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 $𝔞$ 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 $𝔭$ of $A$ is maximal. Choosing any $p\in 𝔭\setminus \{0\}$ there is a nonzero polynomial

$$f=\sum _{k=0}^n{c}_k{X}^k$$

for $c_k\in R$, $c_0\ne 0$ and such that $f(p)=0$. Then

$$c_0=-p\sum _{k=1}^n{c}_k{p}^{k-1}\in 𝔭\cap R,$$

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

$$R/(𝔭\cap R)\to A/𝔭$$

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