Dedekind Domain - Some Examples of Dedekind Domains

Some Examples of Dedekind Domains

All principal ideal domains and therefore all discrete valuation rings are Dedekind domains.

The ring of algebraic integers in a number field K is Noetherian, integrally closed, and of dimension one (to see the last property, observe that for any nonzero ideal I of R, R/I is finite and recall that a finite integral domain is a field), so by (DD4) R is a Dedekind domain. As above, this includes all the examples considered by Kummer and Dedekind and was the motivating case for the general definition, and these remain among the most studied examples.

The other class of Dedekind rings which is arguably of equal importance comes from geometry: let C be a nonsingular geometrically integral affine algebraic curve over a field k. Then the coordinate ring k of regular functions on C is a Dedekind domain. Indeed, this is essentially an algebraic translation of these geometric terms: the coordinate ring of any affine variety is, by definition, a finitely generated k-algebra, so Noetherian; moreover curve means dimension one and nonsingular implies (and, in dimension one, is equivalent to) normal, which by definition means integrally closed.

Both of these constructions can be viewed as special cases of the following basic result:

Theorem: Let R be a Dedekind domain with fraction field K. Let L be a finite degree field extension of K and denote by S the integral closure of R in L. Then S is itself a Dedekind domain.

Applying this theorem when R is itself a PID gives us a way of building Dedekind domains out of PIDs. Taking R = Z this construction tells us precisely that rings of integers of number fields are Dedekind domains. Taking R = k gives us the above case of nonsingular affine curves.

Zariski and Samuel were sufficiently taken by this construction to pose as a question whether every Dedekind domain arises in such a fashion, i.e., by starting with a PID and taking the integral closure in a finite degree field extension. A surprisingly simple negative answer was given by L. Claborn.

If the situation is as above but the extension L of K is algebraic of infinite degree, then it is still possible for the integral closure S of R in L to be a Dedekind domain, but it is not guaranteed. For example, take again R = Z, K = Q and now take L to be the field of all algebraic numbers. The integral closure is nothing else than the ring of all algebraic integers. Since the square root of an algebraic integer is again an algebraic integer, it is not possible to factor any nonzero nonunit algebraic integer into a finite product of irreducible elements, which implies that is not Noetherian! In general, the integral closure of a Dedekind domain in an infinite algebraic extension is a Prüfer domain; it turns out that the ring of algebraic integers is slightly more special than this: it is a Bézout domain.