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Suppose $f(x_1,\dots,x_n)$ is a polynomial in $n$ indeterminates over an infinite field $F$. Suppose $f((a_i))=0$ for all $n$-tuples $(a_i)$ such that $g((a_i))\neq 0$, where $g(x_1,\dots,x_n)$ is another nonzero polynomial over $F$. I know that the set of $n$-tuples such that $g$ is nonzero is nonempty.

Does this imply $f=0$?

I'm curious, because it's true when $n=1$. Since $g$ has only finitely many roots, the set of values on which $g$ is nonzero is infinite since $F$ is infinite, but then $f=0$ as $f$ has infinitely many roots. Does the same argument work in arbitrarily many indeterminates, or does more care need to be taken? Thanks.

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    Oh yes, sorry I missed that.2012-07-16

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Yes, it's true. There are two parts to the result. Both are "well-known":

1) Over an infinite field $F$, a polynomial $f \in F[X_1,\ldots,X_n]$ is $0$ if and only if the corresponding function from $F^n \to F$ is $0$.

2) Over any field $F$, $F[X_1,\ldots,X_n]$ is an integral domain (i.e. if $f g = 0$ then $f=0$ or $g=0$).

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    Thanks Robert, thanks @$J$onas, I believe I can put this together now.2012-07-16