## Dedekind Domains

A Dedekind domain is an integral domain $$R$$ such that

1. Every ideal is finitely generated.

2. Every nonzero prime ideal is a maximal ideal.

3. $$R$$ is integrally closed in its field of fractions.

The last condition means that if $$\alpha/\beta \in K$$ is a root of a monic polynomial over $$R$$, then $$\alpha/\beta \in R$$, that is, $$\beta | \alpha$$ in $$R$$.

The first condition is equivalent to both of the following:

• Every increasing sequence of ideals is eventually constant, that is, given $$I_1 \subset I_2 \subset I_3 ...$$, there exists some $$k$$ for which $$I_k = I_n$$ for all $$n \ge k$$.

• Every nonempty set $$S$$ of ideals has a maximal member, that is, there exists $$M \in S$$ such that $$M\subset I \in S \implies M = I$$.

A ring satisfying these conditions is called a Noetherian ring.

It is not difficult to show that these three conditions are equivalent. First suppose we are given some sequence $$I_1 \subset I_2 \subset I_3 ...$$. Then consider the ideal generated by all the ideals in the sequence. If it is finitely generated, then for some $$k$$, $$I_k$$ contains all the generators, and thus $$I_k = I_n$$ for all $$n \ge k$$.

Now suppose we are given a nonempty set $$S$$ of ideals. Then take any ideal $$I_1 \in S$$, and look for an ideal $$I_2$$ that strictly contains $$I_1$$. Iterating this procedure produces a sequence $$I_1 \subset I_2 \subset ...$$. If there exists some $$k$$ for which $$n \ge k$$ implies $$I_k = I_n$$, then $$I_k$$ is a maximal ideal in $$S$$.

Next suppose we are given some ideal $$I$$. Consider the set $$S$$ of subideals of $$I$$ that are finitely generated. If it has a maximal element $$I'$$, then we must have $$I = I'$$.

Theorem: Every number ring is a Dedekind domain.

Proof: Since a number ring is a free abelian group of finite rank, any ideal must also be a free abelian group of finite rank (because it is a additive subgroup) thus every ideal is finitely generated.

For the second condition, it suffices to show that for every nonzero prime ideal $$P$$, the integral domain $$R/P$$ is a field. We do so by showing $$R/P$$ is finite, which implies it is a field because every finite integral domain is a field (because for any element $$a$$, we must have $$a^n = 1$$ for some $$n$$, thus $$a^{n-1}$$ is the inverse of $$a$$).

We shall show that $$R/I$$ is finite for any nonzero ideal $$I$$. Take any nonzero $$\alpha\in I$$, and let $$m = N^K(\alpha)$$ where $$K$$ is the number field of $$R$$. So $$m = \alpha\beta$$ where $$\beta$$ is the product of the conjugates of $$\alpha$$. Hence $$\beta$$ is also be an algebraic integer. Now $$m$$ is a nonzero integer, so $$\beta= m /\alpha \in K$$, so we deduce $$\beta\in R$$ which means $$m \in I$$. Now $$R/(m)$$ is finite (it contains $$m^n$$ elements), hence $$R/I$$ is finite (its order divides $$m^n$$).

Lastly, suppose $$\alpha$$ is a root of a monic polynomial $$f(x) = a_0 +...+ a_{n-1}x^{n-1} +x^n$$ over $$R$$. Now $$R=\mathbb{Z}[a_0,...,a_{n-1}]$$ is clearly finitely generated, and since $$\mathbb{Z}[a_0,...,a_{n-1},\alpha]$$ is also finitely generated since any expression containing $$\alpha^m$$ for $$m \ge n$$ can be rewritten using smaller powers of $$\alpha$$. Hence $$R$$ is integrally closed.

Ben Lynn blynn@cs.stanford.edu 💡