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. $\square$

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