Lemma 10.45.4. Let $\varphi : R \to S$ be a ring map. Assume

$S$ is generated as an $R$-algebra by elements $x$ such that $x^2, x^3 \in \varphi (R)$, and

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

Then $\varphi $ induces isomorphisms on residue fields and a homeomorphism of spectra. 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). The map on spectra is closed as $S$ is integral over $R$, see Lemmas 10.40.6 and 10.35.22. The image is dense by Lemma 10.29.6. Thus $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is surjective. If $\mathfrak q \subset S$ is a prime lying over $\mathfrak p \subset R$ then the field extension $\kappa (\mathfrak q)/\kappa (\mathfrak p)$ is generated by elements $\alpha \in \kappa (\mathfrak q)$ whose square and cube are in $\kappa (\mathfrak p)$. Thus clearly $\alpha \in \kappa (\mathfrak p)$ and we find that $\kappa (\mathfrak q) = \kappa (\mathfrak p)$. If $\mathfrak q, \mathfrak q'$ were two distinct primes lying over $\mathfrak p$, then at least one of the generators $x$ of $S$ as in (a) would have distinct images in $\kappa (\mathfrak q) = \kappa (\mathfrak p)$ and $\kappa (\mathfrak q') = \kappa (\mathfrak p)$. This would contradict the fact that both $x^2$ and $x^3$ do have the same image. This proves that $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is injective hence a homeomorphism (by what was already shown).

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.
$\square$

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