2
$\begingroup$

I am having trouble understanding a step in my Professor's Lecture notes

She shows that

Lemma 2.2.4 Let $P_1,\ldots,P_5$ be distinct points in $\mathbb{P}_k^2$. There exists a conic in $\mathbb{P}_k^2$ containing these

points. If no four of these points are on a line, the conic is unique.

What I am having trouble with: In the case where 3 of the points $P_1, P_2, P_3$ are on a line, she asserts that the line itself is in the conic.

Is this just a general fact?: Suppose you have a plane curve given by $\sum_{|\alpha|=n} a_\alpha x^{\alpha_1}y^{\alpha_2}z^{\alpha_3}$ with 3 collinear points say(without loss of generality) $P_1=[0:0:1],P_2=[x_0:y_0:z_0],P_3$ on it. The plane curve will be of the form $\sum_{|\alpha|=n, \alpha_3 <n} a_\alpha x^{\alpha_1}y^{\alpha_2}z^{\alpha_3}$ and the line going through these points is given by the morphism $[t:s] \to [tx_0,ty_0,s]$. The question of whether the $\overline{P_1P_2P_3}$ is automatically contained in the plane surface is posed as, "suppose there an $[s:t]\neq [1:z_0]$ or $[0:0:1]$ such that $\sum_{|\alpha|=n} t^{\alpha_1+\alpha_2}s^{\alpha_3} a_\alpha {x_0}^{\alpha_1}{y_0}^{\alpha_2}{z_0}^{\alpha_3}=0$. Then this is zero for all $[t:s]$.

$\endgroup$
  • $\begingroup$ Wait, where did the surface come from? The question is about curves in the plane, right? $\endgroup$ – Schemer Oct 23 '15 at 21:46
  • $\begingroup$ A conic has degree $2$, a line degree $1$. If the line is not contained in the conic there are $2 \cdot 1 = 2$ points of intersection, counted with multiplicity by Bezout's theorem. $\endgroup$ – Jürgen Böhm Oct 23 '15 at 22:09
  • $\begingroup$ oh okay I get the point now. And this special case of Bezout's theorem, with a line intersecting a curve defined by a homogeneous polynomial of degree d, can be done by hand. $\endgroup$ – Hari Rau-Murthy Oct 23 '15 at 23:19
2
$\begingroup$

It equally applies to quadric surfaces and lines in $\mathbb P^3$. The point is that the notion of degree is well defined and well behaved for projective varieties. In this particular case, let's look on an affine patch; we can assume without loss of generality that the line in question is the $x$-axis (defined by $y=0$). If the equation defining the conic is not a multiple of $y$, then restricting the function to the $x$-axis, that is, setting $y$ equal to $0$, gives a degree-$2$ polynomial in $x$, which has exactly $2$ solutions (counted with multiplicities). So if there are more than two points, the quadric is a multiple of $y$, that is, the $x$-axis lies inside the conic.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.