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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