Suppose that we have a map ${\bf f}:=(f_1,f_2,\cdots ,f_n):\mathbb{C}^n\rightarrow \mathbb{C}^n$ given by $$ \mathbb{C}^n\ni {\bf z}:=(z_1,z_1,\cdots,z_n)\rightarrow \big(f_1({\bf z}),f_2({\bf z}),\cdots,f_n({\bf z})\big) \in \mathbb{C}^n$$ such that the $f_i$'s are homogeneous polynomials of degree $d_i$ .
I want to understand why the following is true:
Theorem: The morphism ${\bf f}$ has finite fibers if and only if ${\bf f}^{-1}({\bf 0})={\bf 0}$.
(where I am refering to the cardinality of ${\bf f}^{-1}({\bf y})$ for any ${\bf y}\in \mathbb{C}^n$ and where ${\bf 0}=(0,0,\cdots,0)$ )
Now, I know at least one way to do it: The condition ${\bf f}^{-1}(0)=0$ is equivalent with the fact that $(f_1,f_2,\cdots,f_n)$ is a homogeneous system of parameters for $\mathbb{C}[z_1,z_2,\cdots, z_n]$. This is easy to see, since $${\bf f}^{-1}(0)=0 \iff \sqrt{(f_1,f_2,\cdots,f_n)}=(z_1,z_2,\cdots,z_n)\iff (z_1,z_2,\cdots,z_n)^N\subset (f_1,f_2,\cdots,f_n) $$ for a suitably big $N$, which is, according to Eisenbud for instance, the definition of a system of parameters. Now being a system of parameters implies $\mathbb{C}[f_1,f_2,\cdots,f_n]$ is a finitely generated $\mathbb{C}[z_1,z_2,\cdots,z_n]$-module and this in turn implies ${\bf f}$ is a finite morphism and hence has finite fibers.
However, in the paper I am studying, the theorem is actually used to show the finiteness of ${\bf f}$ (via proper+quasifinite $\Rightarrow$ finite..). I believe there must be another way to show it then, without going through finiteness first. I am not sure if it is going to be a more elementary method or not.
In particular, I think Bezout's theorem is involved, but I cannot figure it out. Say for instance that the fiber ${\bf f}^{-1}({\bf y})$ is an infinite set for some ${\bf y}\in\mathbb{C}^n$. We can translate that into the following situation in projective space:
Consider $f_i'=f_i({\bf z})-y_i\cdot z_{n+1}^{d_i}$; they are homogeneous in $\mathbb{C}[z_1,z_2,\cdots,z_{n+1}]$) and therefore define $n$ hypersurfaces in $\mathbb{CP}^n$. Now, Bezout's theorem says that if the have finitely many intersection points, those would be at most $d_1d_2\cdots d_n$-many. Evenmore, if an intersection point $[z_1:z_2:\cdots:z_{n+1}]$ has $z_{n+1}\neq 0$ it corresponds (by setting $z_{n+1}=1$ and returning to the affine setting) to a solution of ${\bf f}({\bf x})={\bf y}$.
Does an infinitude of intersections imply a common component for the hypersurfaces $f_i'=0$? Would this easily mean that there are nontrivial solutions of ${\bf f}({\bf x})={\bf 0}$? Even worse, I am afraid this isn't true in general (because not everything is a complete intersection...)
Remark: Clearly, the direction (has finite fibers) $\Rightarrow\ {\bf f}^{-1}({\bf 0})={\bf 0}$ is obvious, since otherwise, because of homogeneity, ${\bf 0}$ would have an infinite fiber (a whole line at least). I am looking for the less straight-forward direction.