Duality and Serre's criterion

Question

Let $X$ be a projective scheme, $\mathcal{F}$ a coherent sheaf on $X$ which is $S_2$. Then under what additional conditions is its dual, $\mathcal{H}om_X(\mathcal{F},\mathcal{O}_X)$ also $S_2$?

Answer 1

If $X$ is noetherian and integral, the the dual of a coherent sheaf is always reflexive. If $X$ is even normal then a coherent sheaf is reflexive if and only if it is $S_2$.

You can read about that in Hartshornes paper about reflexive sheaves. Or there is one by Karl Schwede titled "Generalized divisors and reflexive sheaves" which can be found with Google.

Answer 2

To me it seems that whether $F$ satisfies $S_2$ or not doesn't have much of an influence on the question if $\mathrm{\mathcal{H}\it{om}}_{\mathcal{O}_X}(F,\mathcal{O}_X)$ satisfies $S_2$. If $X$ satisfies $S_2$, then the dual of any coherent sheaf on $X$ also satisfies $S_2$. On the other hand, I have no idea how to check whether $\mathrm{\mathcal{H}\it{om}}_{\mathcal{O}_X}(F,\mathcal{O}_X)$ is $S_2$ unless I have information on the depth-behavoiur of $\mathcal{O}_X$. The most general thing I know about the depth of a $\mathrm{Hom}$-module is the following:

Let $A$ be a Noetherian local ring and $M,M'$ two finitely generated $A$-modules. Then $$\mathrm{depth}(\mathrm{Hom}_A(M,M'))\geq \min(\mathrm{depth}(M'),2).\quad(\ast)$$ This easily gives that $\mathrm{Hom}_A(M,M')$ satisfies $S_2$ if so does $M'$.

Since you're asking for the dual, I'll simplify and take $M' = A$. The argument, however, is the same. We take a finite presentation of $M$, $$A^k\to A^m\to M\to 0,$$ and apply $\mathrm{Hom}_A(-,A)$ to get a short exact sequence $$0\to \mathrm{Hom}_A(M,A)\to \mathrm{Hom}_A(A^m,A)\to N\to 0,$$ where $N\subset \mathrm{Hom}_A(A^k,A)$ is the image of $\mathrm{Hom}_A(M,A)\to \mathrm{Hom}_A(A^m,A)$.

From the long exact $\mathrm{Ext}$-sequence we see that $$\mathrm{depth}(\mathrm{Hom}_A(M,A))\geq \min(\mathrm{depth}(A),\mathrm{depth}(N)+1).$$

Suppose $\mathrm{depth}(N) = 0$. Then, since $N\subset \mathrm{Hom}_A(A^k,A)\cong A^k$, also $\mathrm{depth}(A) = 0$, hence $(\ast)$ is trivially true (where $M' = A$).

Consequently, in any case, $\mathrm{depth}(\mathrm{Hom}_A(M,A))\geq \min(\mathrm{depth}(A),2)$. Finally, if $A$ is $S_2$, then $$\mathrm{depth}(\mathrm{Hom}_A(M,A))\geq \min(\dim(A),2)\geq \min(\dim(\mathrm{Hom}_A(M,A)),2),$$ so $\mathrm{Hom}_A(M,A)$ is $S_2$.