# Dedekind domain

A **Dedekind domain** is a integral domain satisfying the following properties:

- is a noetherian ring.
- Every prime ideal of is a maximal ideal.
- is integrally closed in its field of fractions.

Dedekind domains are very important in abstract algebra and number theory. For example, the ring of integers of any number field is a Dedekind domain.

There are several very nice properties of Dedekind domains:

- Dedekind domains have unique prime factorizations of ideals (but not necessarily of elements).

There are also various properties of homological importance that Dedekind domains satisfy.

## Invertibility of Ideals

Let be a Dedekind domain with field of fractions , and let be any nonzero fractional ideal of . We call **invertible** if there is a fractional ideal such that . We shall show that all fractional ideals of are invertible.

Given any nonzero fractional ideal of define . is clearly an -module. Moreover, for any nonzero (such an alpha clearly exists, if for then ) we have by the definition of , and so must be a fractional ideal of . It follows that is a fractional ideal of as well, let . By definition, , and so is an integral ideal. We claim that in fact , and so is invertible.

We will need the following lemmas.

**Lemma 1:** Every nonzero integral ideal of contains a product of prime ideals (counting as the empty product).

*Proof:* Assume that this is not the case. Let be the collection of integral ideals of not containing a product of prime ideals, so is nonempty and . As is noetherian, must have a maximal element, say . Clearly cannot be prime (otherwise it would contain itself), so there must be with but . But then , and so and contain products of prime ideals. But then also contains a product of prime ideals, contradicting the choice of .

**Lemma 2:** For any proper integral ideal , there is some $\gamma\in K\sm R$ (Error compiling LaTeX. ! Undefined control sequence.) for which .

*Proof:* Take any nonzero . By Lemma 1, contains a product of prime ideals, say with minimal (i.e. does not contain a product of prime ideals). As , . As is a proper ideal, it must be contained in some maximal ideal, . Since maximal ideals are prime in commutative rings, is prime. But now . Thus as is prime, for some (if is prime and are ideals with then either or ). But as is a Dedekind domain, must be maximal, so . Now assume WLOG that . By the minimality of , . Take any $b\in P_1\cdots P_{n-1}\sm (a)$ (Error compiling LaTeX. ! Undefined control sequence.) let . We claim that this is the desired .

First if then , a contradiction, so . Now for any , , and so for some . But now , and so , as required.

Now we return to the main proof. Assume that . Then by Lemma 2, there is some $\gamma\in K\sm R$ (Error compiling LaTeX. ! Undefined control sequence.) for which . By the definition of , for any we have and so . It follows that . We claim that this implies (contradicting the choice of ).

Indeed, the map is an -linear map from . As is noetherian, must be a finitely generated -module. Indeed, for some nonzero , must be an integral ideal of , which is finitely generated by the definition of noetherian rings. But if then , so is finitely generated as well. Now take , and let be the matrix representation of with respect to . Then is an matrix with coefficients in and we have: and so is an eigenvalue of . But then is a root of the characteristic polynomial, of . But as has all of its entries in , is a monic polynomial in . Thus as is integrally closed in , .

This is a contradiction, and so we must have , and so is indeed invertible.

In fact, the converse is true as well: if all nonzero ideals are invertible, then is a Dedekind domain. This is sometimes used as a definition of Dedekind domains.