Show that if $f \in \mathbb{C} \left[x_1, \dots, x_n \right]$ vanishes at every point of $\mathbb{Z}^n$, then $f$ is the zero polynomial.

Question

I am working on a problem from Ideas, Varieties, and Algorithms:

Show that if $f \in \mathbb{C} \left[x_1, \dots, x_n \right]$ vanishes at every point of $\mathbb{Z}^n$, then $f$ is the zero polynomial.

My attempt so far:

This proof is by induction on $n$. Let $f \in \mathbb{C}\left[x\right]$ vanish at every point in $\mathbb{Z}$. Then, it must be identically the zero polynomial, as a polynomial of degree $d$ has at most $d$ roots. Hence, $f$ vanishes at every point at $\mathbb{C}$. Now, assume that $g \in \mathbb{C} \left[x_1, \dots, x_{n-1} \right]$ vanishes at every point in $\mathbb{Z}$. We may collect terms and rewrite $f$ as

$$ f(x_1, \dots, x_n) = \sum_{i=0}^m x^i_n g(x_1, \dots, x_{n-1}), $$ for some functions $g_1,\dots,g_n\in \Bbb C[x_1,\dots,x_{n-1}]$.

How do I proceed from here? I was trying to adapt an earlier proof with no success.

Answer 1

This can be turned into a proof by induction on $n$. You have noted that it is true for $n=1$. Then you write $$f=\sum_{i}{x_n^ig_i(x_1,x_2,\ldots,x_{n-1})}$$ for some polynomials $g_i$. If every $g_i$ vanishes at every point of $\mathbb{Z}^{n-1}$, then by induction they are all 0. Hence we may assume there is an integer vector $(k_1,\ldots,k_{n-1})$ for which we can specialize the first $n-1$ variables and obtain a nonzero polynomial $$f=\sum_i{x_n^ig_i(k_1,\ldots,k_{n-1})}$$ However, this polynomial has infinitely many $0$'s since it vanishes at every point of $\mathbb{Z}$, hence it is $0$.

Answer 2

Let $f(x_1, \dots, x_n) = \sum_{i=0}^m g_i(x_1, \dots, x_{n-1}) x^i_n$. Let $h(x_n) = f(a_1, \ldots, a_{n-1}, x_n)$ where $a_1, \ldots, a_{n-1}$ are fixed integers. Since $h$ vanishes on $\mathbb Z$, it must be the zero polynomial. Thus, each $g_i(a_1, \ldots, a_{n-1}) = 0$. Since each $g_i$ vanishes on $\mathbb Z^{n-1}$, it must be zero by induction. We conclude that $f = 0$ as desired.