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
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
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
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
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].
First reduce to the case where R is Noetherian as above.
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).
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.
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.
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.
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