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
Canonical name	ProofOfFiniteSeparableExtensionsOfDedekindDomainsAreDedekind
Date of creation	2013-03-22 18:35:36
Last modified on	2013-03-22 18:35:36
Owner	gel (22282)
Last modified by	gel (22282)
Numerical id	4
Author	gel (22282)
Entry type	Proof
Classification	msc 13F05
Classification	msc 13A15