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Suppose $p(x,y,z)$ is a homogeneous polynomial of degree 2. My question is:

Can I have three distinct lines $L_1,L_2,L_3$ in the projective space $\mathbb{P}^2$ such that $p$ vanishes on every point of the three lines?

My intuition says that this is not possible, but I'm just starting to learn about projective geometry so I do not know how to prove this. Any help?

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  • $\begingroup$ Are you sure you want $p$ to be a homogeneous polynomial in 3 variables? Projective 3-space has 4 co-ordinates. If 3 variables, it vanishes along infinitely many lines. $\endgroup$ – Mohan Oct 15 '15 at 22:44
  • $\begingroup$ Of course, I meant $\mathbb{P}^2$, thanks! $\endgroup$ – u1571372 Oct 15 '15 at 22:47
  • $\begingroup$ If $p$ vanishes along a line defined by $l(x,y,z)=0$, then, $l$ divides $p$. $\endgroup$ – Mohan Oct 15 '15 at 22:52
  • $\begingroup$ Nullstellensatz? $\endgroup$ – Hoot Oct 15 '15 at 23:01
  • $\begingroup$ @Hoot, how can Nullstellensatz be applied here? $\endgroup$ – u1571372 Oct 15 '15 at 23:02
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If $l(x,y,z)=0$ defines a line and $p(x,y,z)$ vanishes on it, then changing variables, you may assume that $l=x$. Thus the restriction of $p$ to $x=0$ is just $p(0,y,z)$. If this polynomial is not identically zero, it is clear that $p$ does not vanish at some point on the line. So, it follows that $p(0,y,z)=0$, which is same as saying $x$ divides $p$. Now changing back to the original coordinates, this means $l$ divides $p$.

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