Some projective bundle

I have stopped studying algebraic geometry a while ago because I got scared by the following exercise.

Exercise. Say we have the coordinate $[a_{00} : a_{01} : a_{02} : a_{11} : a_{12} : a_{22}]$ for $\mathbb{P}^{5}$ and $[x_{0} : x_{1} : x_{2}]$ for $\mathbb{P}^{2}$ over a fixed field $k$. Let $X$ be the closed subscheme of $\mathbb{P}^{5} \times\mathbb{P}^{2}$ cut out by the equation $$a_{00}x_{0}^{2} + a_{01}x_{0}x_{1} + a_{02}x_{0}x_{2} + a_{11}x_{1}^{2} + a_{12}x_{1}x_{2} + a_{22}x_{2}^{2} = 0.$$ Realizing $X$ as a $\mathbb{P}^{4}$-bundle over $\mathbb{P}^{2}$, show that $X$ is a smooth sixfold, and that $\mathrm{Pic}(X) \simeq \mathbb{Z}^{2}$.

As a primitive human being, I usually just take partial derivatives and check whether the non-full-rank-Jacobian locus of them intersects the locus of original equation on every affine open set of an affine open cover. However, I could not possibly think about spending my time on doing so, which is why I stopped studying. I just realized that the author was giving me an hint all along. Once we notice that $X$ is a $\mathbb{P}^{4}$ bundle over $\mathbb{P}^{2}$ then it means that $X$ can be covered by open subsets of the form $\mathbb{P}^{4} \times U$ where $U$ are some open subsets of $\mathbb{P}^{2}$. Thus, once we can check that $X$ is such object, we would get its smoothness for free!

Q. How can we see $X$ as a $\mathbb{P}^{4}$-bundle over $\mathbb{P}^{2}$?

Ad-hoc argument. If we fix any $k$-point $[x_{0} : x_{1} : x_{2}] \in \mathbb{P}^{2}(k)$, then denoting $z_{ij} = x_{i}x_{j}$, we see that any $\boldsymbol{a} = [a_{00} : a_{01} : a_{02} : a_{11} : a_{12} : a_{22}] \in \mathbb{P}^{5}(k)$ such that $$z_{00}a_{00} + z_{01}a_{01} + z_{02}a_{02} + z_{11}a_{11} + z_{12}a_{12} + z_{22}a_{22} = 0$$ will form the points $(\boldsymbol{a}, [x_{0} : x_{1} : x_{2}]) \in X(k)$. Since $z_{ij} = x_{i}x_{j} \in k$ are fixed, these $\boldsymbol{a}$ cuts out a hyperplane of $\mathbb{P}^{5}$, which is isomorphic to $\mathbb{P}^{4}$. Now, as an ad-hoc geometor, I can imagine $\boldsymbol{x} = [x_{0} : x_{1} : x_{2}]$ "continuously" moving in $\mathbb{P}^{2}$, so we would get what we want.

Hopefully legit argument. First off, we need to understand why $X$ is a closed subscheme of $\mathbb{P}^{5} \times \mathbb{P}^{2}$. We first apply the Veronese embedding to the second component $\mathbb{P}^{2} \hookrightarrow \mathbb{P}^{5}$, which restricts to $$[x_{0} : x_{1} : x_{2}] \mapsto [x_{0}^{2} : x_{0}x_{1} : x_{0}x_{2} : x_{1}^{2} : x_{1}x_{2} : x_{2}^{2}]$$ on the sets of $k$-points. We may form such $X$ in $\mathbb{P}^{5} \times \mathbb{P}^{5}$, and this sits in $\mathbb{P}^{5} \times \mathbb{P}^{2}$ as a closed subscheme. Writing $$\boldsymbol{z} = [z_{00} : z_{01} : z_{02} : z_{11} : z_{12} : z_{22}]$$ for the coordinates for the second $\mathbb{P}^{5}$, our $X$ is cut out by the equation $$a_{00}z_{00} + a_{01}z_{01} + a_{02}z_{02} + a_{11}z_{11} + a_{12}z_{12} + a_{22}z_{22} = 0$$ and another degree $2$ homogeneous equation in $z_{ij}$, which records the image of $\mathbb{P}^{2}$ under the Veronese embedding. We are implicitly using the Segre embedding $\mathbb{P}^{5} \times \mathbb{P}^{2} \hookrightarrow \mathbb{P}^{5} \times \mathbb{P}^{5}.$ Anyways, we now understand that $X$ is a legit projective scheme.

To show that $X$ is a $\mathbb{P}^{4}$ bundle over $\mathbb{P}^{2}$, we need to find a locally free sheaf $\mathcal{E}$ of rank $5$ on $\mathbb{P}^{2}$ such that $$X \simeq \textbf{Proj}(\mathrm{Sym}^{\bullet}(\mathcal{E})).$$ This seems very scary, but let's see what I can do. The standard affine open cover for $\mathbb{P}^{2}$ is given by the three affine open subsets $$U_{i} := \mathrm{Spec}(k[x_{0}/x_{i}, x_{1}/x_{i}, x_{2}/x_{i}]) \simeq \mathbb{A}^{2}$$ for $i = 0, 1, 2$. Over each $U_{i}$, we should get an open subset $X_{i}$ of $X$ such that $$X_{i} \simeq \mathrm{Proj}(\mathrm{Sym}^{\bullet}(\mathcal{E}(U_{i}))).$$ Okay. This is less scary because we ran away from the relative stuff. We would be done, if we could magically make this happen so that $X_{i} \simeq \mathbb{P}^{4}$. The defining equation for $X$ is $$a_{00}x_{0}^{2} + a_{01}x_{0}x_{1} + a_{02}x_{0}x_{2} + a_{11}x_{1}^{2} + a_{12}x_{1}x_{2} + a_{22}x_{2}^{2} = 0,$$ so for $X_{i}$, we would need $$a_{00}\left(\frac{x_{0}}{x_{i}}\right)^{2} + a_{01}\left(\frac{x_{0}}{x_{i}}\right)\left(\frac{x_{1}}{x_{i}}\right) + a_{02}\left(\frac{x_{0}}{x_{i}}\right)\left(\frac{x_{2}}{x_{i}}\right) + a_{11}\left(\frac{x_{1}}{x_{i}}\right)^{2} \\ + a_{12}\left(\frac{x_{1}}{x_{i}}\right)\left(\frac{x_{2}}{x_{i}}\right) + a_{22}\left(\frac{x_{2}}{x_{i}}\right)^{2} = 0.$$ We may look at $x_{0}/x_{i}, x_{1}/x_{i}, x_{2}/x_{i} \in k[x_{0}/x_{i}, x_{1}/x_{i}, x_{2}/x_{i}]$ as if they are scalars. With this viewpoint, our $X_{i}$ can be defined by taking $\mathrm{Proj}$ of $k[x_{0}/x_{1}, x_{1}/x_{i}, x_{2}/x_{i}, \boldsymbol{a}]$ modulo the equation above, which we may denote as $$f(x_{0}/x_{i}, x_{1}/x_{i}, x_{2}/x_{i}, \boldsymbol{a}) = 0.$$ Continuing our reverse-engineering, we want $$\mathrm{Sym}^{\bullet}(\mathcal{E}(U_{i})) = k\left[\frac{x_{0}}{x_{i}}, \frac{x_{1}}{x_{i}}, \frac{x_{2}}{x_{i}}, \boldsymbol{a}\right]/\left(f\left(\frac{x_{0}}{x_{i}}, \frac{x_{1}}{x_{i}}, \frac{x_{2}}{x_{i}}, \boldsymbol{a}\right)\right).$$ Hence, it would be nice if we can find a rank $5$ free module $\mathcal{E}(U_{i})$ over $k[x_{0}/x_{i}, x_{1}/x_{i}, x_{2}/x_{i}]$, whose symmetric powers give the above ring. This seems like a cheating, but I think we can just take degree $1$ part (in $\boldsymbol{a} = (a_{ij})$) of the above ring, namely $$\mathcal{E}(U_{i}) := \frac{ R_{i}a_{00} \oplus R_{i}a_{01} \oplus R_{i}a_{02} \oplus R_{i}a_{11} \oplus R_{i}a_{12} \oplus R_{i}a_{22} }{ R_{i} ( x_{0/i}^{2}a_{00} + x_{0/i}x_{1/i}a_{01} + x_{0/i}x_{2/i}a_{02} + x_{1/i}^{2}a_{11} + x_{1/i}x_{2/i}a_{12} + x_{2/i}^{2}a_{22})},$$ where $x_{j/i} := x_{j}/x_{i}$ and $R_{i} := k[x_{0/i}, x_{1/i}, x_{2/i}]$ and call it a day. Gluing over different $U_{i}$'s should work out to construct $\mathcal{E}$.

Q. What about the Picard group?

Let's do this later. Exercise II.7.9 of Hartshorne's book says that for a very general scheme $Y$, which includes $\mathbb{P}^{2}$, for any locally free sheaf $\mathcal{E}$ of $Y$ with finite rank $\geq 2$, we have $$\mathrm{Pic}(\textbf{Proj}(\mathrm{Sym}^{\bullet}(\mathcal{E}))) \simeq \mathrm{Pic}(Y) \times \mathbb{Z}.$$ Since $\mathrm{Pic}(\mathbb{P}^{n}) \simeq \mathbb{Z}$ for any $n \geq 1$ (over any field), taking $Y = \mathbb{P}^{2}$, this general statement will finish the exercise we started.