The Stacks Project


Tag 023F

34.3. 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 34.3.1. 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 a 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 34.3.2. 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 34.3.3. 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 34.3.4. 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 \begin{equation} \tag{34.3.4.1} 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 34.3.5. Suppose that $R \to A$ has a section. Then for any $R$-module $M$ the extended cochain complex (34.3.4.1) is exact.

Proof. By Simplicial, Lemma 14.28.4 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 14.28.3. This implies that the induced map of associated complexes is a homotopy equivalence, see Simplicial, Lemma 14.28.5. 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 34.3.6. Suppose that $R \to A$ is faithfully flat, see Algebra, Definition 10.38.1. Then for any $R$-module $M$ the extended cochain complex (34.3.4.1) 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 34.3.5 applies. $\square$

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

Lemma 34.3.7. 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 34.3.3 and 34.3.6. 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 34.3.8. Let $R \to A$ be a 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 34.3.7. $\square$

Here is the main result of this section. Its proof may seem a little clumsy; for a more highbrow approach see Remark 34.3.11 below.

Proposition 34.3.9. 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 34.3.6). Hence, using Lemma 34.3.8 we may assume that $R \to A$ as 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 34.3.7 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 34.3.2 we have $n = \sum a_i x_i$ in $N$ (because $\sigma^0_0 \circ \delta^1_0 = \text{id}$ 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 34.3.10. 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 10.22.1, 10.22.2 and 10.23.4. 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 29.2.1.

Remark 34.3.11. 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 34.3.9 can be split in the following steps

  1. If $R \to R'$ is faithfully flat, $R \to A$ any ring map, 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 34.3.8. Part (2) uses Lemma 34.3.2. Part (3) we have seen in the proof of Lemma 34.3.5 (it relies on Simplicial, Lemma 14.28.4). 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 $(N, \varphi)_\bullet$.

The code snippet corresponding to this tag is a part of the file descent.tex and is located in lines 205–761 (see updates for more information).

\section{Descent for modules}
\label{section-descent-modules}

\noindent
Let $R \to A$ be a ring map.
By Simplicial, Example \ref{simplicial-example-push-outs-simplicial-object}
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 \ref{simplicial-section-cosimplicial-object}, 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.

\medskip\noindent
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
$M \otimes_R (A/R)_\bullet$.

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

\begin{definition}
\label{definition-descent-datum-modules}
Let $R \to A$ be a ring map.
\begin{enumerate}
\item A {\it descent datum $(N, \varphi)$ for modules
with respect to $R \to A$}
is given by an $A$-module $N$ and a isomorphism of
$A \otimes_R A$-modules
$$
\varphi : N \otimes_R A \to A \otimes_R N
$$
such that the {\it 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).
\item A {\it 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.
\end{enumerate}
\end{definition}

\noindent
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\footnote{Note that
$\tau^2_{ij} = \delta^2_k$, if $\{i, j, k\} = [2] = \{0, 1, 2\}$,
see Simplicial, Definition \ref{simplicial-definition-face-degeneracy}.}.

\medskip\noindent
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.

\begin{lemma}
\label{lemma-descent-datum-cosimplicial}
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 $N_\bullet$\footnote{We should really
write $(N, \varphi)_\bullet$.} 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.
\end{lemma}

\begin{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 \ref{simplicial-section-cosimplicial-object}.
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.

\medskip\noindent
To prove fully that $N_\bullet$ is a cosimplicial module we have to check
all 5 types of relations of
Simplicial, Remark \ref{simplicial-remark-relations-cosimplicial}.
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.
\end{proof}

\noindent
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:
$(m \otimes a) \otimes a' \mapsto a \otimes (m \otimes a')$.

\begin{lemma}
\label{lemma-canonical-descent-datum-cosimplicial}
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 $(A/R)_\bullet \otimes_R M$.
\end{lemma}

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

\begin{definition}
\label{definition-descent-datum-effective-module}
Let $R \to A$ be a ring map.
We say a descent datum $(N, \varphi)$ is {\it effective}
if there exists an $R$-module $M$ and an isomorphism
of descent data from $(M \otimes_R A, can)$ to
$(N, \varphi)$.
\end{definition}

\noindent
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 \ref{simplicial-section-dold-kan-cosimplicial}).
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.

\medskip\noindent
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
\begin{equation}
\label{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 $M = R$.

\begin{lemma}
\label{lemma-with-section-exact}
Suppose that $R \to A$ has a section.
Then for any $R$-module $M$ the extended cochain complex
(\ref{equation-extended-complex}) is exact.
\end{lemma}

\begin{proof}
By
Simplicial, Lemma \ref{simplicial-lemma-push-outs-simplicial-object-w-section}
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 \ref{simplicial-lemma-functorial-homotopy}.
This implies that the induced map of associated
complexes is a homotopy equivalence, see
Simplicial, Lemma \ref{simplicial-lemma-homotopy-s-Q}.
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).
\end{proof}

\begin{lemma}
\label{lemma-ff-exact}
Suppose that $R \to A$ is faithfully flat, see
Algebra, Definition \ref{algebra-definition-flat}.
Then for any $R$-module $M$ the extended cochain complex
(\ref{equation-extended-complex}) is exact.
\end{lemma}

\begin{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).

\medskip\noindent
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 \ref{lemma-with-section-exact}
applies.
\end{proof}

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

\begin{lemma}
\label{lemma-recognize-effective}
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.
\end{lemma}

\begin{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 \ref{lemma-canonical-descent-datum-cosimplicial}
and \ref{lemma-ff-exact}. 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.
\end{proof}

\begin{lemma}
\label{lemma-descent-descends}
Let $R \to A$ be a 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 $R \to A$.
\end{lemma}

\begin{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 \ref{lemma-recognize-effective}.
\end{proof}

\noindent
Here is the main result of this section.
Its proof may seem a little clumsy; for a more highbrow approach see
Remark \ref{remark-homotopy-equivalent-cosimplicial-algebras} below.

\begin{proposition}
\label{proposition-descent-module}
\begin{slogan}
Effective descent for modules along faithfully flat ring maps.
\end{slogan}
Let $R \to A$ be a faithfully flat ring map.
Then
\begin{enumerate}
\item any descent datum on modules with respect to $R \to A$
is effective,
\item the functor $M \mapsto (A \otimes_R M, can)$ from $R$-modules
to the category of descent data is an equivalence, and
\item the inverse functor is given by $(N, \varphi) \mapsto H^0(s(N_\bullet))$.
\end{enumerate}
\end{proposition}

\begin{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 \ref{lemma-ff-exact}). Hence, using
Lemma \ref{lemma-descent-descends}
we may assume that $R \to A$ as 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 \ref{lemma-recognize-effective} it suffices to show that
$A \otimes_R M \to N$ is an isomorphism.

\medskip\noindent
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 \ref{lemma-descent-datum-cosimplicial}
we have $n = \sum a_i x_i$ in $N$ (because
$\sigma^0_0 \circ \delta^1_0 = \text{id}$ 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.
\end{proof}

\begin{remark}
\label{remark-standard-covering}
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 \ref{algebra-lemma-standard-covering},
\ref{algebra-lemma-cover-module} and \ref{algebra-lemma-glue-modules}.
But the results above actually say more because of exactness
in higher degrees. Namely, it implies that {\v C}ech cohomology of
quasi-coherent sheaves on affines is trivial. Thus we get a second
proof of Cohomology of Schemes, Lemma
\ref{coherent-lemma-cech-cohomology-quasi-coherent-trivial}.
\end{remark}

\begin{remark}
\label{remark-homotopy-equivalent-cosimplicial-algebras}
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 {\it cartesian module}.
In this setting, the proof of Proposition \ref{proposition-descent-module}
can be split in the following steps
\begin{enumerate}
\item If $R \to R'$ is faithfully flat, $R \to A$ any ring map,
then descent data for $A/R$ are effective if
descent data for $(R' \otimes_R A)/R'$ are effective.
\item Let $A$ be an $R$-algebra. Descent data for $A/R$ correspond
to cartesian $(A/R)_\bullet$-modules.
\item If $R \to A$ has a section then $(A/R)_\bullet$ is homotopy
equivalent to $R$, the constant cosimplicial
$R$-algebra with value $R$.
\item 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.
\end{enumerate}
For (1) see Lemma \ref{lemma-descent-descends}.
Part (2) uses Lemma \ref{lemma-descent-datum-cosimplicial}.
Part (3) we have seen in the proof of Lemma \ref{lemma-with-section-exact}
(it relies on Simplicial,
Lemma \ref{simplicial-lemma-push-outs-simplicial-object-w-section}).
Moreover, part (4) is a triviality if you think about it right!
\end{remark}

Comments (2)

Comment #1867 by Laurent Moret-Bailly on March 27, 2016 a 4:48 pm UTC

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

Comment #1902 by Johan (site) on April 2, 2016 a 12:23 am UTC

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.

Add a comment on tag 023F

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 lower-right corner).

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 box. So in case this where tag 0321 you just have to write 0321. Beware of the difference between the letter 'O' and the digit 0.

This captcha seems more appropriate than the usual illegible gibberish, right?