Amitsur complex
In algebra, the Amitsur complex is a natural complex associated to a ring homomorphism. It was introduced by {{harvs|txt|author-link=Shimshon Amitsur|last=Amitsur|first=Shimshon|year=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.{{sfn|Artin|1999|loc=III.7|ps=none}}
Definition
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 at the th spot:{{efn|The reference (M. Artin) seems to have a typo, and this should be the correct formula; see the calculation of and in the note.}}
:
Define the degeneracies by multiplying out the th and 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:{{sfn|Artin|1999|loc=III.6|ps=none}}
:
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 -module ,
:
is exact.{{sfn|Artin|1999|loc=Theorem III.6.6|ps=none}}
Proof:
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
:
& 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 ,
: .
This is to say that 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.
= Arc topology case =
{{harvs|txt|last1=Bhatt|first1=Bhargav|author1-link=Bhargav Bhatt (mathematician)|author2-link=Peter Scholze|last2=Scholze|first2=Peter|year=2019|loc=§8}} show that the Amitsur complex is exact if and 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
{{notelist}}
Citations
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References
{{refbegin}}
- {{citation|author-link=Michael Artin|last1=Artin|first1=Michael|url=http://www-math.mit.edu/~etingof/artinnotes.pdf|title=Noncommutative rings (Berkeley lecture notes)|year=1999}}
- {{citation|author-link=Shimshon Amitsur|last=Amitsur|first=Shimshon|title=Simple algebras and cohomology groups of arbitrary fields|journal=Transactions of the American Mathematical Society|volume=90|issue=1|year=1959|pages=73–112}}
- {{citation|last1=Bhatt|first1=Bhargav|author1-link=Bhargav Bhatt (mathematician)|author2-link=Peter Scholze|last2=Scholze|first2=Peter|title=Prisms and Prismatic Cohomology|year=2019|arxiv=1905.08229}}
- {{nlab|id=Amitsur+complex|title=Amitsur complex}}
{{refend}}