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 $\mathbb C^n$?

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 Zariskiclosed and of the same dimension. So were done. $\endgroup$– PVALJul 26, 2014 at 20:14
1 Answer
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 $Y\subset\mathbb{P}^{n1}$ 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.