The Stacks project

Proof. Assume (a) and (b). Note that (b) is equivalent to condition (2) of Lemma 10.46.3. Let $T \subset S$ be the set of elements $x \in S$ such that there exists an integer $n > 0$ such that $x^{p^ n} , p^ n x \in \varphi (R)$. We claim that $T = S$. This will prove that condition (1) of Lemma 10.46.3 holds and hence $\varphi $ induces a homeomorphism on spectra. By assumption (a) it suffices to show that $T \subset S$ is an $R$-sub algebra. If $x \in T$ and $y \in R$, then it is clear that $yx \in T$. Suppose $x, y \in T$ and $n, m > 0$ such that $x^{p^ n}, y^{p^ m}, p^ n x, p^ m y \in \varphi (R)$. Then $(xy)^{p^{n + m}}, p^{n + m}xy \in \varphi (R)$ hence $xy \in T$. We have $x + y \in T$ by Lemma 10.46.5 and the claim is proved.

Since $\varphi $ induces a homeomorphism on spectra, it is in particular surjective on spectra which is a property preserved under any base change, see Lemma 10.30.3. Therefore for any $R \to R'$ the kernel of the ring map $R' \to R' \otimes _ R S$ consists of nilpotent elements, see Lemma 10.30.6, in other words (b) holds for $R' \to R' \otimes _ R S$. It is clear that (a) is preserved under base change. Finally, the condition on residue fields follows from (a) as generators for $S$ as an $R$-algebra map to generators for the residue field extensions. $\square$