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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