# Dedekind Domains

Recall (Corollary 5.2.5) that we proved that the ring of integers of a number field is Noetherian. As we saw before using norms, the ring is finitely generated as a module over  , so it is certainly finitely generated as a ring over  . By the Hilbert Basis Theorem, is Noetherian.

If is an integral domain, the of is the field of all elements , where . The field of fractions of is the smallest field that contains . For example, the field of fractions of is and of is .

Definition 6.1.1 (Integrally Closed)   An integral domain is if whenever is in the field of fractions of and satisfies a monic polynomial , then .

Proposition 6.1.2   If is any number field, then is integrally closed. Also, the ring of all algebraic integers is integrally closed.

Proof. We first prove that  is integrally closed. Suppose is integral over  , so there is a monic polynomial with and . The all lie in the ring of integers of the number field , and is finitely generated as a -module, so is finitely generated as a -module. Since , we can write as a -linear combination of for , so the ring is also finitely generated as a -module. Thus is finitely generated as -module because it is a submodule of a finitely generated -module, which implies that  is integral over  .

Suppose is integral over . Then since is integrally closed,  is an element of , so , as required.

Definition 6.1.3 (Dedekind Domain)   An integral domain  is a if it is Noetherian, integrally closed in its field of fractions, and every nonzero prime ideal of is maximal.

The ring is Noetherian, integrally closed in its field of fractions, and the two prime ideals are maximal. However, it is not a Dedekind domain because it is not an integral domain. The ring is not a Dedekind domain because it is not integrally closed in its field of fractions, as is integrally over and lies in , but not in . The ring is a Dedekind domain, as is any ring of integers of a number field, as we will see below. Also, any field is a Dedekind domain, since it is a domain, it is trivially integrally closed in itself, and there are no nonzero prime ideals so that condition that they be maximal is empty.

Proposition 6.1.4   The ring of integers of a number field is a Dedekind domain.

Proof. By Proposition 6.1.2, the ring is integrally closed, and by Proposition 5.2.5 it is Noetherian. Suppose that  is a nonzero prime ideal of . Let be a nonzero element, and let be the minimal polynomial of . Then

so . Since is irreducible, is a nonzero element of that lies in  . Every element of the finitely generated abelian group is killed by , so is a finite set. Since  is prime, is an integral domain. Every finite integral domain is a field, so is maximal, which completes the proof.

If and are ideals in a ring , the product is the ideal generated by all products of elements in with elements in :

Note that the set of all products , with and , need not be an ideal, so it is important to take the ideal generated by that set. (See the homework problems for examples.)

Definition 6.1.5 (Fractional Ideal)   A is an -submodule of that is finitely generated as an -module.

To avoid confusion, we will sometimes call a genuine ideal an . Also, since fractional ideals are finitely generated, we can clear denominators of a generating set to see that every fractional ideal is of the form for some and ideal .

For example, the collection of rational numbers with denominator or is a fractional ideal of .

Theorem 6.1.6   The set of nonzero fractional ideals of a Dedekind domain  is an abelian group under ideal multiplication.

Before proving Theorem 6.1.6 we prove a lemma. For the rest of this section is the ring of integers of a number field .

Definition 6.1.7 (Divides for Ideals)   Suppose that are ideals of . Then    if .

To see that this notion of divides is sensible, suppose , so . Then and for some integer and , and divides means that , i.e., that there exists an integer  such that , which exactly means that divides , as expected.

Lemma 6.1.8   Suppose  is an ideal of . Then there exist prime ideals such that . In other words,  divides a product of prime ideals. (By convention the empty product is the unit ideal. Also, if , then we take , which is a prime ideal.)

Proof. The key idea is to use that is Noetherian to deduce that the set  of ideals that do not satisfy the lemma is empty. If  is nonempty, then because is Noetherian, there is an ideal that is maximal as an element of . If  were prime, then  would trivially contain a product of primes, so  is not prime. By definition of prime ideal, there exists such that but and . Let and . Then neither nor is in , since is maximal, so both and contain a product of prime ideals. Thus so does , since

which is a contradiction. Thus  is empty, which completes the proof.

We are now ready to prove the theorem.

Proof. [Proof of Theorem 6.1.6] The product of two fractional ideals is again finitely generated, so it is a fractional ideal, and for any nonzero ideal , so to prove that the set of fractional ideals under multiplication is a group it suffices to show the existence of inverses. We will first prove that if is a prime ideal, then has an inverse, then we will prove that nonzero integral ideals have inverses, and finally observe that every fractional ideal has an inverse.

Suppose is a nonzero prime ideal of . We will show that the -module

is a fractional ideal of such that , so that is an inverse of .

For the rest of the proof, fix a nonzero element . Since  is an -module, is an ideal, hence  is a fractional ideal. Since we have , hence either or . If , we are done since then  is an inverse of  . Thus suppose that . Our strategy is to show that there is some not in ; such a  would leave  invariant (i.e., ), so since is an -module it will follow that , a contradiction.

By Lemma 6.1.8, we can choose a product , with  minimal, such that

If no is contained in , then we can choose for each an with ; but then , which contradicts that is a prime ideal. Thus some , say , is contained in , which implies that since every nonzero prime ideal is maximal. Because  is minimal, is not a subset of , so there exists that does not lie in . Then , so by definition of we have . However, , since if it were then would be in . We have thus found our element that does not lie in . To finish the proof that has an inverse, we observe that preserves the -module , and is hence in , a contradiction. More precisely, if is a basis for as a -module, then the action of  on  is given by a matrix with entries in , so the minimal polynomial of has coefficients in . This implies that is integral over , so , since is integrally closed by Proposition 6.1.2. (Note how this argument depends strongly on the fact that is integrally closed!)

So far we have proved that if is a prime ideal of , then is the inverse of in the monoid of nonzero fractional ideals of . As mentioned after Definition 6.1.5 [on Tuesday], every nonzero fractional ideal is of the form for and an integral ideal, so since has inverse , it suffices to show that every integral ideal  has an inverse. If not, then there is a nonzero integral ideal  that is maximal among all nonzero integral ideals that do not have an inverse. Every ideal is contained in a maximal ideal, so there is a nonzero prime ideal such that . Multiplying both sides of this inclusion by and using that , we see that . If , then arguing as in the proof that is the inverse of , we see that each element of preserves the finitely generated -module  and is hence integral. But then , which implies that , a contradiction. Thus . Because is maximal among ideals that do not have an inverse, the ideal does have an inverse, call it . Then is the inverse of , since .

We can finally deduce the crucial Theorem 6.1.10, which will allow us to show that any nonzero ideal of a Dedekind domain can be expressed uniquely as a product of primes (up to order). Thus unique factorization holds for ideals in a Dedekind domain, and it is this unique factorization that initially motivated the introduction of rings of integers of number fields over a century ago.

Theorem 6.1.9   Suppose is an integral ideal of . Then can be written as a product

of prime ideals of , and this representation is unique up to order. (Exception: If , then the representation is not unique.)

Proof. Suppose is an ideal that is maximal among the set of all ideals in that can not be written as a product of primes. Every ideal is contained in a maximal ideal, so is contained in a nonzero prime ideal . If , then by Theorem 6.1.6 we can cancel from both sides of this equation to see that , a contradiction. Thus is strictly contained in , so by our maximality assumption on there are maximal ideals such that . Then , a contradiction. Thus every ideal can be written as a product of primes.

Suppose . If no is contained in , then for each there is an such that . But the product of the is in the , which is a subset of , which contradicts the fact that is a prime ideal. Thus for some . We can thus cancel and from both sides of the equation. Repeating this argument finishes the proof of uniqueness.

Corollary 6.1.10   If is a fractional ideal of then there exists prime ideals and , unique up to order, such that

Proof. We have for some and integral ideal . Applying Theorem 6.1.10 to , , and gives an expression as claimed. For uniqueness, if one has two such product expressions, multiply through by the denominators and use the uniqueness part of Theorem 6.1.10

Example 6.1.11   The ring of integers of is . In , we have

If , with and neither a unit, then , so without loss and . If , then ; since the equation has no solution with , there is no element in with norm , so is irreducible. Also, is not a unit times  or times , since again the norms would not match up. Thus can not be written uniquely as a product of irreducibles in . Theorem 6.1.9, however, implies that the principal ideal can, however, be written uniquely as a product of prime ideals. Using we find such a decomposition:
> R<x> := PolynomialRing(RationalField());
> K := NumberField(x^2+6);
> OK := MaximalOrder(K);
> [K!b : b in Basis(OK)];
[
1,
K.1    // this is sqrt(-6)
]
> Factorization(6*OK);
[
<Prime Ideal of OK
Two element generators:
[2, 0]
[2, 1], 2>,
<Prime Ideal of OK
Two element generators:
[3, 0]
[3, 1], 2>
]

The output means that

where each of the ideals and is prime. I will discuss algorithms for computing such a decomposition in detail, probably next week. The first idea is to write , and hence reduce to the case of writing the , for prime, as a product of primes. Next one decomposes the Artinian ring as a product of local Artinian rings.

William Stein 2004-05-06