# Computing Ext sheaves over complex projective plane

Let $$X:=\mathbb{P}^2_K$$ with $$K$$ algebraically closed field. Take $$p\in X$$ a point and $$\mathcal{I}_p$$ its ideal sheaf. One can prove (using Serre Duality and the exact sequence defining $$\mathcal{I}_p$$) that $$Ext^1((\mathcal{I}_p)_p,\mathcal{O}_{X,p})\cong K$$, so there exists a non trivial extension $$\begin{equation*}0\to\mathcal{O}_X\to\mathcal{E}\to\mathcal{I}_p\to 0 \end{equation*}$$where $$\mathcal{E}$$ is a coherent sheaf over $$X$$. Now I would like to prove that $$\mathcal{Ext}^i(\mathcal{E},\mathcal{O}_X)\cong 0$$ for all $$i>0$$. The case $$i=1$$ is clear since if $$\mathcal{Ext}^1(\mathcal{E},\mathcal{O}_X)$$ were not be zero, then the connecting homomorphism from $$\mathcal{Hom}(\mathcal{O}_X,\mathcal{O}_X)$$ to $$\mathcal{Ext}^1(\mathcal{I}_p, \mathcal{O}_X)$$ would be surjective and this is not possible. So my first question

How can I prove that $$\mathcal{Ext}^i(\mathcal{E},\mathcal{O}_X)\cong 0$$ for all $$i\ge 2$$?

My first attempt was to take $$\mathcal{Ext}^i(-,\mathcal{O}_X)$$ to the exact sequence above, but then I've realized that I should compute $$\mathcal{Ext}^i(\mathcal{I}_p,\mathcal{O}_X)$$ first. Here comes my second question

How can I compute $$\mathcal{Ext}^i(\mathcal{I}_p,\mathcal{O}_X)$$ for all $$i\ge 0$$?

• For the second question use the exact sequence $$0 \to I_p \to \mathcal{O}_{\mathbb{P}^2} \to \mathcal{O}_p \to 0$$ and the isomorphisms $$\mathcal{E}\mathit{xt}^i(\mathcal{O}_p,\mathcal{O}_X) \cong \mathcal{O}_p$$ for $i = 2$ and zero otherwise. – Sasha Mar 11 at 14:51
• @Sasha This is my argument: I have to use Serre duality $Ext^i(E,F)\cong Ext^{2-i}(F,E\otimes\omega_X)^{\vee}$ to the sheaves $E=\mathcal{O}_p, F=\mathcal{O}_X$ and $\omega_X=\mathcal{O}_X(-3)$. So I obtain $Ext^i(\mathcal{O}_p, \mathcal{O}_X)\cong Ext^{2-i}(\mathcal{O}_X, \mathcal{O}_p)^{\vee}$. The last sheaf I wrote is easy to compute since $\mathcal{O}_X$ is locally free and all the Ext groups with index greater or equal to 1 vanish. Finally $Hom(\mathcal{O}_X, \mathcal{O}_p)\cong (\mathcal{O}_p)^{\vee}$. Is it right? Thanks in advance. – John117 Mar 11 at 15:17
• Serre duality tells you about global $Ext$-spaces, while here you need local $Ext$-sheaves. In fact, for $\mathcal{O}_p$ it is not hard to extract the local result from the global. Alternatively, you can use the Grothendieck duality instead of Serre duality. Or you can use the standard Koszul resolution of $\mathcal{O}_p$ to compute the $Ext$-sheaves. – Sasha Mar 11 at 15:23
• Ok, I don't know the Koszul resolution of $\mathcal{O}_p$, moreover I have to think about your comment. Perhaps I'm missing something. Thank you :) – John117 Mar 11 at 15:33

As stated above, from the LES in $$\mathcal{E}xt$$ we have $$\mathcal{E}xt^i(\mathcal{E}, \mathcal{O}_X) \simeq \mathcal{E}xt^i(I_p, \mathcal{O}_X)$$ for all $$i \geq 2$$. How can we compute the term on the left hand side? The key is to observe the following fact:

A point in $$\mathbf{P}^2$$ is a complete intersection of two lines.

What this means is that if $$I_p = \widetilde{I}$$ for $$I$$ an ideal of $$R := \mathbf{C}[x,y,z]$$ generated by $$f$$ and $$g$$, we have a resolution (at the level of graded modules) given by $$0 \to R(-2) \stackrel{c\mapsto (c,-c)}{\to} R(-1) \oplus R(-1) \stackrel{(a,b) \mapsto (af + bg)}{\to} I \to 0.$$ On the level of sheaves, this is $$0 \to \mathcal{O}_X(-2) \to \mathcal{O}_X(-1) \oplus \mathcal{O}_X(-1) \to I_p \to 0.$$ It is now clear that $$\mathcal{E}xt^i(I_p, \mathcal{O}_X) = 0$$ for all $$i \geq 2$$, since in the LES it is sandwiched between sheafy exts of vector bundles (which are zero).