Lemma 38.9.2. Let $R$ be a ring. Let $R \to S$ be a ring map. Assume

1. $R$ is Noetherian,

2. $R \to S$ is of finite type and flat, and

3. 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. Consider the set

$\{ I \subset R \mid S/IS\text{ not projective as }R/I\text{-module}\}$

We have to show this set is empty. To get a contradiction assume it is nonempty. Then it contains a maximal element $I$. Let $J = \sqrt{I}$ be its radical. If $I \not= J$, then $S/JS$ is projective as a $R/J$-module, and $S/IS$ is flat over $R/I$ and $J/I$ is a nilpotent ideal in $R/I$. Applying Algebra, Lemma 10.77.7 we see that $S/IS$ is a projective $R/I$-module, which is a contradiction. Hence we may assume that $I$ is a radical ideal. In other words we are reduced to proving the lemma in case $R$ is a reduced ring and $S/IS$ is a projective $R/I$-module for every nonzero ideal $I$ of $R$.

Assume $R$ is a reduced ring and $S/IS$ is a projective $R/I$-module for every nonzero ideal $I$ of $R$. By generic flatness, Algebra, Lemma 10.118.1 (applied to a localization $R_ g$ which is a domain) or the more general Algebra, Lemma 10.118.7 there exists a nonzero $f \in R$ such that $S_ f$ is free as an $R_ f$-module. Denote $R^\wedge = \mathop{\mathrm{lim}}\nolimits R/(f^ n)$ the $(f)$-adic completion of $R$. Note that the ring map

$R \longrightarrow R_ f \times R^\wedge$

is a faithfully flat ring map, see Algebra, Lemma 10.97.2. Hence by faithfully flat descent of projectivity, see Algebra, Theorem 10.95.6 it suffices to prove that $S \otimes _ R R^\wedge$ is a projective $R^\wedge$-module. To see this we will use the criterion of Lemma 38.9.1. First of all, note that $S/fS = (S \otimes _ R R^\wedge )/f(S \otimes _ R R^\wedge )$ is a projective $R/(f)$-module and that $S \otimes _ R R^\wedge$ is flat and of finite type over $R^\wedge$ as a base change of such. Next, suppose that $\mathfrak p^\wedge$ is a prime ideal of $R^\wedge$. Let $\mathfrak p \subset R$ be the corresponding prime of $R$. As $R \to S$ has geometrically integral fibre rings, the same is true for the fibre rings of any base change. Hence $\mathfrak q^\wedge = \mathfrak p^\wedge (S \otimes _ R R^\wedge )$, is a prime ideals lying over $\mathfrak p^\wedge$ and it is the unique associated prime of $S \otimes _ R \kappa (\mathfrak p^\wedge )$. Thus we win if $f(S \otimes _ R R^\wedge ) + \mathfrak q^\wedge \not= S \otimes _ R R^\wedge$. This is true because $\mathfrak p^\wedge + fR^\wedge \not= R^\wedge$ as $f$ lies in the Jacobson radical of the $f$-adically complete ring $R^\wedge$ and because $R^\wedge \to S \otimes _ R R^\wedge$ is surjective on spectra as its fibres are nonempty (irreducible spaces are nonempty). $\square$

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