## 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.

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).

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 others^{1}.

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 hold 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
\begin{equation} \label{descent-equation-extended-complex} 0 \to M \to A \otimes _ R M \to A \otimes _ R A \otimes _ R M \to \ldots \end{equation}

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.

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

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

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

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:

applying $\sigma $ to the first $A$ we get $\sum \sigma (a_ i) \varphi (x_ i \otimes 1) = \sum \sigma (a_ i) \otimes x_ i$,

applying $\sigma $ to the middle $A$ we get $\sum _ i a_ i \otimes \sum _ j \sigma (a_{ij}) y_ j = \sum a_ i \otimes x_ i$.

Part (1) shows that $\sum \sigma (a_ i) x_ i \in M$. Applying this to $x_ i$ we see that $\sum \sigma (a_{ij})y_ i \in M$ for all $i$. Multiplying out the equation in (2) we conclude that $\sum _ i a_ i (\sum _ j \sigma (a_{ij}) y_ j) = \sum a_ i x_ i = n$. Hence $A \otimes _ R M \to N$ 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.
$\square$

## Comments (4)

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