# If three points on a quadric surface then the line going through them is contained in the quadric

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 $[t:s]$.

• Wait, where did the surface come from? The question is about curves in the plane, right? – Schemer Oct 23 '15 at 21:46
• 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. – Jürgen Böhm Oct 23 '15 at 22:09
• 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. – Hari Rau-Murthy Oct 23 '15 at 23:19

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.