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. $\square$

## Comments (2)

Comment #2154 by Katha on

In the last line it should be $A \otimes_R A$ instead of $A \otimes_A A$.

Comment #2188 by on

Thanks. Fixed here. I removed the other comments which were on the wrong page.

There are also:

• 4 comment(s) on Section 35.3: Descent for modules

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