Square Polynomial system has finite solutions Let $S =$ {$f_{1}, \ldots , f_{n}$} be a set of $n$ polynomials in $n$ variables. Call this set independent if $\forall$ $1 \leq i \leq n, f_i \notin  < S - f_i >$, the ideal generated by the set $ S - f_i $. Is it true that this system always possesses a finite number of solutions?
 A: Sorry, but no: this necessary(*) condition is far from sufficient
except when $n=1$, or when $n=2$ and $f_1$ or $f_2$ is irreducible.
(*) I assume throughout that the unnamed ground field is algebraically
closed.  If not then the condition is not necessary either: consider
$n=2$ and $f_1 = f_2 = x^2+y^2$ over ${\bf R}$ [the only solution is
$(x,y)=(0,0)]$, or indeed a finite field, on which every system of equations
has only finitely many solutions.
For $n=1$ the condition just means that $f_1$ is a nonzero polynomial,
in which case it has finitely many roots.
For $n=2$ the condition says that neither $f_1$ nor $f_2$ is
a multiple of the other; that's sufficient if either polynomial is
irreducible, but if not they could still have a common factor $g$
and vanish simultaneously on the curve $g=0$.  A simple example is
$g=x$, $f_1=xy$, $f_2=x(y+1)$.  Another example,
$(f_1,f_2) = (x y^2, x^2 y)$ [from my Dec.26 comment,
simplifying Mariano Suárez-Alvarez's counterexample for $n=3$],
shows it's even possible for the (set-theoretic) zero-loci of
$f_1$ and $f_2$ to coincide.
Once $n>2$ it is not even enough to assume that each $f_i$ is
irreducible, because two or more of the hypersurfaces $\{f_i = 0\}$
might have a reducible intersection in dimension strictly between
$0$ and $n-1$.  For example, take $n=3$, and consider the vector space
$\Gamma$ of polynomials of degree at most $2$ in $x,y,z$ that vanish
on the line $l: x=y=0$.  Then $\dim(\Gamma) = 7$ (the monomials
$x$, $y$, $x^2$, $xy$, $xz$, $y^2$, and $yz$ constitute a basis),
and the general $f\in\Gamma$ is irreducible; so we can choose
irreducible and linearly independent $f_1,f_2,f_3 \in \Gamma$,
and then no $f_i$ is contained in the ideal generated by the other two,
but the solutions of $f_1=f_2=f_3=0$ always include $l$ and are thus
infinite in number.
