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 $W(w\_1, w\_2, w\_3) \deg(f) \deg(g)$?



Bezout's Theorem for weighted homogeneous polynomials
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](http://riemann.unizar.es/~jortigas/Pau-zgz.pdf), 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 $L\_1L\_2=\frac{1}{2}$.


