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I am reading "Ample divisors on fine moduli spaces on the Projective plane" by Stromme. In the proof of Proposition 2.4, he seems to claim that if $E$ is a torsion free sheaf of projective dimension at most 1 on a smooth variety and the singular locus of $E$ has codimension at least $3$, then $E$ is reflexive. Of course the converse is true without the projective dimension assumption. 

Does anyone know an explanation or reference for this fact?

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    $\begingroup$ I would guess that the condition $pd(F) \leq 1$ is important. Without it you can get a counterexample by considering the ideal sheaf of a point on a variety of dimension $\geq 3$. $\endgroup$
    – naf
    Dec 20, 2013 at 11:33
  • $\begingroup$ Ah yes, of course. Edited. $\endgroup$ Dec 20, 2013 at 11:49

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In general the converse is not true --- take for example the ideal of a point on 3-fold (its reflexive hull is the structure sheaf). On the other hand, if you know that $pd(F) \le 1$ then there is an exact sequence $$ 0 \to E_1 \stackrel{f}\to E_2 \to F \to 0 $$ with $E_1$ and $E_2$ locally free. Taking the dual one gets $$ 0 \to F^* \to E_2^* \stackrel{f^*}\to E_1^* \to F_1 \to 0, $$ where the last sheaf $F_1 := \mathcal{Ext}^1(F,O)$ is supported in codimension 3. In particular, $\mathcal{Ext}^i(F_1,O) = 0$ for $i \le 2$. It follows that $\mathcal{Ext}^i(F_1,O)$ do not contribute into $F^{**}$ and so there is an exact sequence $$ 0 \to E_1^{**} \stackrel{f^{**}}\to E_2^{**} \to F^{**} \to 0, $$ and so since $E_i$ are locally free, it follows that $F^{**} = F$.

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  • $\begingroup$ Great, thanks! This seems to do the trick. $\endgroup$ Dec 20, 2013 at 16:06

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