Lemma 10.32.3 (0544)—The Stacks project

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Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.32: Locally nilpotent ideals
Lemma 10.32.3 (cite)

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Lemma 10.32.3. Let $R \to R'$ be a ring map and let $I \subset R$ be a locally nilpotent ideal. Then $IR'$ is a locally nilpotent ideal of $R'$ .

Proof. This follows from the fact that if $x, y \in R'$ are nilpotent, then $x + y$ is nilpotent too. Namely, if $x^ n = 0$ and $y^ m = 0$ , then $(x + y)^{n + m - 1} = 0$ . $\square$

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