# proof of finite separable extensions of Dedekind domains are Dedekind

Let $R$ be a Dedekind domain with field of fractions $K$ and $L/K$ be a finite (http://planetmath.org/FiniteExtension) separable extension of fields. We show that the integral closure $A$ of $R$ in $L$ is also a Dedekind domain. That is, $A$ is Noetherian (http://planetmath.org/Noetherian), integrally closed and every nonzero prime ideal is maximal (http://planetmath.org/MaximalIdeal).

First, as integral closures are themselves integrally closed, $A$ is integrally closed. Second, as integral closures in separable extensions are finitely generated, $A$ is finitely generated as an $R$-module. Then, any ideal $\mathfrak{a}$ of $A$ is a submodule of $A$, so is finitely generated as an $R$-module and therefore as an $A$-module. So, $A$ is Noetherian.

It only remains to show that a nonzero prime ideal $\mathfrak{p}$ of $A$ is maximal. Choosing any $p\in\mathfrak{p}\setminus\{0\}$ there is a nonzero polynomial

 $f=\sum_{k=0}^{n}c_{k}X^{k}$

for $c_{k}\in R$, $c_{0}\not=0$ and such that $f(p)=0$. Then

 $c_{0}=-p\sum_{k=1}^{n}c_{k}p^{k-1}\in\mathfrak{p}\cap R,$

so $\mathfrak{p}\cap R$ is a nonzero prime ideal in $R$ and is therefore a maximal ideal. So,

 $R/(\mathfrak{p}\cap R)\rightarrow A/\mathfrak{p}$

gives an algebraic extension of the field $R/(\mathfrak{p}\cap R)$ to the integral domain $A/\mathfrak{p}$. Therefore, $A/\mathfrak{p}$ is a field (see a condition of algebraic extension) and $\mathfrak{p}$ is a maximal ideal.

Title proof of finite separable extensions of Dedekind domains are Dedekind ProofOfFiniteSeparableExtensionsOfDedekindDomainsAreDedekind 2013-03-22 18:35:36 2013-03-22 18:35:36 gel (22282) gel (22282) 4 gel (22282) Proof msc 13F05 msc 13A15