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Given a smooth projective variety $X$ over some algebraically closed field $k$ and a locally free sheaf $R$ of $O_X$-algebras, e.g. central simple algebras or orders.

If $M$ is a left $R$-module which is locally free over $O_X$, is it true that $M$ is locally projective over $R$? For example if $X$ is a curve a torsion free $O_X$-module would be locally projective over $R$. Or do we need more conditions for $R$ and $M$ for this to be true?

Why can one compute $Ext_R^1(M,M)$ via $H^1(\mathcal{H}om_R(M,M))$ only in the case $M$ is locally projective over $R$?

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3 Answers 3

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It is true for locally free sheaves of algebras that are central simple algebras at every point, though. These are known as sheaves of Azumaya algebras; I mention them since they were brought up in the question. I don't have a reference here, but the proof is not hard.

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  • $\begingroup$ Yes i am especially/only interested in the case where the $O_X$-algebra $R$ lives in a given central simple $k(X)$-algebra $A$. This includes the Azumaya algebra case. I would really like to have a proof, an idea how to prove or a reference for this statement. According to the following paper, page 2, it could also hold more generally for "hereditary orders", although there $X$ is just a curve. web.maths.unsw.edu.au/~danielch/paper/moduli.pdf $\endgroup$
    – TonyS
    Commented May 22, 2010 at 13:43
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    $\begingroup$ Yes, it is true for hereditary orders but unless I am mistaken if the order is hereditary then $X$ is necessarily a smooth curve. More interesting perhaps is the question for a maximal order (maximal orders over curves are hereditary). I am somewhat doubtful however if a maximal order in general has finite global dimension (which it must if your property is to hold). Artin has some results on the local form for maximal orders over surfaces which might give counterexamples (or support for that problem). $\endgroup$
    – Torsten Ekedahl
    Commented May 22, 2010 at 14:01
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    $\begingroup$ In the case of an Azumaya algebra, the statement is local in the étale topology, so it is enough to prove it for matrix algebras. In this case it follows immediately from Morita equivalence. In the case of hereditary orders, I am not sure, as I am really rusty on this stuff. $\endgroup$
    – Angelo
    Commented May 22, 2010 at 14:09
  • $\begingroup$ I am not so good with Morita equivalence, is the argument the following: If we look locally at a point $p \in X$ the $M_n(R)$-module $M_p$ is of the form $V^n$ for some $R$-module $V$. Since it is free over $R$, we have $M=R^(nk)$ for some k. Then the Morita equivalent $R$-module to $M_p$ is $R^k$, which is free, hence projective over $R$. Since Morita equivalence preserves projectivity we have that $M_p$ is projective. So $M$ is locally projective. $\endgroup$
    – TonyS
    Commented May 22, 2010 at 14:35
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    $\begingroup$ Yes, something like this. You don't need to localize. I would argue as follows. Suppose that $M$ is an $M_n(R)$-module which is projective as an $R$-module. Let $N$ the Morita-equivalent $R$-module; then $M$ is $N^n$, hence $N$ is projective. Since Morita equivalence preserves projectivity, $M$ is projective as an $M_n(R)$-module. $\endgroup$
    – Angelo
    Commented May 22, 2010 at 14:47
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The answer to your first question is a resounding no. An example (among many) is given by $X=\mathrm{Spec} k$, $R=k[x]/(x^2)$ and $M=k$ considered as an $R$-module through the $k$-algebra homomorphism given by $x\mapsto 0$.

As for the second question, the reason is that in general there are correction terms to this formula coming from the failure of $M$ being locally projective. In fact there is a long exact sequence $$ 0\rightarrow H^1(X,\mathcal{H}\mathrm{om}_R(M,M))\rightarrow \mathrm{Ext}^1_R(M,M)\rightarrow H^0(X,\mathcal{E}\mathrm{xt}^1_R(M,M)), $$ where the $\mathcal{E}\mathrm{xt}^i_R(M,M)$ are the sheaves of Ext-classes (their stalks at $x$ are the $\mathrm{Ext}^i_{R_x}(M_x,M_x)$). Hence, you need something like (possibly something a little bit weaker) the vanishing $\mathrm{Ext}^i_{R_x}(M_x,M_x)$ for $i=1$ which in turn are implied by (though not implying) the local $R$-projectivity of $M$. This sequence is most easily obtained by the right hand map taking a sequence to the isomorphism clases of local extensions and the second map is obtained by twisting the trivial sequence by a torsor over its automorphism group. the exactness in the middle comes from the fact that any locally trivial sequence comes from such a twisting.

Addendum: Corrected a typo (stacks) and changed the exact sequence to be correct (one way to get it is from the local to global spectral sequence and I had, as I unfortunately do too often, flipped it 45 degrees).

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  • $\begingroup$ Thank you, this is an easy counterexample and answers my second question. $\endgroup$
    – TonyS
    Commented May 22, 2010 at 13:10
  • $\begingroup$ Is "stacks" supposed to be "stalks"? $\endgroup$
    – Charles Staats
    Commented May 22, 2010 at 16:01
  • $\begingroup$ Yes, i think it is supposed to be "stalks". But I have one last question: Does this long exact sequence come from the local-to-global spectral sequence? Doesn't this sequence look like this: $0 \rightarrow H^1(\mathcal{H}om(M,M) \rightarrow Ext_R^1(M,M) \rightarrow H^0(\mathxcal{E}xt_R^1(M,M)) \rightarrow\ldots$ so we only need vanishing of $Ext_{R_x}^1(M_x,M_x)$ or am I missing something? $\endgroup$
    – TonyS
    Commented May 22, 2010 at 20:27
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Let $R$ be an Azumaya algebra and $M$ an $R$-module which is locally free over $O_X$. Note that $R$ is a locally projective $R^{opp}\otimes_{O_X} R$-module, hence a direct summand of $R^{opp}\otimes_{O_X} R$. It follows that $M = M \otimes_R R$ is a direct summand of $M\otimes_R (R^{opp}\otimes_{O_X} R) = M\otimes_{O_X} R$ which is a locally free $R$-module. Being a direct summand of a locally free $R$-module, it is certainly locally projective. I guess this argument can be found in the Miln book on etale cohomology.

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