If there are complex points that don't annulate a polynomial, could I find integer ones? For simplicity, let's take a linear polynomial with complex coefficients in $n$ variables, $P\in\mathbb{C}[X_{1},\dots,X_{n}]$ (by linear, I mean $P(X_{1},\dots,X_{n})=\alpha_{1} X_{1}+\dots + \alpha_{n} X_{n}$, with $\alpha_{1},\dots,\alpha_{n}\in\mathbb{C}$). Suppose I know that there exists a $\underline{\lambda}=(\lambda_{1},\dots,\lambda_{n})\in \mathbb{C}^{n}$ such that $P(\underline{\lambda})\neq 0$. Can I assure that there exists $\underline{a}=(a_{1},\dots,a_{n})\in\mathbb{Z}^{n}$ such that $P(\underline{a})\neq 0$?.
I have some partial results: it is quite direct to show that, if $P(\underline{\lambda})\neq 0$, then $P((|\lambda_{1}|,\dots,|\lambda_{n}|))\neq 0$, so I can really think the problem considering that $\underline{\lambda}\in\mathbb{R}^{n}$. Taking the latter on account, if $\underline{\lambda}\in\mathbb{Q}^{n}$, the result would be obvious (it would be sufficient to make a "common denominator"). But I don't really get to the point of assuring that, if this is not the case, I am able to find integers that don't annulate the polynomial...
Any hint will be highly appreciated! Thanks in advanced!
 A: It's a consequence of the following general fact:

Let $F$ be a field, let $p\in F[X_1,\cdots, X_n]$ be a polynomial and let $S_1,\cdots, S_n\subseteq F$ be subsets of infinite cardinality such that $p(s)=0$ for all $s\in\prod_{j=1}^n S_j$. Then, $p=0$.

Proof: By induction. Case $n=1$ is just the well-known fact that a non-zero polynomial in one variable has finitely many roots. If it's true for $n$, then let $S_1,\cdots, S_{n+1}\subseteq F$ be infinite subsets and let $p\in F[X_1,\cdots, X_{n+1}]\setminus\{0\}$. Write $p$ as $p(X_1,\cdots, X_n,X_{n+1})=\sum_{j=0}^m f_j(X_1,\cdots, X_n)X_{n+1}^j$, with $f_m\ne 0$. By hypothesis, there is some $(s_1,\cdots,s_n)\in \prod_{j=1}^n S_j$ such that $f_m(s_1,\cdots, s_n)\ne 0$. But then, $p(s_1,\cdots, s_n,X_{n+1})$ is a polynomial in $X_{n+1}$ of degree $m$, therefore it has at most $m$ roots in $F$. Therefore there is some $s_{n+1}\in S_{n+1}$ such that $p(s_1,\cdots, s_n,s_{n+1})\ne 0$. This provides an element $s=(s_1,\cdots,s_{n+1})\in\prod_{j=1}^{n+1}S_j$ such that $p(s)\ne 0$.
