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## 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?

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COnsider in $k[X,Y,Z]$ the three polynomials $X^2YZ$, $XY^2Z$ and $XYZ^2$, which have infinitely many common zeroes. – Mariano Suárez-Alvarez Dec 26 2011 at 4:55
Or just $XY^2$ and $X^2 Y$ in $k[X,Y]$, right? – Noam D. Elkies Dec 26 2011 at 5:17
Thank you Mariano and Noam. It really helped clear some lingering questions. – Harish Dec 26 2011 at 8:52
This is a meta question: Mariano or Noam should maybe make their example an answer -- otherwise, there is no "closure" (we could close the question, but that still does not give closure) – Igor Rivin Dec 26 2011 at 12:02
OK, see below. – Noam D. Elkies Jan 3 2012 at 0:42

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.

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