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 $|\mathcal{O}\_{\mathbb{P}^1\times\mathbb{P}^1}(1,2)|$).



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> 1/ Why does every point on $\mathbb{P}^5$ lie on a unique secant line or tangent line to $S$?
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> 2/ Suppose a cubic fourfold $X\subset \mathbb{P}^5$ contains $S$. Why does $X$ necessarily contain two skewed 2-planes?
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It's fine if you just leave me a reference. Thank you very much.



Elementary question: Cubic 4-fold and rational quartic scroll
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.


