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 $i\ge 0$?

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    $\begingroup$ 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. $\endgroup$ – Sasha Mar 11 at 14:51
  • $\begingroup$ @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. $\endgroup$ – John117 Mar 11 at 15:17
  • $\begingroup$ 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. $\endgroup$ – Sasha Mar 11 at 15:23
  • $\begingroup$ 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 :) $\endgroup$ – 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).


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