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As I understand, the first secant variety of a variety $\mathbb{V}$ is constructed in the following way: Take two arbitrary points on the variety, construct the line passing through them, the closure of all such lines is then the first secant variety.

It is not obvious why this will yield a variety, is it?

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    $\begingroup$ It is obvious that this is a variety because you take the closure. By definition, the result is a Zariski closed subset of projective space. $\endgroup$ – Jesko Hüttenhain Nov 12 '15 at 10:19
  • $\begingroup$ Is the question whether this thing is irreducible? $\endgroup$ – Hoot Nov 12 '15 at 13:22
  • $\begingroup$ @Hoot No, I just want to prove that this is indeed a variety, namely, it is the zero locus of some polynomials. $\endgroup$ – kaiser Nov 12 '15 at 15:36
  • $\begingroup$ @JeskoHüttenhain Really? Not every closed set is a variety, right? $\endgroup$ – kaiser Nov 12 '15 at 15:37
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    $\begingroup$ The Zariski topology is defined such that the closed sets are exactly zero loci of collections of polynomials. $\endgroup$ – Hoot Nov 12 '15 at 16:25
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Let $\mathbb V\subseteq\mathbb P^n$ be a projective irreducible complex variety. Consider the rational map

\begin{align*} \varphi\colon\mathbb V \times \mathbb V \times \mathbb P^1 &\longrightarrow \mathbb P^n\\ ([x_0:\ldots:x_n],[y_0:\ldots:y_n],[t_0:t_1]) & \longmapsto [t_0x_0+t_1y_0:\ldots:t_0x_n+t_1y_n] \end{align*}

which is defined on the open set $$U=\{ (x,y,t) \mid x\ne y \}\subseteq \mathbb V \times \mathbb V \times \mathbb P^1.$$

Then, since $\mathbb V \times \mathbb V \times \mathbb P^1$ is irreducible, the open set $U$ is also irreducible. Therefore, $\varphi(U)$ is also an irreducible subset of $\mathbb P^n$. Note that the Secant variety of $\mathbb V$ is precisely $\overline{\varphi(U)}$. Since $\varphi(U)$ is irreducible, the Secant is therefore also irreducible and closed.

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