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I am trying to show that two distinct irreducible conics $X$ and $Y$ in $\mathbb{A}^2$ intersect in at most four points without relying on Bézout's Theorem. I would appreciate any insight into elementary methods that could be used to achieve this result.

Here's what I attempted:

  1. Let $X = Z(f)$ and $Y = Z(g)$ for irreducible polynomials $f, g \in k[x,y]$, where $f$ and $g$ define the conics $X$ and $Y$ respectively. Since $X \neq Y$ and both $f$ and $g$ are irreducible, it follows that $X \cap Y$ is a finite set.
  2. Then, letting $b$ denote the second coordinate of a point in $X \cap Y$, I considered the system \begin{align*} \begin{cases} f(x,b) = 0, \\ g(x,b) = 0. \end{cases} \end{align*}

My idea was that the finiteness of $X \cap Y$, together with this system, would somehow lead to an upper bound of four points, but I don't know how to make the argument work.

Could someone suggest a more solid, elementary approach that might lead to the desired conclusion of at most four intersection points? Any guidance would be greatly appreciated.

Matheus Frota
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1 Answers1

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There are other ways to do this, but there's no free lunch here - working around Bezout means that you have to spend basically the same amount of effort in developing something else. I'll lay out some of these methods here and leave links to some of the details underlying them.

  1. Any conic in $\Bbb A^2$ is equivalent, up to a change of coordinates, to $y-x^2=0$ or $xy-1=0$ (ref - this proves it in $\Bbb P^2$, but the same proof can be adapted to the affine case by just setting $z=1$). If we can change coordinates so that $f=y-x^2$, then we can plug in $y=x^2$ to $g$, so that we get a polynomial of degree at most 4 in $x$, and any solution to this degree-four polynomial in $x$ has a unique corresponding $y$. If we can change coordinates so that $f=xy-1$, then multiply $g$ by $x$ to the maximal power of $y$ appearing in $g$ and rewrite $xy=1$ to again reach the conclusion that all the $x$-coordinates of points on $V(f,g)$ satisfy a polynomial of degree at most 4, and there is again a unique $y$ for every value of $x$.

  2. Consider $\operatorname{res}_y(f,g)$, the resultant of $f$ and $g$ with respect to $y$. This is a polynomial of degree at most 4 in $x$, and by the general properties of the resultant, each factor $(x-a)$ of the resultant corresponds to a line $x=a$ with an intersection of $f,g$ on it.

  3. In characteristic not two, consider the matrices $Q_f,Q_g$ of the quadratic forms associated to $f,g$. Then $\det(Q_f+tQ_g)$ is a nonconstant polynomial in $t$, so it has a zero, which implies $f+tg$ factors as a product of two linear terms $L_1,L_2$. Then $V(f,g)=V(g,f+tg)=V(g,L_1)\cup V(g,L_2)$, and the intersection of a conic and a line has at most two points. More details here.

KReiser
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