Linear section of a smooth variety Let $X \subset \mathbb{P}^{N}$ be a non-degenerate smooth variety with positive dimension. Take $x_{1}, \ldots, x_{n}$ general points on $X$, with $\mathrm{codim}(X) \geq n - 1$. Denote by $P$ the $(n - 1)$-plane generated by the points $x_{1}, \ldots, x_{n}$. It seems that, if the intersection $P \cap X$ is a curve, then it must be irreducible. Is it true?
 A: First, if $\mathrm{codim}(X) > n - 1$, then by the Trisecant Lemma $P \cap X = \{ x_{1}, \ldots, x_{n} \}$, and therefore, $P \cap X$ is not even a curve.
Assume $\mathrm{codim} = n - 1$. In this case, we can apply the Trisecant Lemma only for $(n - 2)$-planes generated by $n - 1$ of the general points $x_{1}, \ldots, x_{n}$. Let $E$ be an irreducible component of $P \cap X$. Let $P'$ be the $(n - 1)$-plane generated by $x_{1}, \ldots, x_{n-1}$. By Bézout Theorem, $E$ intersects $P'$. But, by the Trisecant Lemma, $P' \cap X = \{ x_{1}, \ldots, x_{n-1} \}$. Thus, we can assume that $x_{1} \in E$. Let $P''$ be the $(n - 2)$-plane generated by $x_{2}, \ldots, x_{n}$. By Bézout Theorem, $E$ intersects $P''$. But, by the Trisecant Lemma, $E \cap P'' = \{ x_{2}, \ldots, x_{n} \}$. Thus, we can assuume that $x_{2} \in E$. If $\deg(E) = 1$, that is, $E$ is a line (joining two general points), then $X \cong \mathbb{P}^{N}$, and in this is not the case by the assumption on the codimension of $X$. Assume $X \not\cong \mathbb{P}^{N}$. Then we have $\deg(E) \geq 2$. We repeat the reason (applying Bézout and Trisecant Lemma), and conclude $x_{3} \in E$. Recalling that a non-degenerate irreducible curve $C$ in $\mathbb{P}^{d}$ of degree $d$ is a rational normal curve, we will have the degree of $E$ at least the number of points $x_{i}'s$ in $E$, unless $X$ is a quadric hypersurface (this is a consequence of Kobayashi-Ochiai's Theorem). Therefore, we conclude that $E$ contains all the points $x_{1}, \ldots, x_{n}$; in particular, $P \cap X$ is irreducible. Now what is important: $E$ is also a normal rational curve, by the previous result that I just mentioned. And again by the Kobayashi-Ochiai's Theorem, $X$ is isomorphic to a quadric hypersurface. Conclusion: the hypothesis $P \cap X$ curve happens only for projective spaces and quadric hypersurfaces, in which cases the result holds.
A: The Segre threefold in $\mathbb{P}^5$, given by the image of $\mathbb{P}^1\times\mathbb{P}^2$ under $([a:b],[x:y:z]) \mapsto [ax:ay:az:bx:by:bz]$, is a counterexample.
It has codimension $2$, but it meets the $2$-plane $T_0=T_1=T_3=0$ in the union of two lines, $T_0=T_1=T_2=T_3=0$ and $T_0=T_1=T_3=T_4=0$.
At this time, I'm unsure about surfaces in $\mathbb{P}^4$, but there might be an example in similar spirit.
