Question:

Curvilinear subschemes and Ext functors.

Easton: 3 days ago

Let $S$ be a smooth, projective surface over $\mathbb{C}$ and let $\xi$ be a $0$-dimensional subscheme. I look at the sheaf $\underline{Ext}^2(O_\xi,O_S)$. Is it true that, if $\xi$ is curvilinear, then $\underline{Ext}^2(O_\xi,O_S)=O_\xi$? What happens instead if $\xi$ is not curvilinear?

Answer:
Miles: 3 days ago

Yes, it is true. Indeed, if $\xi$ is curvilinear then it is lci, so locally there is a resolution $$ 0 \to O_S \xrightarrow{ (g,-f) } O_S \oplus O_S \xrightarrow{ (f,g) } O_S \to O_\xi \to 0. $$ Applying local $Hom$ to $O_S$ one obtains an exact sequence $$ 0 \to O_S \xrightarrow{ (f,g) } O_S \oplus O_S \xrightarrow{ (g,-f) } O_S \to \underline{Ext}^2(O_\xi,O_S) \to 0. $$ It is clear that the second complex is isomorphic to the first, so $\underline{Ext}^2(O_\xi,O_S) \cong O_\xi$.

On the other hand, the simplest example of a subscheme which is not curvilinear is $\xi = Spec\ k[x,y]/(x^2,xy,y^2) \subset Spec\ k[x,y]$. There is a resolution $$ 0 \to O_S^2 \xrightarrow{ \left(\begin{smallmatrix} y & 0 \cr -x & y \cr 0 & -x \end{smallmatrix}\right) } O_S^3 \xrightarrow{ (x^2,xy,y^2) } O_S \to O_\xi \to 0. $$ Applying local $Hom$ to $O_S$ one deduces that $$ \underline{Ext}^2(O_\xi,O_S) = Coker(O_S^3 \xrightarrow{ \left(\begin{smallmatrix} y & -x & 0 \cr 0 & y & -x \end{smallmatrix}\right) } O_S^2). $$ It follows that as a $k[x,y]$-module it is a vector space of dimension 3 with the action of $x$ and $y$ given by $$ x= \left(\begin{smallmatrix} 0 & 0 & 0 \cr 0 & 0 & 0 \cr 1 & 0 & 0 \end{smallmatrix}\right), \qquad y= \left(\begin{smallmatrix} 0 & 0 & 0 \cr 0 & 0 & 0 \cr 0 & 1 & 0 \end{smallmatrix}\right). $$ In particular, it is not generated by 1 vector (as a $k[x,y]$-module), so it is not isomorphic to $O_\xi$.