Let $[X_0:X_1:X_2:X_3]$ be coordinates on $\mathbb{P}^3(\mathbb{C})$. I want to show that given a point $[X_0:X_1:X_2:X_3]\in Z(X_0X_3-X_1X_2)\subset \mathbb{P}^3(\mathbb{C})$, there are precisely two lines in $Z(X_0X_3-X_1X_2)$ containing $[X_0:X_1:X_2:X_3]$ and that any two lines in $Z(X_0X_3-X_1X_2)$ have either $0$ or $1$ point in common.

For the first part, I take a point $[w:x:y:z]\in Z(X_0X_3-X_1X_2)$, a distinct point $[a:b:c:d]\in \mathbb{P}^3$ and a line through these points, namely $$[sw+at:sx+bt:sy+ct:sz+dt].$$ Then I write $$(sz+dt)(sw+at)=(sx+bt)(sy+ct)$$ and by equating coefficients I have $$wd+az=xc+by,\ ad=bc,\ xy=wz.$$

As @GeorgesElencwajg ponted out, I can write $[w:x:y:z]=[a_0b_1:a_0b_1:a_1b_0:a_1b_1]$ for a unique pair of points $[a_0:a_1]$, $[b_0:b_1]\in\mathbb{P}^1$. Then the above system of three equations is equivalent to the following system of two equations: $$a_0b_0d+a_1b_1a=a_0b_1c+a_1b_0b,\ ad=bc.$$

It suffices to consider 4 cases: $a_0,b_0\ne 0$, $a_0,b_1\ne 0$, $a_1,b_0\ne 0$, $a_1,b_1\ne 0$. They are all analogous to one another. Let me consider the first case.

Case 1. $a_0\ne 0, b_0\ne 0$. Set $a_1/a_0=\alpha$ and $b_1/b_0=\beta$. Then $[a_0:a_1]=[1:\alpha],\ [b_0:b_1]=[1:\beta]$. The first equation implies $d=\beta c + \alpha b - \alpha \beta a$, and the second one says $a\beta c + a \alpha b - \alpha \beta a^2 - bc=0$ or equivalently $(c-\alpha a)(b - a\beta)=0$. That is, either $c=a\alpha$ or $b=a\beta$.

But why does the above imply that (in this case) there are precisely two lines through $[w:x:y:z]$? Specifically, consider the case $c=a\alpha$. Then $d=b\alpha$ and $[a:b:c:d]=[a:b:a\alpha:b\alpha]$. Three cases:

  • $a=0,b\ne 0\implies [a:b:c:d]=[0:1:0:\alpha]$, this point determines a unique line
  • $a\ne 0, b=0 \implies [a:b:c:d]=[1:0:\alpha:0]$, this point determines a unique line
  • $a\ne 0, b\ne 0 \implies c\ne 0, d\ne 0$. Why does the point $[a:b:c:d]=[a:b:a\alpha:b\alpha]$ determine a unique line in this case?

For the second part, how should I start at least?

  • $\begingroup$ I think your confusion stems from using the same letters for a special point and the variables. Denote the point by $ [a_0:a_1:a_2:a_4]$ and see what happens. $\endgroup$ – Mohan Mar 27 '17 at 8:12
  • $\begingroup$ @user428554: The second part is exactly Bézout's theorem, have you studied it? (I guess that the problem should say "$0$ or $1$ points in common in $Z(X_0X_3 - X_1X_2)$", because otherwise any two projective lines intersect in a single point if they do not coincide.) $\endgroup$ – Alex M. Mar 27 '17 at 17:23
  • $\begingroup$ Your 6th line does not make sense in projective space. You must write the parametric equation of a line through $[w:x:y:z]$ as $$[sw+at:sx+bt:sy+ct:sz+dt]$$ (and assume $[w:x:y:z]\neq [a:b:c:d]$) $\endgroup$ – Georges Elencwajg Mar 27 '17 at 20:05
  • $\begingroup$ @GeorgesElencwajg, could you please clarify why the 6th line does not make sense in projective space? $\endgroup$ – user428554 Mar 27 '17 at 20:15
  • $\begingroup$ Moreover, if I write the parametric equation of a line as $$[sw+at:sx+bt:sy+ct:sz+dt]$$ then this will yield the same system of three equations. $\endgroup$ – user428554 Mar 27 '17 at 20:32

Use the Segre embedding $\mathbb P^1\times \mathbb P^1 \hookrightarrow \mathbb P^3$, which implies that :

1) Every point $P\in Z=Z(X_0X_3-X_1X_2)$ can (and should!) be written $P=(a_0b_0:a_0b_1:a_1b_0:a_1b_1)$ for a unique pair $(a_0:a_1), (b_0:b_1)\in \mathbb P^1$
2) The two lines lying on the quadric $Z$ and passing through $P$ are given parametrically by: $$(a_0u:a_0v:a_1u:a_1v)_{(u:v)\in \mathbb P^1} \quad \operatorname {and}\quad (b_0s:b_1s:b_0t:b_1t)_{(s:t)\in \mathbb P^1}$$


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