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 💡