4
$\begingroup$

I apologize if this is a trivial question. Let $X$ be a closed affine subvariety of $\mathbb C^n$ given by the union of lines trough the origin. Does $X$ smooth implies $X$ is a subspace of $\mathbb C^n$?

$\endgroup$
1
  • 7
    $\begingroup$ Thinking purely in geometric terms, $X \subseteq T_0X$ for somewhat obvious reasons (basic differential geometry say). So we have $X \subseteq T_0X$ both Zariski-closed and of the same dimension. So were done. $\endgroup$
    – PVAL
    Jul 26, 2014 at 20:14

1 Answer 1

1
$\begingroup$

I do not know if I got you question. Anyway, if $X$ is the union of lines through the origin then the projective closure $\overline{X}\subset\mathbb{P}^n$ is a cone over a subvariety $Y\subset\mathbb{P}^{n-1}$ with vertex $p=(1:0:...:0)$. If $\overline{X}$ is smooth follwing the comment of PVAL che can consider $\mathbb{T}_p\overline{X}$. Now, any line through $p$ contained in $\overline{X}$ is contained in $\mathbb{T}_p\overline{X}$ as well. Therefore $\overline{X} = \mathbb{T}_p\overline{X}$. Finally $\overline{X}$ and $X$ are linear subspaces.

$\endgroup$
0

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.