
Lemma 10.135.17. Let $R \to S$ be a ring map of finite presentation. Let $R \to R'$ be a flat ring map. Denote $S' = R' \otimes _ R S$ the base change. Let $U \subset \mathop{\mathrm{Spec}}(S)$ be the set of primes at which $R \to S$ is smooth. Let $V \subset \mathop{\mathrm{Spec}}(S')$ the set of primes at which $R' \to S'$ is smooth. Then $V$ is the inverse image of $U$ under the map $f : \mathop{\mathrm{Spec}}(S') \to \mathop{\mathrm{Spec}}(S)$.

Proof. By Lemma 10.132.8 we see that $\mathop{N\! L}\nolimits _{S/R} \otimes _ S S'$ is homotopy equivalent to $\mathop{N\! L}\nolimits _{S'/R'}$. This already implies that $f^{-1}(U) \subset V$.

Let $\mathfrak q' \subset S'$ be a prime lying over $\mathfrak q \subset S$. Assume $\mathfrak q' \in V$. We have to show that $\mathfrak q \in U$. Since $S \to S'$ is flat, we see that $S_{\mathfrak q} \to S'_{\mathfrak q'}$ is faithfully flat (Lemma 10.38.17). Thus the vanishing of $H_1(L_{S'/R'})_{\mathfrak q'}$ implies the vanishing of $H_1(L_{S/R})_{\mathfrak q}$. By Lemma 10.77.5 applied to the $S_{\mathfrak q}$-module $(\Omega _{S/R})_{\mathfrak q}$ and the map $S_{\mathfrak q} \to S'_{\mathfrak q'}$ we see that $(\Omega _{S/R})_{\mathfrak q}$ is projective. Hence $R \to S$ is smooth at $\mathfrak q$ by Lemma 10.135.12. $\square$

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