Dedekind Domains, 11 Definitions

11 Definitions

The following conditions on an integral domain R, with a fraction field F, are equivalent.

R is dedekind. That is, every nonzero proper ideal in R is uniquely a product of finitely many prime ideals.
Every nonzero proper fractional ideal in R is uniquely a product of finitely many prime ideals or their inverses.
R is noetherian and every nonzero maximal ideal is invertible.
Every nonzero prime ideal in R is invertible.
Every nonzero ideal in R is invertible.
Every nonzero fractional ideal of R is invertible.
Every ideal in R is projective.
Every fractional ideal of R is projective.
R is noetherian and integrally closed, and every nonzero prime ideal is maximal.
R is noetherian, and each localization R_P is a discrete valuation ring.
R is noetherian, and every nonzero prime ideal is maximal, and every nonzero primary ideal is a prime power.

An earlier theorem takes care of 1 → 2, and 2 → 1 is obvious.

An earlier theorem takes care of: 1 → 10 → 5 → 4 → 3 → 1, so these are all equivalent.

Since you can always flip the exponents on the prime factors of a fractional ideal, 2 → 6. And 6 → 5 is trivial, so 6 is brought into the fold.

Invertible is the same as projective, so 8 ⇔ 6 and 7 ⇔ 5.

5 implies 9

every ideal is invertible, and finitely generated, hence R is noetherian. We already know the prime ideals are maximal, so we only need show R is integrally closed.

If x is integral over R, then R[x] is a finitely generated R module inside F. Thus R[x] is a fractional ideal, and inside a dedekind domain, every fractional ideal is invertible.

Consider R[x]*R[x], the product of the two fractional ideals, and remember that R[x] contains 1. The product is simply R[x]. If R[x]′ is the inverse of R[x], we can write the following.

R[x] = R*R[x] = R[x]′*R[x]*R[x] = R[x]′*R[x] = R

This shows x lies in R, and R is integrally closed.

9 implies 10

Show 9 → 10, and that will make 9 equivalent to the others.

Let P be a prime ideal and let R_P be the localization of R about P. By preimage correspondence, an ascending chain in R_P pulls back to an ascending chain in R, therefore R_P is noetherian.

The property of being integrally closed also survives localization, so R_P is integrally closed.

Prime and maximal ideals are synonymous, so there are no prime ideals properly contained in P. Apply prime correspondence, and there are no prime ideals of the ring R_P inside the maximal ideal P_P. In other words, R_P has but one prime ideal. This satisfies the conditions of an earlier theorem, hence every R_P is a valuation ring.

That completes 5 → 9 → 10.

11

condition 11 deals with primary ideals and laskerian rings. If you are not familiar with this topic, you might want to skip ahead to the next section.

Let R be a noetherian integral domain where all nonzero prime ideals are maximal. If R were integrally closed it would be dedekind by condition 9. Instead, we are given condition 11: every nonzero primary ideal is a prime power.

Since R is noetherian it is laskerian. Let J be a nonzero proper ideal and write it as the finite intersection of primary ideals. Since all primes are maximal, the intersection equals the product. Each primary ideal is a prime power, hence J is a finite product of prime ideals.

Suppose the representation is not unique. Some other product yields J, hence some other intersection yields J. By first uniqueness, the prime ideals used to build J have to be the same. Since these primes are maximal they are all isolated in σ(J). Apply second uniqueness, and the primary ideals are fixed. The representation of J is unique, and R is dedekind.

Now for 11 → 1. Let H be a P primary ideal in a dedekind domain R. Of course H is a finite product of prime ideals. The radical of this product is the radical of the intersection. This in turn is the intersection of the radicals. Now rad(H) = P, which is the intersection of all the primes that divide H. Thus P lies in each prime factor of H, and since P is maximal, H is a power of P.

Conversely let H = Pⁿ. With no zeero divisors, the image of Pⁿ in the local ring R_P is the image of P, raised to the n^th power. There are no other ideals in R that lead to proper ideals in R_P. Thus the preimage of the image of Pⁿ, i.e. the saturation of Pⁿ, is Pⁿ. In addition, the saturation of Pⁿ is P primary. Therefore H is P primary iff H = Pⁿ.

That takes care of condition 11.

Fraction Ring

If R is dedekind and S is a multiplicatively closed set, R/S is dedekind.

Let H/S be a proper ideal in R/S. Since H is invertible in R, H/S is invertible in R/S. All ideals are invertible, and by condition 5, R/S is dedekind.

The prime ideals of R/S correspond to the primes of R that miss S.