Elementary question: Cubic 4-fold and rational quartic scroll Forgive me to ask an elementary question, because I really need the answer to this today (I already asked this in Stackexchange.)
Let $S$ be the rational quartic scroll in $\mathbb{P}^5$ ($S$ is the image of the embedding $\mathbb{P}^1\times\mathbb{P}^1\rightarrow\mathbb{P}^5$ via $|\mathcal{O}_{\mathbb{P}^1\times\mathbb{P}^1}(1,2)|$).

1/ Why does every point on $\mathbb{P}^5$ lie on a unique secant line or tangent line to $S$?
2/ Suppose a cubic fourfold $X\subset \mathbb{P}^5$ contains $S$. Why does $X$ necessarily contain two skewed 2-planes?

It's fine if you just leave me a reference. Thank you very much.
 A: Here is another (sketch of) proof of 1). Let $V$ be a 2-dimensional vector space; you can view your scroll as the image of the embedding $\mathbb{P}(V)\times \mathbb{P}(V)\rightarrow \mathbb{P}(V\otimes \mathsf{S}^2V)$ given by $(v,w)\mapsto v\otimes w^2$.  Let $p$ be an element of  $\mathbb{P}(V\otimes \mathsf{S}^2V) \smallsetminus S$; it can be written  $v_1\otimes t_1+v_2\otimes t_2$. The line $\langle t_1,t_2\rangle$ in $\mathbb{P}(\mathsf{S}^2V)$ is determined by $p$; it contains  2 rank 1 tensors $w_1^2$ and $w_2^2$ (up to scalar), given by the intersection of  $\langle t_1,t_2\rangle$ with the conic of rank 1 tensors. Thus $p$ can be written $v'_1\otimes w_1^2+v'_2\otimes w_2^2$, where the points $(v'_1,w_1)$ and $(v'_2,w_2)$ of $\mathbb{P}(V)\times \mathbb{P}(V)$ are uniquely determined, and span the unique secant line passing through $p$.
A: The question about secant lines reduces to a Chern class computation on the blowing up $\nu:\widetilde{\Sigma \times \Sigma}\to \Sigma \times \Sigma$ along the diagonal, where $\Sigma$ is the cubic quartic scroll.  Your claim is equivalent to the assertion that in $A^4(\widetilde{\Sigma \times \Sigma})$, the following class equals $2$ (overcounting by $2$, because the pair of points is ordered). $$2 = a^4 - 3a^2 b + b^2,\ \ a = \nu^*\text{pr}_1^*c_1(\mathcal{O}(1)) + \nu^*\text{pr}_2^*c_1(\mathcal{O}(1)) - i_*(1),$$ $$b = \nu^*\left( \text{pr}_1^*c_1(\mathcal{O}(1))\cdot \text{pr}_2^*c_1(\mathcal{O}(1))\right) - i_*\pi^*c_1(\mathcal{O}(1)) $$ where $i:E\to \widetilde{\Sigma\times \Sigma}$ is the exceptional divisor, where $\pi:E\to \Sigma$ is the projection to the diagonal, and where $\mathcal{O}(1)$ is the usual Serre twisting sheaf on $\mathbb{P}^5$.  I have not computed whether or not $a^4-3a^2b+b^2$ does indeed equal $2$ . . .
Edit. The computation by abx seems more robust, since it is not vulnerable to arithmetic mistakes.  Nonetheless, I just got back to the Chern class computation.  I compute that $a^4$ equals $44$, $a^2b$ equals $18$ and $b^2$ equals $12$.  Thus $a^4-3a^2b+b^2$ equals $44-3(18)+12$.  This does equal $2$, as expected.
