# The Stacks Project

## Tag 08WE

### 34.4. 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].

#### 34.4.1. 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 $$\tag{34.4.1.1} \xymatrix{ A \ar[r]^f & B \ar@<1ex>[r]^{g_1} \ar@<-1ex>[r]_{g_2} & C }$$ 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 34.4.2. A split equalizer is a diagram (34.4.1.1) 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 $$\tag{34.4.2.1} h \circ f = 1_A, \quad f \circ h = i \circ g_1, \quad i \circ g_2 = 1_B.$$

The point is that the equalities among arrows force (34.4.1.1) 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.

#### 34.4.3. 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 34.4.4. 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 10.81.1.

Definition 34.4.5. A ring map $f: R \to S$ is universally injective if it is universally injective as a morphism in $\text{Mod}_R$.

Example 34.4.6. Any split injection in $\text{Mod}_R$ is universally injective. In particular, any split injection in $\textit{Rings}$ is universally injective.

Example 34.4.7. 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 34.4.8, 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 34.4.8. Any faithfully flat ring map is universally injective.

Proof. This is a reformulation of Algebra, Lemma 10.81.11. $\square$

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 34.4.9. Let $R$ be a ring. Define the contravariant functor $C$ $: \text{Mod}_R \to \text{Mod}_R$ by setting $$C(M) = \mathop{\rm Hom}\nolimits_{\textit{Ab}}(M, \mathbf{Q}/\mathbf{Z}),$$ 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 15.51.

Lemma 34.4.10. 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 15.51.6 and the other properties follow from this, see Remark 34.4.4. $\square$

Remark 34.4.11. We will use frequently the standard adjunction between $\mathop{\rm Hom}\nolimits$ and tensor product, in the form of the natural isomorphism of contravariant functors $$\tag{34.4.11.1} C(\bullet_1 \otimes_R \bullet_2) \cong \mathop{\rm Hom}\nolimits_R(\bullet_1, C(\bullet_2)): \text{Mod}_R \times \text{Mod}_R \to \text{Mod}_R$$ 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 10.13.5. 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 34.4.12. 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 (34.4.11.1), for any $P \in \text{Mod}_R$ we have a commutative diagram $$\xymatrix@C=9pc{ \mathop{\rm Hom}\nolimits_R( P, C(N)) \ar[r]_{\mathop{\rm Hom}\nolimits_R(P,C(f))} \ar[d]^{\cong} & \mathop{\rm Hom}\nolimits_R(P,C(M)) \ar[d]^{\cong} \\ C(P \otimes_R N ) \ar[r]^{C(1_{P} \otimes f)} & C(P \otimes_R M ). }$$ 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 $1_{C(M)} \in \mathop{\rm Hom}\nolimits_R(C(M), C(M))$ to some $g \in \mathop{\rm Hom}\nolimits_R(C(N), C(M))$ splitting $C(f)$. Conversely, if $C(f)$ is a split surjection, then both rows in the above diagram are surjective, so by Lemma 34.4.10, $1_{P} \otimes f$ is injective. $\square$

Remark 34.4.13. 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 (34.4.11.1) this amounts to splitting $\mathop{\rm Hom}\nolimits_R(P, C(f))$ functorially in $P$, and this is achieved by the map $g \circ \bullet$.

#### 34.4.14. Descent for modules and their morphisms

Throughout this subsection, fix a ring map $f: R \to S$. As seen in Section 34.3 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 34.3.)

Recall1 from Definition 34.3.1 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 $$\tag{34.4.14.1} (\theta \otimes \delta_2^2) \circ (\theta \otimes \delta_2^0) = (\theta \otimes \delta_2^1): M \otimes_{S, \delta^1_{01}} S_3 \to M \otimes_{S,\delta^1_{12}} S_3.$$ 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 34.4.15. 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 34.4.16. For $(M,\theta) \in DD_{S/R}$, the diagram $$\tag{34.4.16.1} \xymatrix@C=8pc{ M \ar[r]^{\theta \circ (1_M \otimes \delta_0^1)} & M \otimes_{S, \delta_1^1} S_2 \ar@<1ex>[r]^{(\theta \otimes \delta_2^2) \circ (1_M \otimes \delta^2_0)} \ar@<-1ex>[r]_{1_{M \otimes S_2} \otimes \delta^2_1} & M \otimes_{S, \delta_{12}^1} S_3 }$$ 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 (34.4.2.1), the first follows from tensoring the cocycle condition (34.4.14.1) with $\sigma_1^1$ and the others are immediate. $\square$

Lemma 34.4.17. For $(M, \theta) \in DD_{S/R}$, the diagram $$\tag{34.4.17.1} \xymatrix@C=8pc{ C(M \otimes_{S, \delta_{12}^1} S_3) \ar@<1ex>[r]^{C((\theta \otimes \delta_2^2) \circ (1_M \otimes \delta^2_0))} \ar@<-1ex>[r]_{C(1_{M \otimes S_2} \otimes \delta^2_1)} & C(M \otimes_{S, \delta_1^1} S_2 ) \ar[r]^{C(\theta \circ (1_M \otimes \delta_0^1))} & C(M). }$$ obtained by applying $C$ to (34.4.16.1) is a split coequalizer.

Proof. Omitted. $\square$

Lemma 34.4.18. The diagram $$\tag{34.4.18.1} \xymatrix@C=8pc{ S_1 \ar[r]^{\delta^1_1} & S_2 \ar@<1ex>[r]^{\delta^2_2} \ar@<-1ex>[r]_{\delta^2_1} & S_3 }$$ is a split equalizer.

Proof. In Lemma 34.4.16, take $(M, \theta) = f^*(S)$. $\square$

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

Definition 34.4.19. 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 $$\tag{34.4.19.1} \xymatrix@C=8pc{f_*(M,\theta) \ar[r] & M \ar@<1ex>^{\theta \circ (1_M \otimes \delta_0^1)}[r] \ar@<-1ex>_{1_M \otimes \delta_1^1}[r] & M \otimes_{S, \delta_1^1} S_2 }$$ is an equalizer.

Using Lemma 34.4.16 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 34.4.21), but in the general case some argument is needed.

Lemma 34.4.20. If $f$ is universally injective, then the diagram $$\tag{34.4.20.1} \xymatrix@C=8pc{ f_*(M, \theta) \otimes_R S \ar[r]^{\theta \circ (1_M \otimes \delta_0^1)} & M \otimes_{S, \delta_1^1} S_2 \ar@<1ex>[r]^{(\theta \otimes \delta_2^2) \circ (1_M \otimes \delta^2_0)} \ar@<-1ex>[r]_{1_{M \otimes S_2} \otimes \delta^2_1} & M \otimes_{S, \delta_{12}^1} S_3 }$$ obtained by tensoring (34.4.19.1) over $R$ with $S$ is an equalizer.

Proof. By Lemma 34.4.12 and Remark 34.4.13, 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 (34.4.17.1); 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 34.4.11, 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 34.4.10, we conclude (34.4.20.1) must also be an equalizer. $\square$

Remark 34.4.21. 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 34.4.20 is immediate because tensoring over $R$ with $S$ preserves all equalizers.

Theorem 34.4.22. The following conditions are equivalent.

1. (a)    The morphism $f$ is a descent morphism for modules.
2. (b)    The morphism $f$ is an effective descent morphism for modules.
3. (c)    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 34.4.20, 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 (34.4.18.1) 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. $\square$

#### 34.4.23. 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 34.4.24. 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 34.4.10, $$0 \to C(Q \otimes_R M) \to C(P \otimes_R M) \to C(N \otimes_R M) \to 0$$ is exact. By (34.4.11.1), this last sequence can be rewritten as $$0 \to \mathop{\rm Hom}\nolimits_R(Q, C(M)) \to \mathop{\rm Hom}\nolimits_R(P, C(M)) \to \mathop{\rm Hom}\nolimits_R(N, C(M)) \to 0.$$ Hence $C(M)$ is an injective object of $\text{Mod}_R$. $\square$

Theorem 34.4.25. If $M \otimes_R S$ has one of the following properties as an $S$-module

1. (a)    finitely generated;
2. (b)    finitely presented;
3. (c)    flat;
4. (d)    faithfully flat;
5. (e)    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 $\mathop{\rm Coker}(F \to M) \otimes_R S$ is zero and hence $\mathop{\rm Coker}(F \to M)$ 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 34.4.10 the map $C(M \otimes_R S) \to C(M'' \otimes_R S)$ is injective. If $M \otimes_R S$ is flat, then Lemma 34.4.24 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 34.4.12, 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 (34.4.11.1) 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 34.4.10, $$0 \to M' \otimes_R N \to M'' \otimes_R N \to M \otimes_R N \to 0$$ 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 10.74.8. 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 10.77.2. $\square$

There is a variant for $R$-algebras.

Theorem 34.4.26. If $A \otimes_R S$ has one of the following properties as an $S$-algebra

1. (a)    of finite type;
2. (b)    of finite presentation;
3. (c)    formally unramified;
4. (d)    unramified;
5. (e)    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 $\mathop{\rm Coker}(F \to A) \otimes_R S$ is zero and hence $\mathop{\rm Coker}(F \to A)$ 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 10.144.2 or [EGA4, Proposition 17.2.1]. Since $\Omega_{(A \otimes_R S)/S} = \Omega_{A/R} \otimes_R S$, the vanishing descends by Theorem 34.4.22.

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 10.147.2.

To prove (e), recall that by Algebra, Lemma 10.147.8 or [EGA4, Théorème 17.6.1] the algebra $A$ is étale over $R$ if and only if $A$ is flat, unramified, and of finite presentation over $R$. $\square$

Remark 34.4.27. 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 34.3.1.

The code snippet corresponding to this tag is a part of the file descent.tex and is located in lines 762–1531 (see updates for more information).

\section{Descent for universally injective morphisms}
\label{section-descent-universally-injective}

\noindent
Numerous constructions in algebraic geometry are made using techniques of
{\it 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 {\it effective descent morphisms}.
Early in the Grothendieckian development of modern algebraic geometry,
the class of morphisms which are {\it quasi-compact} and {\it faithfully flat}
was shown to be effective for descending objects, morphisms, and many
properties thereof.

\medskip\noindent
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.

\medskip\noindent
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
{\it 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.

\medskip\noindent
The history of this result is a bit involved: it was originally asserted
by Olivier \cite{olivier}, who  called universally injective morphisms
{\it pure}, but without a clear indication of proof. One can extract the
result from the work of Joyal and Tierney \cite{joyal-tierney}, but to the
best of our knowledge, the first free-standing proof to appear in the
literature is that of Mesablishvili \cite{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 \cite{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 \cite{janelidze-tholen}.

\medskip\noindent
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 \cite{mesablishvili2}.

\subsection{Category-theoretic preliminaries}
\label{subsection-category-prelims}

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

\medskip\noindent
For two morphisms $g_1, g_2: B \to C$, recall that an {\it 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

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

is an equalizer. Reversing arrows gives the definition of a {\it coequalizer}.
See Categories, Sections \ref{categories-section-equalizers} and
\ref{categories-section-coequalizers}.

\medskip\noindent
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.

\begin{definition}
\label{definition-split-equalizer}
A {\it split equalizer} is a diagram (\ref{equation-equalizer}) 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

\label{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{definition}

\noindent
The point is that the equalities among arrows force (\ref{equation-equalizer})
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 {\it split coequalizer}, whose definition is apparent.

\subsection{Universally injective morphisms}
\label{subsection-universally-injective}

\noindent
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.

\begin{remark}
\label{remark-reflects}
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 \ref{homology-section-functors}).
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.
\end{remark}

\noindent
Let $R$ be a ring. Recall that a morphism $f : M \to N$ in $\text{Mod}_R$
is {\it 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 \ref{algebra-definition-universally-injective}.

\begin{definition}
\label{definition-universally-injective}
A ring map $f: R \to S$ is {\it universally injective}
if it is universally injective as a morphism in $\text{Mod}_R$.
\end{definition}

\begin{example}
\label{example-split-injection-universally-injective}
Any split injection in $\text{Mod}_R$ is universally injective. In particular,
any split injection in $\textit{Rings}$ is universally injective.
\end{example}

\begin{example}
\label{example-cover-universally-injective}
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 \ref{lemma-faithfully-flat-universally-injective},
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.
\end{example}

\begin{lemma}
\label{lemma-faithfully-flat-universally-injective}
Any faithfully flat ring map is universally injective.
\end{lemma}

\begin{proof}
This is a reformulation of Algebra, Lemma
\ref{algebra-lemma-faithfully-flat-universally-injective}.
\end{proof}

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

\begin{definition}
\label{definition-C}
Let $R$ be a ring. Define the contravariant functor
{\it $C$} $: \text{Mod}_R \to \text{Mod}_R$ by setting
$$C(M) = \Hom_{\textit{Ab}}(M, \mathbf{Q}/\mathbf{Z}),$$
with the $R$-action on $C(M)$ given by $rf(s) = f(rs)$.
\end{definition}

\noindent
This functor was denoted $M \mapsto M^\vee$ in
More on Algebra, Section \ref{more-algebra-section-injectives-modules}.

\begin{lemma}
\label{lemma-C-is-faithful}
For a ring $R$, the functor $C : \text{Mod}_R \to \text{Mod}_R$ is
exact and reflects injections and surjections.
\end{lemma}

\begin{proof}
Exactness is More on Algebra, Lemma \ref{more-algebra-lemma-vee-exact}
and the other properties follow from this, see
Remark \ref{remark-reflects}.
\end{proof}

\begin{remark}
We will use frequently the standard adjunction between $\Hom$ and tensor
product, in the form of the natural isomorphism of contravariant functors

C(\bullet_1 \otimes_R \bullet_2) \cong \Hom_R(\bullet_1, C(\bullet_2)):
\text{Mod}_R \times \text{Mod}_R \to \text{Mod}_R

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 \ref{algebra-lemma-hom-from-tensor-product-variant}.
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 $Q \in \text{Mod}_R$.
\end{remark}

\begin{lemma}
\label{lemma-split-surjection}
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.
\end{lemma}

\begin{proof}
By (\ref{equation-adjunction}), for any $P \in \text{Mod}_R$ we have a
commutative diagram
$$\xymatrix@C=9pc{ \Hom_R( P, C(N)) \ar[r]_{\Hom_R(P,C(f))} \ar[d]^{\cong} & \Hom_R(P,C(M)) \ar[d]^{\cong} \\ C(P \otimes_R N ) \ar[r]^{C(1_{P} \otimes f)} & C(P \otimes_R M ). }$$
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
$1_{C(M)} \in \Hom_R(C(M), C(M))$ to some $g \in \Hom_R(C(N), C(M))$ splitting
$C(f)$.
Conversely, if $C(f)$ is a split surjection, then
both rows in the above diagram are surjective,
so by Lemma \ref{lemma-C-is-faithful}, $1_{P} \otimes f$ is injective.
\end{proof}

\begin{remark}
\label{remark-functorial-splitting}
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 (\ref{equation-adjunction}) this amounts to splitting $\Hom_R(P, C(f))$  functorially in $P$,
and this is achieved by the map $g \circ \bullet$.
\end{remark}

\subsection{Descent for modules and their morphisms}
\label{subsection-descent-modules-morphisms}

\noindent
Throughout this subsection, fix a ring map $f: R \to S$. As seen in
Section \ref{section-descent-modules} 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.

\medskip\noindent
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 \ref{section-descent-modules}.)

\medskip\noindent
Recall\footnote{To be precise, our $\theta$ here is the inverse of
$\varphi$ from Definition \ref{definition-descent-datum-modules}.}
from Definition \ref{definition-descent-datum-modules} that for
$M \in \text{Mod}_S$, a {\it 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 {\it cocycle condition}

\label{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.

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

\medskip\noindent
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 $f^*: \text{Mod}_R \to DD_{S/R}$.

\begin{definition}
\label{definition-effective-descent}
The functor $f^*: \text{Mod}_R \to DD_{S/R}$
is called {\it base extension along $f$}. We say that $f$ is a
{\it descent morphism for modules} if $f^*$ is fully
faithful. We say that $f$ is an {\it effective descent morphism for modules}
if $f^*$ is an equivalence of categories.
\end{definition}

\noindent
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.

\begin{lemma}
\label{lemma-equalizer-M}
For $(M,\theta) \in DD_{S/R}$, the diagram

\label{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
}

is a split equalizer.
\end{lemma}

\begin{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 (\ref{equation-split-equalizer-conditions}),
the first follows from tensoring the cocycle condition
(\ref{equation-cocycle-condition}) with $\sigma_1^1$
and the others are immediate.
\end{proof}

\begin{lemma}
\label{lemma-equalizer-CM}
For $(M, \theta) \in DD_{S/R}$, the diagram

\label{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).
}

obtained by applying $C$ to (\ref{equation-equalizer-M}) is a split
coequalizer.
\end{lemma}

\begin{proof}
Omitted.
\end{proof}

\begin{lemma}
\label{lemma-equalizer-S}
The diagram

\label{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
}

is a split equalizer.
\end{lemma}

\begin{proof}
In Lemma \ref{lemma-equalizer-M}, take $(M, \theta) = f^*(S)$.
\end{proof}

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

\begin{definition}
\label{definition-pushforward}
Define the functor {\it $f_*$} $: DD_{S/R} \to \text{Mod}_R$ by taking
$f_*(M, \theta)$ to be the $R$-submodule of $M$ for which the diagram

\label{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
}

is an equalizer.
\end{definition}

\noindent
Using Lemma \ref{lemma-equalizer-M} 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 $f^*$.

\medskip\noindent
We are ready for the key lemma. In the faithfully flat case this is a
triviality (see Remark \ref{remark-descent-lemma}),
but in the general case some argument is needed.

\begin{lemma}
\label{lemma-descent-lemma}
If $f$ is universally injective, then the diagram

\label{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
}

obtained by tensoring (\ref{equation-equalizer-f}) over $R$ with $S$ is an
equalizer.
\end{lemma}

\begin{proof}
By
Lemma \ref{lemma-split-surjection} and
Remark \ref{remark-functorial-splitting},
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
(\ref{equation-coequalizer-CM}); 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 \ref{remark-adjunction}, 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 \ref{lemma-C-is-faithful}, we conclude
(\ref{equation-equalizer-f2}) must also be an equalizer.
\end{proof}

\begin{remark}
\label{remark-descent-lemma}
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 \ref{lemma-descent-lemma}
is immediate because tensoring over $R$ with $S$ preserves all equalizers.
\end{remark}

\begin{theorem}
\label{theorem-descent}
The following conditions are equivalent.
\begin{enumerate}
\item[(a)] The morphism $f$ is a descent morphism for modules.
\item[(b)] The morphism $f$ is an effective descent morphism for modules.
\item[(c)] The morphism $f$ is universally injective.
\end{enumerate}
\end{theorem}

\begin{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.

\medskip\noindent
We finally check that (c) implies (b). By Lemma \ref{lemma-descent-lemma}, 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 (\ref{equation-equalizer-S}) 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.
\end{proof}

\subsection{Descent for properties of modules}
\label{subsection-descent-properties-modules}

\noindent
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
\cite{mesablishvili2}.

\begin{lemma}
\label{lemma-flat-to-injective}
If $M \in \text{Mod}_R$ is flat, then $C(M)$ is an injective $R$-module.
\end{lemma}

\begin{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 \ref{lemma-C-is-faithful},
$$0 \to C(Q \otimes_R M) \to C(P \otimes_R M) \to C(N \otimes_R M) \to 0$$
is exact. By (\ref{equation-adjunction}), this last sequence can be rewritten
as
$$0 \to \Hom_R(Q, C(M)) \to \Hom_R(P, C(M)) \to \Hom_R(N, C(M)) \to 0.$$
Hence $C(M)$ is an injective object of $\text{Mod}_R$.
\end{proof}

\begin{theorem}
\label{theorem-descend-module-properties}
If $M \otimes_R S$ has one of the following properties as an $S$-module
\begin{enumerate}
\item[(a)]
finitely generated;
\item[(b)]
finitely presented;
\item[(c)]
flat;
\item[(d)]
faithfully flat;
\item[(e)]
finite projective;
\end{enumerate}
then so does $M$ as an $R$-module (and conversely).
\end{theorem}

\begin{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
$\Coker(F \to M) \otimes_R S$ is zero and hence $\Coker(F \to M)$
is zero. This proves (a).

\medskip\noindent
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 \ref{algebra-lemma-extension}
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.

\medskip\noindent
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 \ref{lemma-C-is-faithful} the map
$C(M \otimes_R S) \to C(M'' \otimes_R S)$ is injective.
If $M \otimes_R S$ is flat, then
Lemma \ref{lemma-flat-to-injective} 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 \ref{lemma-split-surjection}, 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 (\ref{equation-adjunction}) 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 \ref{lemma-C-is-faithful},
$$0 \to M' \otimes_R N \to M'' \otimes_R N \to M \otimes_R N \to 0$$
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 \ref{algebra-lemma-characterize-flat}.
This proves (c).

\medskip\noindent
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.

\medskip\noindent
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 \ref{algebra-lemma-finite-projective}.
\end{proof}

\noindent
There is a variant for $R$-algebras.

\begin{theorem}
\label{theorem-descend-algebra-properties}
If $A \otimes_R S$ has one of the following properties as an $S$-algebra
\begin{enumerate}
\item[(a)]
of finite type;
\item[(b)]
of finite presentation;
\item[(c)]
formally unramified;
\item[(d)]
unramified;
\item[(e)]
\'etale;
\end{enumerate}
then so does $A$ as an $R$-algebra (and of course conversely).
\end{theorem}

\begin{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
$\Coker(F \to A) \otimes_R S$ is zero and hence $\Coker(F \to A)$
is zero. This proves (a).

\medskip\noindent
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 \ref{algebra-lemma-finite-presentation-independent}
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.

\medskip\noindent
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 \ref{algebra-lemma-characterize-formally-unramified} or
\cite[Proposition~17.2.1]{EGA4}.
Since $\Omega_{(A \otimes_R S)/S} = \Omega_{A/R} \otimes_R S$,
the vanishing descends by Theorem \ref{theorem-descent}.

\medskip\noindent
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 \ref{algebra-lemma-formally-unramified-unramified}.

\medskip\noindent
To prove (e), recall that by
Algebra, Lemma \ref{algebra-lemma-etale-flat-unramified-finite-presentation}
or \cite[Th\'eor\eme~17.6.1]{EGA4} the algebra
$A$ is \'etale over $R$ if and only if
$A$ is flat, unramified, and of finite presentation over $R$.
\end{proof}

\begin{remark}
\label{remark-when-locally-split}
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.
\end{remark}

Comment #1688 by Aravind Asok on November 11, 2015 a 1:43 pm UTC

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

Comment #1736 by Johan (site) on December 15, 2015 a 6:46 pm UTC

Thanks, fixed here.

Comment #2458 by Matthieu Romagny on March 15, 2017 a 3:14 pm UTC

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

Comment #2497 by Johan (site) on April 13, 2017 a 11:19 pm UTC

Thanks, fixed here.

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