37.42 Étale localization of integral morphisms
Some variants of the results of Section 37.41 for the case of integral morphisms.
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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