Today’s (quick) topic focuses on Dedekind domains. These come up when you take the ring of integers in any finite extension of ${\mathbb{Q}}$ (i.e. number fields). In these, you don’t necessarily have unique factorization. But you do have something close, which makes these crucial.

Definition 1 A Dedekind domain is a Noetherian integral domain ${A}$ that is integrally closed, and of Krull dimension one—that is, each nonzero prime ideal is maximal.

A DVR is a Dedekind domain, and the localization of a Dedekind domain at a nonzero prime is a DVR by this. Another example (Serre) is to take a nonsingular affine variety ${V}$ of dimension 1 and consider the ring of globally regular functions ${k[V]}$; the localizations at closed points are DVRs, so the ring is a Dedekind domain.

Now assume ${A}$ is Dedekind.

A f.g. ${A}$-submodule of the quotient field ${F}$ is called a fractional ideal; by multiplying by some element of ${A}$, we can always pull a fractional ideal into ${A}$, when it becomes an ordinary ideal. The sum and product of two fractional ideals is a fractional ideal.

Theorem 2 (Invertibility) If ${I}$ is a nonzero fractional ideal and ${ I^{-1} := \{ x \in F: xI \subset A \}}$, then ${I^{-1}}$ is a fractional ideal and ${I I^{-1} = A}$

Thus, the nonzero fractional ideals are an abelian group under multiplication.

To see this, note that invertibility is preserved under localization: for a multiplicative set ${S}$, we have ${S^{-1} ( I^{-1} ) = (S^{-1} I)^{-1}}$, where the second ideal inverse is with respect to ${S^{-1}A}$; this follows from the fact that ${I}$ is finitely generated. Note also that invertibility is true for discrete valuation rings.

So for all primes ${\mathfrak{p}}$, we have ${(I I^{-1})_{\mathfrak{p}} = A_{\mathfrak{p}}}$, which means the inclusion of ${A}$-modules ${I I^{-1} \rightarrow A}$ is an isomorphism at each localization. Therefore it is an isomorphism, by general algebra.

The next result says we have unique factorization of ideals:

Theorem 3 (Factorization) Each ideal ${I \subset A}$ can be written uniquely as a product of powers of prime ideals.

Let’s use the pseudo-inductive argument. Let ${I}$ be the maximal ideal which can’t be written in such a manner, since ${A}$ is Noetherian. Then ${I}$ isn’t prime, so it’s contained in some prime ${\mathfrak{p}}$. But ${I = (I\mathfrak{p}^{-1})\mathfrak{p}}$, and ${I\mathfrak{p}^{-1} \neq I}$ can be written as a product of primes, contradiction.

Uniqueness follows by localizing at each prime.