Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
I assume you suppose $g\neq 0$. This is certainly true for $\mathbb R^n$ and $\mathbb C^n$, given that the set of points at which $g$ does not vanish is dense in the metric topology. I haven't thought of an approach for general infinite fields though. I'm tempted to use the dimensions of the affine varieties that make up the zero sets of $f$, but I'm not sure this works when $F$ is not algebraically closed, because I usually assume closure for this type of thing. – Alex Becker Jul 16 '12 at 4:10
@AlexBecker Yes, I am assuming $g\neq 0$. I mention so in the second sentence, apologies if was not communicated well. – Son Bi Jul 16 '12 at 4:12
Oh yes, sorry I missed that. – Alex Becker Jul 16 '12 at 4:13
up vote 6 down vote accepted

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 $g=0$).

share|cite|improve this answer
Suggestion to Son Bi: The identification $F[X_1,\ldots,X_n]=F[X_1,\ldots,X_{n-1}][X_n]$ is useful. 1) can be proved by induction, and you already have the base case. For 2) you can show that $R[X]$ is an integral domain if $R$ is. – Jonas Meyer Jul 16 '12 at 4:27
Thanks Robert, thanks @Jonas, I believe I can put this together now. – Son Bi Jul 16 '12 at 4:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.