If an Azumaya algebra $A$ on a scheme $X$ is trivialised by $A = \mathcal{End}(\mathcal{O})$, $\mathcal{O}$ locally free on $X$, why is $\mathcal{O}$ uniquely determined up to tensoring with a line bundle?
2 Answers
Here's an easier answer. If $A=End_R(E)\cong End_R(F)$ with center a commutative ring $R$, then $F$ is an $End(E)$ module via the isomorphism. Now Morita theory says that $L:=Hom_A(E,F)$ is a rank 1 projective module over $R$ which you can check by localizing at any prime ideal and computing in matrices. Moreover $E\otimes_R L\to F$ is similarly an isomorphism. This argument generally goes under the name of the SkolemNoether theorem.
One way to see this is to look at the sequence in cohomology
$H^1(X,\mathbb{G}_m) \to H^1(X,\mathbf{GL}_n) \to H^1(X,\mathbf{PGL}_n) \to \mathbf{Br} X$
which comes from the obvious short exact sequence of algebraic groups. Now $H^1(X,\mathbf{PGL}_n)$ classifies Azumaya algebras on $X$ of degree $n$. The trivial ones are those in the image of $H^1(X,\mathbf{GL}_n)$, which classifies vector bundles of rank $n$; the trick is to convince yourself that the arrow between those groups is the one taking a vector bundle to its sheaf of endomorphisms. There's a brief proof about a third of the way down p.143 of Milne's Etale cohomology. So the exact sequence says that two vector bundles trivialise the same Azumaya algebra if they differ by an element of $H^1(X,\mathbb{G}_m)$, that is, by the class of a line bundle.

$\begingroup$ Wait: "two vector bundles trivialise the same Azumaya algebra if they differ by an element of $H^1(X,\mathbb{G}_m)$, that is, by the class of a line bundle." how do you conclude this? This is only an exact sequence of pointed sets. $\endgroup$– user19475Jun 5, 2013 at 14:02

$\begingroup$ Timo  you're right, there's a bit more to say here, and it's not obvious that my answer is correct. Unfortunately I am travelling and don't have the right reference to hand. The usual thing in this situation is to wave one's hands and say "by twisting". I believe Serre goes through the argument in some detail in the chapter on nonAbelian cohomology in "Cohomologie Galoisienne". $\endgroup$ Jun 7, 2013 at 10:49

$\begingroup$ Yes, thank you. In fact, I just looked into "Coh. Gal." before reading your comment. ;) It seems to follow from p. 50, Section 5.4 Corollary 1. I have to check it. $\endgroup$– user19475Jun 7, 2013 at 18:03

$\begingroup$ resp. p. 54, Section 5.7, Proposition 42. One also has to check that this result transfers from group cohomology to (Cech) etale cohomology. $\endgroup$– user19475Jun 7, 2013 at 18:12