In algebra, the Amitsur complex is a natural complex associated to a ring homomorphism. It was introduced by Shimshon 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.[1]
Let be a homomorphism of (not-necessary-commutative) rings. First define the cosimplicial set (where refers to , not ) as follows. Define the face maps by inserting 1 at the i-th spot:[note 1]
Define the degeneracies by multiplying out the i-th and (i + 1)-th spots:
They satisfy the "obvious" cosimplicial identities and thus is a cosimplicial set. It then determines the complex with the augumentation , the Amitsur complex:[2]
where
Exactness of the Amitsur complex[]
Faithfully flat case[]
In the above notations, if is right faithfully flat, then a theorem of Alexander Grothendieck states that the (augmented) complex is exact and thus is a resolution. More generally, if is right faithfully flat, then, for each left R-module M,
Step 1: The statement is true if splits as a ring homomorphism.
That " splits" is to say for some homomorphism ( is a retraction and a section). Given such a , define
by
An easy computation shows the following identity: with ,
.
This is to say that h is a homotopy operator and so determines the zero map on cohomology: i.e., the complex is exact.
Step 2: The statement is true in general.
We remark that is a section of . Thus, Step 1 applied to the split ring homomorphism implies:
where , is exact. Since , etc., by "faithfully flat", the original sequence is exact.
The case of the arc topology[]
Bhargav Bhatt and Peter 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[]
^Note the reference (M. Artin) seems to have a typo, and this should be the correct formula; see the calculation of and in the note.