# Polynomials vanishing on an infinite set

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 $0$.

• The argument is already fine. Commented Jan 25, 2012 at 1:24
• Does it help you to think of $p$ as a univariate polynomial in $x_n$ with coefficients in the field $k(x_1,\ldots,x_{n-1})$? Commented Jan 25, 2012 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 $0$. Commented Jan 25, 2012 at 1:51
• @smackcrane - The missing idea is to do something like: think about subbing in $(a_1,\dots,a_{n-1})\in A_1\times\dots\times A_{n-1}$ separately from subbing in $a_n\in A_n$. I.e. fix a specific $(a_1,\dots,a_{n-1})$ and sub it in for $x_1,\dots,x_{n-1}$ in your original polynomial. (Call it $F$.) This gives you a polynomial $f\in k[x_n]$. You know that when you sub in any $a_n\in A_n$, you'll get zero, since $f$ is already the specialization of $F$ at $(a_1,\dots,a_{n-1})$. Since there are infinitely many $a_n$'s, this means $f=0$. Cont'd... Commented Jun 16, 2017 at 15:33
• This tells you that, now viewing $F$ as a polynomial in $x_n$ with coefficients $g_i\in k[x_1,\dots,x_{n-1}]$, that each $g_i$ is a polynomial that will become zero when you sub in any $(a_1,\dots,a_n)$. Now you can invoke the induction hypothesis to conclude each $g_i$ is zero and therefore $F$ is zero. Commented Jun 16, 2017 at 15:34