Let $X$ be a complex algebraic variety. Its Hilbert scheme represents the functor $G$ from schemes to sets given by $$G(S)=\{Z\subset X\times S|\, Z \mbox{ is a closed subscheme, flat and proper over } S\}.$$

Consider the scheme $\mathcal{F}^X$ which parameterizes pairs: closed proper subscheme of $X$ and a point on it. More precisely, $\mathcal{F}^X$ represents the functor $H$ from schemes to sets $$H(S)=\{Z\subset X\times S \mbox{ is as above }, f\colon S\to Z \mbox{ is a section of the projection } Z\to S\}.$$

I was told in another post Basic questions on the Hilbert scheme/ Douady space the following facts:

1) $\mathcal{F}^X$ does exits.

2) We have the obvious morphisms $\mathcal{F}^X\to X$, $\mathcal{F}^X\to Hilb(X)$ by forgetting either subvariety of a point on it. Then the morphism $$\mathcal{F}^X\to X\times Hilb(X)$$ is a closed imbedding.

Let now $Y\subset X$ be a closed subscheme. I was told in the same above mentioned post that $Hilb(Y)$ is a closed subscheme of $Hilb(X)$. Thus we have three pairs of closed subschemes $$\mathcal{F}^X\subset X\times Hilb(X),$$ $$\mathcal{F}^Y\subset Y\times Hilb(Y)\subset X\times Hilb(X).$$

Question. Is it true that $$\mathcal{F}^Y=\mathcal{F}^X\times_{(X\times Hilb(X))} (Y\times Hilb(Y))?$$

I am particularly interested in a modification of this question when $X$ is a complex analytic (rather than algebraic) scheme; in that case an analogue of Hilbert scheme is called Douady space.

A reference would be very helpful.


1 Answer 1


This is true. The condition defining $F^X$ in $X\times Hilb(X)$ is the vanishing of the morphism $I_{Z,X} \subset O_X \to O_x$ (where $Z$ is a subscheme and $x$ is a point). It follows that the fiber product embedds into $Y \times Hilb(Y)$ and is given there by the vanishing of the above morphism. It remains to note that for $(x,Z) \in Y \times Hilb(Y)$ the above composition factors as $$ I_{Z,X} \to I_{Z,Y} \to O_Y \to O_y, $$ where the first map is the restiction to $Y$. This first map is surjective, so the composition vanishes if and only if the composition $I_{Z,Y} \to O_Y \to O_y$ does, which is precisely the property defining $F^Y$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy