On the place where $\mathrm{Hilb}_{lines}^{x}(X)$ is smooth. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T01:18:00Z http://mathoverflow.net/feeds/question/69002 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/69002/on-the-place-where-mathrmhilb-linesxx-is-smooth On the place where $\mathrm{Hilb}_{lines}^{x}(X)$ is smooth. gio 2011-06-28T06:48:03Z 2011-06-28T14:48:29Z <p>Let <code>$X\subset \mathbb{P}_{\mathbb{C}}^N$</code> be irreducible generically smooth closed subscheme and let <code>$\mathrm{Hilb}_{lines}^{x}(X)$</code> denote the Hilbert scheme of lines contained in $X$ and passing through the point $x\in X$. Is it true that the set <code>$$\{ x\in X : \mathrm{Hilb}_{lines}^{x}(X) \mbox{ is smooth } \}$$</code> is constructible?</p> <p>Thanks in advance.</p> http://mathoverflow.net/questions/69002/on-the-place-where-mathrmhilb-linesxx-is-smooth/69023#69023 Answer by ulrich for On the place where $\mathrm{Hilb}_{lines}^{x}(X)$ is smooth. ulrich 2011-06-28T14:48:29Z 2011-06-28T14:48:29Z <p>Yes. This can be seen as follows:</p> <p>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:</p> <blockquote> <p>Let $f:Y \to X$ be a proper morphism of finite type schemes over a field. Then the set <code>$\{x \in X | f^{-1}(x) \mbox{ is smooth } \}$</code> is constructible.</p> </blockquote> <p>To prove this, by replacing $X$ by $X_{red}$ and $Y$ by <code>$Y \times_X X_{red}$</code> 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.</p>