I am reading some texts about the tangent space to the Hilbert scheme. Apparently, $T_{[Z]}(Hilb (X)) = H^0(Z, N_{Z/X})$ for any $Z$ is a consensus which I agree with, but then regarding the hilbert scheme of points, people generally use that $T_{[Z]}(Hilb_r (X)) = \text{Hom}_{O_X}(I_Z, i_*O_Z)$, and there are a lot of proofs of this last equality but I don't really see where they use that $Z$ is finite.

I know that $ H^0(Z, N_{Z/X}) = \text{Hom}_{O_Z}(I_Z/I_Z^2, O_Z)$, so maybe I am missing something and the equality $\text{Hom}_{O_Z}(I_Z/I_Z^2, O_Z) = \text{Hom}_{O_X}(I_Z, i_*O_Z)$ is obvious, or is obvious when $Z$ is a finite scheme.

The usual proofs of $T_{[Z]}(Hilb_r (X)) = \text{Hom}_{O_X}(I_Z, i_*O_Z)$ basically try to characterize all quotients $Q$ of $O_X [ \varepsilon ]$ that are flat over $k[ \varepsilon ]$ and such that, $Q/\varepsilon Q = O_Z$. Chasing a bit in some diagram they show that this is equivalent as giving a $O_X$-morphism from $I_Z$ to $i_*O_Z$. For some links, see for example page 22 here. The proof convinces me but it doesn't say where thye use that $Z$ is a finite scheme.

Can someone shed some light on this topic? Thank you

Notes: Here $X$ is a projective variety, and $F[\varepsilon] = F\otimes k(t)/(t^2)$. When I say a finite scheme a mean a scheme of finite type over a field whose topological space is finite. but it can be non-reduced


1 Answer 1


First, let's just set $i :Z \to X$ for the closed immersion.

Next, a small correction: the the second formula for the tangent space should be $T_{[Z]} \mathrm{Hilb}_r(X) = \mathrm{Hom}_{\mathcal{O}_X}(I_Z,\mathcal{O}_Z)$.

In this formula, we are viewing $\mathcal{O}_Z$ as a $\mathcal{O}_X$-module, so it would be more proper to write $T_{[Z]} \mathrm{Hilb}_r(X) = \mathrm{Hom}_{\mathcal{O}_X}(I_Z,i_* \mathcal{O}_Z)$. By adjunction, we have $\mathrm{Hom}_{\mathcal{O}_X}(I_Z,i_* \mathcal{O}_Z) = \mathrm{Hom}_{\mathcal{O}_Z}(i^* I_Z,\mathcal{O}_Z)$. Finally, noting that $i^* I_Z$ is the conormal sheaf $I_Z /I_Z^2$, we have that the two formulas agree.

Indeed, none of this really depends on $Z$ being finite.

  • $\begingroup$ Thank you very much, I will correct the mistakes :) $\endgroup$ Apr 24 at 12:41

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