Criterions for Reflexiveness of sheaves and a special case

Question

A coherent sheaf $V$ on a say noetherian scheme is called reflexive if the canonical map $V \rightarrow \mathcal Hom_{\mathcal O_X}(\mathcal Hom_{\mathcal O_X}(V,\mathcal O_X),\mathcal O_X)$ is an isomorphism of sheaves.

In principle, one can define this notion also for quasicoherent sheaves, and this is what my question is about: does one have criteria about when a quasicoherent module is reflexive? Or is the question of reflexiveness in general very hard to answer?

What I am particularly interested in as a special case:

If one has an infinite chain of inclusions $V_1 \subset V_2 \subset ...$ reflexive sheaves, then one knows of course that each $V_i$ is reflexive. But is the union of all these sheaves also reflexive?

Answer 1

I'll add another example to the mix even in the ring setting (as opposed to the sheaf setting). Fix a DVR $(R, \langle x \rangle)$. Then the fraction field $K(R) = \bigcup_n (x^{-n}R)$, is an ascending union of free (reflexive) modules. But clearly $K(R)$ is not itself reflexive.

On the other hand, under mild conditions on the scheme $X$, for example if it is normal, then for a coherent module being reflexive is equivalent to being S2 (Serre's second condition). The S2 condition behaves fairly well for quasi-coherent modules and might be useful for you. For example, see the paper(s) of Hartshorne: "Generalized divisors ..."

Answer 2

Let $R$ be a discrete valuation ring and $V$ a free $\mathcal O_X$-module of infinite rank on $X=\mathrm{Spec}(R)$. Then, neither $\mathcal Hom(V,\mathcal O_X)$ nor $\mathcal Hom(\mathcal Hom(V,\mathcal O_X),\mathcal O_X)$ is quasi-coherent. In particular, $V$ is not reflexive.

Answer 3

Let $O$ be the valuation ring of a valuation $v$ having a subgroup of the reals different from the integers as its value group. Let $v_1>v_2>\ldots >0$ be a monotonously decreasing sequence of values converging to $0$ and chose elements $x_i\in O$ such that $v(x_i)=v_i$. Then $O x_1\subset O x_2\subset\ldots \subset M$, where $M$ is the maximal ideal of $O$. Moreover

$M=\bigcup\limits_{i}Ox_i$.

But $M$ is not reflexive: $(O:M)=O$, hence $(O:(O:M))=(O:O)=O$.

However: this may only tell us that we should define reflexivity for non-coherent modules / sheaves in a different way.

Answer 4

The problem with this attempt is that the dual of a direct sum is a direct product. So let $X=\mathrm{Spec} k$ and $V$ an infinite dimensional vector space over $k$. Then the (double) dual of $V$ is much larger than $V$ so they cannot be isomorphic.