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Consider over $\mathbb{C}$. Let $(X,\mathcal{O}(1))$ be a smooth projective scheme with an ample polarisation. Let $P(t):=\chi(X,\mathcal{O}(t))$ denote the Hilbert polynomial of $\mathcal{O}_X$. Choose a decomposition $P(t)=I(t)+Q(t)$ such that $I(t),Q(t)$ are also Hilbert polinomials. If a sheaf of ideals $\mathcal{I}\subseteq\mathcal{O}_X$ has Hilbert polynomial $I(t)$, then the sheaf $\mathcal{O}_X/\mathcal{I}$ has Hilbert polynomial $Q(t)$.

By smoothness, we have that $\mathcal{O}_X$ is Gieseker stable. Then any ideal $\mathcal{I}\subseteq\mathcal{O}_X$ is stable, because any subsheaf of a rank one pure sheaf is Gieseker stable.

Consider the following moduli scheme of stable sheaves and Hilbert scheme $$\mathrm{M}^{I(t),\mathrm{s}}:=\{\textrm{stable sheaves on }X\textrm{ of Hilbert polynomial }I(t)\}\\ \mathrm{Hilb}_X^{Q(t)}:=\{\textrm{subschemes }Z\hookrightarrow X\textrm{ s.t. }\mathcal{O}_Z\textrm{ has Hilbert polynomial }Q(t)\}.$$

For a closed subschem $Z\hookrightarrow X$, the associated ideal $\mathcal{I}_Z:=\ker(\mathcal{O}_X\to \mathcal{O}_Z)$ is stable of Hilbert polynomail $I(t)$, defining a point of $\mathrm{M}^{I(t),\mathrm{s}}$. This defines a morphism $$\mathrm{Hilb}_X^{Q(t)}\to \mathrm{M}^{I(t),\mathrm{s}},\quad (Z\hookrightarrow X)\mapsto \mathcal{I}_Z. $$

The question is when is the morphism an isomorphism? It seems that it is an isomorphism when $X=\mathbb{P}^3$ and the Hilbert scheme parametrises curves according to discussions in this MO. Does it suffices to assume that $X$ is smooth, and $Z\hookrightarrow X$ has higher codimension?

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This works for any smooth projective variety $X$ under the assumption $$ \mathrm{Pic}^0(X) = 0 $$ and any $Z$ of codimension at least 2. For the proof see Lemma B.5.6 in Kuznetsov, Alexander G.; Prokhorov, Yuri G.; Shramov, Constantin A. Hilbert schemes of lines and conics and automorphism groups of Fano threefolds. Jpn. J. Math. 13 (2018), no. 1, 109--185.

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  • $\begingroup$ I find this MO answer saying that if we replace $\mathrm{M}^{I(t),\mathrm{s}}$ by the moduli scheme of stable sheaves with trivial determinant, then we have an isomorphism with the Hilbert scheme. $\endgroup$ Aug 15, 2023 at 3:06
  • $\begingroup$ I agree, the proof of Lemma B.5.6, indeed, proves this. $\endgroup$
    – Sasha
    Aug 15, 2023 at 4:23

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