# 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 $\U0001d51e$ 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 $\U0001d52d$ of $A$ is maximal. Choosing any $p\in \U0001d52d\setminus \{0\}$ 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

$${c}_{0}=-p\sum _{k=1}^{n}{c}_{k}{p}^{k-1}\in \U0001d52d\cap R,$$ |

so $\U0001d52d\cap R$ is a nonzero prime ideal in $R$ and is therefore a maximal ideal^{}. So,

$$R/(\U0001d52d\cap R)\to A/\U0001d52d$$ |

gives an algebraic extension^{} of the field $R/(\U0001d52d\cap R)$ to the integral domain^{} $A/\U0001d52d$. Therefore, $A/\U0001d52d$ is a field (see a condition of algebraic extension) and $\U0001d52d$ is a maximal ideal.

Title | proof of finite separable extensions of Dedekind domains are Dedekind |
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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 |