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 $(x + y)^{n + m - 1} = 0$. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$
). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)
There are also: