Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $X\subset \mathbb{P}_{\mathbb{C}}^N$ be irreducible generically smooth closed subscheme and let $\mathrm{Hilb}_{lines}^{x}(X)$ denote the Hilbert scheme of lines contained in $X$ and passing through the point $x\in X$. Is it true that the set $$ \{ x\in X : \mathrm{Hilb}_{lines}^{x}(X) \mbox{ is smooth } \} $$ is constructible?

Thanks in advance.

share|improve this question

1 Answer 1

up vote 4 down vote accepted

Yes. This can be seen as follows:

Let $Hilb_{lines}(X)$ denote the Hilbert scheme of lines in $X$ and let $\Gamma \subset X \times Hilb_{lines}(X)$ be the correponding universal family. Then $Hilb_{lines}^x(X) = p^{-1}(x)$ where $p:\Gamma \to X$ is induced by the first projection. So we are reduced to the folowing (well-known) statement:

Let $f:Y \to X$ be a proper morphism of finite type schemes over a field. Then the set $\{x \in X | f^{-1}(x) \mbox{ is smooth } \}$ is constructible.

To prove this, by replacing $X$ by $X_{red}$ and $Y$ by $Y \times_X X_{red}$ we may assume $X$ is reduced (since we only care about the fibres). By generic flatness, we may find a finite stratification of $X$ by locally closed reduced subschemes $X_i$ so that the induced morphisms $Y \times_X X_i \to X_i$ are all flat. For a flat proper morphism the locus of points in the base so that the fibres are smooth is open. It follows that the set we are interested in is a finite union of open subsets of closed subsets of $X$, so is constructible.

share|improve this answer
    
Thank you very much. –  gio Jun 28 '11 at 17:07

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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