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 $$\theta: R \to S$$ 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 $$\otimes_{\Z}$$) 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 $$\delta^n = \sum_{i=0}^{n+1} (-1)^i d^i.$$

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{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}$$ An easy computation shows the following identity: with $$\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
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).