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A foundational result in Grothendieck's descent theory and in his ├ętale cohomology is the exactness of Amitsur's complex. More precisely, suppose we have an $A$-algebra $A\to B$; then there is a cosimplicial complex associated to it whose $n-$cosimplices are $B\ ^ {\otimes {(n+1)} }$, and from there a complex obtains $$ 0\to A \to B \to B \otimes B \to \ldots \quad (AMITSUR) $$
For example the map $B \to B \otimes B$ is $b\mapsto 1\otimes b -b \otimes 1$ and the following maps are obtained similarly by inserting $1$'s in tensor products of copies of $B$ and taking alternating sums. The key result is that this Amitsur complex is exact if the initial algebra $A \to B$ is faithfully flat.

The proof is splendid: "one" (ah, that's the point!) remarks that if the structural map has an $A$- linear retraction, then it is easy to conclude by constructing a homotopy. And then one reduces to this case by a bold gambit: since one doesn't know how to prove exactness of $(AMITSUR)$ one tensors with $B$ and gets the even more complicated complex $(AMITSUR)\otimes B$ . But now the initial map $A\to B$ has become $B \to B\otimes B: b \mapsto 1\otimes b$ , which HAS a retraction: just take the product $B\otimes B \to B: b\otimes b' \mapsto bb'$ . So the tensored complex is exact and the initial complex was necessarily exact by faithful flatness.

Question: who proved this? I suspect the argument I sketched is due to Grothendieck since I couldn't find a reference to Amitsur in EGA nor in SGA. So, what exactly did Amitsur prove in this context and how did he do it? I have a vague intuition that he didn't express himself in terms of faithful flatness, but my Internet search failed miserably. So, dear mathoverflow participants, you are my last hope...

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Amitsur introduced the concept here: jstor.org/pss/1993268 right? –  Gjergji Zaimi Mar 4 '10 at 19:51
    
And then his results were simplified by Rosenberg and Zelinsky, jstor.org/pss/1993305 –  Gjergji Zaimi Mar 4 '10 at 20:09
    
Thanks for the references, Gjergji. By the way, isn't Gjergji the equivalent of Georges in a language which I confess completely eludes me? It would be nice if we were namesakes :-) –  Georges Elencwajg Mar 4 '10 at 20:16
    
The language you are looking for is Albanian, and yes, Gjergji is one of the equivalent versions of George/Georges :-) . Unfortunately it comes with an unfriendly pronunciation. –  Gjergji Zaimi Mar 4 '10 at 21:30
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Grothendieck describes a version of this argument in the FGA on flat descent (I think FGA I). You might want to look in there to see if he mentions Amitsur. (I can't remember now whether or not he does, but it is conceivable, since FGA does have more references to earlier work than some of Grothendieck's subsequent writings.) –  Emerton Mar 5 '10 at 2:31
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