Bezout's Theorem concludes that if $f\_1,f\_2,\cdots,f\_n\in k[x\_1,x\_2,\cdots,x\_n]$ have finite intersection points, then they have at most $d\_1d\_2\cdots d\_n$ intersection points, where $d\_i$ is the degree of $f\_i$.


When $n=2$, if $f\_1$ and $f\_2$ don't have a common divisor of positive degree, then they have finite intersection points.


When $n\geq 3$, is there any condition that can guarantee that $f\_1,f\_2,\cdots,f\_n\in k[x\_1,x\_2,\cdots,x\_n]$ have finite intersection points?



Sufficient conditions to guarantee finite intersection points in Bezout's Theorem
The condition you want is
$$
{\rm Res}(h\_1,\ldots,h\_n)\neq 0\ .
$$
Here $h\_i$ is the (leading/highest degree) homogeneous part of $f\_i$ of degree $d\_i$.
The expression on the left is the homogeneous resultant of the given collection of forms $h\_1,\ldots,h\_n$. It is a polynomial in the coefficients of these forms, which is separately homogeneous of degree $\prod\_{j\neq i}d\_j$ in the coefficients of each form $h\_i$. The condition essentially says there is no solution of the $f$ vanishing system at infinity, but that also rules out higher dimensional components in the common zero locus.
Note that this is related to the notion of residue:
$$
\sum\_{x}\frac{g(x)}{J\_f(x)}\ .
$$
Here $J\_f(x)$ is the Jacobian determinant of $f\_1,\ldots,f\_n$ at the point $x$, and $g$ is some other polynomial.
The sum is over all $x$ which solve the system $f\_1(x)=0,\ldots,f\_n(x)=0$, counting solutions with their multiplicities.


The residue is a linear form in the coefficients of $G$ and the coefficients of that linear form are rational functions of the coefficients of the $f$ polynomials.
The denominator of these rational functions is given by the homogeneous resultant featuring in the nonvanishing condition above.


For more details, see the reference by Cattani and Dickenstein (Theorem 1.7.2) that I mentioned in this other MO answer:


[Residues in several complex variables](https://mathoverflow.net/questions/365866/residues-in-several-complex-variables/365880#365880)


