# 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.[1]

## Definition

Let $\displaystyle{ \theta\colon R \to S }$ 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 $\displaystyle{ \otimes_{\Z} }$) as follows. Define the face maps $\displaystyle{ d^i\colon S^{\otimes {n+1}} \to S^{\otimes n+2} }$ by inserting 1 at the i-th spot:[lower-alpha 1]

$\displaystyle{ d^i(x_0 \otimes \cdots \otimes x_n) = x_0 \otimes \cdots \otimes x_{i-1} \otimes 1 \otimes x_i \otimes \cdots \otimes x_n. }$

Define the degeneracies $\displaystyle{ s^i\colon S^{\otimes n+1} \to S^{\otimes n} }$ by multiplying out the i-th and (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 $\displaystyle{ \delta^n = \sum_{i=0}^{n+1} (-1)^i d^i. }$

## Exactness of the Amitsur complex

### Faithfully flat case

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 R-module 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 $\displaystyle{ \theta\colon R \to S }$ splits as a ring homomorphism.

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

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

by

\displaystyle{ \begin{align} & 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{align} }

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

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

This is to say that 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. $\displaystyle{ \square }$

### The case of the arc topology

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).

## Notes

1. Note the reference (M. Artin) seems to have a typo, and this should be the correct formula; see the calculation of $\displaystyle{ s_0 }$ and $\displaystyle{ d^2 }$ in the note.

## References

1. Artin 1999, III.7.
2. Artin 1999, III.6.
3. Artin 1999, Theorem III.6.6.