Let $\mathbf{P} = \mathbf{P}^n(k)$ be the $n$-dimensional projective space over a field $k$, let $A, B$ be projective varieties in $\mathbf{P}$ such that $A \subset B$. Now define $V(A,B)$ to be the set of all projective varieties $C$ such that $A \subseteq C \subseteq B$. What can be said about the structure of $V(A,B)$ ? What about the same question where we allow "quasi-projective'' instead of ''projective" ? Can one see $V(A,B)$ as a variety (and what is its dimension) ?


1 Answer 1


This is more reasonable if you insist that $C$ have a given Hilbert polynomial. Otherwise, consider the case $A = \emptyset$, $B = \bf P = \bf P^2$. Then you have curves of every degree, so your $V(A,B)$ will be of infinite type.

With $C$'s Hilbert polynomial fixed, the natural thing to look at is the subschemes $C$ between $A$ and $B$. This defines a closed subscheme $X$ of the Hilbert scheme of $\bf P$ with this chosen Hilbert polynomial, so in particular is projective. (I get the impression from your question that you need to learn about Hilbert schemes, from e.g. Eisenbud and Harris.)

If you only want varieties, then you're thinking about an open subscheme $V(A,B)$ of that $X$. Given how badly behaved Hilbert schemes are, I doubt it $V(A,B)$ will be irreducible, and I suspect figuring out the dimensions of its components will be quite intractable in general.

  • $\begingroup$ Hey - thanks for the nice answer ! Supposing that one considers quasi-projective varieties $A, B, C$ (so that, as you state, it is possible to obtain a thing of infinite type), is there a formal statement somewhere that $V(A,B)$ iself is always a variety, or a scheme ? $\endgroup$
    – THC
    Sep 12, 2014 at 13:39

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