If $E$ is a coherent sheaf on projective space then is it always true that its dual $ E^* $ is also coherent.? Moreover, if $E$ is also torsionfree or reflexive then, is $E^* $ also torsionfree or reflexive?
3 Answers
The dual of any coherent sheaf on a projective space is reflexive (and in particular torsion free). The easiest way to see it is the following. Choose a locally free resolution $\dots \to F_1 \to F_0 \to E \to 0$. After dualizing it gives an exact sequence $0 \to E^* \to F_0^* \to F_1^*$, thus $E^*$ is a kernel of a morphism of locally free sheaves, hence reflexive.

1$\begingroup$ Am I missing something, or is it true that this same argument applies to any smooth projective variety (not just projective space), so that in fact the dual of a coherent sheaf on such a variety is reflexive? $\endgroup$ Dec 29, 2016 at 21:41

If $E$ is a coherent sheaf on a noetherian scheme, the dual $E^*=Hom_{O_X}(E, O_X)$ is always coherent. If $A$ is an affine open subset, then $E^*$ is the sheaf associated to the $A$module $Hom_A( \Gamma(A, E), \Gamma(A, O_X))$. More generally, sheaf hom of any two sheaves preserves coherence.
This is a corollary of the fact that if $M,N$ are finitely presented $A$modules, then for any multiplicative subset $S$, $S^{1}Hom_A(M,N) = Hom_{S^{1}A}(S^{1}M, S^{1}N)$, which can be found in any commutative algebra textbook (e.g. Eisenbud).
A good starting reading on reflexive sheaves is (at least the first section of) Hartshorne: Stable reflexive sheaves.