## 38.9 Projective modules

The following lemma can be used to prove projectivity by Noetherian induction on the base, see Lemma 38.9.2.

Lemma 38.9.1. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $R \to S$ be a ring map, and $N$ an $S$-module. Assume

$R$ is Noetherian and $I$-adically complete,

$R \to S$ is of finite type,

$N$ is a finite $S$-module,

$N$ is flat over $R$,

$N/IN$ is projective as a $R/I$-module, and

for any prime $\mathfrak q \subset S$ which is an associated prime of $N \otimes _ R \kappa (\mathfrak p)$ where $\mathfrak p = R \cap \mathfrak q$ we have $IS + \mathfrak q \not= S$.

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

**Proof.**
By Lemma 38.8.4 the map $N \to N^\wedge $ is universally injective. By Lemma 38.8.2 the module $N^\wedge $ is Mittag-Leffler. By Algebra, Lemma 10.89.7 we conclude that $N$ is Mittag-Leffler. Hence $N$ is countably generated, flat and Mittag-Leffler as an $R$-module, whence projective by Algebra, Lemma 10.93.1.
$\square$

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

$R$ is Noetherian,

$R \to S$ is of finite type 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.**
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).
$\square$

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.
$\square$

## Comments (0)