Prove that $g(x)=0$ for all $x$.

Question

Let $f$ and $g$ be two polynomials on $\Bbb C^n$ s.t $fg=0$ clearly either $f=0$ or $g=0$ as $\Bbb C[x_1, \cdots, x_n]$ is an integral domain.

Suppose $ g$ has the property that if $f(x)≠0$, then $g(x)=0$. Then prove that $g(x)=0$ for all $x$.

I have come to this step while proving every Zariski open set is dense so please don't use it and tag that question here. I was searching for the answer in math stack but didn't get it. Please help

I asked this question here:Suppose $ g$ has the property that if $f(x)≠0$, then $g(x)=0$. Then prove that $g(x)=0$ for all $x$.

For $n=1$ this result is obvious as both $f,g$ have finitely many zeros. So we are done.

From the comment, I tried to prove it in this way. $f(x)$ will be non-zero for infinitely many $x∈\Bbb C^n$, so $g(x)$ will have infinitely zeroes and writing out what this means for its coefficients, can we get $n$ independent equation so that you can make the coefficients zero? In other words, the matrix created by putting the nonzero points will give an invertible matrix. How do we ensure that?

If we get this then it is done. It should be solved with polynomial concepts. I am relatively new in math stack so I can't put it for bounty and I waited for a day to get no further comment and answer in that section but didn't get any reply. I need the answer bit quickly to revise my concepts before algebraic geometry exam. Please help me here.

Answer 1

Your argument here doesn't quite seem to work, because it is not clear how you are supposed to deal with the fact that, while a non-zero polynomial $g\in \Bbb C[x]$ can only have finitely many roots, it is perfectly fine for the zero locus of a non-zero polynomial $g\in\Bbb C[x_1,\cdots, x_n]$ to contain infinitely many points (in fact, it always does).

What cannot happen is that a non-zero polynomial $g\in\Bbb C[x_1,\cdots,x_n]$ evaluates to $0$ for all $x$ in some euclidean ball, due to the following lemma:

Let $k$ be a field, let $g\in k[x_1,\cdots,x_n]$ and let $S_1,\cdots, S_n\subseteq k$ be infinite subsets such that $g(s)=0$ for all $s\in S_1\times \cdots\times S_n$. Then, $g=0$.

This may be proved by induction on $n$.