Lemma 10.143.8. Let $R \to S$ and $R \to S'$ be étale. Then any $R$-algebra map $S' \to S$ is étale.

Proof. First of all we note that $S' \to S$ is of finite presentation by Lemma 10.6.2. Let $\mathfrak q \subset S$ be a prime ideal lying over the primes $\mathfrak q' \subset S'$ and $\mathfrak p \subset R$. By Lemma 10.143.5 the ring map $S'_{\mathfrak q'}/\mathfrak p S'_{\mathfrak q'} \to S_{\mathfrak q}/\mathfrak p S_{\mathfrak q}$ is a map of finite separable extensions of $\kappa (\mathfrak p)$. In particular it is flat. Hence by Lemma 10.128.8 we see that $S'_{\mathfrak q'} \to S_{\mathfrak q}$ is flat. Thus $S' \to S$ is flat. Moreover, the above also shows that $\mathfrak q'S_{\mathfrak q}$ is the maximal ideal of $S_{\mathfrak q}$ and that the residue field extension of $S'_{\mathfrak q'} \to S_{\mathfrak q}$ is finite separable. Hence from Lemma 10.143.7 we conclude that $S' \to S$ is étale at $\mathfrak q$. Since being étale is local (see Lemma 10.143.3) we win. $\square$

Comment #8297 by nhw on

Small typo: "map finite separable extensions" should be "map of finite separable extensions" I think.

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