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

Given a smooth and projective surface $S$ over an algebraically closed field $k$ and a sheaf of Azumaya algebras $R$, i.e. $R$ is a locally free $O_S$-module of finite rank. Let $M$ be a coherent and torsion free $O_S$-module, which is also a left $R$-module, such that generically $M_\eta$ is a simple $R_\eta$-module. Then we have $Hom_R(M,M)=k$.

Now $M^\*:=Hom_{O_S}(M,O_S)$ is a right $R$-module and $M^{\*\*}$ is a left $R$-module. We have the canonical map $\iota: M \rightarrow M^{\*\*}$.

Is it true that $Hom_R(M,M^{\*\*})$ just consists of the muliples of $\iota$, i.e. is it a one dimensional $k$-vector space?

I tried to use the sequence $0\rightarrow M\rightarrow M^{\*\*} \rightarrow Q\rightarrow 0$. Since $M$ is torsion free $Q$ has support in codimension 2. Then apply $Hom_R(M, - )$, which is left exact, so we get, with $Hom_R(M,M)=k$: $0\rightarrow k\rightarrow Hom_R(M,M^{\*\*}) \rightarrow Hom_R(M,Q)$. But here i am stuck.

Or is this assertion wrong, i.e. are there more morphisms? If it is right, can it be generalized to a bigger class of algebras $R$?

share|improve this question
1  
It is better to apply $Hom_R(-,M^{**})$. Since $Q$ is in codimension 2 one has $Hom(Q,M^{**}) = Ext^1(Q,M^{**}) = 0$, so $Hom(M,M^{**}) = Hom(M^{**},M^{**})$. –  Sasha Sep 22 '10 at 15:36
    
Okay, i see this long exact sequence. But why do these groups vanish? Just because $Q$ live in codimension 2? I cannot see this. We still have all $H^0$ groups, e.g. $H^0(\mathcal{E}xt^1 (Q,M^{\*\*}))$ which shows up in the local to global spectral sequence for $Ext^1$. Or am i missing the point here? –  TonyS Sep 22 '10 at 16:58
1  
Local $Hom$ and local $Ext^1$ vanish because $Q$ lives in codimension 2. Then local-to-global spectral sequence shows that global $Hom$ and $Ext^1$ vanish as well. –  Sasha Sep 22 '10 at 17:29
1  
It may be not too obvious, but it is a standard fact. Usually it is proved using the notion of depth e.t.c. But an easy way to explain this is the following. First, it is clear that local $Hom$ is supported at the support of $Q$. On the other hand, local $Hom$ is torsion free. Hence it is zero. Now locally we can choose a pair of Cartier divisors $D_1,D_2$ such that $Q$ is supported on $Z = D_1 \cap D_2$ (scheme theoretically) and $codim Z = 2$. Further, locally we can choose a surjection $O_D^n \to Q$. Let $Q′$ be the kernel. Then local $Hom$ for $Q′$ vanishes by the same reason. –  Sasha Sep 22 '10 at 18:25
1  
Sorry, $O_D$ should be $O_Z$. Continue. Hence local $Ext^1$ for $Q$ injects into local $Ext^1$ for $O_Z^n$. The latter can be computed explicitly using the Koszul resolution $$ 0 \to O(-D_1 - D_2) \to O(-D_1) \oplus O(-D_2) \to O \to O_Z \to 0. $$ The result is zero. –  Sasha Sep 22 '10 at 18:27

1 Answer 1

up vote 3 down vote accepted

Any $R$-homomorphism (in fact any $\mathcal O_S$-homomorphism) $M \to M^{**}$ extends to a morphism $M^{**}\to M^{**}$ (as $M$ is locally free in codimension $1$ and $M^{**}$ is the maximal extension from outside codimension $2$. This gives what you want. as $Hom_R(M^{**},M^{**})=k$ for the same reason as it is true of $M$.

share|improve this answer
    
Thanks for your answer, i have some questions: what does extension mean here? Something like: given $f: M\rightarrow M^{\*\*}$, there is a unique $g: M^{\*\*} \rightarrow M^{\*\*}$ with $f=g \iota$. Since $End_R(M^{\*\*})=k$ $g$ must be a multiple of the identity and so f is multiple of $\iota$? How do I find this g exactly? –  TonyS Sep 22 '10 at 15:54
1  
Your interpretation is correct. You can find $g$ either by finding a codimension $2$ subset outside of which $\iota$ is an isomorphism and then note that (local) sections of $M^{**}$ defined outside of a codimension $2$ subset extend over the subset (algebraic version of Hartog's theorem) or you can look at $g^{**}\colon M^{**}\to M^{****}=M^{**}$. –  Torsten Ekedahl Sep 22 '10 at 16:23
    
So we actually don't need $R$ to be an Azumaya algebra. Nice! –  TonyS Sep 22 '10 at 16:54

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.