s : x1 + . . . + xr+s = 0 .
The first isomorphism theorem yields a group isomomorphism
ψ(U ) ∼
=
U
.
ker(ψ|U )
(2.204)
Since ker(ψ|U ) = µ(K) is finite (by 2.74),
rank(U ) = rank(ψ(U ))
(2.205)
Since ψ(U ) is a lattice in H, let
ψ(U ) = Ze1 + . . . + Zek ,
k ∈ Z>0 ,
where e1 , . . . , ek are independent vectors in H. Now
rank(U ) = rank(ψ(U )) = k ≤ dim(H) = r + s − 1.
(2.206)
Moreover, the above proof shows that rank(U ) = rank(ψ(U )). Thus, to complete the proof of 2.73, it remains
to show that ψ(U ) is a full lattice in H.
Lemma 2.79. Assume (aij ) ∈ Mm (R), m ∈ Z>0 satisfies
• if i, j ∈ {1, 2, . . . , m} and i 6= j then aij < 0, and
Pm
• if i ∈ {1, 2, . . . , m} then j=1 aij > 0.
Then (aij ) is invertible.
Proof. Proof by contradiction: assume that (aij ) is not invertible. Then there exists (x1 , . . . , xm ) ∈ Rm \ {0}
satisfying
m
X
aij xj = 0,
for i = 1, 2, . . . , m.
(2.207)
j=1
Let k ∈ {1, 2, . . . , m} satisfy |xk | = max{|xj | : j ∈ {1, 2, . . . , m}}. Without loss of generality, xk = 1. Then
|xj | ≤ 1 for j = 1, 2, . . . , m, so
0=
m
X
akj xj = akk +
j=1
X
akj xj ≥ akk +
j6=k
X
j6=k
akj =
m
X
akj > 0,
j=1
contradiction. Hence (aij ) is invertible.
Proposition 2.80. Let K be a number field with r real embeddings and 2s non-real complex embeddings,
×
and let U = OK
. Let
{σ1 , . . . , σr , σr+1 , σr+1 , . . . , σr+s , σr+s }
77
be the set of embeddings K ,→ C. Define a homomorphism
ψ : K × → (Rr+s , +)
α 7→ (log |σ1 α|, . . . , log |σr α|, 2 log |σr+1 α|, . . . , 2 log |σr+s α|)
and a hyperplane
H := (x1 , . . . , xr+s ) ∈ Rr+s : x1 + . . . + xr+s = 0 .
Then ψ(U ) is a full lattice in H.
Proof. By 2.77, ψ(U ) is a lattice in H. Thus, it remains to find r + s − 1 vectors in L(U ) that are linearly
independent over R. Define a ring embedding
σ : K ,→ Rr × Cs
α 7→ (σ1 α, . . . , σr+s α)
and a multiplicative map (multiplication in Rs × Cs is componentwise)
Nm : Rr × Cs → R
(x1 , . . . , xr , xr+1 , . . . , xr+s ) 7→ x1 · · · xr · |xr+1 |2 · · · |xr+s |2 .
Note that Nm(σ(α)) = NmK/Q (α), for α ∈ K (by 2.14).
∼ R2n , with basis {σ(a1 ), . . . , σ(an )}, where
From the proof of 2.72, σ(OK ) is a full lattice in Rr × Cs =
{a1 , . . . , an } is an integral basis for K. Let x ∈ S := {y ∈ Rr × Cs : |Nm(y)| = 1}. With componentwise
multiplication, define
x · σ(OK ) := {x · σ(α) : α ∈ OK }.
(2.208)
This is a full lattice in Rr × Cs , with basis {x · σ(a1 ), . . . , x · σ(an )}. The volume of a fundamental domain
for x · σ(OK ) is | det(M1 )|, where the ith row (for i = 1, 2, . . . , n) of M1 ∈ Mn (R) is
(x1 σ1 (αi ), . . . , xr σr (αi ), Re(xr+1 σr+1 (ai )), Im(xr+1 σr+1 (ai )), . . . , Re(xr+s σr+1 (ai )), Im(xr+1 σr+s (ai ))).
Let M2 ∈ Mn (C) be the matrix whose ith row (for i = 1, 2, . . . , n) is
(x1 σ1 (ai ), . . . , xr σr (ai ), xr+1 σr+1 (ai ), xr+1 · σr+1 (ai ), . . . , xr+s σr+s (ai ), xr+s · σr+s (ai )).
The matrix M1 is obtained from the matrix M2 via the following (ordered) column operations: for t =
1, . . . , s,
• Add column r + 2t to column r + 2t − 1 (doesn’t change the determinant).
• Multiply column r + 2t − 1 by
1
2
(halves the determinant).
• Subtract column r + 2t − 1 from column r + 2t (doesn’t change the determinant).
• Multiply column r + 2t by i (multiplies the determinant by i).
78
Hence det(M1 ) = ( 2i )s det(M2 ), so
| det(M1 )| = 2−s | det(M2 )|.
(2.209)
Further, M2 is obtained from M3 by column multiplication, where the ith row (for i = 1, 2, . . . , n) of
M3 ∈ Mn (C) is
(σ1 (ai ), . . . , σr (ai ), σr+1 (ai ), σr+1 (ai ), . . . , σr+s (ai ), σr+s (ai )).
Thus,
det(M2 ) = Nm(x) det(M3 ).
(2.210)
From the proof of 2.72,
| det(M3 )| =
p
|∆|,
(2.211)
where ∆ is the discriminant of K. Finally, the volume of a fundamental domain for the lattice x · σ(OK ) is
p
p
| det(M1 )| = 2−s |Nm(x)| |∆| = 2−s |∆|.
(2.212)
Let T be a compact, convex subset of Rr × Cs such that
√
µ(T ) ≥ 2n−s ∆.
(2.213)
As T is bounded, there exists M ∈ R>0 such that if τ ∈ T then |Nm(τ )| ≤ M . Let P be the set of γ ∈ OK
such that there exists x ∈ S for which x · σ(γ) ∈ T . We now show that the set {γ · OK : γ ∈ P } is finite.
Assume that γ ∈ OK , x ∈ S and x · σ(γ) ∈ T . Then |Nm(x · σ(γ))| ≤ M (as x · σ(γ) ∈ T ), so
N (γ · OK ) = |NmK/Q (γ)| = |Nm(σ(γ))| = |Nm(x)| · |Nm(σ(γ))| = |Nm(x · σ(γ))| ≤ M.
(2.214)
By 2.58 (b), there are only finitely many possible ideals γ · OK . Hence, the set {γ · OK : γ ∈ P } is finite. Let
{γ · OK : γ ∈ P } = {γ1 · OK , . . . , γt OK },
t ∈ Z≥0 ,
γ1 , . . . , γt ∈ OK .
(2.215)
The set
T 0 := σ(γ1−1 ) · T ∪ . . . ∪ σ(γt−1 ) · T
(2.216)
is a bounded subset of Rr × Cs , so there exists K ∈ R>0 such that if u ∈ T 0 then each coordinate of u has
absolute value less than K. We now show that if x ∈ S then there exists ε ∈ U such that each coordinate of
x · σ(ε) has absolute value less than K. Let x ∈ S. By Minkowski’s theorem (2.71), there exists γ ∈ OK \ {0}
such that x · σ(γ) ∈ T (note that γ 6= 0 because σ : K ,→ Rr × Cs is injective). There exists i ∈ {1, 2, . . . , t}
such that γ · OK = γi · OK . Hence there exists ε ∈ U such that γ = γi · ε. Since σ : K ,→ Rr × Cs is a ring
homomorphism,
x · σ() = σ(γi−1 ) · x · σ(γ) ∈ T 0 .
(2.217)
Hence, each coordinate of x · σ(ε) has absolute value less than K. We have shown that if x ∈ S then there
exists ε ∈ U such that each coordinate of x · σ(ε) has absolute value less than K.
For i ∈ {1, 2, . . . , r + s − 1}, define εi as follows. Choose x = (x1 , . . . , xr , xr+1 , . . . , xr+s ) ∈ S by letting
xj = K for j ∈ {1, 2, . . . , r + s} \ {i}, and then choosing xi such that |Nm(x)| = 1. Let εi ∈ U be such that
each coordinate of x · σ(εi ) has absolute value less than K.
79
It remains to show that ψ(ε1 ), . . . , ψ(εr+s−1 ) ∈ ψ(U ) are linearly independent over R. Let L ∈ Mr+s−1 (R)
be the matrix whose ith row (for i = 1, 2, . . . , r + s − 1) is
(ψ1 (εi ), . . . , ψr+s−1 (εi )),
where, for x ∈ U ,

log |σ x|,
j
ψj (x) =
2 log |σj x|,
j = 1, 2, . . . , r
(2.218)
j = r + 1, . . . , r + s.
It remains to show that L is invertible. If i, j ∈ {1, 2, . . . , r + s − 1} and i 6= j, then
K > |xj · σj (εi )| = |xj | · |σj (εi )| = K · |σj (εi )|
∴ |σj (εi )| < 1
∴ log |σj (εi )| < 0,
2 log |σj (εi )| < 0.
For i = 1, 2, . . . , r + s − 1, ψ(εi ) ∈ H, so
ψ1 (εi ) + . . . + ψr+s (εi ) = 0 =⇒ ψ1 (εi ) + . . . + ψr+s−1 (εi ) = −ψr+s (εi ) > 0.
(2.219)
Thus, by 2.79, L is invertible. Hence ψ(ε1 ), . . . , ψ(εr+s−1 ) ∈ ψ(U ) are linearly independent over R, so ψ(U )
is a full lattice in H.
Now rank(U ) = rank(ψ(U )) = r + s − 1, which completes the proof of 2.73.
2.11
Fundamental units and regulators
References: Milne [8] and Borevich-Shafarevich [2].
×
Let K be a number field with r real embeddings and 2s non-real complex embeddings, and let U = OK
. A
set {u1 , . . . , ur+s−1 } ⊆ U is a system of fundamental units if each u ∈ U can be written uniquely in the form
m
r+s−1
1
u = ζum
1 · · · ur+s−1 ,
ζ ∈ µ(K),
m1 , . . . , mr+s−1 ∈ Z.
The unit theorem (2.73) implies that a system of fundamental units exists. Let {u1 , . . . , ur+s−1 } ⊆ U be a
system of fundamental units. Let
{σ1 , . . . , σr , σr+1 , σr+1 , . . . , σr+s , σr+s }
be the set of embeddings K ,→ C. Define a homomorphism
ψ = (ψ1 , . . . , ψr+s ) : K × → (Rr+s , +)
α 7→ (log |σ1 α|, . . . , log |σr α|, 2 log |σr+1 α|, . . . , 2 log |σr+s α|)
and a hyperplane
H := (x1 , . . . , xr+s ) ∈ Rr+s : x1 + . . . + xr+s = 0 .
80
For i = 1, 2, . . . , r + s − 1, let bi = ψ(ui ). As ψ : K × → (Rr+s , +) is a homomorphism,
( r+s−1
) (
!
)
r+s−1
r+s−1
X
X
Y
ai
Zbi =
ai bi : a1 , . . . , ar+s−1 ∈ Z = ψ
ui
: a1 , . . . , ar+s−1 ∈ Z = ψ(U ), (2.220)
i=1
i=1
i=1
because, by 2.77, ker(ψ|U ) = µ(K).
Pr+s−1
(b1 , . . . , br+s−1 are independent over R:) Assume that a1 , . . . , ar+s−1 ∈ Z and i=1 ai bi = 0. Then
!
r+s−1
r+s−1
Y
Y
ψ
uai i = 0 =⇒
(2.221)
uai i ∈ ker(ψ|U ) = µ(K),
i=1
i=1
by 2.77. Since {u1 , . . . , ur+s−1 } is a system of fundamental units, a1 = . . . = ar+s−1 = 0.
b1 , . . . , br+s−1 are independent over R.
Hence
Thus, {b1 , . . . , br+s−1 } is a basis for ψ(U ) as a full lattice in H, and therefore also a basis for H ⊆ Rr+s .
1
(1, 1, . . . , 1) ∈ Rr+s is orthogonal to H and has unit length. Thus, the (r + s − 1)The vector l0 := √r+s
62
dimensional volume of a fundamental domain for ψ(U ) is equal to the (r + s)-dimensional volume of the
1
parallelopiped determined by the vectors l0 , b1 , . . . , br . The regulator, R, of the number field K is √r+s
times
this volume. Then (r + s)R is equal to the absolute value of the determinant of the (r + s) × (r + s) matrix


1 ψ1 (u1 ) . . . ψ1 (ur+s−1 )
.

..
..
.
.
(2.222)
.
.

.
1 ψr+s (u1 ) . . . ψr+s (ur+s−1 )
Pr+s
Let i ∈ {1, 2, . . . , r + s − 1}. Then j=1 ψj (ui ) = 0, since ψ(ui ) ∈ ψ(U ) ⊆ H. Hence, adding each other
row to the ith row tells us that the regulator R is equal to the absolute value of the determinant of any
(r + s − 1) × (r + s − 1) minor of the (r + s) × (r + s − 1) matrix


ψ1 (u1 ) . . . ψ1 (ur+s−1 )


..
..
..

.
(2.223)
.
.
.


ψr+s (u1 ) . . . ψr+s (ur+s−1 )
1
times the volume of a fundamental domain for the lattice ψ(U ) in H, the regulator
Since the regulator is √r+s
is an algebraic invariant of the number field K (also, R > 0). In particular, it is independent of the choice
of system of fundamental units. Loosely speaking, the regulator measures the ‘density’ of units: a lower
regulator corresponds to a higher density of units.63
3
The class number formula
The Riemann zeta function is given by
ζ(z) =
∞
X
1
.
kz
k=1
62 Recall
63 The
that the volume measure is Haar measure. In Euclidean space, this is Lebesgue measure.
lattice points in a lattice of small fundamental volume are dense.
81
Theorem 3.1.
(a) If z ∈ C and Re(z) > 1, then ζ(z) converges.
(b)
lim (z − 1)ζ(z) = 1.
(3.1)
z→1+
In fact ζ can be analytically continued,64 so it has a simple pole at z = 1, with residue equal to
lim (z − 1)ζ(z) = 1.
z→1
Corollary 3.2 (Euler product formula). If z ∈ C and Re(z) > 1, then
−1
Y
1
ζ(z) =
1− z
,
p
p
(3.2)
where the product is over all nonzero prime numbers p.
Let K be a number field. The Dedekind zeta function, ζK , for the number field K, is given by
ζK (z) =
X
I
1
,
N (I)z
(3.3)
summed over all nonzero ideals I C OK . The Dedekind zeta function is a generalization of the Riemann zeta
function, since ζQ ≡ ζ (use 2.61 with K = Q).
The (analytic) class number formula generalizes the formula in 3.1 to Dedekind zeta functions:
Theorem 3.3. Let K be a number field with r real embeddings and 2s non-real complex embeddings. Let
h, R, and D be the class number, regulator, and discriminant of K, respectively, and let m be the number of
roots of unity in K. Then
(a) If z ∈ C and Re(z) > 1, then ζK (z) converges.
(b)
lim (z − 1)ζK (z) =
z→1+
2r+s π s hR
p
.
m |D|
In fact, for any number field K, ζK can be analytically continued,
simple pole at z = 1, with residue equal to
lim (z − 1)ζK (z) =
z→1
65
(3.4)
, so the theorem implies that ζK has a
2r+s π s hR
p
.
m |D|
The class number formula is an important bridge between algebraic and analytic number theory. Indeed, in
equation (3.4), the left hand side is purely analytic, and the right hand side is purely algebraic. It is amazing
that there should be an equation connecting so many different algebraic invariants of number fields.
64 See
Lang [5], Chapter VIII.
see Lang [5], Chapter VIII.
65 Again,
82
The formula sometimes enables us to compute the class number of a number field without explicitly computing the ideal class group, for instance, for quadratic number fields (see Rincón [10]). Another application
of 3.3 is to prove Dirichlet’s prime number theorem (again, see Rincón [10]).
Proof of 3.1.
(a) Let z ∈ C be such that Re(z) > 1. Let z = x + iy, where x ∈ R>1 and y ∈ R. We will show that
P∞
ζ(z) = k=1 k1z converges absolutely.
∞
X
1
|k x+iy |
k=1
=
∞
X
1
,
kx
(3.5)
k=1
which converges, since x ∈ R>1 . Hence ζ(z) converges absolutely, and therefore converges.
(b) Let z ∈ R>1 . For k ∈ Z>0 ,
1
<
(k + 1)z
k+1
Z
k
dx
1
< z.
z
x
k
Summing over all k ∈ Z>0 yields
Z
∞
ζ(z) − 1 <
1
1
dx
=
< ζ(z)
xz
z−1
=⇒ 1 < (z − 1)ζ(z) < z
=⇒ lim+ (z − 1)ζ(z) = 1.
z→1
Proof of 3.2. Let z ∈ C be such that Re(z) > 1. By 3.1, ζ(Re(z)) converges, so
lim
∞
X
N →∞
Thus, in order to prove that ζ(z) =
p 1−
Q
1
pz
k=N +1
1
= 0.
|k z |
(3.6)
−1
, it suffices to prove that
∞
Y X
1 −1 1
ζ(z) −
1
−
≤
,
pz
|k z |
p≤N
k=N +1
for all N ∈ Z>0 , where the product is over all prime numbers p ≤ N .
Let N ∈ Z>0 . Let p1 , . . . , pr be the prime numbers that are less than or equal to N .
Y p≤N
1−
r
X 1
1 −1 Y 1
1
=
1
+
+
+
.
.
.
=
,
pz
pzi
p2z
ks
i
i=1
where S is the set of m ∈ Z>0 such that m has no prime divisors that are greater than N . Now
∞
Y X 1
X
X
1 −1 1
1
ζ(z) −
=
≤
≤
.
1
−
pz
kz
|k z |
|k z |
p≤N
(3.7)
k∈S
k∈Z>0 \S
83
k∈Z>0 \S
k=N +1
(3.8)
Hence,
−1
Y
1
.
ζ(z) =
1− z
p
p
Now we prove 3.3. Our approach follows Osserman [9] and Sivek [11], supplemented by Borevich-Shafarevich
[2] and Milne [8].
For each ideal class C of OK , define
ζK,C (z) =
Y
I∈C
1
1−
N (I)z
!−1
,
(3.9)
where the product is over all nonzero ideals I C OK in the ideal class C. For all z ∈ C such that Re(z) > 1,
X
ζK (z) =
ζK,C (z),
(3.10)
C
summed over all ideal classes C. There are h ∈ Z>0 ideal classes, so it suffices to prove the following:
Proposition 3.4. Let C be an ideal class of OK .
(a) If z ∈ C and Re(z) > 1, then ζK,C (z) converges.
(b)
lim+ (z − 1)ζK,C (z) =
z→1
2r+s π s R
p
.
m |D|
(3.11)
Fix I0 ∈ C −1 . Then an ideal I C OK is in C if and only if there exists α ∈ I0 such that I0 I = (α). Using
2.61,
X
X N (I0 )z
1
= N (I0 )z
.
(3.12)
ζK,C (z) =
z
N ((α))
|Nm(α)|z
(α):α∈I0
(α):α∈I0
We would like to sum over elements rather than ideals, since associates generate the same ideal. The next
proposition allows us to achieve this, but first we need some preliminary definitions.
Let n := r+2s, and identify Rr ×Cs with Rn using the basis {1, i} for C (we will often make this identification).
Let {σ1 , . . . , σr , σr+1 , σr+1 , . . . , σr+s } be the set of embeddings K ,→ C. Define an embedding
φ : K ,→ Rr × Cs
α 7→ (σ1 α, . . . , σr+s α),
a homomorphism
N : Rr × Cs → R≥0
(x1 , . . . , xr ; xr+1 , . . . , xr+s ) 7→
r
Y
i=1
84
|xi |
s
Y
j=1
|xr+j |2
and a homomorphism66
χ : (R× )r × (C× )s → (Rr+s , +)
(x1 , . . . , xr ; xr+1 , . . . , xr+s ) 7→ (log |x1 |, . . . , log |xr |, 2 log |xr+1 |, . . . , 2 log |xr+s |).
Also define a homomorphism
ψ = (ψ1 , . . . , ψr+s ) : K × → (Rr+s , +)
α 7→ (log |σ1 α|, . . . , log |σr α|, 2 log |σr+1 α|, . . . , 2 log |σr+s α|)
and a hyperplane
H := (x1 , . . . , xr+s ) ∈ Rr+s : x1 + . . . + xr+s = 0 .
Note that ψ = χ ◦ φ
if x ∈ X and t ∈
, and that if x ∈ Rn then log N (x) is the sum of the coordinates of χ(x). Crucially,
K×
R>0 then
N (tx) = tn N (x).
×
Let {u1 , . . . , ur+s−1 } be a system of fundamental units for U := OK
, and let
i = 1, 2, . . . , r + s − 1.
bi = ψ(ui ),
Recall, from the section 2.11, that {b1 , . . . , br+s−1 } is a basis for ψ(U ) as a full lattice in H, and therefore also a
basis for the hyperplane H ⊆ Rr+s . Define λ = (λ1 , . . . , λr ; λr+1 , . . . , λr+s ) = (1, . . . , 1; 2, . . . , 2) ∈ Rr+s \ H.
As λ ∈
/ H, it follows that {λ, b1 , . . . , br+s−1 } is a basis for Rr+s .
Proposition 3.5. Let X be the set of x ∈ Rn such that
(i) N (x) 6= 0.
(ii) χ(x) = cλ +
Pr+s−1
(iii) 0 ≤ arg(x1 ) <
i=1
2π
m,
c i bi ,
c ∈ R,
0 ≤ c1 , . . . , cr+s−1 < 1.
where x1 is the first coordinate of x ∈ (R× )r × (C× )s .67
Then
(a) X is a cone in Rn , i.e. if x ∈ X and t ∈ R>0 then tx ∈ X.
(b) Let y ∈ Rn be such that N (y) 6= 0. Then there exists a unique element (unit) ε ∈ U such that φ(ε)·y ∈ X,
with componentwise multiplication in Rr × Cs .
(c) The set S := {x ∈ X : N (x) ≤ 1} is bounded.
(d) The volume of S is
vol(S) =
2r π s R
.
m
(3.13)
In order to prove 3.5, we need a lemma:
in (R× )r × (C× )s is componentwise.
r > 0, then x1 ∈ R, and the condition is that x1 > 0. Note that the first condition, that N (x) 6= 0, is equivalent to the
condition that x ∈ (R× )r × (C× )s .
66 Multiplication
67 If
85
Lemma 3.6. Let x = (x1 , . . . , xr ; xr+1 , . . . , xr+s ) ∈ (R× )r × (C× )s . As {λ, b1 , . . . , br+s−1 } is a basis for
Rr+s , let
χ(x) = cλ + c1 b1 + . . . + cr+s−1 br+s−1 ,
c, c1 , . . . , cr+s−1 ∈ R.
Then
c=
1
log N (x).
n
(3.14)
Proof. For j = 1, . . . , r + s, equating two expressions for the jth coordinate of χ(x) yields
λj log |xj | = cλj +
r+s−1
X
ci ψj (ui ).
(3.15)
i=1
Summing over j yields log N (x) = cn, as ψ(
Qr+s−1
i=1
c=
uci i ) ∈ ψ(U ) ⊆ H. Hence,
1
log N (x).
n
Proof of 3.5.
(a) Let x ∈ X, t ∈ R>0 . To show: tx ∈ X.
(i) N (tx) = tn N (x) 6= 0, since t 6= 0 and N (x) 6= 0.
Pr+s−1
(ii) Let χ(x) = cλ + i=1 ci bi , where c ∈ R and 0 ≤ c1 , . . . , cr+s−1 < 1. Then
χ(tx) = log(t) · λ + χ(x) = (c0 + log t)λ +
r+s−1
X
c i bi ,
0 ≤ c1 , . . . , cr+s−1 < 1.
i=1
(iii) Let x1 be the first coordinate of x ∈ (R× )r ×(C× )s . Then the first coordinate of tx ∈ (R× )r ×(C× )s
is tx1 , which has argument equal to arg(x1 ) ∈ [0, 2π
m ).
(b) Let y ∈ Rn be such that N (y) 6= 0. To show: there exists a unique element (unit) ε ∈ U such that
φ(ε) · y ∈ X, with componentwise multiplication in Rr × Cs .
• (Existence:) First we address condition (ii) for X. As {λ, b1 , . . . , br+s−1 } is a basis for Rr+s , let
χ(y) = cλ + c1 b1 + . . . + cr+s−1 br+s−1 ,
c, c1 , . . . , cr+s−1 ∈ R.
(3.16)
For i = 1, 2, . . . , r + s − 1, let
mi ∈ Z,
ci = mi + µi ,
0 ≤ µi < 1.
(3.17)
m
r+s−1
−1
1
Let u = um
). Then
1 · · · ur+s−1 ∈ U , and let z = y · φ(u
χ(z) = χ(y) + ψ(u−1 ) = cλ + µ1 b1 + . . . + µr+s−1 br+s−1 ,
0 ≤ µi < 1.
(3.18)
Now that condition (ii) is satisfied, we address condition (iii). Let z1 be the first coordinate of
z ∈ Rr × Cs . Let r be the unique integer such that
0 ≤ arg(z1 ) −
86
2π
2πr
<
.
m
m
(3.19)
2πi
Choose w ∈ µ(K) such that σ1 (w) = e m . Note that ε := u−1 w−r ∈ U . We will show that
y · φ(ε) ∈ X. Note that
y · φ(ε) = y · φ(u−1 w−r ) = z · φ(w−r ).
(3.20)
(i) As u−1 w−r ∈ U , 2.15 yields Nm(u−1 w−r ) = ±1. Now
N (y · φ(u−1 w−r )) = N (y) · N (φ(u−1 wr )) = N (y) · |Nm(u−1 w−r )| = N (y) 6= 0.
(ii) Recall that χ(z) = cλ + µ1 b1 + . . . + µr+s−1 br+s−1 , where 0 ≤ µ1 , . . . , µr+s−1 < 1. Now
χ(z · φ(w−r )) = χ(z) + r · ψ(w) = χ(z),
(3.21)
as w ∈ µ(K) = ker(ψ|U ), by 2.77. Now, using equation (3.18),
χ(z · φ(w−r )) = χ(z) = χ(y) · ψ(u−1 )
0 ≤ µi < 1.
= cλ + µ1 b1 + . . . + µr+s−1 br+s−1 ,
(iii) The first coordinate of z · φ(w−r ) is z1 · σ1 (w)−r , which has argument equal to
2πr h 2π arg(z1 ) −
∈ 0,
,
m
m
directly from the inequality (3.19).
• (Uniqueness:) Let u ∈ U be such that z := φ(u) · y ∈ X. Let v ∈ U be such that φ(v) · z ∈ X.
To show: φ(v) = 1. Let
0 ≤ c, c1 , . . . , cr+s−1 < 1.
χ(z) = cλ + c1 b1 + . . . + cr+s−1 br+s−1 ,
(3.22)
As v ∈ U , there exist ρ ∈ µ(K) and m1 , . . . , mr+s−1 ∈ Z such that
m
r+s−1
1
v = ρ · um
1 · · · ur+s−1 .
Now
χ(φ(v) · z) = ψ(v) + χ(z) = ψ(ρ) +
(3.23)
r+s−1
X
r+s−1
X
i=1
i=1
(mi + ci )bi =
(mi + ci )bi ,
(3.24)
since ρ ∈ µ(K) = ker(ψ|U ), by 2.77. For i = 1, 2, . . . , r + s − 1,
mi = (mi + ci ) − ci ∈ (−1, 1) ∩ Z = {0},
(3.25)
since φ(v) · z ∈ X implies that 0 ≤ mi + ci < 1. Hence m1 = . . . = mr+s−1 = 0, so v = ρ ∈ µ(K).
Let z1 be the first coordinate of z ∈ Rr × Cs . Then 0 ≤ arg(z1 ) <
φ(v) · z is equal to (modulo 2π)
arg(φ(v)) + arg(z1 ),
2π
m.
The first coordinate of
which also lies in the interval [0, 2π
m ), since φ(v) · z ∈ X. Now
arg(φ(v)) = (arg(φ(v)) + arg(z1 )) − arg(z1 ) ∈
As µ(K) = he
2πi
m
−
i, the argument of φ(v) is also an integer multiple of
87
2π 2π ,
.
m m
2π
m.
(3.26)
Hence, arg(φ(v)) = 0.
Hence, there exists a unique element (unit) ε ∈ U such that φ(ε) · y ∈ X.
(c) Define S0 := {x ∈ X : N (x) = 1}. Then S = {tS0 : 0 < t ≤ 1}. Thus, in order to show that S is
bounded, it suffices to show that S0 is bounded.
(χ(S0 ) ⊆ H :) Let x ∈ S0 . The sum of the coordinates of χ(x) is log N (x) = log 1 = 0, so x ∈ H. Hence,
(χ(S0 ) ⊆ H).
(χ(S0 ) ⊆ Rr+s is bounded:) As {λ, b1 , . . . , br+s−1 } is a basis for Rr+s , it suffices to prove that there
exists B ∈ R>0 such that if x = (x1 , . . . , xr+s ) ∈ S0 and
χ(x) = cλ +
r+s−1
X
ci bi ,
c, c1 , . . . , cr+s−1 ∈ R,
(3.27)
i=1
then |c|, |c1 |, . . . , |cr+s−1 | ≤ B. Let B := 1. Let x = (x1 , . . . , xr+s ) ∈ S0 ⊆ X, and let χ(x) =
Pr+s−1
cλ + i=1 ci bi , where c, c1 , . . . , cr+s−1 ∈ R. As x ∈ S0 , 3.6 yields
c=
1
1
log N (x) = log 1 = 0.
n
n
(3.28)
For i = 1, 2, . . . , r + s − 1, x ∈ X implies that 0 ≤ ci < 1. Thus |c|, |c1 |, . . . , |cr+s−1 | ≤ B. Hence,
χ(S0 ) ⊆ Rr+s is bounded.
Now we finish proving that S0 (and hence S) is bounded. It suffices to prove that there exists R ∈ R>0
such that if x = (x1 , . . . , xr ; xr+1 , . . . , xr+s ) ∈ S0 then |x1 |, . . . , |xr+s | ≤ R. As χ(S0 ) is bounded, there
exists M ∈ R>0 such that if x = (x1 , . . . , xr ; xr+1 , . . . , xr+s ) ∈ S0 , then
λi log |xi | ≤ M,
i = 1, 2, . . . , r + s.
(3.29)
Let R := eM . Let x = (x1 , . . . , xr ; xr+1 , . . . , xr+s ) ∈ S0 . For i = 1, 2, . . . , r + s,
λi log |xi | ≤ M =⇒ |xi | ≤ eM = R.
(3.30)
Hence S0 is bounded, so S is bounded.
(d) We need to show that the volume of S := {x ∈ X : N (x) ≤ 1} is
vol(S) =
2r π s R
.
m
For k = 0, 1, . . . , m − 1, let Sk := e2πki/m S. Recall that the volume measure is Haar measure, and
therefore Lebesgue measure on Rn . Multiplication by a unit68 is volume-preserving, therefore
k = 0, 1, . . . , m − 1.
vol(S) = vol(Sk ),
Let
T := {x = (x1 , . . . , xr ; xr+1 , . . . , xr+s ) ∈ tm−1
k=0 Sk : x1 > 0, . . . , xr > 0},
68 That
is, an element u ∈ C such that |u| = 1.
88
(3.31)
where t denotes disjoint union.
r
(vol(S) = 2m vol(T ):) Consider the 2r volume-preserving linear transformations on Rr × Cs defined by
multiplication by (±1, . . . , ±1; 1, . . . , 1). Applying these to T yields 2r pairwise-disjoint sets, whose union
is tm−1
k=0 Sk . Hence,
2r
(3.32)
vol(S) = vol(T ).
m
It remains to show that vol(T ) = π s R. We can replace the coordinate system (x1 , . . . , xr ; xr+1 , . . . , xr+s )
for Rr ×Cs ∼
= Rn with the coordinate system (x1 , . . . , xr , yr+1 , zr+1 , . . . , yr+s , zr+s ) by letting xj = yj + izj
(where yj , zj ∈ R) for j = r1 , . . . , r + s. Now
Z
vol(T ) =
dx1 · · · dxr dyr+1 dzr+1 · · · dyr+s dzr+s .
(3.33)
T
Now we change to polar coordinates for each copy of C in the product Rr × Cs ; that is, let
ρj = |xj |,
yj + izj = ρj eiθj ,
ρj > 0,
for j = 1, 2, . . . , r
0 ≤ θj < 2π,
for j = r + 1, . . . , r + s,
which gives a new coordinate system (ρ1 , . . . , ρr , ρr+1 , θr+1 , . . . , ρr+s , θr+s ) for Rn . Note that ρj = xj
on T , for j = 1, 2, . . . , r, so the s changes to polar coordinates mean that the Jacobian determinant of
the transformation is ρr+1 · · · ρr+s > 0. Hence,
Z
vol(T ) =
ρr+1 · · · ρr+s dρ1 · · · dρr+s dθr+1 dθr+s .
(3.34)
T
Now we determine the conditions describing T . Let x ∈ T .
• ρ1 > 0, . . . , ρr+s > 0.
• As x ∈ T , there exists k ∈ {0, 1, . . . , m − 1} such that e−2πki/m x ∈ X. Then
χ(x) = χ(e−2πki/m ) + χ(x) = χ(e−2πki/m x),
(3.35)
so let (by 3.6)
λ
χ(x) = log
n
r+s
Y
k=1
!
ρλk k
+
r+s−1
X
ξk bk ,
(3.36)
k=1
and the condition is that 0 ≤ ξk < 1 for k = 1, 2, . . . , r + s − 1.
• 0 ≤ θj < 2π for j = r + 1, . . . , r + s.
In order to show that (for j = r + 1, . . . , r + s) θj can take any value in [0, 2π) independently, we
need to show that if j ∈ {r + 1, . . . , r + s} then T is closed under the action of multiplying the jth
coordinate by an element of G := {z ∈ C : |z| = 1}. Let j ∈ {r +1, . . . , r +s}, g ∈ G, t ∈ T . Let g(t)
be the result of multiplying the jth coordinate of t by g. To show: g(t) ∈ T . The first r coordinates
of g(t) are the same as for t, therefore it remains to show that there exists k ∈ {0, 1, . . . , m − 1}
such that g(t) ∈ Sk . However,
g(t) ∈ Sk ⇔ e2πki/m t ∈ S ⇔ e−2πki/m g(t) ∈ X,
89
(3.37)
since N (g(t)) = N (t) ≤ 1. Since acting by g and multiplying the jth coordinate by e−2πki/m ∈ G
does not affect conditions (i) and (ii) for X, it remains to show that there exists k ∈ {0, 1, . . . , m−1}
such that the argument of the first coordinate of e−2πki/m g(t) ∈ Rr × Cs is in the interval [0, 2π
m ),
which is true. Hence (for j = r + 1, . . . , r + s) θj can take any value in [0, 2π) independently.
We change to the coordinate system (θr+1 , . . . , θr+s , ξ, ξ1 , . . . , ξr+s−1 ), with ξ1 , . . . , ξr+s−1 as in equation
Qr+s
(3.36) and ξ := k=1 ρλk k . Now T is described by a rectangular region:
• 0 < ξ ≤ 1.
• 0 ≤ ξk < 1 for k = 1, 2, . . . , r + s.
Now we compute the Jacobian determinant of the transformation. Equating jth coordinates in equation
(3.36) yields (1.15) of Borevich-Shafarevich [2]:
λj log ρj =
r+s−1
X
λj
log ξ +
ξk ψj (uk ),
n
j = 1, 2, . . . , r + s.
(3.38)
k=1
Thus, for j = 1, 2, . . . , r + s and k = 1, 2, . . . , r + s − 1,
ρj
∂ρj
=
,
∂ξ
nξ
∂ρj
ρj
=
ψj (uk ).
∂ξk
λj
Hence, the Jacobian determinant of the transformation is
ρ
ρ1
1
...
nξ
λ1 ψ1 (u1 )
.
.
..
..
J = ..
.
ρr+s ρr+s
nξ
ψr+s (u1 ) . . .
λ
r+s
=
ρ1 · · · ρr+s
nξ2s
λ
1
.
.
.
λr+s
(3.39)
..
.
ρr+s
ψ
(u
)
r+s
r+s−1
λr+s
ρ1
λ1 ψ1 (ur+s−1 )
ψ1 (u1 ) . . .
..
..
.
.
ψr+s (u1 ) . . .
ψ1 (ur+s−1 )
..
.
ψr+s (ur+s−1 )
(3.40)
As ψ(U ) ⊆ H, adding rows 2, 3, . . . , r + s to row 1 yields
n
0
0
0
ψ2 (u1 ) . . . ψ2 (ur+s−1 ) ρ1 · · · ρr+s λ2
J=
..
..
..
..
nξ2s
.
.
.
.
λr+s ψr+s (u1 ) . . . ψr+s (ur+s−1 ) ψ (u ) . . . ψ (u
2 1
2 r+s−1 ) ρ1 · · · ρr+s
ρ1 · · · ρr+s ..
..
..
=
=
R
.
.
.
s
ξ2
ξ2s
ψr+s (u1 ) . . . ψr+s (ur+s−1 ) =
R
ρ1 · · · ρr+s
R= s
.
ρ1 · · · ρr ρ2r+1 · · · ρ2r+s 2s
2 ρr+1 · · · ρr+s
Continuing from equation (3.34), and using equation (3.41),
Z
vol(T ) =
ρr+1 · · · ρr+s dρ1 · · · dρr+s dθr+1 dθr+s
T
90
(3.41)
Z
2π
Z
2π
dθr+1 · · ·
=
= (2π)
Z
1
Z
ρr+1 · · · ρr+s |J|dξdξ1 · · · dξr+s−1
0
0
s
1
...
0
0
1
Z
dθr+s
1
Z
...
0
0
R
dξdξ1 · · · dξr+s−1 = π s R.
2s
(3.42)
Thus, from equations (3.32) and (3.42),
vol(S) =
2r
2r π s R
vol(T ) =
.
m
m
Proposition 3.7. Let X ⊆ Rn be a cone such that
• If x ∈ X then N (x) 6= 0.
• The set {x ∈ X : N (x) ≤ 1} is bounded, with volume vX > 0.
Let L be a full lattice in Rn , such that the volume of a fundamental domain in L is vL . Define
X
1
.
ζX,L (z) =
N (x)z
(3.43)
x∈X∩L
Then
(a) If z ∈ C and Re(z) > 1, then ζX,L (z) converges.
(b)
lim (z − 1)ζX,L (z) =
z→1+
vX
.
vL
(3.44)
Proof. Let S = {x ∈ X : N (x) ≤ 1}. For t ∈ R>0 , let Lt be the lattice 1t L, which has fundamental volume
1
tn vL , and let P (t) be the number of points in Lt ∩ S (which is finite, by 2.69, as S is bounded).
We can fill S with smaller and smaller parallelopipeds, i.e. fundamental domains for Lt (letting t → ∞).
The number of such parallelopipeds is approximated by P (t). Thus,
vL
(3.45)
vX = vol(S) = lim P (t) n ,
t→∞
t
so
vX
P (t)
(3.46)
= lim n .
t→∞ t
vL
Hence,
lim P (t) = ∞.
(3.47)
t→∞
Let t ∈ R>0 . Note that P (t) is also the number of points in L ∩ tS, since L ∩ tS = t · (Lt ∩ S). Moreover,
L ∩ tS = {x ∈ X : N (z) ≤ tn }, since N (tx) = tn N (x) for all x ∈ X and for all t ∈ R>0 . Hence, for all
t ∈ R>0 , the number of points in {x ∈ X ∩ L : N (z) ≤ tn } is P (t), which is finite. The norms of points in
X ∩ L are therefore discrete. Moreover, |X ∩ L| = limt→∞ P (t) = ∞. Let
X ∩ L = {x1 , x2 , . . .},
Claim.
lim
k→∞
N (x1 ) ≤ N (x2 ) ≤ . . . .
k
vX
=
.
N (xk )
vL
91
(3.48)
(3.49)
Proof of claim.
P (t1/n )
vX
P (t)
#{x ∈ X ∩ L : N (x) ≤ t}
= lim n = lim
= lim
.
t→∞
t→∞ t
t→∞
vL
t
t
(3.50)
For k ∈ Z>0 and ε ∈ R>0 ,
#{x ∈ X ∩ L : N (x) ≤ N (xk ) − ε} < k ≤ #{x ∈ X ∩ L : N (x) ≤ N (xk )}.
(3.51)
Note that limk→∞ N (xk ) = ∞, since #{x ∈ X ∩ L : N (x) ≤ tn } = P (t) is finite for each t ∈ R>0 .
•
#{x ∈ X ∩ L : N (x) ≤ N (xk ) − ε}
k
N (xk )
<
·
.
N (xk ) − ε
N (xk ) N (xk ) − ε
Now
(3.52)
#{x ∈ X ∩ L : N (x) ≤ t}
#{x ∈ X ∩ L : N (x) ≤ N (xk ) − ε}
vX
= lim
= lim
t→∞
k→∞
vL
t
N (xk ) − ε
•
≤ lim
k
N (xk )
k
·
= lim
.
k→∞ N (xk ) N (xk ) − ε
k→∞ N (xk )
(3.53)
#{x ∈ X ∩ L : N (x) ≤ N (xk )}
k
≤
.
N (xk )
N (xk )
(3.54)
Now
lim
k→∞
k
#{x ∈ X ∩ L : N (x) ≤ N (xk )}
#{x ∈ X ∩ L : N (x) ≤ t}
vX
≤ lim
= lim
=
. (3.55)
t→∞
N (xk ) k→∞
N (xk )
t
vL
Hence limk→∞
k
N (xk )
=
vX
vL ,
so the claim is proven.
Now we complete the proof of 3.7.
P
1
(a) Let z ∈ C be such that Re(z) > 1. In order to show that ζX,L (z) = x∈X∩L N (x)
z converges, we prove
that it converges absolutely.
∞
∞
∞
kz X
X
1 k z X 1
1
=
·
·
(3.56)
=
.
z
z
z
z
Re(z)
|N (xk ) |
|k | N (xk )
N (xk ) k
k=1
The series
P∞
k=1
k=1
1
k=1 kRe(z)
converges, as Re(z) > 1. Thus, we can apply the limit comparison test.
k z k z vX z lim by the claim
= lim
=
,
k→∞ N (xk ) vL k→∞ N (xk )z ∈ R>0 .
Thus, by the limit comparison test, the series
∞
X
k=1
∞
X 1
1
=
z
|N (xk ) |
k Re(z)
k=1
kz ·
N (xk )z converges, i.e. the series ζX,L (z) converges absolutely. Hence, ζX,L (z) converges.
92
(3.57)
(b) Let ε ∈ R>0 . The claim implies that there exists k0 ∈ Z>0 such that if k ≥ k0 then
v
1
v
1
1
X
X
−ε
<
<
+ε .
vL
k
N (xk )
vL
k
(3.58)
Let z ∈ R>1 . Then
v
X
vL
∞
∞
∞
v
z
z
X
X
X
1
1
1
X
<
(z
−
1)
<
,
− ε (z − 1)
+
ε
(z
−
1)
z
z
k
N (xk )
vL
kz
k=k0
k=k0
(3.59)
k=k0
so
v
X
vL
−ε
∞
∞
∞
v
X
X
X
1
1
1
X
<
lim
(z
−
1)
<
+
ε
lim
(z
−
1)
z
z
+
+
k
N (xk )
vL
kz
z→1
z→1
lim+ (z − 1)
z→1
k=k0
k=k0
As
lim+ (z − 1)
kX
0 −1
z→1
k=1
(3.60)
k=k0
kX
0 −1
1
1
=
0
=
lim
(z
−
1)
,
kz
N (xk )z
z→1+
(3.61)
k=1
equation (3.60) is equivalent to
v
v
X
X
− ε lim+ (z − 1)ζ(z) < lim+ (z − 1)ζX,L (z) <
+ ε lim+ (z − 1)ζ(z).
vL
vL
z→1
z→1
z→1
(3.62)
As limz→1+ (z − 1)ζ(z) = 1 (from 3.1),
vX
vX
− ε < lim+ (z − 1)ζX,L (z) <
+ ε.
vL
vL
z→1
Taking the limit as ε → 0+ yields
lim (z − 1)ζX,L (z) =
z→1+
(3.63)
vX
.
vL
Finally, we can complete the proof of 3.4, and hence of 3.3.
Proof. Let α ∈ I0 . By 3.5, α has a unique associate α0 ∈ I0 such that x := φ(α0 ) ∈ φ(I0 ) ∩ X. Let α0 = α · ε,
where ε ∈ U . Using 2.15 and 2.14,
|Nm(α)| = |Nm(ε)| · |Nm(α)| = |Nm(α0 )| = N (φ(α)) = N (x).
(3.64)
By 3.5, X ⊆ Rn is a cone such that
• If x ∈ X then N (x) 6= 0.
• The set {x ∈ X : N (x) ≤ 1} is bounded, with volume
vX =
2r π s R
> 0.
m
By 2.72, L := φ(I0 ) is a full lattice in Rr+s , and the volume of a fundamental domain for L is
p
vL := 2−s · N (I0 ) · |D|.
93
(3.65)
(3.66)
(a) Let z ∈ C be such that Re(z) > 1. Then
X
ζK,C (z) = N (I0 )z
(α):α∈I0
X
1
1
= N (I0 )z
,
z
|Nm(α)|
N (x)z
(3.67)
x∈X∩L
which converges, by 3.7.
(b) Using 3.7,
X
lim+ (z − 1)ζK,C (z) = lim+ (z − 1)N (I0 )z
z→1
z→1
= N (I0 )
(α):α∈I0
X
1
1
= N (I0 ) lim+ (z − 1)
z
|Nm(α)|
N (x)z
z→1
x∈X∩L
p 2r+s π s R
vX
2r π s R −s
.
= N (I0 ) ·
÷ 2 · N (I0 ) · |D| = p
vL
m
m |D|
Summing over the h ideal classes C of OK yields
(a) If z ∈ C and Re(z) > 1, then ζK (z) converges.
(b)
lim+ (z − 1)ζK (z) =
z→1
2r+s π s hR
p
,
m |D|
which completes the proof of 3.3.
There is also an analagous Euler product formula:
Corollary 3.8. Let K be a number field. If z ∈ C and Re(z) > 1, then
−1
Y
1
ζK (z) =
1−
,
N (p)z
p
(3.68)
where the product is over all nonzero prime ideals p C OK .
Proof. Let z ∈ C be such that Re(z) > 1. By 3.3, ζK (Re(z)) converges, so
X
1
lim
= 0.
N →∞
|N (I)z |
(3.69)
I:N (I)>N
Q
Thus, in order to prove that ζK (z) = p 1 −
ζK (z) −
Y
p:N (p)≤N
1
N (p)z
−1
, it suffices to prove that
1 −1 1−
≤
N (p)z
X
I:N (I)>N
1
,
|N (I)z |
for all N ∈ Z>0 , where the product is over all nonzero prime ideals p C OK such that N (p) ≤ N .
Let N ∈ Z>0 . By 2.58, there are only finitely many nonzero prime ideals p C OK such that N (p) ≤ N ;
denote these by p1 , . . . , pr .
Y
p:N (p)≤N
r
1−
X 1
1 −1 Y 1
1
=
1+
+
+ ... =
,
z
z
2z
N (p)
N (pi )
N (pi )
N (I)s
i=1
I∈S
94
(3.70)
where S is the set of nonzero ideals I C OK satisfying: if p is a nonzero prime ideal in OK then N (p) ≤ N .
Now
Y X
X
1 −1 X 1
1
1
1
−
=
≤
≤
.
(3.71)
z
z
z
N (p)
N (I)
|N (I) |
|N (I)z |
p:N (p)≤N
I ∈S
/
I ∈S
/
I:N (I)>N
Hence,
ζK (z) =
Y
p
1
1−
N (p)z
95
−1
.
References
[1] Michael Atiyah and Ian G. Macdonald. Introduction to Commutative Algebra. Addison-Wesley, 1969.
[2] Zenon Borevich and Igor Shafarevich. Number Theory. Academic Press, 1966.
[3] Rodney Coleman. On the image of Euler’s totient function.
http://arxiv.org/abs/0910.2223v1, 2009.
[4] Serge Lang. Algebra, revised 3rd ed. Springer-Verlag, 2002.
[5] Serge Lang. Algebraic Number Theory, 2nd ed. Springer-Verlag, 1994.
[6] Kimball Martin. Nonunique factorization and principalization in number fields. Proc. Amer. Math. Soc.
139, No. 9: 3025-3038, 2011.
[7] James Milne. Course notes on field theory.
http://www.jmilne.org/math/CourseNotes/ft.html, 2008.
[8] James Milne. Course notes on algebraic number theory.
http://www.jmilne.org/math/CourseNotes/ant.html, 2009.
[9] Brian Osserman. Course notes on number theory.
http://www.math.ucdavis.edu/~osserman/classes/254A/, 2005.
[10] José Alfredo Cañizo Rincón. The Dedekind Zeta Function and the Class Number Formula. Final paper.
http://www-math.mit.edu/~kedlaya/Math254B/zetafunction.pdf, 2002.
[11] Gary Sivek. The analytic class number formula. Final paper.
http://wstein.org/129-05/final_papers/Gary_Sivek.pdf, 2005.
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