Let $\mathcal{C},\Gamma\subseteq \mathbb{P}^2(\mathbb{C})$ be two smooth algebraic curves defined by: $$\mathcal{C}:F(X_0,X_1,X_2)=0 \ \ \ \ \ \ \Gamma:G(X_0,X_1,X_2)=0$$ such that none of them contains the other one. We already know that $\mathcal{C}$ can be canonically equipped with a compact Riemann surface structure. We've defined the intersection divisor of $\Gamma$ with $\mathcal{C}$ as: $$\Gamma\cdot \mathcal{C}:=\sum_{p\in\mathcal{C}} I(p)p$$ where: $$I(p):=\nu_p\left(\frac{G}{H}\right)$$ where $H=H(X_0,X_1,X_2)$ is an homogenous polynomial of degree $\deg(G)$ that doesn't vanish in $p$ (this definition doesn't depend on the choice of $H$ and it's meaningful since $G/H$ is a meromorphic function of $\mathcal{C}$).
Now I'd like to prove that:
$$\deg(\Gamma\cdot \mathcal{C})=\deg(G)\deg(F)$$
The book that I'm using proves this when $$G(X_0,X_1,X_2)=X_1^m$$ and then states that this is not a loss of generality. Why is it so? Why is it sufficient to prove this statement just for monomials?
That's the way the geometers of the old Italian school proved Bezout's Theorem.
The basic idea is that you can deform the curve of degree $m$ to make it become $m$ copies of a line, and in fact the $X_1=0$ line up to a change of coordinates (concretely, deforming a curve means changing the coefficients of its equation in a continous way). One then notices that as the curve is deformed the intersection points vary continously as well, thus the degree of the intersection divisor won't change.
Of course this approach needs some finer formalization since it is not entirely obvious that this deformation can be done in such a way at every step except the final the curve remains smooth (this is possibly irrelevant) and also that when intersection points collide the index $I(P)$ behaves as one would expect.