# What does the tangent space of $V -\{ 0\}$ look like, for a finite dimensional vector space $V$?

Let $K$ be an algebraically closed field, and $V$ a finite dimensional vector space over it, $x$ a point in $V -\{ 0\}$. What does the tangent space of $V -\{ 0\}$ at $x$ look like?

When $X \in \mathbb{A}^n$ is a closed subset, with $\mathscr{I}(X) = <f_1, \cdots, f_m>$ for $f_i \in K[T_1, \cdots, T_n]$, the geometric tangent space of $X = (x_1, \cdots, x_n)$ at $x$ is defined to be $\{ a \in K^n | (\sum\limits_{i=1}^n \frac{\partial f_j}{\partial T_i}(x)(T_i - x_i)) (a) = 0, 1 \leq j \leq m \}$. But when $X$ is not closed, what happens?

Further, in order to prove

The canonical morphism $\phi: V- \{0 \} \rightarrow \mathbb{P}(V)$ is separable,

do I have to prove $d \phi_x : \mathscr{T}(V- \{0 \})_x \rightarrow \mathscr{T}(\mathbb{P}(V))_y$ is surjective for a point $x \in V- \{0 \}$ and $y = \phi (x) \in \mathbb{P}(V)$?

If I have to prove the surjetivity of differential, how can I identify the tangent spaces and the differential?

(Page 42 on Linear Algebraic Groups, GTM21, written by James E. Humphreys)

Thanks a lot.

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## 1 Answer

Unless I don't understand your question, tangent space can be defined in terms of the local ring $O_x$ at $x$, as $(m/m^2)*$ where $m$ is maximal in $O_x$.

In particular, it doesn't change when you restrict yourself to an open subset. So tangent space at each point of your variety is just $\mathbb{A}^n$.

You have a map from $O_y \rightarrow O_x$, and it induces a map $m_y \rightarrow m_x$. Passing to quotient and taking $*$ gives you the differential.

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Thank you very much for the answer. I have seen the definition of tangent space using the local ring. This is clear in theory, but not quite clear to me in practice. I am wondering when $K = \mathbb{C}$ and $\mathrm{dim}V = 2$, $x = (0,1)$, $y = [0,1]$, what $\mathscr{T}(V - \{ 0 \})$ and $\mathscr{T}( \mathbb{P} (V))$ consist of and how $d \phi_x$ takes place. Would you please give me some hints, if it does not cause you too much trouble? Sorry for the presumptuous request. Thank again. –  ShinyaSakai Sep 27 '11 at 2:26
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