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Let $R$ be a Noetherian local ring and $S$ a regular local subring of $R$ such that $R$ is a finite $S$-module.

Question: Why is it true that every regular system of parameters of $S$ is a system of parameters of $R$?

Motivation: Proof of Prp. 2.2.11 in Bruns and Herzog, CMR.

Manos
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1 Answers1

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If $R$ is a finite $S$-module, then the ring extension $S \subseteq R$ is integral. This already implies that $\dim S = \dim R$ (this doesn't use $R, S$ local, or $S$ regular), and also implies the result about systems of parameters (and the hypothesis that $S$ is regular is not needed), as follows: $\newcommand{\n}{\mathfrak{n}}$ $\newcommand{\m}{\mathfrak{m}}$ $\newcommand{\ht}{\text{ht}}$

Suppose $(S, \n) \subseteq (R, \m)$, $\dim S = d$. If $a_1, ..., a_d \in \n$ is an sop in $S$, then $I := (a_1,...,a_d)S$ is $\n$-primary. Let $p$ be a minimal prime of $IR$. Now $p$ contracts to a prime of $S$ containing $a_1,...,a_d$, so $p \cap S = \n$. By incomparability of primes in an integral extension, $p = \m$, so $a_1, ..., a_d$ is an sop of $R$.

zcn
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  • Nice answer! So when we say that $(S,n)$ is a local subring of $(R,m)$ we mean that $m \cap S =n$? – Manos Jan 28 '14 at 16:30
  • Yes, it means the inclusion $S \subseteq R$ is a morphism of local rings, i.e. $nR \subseteq m$ (which is equivalent to what you said). – zcn Jan 28 '14 at 18:05