Amitsur complex

In algebra, the Amitsur complex is a natural complex associated to a ring homomorphism. It was introduced in (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 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 $\delta ^{-1}:M{\overset {\theta \otimes \operatorname {id} _{M}}{\to }}S\otimes _{R}M$,

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

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. $\square$

The case of the arc topology

Bhatt & 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).

References

• Artin, Michael (1999), Noncommutative rings (Berkeley lecture notes) (PDF)
• Amitsur, Shimshon (1959), "Simple algebras and cohomology groups of arbitrary fields", Transactions of the American Mathematical Society, 90 (1): 73–112
• Bhatt, Bhargav; Scholze, Peter (2019), Prisms and Prismatic Cohomology, arXiv:1905.08229
• Amitsur complex at the nLab