# The Stacks Project

## Tag 0544

Lemma 10.31.2. 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 $(x + y)^{n + m - 1} = 0$. $\square$

The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 5913–5917 (see updates for more information).

\begin{lemma}
\label{lemma-locally-nilpotent}
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 $R'$.
\end{lemma}

\begin{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
$(x + y)^{n + m - 1} = 0$.
\end{proof}

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