I am reading Nakajima's Lectures on Hilbert Schemes of Points on Surfaces, and I am confused about a detail at the top of page 13. He is sketching a proof that $X^{[n]}(=$ Hilbert scheme of $n$ points on a nonsingular surface $X$) is nonsingular. The strategy is to show that for $[Z]\in X^{[n]}$, $\dim T_{[Z]}X^{[n]}=2n$. The method of doing this is a fairly straightforward argument using $$0\to \mathcal{I}_Z\to\mathcal{O}_X\to \mathcal{O}_X/\mathcal{I}_Z\to 0$$ and the long exact sequence for $\operatorname{Ext}^\bullet_{\mathcal{O}_X}(-,\mathcal{O}_Z)$. I need help understanding the following claim:
The Euler characteristic $\sum_{i=0}^2 (-1)^i \dim \operatorname{Ext}^i(\mathcal{I}_Z,\mathcal{O}_Z)$ is independent of $Z$.
I believe Nakajima means $\chi(X,\mathcal{O}_Z)=\sum_{i=1}^2(-1)^i\dim \operatorname{Ext}^i(\mathcal{I}_Z,\mathcal{O}_Z).$ I agree that for points $[Z]\in X^{[n]}$, $\chi(X,\mathcal{O}_Z)$ should be the same; namely equal to $n$. I just don't understand why the Euler Characteristic can be written in this fashion. It's probably easy, but escapes me!
It is not true that the alternating sum of dimensions of $\mathrm{Ext}^i$ is equal to the Euler characteristic of $\mathcal{O}_Z$, but it is true that it is independent of $Z$. The point is that it only depends on the classes of arguments in the numerical Grothendieck group $K_0(X)$; indeed, this follows from the long exact sequence of $\mathrm{Ext}$-spaces.
Now, it remains to note that the classes of $\mathcal{O}_Z$ and $\mathcal{I}_Z$ do not depend on $Z$. Indeed, $\mathcal{O}_Z$ has a filtration with $n$ factors isomorphic to $\mathcal{O}_{x_i}$, the structure sheaves of points, hence its class is equal to $n[\mathcal{O}_x]$ for any point $x$ on $X$. Similarly, the class of $\mathcal{I}_Z$ is equal to $[\mathcal{O}_X] - n[\mathcal{O}_x]$.