# Dedekind Domains

Recall (Corollary 2.4.6) that we proved that the ring of integers of a number field is noetherian, as follows. 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 field of fractions of  is the field of all equivalence classes of formal quotients , where with , and if . For example, the field of fractions of  is (canonically isomorphic to) and the field of fractions of is . The field of fractions of the ring of integers of a number field  is just the number field .

Example 3.1.1   We compute the fraction fields mentioned above.
sage: Frac(ZZ)
Rational Field

In Sage the Frac command usually returns a field canonically isomorphic to the fraction field (not a formal construction).
sage: K.<a> = QuadraticField(5)
sage: OK = K.ring_of_integers(); OK
Order with module basis 1/2*a + 1/2, a in Number Field
in a with defining polynomial x^2 - 5
sage: Frac(OK)
Number Field in a with defining polynomial x^2 - 5


The fraction field of an order - i.e., a subring of of finite index - is also the number field again.

sage: O2 = K.order(2*a); O2
Order with module basis 1, 2*a in Number Field
in a with defining polynomial x^2 - 5
sage: Frac(O2)
Number Field in a with defining polynomial x^2 - 5


Definition 3.1.2 (Integrally Closed)   An integral domain is integrally closed in its field of fractions if whenever is in the field of fractions of and satisfies a monic polynomial , then .

Proposition 3.1.3   If is any number field, then is integrally closed. Also, the ring of all algebraic integers (in a fixed choice of ) 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  .

Without loss we may assume that , so that . Suppose is integral over . Then since is integrally closed,  is an element of , so , as required.

Definition 3.1.4 (Dedekind Domain)   An integral domain  is a Dedekind domain if it is noetherian, integrally closed in its field of fractions, and every nonzero prime ideal of is maximal.

The ring 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 an integral domain, it is trivially integrally closed in itself, and there are no nonzero prime ideals so the condition that they be maximal is empty. The ring is not noetherian, but it is integrally closed in its field of fraction, and every nonzero prime ideal is maximal.

Proposition 3.1.5   The ring of integers of a number field is a Dedekind domain.

Proof. By Proposition 3.1.3, the ring is integrally closed, and by Proposition 2.4.6 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 (see Exercise 10), 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 Exercise 11).

Definition 3.1.6 (Fractional Ideal)   A fractional ideal is a nonzero -submodule  of  that is finitely generated as an -module.

We will sometimes call a genuine ideal an integral ideal. The notion of fractional ideal makes sense for an arbitrary Dedekind domain - it is an -module that is finitely generated as an -module.

Example 3.1.7   We multiply two fractional ideals in SAGE:
sage: K.<a> = NumberField(x^2 + 23)
sage: I = K.fractional_ideal(2, 1/2*a - 1/2)
sage: J = I^2
sage: I
Fractional ideal (2, 1/2*a - 1/2) of Number Field ...
sage: J
Fractional ideal (4, 1/2*a + 3/2) of Number Field ...
sage: I*J
Fractional ideal (-1/2*a - 3/2) of Number Field ...


Since fractional ideals are finitely generated, we can clear denominators of a generating set to see that there is some nonzero such that

with an integral ideal. Thus dividing by , we see that every fractional ideal is of the form

for some and integral ideal .

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

Theorem 3.1.8   The set of fractional ideals of a Dedekind domain  is an abelian group under ideal multiplication with identity element .

Note that fractional ideals are nonzero by definition, so it is not necessary to write nonzero fractional ideals'' in the statement of the theorem. We will only prove Theorem 3.1.8 in the case when is the ring of integers of a number field . Before proving Theorem 3.1.8 we prove a lemma. For the rest of this section is the ring of integers of a number field .

Definition 3.1.9 (Divides for Ideals)   Suppose that are ideals of . Then we say that  divides  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 3.1.10   Suppose  is a nonzero ideal of . Then there exist prime ideals such that , i.e.,  divides a product of prime ideals.

Proof. Let be the set of nonzero ideals of that do not satisfy the conclusion of the lemma. The key idea is to use that is noetherian to show that is the empty set. If  is nonempty, then since 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 we may assume that  is not prime. Thus 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, say and . Then

so contains a product of primes. This is a contradiction, since we assumed . Thus  is empty, which completes the proof.

We are now ready to prove the theorem.

Proof. [Proof of Theorem 3.1.8] Note that we will only prove Theorem 3.1.8 in the case when is the ring of integers of a number field .

The product of two fractional ideals is again finitely generated, so it is a fractional ideal, and for any 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 all nonzero integral ideals have inverses, and finally observe that every fractional ideal has an inverse. (Note: Once we know that the set of fractional ideals is a group, it will follows that inverses are unique; until then we will be careful to write an'' instead of the''.)

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 since  is maximal, either or . If , we are done since then  is an inverse of  . Thus suppose that . Our strategy is to show that there is some , with . Since , such a  would leave  invariant, i.e., . Since is an -module we will see that it will follow that , a contradiction.

By Lemma 3.1.10, we can choose a product , with  minimal, with

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  (because satisfies the minimal polynomial of , by the Cayley-Hamilton theorem - here we also use that , since is a finite set). This implies that  is integral over  , so , since  is integrally closed by Proposition 3.1.3. (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 3.1.6, 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 an inverse of  , we see that each element of preserves the finitely generated -module  and is hence integral. But then , which, upon multiplying both sides by , implies that , a contradiction. Thus . Because is maximal among ideals that do not have an inverse, the ideal does have an inverse . Then is an inverse of , since

We can finally deduce the crucial Theorem 3.1.11, 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 ideals to mathematics over a century ago.

Theorem 3.1.11   Suppose is a nonzero integral ideal of . Then can be written as a product

of prime ideals of , and this representation is unique up to order.

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 3.1.8 we can cancel from both sides of this equation to see that , a contradiction. Since , we have , and by the above observation  is strictly contained in . 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 , which is a subset of , which contradicts that is a prime ideal. Thus for some . We can thus cancel and from both sides of the equation by multiplying both sides by the inverse. Repeating this argument finishes the proof of uniqueness.

Theorem 3.1.12   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 3.1.11 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 3.1.11

Example 3.1.13   The ring of integers of is . 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 3.1.12, however, implies that the principal ideal can, however, be written uniquely as a product of prime ideals. An explicit decomposition is

 (3.1)

where each of the ideals and is prime. We will discuss algorithms for computing such a decomposition in detail in Chapter 4. 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 finite (as a set) ring .

The factorization (3.1.1) can be compute using SAGE as follows:

sage: K.<a> = NumberField(x^2 + 6); K
Number Field in a with defining polynomial x^2 + 6
sage: K.factor_integer(6)
(Fractional ideal (2, a) of Number Field ...)^2 *
(Fractional ideal (3, a) of Number Field ...)^2


William Stein 2012-09-24