# The first order deformations of Hilbert Scheme

Let $X$ be a complex smooth variety. We know for $Z\in Hilb^n(X)$, $T_Z(Hilb^n)=Hom(I_Z, O_Z)$. Hence there is a 1-1 correspodence between $f\in Hom(I_Z, O_Z)$ and the set of first-order deformations of Z. My question is can we express the correspodence explicitly. For example, let $X=A^2$ with coordinates x,y and $I_Z=(x,y)$. Take an element $f\in Hom(I_Z, O_Z)$. What is the corresponding first-order deformations of Z. Is it $(x+sf(x), y+sf(y))$?

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You are right. And the same is true in general. Assume $X$ is affine. Choose a collection of generators $g_1,\dots,g_r$ of $I_Z$ --- then $I_Z = Im(O_X^r \stackrel{(g_1,\dots,g_r)}\to O_X)$. Given a map $f:I_Z \to O_Z$ consider the composition $O_X^r \to I_Z \to O_Z$ and choose its lift $\tilde{f} = (\tilde{f}_1,\dots,\tilde{f}_r):O_X^r \to O_X$. Then the first order deformation is given by the ideal $$Im(O_X^r \stackrel{(g_1 + s\tilde{f}_1,\dots,g_r + s\tilde{f}_r)}\to O_X).$$

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Can you give us a proof? Thank you! – Diego Maradona Sep 21 '10 at 12:00
If you replace $X$ by $X \times Spec k[s]/s^2$ and define an ideal by the same formula, you will get a flat family over $Spec k[s]/s^2$ which is the same (since $Hilb$ represents a functor!) as a map $Spec k[s]/s^2 \to Hilb$, and this is the same as a point in the tangent space. Vice versa, any flat ideal in $X \times Spec k[s]/s^2$ can be written in this form. That's all. – Sasha Sep 21 '10 at 18:34