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.



On the place where $\mathrm{Hilb}_{lines}^{x}(X)$ is smooth.
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.


