Lemma 23.2.6 (07GR)—The Stacks project

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Table of contents
Part 1: Preliminaries
Chapter 23: Divided Power Algebra
Section 23.2: Divided powers
Lemma 23.2.6 (cite)

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Lemma 23.2.6. Let $p$ be a prime number. Let $A$ be a ring, let $I \subset A$ be an ideal, and let $\gamma$ be a divided power structure on $I$ . Assume $p$ is nilpotent in $A/I$ . Then $I$ is locally nilpotent if and only if $p$ is nilpotent in $A$ .

Proof. If $p^ N = 0$ in $A$ , then for $x \in I$ we have $x^{pN} = (pN)!\gamma _{pN}(x) = 0$ because $(pN)!$ is divisible by $p^ N$ . Conversely, assume $I$ is locally nilpotent. We've also assumed that $p$ is nilpotent in $A/I$ , hence $p^ r \in I$ for some $r$ , hence $p^ r$ nilpotent, hence $p$ nilpotent. $\square$

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