# Amitsur complex

In algebra, the Amitsur complex is a natural complex associated to a ring homomorphism. It was introduced by Shimshon Amitsur (1959). When the homomorphism is faithfully flat, the Amitsur complex is exact (thus determining a resolution), which is the basis of the theory of faithfully flat descent.
The notion should be thought of as a mechanism to go beyond the conventional localization of rings and modules.

## Definition

Let $\theta :R\to S$ be a homomorphism of (not-necessary-commutative) rings. First define the cosimplicial set $C^{\bullet }=S^{\otimes \bullet +1}$ (where $\otimes$ refers to $\otimes _{R}$, not $\otimes _{\mathbb {Z} }$) as follows. Define the face maps $d^{i}:S^{\otimes {n+1}}\to S^{\otimes n+2}$ by inserting $1$ at the $i$th spot:
    $d^{i}(x_{0}\otimes \cdots \otimes x_{n})=x_{0}\otimes \cdots \otimes x_{i-1}\otimes 1\otimes x_{i}\otimes \cdots \otimes x_{n}.$
Define the degeneracies $s^{i}:S^{\otimes n+1}\to S^{\otimes n}$ by multiplying out the $i$th and $(i+1)$th spots:
    $s^{i}(x_{0}\otimes \cdots \otimes x_{n})=x_{0}\otimes \cdots \otimes x_{i}x_{i+1}\otimes \cdots \otimes x_{n}.$
They satisfy the "obvious" cosimplicial identities and thus $S^{\otimes \bullet +1}$ is a cosimplicial set. It then determines the complex with the augumentation $\theta$, the Amitsur complex:
    $0\to R\,{\overset {\theta }{\to }}\,S\,{\overset {\delta ^{0}}{\to }}\,S^{\otimes 2}\,{\overset {\delta ^{1}}{\to }}\,S^{\otimes 3}\to \cdots$
where $\delta ^{n}=\sum _{i=0}^{n+1}(-1)^{i}d^{i}.$

## Exactness of the Amitsur complex
### Faithfully flat case

In the above notations, if $\theta$ is right faithfully flat, then a theorem of Alexander Grothendieck states that the (augmented) complex $0\to R{\overset {\theta }{\to }}S^{\otimes \bullet +1}$ is exact and thus is a resolution. More generally, if $\theta$ is right faithfully flat, then, for each left $R$-module $M$,
    $0\to M\to S\otimes _{R}M\to S^{\otimes 2}\otimes _{R}M\to S^{\otimes 3}\otimes _{R}M\to \cdots$
is exact.
Proof:
Step 1: The statement is true if $\theta :R\to S$ splits as a ring homomorphism.
That "$\theta$ splits" is to say $\rho \circ \theta =\operatorname {id} _{R}$ for some homomorphism $\rho :S\to R$ ($\rho$ is a retraction and $\theta$ a section). Given such a $\rho$, define
    $h:S^{\otimes n+1}\otimes M\to S^{\otimes n}\otimes M$
by
    ${\begin{aligned}&h(x_{0}\otimes m)=\rho (x_{0})\otimes m,\\&h(x_{0}\otimes \cdots \otimes x_{n}\otimes m)=\theta (\rho (x_{0}))x_{1}\otimes \cdots \otimes x_{n}\otimes m.\end{aligned}}$
An easy computation shows the following identity: with $\delta ^{-1}=\theta \otimes \operatorname {id} _{M}:M\to S\otimes _{R}M$,
    $h\circ \delta ^{n}+\delta ^{n-1}\circ h=\operatorname {id} _{S^{\otimes n+1}\otimes M}$.
This is to say that $h$ is a homotopy operator and so $\operatorname {id} _{S^{\otimes n+1}\otimes M}$ determines the zero map on cohomology: i.e., the complex is exact.
Step 2: The statement is true in general.
We remark that $S\to T:=S\otimes _{R}S,\,x\mapsto 1\otimes x$ is a section of $T\to S,\,x\otimes y\mapsto xy$. Thus, Step 1 applied to the split ring homomorphism $S\to T$ implies:
    $0\to M_{S}\to T\otimes _{S}M_{S}\to T^{\otimes 2}\otimes _{S}M_{S}\to \cdots ,$
where $M_{S}=S\otimes _{R}M$, is exact. Since $T\otimes _{S}M_{S}\simeq S^{\otimes 2}\otimes _{R}M$, etc., by "faithfully flat", the original sequence is exact. $\square$

### Arc topology case

Bhargav Bhatt and Peter Scholze (2019, §8) show that the Amitsur complex is exact if $R$ and $S$ 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).