Consider the Veronese surface $S$ defined by the map: $$\phi: \mathbb{P}^2\rightarrow \mathbb{P}^5$$ where $\phi(x_0,x_1,x_2)=(x_0^2,x_0x_1,x_0x_2,x_1^2,x_1x_2,x_2^2)$.
The problem asks to show that a line in $\mathbb{P}^2$ is mapped to a conic in $\mathbb{P^5}$
I will use the following notation The definition of the Veronese map will be taken from the question above. I will denote the coordinates in $\mathbb{P}^{5}$ by $[z_{0}:\ldots:z_{5}]$ and the coordinates in $\mathbb{P}^{2}$ by $[x_{0}:x_{1}: x_{2}]$.
Claim The Veronese surface is cut out by three quadrics: $$C_{1}: z_{0}z_{3} - z_{1}^{2} = 0$$ $$C_{2}: z_{0}z_{5} -z_{2}^{2}= 0 $$ $$C_{3}: z_{3}z_{5} - z_{4}^{2} = 0. $$
Suppose a line in $\mathbb{P}^2$ is defined by $ax_0+bx_1+cx_2=0$. So can the image of the line in $\mathbb{P}^5$ be thought as the intersection of $S$ and something else? I have the following questions:
- How do we show that it is a conic?