## 35.3 Descent for modules

Let $R \to A$ be a ring map. By Simplicial, Example 14.5.5 this gives rise to a cosimplicial $R$-algebra

$\xymatrix{ A \ar@<1ex>[r] \ar@<-1ex>[r] & A \otimes _ R A \ar@<0ex>[l] \ar@<2ex>[r] \ar@<0ex>[r] \ar@<-2ex>[r] & A \otimes _ R A \otimes _ R A \ar@<1ex>[l] \ar@<-1ex>[l] }$

Let us denote this $(A/R)_\bullet$ so that $(A/R)_ n$ is the $(n + 1)$-fold tensor product of $A$ over $R$. Given a map $\varphi : [n] \to [m]$ the $R$-algebra map $(A/R)_\bullet (\varphi )$ is the map

$a_0 \otimes \ldots \otimes a_ n \longmapsto \prod \nolimits _{\varphi (i) = 0} a_ i \otimes \prod \nolimits _{\varphi (i) = 1} a_ i \otimes \ldots \otimes \prod \nolimits _{\varphi (i) = m} a_ i$

where we use the convention that the empty product is $1$. Thus the first few maps, notation as in Simplicial, Section 14.5, are

$\begin{matrix} \delta ^1_0 & : & a_0 & \mapsto & 1 \otimes a_0 \\ \delta ^1_1 & : & a_0 & \mapsto & a_0 \otimes 1 \\ \sigma ^0_0 & : & a_0 \otimes a_1 & \mapsto & a_0a_1 \\ \delta ^2_0 & : & a_0 \otimes a_1 & \mapsto & 1 \otimes a_0 \otimes a_1 \\ \delta ^2_1 & : & a_0 \otimes a_1 & \mapsto & a_0 \otimes 1 \otimes a_1 \\ \delta ^2_2 & : & a_0 \otimes a_1 & \mapsto & a_0 \otimes a_1 \otimes 1 \\ \sigma ^1_0 & : & a_0 \otimes a_1 \otimes a_2 & \mapsto & a_0a_1 \otimes a_2 \\ \sigma ^1_1 & : & a_0 \otimes a_1 \otimes a_2 & \mapsto & a_0 \otimes a_1a_2 \end{matrix}$

and so on.

An $R$-module $M$ gives rise to a cosimplicial $(A/R)_\bullet$-module $(A/R)_\bullet \otimes _ R M$. In other words $M_ n = (A/R)_ n \otimes _ R M$ and using the $R$-algebra maps $(A/R)_ n \to (A/R)_ m$ to define the corresponding maps on $M \otimes _ R (A/R)_\bullet$.

The analogue to a descent datum for quasi-coherent sheaves in the setting of modules is the following.

Definition 35.3.1. Let $R \to A$ be a ring map.

1. A descent datum $(N, \varphi )$ for modules with respect to $R \to A$ is given by an $A$-module $N$ and an isomorphism of $A \otimes _ R A$-modules

$\varphi : N \otimes _ R A \to A \otimes _ R N$

such that the cocycle condition holds: the diagram of $A \otimes _ R A \otimes _ R A$-module maps

$\xymatrix{ N \otimes _ R A \otimes _ R A \ar[rr]_{\varphi _{02}} \ar[rd]_{\varphi _{01}} & & A \otimes _ R A \otimes _ R N \\ & A \otimes _ R N \otimes _ R A \ar[ru]_{\varphi _{12}} & }$

commutes (see below for notation).

2. A morphism $(N, \varphi ) \to (N', \varphi ')$ of descent data is a morphism of $A$-modules $\psi : N \to N'$ such that the diagram

$\xymatrix{ N \otimes _ R A \ar[r]_\varphi \ar[d]_{\psi \otimes \text{id}_ A} & A \otimes _ R N \ar[d]^{\text{id}_ A \otimes \psi } \\ N' \otimes _ R A \ar[r]^{\varphi '} & A \otimes _ R N' }$

is commutative.

In the definition we use the notation that $\varphi _{01} = \varphi \otimes \text{id}_ A$, $\varphi _{12} = \text{id}_ A \otimes \varphi$, and $\varphi _{02}(n \otimes 1 \otimes 1) = \sum a_ i \otimes 1 \otimes n_ i$ if $\varphi (n \otimes 1) = \sum a_ i \otimes n_ i$. All three are $A \otimes _ R A \otimes _ R A$-module homomorphisms. Equivalently we have

$\varphi _{ij} = \varphi \otimes _{(A/R)_1, \ (A/R)_\bullet (\tau ^2_{ij})} (A/R)_2$

where $\tau ^2_{ij} : [1] \to [2]$ is the map $0 \mapsto i$, $1 \mapsto j$. Namely, $(A/R)_{\bullet }(\tau ^2_{02})(a_0 \otimes a_1) = a_0 \otimes 1 \otimes a_1$, and similarly for the others1.

We need some more notation to be able to state the next lemma. Let $(N, \varphi )$ be a descent datum with respect to a ring map $R \to A$. For $n \geq 0$ and $i \in [n]$ we set

$N_{n, i} = A \otimes _ R \ldots \otimes _ R A \otimes _ R N \otimes _ R A \otimes _ R \ldots \otimes _ R A$

with the factor $N$ in the $i$th spot. It is an $(A/R)_ n$-module. If we introduce the maps $\tau ^ n_ i : [0] \to [n]$, $0 \mapsto i$ then we see that

$N_{n, i} = N \otimes _{(A/R)_0, \ (A/R)_\bullet (\tau ^ n_ i)} (A/R)_ n$

For $0 \leq i \leq j \leq n$ we let $\tau ^ n_{ij} : [1] \to [n]$ be the map such that $0$ maps to $i$ and $1$ to $j$. Similarly to the above the homomorphism $\varphi$ induces isomorphisms

$\varphi ^ n_{ij} = \varphi \otimes _{(A/R)_1, \ (A/R)_\bullet (\tau ^ n_{ij})} (A/R)_ n : N_{n, i} \longrightarrow N_{n, j}$

of $(A/R)_ n$-modules when $i < j$. If $i = j$ we set $\varphi ^ n_{ij} = \text{id}$. Since these are all isomorphisms they allow us to move the factor $N$ to any spot we like. And the cocycle condition exactly means that it does not matter how we do this (e.g., as a composition of two of these or at once). Finally, for any $\beta : [n] \to [m]$ we define the morphism

$N_{\beta , i} : N_{n, i} \to N_{m, \beta (i)}$

as the unique $(A/R)_\bullet (\beta )$-semi linear map such that

$N_{\beta , i}(1 \otimes \ldots \otimes n \otimes \ldots \otimes 1) = 1 \otimes \ldots \otimes n \otimes \ldots \otimes 1$

for all $n \in N$. This hints at the following lemma.

Lemma 35.3.2. Let $R \to A$ be a ring map. Given a descent datum $(N, \varphi )$ we can associate to it a cosimplicial $(A/R)_\bullet$-module $N_\bullet$2 by the rules $N_ n = N_{n, n}$ and given $\beta : [n] \to [m]$ setting we define

$N_\bullet (\beta ) = (\varphi ^ m_{\beta (n)m}) \circ N_{\beta , n} : N_{n, n} \longrightarrow N_{m, m}.$

This procedure is functorial in the descent datum.

Proof. Here are the first few maps where $\varphi (n \otimes 1) = \sum \alpha _ i \otimes x_ i$

$\begin{matrix} \delta ^1_0 & : & N & \to & A \otimes N & n & \mapsto & 1 \otimes n \\ \delta ^1_1 & : & N & \to & A \otimes N & n & \mapsto & \sum \alpha _ i \otimes x_ i \\ \sigma ^0_0 & : & A \otimes N & \to & N & a_0 \otimes n & \mapsto & a_0n \\ \delta ^2_0 & : & A \otimes N & \to & A \otimes A \otimes N & a_0 \otimes n & \mapsto & 1 \otimes a_0 \otimes n \\ \delta ^2_1 & : & A \otimes N & \to & A \otimes A \otimes N & a_0 \otimes n & \mapsto & a_0 \otimes 1 \otimes n \\ \delta ^2_2 & : & A \otimes N & \to & A \otimes A \otimes N & a_0 \otimes n & \mapsto & \sum a_0 \otimes \alpha _ i \otimes x_ i \\ \sigma ^1_0 & : & A \otimes A \otimes N & \to & A \otimes N & a_0 \otimes a_1 \otimes n & \mapsto & a_0a_1 \otimes n \\ \sigma ^1_1 & : & A \otimes A \otimes N & \to & A \otimes N & a_0 \otimes a_1 \otimes n & \mapsto & a_0 \otimes a_1n \end{matrix}$

with notation as in Simplicial, Section 14.5. We first verify the two properties $\sigma ^0_0 \circ \delta ^1_0 = \text{id}$ and $\sigma ^0_0 \circ \delta ^1_1 = \text{id}$. The first one, $\sigma ^0_0 \circ \delta ^1_0 = \text{id}$, is clear from the explicit description of the morphisms above. To prove the second relation we have to use the cocycle condition (because it does not holds for an arbitrary isomorphism $\varphi : N \otimes _ R A \to A \otimes _ R N$). Write $p = \sigma ^0_0 \circ \delta ^1_1 : N \to N$. By the description of the maps above we deduce that $p$ is also equal to

$p = \varphi \otimes \text{id} : N = (N \otimes _ R A) \otimes _{(A \otimes _ R A)} A \longrightarrow (A \otimes _ R N) \otimes _{(A \otimes _ R A)} A = N$

Since $\varphi$ is an isomorphism we see that $p$ is an isomorphism. Write $\varphi (n \otimes 1) = \sum \alpha _ i \otimes x_ i$ for certain $\alpha _ i \in A$ and $x_ i \in N$. Then $p(n) = \sum \alpha _ ix_ i$. Next, write $\varphi (x_ i \otimes 1) = \sum \alpha _{ij} \otimes y_ j$ for certain $\alpha _{ij} \in A$ and $y_ j \in N$. Then the cocycle condition says that

$\sum \alpha _ i \otimes \alpha _{ij} \otimes y_ j = \sum \alpha _ i \otimes 1 \otimes x_ i.$

This means that $p(n) = \sum \alpha _ ix_ i = \sum \alpha _ i\alpha _{ij}y_ j = \sum \alpha _ i p(x_ i) = p(p(n))$. Thus $p$ is a projector, and since it is an isomorphism it is the identity.

To prove fully that $N_\bullet$ is a cosimplicial module we have to check all 5 types of relations of Simplicial, Remark 14.5.3. The relations on composing $\sigma$'s are obvious. The relations on composing $\delta$'s come down to the cocycle condition for $\varphi$. In exactly the same way as above one checks the relations $\sigma _ j \circ \delta _ j = \sigma _ j \circ \delta _{j + 1} = \text{id}$. Finally, the other relations on compositions of $\delta$'s and $\sigma$'s hold for any $\varphi$ whatsoever. $\square$

Note that to an $R$-module $M$ we can associate a canonical descent datum, namely $(M \otimes _ R A, can)$ where $can : (M \otimes _ R A) \otimes _ R A \to A \otimes _ R (M \otimes _ R A)$ is the obvious map: $(m \otimes a) \otimes a' \mapsto a \otimes (m \otimes a')$.

Lemma 35.3.3. Let $R \to A$ be a ring map. Let $M$ be an $R$-module. The cosimplicial $(A/R)_\bullet$-module associated to the canonical descent datum is isomorphic to the cosimplicial module $(A/R)_\bullet \otimes _ R M$.

Proof. Omitted. $\square$

Definition 35.3.4. Let $R \to A$ be a ring map. We say a descent datum $(N, \varphi )$ is effective if there exists an $R$-module $M$ and an isomorphism of descent data from $(M \otimes _ R A, can)$ to $(N, \varphi )$.

Let $R \to A$ be a ring map. Let $(N, \varphi )$ be a descent datum. We may take the cochain complex $s(N_\bullet )$ associated with $N_\bullet$ (see Simplicial, Section 14.25). It has the following shape:

$N \to A \otimes _ R N \to A \otimes _ R A \otimes _ R N \to \ldots$

We can describe the maps. The first map is the map

$n \longmapsto 1 \otimes n - \varphi (n \otimes 1).$

The second map on pure tensors has the values

$a \otimes n \longmapsto 1 \otimes a \otimes n - a \otimes 1 \otimes n + a \otimes \varphi (n \otimes 1).$

It is clear how the pattern continues.

In the special case where $N = A \otimes _ R M$ we see that for any $m \in M$ the element $1 \otimes m$ is in the kernel of the first map of the cochain complex associated to the cosimplicial module $(A/R)_\bullet \otimes _ R M$. Hence we get an extended cochain complex

35.3.4.1
$$\label{descent-equation-extended-complex} 0 \to M \to A \otimes _ R M \to A \otimes _ R A \otimes _ R M \to \ldots$$

Here we think of the $0$ as being in degree $-2$, the module $M$ in degree $-1$, the module $A \otimes _ R M$ in degree $0$, etc. Note that this complex has the shape

$0 \to R \to A \to A \otimes _ R A \to A \otimes _ R A \otimes _ R A \to \ldots$

when $M = R$.

Lemma 35.3.5. Suppose that $R \to A$ has a section. Then for any $R$-module $M$ the extended cochain complex (35.3.4.1) is exact.

Proof. By Simplicial, Lemma 14.28.5 the map $R \to (A/R)_\bullet$ is a homotopy equivalence of cosimplicial $R$-algebras (here $R$ denotes the constant cosimplicial $R$-algebra). Hence $M \to (A/R)_\bullet \otimes _ R M$ is a homotopy equivalence in the category of cosimplicial $R$-modules, because $\otimes _ R M$ is a functor from the category of $R$-algebras to the category of $R$-modules, see Simplicial, Lemma 14.28.4. This implies that the induced map of associated complexes is a homotopy equivalence, see Simplicial, Lemma 14.28.6. Since the complex associated to the constant cosimplicial $R$-module $M$ is the complex

$\xymatrix{ M \ar[r]^0 & M \ar[r]^1 & M \ar[r]^0 & M \ar[r]^1 & M \ldots }$

we win (since the extended version simply puts an extra $M$ at the beginning). $\square$

Lemma 35.3.6. Suppose that $R \to A$ is faithfully flat, see Algebra, Definition 10.39.1. Then for any $R$-module $M$ the extended cochain complex (35.3.4.1) is exact.

Proof. Suppose we can show there exists a faithfully flat ring map $R \to R'$ such that the result holds for the ring map $R' \to A' = R' \otimes _ R A$. Then the result follows for $R \to A$. Namely, for any $R$-module $M$ the cosimplicial module $(M \otimes _ R R') \otimes _{R'} (A'/R')_\bullet$ is just the cosimplicial module $R' \otimes _ R (M \otimes _ R (A/R)_\bullet )$. Hence the vanishing of cohomology of the complex associated to $(M \otimes _ R R') \otimes _{R'} (A'/R')_\bullet$ implies the vanishing of the cohomology of the complex associated to $M \otimes _ R (A/R)_\bullet$ by faithful flatness of $R \to R'$. Similarly for the vanishing of cohomology groups in degrees $-1$ and $0$ of the extended complex (proof omitted).

But we have such a faithful flat extension. Namely $R' = A$ works because the ring map $R' = A \to A' = A \otimes _ R A$ has a section $a \otimes a' \mapsto aa'$ and Lemma 35.3.5 applies. $\square$

Here is how the complex relates to the question of effectivity.

Lemma 35.3.7. Let $R \to A$ be a faithfully flat ring map. Let $(N, \varphi )$ be a descent datum. Then $(N, \varphi )$ is effective if and only if the canonical map

$A \otimes _ R H^0(s(N_\bullet )) \longrightarrow N$

is an isomorphism.

Proof. If $(N, \varphi )$ is effective, then we may write $N = A \otimes _ R M$ with $\varphi = can$. It follows that $H^0(s(N_\bullet )) = M$ by Lemmas 35.3.3 and 35.3.6. Conversely, suppose the map of the lemma is an isomorphism. In this case set $M = H^0(s(N_\bullet ))$. This is an $R$-submodule of $N$, namely $M = \{ n \in N \mid 1 \otimes n = \varphi (n \otimes 1)\}$. The only thing to check is that via the isomorphism $A \otimes _ R M \to N$ the canonical descent data agrees with $\varphi$. We omit the verification. $\square$

Lemma 35.3.8. Let $R \to A$ be a faithfully flat ring map, and let $R \to R'$ be faithfully flat. Set $A' = R' \otimes _ R A$. If all descent data for $R' \to A'$ are effective, then so are all descent data for $R \to A$.

Proof. Let $(N, \varphi )$ be a descent datum for $R \to A$. Set $N' = R' \otimes _ R N = A' \otimes _ A N$, and denote $\varphi ' = \text{id}_{R'} \otimes \varphi$ the base change of the descent datum $\varphi$. Then $(N', \varphi ')$ is a descent datum for $R' \to A'$ and $H^0(s(N'_\bullet )) = R' \otimes _ R H^0(s(N_\bullet ))$. Moreover, the map $A' \otimes _{R'} H^0(s(N'_\bullet )) \to N'$ is identified with the base change of the $A$-module map $A \otimes _ R H^0(s(N)) \to N$ via the faithfully flat map $A \to A'$. Hence we conclude by Lemma 35.3.7. $\square$

Here is the main result of this section. Its proof may seem a little clumsy; for a more highbrow approach see Remark 35.3.11 below.

Proposition 35.3.9. Let $R \to A$ be a faithfully flat ring map. Then

1. any descent datum on modules with respect to $R \to A$ is effective,

2. the functor $M \mapsto (A \otimes _ R M, can)$ from $R$-modules to the category of descent data is an equivalence, and

3. the inverse functor is given by $(N, \varphi ) \mapsto H^0(s(N_\bullet ))$.

Proof. We only prove (1) and omit the proofs of (2) and (3). As $R \to A$ is faithfully flat, there exists a faithfully flat base change $R \to R'$ such that $R' \to A' = R' \otimes _ R A$ has a section (namely take $R' = A$ as in the proof of Lemma 35.3.6). Hence, using Lemma 35.3.8 we may assume that $R \to A$ has a section, say $\sigma : A \to R$. Let $(N, \varphi )$ be a descent datum relative to $R \to A$. Set

$M = H^0(s(N_\bullet )) = \{ n \in N \mid 1 \otimes n = \varphi (n \otimes 1)\} \subset N$

By Lemma 35.3.7 it suffices to show that $A \otimes _ R M \to N$ is an isomorphism.

Take an element $n \in N$. Write $\varphi (n \otimes 1) = \sum a_ i \otimes x_ i$ for certain $a_ i \in A$ and $x_ i \in N$. By Lemma 35.3.2 we have $n = \sum a_ i x_ i$ in $N$ (because $\sigma ^0_0 \circ \delta ^1_1 = \text{id}$ in any cosimplicial object). Next, write $\varphi (x_ i \otimes 1) = \sum a_{ij} \otimes y_ j$ for certain $a_{ij} \in A$ and $y_ j \in N$. The cocycle condition means that

$\sum a_ i \otimes a_{ij} \otimes y_ j = \sum a_ i \otimes 1 \otimes x_ i$

in $A \otimes _ R A \otimes _ R N$. We conclude two things from this. First, by applying $\sigma$ to the first $A$ we conclude that $\sum \sigma (a_ i) \varphi (x_ i \otimes 1) = \sum \sigma (a_ i) \otimes x_ i$ which means that $\sum \sigma (a_ i) x_ i \in M$. Next, by applying $\sigma$ to the middle $A$ and multiplying out we conclude that $\sum _ i a_ i (\sum _ j \sigma (a_{ij}) y_ j) = \sum a_ i x_ i = n$. Hence by the first conclusion we see that $A \otimes _ R M \to N$ is surjective. Finally, suppose that $m_ i \in M$ and $\sum a_ i m_ i = 0$. Then we see by applying $\varphi$ to $\sum a_ im_ i \otimes 1$ that $\sum a_ i \otimes m_ i = 0$. In other words $A \otimes _ R M \to N$ is injective and we win. $\square$

Remark 35.3.10. Let $R$ be a ring. Let $f_1, \ldots , f_ n\in R$ generate the unit ideal. The ring $A = \prod _ i R_{f_ i}$ is a faithfully flat $R$-algebra. We remark that the cosimplicial ring $(A/R)_\bullet$ has the following ring in degree $n$:

$\prod \nolimits _{i_0, \ldots , i_ n} R_{f_{i_0}\ldots f_{i_ n}}$

Hence the results above recover Algebra, Lemmas 10.24.2, 10.24.1 and 10.24.5. But the results above actually say more because of exactness in higher degrees. Namely, it implies that Čech cohomology of quasi-coherent sheaves on affines is trivial. Thus we get a second proof of Cohomology of Schemes, Lemma 30.2.1.

Remark 35.3.11. Let $R$ be a ring. Let $A_\bullet$ be a cosimplicial $R$-algebra. In this setting a descent datum corresponds to an cosimplicial $A_\bullet$-module $M_\bullet$ with the property that for every $n, m \geq 0$ and every $\varphi : [n] \to [m]$ the map $M(\varphi ) : M_ n \to M_ m$ induces an isomorphism

$M_ n \otimes _{A_ n, A(\varphi )} A_ m \longrightarrow M_ m.$

Let us call such a cosimplicial module a cartesian module. In this setting, the proof of Proposition 35.3.9 can be split in the following steps

1. If $R \to R'$ and $R \to A$ are faithfully flat, then descent data for $A/R$ are effective if descent data for $(R' \otimes _ R A)/R'$ are effective.

2. Let $A$ be an $R$-algebra. Descent data for $A/R$ correspond to cartesian $(A/R)_\bullet$-modules.

3. If $R \to A$ has a section then $(A/R)_\bullet$ is homotopy equivalent to $R$, the constant cosimplicial $R$-algebra with value $R$.

4. If $A_\bullet \to B_\bullet$ is a homotopy equivalence of cosimplicial $R$-algebras then the functor $M_\bullet \mapsto M_\bullet \otimes _{A_\bullet } B_\bullet$ induces an equivalence of categories between cartesian $A_\bullet$-modules and cartesian $B_\bullet$-modules.

For (1) see Lemma 35.3.8. Part (2) uses Lemma 35.3.2. Part (3) we have seen in the proof of Lemma 35.3.5 (it relies on Simplicial, Lemma 14.28.5). Moreover, part (4) is a triviality if you think about it right!

[1] Note that $\tau ^2_{ij} = \delta ^2_ k$, if $\{ i, j, k\} = [2] = \{ 0, 1, 2\}$, see Simplicial, Definition 14.2.1.
[2] We should really write $(N, \varphi )_\bullet$.

Comment #1867 by Laurent Moret-Bailly on

Lemma 34.3.7: I think faithful flatness is not needed for the "if" part.

Comment #1902 by on

Sorry, I tried to figure out what you are saying, but I could not. In any case you do not seem to be saying the lemma is wrong, only that it could be improved upon, so I'll leave it for now.

Comment #5940 by Friedrich Knop on

Typo: "because it does not holds"

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