Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'd like some help making this argument complete and rigorous (if it's correct - if not, help with that would be nice).

Here $k$ is a field.

Let $A_1,\ldots,A_n \subseteq k$ be infinite subsets. Then any polynomial in $k[x_1,\ldots,x_n]$ that vanishes on $A_1\times\cdots\times A_n\subseteq k^n$ must be $0$ (as a polynomial).

This is what I have ...

For the case $n=1$, a non-constant polynomial can only have as many roots as its degree, and in particular, it must have a finite number of roots. The only polynomial in one variable that has an infinite number of roots is $0$, so if a polynomial in $k[x_1,\ldots,x_n]$ vanishes on an infinite subset then it must be $0$.

For the inductive step, suppose the proposition is true for less than $n$ subsets and variables. Let $p\in k[x_1,\ldots,x_n]$ vanish on $A_1\times\cdots\times A_n$. Fix $x_n$ as some $a\in A_n$, and we have a polynomial in $n-1$ variables that vanishes on the set $A_1\times\cdots\times A_{n-1}$, so by the inductive hypothesis it must be identically $0$. (Now it gets sketchy). Since this is true for any of the infinite values in $A_n$, and , $p$ must be $0$.

share|improve this question
    
The argument is already fine. –  Qiaochu Yuan Jan 25 '12 at 1:24
5  
Does it help you to think of $p$ as a univariate polynomial in $x_n$ with coefficients in the field $k(x_1,\ldots,x_{n-1})$? –  Pete L. Clark Jan 25 '12 at 1:25
    
@PeteL.Clark i see ... regarding $p$ as a polynomial in $x_n$, there must be a finite number of values for $x_n$ that make $p$ identically $0$. But in fact there are an infinite number of such values, so $p$ must be $0$. –  smackcrane Jan 25 '12 at 1:51

1 Answer 1

up vote 1 down vote accepted

Answered satisfactorily in the comments.

share|improve this answer

Your Answer

 
discard

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.