Lemma 10.137.18. 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 10.134.8 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 10.39.17). Thus the vanishing of $H_1(L_{S'/R'})_{\mathfrak q'}$ implies the vanishing of $H_1(L_{S/R})_{\mathfrak q}$. By Lemma 10.78.6 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 10.137.12. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)