Amitsur Complex

In algebra, the Amitsur complex is a natural complex associated to a ring homomorphism. It was introduced by . 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 $\theta\colon R \to S$ be a homomorphism of (not-necessary-commutative) rings. First define the cosimplicial set $C^\bullet = S^$ (where $\otimes$ refers to $\otimes_R$, not $\otimes_$) as follows. Define the face maps $d^i\colon S^ \to S^$ by inserting 1 at the ''i''-th spot:
:$d^i(x_0 \otimes \cdots \otimes x_n) = x_0 \otimes \cdots \otimes x_ \otimes 1 \otimes x_i \otimes \cdots \otimes x_n.$
Define the degeneracies $s^i\colon S^ \to S^$ by multiplying out the ''i''-th and (''i'' + 1)-th spots:
:$s^i(x_0 \otimes \cdots \otimes x_n) = x_0 \otimes \cdots \otimes x_i x_ \otimes \cdots \otimes x_n.$
They satisfy the "obvious" cosimplicial identities and thus $S^$ is a cosimplicial set. It then determines the complex with the augumentation $\theta$, the Amitsur complex:
:$0 \to R \,\overset\to\, S \,\overset\to\, S^ \,\overset\to\, S^ \to \cdots$
where $\delta^n = \sum_^ (-1)^i d^i.$

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 $0 \to R \overset\to S^$ is exact and thus is a resolution. More generally, if $\theta$ is right faithfully flat, then, for each left ''R''-module ''M'',
:$0 \to M \to S \otimes_R M \to S^ \otimes_R M \to S^ \otimes_R M \to \cdots$
is exact.

''Proof'':

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

That "$\theta$ splits" is to say $\rho \circ \theta = \operatorname_R$ for some homomorphism $\rho\colon S \to R$ ($\rho$ is a retraction and $\theta$ a section). Given such a $\rho$, define
:$h\colon S^ \otimes M \to S^ \otimes M$
by
:$\begin & h(x_0 \otimes m) = \rho(x_0) \otimes m, \\ & h(x_0 \otimes \cdots \otimes x_n \otimes m) = \theta(\rho(x_0)) x_1 \otimes \cdots \otimes x_n \otimes m. \end$
An easy computation shows the following identity: with $\delta^=\theta \otimes \operatorname_M\colon M \to S \otimes_R M$,
:$h \circ \delta^n + \delta^ \circ h = \operatorname_$.
This is to say that ''h'' is a homotopy operator and so $\operatorname_$ 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:
:$0 \to M_S \to T \otimes_S M_S \to T^ \otimes_S M_S \to \cdots,$
where $M_S = S \otimes_R M$, is exact. Since $T \otimes_S M_S \simeq S^ \otimes_R M$, etc., by "faithfully flat", the original sequence is exact. $\square$

The case of the arc topology

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

References

Bibliography

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* {{nlab, id=Amitsur+complex, title=Amitsur complex

Algebra