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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 $R$?

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    $\begingroup$ 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 $Hom(M,M^{**}) = Hom(M^{**},M^{**})$. $\endgroup$
    – Sasha
    Commented Sep 22, 2010 at 15:36
  • $\begingroup$ 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? $\endgroup$
    – TonyS
    Commented Sep 22, 2010 at 16:58
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    $\begingroup$ 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. $\endgroup$
    – Sasha
    Commented Sep 22, 2010 at 17:29
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    $\begingroup$ 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. $\endgroup$
    – Sasha
    Commented Sep 22, 2010 at 18:25
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    $\begingroup$ 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. $\endgroup$
    – Sasha
    Commented Sep 22, 2010 at 18:27

1 Answer 1

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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 $M$.

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  • $\begingroup$ 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? $\endgroup$
    – TonyS
    Commented Sep 22, 2010 at 15:54
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    $\begingroup$ 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 $g^{**}\colon M^{**}\to M^{****}=M^{**}$. $\endgroup$
    – Torsten Ekedahl
    Commented Sep 22, 2010 at 16:23
  • $\begingroup$ So we actually don't need $R$ to be an Azumaya algebra. Nice! $\endgroup$
    – TonyS
    Commented Sep 22, 2010 at 16:54

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