Download MATH 254A: RINGS OF INTEGERS AND DEDEKIND DOMAINS 1

MATH 254A: RINGS OF INTEGERS AND DEDEKIND
DOMAINS
BRIAN OSSERMAN
1. First properties of rings of integers
The discriminant gives us a somewhat constructive proof of the following fact:
Proposition 1.1. Let I be a non-zero ideal in a ring of integers OK , where [K :
Q] = n. Then I is a free Z-module of rank n, and generated by a Q-basis for K.
More precisely, I contains a Q-basis for K, and among such bases, any basis
whose discriminant has minimal absolute value freely generates I as a Z-module.
Proof. Let (α1 , . . . , αn ) be any Q-basis for K; we first claim there for each i, there
exists a non-zero di ∈ Z such that di αi ∈ OK . Indeed, it is easy to check that it
is enough to let di be the leading term of any integer polynomial satisfied by αi .
Thus, for any non-zero β ∈ I, we find that (βd1 α1 , . . . , βdn αn ) is a Q-basis for K
contained in I.
It remains to show that such a basis with minimal discriminant over Q is in
fact a Z-basis for I. Note that because the discriminant of elements
Pnin OK is an
integer, there is a minimal one. Take any α ∈ I, and write α = i=1 ai αi , for
ai ∈ Q. We want to show that ai ∈ Z for all i. Suppose not; without loss of
generality, we may assume a1 6∈ Z. Write a1 = m + ǫ, with 0 < ǫ < 1 and m = ⌊a1 ⌋,
and replace α1 by α′1 = α − mα1 , which is still in I. We then obtain a new
basis inside I, and the determinant of the transition matrix is ǫ, so by the changeof-basis formula of the previous section, this new basis has smaller discriminant,
contradicting minimality.
We can immediately conclude:
Corollary 1.2. We can define the discriminant DK of a number field K in terms
of any generators of OK over Z as in the proposition. It is a well-defined, non-zero
integer.
Proof. The existence of such a basis follows from the proposition, and the fact
that it is a well-defined, non-zero integer follows from the last three results of the
previous section, noting that the transition matrix between any two Z-bases must
have determinant ±1.
The discriminant will be a basic and important tool for studying number fields.
We also find:
Corollary 1.3. The field of fractions of OK is K.
The proposition also allows us to conclude the following important fact:
Corollary 1.4. Let I be a non-zero ideal in a ring of integers OK . Then OK /I is
finite.
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BRIAN OSSERMAN
Proof. Suppose we know that I contains some integer m ∈ Z. Then OK /I is a
quotient of OK /(m), which from the proposition is isomorphic as a Z-module to
(Z/mZ)n , hence both are finite. The corollary then follows from the easy lemma
below.
Lemma 1.5. Let R be an integral domain, and S an integral domain containing
R. Let I be an ideal of S, and suppose I contains a non-zero element α satisfying
a non-zero polynomial f(x) ∈ R[x]. Then I ∩ R 6= 0.
Proof. Because S is an integral domain, we may factor out any powers of x dividing
f(x), and can therefore assume that f(0) 6= 0. But α|(f(α) − f(0)), so −f(0) =
f(α) − f(0) ∈ I ∩ R, as desired.
This finiteness result will play a key role in the following lectures, and indeed in
all of algebraic number theory.
2. Dedekind domains
We are now ready to prove the following theorem:
Theorem 2.1. A ring of integers OK is a Dedekind domain: i.e., it satisfies
(i) OK is Noetherian;
(ii) Every non-zero prime ideal of OK is maximal;
(iii) OK is integrally closed in its field of fractions.
Algebraic Geometry Remark 2.2. From an algebraic geometry perspective, we see
that the definition of OK means that it has dimension one
√ and is normal; i.e., we
can think of it as a nonsingular curve. Subrings such as Z[ −3] are still curves, but
have singularities, and their containment in the ring of integers corresponds to the
normalization map. As in the geometric situation, in order to study the singular
curve it is frequently helpful to start by studying the normalization, so we focus
primarily on the rings of integers themselves.
Proof. The first two properties follow easily from Corollary 1.4. Recall that for OK
to be Noetherian means that any ascending chain of ideals stabilizes. But if I is a
non-zero ideal, OK /I is finite, and the ideals containing I in OK are in bijection
with the ideals of OK /I, hence there can only be finitely many.
Similarly, if p is a non-zero prime ideal of OK , then OK /p is a finite integral
domain, and it is a general fact that any finite integral domain is a field, so that
p is maximal. Indeed, let R be a finite integral domain, and a ∈ R a non-zero
element. Then we must have ak1 = ak2 for some k1 > k2 , by finiteness. Because R
is an integral domain, we find ak1−k2 = 1, so a is invertible, and R is a field.
Finally, we need to know that because OK is integrally closed over Z, it must be
integrally closed in K as well. Recall the lemma from the previous lecture stating
that α ∈ K is integral over Z if and only if there exists a non-zero finitely-generated
Z-module W ⊂ K such that αW ⊂ W , and that if α is integral, then Z[α] is such a
module. Now, suppose α ∈ K satisfies a polynomial xm +am−1 xm−1 +· · ·+a1 x+a0 ,
with ai ∈ OK for all i. Then consider the ring W := Z[{ai }i , α] ⊂ K as a Z-module.
Z[{ai }i ] is a finitely-generated Z-module because the ai are integral over Z, and W
is a finitely-generated Z[{ai }i ]-module for the same reason. It easily follows that W
is a finitely-generated Z-module by taking products of generators. Since αW ⊂ W ,
we find that α is integral over Z, as desired.
MATH 254A: RINGS OF INTEGERS AND DEDEKIND DOMAINS
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3. Main Properties of Dedekind domains
Dedekind domains have two main properties which we will exploit. The first is:
Theorem 3.1. Every non-zero ideal of a Dedekind domain may be uniquely factored
as a product of prime ideals, up to reordering.
The second requires a definition:
Definition 3.2. Let R be an integral domain with fraction field K. We say that
I ⊂ K is a fractional ideal of R if it is closed under addition and under scalar
multiplication by elements of R, and if there exists a non-zero d ∈ R such that
dI ⊂ R. A fractional ideal is principal if it is of the form αR, for some α ∈ K.
Given fractional ideals I, J of R, the product IJ is defined to be
X
{α ∈ L : α =
iℓ jℓ , iℓ ∈ I, jℓ ∈ J}
ℓ
The product of two fractional ideals is easily seen to be a fractional ideal.
Theorem 3.3. The set of fractional ideals of a Dedekind domain form a group
under multiplication.
We will prove these two theorems next week, and give applications.