The Stacks project

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
\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
\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 (, the first follows from tensoring the cocycle condition ( with $\sigma _1^1$ and the others are immediate. $\square$

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

Proof. Omitted. $\square$

Lemma 35.4.18. The diagram
\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
\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
\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 ( 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 (; 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 ( 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 ( 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. $\square$

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

Comments (0)

There are also:

  • 4 comment(s) on Section 35.4: Descent for universally injective morphisms

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 08WW. Beware of the difference between the letter 'O' and the digit '0'.