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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

Umberto P.
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Ri-Li
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    This is not my field of expertise, but can you say something like: $f(x)$ will be non-zero for infinitely many $x\in \mathbb{C}^n$, so $g(x)$ will have infinitely zeroes and writing out what this means for its coeffcients, they must all be zero, hence $g=0$? – ThePuix Jul 19 '19 at 15:05
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    I agree with @ThePuix, only that one has to assume that $f \neq 0$. In fact, if $f=0$, then your condition is useless (since it is only valid on $x$ such that $f(x) \neq 0$). Also, if $f \neq 0$, you can use the fact at the beginning to derive that $g=0$. – Pirer Jul 19 '19 at 15:38
  • Can you please elaborate the answer, I was trying in the way putting $x_1=\cdots=x_{n-1}=0$ and then considering it as a polynomial in $x_n$(WLOG) but I was not successful. Can you please teach me the tricks that you are saying? – Ri-Li Jul 19 '19 at 18:24
  • @ThePuix can you help me? – Ri-Li Jul 19 '19 at 19:16
  • How do you ensure that you have $n$ 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 you ensure that? – Ri-Li Jul 19 '19 at 20:41

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It can be shown that for any infinite field $k$ we have $I(k^n)=(0)$, where for $A\subseteq k^n$ we define $$I(A)=\{f\in\mathbb{C}[x_1,\ldots,x_n]:f(a)=0\text{ for all }a\in A\}$$ in case you don't use this notation.

We can deduce this quickly for $\mathbb{C}$ in particular by the Nullstellensatz: $I(\mathbb{C}^n)=I(V((0)))=\sqrt{(0)}=(0)$ as $\mathbb{C}[x_1,\ldots,x_n]$ is an integral domain.

Then we have shown that the only polynomial which vanishes everywhere on $\mathbb{C}^n$ is $0$. Clearly $fg$ vanishes on all of $\mathbb{C}^n$, so $fg=0$ and then, as you said, either $f=0$ or $g=0$.

Note that Zariski open sets are only required to be dense in irreducible varieties (which $\mathbb{C}^n$ is, since $(0)$ is prime as $\mathbb{C}[x_1,\ldots,x_n]$ is an integral domain).

For example, consider the variety $A=V(xy)\subsetneq\mathbb{C}^2$. Take the open set $D_A(x)\subseteq A$, then this is contained in $V(y)\subsetneq A$, and so is not dense. This is because $A$ is not irreducible, we can write $A=V(x)\cup V(y)$, with $V(x)\nsubseteq V(y)$ and $V(y)\nsubseteq V(x)$.

Dave
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