Veronese surface contains no lines

Question

Why does Veronese surface contain no lines? Can you give me a reference about this fact? Thank you for your answers.

Answer 1

We reason by contradiction. Let $\phi : \mathbb{P}^2 \to \mathbb{P^5}$ be the Veronese embedding and $C'$ be the preimage of the given $C \cong \mathbb{P}^1$. Then $1 = C.O_{\mathbb{P}^5}(1) = \phi_*C'.O_{\mathbb{P}^5}(1) = C'.\phi^*O_{\mathbb{P}^5}(1) = C'.O_{\mathbb{P}^2}(2) = 2d$ where $d$ is the degree of $C'$ inside $\mathbb{P}^2$. This is a contradiction and therefore we are done.

Maybe I'll try to express the above solution in simple terms of Bezout theorem.

Assume the line $C \cong \mathbb{P}^1$ exists. Take a hyperplane $H$ in $\mathbb{P}^5$ which intersects $C$ in one point. Then, on one hand the preimage $\phi^{-1}(C \cap H)$ consists of a single point. On the other hand, it's an intersection $C' \cap \phi^{-1}H$ which is by Bezout theorem equal to $deg(C')deg(\phi^{-1}H)$ and $\phi^{-1}H$ is a quadric, i.e., a curve of degree $2$. Contradiction:>

You could probably look to Harris and Eisenbud "3264 and all that..".