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A line in $\mathbb{P}^3$ through points $[u_1:u_2:u_3:u_4]$ and $[v_1:v_2:v_3:v_4]$ is $\{[ku_1+lv_1:\dots:ku_3+lv_3]\}$ where either $k\ne 0$ or $l\ne 0$. But there's something worrying me, and it comes from the following problem.

Let $Z_1=Z(X_1^2-X_0X_2)$, $Z_2=Z(X_1X_2-X_0X_3)$, $Z_3=Z(X_2^2-X_1X_3)$. I want to show that $Z_1$ intersects $Z_2$ in the union of $Z_1\cap Z_2 \cap Z_3$ and a line in $\mathbb{P}^3(\mathbb{C})$.

I consider 3 cases: take $[X_0:X_1:X_2:X_3]\in Z_1\cap Z_2$

  • if $X_1\ne 0$, then $X_0,X_2,X_3\ne 0$ and therefore $X_0=X_1X_2/X_3\implies X_3X_1^2=X_1X_2^2\implies X_3X_1=X_2^2$, so $[X_0:X_1:X_2:X_3]\in Z_1\cap Z_2 \cap Z_3$
  • if $X_1=X_2=0$, then clearly $[X_0:X_1:X_2:X_3]\in Z_1\cap Z_2 \cap Z_3$
  • if $X_1=0, X_2\ne 0$, then $X_0=0$ and I want to conclude that $[X_0:X_1:X_2:X_3]$ belongs to the line $\{[0:0:X_2:X_3]\}$

But in the latter case $X_3$ may be zero or nonzero. If it is nonzero, then by what I said at the beginning, $\{[0:0:X_2:X_3]\}$ is the line thorugh the points $[0:0:1:0]$ and $[0:0:0:1]$. However if $X_3$ is zero then $\{[0:0:X_2:X_3]\}=\{[0:0:X_2:0]\}$, and this is not a line because $[0:0:0:0]$ is not a point of $\mathbb{P}^3$.

So I have 2 questions:

  1. What should I do with the case $X_3=0$?
  2. Is the rest of my reasoning correct?
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  • $\begingroup$ The third bullet, why does $X_1 = 0,X_2\neq 0$ imply that $X_0 = 0$? $\endgroup$ – Tanner Strunk Mar 26 '17 at 20:00
  • $\begingroup$ @TannerStrunk: thanks, there was a typo in the definition of $Z_1$. $\endgroup$ – user428554 Mar 26 '17 at 20:05
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    $\begingroup$ If $X_0=X_1=X_3=0$ you have just a point (which is indeed on the line passing through $[0:0:1:0]$ and $[0:0:0:1]$). $\endgroup$ – Roland Mar 26 '17 at 20:09
  • $\begingroup$ Your reasoning was correct except that you were treating the third case as kinda...separate from the case of $[X_0:X_1:X_2:X_3]\in Z_1\cap Z_2\cap Z_3$, and...it kinda is. But then you have to throw in a point at infinity to get a line. $\endgroup$ – Tanner Strunk Mar 26 '17 at 20:14
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    $\begingroup$ Yes it is true, you proved it ! You just didn't need to consider the two cases $X_3\neq 0$ and $X_3=0$. Both of these cases produce points on the line $[0:0:U:V]$. The case $X_3=0$ produces only one point of this line (namely $[0:0:1:0]$) whereas the case $X_3\neq 0$ produces the others. $\endgroup$ – Roland Mar 26 '17 at 20:26
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So...you're right in how you're thinking about this problem. (I actually just did this problem last semester in my first alg. geo. class in grad. school and pulled up my solution set.) Ok so you have the line given by $[0:0:1:a]$ union with a point at infinity, $[0:1:0:0]$, a point lying in the set $Z_1\cap Z_2$. Sort of the point here is that you have to have both the line and the variety $Z_1\cap Z_2\cap Z_3$ in your decomposition of $Z_1\cap Z_2$ and the variety and the line don't have trivial intersection.

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