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## $Hilb_{lines}^{x}(X)$ and $Hilb_{lines}^{x}(X_{red})$

Let $X$ be a irreducible closed subscheme of $\mathbb{P}^N_{\mathbb{C}}$, and $U$ is a nonempty open where $X$ is smooth and moreover for every $x\in U$ and for every line $l\subseteq X$ with $x\in l$ assume $l \subseteq X\setminus \mathrm{sing}(X)$. If $x\in U$, clearly $$Hilb_{lines}^{x}(X)=Hilb_{lines}^{x}(X_{red})$$ as the identity of sets, but is it true as identity of schemes?

Thanks.

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## 1 Answer

If you let $V = X \setminus \mathrm{sing}(X)$, then assuming $x\in U$ we see that the Hilbert scheme of lines in $X$ passing through $x$ is the same as the Hilbert scheme of lines in $V$ passing through $x$. Indeed, the flat families of lines through $x$ in $X$ are exactly the same as the flat families of lines through $x$ in $V$, so the representing schemes are the same as well (if somebody has a better way of saying this please comment!). Furthermore, $V$ is reduced, so this is also the same as the Hilbert scheme of lines in $X_{\mathrm{red}}$ passing through $x$.

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 I thought what you said, but object $Hilb_{lines}^x(V)$ is not nice because $V$ is not projective... – gio Jul 2 2011 at 23:24 It's not a priori nice, but it does exist and in fact is the same as the two schemes you're considering. Since the lines you are considering are all in fact contained in V, things behave much better than they normally would. I suppose the real point is that any infinitesimal deformation of a line in X through x will continue to lie in V. – Jack Huizenga Jul 2 2011 at 23:46