Lemma 37.42.1. Let $R \to S$ be an integral ring map. Let $\mathfrak p \subset R$ be a prime ideal. Assume

there are finitely many primes $\mathfrak q_1, \ldots , \mathfrak q_ n$ lying over $\mathfrak p$, and

for each $i$ the maximal separable subextension $\kappa (\mathfrak q)/\kappa (\mathfrak q_ i)_{sep}/\kappa (\mathfrak p)$ (Fields, Lemma 9.14.6) is finite over $\kappa (\mathfrak p)$.

Then there exists an étale ring map $R \to R'$ and a prime $\mathfrak p'$ lying over $\mathfrak p$ such that

\[ S \otimes _ R R' = A_1 \times \ldots \times A_ m \]

with $R' \to A_ j$ integral having a unique prime $\mathfrak r_ j$ over $\mathfrak p'$ such that $\kappa (\mathfrak r_ j)/\kappa (\mathfrak p')$ is purely inseparable.

**First proof.**
This proof uses Algebra, Lemma 10.145.4. Namely, choose a generator $\theta _ i \in \kappa (\mathfrak q_ i)_{sep}$ of this field over $\kappa (\mathfrak p)$ (Fields, Lemma 9.19.1). The spectrum of the fibre ring $S \otimes _ R \kappa (\mathfrak p)$ is finite discrete with points corresponding to $\mathfrak q_1, \ldots , \mathfrak q_ n$. By the Chinese remainder theorem (Algebra, Lemma 10.15.4) we see that $S \otimes _ R \kappa (\mathfrak p) \to \prod \kappa (\mathfrak q_ i)$ is surjective. Hence after replacing $R$ by $R_ g$ for some $g \in R$, $g \not\in \mathfrak p$ we may assume that $(0, \ldots , 0, \theta _ i, 0, \ldots , 0) \in \prod \kappa (\mathfrak q_ i)$ is the image of some $x_ i \in S$. Let $S' \subset S$ be the $R$-subalgebra generated by our $x_ i$. Since $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(S')$ is surjective (Algebra, Lemma 10.36.17) we conclude that $\mathfrak q_ i' = S' \cap \mathfrak q_ i$ are the primes of $S'$ over $\mathfrak p$. By our choice of $x_ i$ we conclude these primes are distinct that and $\kappa (\mathfrak q'_ i)_{sep} = \kappa (\mathfrak q_ i)_{sep}$. In particular the field extensions $\kappa (\mathfrak q_ i)/\kappa (\mathfrak q'_ i)$ are purely inseparable. Since $R \to S'$ is finite we may apply Algebra, Lemma 10.145.4. and we get $R \to R'$ and $\mathfrak p'$ and a decomposition

\[ S' \otimes _ R R' = A'_1 \times \ldots \times A'_ m \times B' \]

with $R' \to A'_ j$ integral having a unique prime $\mathfrak r'_ j$ over $\mathfrak p'$ such that $\kappa (\mathfrak r'_ j)/\kappa (\mathfrak p')$ is purely inseparable and such that $B'$ does not have a prime lying over $\mathfrak p'$. Since $R' \to B'$ is finite (as $R \to S'$ is finite) we can after localizing $R'$ at some $g' \in R'$, $g' \not\in \mathfrak p'$ assume that $B' = 0$. Via the map $S' \otimes _ R R' \to S \otimes _ R R'$ we get the corresponding decomposition for $S$.
$\square$

**Second proof.**
This proof uses strict henselization. First, assume $R$ is strictly henselization with maximal ideal $\mathfrak p$. Then $S/\mathfrak p S$ has finitely many primes corresponding to $\mathfrak q_1, \ldots , \mathfrak q_ n$, each maximal, each with purely inseparable residue field over $\kappa (\mathfrak p)$. Hence $S/\mathfrak p S$ is equal to $\prod (S/\mathfrak p S)_{\mathfrak p_ i}$. By More on Algebra, Lemma 15.11.6 we can lift this product decomposition to a product composition of $S$ as in the statement.

In the general case, let $R^{sh}$ be the strict henselization of $R_\mathfrak p$. Then we can apply the result of the first paragraph to $R^{sh} \to S \otimes _ R R^{sh}$. Consider the $m$ mutually orthogonal idempotents in $S \otimes _ R R^{sh}$ corresponding to the product decomposition. Since $R^{sh}$ is a filtered colimit of étale ring maps $(R, \mathfrak p) \to (R', \mathfrak p')$ by Algebra, Lemma 10.155.11 we see that these idempotents descend to some $R'$ as desired.
$\square$

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