Surface of type $(2,2)$ on the Segre cubic scroll $\mathbb{P}^1 \times \mathbb{P}^2 \subset \mathbb{P}^5$ Let $S=\mathbb{P}^1 \times \mathbb{P}^2 \subset \mathbb{P}^5$ embedded with the Segre embedding given by $\mathcal{O}_S(1,1)$.
If we intersect $S$ with a general smooth quadric $Q \subset \mathbb{P}^5$ we get a smooth surface $X \subset S$ of type $(2,2)$. Since $deg(S)=3$ by Bertini we have that $deg(X)=6$.
By construction we have that (up to identifying every $\mathbb{P}^2$ in the ruling by projecting from the $\mathbb{P}^1$ factor) for every point $p \in \mathbb{P}^2$ there exist only $2$ conics $C_1,C_2$ passing through $p$, so this is not a surface of the type $Y=\mathbb{P}^1 \times \mathbb{P}^1$ embedded with bidegree $(2,1)$. In fact we have that $Y \subset \mathbb{P}^5$ but $deg(Y)=4 \neq deg(X)$.
My question now is how to prove that $X$ is rational? (I've the feeling that it must be so but I'm currently not able to figure it out).
Also I had the feeling that $X$ could be viewed as the total space of a conic $\mathcal{C}$ in the $\mathbb{P}^5=\mathbb{P}(\mathbb{C}[x^2,y^2,z^2,xy,xz,yz])$ where $[x,y,z]$ are homogeneous coordinates of $\mathbb{P}^2$, where by total space I mean the space consisting of the union of the conics $C_x$ corresponding to the point $x \in \mathcal{C}$. This would imply that there are $6$ singular conics among the conics in the $\mathbb{P}^2$ fibers of $S$, since $deg(\mathcal{C} \cap \mathbb{S}ec_2(v_2(\mathbb{P}^2)))=2 \cdot 3=6$.
Anyone knows if $6$ is the correct number of singular conics in a smooth quadric section of the Segre cubic scroll in $\mathbb{P}^5$? The fact that $6$ is also the degree is a coincidence?
Thanks in advance.
 A: This surface $X$ satisfies $p_g(X)=q(X)=0$. Moreover, if $H_X$ is the hyperplane section, by adjunction we find $$K_X^2=2, \quad K_X H_X=-4.$$
Therefore no multiple of the canonical divisor can be effective, in particular $P_2(X)=q(X)=0$ and $X$ is rational by Castelnuovo criterion.
In fact, $X$ is a conic bundle with six singular fibres, that turns out to be the blow-up at six points of the quadric $\mathbb{F}_0=\mathbb{P}^1 \times \mathbb{P}^1$. This is Example 1.9, p. 236 in
S. Di Rocco, K. Ranestad: On surfaces in $\mathbb{P}^{6}$ with no trisecant lines (http://dx.doi.org/10.1007/BF02384319), Ark. Mat. 38, No. 2, 231-261 (2000). ZBL1035.14011.
A: Here is another approach. Take a smooth surface $X$ of type $(d,2)$ in $S$. The general fiber of the projection $\pi:X\rightarrow \mathbb{P}^1$ is a smooth conic. So $X$ is rationally connected and since $X$ has dimension $2$ it is rational. This does not depend on $d$.
The bi-homogeneous polynomial cutting out $X$ in $S$ of the the following form:
$$a_{0,0}(u,v)x^2+a_{0,1}(u,v)xy+a_{0,2}(u,v)xz+a_{1,1}(u,v)y^2+a_{1,2}(u,v)yz+a_{2,2}(u,v)z^2$$
where $[u:v]$ and $[x:y:z]$ are homogeneous coordinates on $\mathbb{P}^1$ and $\mathbb{P}^2$ respectively, and the $a_{i,j}(u,v)$ are homogeneous polynomials of degree $d$. The matrix of the conic $C_{u,v}$ over the point $[u:v]\in\mathbb{P}^1$ is a $3\times 3$ matrix whose entries are homogeneous polynomials of degree $d$. So its determinant is a homogeneous polynomial of degree $3d$. The singular conics in the conic bundle correspond to the zeros of the determinat. So you have $3d$ singular conics.
