Can the method of resolvents be used to give a proof of Bezout's Theorem? It seems to me like it should but I am unable to finish the proof. Here is what I have so far.
Take two homogeneous polynomials $F(X,Y,Z)$ and $G(X,Y,Z)$, of order $N$ and $M$, respectively, and treat $X$ and $Y$ as constants for the moment. Then the resolvent will be equal to zero iff the two polynomials share a factor. But the resolvent is an homogeneous polynomial in two variables of order $M+N$, hence is a projectivisation of a one variable polynomial. If the field is algebraically closed then this gives us exactly $M+N$ projective solutions to res$(X,Y) = 0$, counting with multiplicity, call them $[X_i,Y_i]$. But then if we consider $F[X_i,Y_i,Z]$ and $G[X_i,Y_i,Z]$ and look to solve for $Z$ we can have up to $\max(M,N)$ solutions for each $i$. If this method is to work we must apply some other fact to make sure there is exactly one $Z_i$ for each of the $[X_i,Y_i]$. Does this method work at all?
As Daniel McLaury pointed out, I believe you are talking about the resolutant, not the resolvent. That resultant can indeed be used to proove Bezout's theorem. For the moment, assume that all $Y_i\neq0$. Then you can w.l.o.g. assume $Y_i=1$ (i.e. fix one representant, or a specific “drawing plane”), so the resultant will basically tell you the $X$ coordinate for a point of intersection. Now you can take the line $X=X_i$ and intersect it with either curve. You will obtain $M$ intersections for one and $N$ intersections for the other curve.
But the claim is that (for a root with multiplicity one) only one of these will be common to both curves. To see this, assume there were two common intersections for a given $X$ value. Then you could perturb your curves slightly, so that these two intersections end up in two different $X$ positions. In that case, the resultant would clearly have two distinct zeros. Now as you smootly reduce your perturbation, the two zeros of the resultant would move towards one another, and in the end would result in a double root of that polynomial.
So the general rule is that the multiplicity of a root of the resultant will indicate the number of intersections associated with it. Hence knowing the number of zeros, and taking multiplicities into account, you already know the number of intersections.
The above assumed $Y_i\neq 0$, but since the argument holds for any projective coordinate system, it must hold in general, and the assumption can be dropped.