The Stacks project

35.4 Descent for universally injective morphisms

Numerous constructions in algebraic geometry are made using techniques of descent, such as constructing objects over a given space by first working over a somewhat larger space which projects down to the given space, or verifying a property of a space or a morphism by pulling back along a covering map. The utility of such techniques is of course dependent on identification of a wide class of effective descent morphisms. Early in the Grothendieckian development of modern algebraic geometry, the class of morphisms which are quasi-compact and faithfully flat was shown to be effective for descending objects, morphisms, and many properties thereof.

As usual, this statement comes down to a property of rings and modules. For a homomorphism $f: R \to S$ to be an effective descent morphism for modules, Grothendieck showed that it is sufficient for $f$ to be faithfully flat. However, this excludes many natural examples: for instance, any split ring homomorphism is an effective descent morphism. One natural example of this even arises in the proof of faithfully flat descent: for $f: R \to S$ any ring homomorphism, $1_ S \otimes f: S \to S \otimes _ R S$ is split by the multiplication map whether or not it is flat.

One may then ask whether there is a natural ring-theoretic condition implying effective descent for modules which includes both the case of a faithfully flat morphism and that of a split ring homomorphism. It may surprise the reader (at least it surprised this author) to learn that a complete answer to this question has been known since around 1970! Namely, it is not hard to check that a necessary condition for $f: R \to S$ to be an effective descent morphism for modules is that $f$ must be universally injective in the category of $R$-modules, that is, for any $R$-module $M$, the map $1_ M \otimes f: M \to M \otimes _ R S$ must be injective. This then turns out to be a sufficient condition as well. For example, if $f$ is split in the category of $R$-modules (but not necessarily in the category of rings), then $f$ is an effective descent morphism for modules.

The history of this result is a bit involved: it was originally asserted by Olivier [olivier], who called universally injective morphisms pure, but without a clear indication of proof. One can extract the result from the work of Joyal and Tierney [joyal-tierney], but to the best of our knowledge, the first free-standing proof to appear in the literature is that of Mesablishvili [mesablishvili1]. The first purpose of this section is to expose Mesablishvili's proof; this requires little modification of his original presentation aside from correcting typos, with the one exception that we make explicit the relationship between the customary definition of a descent datum in algebraic geometry and the one used in [mesablishvili1]. The proof turns out to be entirely category-theoretic, and consequently can be put in the language of monads (and thus applied in other contexts); see [janelidze-tholen].

The second purpose of this section is to collect some information about which properties of modules, algebras, and morphisms can be descended along universally injective ring homomorphisms. The cases of finite modules and flat modules were treated by Mesablishvili [mesablishvili2].

35.4.1 Category-theoretic preliminaries

We start by recalling a few basic notions from category theory which will simplify the exposition. In this subsection, fix an ambient category.

For two morphisms $g_1, g_2: B \to C$, recall that an equalizer of $g_1$ and $g_2$ is a morphism $f: A \to B$ which satisfies $g_1 \circ f = g_2 \circ f$ and is universal for this property. This second statement means that any commutative diagram

\[ \xymatrix{A' \ar[rd]^ e \ar@/^1.5pc/[rrd] \ar@{-->}[d] & & \\ A \ar[r]^ f & B \ar@<1ex>[r]^{g_1} \ar@<-1ex>[r]_{g_2} & C } \]

without the dashed arrow can be uniquely completed. We also say in this situation that the diagram

35.4.1.1
\begin{equation} \label{descent-equation-equalizer} \xymatrix{ A \ar[r]^ f & B \ar@<1ex>[r]^{g_1} \ar@<-1ex>[r]_{g_2} & C } \end{equation}

is an equalizer. Reversing arrows gives the definition of a coequalizer. See Categories, Sections 4.10 and 4.11.

Since it involves a universal property, the property of being an equalizer is typically not stable under applying a covariant functor. Just as for monomorphisms and epimorphisms, one can get around this in some cases by exhibiting splittings.

Definition 35.4.2. A split equalizer is a diagram (35.4.1.1) with $g_1 \circ f = g_2 \circ f$ for which there exist auxiliary morphisms $h : B \to A$ and $i : C \to B$ such that

35.4.2.1
\begin{equation} \label{descent-equation-split-equalizer-conditions} h \circ f = 1_ A, \quad f \circ h = i \circ g_1, \quad i \circ g_2 = 1_ B. \end{equation}

The point is that the equalities among arrows force (35.4.1.1) to be an equalizer: the map $e$ factors uniquely through $f$ by writing $e = f \circ (h \circ e)$. Consequently, applying a covariant functor to a split equalizer gives a split equalizer; applying a contravariant functor gives a split coequalizer, whose definition is apparent.

35.4.3 Universally injective morphisms

Recall that $\textit{Rings}$ denotes the category of commutative rings with $1$. For an object $R$ of $\textit{Rings}$ we denote $\text{Mod}_ R$ the category of $R$-modules.

Remark 35.4.4. Any functor $F : \mathcal{A} \to \mathcal{B}$ of abelian categories which is exact and takes nonzero objects to nonzero objects reflects injections and surjections. Namely, exactness implies that $F$ preserves kernels and cokernels (compare with Homology, Section 12.7). For example, if $f : R \to S$ is a faithfully flat ring homomorphism, then $\bullet \otimes _ R S: \text{Mod}_ R \to \text{Mod}_ S$ has these properties.

Let $R$ be a ring. Recall that a morphism $f : M \to N$ in $\text{Mod}_ R$ is universally injective if for all $P \in \text{Mod}_ R$, the morphism $f \otimes 1_ P: M \otimes _ R P \to N \otimes _ R P$ is injective. See Algebra, Definition 10.82.1.

Definition 35.4.5. A ring map $f: R \to S$ is universally injective if it is universally injective as a morphism in $\text{Mod}_ R$.

Example 35.4.6. Any split injection in $\text{Mod}_ R$ is universally injective. In particular, any split injection in $\textit{Rings}$ is universally injective.

Example 35.4.7. For a ring $R$ and $f_1, \ldots , f_ n \in R$ generating the unit ideal, the morphism $R \to R_{f_1} \oplus \ldots \oplus R_{f_ n}$ is universally injective. Although this is immediate from Lemma 35.4.8, it is instructive to check it directly: we immediately reduce to the case where $R$ is local, in which case some $f_ i$ must be a unit and so the map $R \to R_{f_ i}$ is an isomorphism.

Lemma 35.4.8. Any faithfully flat ring map is universally injective.

Proof. This is a reformulation of Algebra, Lemma 10.82.11. $\square$

The key observation from [mesablishvili1] is that universal injectivity can be usefully reformulated in terms of a splitting, using the usual construction of an injective cogenerator in $\text{Mod}_ R$.

Definition 35.4.9. Let $R$ be a ring. Define the contravariant functor $C$ $ : \text{Mod}_ R \to \text{Mod}_ R$ by setting

\[ C(M) = \mathop{\mathrm{Hom}}\nolimits _{\textit{Ab}}(M, \mathbf{Q}/\mathbf{Z}), \]

with the $R$-action on $C(M)$ given by $rf(s) = f(rs)$.

This functor was denoted $M \mapsto M^\vee $ in More on Algebra, Section 15.55.

Lemma 35.4.10. For a ring $R$, the functor $C : \text{Mod}_ R \to \text{Mod}_ R$ is exact and reflects injections and surjections.

Proof. Exactness is More on Algebra, Lemma 15.55.6 and the other properties follow from this, see Remark 35.4.4. $\square$

Remark 35.4.11. We will use frequently the standard adjunction between $\mathop{\mathrm{Hom}}\nolimits $ and tensor product, in the form of the natural isomorphism of contravariant functors

35.4.11.1
\begin{equation} \label{descent-equation-adjunction} C(\bullet _1 \otimes _ R \bullet _2) \cong \mathop{\mathrm{Hom}}\nolimits _ R(\bullet _1, C(\bullet _2)): \text{Mod}_ R \times \text{Mod}_ R \to \text{Mod}_ R \end{equation}

taking $f: M_1 \otimes _ R M_2 \to \mathbf{Q}/\mathbf{Z}$ to the map $m_1 \mapsto (m_2 \mapsto f(m_1 \otimes m_2))$. See Algebra, Lemma 10.14.5. A corollary of this observation is that if

\[ \xymatrix@C=9pc{ C(M) \ar@<1ex>[r] \ar@<-1ex>[r] & C(N) \ar[r] & C(P) } \]

is a split coequalizer diagram in $\text{Mod}_ R$, then so is

\[ \xymatrix@C=9pc{ C(M \otimes _ R Q) \ar@<1ex>[r] \ar@<-1ex>[r] & C(N \otimes _ R Q) \ar[r] & C(P \otimes _ R Q) } \]

for any $Q \in \text{Mod}_ R$.

Lemma 35.4.12. Let $R$ be a ring. A morphism $f: M \to N$ in $\text{Mod}_ R$ is universally injective if and only if $C(f): C(N) \to C(M)$ is a split surjection.

Proof. By (35.4.11.1), for any $P \in \text{Mod}_ R$ we have a commutative diagram

\[ \xymatrix@C=9pc{ \mathop{\mathrm{Hom}}\nolimits _ R( P, C(N)) \ar[r]_{\mathop{\mathrm{Hom}}\nolimits _ R(P,C(f))} \ar[d]^{\cong } & \mathop{\mathrm{Hom}}\nolimits _ R(P,C(M)) \ar[d]^{\cong } \\ C(P \otimes _ R N ) \ar[r]^{C(1_{P} \otimes f)} & C(P \otimes _ R M ). } \]

If $f$ is universally injective, then $1_{C(M)} \otimes f: C(M) \otimes _ R M \to C(M) \otimes _ R N$ is injective, so both rows in the above diagram are surjective for $P = C(M)$. We may thus lift $1_{C(M)} \in \mathop{\mathrm{Hom}}\nolimits _ R(C(M), C(M))$ to some $g \in \mathop{\mathrm{Hom}}\nolimits _ R(C(N), C(M))$ splitting $C(f)$. Conversely, if $C(f)$ is a split surjection, then both rows in the above diagram are surjective, so by Lemma 35.4.10, $1_{P} \otimes f$ is injective. $\square$

Remark 35.4.13. Let $f: M \to N$ be a universally injective morphism in $\text{Mod}_ R$. By choosing a splitting $g$ of $C(f)$, we may construct a functorial splitting of $C(1_ P \otimes f)$ for each $P \in \text{Mod}_ R$. Namely, by (35.4.11.1) this amounts to splitting $\mathop{\mathrm{Hom}}\nolimits _ R(P, C(f))$ functorially in $P$, and this is achieved by the map $g \circ \bullet $.

35.4.14 Descent for modules and their morphisms

Throughout this subsection, fix a ring map $f: R \to S$. As seen in Section 35.3 we can use the language of cosimplicial algebras to talk about descent data for modules, but in this subsection we prefer a more down to earth terminology.

For $i = 1, 2, 3$, let $S_ i$ be the $i$-fold tensor product of $S$ over $R$. Define the ring homomorphisms $\delta _0^1, \delta _1^1: S_1 \to S_2$, $\delta _{01}^1, \delta _{02}^1, \delta _{12}^1: S_1 \to S_3$, and $\delta _0^2, \delta _1^2, \delta _2^2: S_2 \to S_3$ by the formulas

\begin{align*} \delta ^1_0 (a_0) & = 1 \otimes a_0 \\ \delta ^1_1 (a_0) & = a_0 \otimes 1 \\ \delta ^2_0 (a_0 \otimes a_1) & = 1 \otimes a_0 \otimes a_1 \\ \delta ^2_1 (a_0 \otimes a_1) & = a_0 \otimes 1 \otimes a_1 \\ \delta ^2_2 (a_0 \otimes a_1) & = a_0 \otimes a_1 \otimes 1 \\ \delta _{01}^1(a_0) & = 1 \otimes 1 \otimes a_0 \\ \delta _{02}^1(a_0) & = 1 \otimes a_0 \otimes 1 \\ \delta _{12}^1(a_0) & = a_0 \otimes 1 \otimes 1. \end{align*}

In other words, the upper index indicates the source ring, while the lower index indicates where to insert factors of 1. (This notation is compatible with the notation introduced in Section 35.3.)

Recall1 from Definition 35.3.1 that for $M \in \text{Mod}_ S$, a descent datum on $M$ relative to $f$ is an isomorphism

\[ \theta : M \otimes _{S,\delta ^1_0} S_2 \longrightarrow M \otimes _{S,\delta ^1_1} S_2 \]

of $S_2$-modules satisfying the cocycle condition

35.4.14.1
\begin{equation} \label{descent-equation-cocycle-condition} (\theta \otimes \delta _2^2) \circ (\theta \otimes \delta _2^0) = (\theta \otimes \delta _2^1): M \otimes _{S, \delta ^1_{01}} S_3 \to M \otimes _{S,\delta ^1_{12}} S_3. \end{equation}

Let $DD_{S/R}$ be the category of $S$-modules equipped with descent data relative to $f$.

For example, for $M_0 \in \text{Mod}_ R$ and a choice of isomorphism $M \cong M_0 \otimes _ R S$ gives rise to a descent datum by identifying $M \otimes _{S,\delta ^1_0} S_2$ and $M \otimes _{S,\delta ^1_1} S_2$ naturally with $M_0 \otimes _ R S_2$. This construction in particular defines a functor $f^*: \text{Mod}_ R \to DD_{S/R}$.

Definition 35.4.15. The functor $f^*: \text{Mod}_ R \to DD_{S/R}$ is called base extension along $f$. We say that $f$ is a descent morphism for modules if $f^*$ is fully faithful. We say that $f$ is an effective descent morphism for modules if $f^*$ is an equivalence of categories.

Our goal is to show that for $f$ universally injective, we can use $\theta $ to locate $M_0$ within $M$. This process makes crucial use of some equalizer diagrams.

Lemma 35.4.16. For $(M,\theta ) \in DD_{S/R}$, the diagram

35.4.16.1
\begin{equation} \label{descent-equation-equalizer-M} \xymatrix@C=8pc{ M \ar[r]^{\theta \circ (1_ M \otimes \delta _0^1)} & M \otimes _{S, \delta _1^1} S_2 \ar@<1ex>[r]^{(\theta \otimes \delta _2^2) \circ (1_ M \otimes \delta ^2_0)} \ar@<-1ex>[r]_{1_{M \otimes S_2} \otimes \delta ^2_1} & M \otimes _{S, \delta _{12}^1} S_3 } \end{equation}

is a split equalizer.

Proof. Define the ring homomorphisms $\sigma ^0_0: S_2 \to S_1$ and $\sigma _0^1, \sigma _1^1: S_3 \to S_2$ by the formulas

\begin{align*} \sigma ^0_0 (a_0 \otimes a_1) & = a_0a_1 \\ \sigma ^1_0 (a_0 \otimes a_1 \otimes a_2) & = a_0a_1 \otimes a_2 \\ \sigma ^1_1 (a_0 \otimes a_1 \otimes a_2) & = a_0 \otimes a_1a_2. \end{align*}

We then take the auxiliary morphisms to be $1_ M \otimes \sigma _0^0: M \otimes _{S, \delta _1^1} S_2 \to M$ and $1_ M \otimes \sigma _0^1: M \otimes _{S,\delta _{12}^1} S_3 \to M \otimes _{S, \delta _1^1} S_2$. Of the compatibilities required in (35.4.2.1), the first follows from tensoring the cocycle condition (35.4.14.1) with $\sigma _1^1$ and the others are immediate. $\square$

Lemma 35.4.17. For $(M, \theta ) \in DD_{S/R}$, the diagram

35.4.17.1
\begin{equation} \label{descent-equation-coequalizer-CM} \xymatrix@C=8pc{ C(M \otimes _{S, \delta _{12}^1} S_3) \ar@<1ex>[r]^{C((\theta \otimes \delta _2^2) \circ (1_ M \otimes \delta ^2_0))} \ar@<-1ex>[r]_{C(1_{M \otimes S_2} \otimes \delta ^2_1)} & C(M \otimes _{S, \delta _1^1} S_2 ) \ar[r]^{C(\theta \circ (1_ M \otimes \delta _0^1))} & C(M). } \end{equation}

obtained by applying $C$ to (35.4.16.1) is a split coequalizer.

Proof. Omitted. $\square$

Lemma 35.4.18. The diagram

35.4.18.1
\begin{equation} \label{descent-equation-equalizer-S} \xymatrix@C=8pc{ S_1 \ar[r]^{\delta ^1_1} & S_2 \ar@<1ex>[r]^{\delta ^2_2} \ar@<-1ex>[r]_{\delta ^2_1} & S_3 } \end{equation}

is a split equalizer.

Proof. In Lemma 35.4.16, take $(M, \theta ) = f^*(S)$. $\square$

This suggests a definition of a potential quasi-inverse functor for $f^*$.

Definition 35.4.19. Define the functor $f_*$ $: DD_{S/R} \to \text{Mod}_ R$ by taking $f_*(M, \theta )$ to be the $R$-submodule of $M$ for which the diagram

35.4.19.1
\begin{equation} \label{descent-equation-equalizer-f} \xymatrix@C=8pc{f_*(M,\theta ) \ar[r] & M \ar@<1ex>^{\theta \circ (1_ M \otimes \delta _0^1)}[r] \ar@<-1ex>_{1_ M \otimes \delta _1^1}[r] & M \otimes _{S, \delta _1^1} S_2 } \end{equation}

is an equalizer.

Using Lemma 35.4.16 and the fact that the restriction functor $\text{Mod}_ S \to \text{Mod}_ R$ is right adjoint to the base extension functor $\bullet \otimes _ R S: \text{Mod}_ R \to \text{Mod}_ S$, we deduce that $f_*$ is right adjoint to $f^*$.

We are ready for the key lemma. In the faithfully flat case this is a triviality (see Remark 35.4.21), but in the general case some argument is needed.

Lemma 35.4.20. If $f$ is universally injective, then the diagram

35.4.20.1
\begin{equation} \label{descent-equation-equalizer-f2} \xymatrix@C=8pc{ f_*(M, \theta ) \otimes _ R S \ar[r]^{\theta \circ (1_ M \otimes \delta _0^1)} & M \otimes _{S, \delta _1^1} S_2 \ar@<1ex>[r]^{(\theta \otimes \delta _2^2) \circ (1_ M \otimes \delta ^2_0)} \ar@<-1ex>[r]_{1_{M \otimes S_2} \otimes \delta ^2_1} & M \otimes _{S, \delta _{12}^1} S_3 } \end{equation}

obtained by tensoring (35.4.19.1) over $R$ with $S$ is an equalizer.

Proof. By Lemma 35.4.12 and Remark 35.4.13, the map $C(1_ N \otimes f): C(N \otimes _ R S) \to C(N)$ can be split functorially in $N$. This gives the upper vertical arrows in the commutative diagram

\[ \xymatrix@C=8pc{ C(M \otimes _{S, \delta _1^1} S_2) \ar@<1ex>^{C(\theta \circ (1_ M \otimes \delta _0^1))}[r] \ar@<-1ex>_{C(1_ M \otimes \delta _1^1)}[r] \ar[d] & C(M) \ar[r]\ar[d] & C(f_*(M,\theta )) \ar@{-->}[d] \\ C(M \otimes _{S,\delta _{12}^1} S_3) \ar@<1ex>^{C((\theta \otimes \delta _2^2) \circ (1_ M \otimes \delta ^2_0))}[r] \ar@<-1ex>_{C(1_{M \otimes S_2} \otimes \delta ^2_1)}[r] \ar[d] & C(M \otimes _{S, \delta _1^1} S_2 ) \ar[r]^{C(\theta \circ (1_ M \otimes \delta _0^1))} \ar[d]^{C(1_ M \otimes \delta _1^1)} & C(M) \ar[d] \ar@{=}[dl] \\ C(M \otimes _{S, \delta _1^1} S_2) \ar@<1ex>[r]^{C(\theta \circ (1_ M \otimes \delta _0^1))} \ar@<-1ex>[r]_{C(1_ M \otimes \delta _1^1)} & C(M) \ar[r] & C(f_*(M,\theta )) } \]

in which the compositions along the columns are identity morphisms. The second row is the coequalizer diagram (35.4.17.1); this produces the dashed arrow. From the top right square, we obtain auxiliary morphisms $C(f_*(M,\theta )) \to C(M)$ and $C(M) \to C(M\otimes _{S,\delta _1^1} S_2)$ which imply that the first row is a split coequalizer diagram. By Remark 35.4.11, we may tensor with $S$ inside $C$ to obtain the split coequalizer diagram

\[ \xymatrix@C=8pc{ C(M \otimes _{S,\delta _2^2 \circ \delta _1^1} S_3) \ar@<1ex>^{C((\theta \otimes \delta _2^2) \circ (1_ M \otimes \delta ^2_0))}[r] \ar@<-1ex>_{C(1_{M \otimes S_2} \otimes \delta ^2_1)}[r] & C(M \otimes _{S, \delta _1^1} S_2 ) \ar[r]^{C(\theta \circ (1_ M \otimes \delta _0^1))} & C(f_*(M,\theta ) \otimes _ R S). } \]

By Lemma 35.4.10, we conclude (35.4.20.1) must also be an equalizer. $\square$

Remark 35.4.21. If $f$ is a split injection in $\text{Mod}_ R$, one can simplify the argument by splitting $f$ directly, without using $C$. Things are even simpler if $f$ is faithfully flat; in this case, the conclusion of Lemma 35.4.20 is immediate because tensoring over $R$ with $S$ preserves all equalizers.

Theorem 35.4.22. The following conditions are equivalent.

  1. The morphism $f$ is a descent morphism for modules.

  2. The morphism $f$ is an effective descent morphism for modules.

  3. The morphism $f$ is universally injective.

Proof. It is clear that (b) implies (a). We now check that (a) implies (c). If $f$ is not universally injective, we can find $M \in \text{Mod}_ R$ such that the map $1_ M \otimes f: M \to M \otimes _ R S$ has nontrivial kernel $N$. The natural projection $M \to M/N$ is not an isomorphism, but its image in $DD_{S/R}$ is an isomorphism. Hence $f^*$ is not fully faithful.

We finally check that (c) implies (b). By Lemma 35.4.20, for $(M, \theta ) \in DD_{S/R}$, the natural map $f^* f_*(M,\theta ) \to M$ is an isomorphism of $S$-modules. On the other hand, for $M_0 \in \text{Mod}_ R$, we may tensor (35.4.18.1) with $M_0$ over $R$ to obtain an equalizer sequence, so $M_0 \to f_* f^* M$ is an isomorphism. Consequently, $f_*$ and $f^*$ are quasi-inverse functors, proving the claim. $\square$

35.4.23 Descent for properties of modules

Throughout this subsection, fix a universally injective ring map $f : R \to S$, an object $M \in \text{Mod}_ R$, and a ring map $R \to A$. We now investigate the question of which properties of $M$ or $A$ can be checked after base extension along $f$. We start with some results from [mesablishvili2].

Lemma 35.4.24. If $M \in \text{Mod}_ R$ is flat, then $C(M)$ is an injective $R$-module.

Proof. Let $0 \to N \to P \to Q \to 0$ be an exact sequence in $\text{Mod}_ R$. Since $M$ is flat,

\[ 0 \to N \otimes _ R M \to P \otimes _ R M \to Q \otimes _ R M \to 0 \]

is exact. By Lemma 35.4.10,

\[ 0 \to C(Q \otimes _ R M) \to C(P \otimes _ R M) \to C(N \otimes _ R M) \to 0 \]

is exact. By (35.4.11.1), this last sequence can be rewritten as

\[ 0 \to \mathop{\mathrm{Hom}}\nolimits _ R(Q, C(M)) \to \mathop{\mathrm{Hom}}\nolimits _ R(P, C(M)) \to \mathop{\mathrm{Hom}}\nolimits _ R(N, C(M)) \to 0. \]

Hence $C(M)$ is an injective object of $\text{Mod}_ R$. $\square$

Theorem 35.4.25. If $M \otimes _ R S$ has one of the following properties as an $S$-module

  1. finitely generated;

  2. finitely presented;

  3. flat;

  4. faithfully flat;

  5. finite projective;

then so does $M$ as an $R$-module (and conversely).

Proof. To prove (a), choose a finite set $\{ n_ i\} $ of generators of $M \otimes _ R S$ in $\text{Mod}_ S$. Write each $n_ i$ as $\sum _ j m_{ij} \otimes s_{ij}$ with $m_{ij} \in M$ and $s_{ij} \in S$. Let $F$ be the finite free $R$-module with basis $e_{ij}$ and let $F \to M$ be the $R$-module map sending $e_{ij}$ to $m_{ij}$. Then $F \otimes _ R S\to M \otimes _ R S$ is surjective, so $\mathop{\mathrm{Coker}}(F \to M) \otimes _ R S$ is zero and hence $\mathop{\mathrm{Coker}}(F \to M)$ is zero. This proves (a).

To see (b) assume $M \otimes _ R S$ is finitely presented. Then $M$ is finitely generated by (a). Choose a surjection $R^{\oplus n} \to M$ with kernel $K$. Then $K \otimes _ R S \to S^{\oplus r} \to M \otimes _ R S \to 0$ is exact. By Algebra, Lemma 10.5.3 the kernel of $S^{\oplus r} \to M \otimes _ R S$ is a finite $S$-module. Thus we can find finitely many elements $k_1, \ldots , k_ t \in K$ such that the images of $k_ i \otimes 1$ in $S^{\oplus r}$ generate the kernel of $S^{\oplus r} \to M \otimes _ R S$. Let $K' \subset K$ be the submodule generated by $k_1, \ldots , k_ t$. Then $M' = R^{\oplus r}/K'$ is a finitely presented $R$-module with a morphism $M' \to M$ such that $M' \otimes _ R S \to M \otimes _ R S$ is an isomorphism. Thus $M' \cong M$ as desired.

To prove (c), let $0 \to M' \to M'' \to M \to 0$ be a short exact sequence in $\text{Mod}_ R$. Since $\bullet \otimes _ R S$ is a right exact functor, $M'' \otimes _ R S \to M \otimes _ R S$ is surjective. So by Lemma 35.4.10 the map $C(M \otimes _ R S) \to C(M'' \otimes _ R S)$ is injective. If $M \otimes _ R S$ is flat, then Lemma 35.4.24 shows $C(M \otimes _ R S)$ is an injective object of $\text{Mod}_ S$, so the injection $C(M \otimes _ R S) \to C(M'' \otimes _ R S)$ is split in $\text{Mod}_ S$ and hence also in $\text{Mod}_ R$. Since $C(M \otimes _ R S) \to C(M)$ is a split surjection by Lemma 35.4.12, it follows that $C(M) \to C(M'')$ is a split injection in $\text{Mod}_ R$. That is, the sequence

\[ 0 \to C(M) \to C(M'') \to C(M') \to 0 \]

is split exact. For $N \in \text{Mod}_ R$, by (35.4.11.1) we see that

\[ 0 \to C(M \otimes _ R N) \to C(M'' \otimes _ R N) \to C(M' \otimes _ R N) \to 0 \]

is split exact. By Lemma 35.4.10,

\[ 0 \to M' \otimes _ R N \to M'' \otimes _ R N \to M \otimes _ R N \to 0 \]

is exact. This implies $M$ is flat over $R$. Namely, taking $M'$ a free module surjecting onto $M$ we conclude that $\text{Tor}_1^ R(M, N) = 0$ for all modules $N$ and we can use Algebra, Lemma 10.75.8. This proves (c).

To deduce (d) from (c), note that if $N \in \text{Mod}_ R$ and $M \otimes _ R N$ is zero, then $M \otimes _ R S \otimes _ S (N \otimes _ R S) \cong (M \otimes _ R N) \otimes _ R S$ is zero, so $N \otimes _ R S$ is zero and hence $N$ is zero.

To deduce (e) at this point, it suffices to recall that $M$ is finitely generated and projective if and only if it is finitely presented and flat. See Algebra, Lemma 10.78.2. $\square$

There is a variant for $R$-algebras.

Theorem 35.4.26. If $A \otimes _ R S$ has one of the following properties as an $S$-algebra

  1. of finite type;

  2. of finite presentation;

  3. formally unramified;

  4. unramified;

  5. étale;

then so does $A$ as an $R$-algebra (and of course conversely).

Proof. To prove (a), choose a finite set $\{ x_ i\} $ of generators of $A \otimes _ R S$ over $S$. Write each $x_ i$ as $\sum _ j y_{ij} \otimes s_{ij}$ with $y_{ij} \in A$ and $s_{ij} \in S$. Let $F$ be the polynomial $R$-algebra on variables $e_{ij}$ and let $F \to M$ be the $R$-algebra map sending $e_{ij}$ to $y_{ij}$. Then $F \otimes _ R S\to A \otimes _ R S$ is surjective, so $\mathop{\mathrm{Coker}}(F \to A) \otimes _ R S$ is zero and hence $\mathop{\mathrm{Coker}}(F \to A)$ is zero. This proves (a).

To see (b) assume $A \otimes _ R S$ is a finitely presented $S$-algebra. Then $A$ is finite type over $R$ by (a). Choose a surjection $R[x_1, \ldots , x_ n] \to A$ with kernel $I$. Then $I \otimes _ R S \to S[x_1, \ldots , x_ n] \to A \otimes _ R S \to 0$ is exact. By Algebra, Lemma 10.6.3 the kernel of $S[x_1, \ldots , x_ n] \to A \otimes _ R S$ is a finitely generated ideal. Thus we can find finitely many elements $y_1, \ldots , y_ t \in I$ such that the images of $y_ i \otimes 1$ in $S[x_1, \ldots , x_ n]$ generate the kernel of $S[x_1, \ldots , x_ n] \to A \otimes _ R S$. Let $I' \subset I$ be the ideal generated by $y_1, \ldots , y_ t$. Then $A' = R[x_1, \ldots , x_ n]/I'$ is a finitely presented $R$-algebra with a morphism $A' \to A$ such that $A' \otimes _ R S \to A \otimes _ R S$ is an isomorphism. Thus $A' \cong A$ as desired.

To prove (c), recall that $A$ is formally unramified over $R$ if and only if the module of relative differentials $\Omega _{A/R}$ vanishes, see Algebra, Lemma 10.148.2 or [Proposition 17.2.1, EGA4]. Since $\Omega _{(A \otimes _ R S)/S} = \Omega _{A/R} \otimes _ R S$, the vanishing descends by Theorem 35.4.22.

To deduce (d) from the previous cases, recall that $A$ is unramified over $R$ if and only if $A$ is formally unramified and of finite type over $R$, see Algebra, Lemma 10.151.2.

To prove (e), recall that by Algebra, Lemma 10.151.8 or [Théorème 17.6.1, EGA4] the algebra $A$ is étale over $R$ if and only if $A$ is flat, unramified, and of finite presentation over $R$. $\square$

Remark 35.4.27. It would make things easier to have a faithfully flat ring homomorphism $g: R \to T$ for which $T \to S \otimes _ R T$ has some extra structure. For instance, if one could ensure that $T \to S \otimes _ R T$ is split in $\textit{Rings}$, then it would follow that every property of a module or algebra which is stable under base extension and which descends along faithfully flat morphisms also descends along universally injective morphisms. An obvious guess would be to find $g$ for which $T$ is not only faithfully flat but also injective in $\text{Mod}_ R$, but even for $R = \mathbf{Z}$ no such homomorphism can exist.

[1] To be precise, our $\theta $ here is the inverse of $\varphi $ from Definition 35.3.1.

Comments (4)

Comment #1688 by Aravind Asok on

Typo: One may then ask whether one there is"; removeone"

Comment #2458 by Matthieu Romagny on

In introductory blurb, the word 'out' is missing in the sentence 'The proof turns out to be entirely category-theoretic'.

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