# Secant variety of a curve

$$\underline {Background}$$: Let,$$X$$ be a curve in $$\mathbb{P}^{2}$$ which is not a line.We denote secant variety of $$X$$ to be $$\sigma_2(X)$$

$$\underline {Question}$$: To prove $$\sigma_2(X)=\mathbb{P}^2$$

$$\underline {Attempt}$$: One inclusion is obvious.

To show the other inclusion it is sufficient to show that every nonzero point in $$\mathbb K^{3}$$ is in one of the spaces spanned by $$2$$ roots of the corresponding degree $$d$$ polynomials.But I cannot see why is that the case?

Also I donot see what is the problem if this is a line

Any help from anyone is welcome.

If $$C\subset \Bbb P^2$$ is a curve which is not a line, either it contains a component of degree $$\geq 2$$ or it contains a line and a point not on that line. In the case when it has a component of degree $$\geq2$$, every line in $$\Bbb P^2$$ intersects the curve in at least two points by Bezout's theorem, which proves the claim. If $$C$$ contains a line $$L$$ and a point $$P$$ not on that line, then every point in $$\Bbb P^2$$ is on a line $$L'$$ through $$P$$, and again by Bezout's theorem $$L'$$ must intersect $$L$$ once and thus $$C$$ twice.
If $$C$$ is a line, then the line through any two points in $$C$$ is just $$C$$ itself. So you don't get any points not on $$C$$ when constructing the secant variety.
• I mean atleast in $\mathbb{P}^2$ – HARRY Apr 29 at 9:59