A very simple question about a Harris' exercise

Question

In the book of Harris Algebraic Geometry, a First Course, exercise 2.29, it is asked to show that the composition of $1\times \nu: \mathbb{P}^1\times\mathbb{P}^1 \to \mathbb{P}^1\times\mathbb{P}^2$ where $\nu$ is the Veronese embedding with the Segre embedding $\sigma:\mathbb{P}^1\times\mathbb{P}^2 \to \mathbb{P}^5$ can be represented (after identification of $\mathbb{P}^1\times\mathbb{P}^1$ with its image the quadric $Q: Z_0Z_3-Z_1Z_2=0$ in $\mathbb{P}^3$ by the Segre mapping), by the map $$\phi: [Z_0,Z_1,Z_2,Z_3]\to [F_0(Z),\cdots,F_1(Z)]$$ where $(Q,F_0,\cdots,F_5)$ is a basis of the 7 dimensional space of homogenous quadratic polynomials that contains the line $L: Z_0=Z_1=0$ of $Q$.

This seems strange to me because $\phi$ is not even defined on the line $L$. What did I miss ?

Answer 1

The composition $\sigma\circ(1\times\nu)$ will map $((x:y),(z:w))\mapsto ((x:y),(z^2:zw:w^2))\mapsto (xz^2:xzw:xw^2:yz^2:yzw:yw^2).$

Identifying $\Bbb P^1\times\Bbb P^1\cong Q\subseteq\Bbb P^3$ takes $((x:y),(z:w))\mapsto (xz:xw:yz:yw)$ or conversely, $(a:b:c:d)\mapsto ((1:c/a),(1:b/a))=((a:c),(a:b))$ where we suppose that $a\neq 0.$ Thus, on the affine patch $a\neq 0,$ the composition is determined by

$$(a:b:c:d)\mapsto (a^3:a^2b:ab^2:ca^2:cab:cb^2)=(a^3:a^2b:ab^2:ca^2:a^2d:adb) =(a^2:ab:b^2:ac:ad:bd),$$ where we've used $ad=bc.$

Thus, we wish to show that $\langle a^2,ab,b^2,ac,ad,bd,ad-bc\rangle$ is a basis of the homogeneous quadratics containing $a=b=0.$ This part should be routine.

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Edit: The identification on the patch $c\neq 0$ is given by $(a:b:c:d)\mapsto ((a/c:1),(1:d/c))=((a:c),(c:d)).$ On this patch the composition is determined by $$(a:b:c:d)\mapsto (ac^2:acd:ad^2:c^3:c^2d:cd^2)=(ac^2:bc^2:bcd:c^3:c^2d:cd^2)=(ac:bc:bd:c^2:cd:d^2).$$

On this patch we find a similar expression, but now our forms vanish on $c=d=0.$

I think the question is not saying that the first expression should be valid everywhere on $Q,$ since it does not make sense to evaluate it on $a=b=0,$ as you say. On $c\neq 0$ or $d\neq 0$ we have this alternate expression, which should coincide with the original one on the intersection $a,c\neq 0$ for example. This way we'll get a globally defined map on $Q.$ Unfortunately, the way the question seems to be stated doesn't make this precise at all.