The Stacks project

Proof. Assume (a) and (b). The map on spectra is closed as $S$ is integral over $R$, see Lemmas 10.41.6 and 10.36.22. The image is dense by Lemma 10.30.6. 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 10.30.3. 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. $\square$