Suppose $X$ is a projective scheme over an algebraically closed field $k$, denote its Hilbert scheme with Hilbert polynomial $p$ by $\text{Hilb}^p_X$, then from section 1.1 of Nakajima's book, Lectures on Hilbert Schemes of Points on Surfaces, it claims that

" Moreover, if we have an open subscheme $Y$ of $X$, then we have the corresponding open subscheme $\text{Hilb}^p_Y$ of $\text{Hilb}^p_X$ parametrizing subschemes in $Y$. In particular, $\text{Hilb}^p_Y$ is defined for a quasi-projective scheme $Y$."

I could see this is true for the Hilbert scheme of points, i.e. $p$ is a constant polynomial. Is it true generally? One basic problem is could the Hilbert polynomial of a quasi-projective variety be defined?

  • $\begingroup$ Not sure if this is good enough for you but there is some discussion in section 9 here $\endgroup$ – Billy O. Mar 14 '18 at 7:51

In the sentence you quoted, "parametrizing subschemes in $Y$" should be interpreted as "parametrizing closed subschemes of $X$ which are entirely contained in $Y$". So these subschemes are still projective, and their Hilbert polynomial is well-defined.

So, for example, if $Y$ is affine, then $Hilb_Y^p=\emptyset$ whenever $p$ has positive degree, as $Y$ does not contain any projective scheme of positive dimension.

Note also that a family of such subschemes in $Y$ over a base $S$ can be defined intrinsecally (without the need of the open immersion $Y\hookrightarrow X$) as a closed subscheme $Z\subset Y\times S$ which is flat and projective over $S$.

  • $\begingroup$ I think that is not correct. Have you considered Lemma 6.2 and Theorem 6.3 of "Construction of Hilbert and Quot Schemes", Nitin Nitsure? I think they are the answer to OP's question. $\endgroup$ – MonLau Mar 28 '19 at 12:25
  • $\begingroup$ I can't see any contradiction, can you be more specific? $\endgroup$ – Andrea Mar 30 '19 at 8:12
  • $\begingroup$ Well, probaly I am wrong, but as I understand, for Hilbert schemes, Lemma 6.2 is saying that every closed subscheme of $Y$ can be prolonged to a closed subscheme of $X$, and Theorem 6.3 says that $Hilb_Y^p$ is actually "parametrizing closed subschemes of $Y$", not "closed subschemes of $X$ which are entirely contained in $Y$" (Regarding OP's question, Lemma 6.2 is telling you how to compute the Hilbert polynomial of a quaisprojective variety). $\endgroup$ – MonLau Mar 30 '19 at 11:46
  • $\begingroup$ I agree with the first part of your sentence (of course that is the same as taking the closure in $X$), but not with the second: look at how the Hilbert scheme is defined between pages 3 and 4 of the notes you linked (in particular page 4, line 5), that's essentially the last sentence of my answer. $\endgroup$ – Andrea Mar 30 '19 at 16:01
  • $\begingroup$ I see, many thanks $\endgroup$ – MonLau Mar 30 '19 at 21:56

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.