Amitsur complex

Let ${\displaystyle \theta :R\to S}$ be a homomorphism of (not-necessary-commutative) rings. First define the cosimplicial set ${\displaystyle C^{\bullet }=S^{\otimes \bullet +1}}$ (where ${\displaystyle \otimes }$ refers to ${\displaystyle \otimes _{R}}$, not ${\displaystyle \otimes _{\mathbb {Z} }}$) as follows. Define the face maps ${\displaystyle d^{i}:S^{\otimes {n+1}}\to S^{\otimes n+2}}$ by inserting ${\displaystyle 1}$ at the ${\displaystyle i}$th spot:^[a]

${\displaystyle d^{i}(x_{0}\otimes \cdots \otimes x_{n})=x_{0}\otimes \cdots \otimes x_{i-1}\otimes 1\otimes x_{i}\otimes \cdots \otimes x_{n}.}$

Define the degeneracies ${\displaystyle s^{i}:S^{\otimes n+1}\to S^{\otimes n}}$ by multiplying out the ${\displaystyle i}$th and ${\displaystyle (i+1)}$th spots:

${\displaystyle s^{i}(x_{0}\otimes \cdots \otimes x_{n})=x_{0}\otimes \cdots \otimes x_{i}x_{i+1}\otimes \cdots \otimes x_{n}.}$

They satisfy the "obvious" cosimplicial identities and thus ${\displaystyle S^{\otimes \bullet +1}}$ is a cosimplicial set. It then determines the complex with the augumentation ${\displaystyle \theta }$, the Amitsur complex:^[2]

${\displaystyle 0\to R\,{\overset {\theta }{\to }}\,S\,{\overset {\delta ^{0}}{\to }}\,S^{\otimes 2}\,{\overset {\delta ^{1}}{\to }}\,S^{\otimes 3}\to \cdots }$

where ${\displaystyle \delta ^{n}=\sum _{i=0}^{n+1}(-1)^{i}d^{i}.}$

In the above notations, if ${\displaystyle \theta }$ is right faithfully flat, then a theorem of Alexander Grothendieck states that the (augmented) complex ${\displaystyle 0\to R{\overset {\theta }{\to }}S^{\otimes \bullet +1}}$ is exact and thus is a resolution. More generally, if ${\displaystyle \theta }$ is right faithfully flat, then, for each left ${\displaystyle R}$-module ${\displaystyle M}$,

${\displaystyle 0\to M\to S\otimes _{R}M\to S^{\otimes 2}\otimes _{R}M\to S^{\otimes 3}\otimes _{R}M\to \cdots }$

is exact.^[3]

Proof:

Step 1: The statement is true if ${\displaystyle \theta :R\to S}$ splits as a ring homomorphism.

That "${\displaystyle \theta }$ splits" is to say ${\displaystyle \rho \circ \theta =\operatorname {id} _{R}}$ for some homomorphism ${\displaystyle \rho :S\to R}$ (${\displaystyle \rho }$ is a retraction and ${\displaystyle \theta }$ a section). Given such a ${\displaystyle \rho }$, define

${\displaystyle h:S^{\otimes n+1}\otimes M\to S^{\otimes n}\otimes M}$

by

${\displaystyle {\begin{aligned}&h(x_{0}\otimes m)=\rho (x_{0})\otimes m,\\&h(x_{0}\otimes \cdots \otimes x_{n}\otimes m)=\theta (\rho (x_{0}))x_{1}\otimes \cdots \otimes x_{n}\otimes m.\end{aligned}}}$

An easy computation shows the following identity: with ${\displaystyle \delta ^{-1}=\theta \otimes \operatorname {id} _{M}:M\to S\otimes _{R}M}$,

${\displaystyle h\circ \delta ^{n}+\delta ^{n-1}\circ h=\operatorname {id} _{S^{\otimes n+1}\otimes M}}$.

This is to say that ${\displaystyle h}$ is a homotopy operator and so ${\displaystyle \operatorname {id} _{S^{\otimes n+1}\otimes M}}$ determines the zero map on cohomology: i.e., the complex is exact.

Step 2: The statement is true in general.

We remark that ${\displaystyle S\to T:=S\otimes _{R}S,\,x\mapsto 1\otimes x}$ is a section of ${\displaystyle T\to S,\,x\otimes y\mapsto xy}$. Thus, Step 1 applied to the split ring homomorphism ${\displaystyle S\to T}$ implies:

${\displaystyle 0\to M_{S}\to T\otimes _{S}M_{S}\to T^{\otimes 2}\otimes _{S}M_{S}\to \cdots ,}$

where ${\displaystyle M_{S}=S\otimes _{R}M}$, is exact. Since ${\displaystyle T\otimes _{S}M_{S}\simeq S^{\otimes 2}\otimes _{R}M}$, etc., by "faithfully flat", the original sequence is exact. ${\displaystyle \square }$

Bhargav Bhatt and Peter Scholze (2019, §8) show that the Amitsur complex is exact if ${\displaystyle R}$ and ${\displaystyle S}$ are (commutative) perfect rings, and the map is required to be a covering in the arc topology (which is a weaker condition than being a cover in the flat topology).