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Let $X$ be a smooth complex manifold. Then the Hilbert scheme of one point is canonically isomorphic to $X$ $$ \mathrm{Hilb}^1(X)\cong X. $$ Is this still true for general scheme $X$ (not necessarily regular, reduced etc) over a field $k$? If not what kind of assumption does one need for the claim to hold?

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You might need a ground field or a base scheme to define the Hilbert scheme. – user18119 Nov 20 '12 at 8:55
You are right. I fixed a base field $k$. – M. K. Nov 20 '12 at 19:11
up vote 3 down vote accepted

For a general ground field $k$ and an algebraic variety $X$ over $k$, the Hilbert scheme of one point should parametrises rational points $X$. But $X$ has exactly the same rational points as $X$ itself ! So $\mathrm{Hilb}^1(X)=X$.

In a more formal setting, and to be really related to Hilbert schemes, let $X\to S$ be a projective scheme with a given ample sheaf $O_X(1)$. For any $S$-scheme $T$, consider the set $H_X(T)$ of closed subschemes $Z$ of $X\times_S T$, flat over $T$, whose fibers have constant Hilbert polynomials $1$ (with respect to $O_X(1)$). The the "Hilbert schemes of one point" for $X$ is the scheme which represents the functor $$ \underline{\mathrm{Sch}}_S \to \underline{\mathrm{Set}}, \quad T \mapsto H_X(T). $$ Now if $Z\in H_X(T)$, then the fibers of $Z\to T$ are projective varieties with constant Hilbert polynomial $1$. So each fiber is just a rational point. Hence $Z\to T$ is an isomorphism (finite and flat, of degree $1$). So $H_X(T)$ is just the set of sections of $X\times_S T\to T$. But this functor is nothing but the functor represented by $X$. So the corresponding Hilbert scheme is $X$.

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