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$1 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. $\square$

[1] We should really write $(N, \varphi )_\bullet$.

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