Amitsur complex explained

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 ^{⊗} (where ⊗ refers to ⊗ _{R} , not ⊗ _{\Z} ) as follows. Define the face maps d ^{i} : S ^{⊗} } \to S^ by inserting 1 at the i th spot:d ^{i(x} _{0} ⊗ … ⊗ x _{n)} = x _{0} ⊗ … ⊗ x _{i-1} ⊗ 1 ⊗ x _{i} ⊗ … ⊗ x _{n.} Define the degeneracies s ^{i} : S ^{⊗} \to S ^{⊗} by multiplying out the i th and (i +1 ) th spots:s ^{i(x} _{0} ⊗ … ⊗ x _{n)} = x _{0} ⊗ … ⊗ x _{i} x _{i+1} ⊗ … ⊗ 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{\theta}\to S \overset{\delta ^{0}\to} S ^{⊗} \overset{\delta ^{1}\to} S ^{⊗} \to … where \delta ^{n} = \sum ^{n +1} _{i =0} ( -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{\theta}\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 ⊗ _{R} M \to S ^{⊗} ⊗ _{R} M \to S ^{⊗} ⊗ _{R} M \to … 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 , defineh : S ^{⊗} ⊗ M \to S ^{⊗} ⊗ M by\begin{align} & h(x _{0} ⊗ m) = \rho(x _{0)} ⊗ m, \\ & h(x _{0} ⊗ … ⊗ x _{n} ⊗ m) = \theta(\rho(x _{0))} x _{1} ⊗ … ⊗ x _{n} ⊗ m. \end{align} An easy computation shows the following identity: with \delta ^{-1} = \theta ⊗ \operatorname{id} _{M} : M \to S ⊗ _{R} M ,h \circ \delta ^{n} + \delta ^{n-1} \circ h = \operatorname{id} _{S ^{⊗} ⊗ M} .This is to say that h is a homotopy operator and so \operatorname{id} _{S ^{⊗} ⊗ 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 ⊗ _{R} S, x \mapsto 1 ⊗ x is a section of T \to S, x ⊗ y \mapsto xy . Thus, Step 1 applied to the split ring homomorphism S \to T implies:0 \to M _{S} \to T ⊗ _{S} M _{S} \to T ^{⊗} ⊗ _{S} M _{S} \to … , where M _{S} = S ⊗ _{R} M , is exact. Since T ⊗ _{S} M _{S} \simeq S ^{⊗} ⊗ _{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)