The Stacks project

Lemma 10.126.9. Let $R$ be a ring. Let $I \subset R$ be a locally nilpotent ideal. Let $S \to S'$ be an $R$-algebra map such that $S \to S'/IS'$ is surjective and such that $S'$ is of finite type over $R$. Then $S \to S'$ is surjective.

Proof. Write $S' = R[x_1, \ldots , x_ m]/K$ for some ideal $K$. By assumption there exist $g_ j = x_ j + \sum \delta _{j, J} x^ J \in R[x_1, \ldots , x_ n]$ with $\delta _{j, J} \in I$ and with $g_ j \bmod K \in \mathop{\mathrm{Im}}(S \to S')$. Hence it suffices to show that $g_1, \ldots , g_ m$ generate $R[x_1, \ldots , x_ n]$. Let $R_0 \subset R$ be a finitely generated $\mathbf{Z}$-subalgebra of $R$ containing at least the $\delta _{j, J}$. Then $R_0 \cap I$ is a nilpotent ideal (by Lemma 10.32.5). It follows that $R_0[x_1, \ldots , x_ n]$ is generated by $g_1, \ldots , g_ m$ (because $x_ j \mapsto g_ j$ defines an automorphism of $R_0[x_1, \ldots , x_ m]$; details omitted). Since $R$ is the union of the subrings $R_0$ we win. $\square$

Comments (0)

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Your email address will not be published.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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 07RD. Beware of the difference between the letter 'O' and the digit '0'.