Let $X$ be a complex manifold. I just recall some notions, for my question below to be clear.
Let $F\in\textrm{Coh}(X)$ be coherent of rank $r$. If it is torsionfree, it has a well-defined determinant line bundle $\det F=(\Lambda^rF)^{\vee\vee}\in\textrm{Pic}(X)$. If $r=1$, then $\det F=F^{\vee\vee}$. Furthermore, $F$ is torsionfree iff the natural map $F\to F^{\vee\vee}$ is injective, and is said to be reflexive iff that map is an isomorphism.
It follows:
Fact. A reflexive rank one sheaf $F$ on $X$, such that $\det F=\mathcal O_X$, is necessarily $\mathcal O_X$ itself.
Let $I\subset \mathcal O_X$ be an ideal sheaf. Then $I$ is torsionfree of rank one, so it embeds into its double dual $0\to I\to I^{\vee\vee}=\det I$.
Question: why is $I^{\vee\vee}=\mathcal O_X$?
I believe I should use the Fact, but I do not see why $I^{\vee\vee}$ should be reflexive with trivial determinant.
It is obviously reflexive because it is a line bundle, and it actually is its own determinant, but why is it trivial?
Thanks for any help.
It's been a while since this was posted, but I'd like to note that the reason it is hard to show is that it is false. If $I$ is the ideal sheaf of a codimension $1$ subvariety (say $X$ is projective), then the determinant is not trivial: $I$ is reflexive, and $\det I = I$.
If $I$ is the ideal sheaf of a subvariety of codimension $\geq 2$, then the result is true.