# The Stacks Project

## Tag 07GR

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. $\square$

The code snippet corresponding to this tag is a part of the file dpa.tex and is located in lines 314–319 (see updates for more information).

\begin{lemma}
\label{lemma-nil}
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 $A$.
\end{lemma}

\begin{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.
\end{proof}

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