Amitsur complex

In algebra, the Amitsur complex is a natural complex associated to a ring homomorphism. It was introduced by Template:Harvs. 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.Template:Sfn

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:Template:Efn

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

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:Template:Sfn

$0\to R\stackrel{\theta }{\to }S\stackrel{{\delta }^0}{\to }{S}^{\otimes 2}\stackrel{{\delta }^1}{\to }{S}^{\otimes 3}\to \cdots$

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 $0\to R\stackrel{\theta }{\to }{S}^{\otimes \bullet +1}$ 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.Template:Sfn

Proof:

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

That "$\theta$ splits" is to say $\rho \circ \theta ={\mathrm{id}}_R$ 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

$\begin{array}{cc}& h(x_0\otimes m)=\rho (x_0)\otimes m,\\ & h(x_0\otimes \cdots \otimes x_n\otimes m)=\theta (\rho (x_0))x_1\otimes \cdots \otimes x_n\otimes m.\end{array}$

An easy computation shows the following identity: with ${\delta }^{-1}=\theta \otimes {\mathrm{id}}_M:M\to S{\otimes }_{R}M$,

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

This is to say that $h$ is a homotopy operator and so ${\mathrm{id}}_{S^{\otimes n+1}\otimes M}$ 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. $◻$

Arc topology case

Template:Harvs 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

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• Template:Nlab