The Stacks project

Remark 38.9.4. Lemma 38.9.3 is a key step in the development of results in this chapter. The analogue of this lemma in [GruRay] is [I Proposition 3.3.1, GruRay]: If $R \to S$ is smooth with geometrically integral fibres, then $S$ is projective as an $R$-module. This is a special case of Lemma 38.9.3, but as we will later improve on this lemma anyway, we do not gain much from having a stronger result at this point. We briefly sketch the proof of this as it is given in [GruRay].

1. First reduce to the case where $R$ is Noetherian as above.

2. Since projectivity descends through faithfully flat ring maps, see Algebra, Theorem 10.95.6 we may work locally in the fppf topology on $R$, hence we may assume that $R \to S$ has a section $\sigma : S \to R$. (Just by the usual trick of base changing to $S$.) Set $I = \mathop{\mathrm{Ker}}(S \to R)$.

3. Localizing a bit more on $R$ we may assume that $I/I^2$ is a free $R$-module and that the completion $S^\wedge $ of $S$ with respect to $I$ is isomorphic to $R[[t_1, \ldots , t_ n]]$, see Morphisms, Lemma 29.34.20. Here we are using that $R \to S$ is smooth.

4. To prove that $S$ is projective as an $R$-module, it suffices to prove that $S$ is flat, countably generated and Mittag-Leffler as an $R$-module, see Algebra, Lemma 10.93.1. The first two properties are evident. Thus it suffices to prove that $S$ is Mittag-Leffler as an $R$-module. By Algebra, Lemma 10.91.4 the module $R[[t_1, \ldots , t_ n]]$ is Mittag-Leffler over $R$. Hence Algebra, Lemma 10.89.7 shows that it suffices to show that the $S \to S^\wedge $ is universally injective as a map of $R$-modules.

5. Apply Lemma 38.7.4 to see that $S \to S^\wedge $ is $R$-universally injective. Namely, as $R \to S$ has geometrically integral fibres, any associated point of any fibre ring is just the generic point of the fibre ring which is in the image of $\mathop{\mathrm{Spec}}(S^\wedge ) \to \mathop{\mathrm{Spec}}(S)$.

There is an analogy between the proof as sketched just now, and the development of the arguments leading to the proof of Lemma 38.9.3. In both a completion plays an essential role, and both times the assumption of having geometrically integral fibres assures one that the map from $S$ to the completion of $S$ is $R$-universally injective.