Line inside the image of the Segre embedding $\sigma_{2,1}$

Question

Why any line in $\mathbb P^5$, which is contained inside the image of the Segre embedding $\sigma_{2,1}$, is also contained inside some $\sigma_{2,1}(x; \mathbb P^1)$ with $x \in \mathbb P^2$ or in $\sigma_{2,1}( \mathbb P^2, y)$ with $y \in \mathbb P^1$ ? I thought that the best way to do that was to show that if the line passes through $\sigma_{2,1}( x, y)$ and $\sigma_{2,1}( x', y')$, then $x=x'$ or $y=y'$. Although I cannot manage to find a way to prove this. Can anyone help me?

Answer 1

The line bundle inducing the Segre embedding is $O(1,1) = O(1) \boxtimes O(1)$. A rational curve $C \subset P^2 \times P^1$ becomes a line in the embedding, if this line bundles restricts to $C$ as $O(1)$. Since both factors of $O(1,1)$ restrict nonnegatively, it follows that one restricts as $O(1)$ and the other as $O$. This means that $C$ is contained in a fiber of one of the projections.