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Bezout's Theorem states that for two homogeneous polynomials $f(x,y,z), g(x,y,z)$ over an algebraically closed field of degrees $m,n$ respectively, such that the two polynomials do not share a common component, then the number of intersections of $f,g$ is equal to $mn$ counting multiplicity. Is there an analogue of this theorem for WEIGHTED homogeneous polynomials? That is, suppose that $w_1, w_2, w_3$ are three coprime positive integers, and for a given polynomial $h(x,y,z)$ let $e_1(h), e_2(h), e_3(h)$ denote the degrees of $x,y,z$ in $h$ respectively. We say that $h$ is weighted homogeneous of degree $d$ with weight $(w_1, w_2, w_3)$ if $h$ satisfies $w_1 e_1(h) + w_2 e_2(h) + w_3 e_3(h) = d$. If we allow $d$ to vary across all positive integers, then the resulting set of polynomials is the set of weighted homogeneous polynomials with weight $(w_1, w_2, w_3)$.

So my question is, is there an analogue to Bezout's Theorem in this setting? That is, is there a constant $W = W(w_1, w_2, w_3)$ which depends on $w_1, w_2, w_3$ such that if $f,g$ are two weighted homogeneous polynomials with weight $(w_1, w_2, w_3)$ with no common componets, then the number of intersections of $f,g$ is bounded by $W(w_1, w_2, w_3) \deg(f) \deg(g)$?

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I would try to estimate, for large $d$, the dimension of the space of homogeneous polynomials of degree $d$ modulo multiples of $f$ and multiples of $g$. You could show that this did not depend on $f$ and $g$ other than their degrees, then find the correct answer for polynomials of the form $e_i^k$. This seems like it would work. I'll think about it and come back to it later. –  Will Sawin Nov 14 '11 at 21:54

1 Answer 1

up vote 15 down vote accepted

The classical Bézout theorem works for curves in the projective space $\mathbb{P}^2$.

In the case of weighted homogeneous polynomials one needs a Bézout theorem in the weighted projective plane $\mathbb{P}^2(w_1, w_2, w_3)$. Such a result can be found, for instance, in the paper by Bartolo, Martin-Morales and Ortigas-Galindo Q-resolutions and intersection numbers, Section 5. It turns out that the intersection number of two curves of equation $f=0$ and $g=0$ in $\mathbb{P}^2(w_1, w_2, w_3)$ is given by $$\frac{1}{w_1 w_2 w_3} \deg_{\omega}(f) \deg_{\omega}(g),$$ where $\deg_{\omega}$ denotes the weighted degree, see also auniket's comment below.

Since $\mathbb{P}^2(w_1, w_2, w_3)$ is a singular variety (with cyclic quotient singularities, hence $\mathbb{Q}$-factorial), this formula makes sense only as an intersection formula for $\mathbb{Q}$-divisors. In fact, if the zero locus of your polynomials intersect the singular locus of the weighted projective plane, it may happen that the corresponding Weil divisors are not Cartier. Consequently, one can obtain a rational intersection number, insted of an integer one.

For instance, let us consider $\mathbb{P}(1,1,2)$, which is isomorphic to a quadric cone in $\mathbb{P}^3$. If $x, y, z$ are the weighted homogeneous coordinates, the ruling of the cone is generated by $x=0$ and $y=0$; furthermore, the intersection numbers for any two curves $L_1$, $L_2 \subset \mathbb{P}(1,1,2)$ of equation $$\lambda_1x+\mu_1y=0, \quad \lambda_2x+\mu_2y=0 \quad (\lambda_i, \mu_i \in \mathbb{C})$$ is equal to $\frac{1 \cdot 1}{ 1 \cdot 1 \cdot 2} = \frac{1}{2}$.

This happens because a line $L$ in the ruling is not a Cartier divisor (since it passes through the vertex $[0:0:1]$) but $2L$ is Cartier, being linearly equivalent to a conic, i.e. to a hyperplane section of the cone. Now two hyperplane sections intersect in two points, so we have $(2L_1)(2L_2)=2$, that is $L_1L_2=\frac{1}{2}$.

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Let me explicitly state, just for clarity, that $W(w_1, w_2, w_3) = \frac{1}{w_1w_2w_3}$. Of course in the OP's formula $\deg(f)$ and $\deg(g)$ should be interpreted as weighted degrees. –  auniket Nov 14 '11 at 22:56
    
Dear auniket, thank you for your comment. I will put the value of $W$ in the answer. –  Francesco Polizzi Nov 14 '11 at 22:59

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