Show that quadratics $f, g\in\mathbb {C}[x, y]$ have at most four common zeroes, unless they have a non-constant factor in common.

Question

Here is a problem from Algebra, Artin:

Suppose $f, g\in\mathbb {C}[x, y]$ are quadratic. Prove that $f$ and $g$ have at most four common zeroes, unless they have a non-constant factor in common.

1. The book gives the general fact that the number of comnon zeros of such $\mathit f, \mathit g$ (no need to be quadratic) is bounded by the Bezout bound and also gives the proof of the finiteness of the number of common zeroes. So with this fact the above problem is obviously solved.

2. I am trying solving the problem without Bezout's Theorem. Are there some special properties of quadratic polynomials help to solve the problem?

Answer 1

First, the question is about intersection of two conics in the affine plane $\mathbb{A}^2$. Clearly, it reduces to the question about intersection of two conics in the projective plane $\mathbb{P}^2$. So, I will think of $f$ and $g$ as of homogeneous quadratic polynomials in three variables.

If $f$ is irreducible, then the conic $C_f = \{f = 0\}$ is isomorphic to $\mathbb{P}^1$. Any isomorphism $\phi \colon \mathbb{P}^1 \to C_f$ is given by quadratic polynomials (it is equivalent to the Veronese embedding). The points of intersection $C_f \cap C_g$ correspond to zeroes of the polynomial $g \circ \phi$, which is a homogeneous polynomial of degree 4 in two variables, hence has 4 zeroes on $\mathbb{P}^1$.

If $f$ is reducible, then $C_f = L_1 \cup L_2$, and you can check that $L_i \cap C_g$ consists of at most 2 points each by the same argument.

So, a short form of this answer is that rationality of a conic helps.

Answer 2

Inspired by @Sasha's answer, let me give a somewhat adventitious proof based on elementary geometry.

Just as @Sasha, I reduce the question to proving that two conics in $\mathbb{P}^2$ have at most $4$ points of intersection unless they are degenerate and have a line (or two) in common.

To prove this, I assume the contrary. So there are two conics $C$ and $D$ that have more than $4$ points of intersection and yet no line in common.

Pick two points of intersection of $C$ and $D$, and call them $A$ and $B$.

Now, one quirk of the complex projective plane is that there are two points $P_1$ and $P_2$ at infinity that have the property that they belong to all circles. More precisely,

(1) the conics that pass through both $P_1$ and $P_2$ are precisely the circles as well as the degenerate conics containing the line at infinity.

These two points are given in homogeneous Cartesian coordinates by $P_1 = \left(1 : i : 0\right)$ and $P_2 = \left(1 : -i : 0\right)$, where $i = \sqrt{-1}$. Proving (1) is fairly straightforward -- just check that a conic given by the equation $ax^2 + by^2 + cz^2 + dyz + ezx + fxy = 0$ contains $P_1$ and $P_2$ if and only if it satisfies $a = b$ and $f = 0$, but the latter two conditions transform the equation of the conic into $\left(ax^2 + ezx\right) + \left(ay^2 + dyz\right) + cz^2 = 0$, which is either a circle or a line united with the line at infinity.

Now, obviously, we can pick any (invertible) projective transformation and apply it to both $C$ and $D$; this does not change the number of points of intersection of $C$ with $D$. In particular, we can pick any projective transformation that sends $A$ and $B$ to $P_1$ and $P_2$. (Such a transformation exists: Indeed, any given three non-collinear points can be mapped to any three given non-collinear points by a projective transformation. To prove this, recall that projective transformations of $\mathbb{P}^2$ are given by invertible $3\times 3$-matrices.) After applying this transformation, the two conics $C$ and $D$ now pass through $P_1$ and $P_2$, and thus are circles or pairs of lines (by (1)). If they are both circles, then we are done: Indeed, we know that two circles have at most $2$ finite points of intersection, which (combined with the at most $2$ infinite points they may have in common) yields that they have at most $4$ points of intersection; thus we get a contradiction to our assumption. If one of them is a pair of lines, then our life is even easier (because each of the two lines intersects the other circle or pair of lines in at most $2$ points, and thus again we get at most $4$ points of intersection).

I'm afraid there are no pictures for this kind of argument, though.