Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Given $\mathcal{O}=k[[u,v]]$ with maximal ideal $\mathfrak{m}$ and an $\mathcal{O}$-algebra $A$, free as an $\mathcal{O}$-module of rank $n^2$. $A$ is genertaed by two elements $x,y$ with $x^n=u$, $y^n=v$ and $xy=\xi_n yx$, where $\xi_n$ is an n-root of unity. (But i think this should be true in more general situations.)

Then we can look at the sheaf $\mathcal{A}$ associated to $A$ on $U=Spec(\mathcal{O})\backslash \lbrace \mathfrak{m} \rbrace$.

Why does $H^1_{zar}(U,\mathcal{A}^{\times})$ classify fractional left $A$-ideals $L$, which are reflexive as $\mathcal{O}$-modules?

Shouldn't this group classify modules, which look locally like $\mathcal{A}$, as some kind of Picard group for $\mathcal{A}$? But then i think this group should correspond to principal left ideals. But this seems wrong, because then this group would be zero, but in the text it is proven seperately that every such ideal is principal. And why are these ideals reflexive?

This is stated in M.Artin's article "Local structure of maximal orders on surfaces".It says on page 26: "And OF COURSE, $H^1_{zar}(U,\mathcal{A}^{\times})$ classifies reflexive fractional left ideals in $A$." Or on page 31 it says: "locally principal $\mathcal{A}$-module (or left ideal)".

Is this so obvious? Am I missing something?

share|improve this question

1 Answer 1

up vote 2 down vote accepted

Let $M$ be a coherent (left) $\mathcal{A}$-module which is Zariski locally isomorphic to $\mathcal{A}$. Then in particular, the stalk at the generic point of $U$ is isomorphic to the stalk of $\mathcal{A}$ at the generic point. Choose such an isomorphism. By adjunction, this is equivalent to an $\mathcal{A}$-module homomorphism from $M$ to the pushforward to $U$ of the stalk of $\mathcal{A}$ at the generic point. This pushforward is just the constant sheaf $\underline{D}$, where $D$ is the central simple algebra $A\otimes_{\mathcal{O}} K$ over the fraction field $K$ of $\mathcal{O}$. So now you have an $\mathcal{A}$-module homomorphism $i:M\rightarrow \underline{D}$ which becomes an isomorphism when you take stalks at the generic point. Pulling $i$ back to the Zariski opens of your trivializing cover, it is easy to see that $i$ is injective (essentially because $\mathcal{A}$ is a torsion-free $\mathcal{O}$-module). So now $i$ identifies $M$ as an $\mathcal{A}$-submodule of $\underline{D}$. Finally take global sections to get an $A$-submodule of $D$.

share|improve this answer
    
Thanks a lot. So it is not that obvious, but with this nice construction it's understandable. –  TonyS Jul 15 '11 at 8:19

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.