This is part of Exercise 5.13 from Undergraduate Algebraic Geometry by Reid:
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 $\mathbb{P^5}$ and a conic in $\mathbb{P}^2$ is mapped to a quartic in $\mathbb{P}^5$.
My attempt:
Suppose a line in $\mathbb{P}^2$ is defined by $ax_0+bx_1+cx_2=0$. Then we have also $V: ay_0+by_1+cy_2=0$ in $\mathbb{P}^5$. So the image of the line in $\mathbb{P}^5$ is the intersection of $V$ and $S$. I have the following questions:
- How do we show that it is a conic?
- How do we decide we don't need more equations? For example, $ay_1+by_3+cy_5=0$ can also define it. So does $ay_2+by_4+cy_5=0$.
I read some pages by Harris. It has some nice description of the Veronese surface, but my questions are not solved. I have similar questions then about a conic mapped to quartic.
Thank you for your help!
Edit:
Let $x_0^2+x_1^2+x_2^2=0$ be a conic in $\mathbb{P}^2$. Its image in $\mathbb{P}^5$ is the intersection of $y_0+y_3+y_5=0$ and the surface $S$. Making a change of variable so $y_5=0$ and plugging $-y_0-y_3$ into $y_5$ of the three defining equations of $S$, I got $$y_1^2=-(y_3^2+y_4^2)\\ y_1^2=-(y_0^2-y_2^2)\\ y_1^2=y_0y_3$$
The pullback of the first two are union of the conic and a line ($x_0=0$ and $x_1=0$, respectively). How to write it as a single quartic so that the pullback does not contain the extra line?