Lemma 10.126.11. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $S \to S'$ be an $R$-algebra map. Assume that

1. $I$ is locally nilpotent,

2. $S/IS \to S'/IS'$ is an isomorphism,

3. $S$ is of finite type over $R$,

4. $S'$ of finite presentation over $R$, and

5. $S'$ is flat over $R$.

Then $S \to S'$ is an isomorphism.

Proof. By Lemma 10.126.9 the map $S \to S'$ is surjective. As $I$ is locally nilpotent, so are the ideals $IS$ and $IS'$ (Lemma 10.32.3). Hence every prime ideal $\mathfrak q$ of $S$ contains $IS$ and (trivially) $S_\mathfrak q/IS_\mathfrak q \cong S'_\mathfrak q/IS'_\mathfrak q$. Thus Lemma 10.126.10 applies and we see that $S_\mathfrak q \to S'_\mathfrak q$ is an isomorphism for every prime $\mathfrak q \subset S$. It follows that $S \to S'$ is injective for example by Lemma 10.23.1. $\square$

Comment #213 by Rex on

Typo: "is an ismorphism"

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