# Hilbert schemes and moduli of ideal sheaves

Let $X$ be a smooth projective variety over $\mathbb{C}$. The Hilbert scheme on $X$ parametrizes quotients $\mathcal{O}_X \to E$ with fixed Hilbert polynomial. Let us fix the Hilbert polynomial to lead to subschemes of codimension at least $2$. Then we can also look at the moduli space parametrizing semistable rank $1$ sheaves with corresponding Hilbert polynomial and trivial determinant, i.e. ideal sheaves.

Whenever we have a family of quotients, we can take the kernel to get a family of ideal sheaves. That should induce a bijective morphism. However, it seems unclear to me how to get an inverse morphisms.

Under what kind of hypotheses are these two moduli spaces the same? I am in particular interested in the case of Hilbert schemes of curves in $\mathbb{P}^3$.

Edit: Since I never wrote it down and it came up in the comments, let me write down what the functors are I am exactly talking about.

The Hilbert scheme represents the functor that maps $S$ to quotients $\mathcal{O}_{X \times S} \to E$ that are flat over the base $S$.

The moduli of semistable sheaves represents the functor that maps $S$ to flat families $E \in \operatorname{Coh}(X \times S)$ of semistable sheaves modulo tensoring with line bundles pulled back from $S$.

We then look at the corresponding connected components coming from fixing the Hilbert polynomial and ask whether they are the same.

It seems to me that $$\mathscr{I}_Z^{\vee\vee}\simeq \mathscr{O}_X,$$ because the left-hand side is reflexive and isomorphic to $\mathscr{O}_X$ away from $Z$, which has codimension at least $2$ by assumption. In other words, the abstract sheaf $\mathscr{I}_Z$ is canonically embedded in $\mathscr{O}_X$ via the double-dual map $$\mathscr{I}_Z\to \mathscr{I}_Z^{\vee\vee}\simeq \mathscr{O}_X.$$
So the inverse map you are looking for is induced by $$\mathscr{I}_Z\mapsto (\mathscr{O}_X\simeq\mathscr{I}_Z^{\vee\vee}\to \text{coker}(\mathscr{I}_Z\to \mathscr{I}_Z^{\vee\vee})\simeq \mathscr{O}_Z).$$
I suppose I used that $X$ is smooth in the identification $$\mathscr{I}_Z^{\vee\vee}\simeq \mathscr{O}_X,$$ but you can probably relax this a bit. (I'm a little worried about making this work in families, which is, strictly speaking, necessary; so this answer is not quite complete.)
• "I'm a little worried about making this work in families, which is, strictly speaking, necessary; so this answer is not quite complete." Please see EGA III_2, Corollaire 7.7.8.2, p. 70: $Y$ is the moduli space of sheaves, $X$ is the product of $Y$ and your fixed variety, $\mathcal{F}$ is the structure sheaf of $X$, and $\mathcal{G}$ is the universal ideal sheaf. Proj of the symmetric algebra of $\mathcal{N}$ is the scheme which naturally maps to the Hilbert scheme. It remains to prove that $\mathcal{N}$ is invertible. This can be checked on closed, affine, Artin subschemes of $Y$. – Jason Starr Nov 15 '15 at 16:30
• By the way, if $X$ is a proper Deligne-Mumford stack (rather than a scheme), then existence of $\mathcal{N}$ is generalized via the following article: M. Lieblich, "Remarks on the stack of coherent algebras", Int. Math. Res. Not. 2006, Art. ID 75273. If $X$ is a proper Artin stack whose diagonal is finite, this follows from the following article: Olsson, M. "Hom-stacks and restriction of scalars." Duke Math. J. 134 (2006), no. 1, 139–164. – Jason Starr Nov 15 '15 at 18:55
• Since formation of $\mathcal{N}$ is compatible with arbitrary base change, by Daniel's argument the base change of $\mathcal{N}$ from $Y$ to each closed point $y$ of $Y$ is a rank $1$ locally free sheaf. Also, by construction $\mathcal{N}$ is coherent. Thus, by Nakayama's Lemma, Zariski locally $\mathcal{N}$ is a quotient of the structure sheaf of $Y$. If the kernel of this quotient is not zero, then Krull's Intersection Theorem implies the same after base change to $\mathcal{O}_{Y,y}/\mathfrak{m}_{Y,y}^N$, which is Artinian. – Jason Starr Nov 20 '15 at 19:21
• Jason's argument shows that $\mathscr{N}$ is a line bundle on $X\times S$ pulled back from the base $S$; to conclude that it is in fact trivial (and hence $\mathscr{N}^{\vee}$ is trivial, which is what we want) use that $\mathscr{N}^\vee\simeq \det(\mathscr{I}_Z)$ b/c it is so away from $Z$ (and its formation commutes w/ base change) and is locally free; but your condition is that $\det(\mathscr{I}_Z)$ is trivial. – Daniel Litt Nov 21 '15 at 3:50