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$\underline {Background}$: 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.

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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.

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  • $\begingroup$ Thank you@KReiser for such a nice comprehensive answer.So just to clarify the last point,secant variety of a line will be the line itself right? $\endgroup$ – HARRY Apr 29 at 9:48
  • $\begingroup$ I mean atleast in $\mathbb{P}^2$ $\endgroup$ – HARRY Apr 29 at 9:59
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    $\begingroup$ Yes, that is correct. $\endgroup$ – KReiser Apr 29 at 17:18

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