Let $X$ be a smooth, projective rational surface and $Z$ be a zerodimensional subscheme of $X$. Denote by $\mathcal{I}_Z$ the ideal sheaf of $Z$ in $X$ and $\mathcal{O}_Z$ the structure sheaf. Is it true that $$\mbox{Hom}_X(\mathcal{I}_Z,\mathcal{O}_Z) \cong \mbox{Ext}^1_X(\mathcal{I}_Z, \mathcal{I}_Z)?$$ Any hint or reference will be most welcome.
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$\begingroup$ Morally Ext1(I,I) is the tangent space to the module of sheaves at I, so it should be true because Pic0(X)={0}. If you take the long exact sequence coming from deriving Hom(I,) applies to the ideal sequence of Z you see that Hom(I,Oz)=Ext1(I,I) exactly when Ext1(I,Ox)=0. Taking the long exact sequence coming from deriving Hom(,Ox) applied to the ideal sequence shows Ext1(Ox,Ox)=Ext1(I,Ox). Ext1(Ox,Ox)=H1(Ox)=0 because Pic0(X)={0} and H1(Ox) parametrizes infinitesimal deformations of Ox. $\endgroup$ – Yosemite Stan Oct 9 at 19:04

2$\begingroup$ @YosemiteStan I just deleted a wrong answer along those lines (because I had made a mistake). Using Serre duality, one can show that $\operatorname{Ext}^1_X(\mathcal I_Z,\mathcal O_X) \neq 0$ (in fact it has dimension equal to the length of $Z$). But it can still be the case that the map to $\operatorname{Ext}^1_X(\mathcal I_Z,\mathcal O_Z)$ is injective, which would also be enough. $\endgroup$ – R. van Dobben de Bruyn Oct 9 at 19:31

1$\begingroup$ Hmmm, I guess I applied the dimension vanishing on the wrong side. $\endgroup$ – Yosemite Stan Oct 9 at 21:00
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See Lemma B.5.6 in A. Kuznetsov, Yu. Prokhorov, C. Shramov, "Hilbert schemes of lines and conics and automorphism groups of Fano threefolds", Japanese Journal of Mathematics, V. 13 (2018), N. 1, pp. 109185.