Lemma 10.45.7. Let $\varphi : R \to S$ be a ring map. Let $p$ be a prime number. Assume

$S$ is generated as an $R$-algebra by elements $x$ such that there exists an $n > 0$ with $x^{p^ n} \in \varphi (R)$ and $p^ nx \in \varphi (R)$, and

$\mathop{\mathrm{Ker}}(\varphi )$ is locally nilpotent,

Then $\varphi $ induces a homeomorphism of spectra and induces residue field extensions satisfying the equivalent conditions of Lemma 10.45.6. For any ring map $R \to R'$ the ring map $R' \to R' \otimes _ R S$ also satisfies (a) and (b).

**Proof.**
Assume (a) and (b). Note that (b) is equivalent to condition (2) of Lemma 10.45.3. 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 10.45.3 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 10.45.5 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 10.29.3. Therefore for any $R \to R'$ the kernel of the ring map $R' \to R' \otimes _ R S$ consists of nilpotent elements, see Lemma 10.29.6, 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.
$\square$

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