The Stacks project

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

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

in $A \otimes _ R A \otimes _ R N$. We conclude two things from this:

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

2. 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. $\square$