# Picard Group of Universal Plane Conic

This is from 18.4.5 of Vakil's book Foundations of Algebraic Geometry.

We work over a fixed field $k$, if $\mathbb{P}^2$ has projective coordinates $x_0,~x_1,~x_2$ and $\mathbb{P}^5$ has coordinates $a_{00},a_{01},a_{02},a_{11},a_{12},a_{22}$, the universal plane conic $\mathcal{C}$ in $\mathbb{P}^2 \times \mathbb{P}^5$ is the scheme cut out by the equation, $$a_{00}x_0^2+a_{01}x_0 x_1+\cdots+a_{22} x_{2}^2=0$$

Then in Exercise 18.4 M, by intepreting $\mathcal{C}$ as a $\mathbb{P}^4$-bundle over $\mathbb{P}^2$, show that $\mathcal{C}$ is a smooth sixfold, and that $\text{Pic}(\mathcal{C})=\mathbb{Z} \times \mathbb{Z}$.

I do not see why by intepreting $\mathcal{C}$ as a $\mathbb{P}^4$-bundle over $\mathbb{P}^2$ could solve this question. I do not know much about computing Picard group.

• Here is a general fact. Let $X$ be a schme, $E$ a vector bundle over $X$ and let $\pi:Y=\mathbb{P}(E)\to X$ be the projectivized bundle and let $\mathcal{O}_Y(1)$ be the tautological line bundle. Then the Picard group of $Y$ is the direct sum of the pull back of that of $X$ via $\pi$ and $\mathbb{Z}\mathcal{O}_Y(1)$. – Mohan Mar 1 '16 at 21:07
• @Mohan Thank you, do you know where I could find a proof for this general fact? – Wenzhe Mar 1 '16 at 22:57
• Fulton's intersection theory has a proof. It appears as an exercise in Hartshorne, 12.5, at the end of Chapter III. – Mohan Mar 1 '16 at 23:18