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 $0$.

  • $\begingroup$ The argument is already fine. $\endgroup$ – Qiaochu Yuan Jan 25 '12 at 1:24
  • 5
    $\begingroup$ Does it help you to think of $p$ as a univariate polynomial in $x_n$ with coefficients in the field $k(x_1,\ldots,x_{n-1})$? $\endgroup$ – Pete L. Clark Jan 25 '12 at 1:25
  • 1
    $\begingroup$ @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 $0$. $\endgroup$ – smackcrane Jan 25 '12 at 1:51
  • $\begingroup$ @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... $\endgroup$ – Ben Blum-Smith Jun 16 '17 at 15:33
  • $\begingroup$ 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. $\endgroup$ – Ben Blum-Smith Jun 16 '17 at 15:34

Answered satisfactorily in the comments.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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