Lemma 10.45.11. Let $\varphi : R \to S$ be a ring map such that

the kernel of $\varphi $ is locally nilpotent, and

$S$ is generated as an $R$-algebra by elements $x$ such that there exist $n > 0$ and a polynomial $P(T) \in R[T]$ whose image in $S[T]$ is $(T - x)^ n$.

Then $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is a homeomorphism and $R \to S$ induces purely inseparable extensions of residue fields. Moreover, conditions (1) and (2) remain true on arbitrary base change.

**Proof.**
We may replace $R$ by $R/\mathop{\mathrm{Ker}}(\varphi )$, see Lemma 10.45.1. Assumption (2) implies $S$ is generated over $R$ by elements which are integral over $R$. Hence $R \subset S$ is integral (Lemma 10.35.7). In particular $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is surjective and closed (Lemmas 10.35.17, 10.40.6, and 10.35.22).

Let $x \in S$ be one of the generators in (2), i.e., there exists an $n > 0$ be such that $(T - x)^ n \in R[T]$. Let $\mathfrak p \subset R$ be a prime. The $\kappa (\mathfrak p) \otimes _ R S$ ring is nonzero by the above and Lemma 10.16.9. If the characteristic of $\kappa (\mathfrak p)$ is zero then we see that $nx \in R$ implies $1 \otimes x$ is in the image of $\kappa (\mathfrak p) \to \kappa (\mathfrak p) \otimes _ R S$. Hence $\kappa (\mathfrak p) \to \kappa (\mathfrak p) \otimes _ R S$ is an isomorphism. If the characteristic of $\kappa (\mathfrak p)$ is $p > 0$, then write $n = p^ k m$ with $m$ prime to $p$. In $\kappa (\mathfrak p) \otimes _ R S[T]$ we have

\[ (T - 1 \otimes x)^ n = ((T - 1 \otimes x)^{p^ k})^ m = (T^{p^ k} - 1 \otimes x^{p^ k})^ m \]

and we see that $mx^{p^ k} \in R$. This implies that $1 \otimes x^{p^ k}$ is in the image of $\kappa (\mathfrak p) \to \kappa (\mathfrak p) \otimes _ R S$. Hence Lemma 10.45.7 applies to $\kappa (\mathfrak p) \to \kappa (\mathfrak p) \otimes _ R S$. In both cases we conclude that $\kappa (\mathfrak p) \otimes _ R S$ has a unique prime ideal with residue field purely inseparable over $\kappa (\mathfrak p)$. By Remark 10.16.8 we conclude that $\varphi $ is bijective on spectra.

The statement on base change is immediate.
$\square$

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