Lemma 10.106.2 (00NP)—The Stacks project

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Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.106: Regular local rings
Lemma 10.106.2 (cite)

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Lemma 10.106.2. Any regular local ring is a domain.

Proof. We will use that $\bigcap \mathfrak m^ n = 0$ by Lemma 10.51.4. Let $f, g \in R$ such that $fg = 0$ . Suppose that $f \in \mathfrak m^ a$ and $g \in \mathfrak m^ b$ , with $a, b$ maximal. Since $fg = 0 \in \mathfrak m^{a + b + 1}$ we see from the result of Lemma 10.106.1 that either $f \in \mathfrak m^{a + 1}$ or $g \in \mathfrak m^{b + 1}$ . Contradiction. $\square$

Comments (2)

Comment #8848 by Et on March 09, 2024 at 10:42

Proposed simpler proof: $\bigcap \mathfrak{m}^n =0$ by lemma 00IP, hence the map $R \rightarrow \bigoplus \mathfrak{m}^n / \mathfrak{m}^{n+1}$ is an inclusion. By lemma 00NO $\bigoplus \mathfrak{m}^n / \mathfrak{m}^{n+1}$ is an integral domain, and hence so is $R$ .

Comment #9238 by Stacks project on May 16, 2024 at 14:52

This doesn't work because there is no ring map $R \to \bigoplus \mathfrak m^n/\mathfrak m^{n + 1}$ . A posteriori we see that there is a map of multiplicative monoids -- but this isn't true in general for Noetherian local rings.

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