Lemma 38.9.3 (05FT)—The Stacks project

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Table of contents
Part 2: Schemes
Chapter 38: More on Flatness
Section 38.9: Projective modules
Lemma 38.9.3 (cite)

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Lemma 38.9.3. Let $R$ be a ring. Let $R \to S$ be a ring map. Assume

$R \to S$ is of finite presentation and flat, and
every fibre ring $S \otimes _ R \kappa (\mathfrak p)$ is geometrically integral over $\kappa (\mathfrak p)$ .

Then $S$ is projective as an $R$ -module.

Proof. We can find a cocartesian diagram of rings

$\xymatrix{ S_0 \ar[r] & S \\ R_0 \ar[u] \ar[r] & R \ar[u] }$

such that $R_0$ is of finite type over $\mathbf{Z}$ , the map $R_0 \to S_0$ is of finite type and flat with geometrically integral fibres, see More on Morphisms, Lemmas 37.34.4, 37.34.6, 37.34.7, and 37.34.11. By Lemma 38.9.2 we see that $S_0$ is a projective $R_0$ -module. Hence $S = S_0 \otimes _{R_0} R$ is a projective $R$ -module, see Algebra, Lemma 10.94.1. $\square$

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