1

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.

Ri-Li
  • 9,038
  • But isn't it so $fg = 0$ for this example ? So it must be $f = 0$ or $g = 0$, but can't be $f = 0$. – Jakobian Jul 20 '19 at 18:01
  • Actually in my last post the commentators raised this question what will happen if $f=0$ that is why I added this. I can edit it now. – Ri-Li Jul 20 '19 at 18:10
  • Dupe of this. – Bill Dubuque Jul 20 '19 at 19:03
  • 1
    I don't understand. You're saying it's clear that if $h_1h_2 = 0$ for polynomials $h_1, h_2$, then must be $h_1 = 0$ or $h_2 = 0$. Well, $f(x)g(x) = 0$ for all $x$. If $f(x) = 0$, then $f(x)g(x) = 0$. If $f(x)\neq 0$, then $g(x) = 0$, so $f(x)g(x) = 0$. Then must be $f = 0$ or $g = 0$. And yes, assumption that $f \neq 0$ is essential, I wasn't denying it – Jakobian Jul 20 '19 at 20:03
  • Yes I understood later on that I got confused in $f(x)g(x)=0$ at a particular $x$ and $fg=0$ in $\Bbb C[x_1,...,x_n]$. Now it is clear. This is just a statement of proving that $\Bbb C[x_1,...,x_n]$ which is clear to me. I have proved it. Thanks anyway. – Ri-Li Jul 20 '19 at 20:32

1 Answers1

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 $n$.

  • I don't know what are you trying to say. I mentioned in the question that I am trying to solve it in that way. Are you giving a contradiction? My question is to prove that $g=0$. – Ri-Li Jul 20 '19 at 18:45
  • For $n=1$ this result is obvious as both $f,g$ have finitely many zeros. So we are done. We are mainly concerned about the higher dimension. – Ri-Li Jul 20 '19 at 18:48
  • The way I'm reading what you wrote is "I think it should go some way and perhaps there is a way to make it go that way". What I have written should be interpreted as: "One must prove the highlighted lemma by induction and then it is all obvious." –  Jul 20 '19 at 18:51