# Tangent space to Hilbert schemes of points

Let $$X$$ be a smooth, projective rational surface and $$Z$$ be a zero-dimensional 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.

• 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. – Yosemite Stan Oct 9 at 19:04
• @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. – R. van Dobben de Bruyn Oct 9 at 19:31
• Hmmm, I guess I applied the dimension vanishing on the wrong side. – Yosemite Stan Oct 9 at 21:00