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.
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 the following
Theorem: All fractional ideals of are invertible.
Proof: 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 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 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 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.
This result has a number of important applications.
First, as we clearly have for all fractional ideals , the set of fractional ideals, , of becomes an abelian group under multiplication. The identity element is and the existence of inverses is guaranteed by the above result. The set of principal ideal fractional ideals, forms a subgroup of this group. The quotient group is is then the ideal class group of , a object of great importance in algebraic number theory.
We may define divisibility of ideals in the "obvious" way. Namely for fractional ideals we say that iff there is some integral ideal for which . The above result gives a useful characterization of divisibility.
Theorem: For fractional ideals of , iff .
Proof: First, if , then there is some for which . But this gives . Conversely assume that . Consider the fractional ideal . We have . Thus is a fractional ideal and . It follows that is an integral ideal of . But now and so .
As an application of this, we get that and for all ideals and .