Is $\mathbb A^n(\mathbb Z)$ Zariski dense in $\mathbb P^n(\mathbb C)$? Since $\mathbb A^1(\mathbb Z)$ is infinite, it is obvious that  $\mathbb A^1(\mathbb Z)$ is dense in $\mathbb P^1(\mathbb C)$ with respect to the Zariski topology. Does the same property hold for  $\mathbb A^n(\mathbb Z)\subset \mathbb P^n(\mathbb C)$? I am considering  the canonical embedding of the affine space.
I think that the answer is positive, otherwise I'd need a system of polynomial equations which is zero on $\mathbb A^n(\mathbb Z)$ but not on the whole $\mathbb P^n(\mathbb C)$. This looks odd to me, but what is the formal argument?
 A: Yes. You should be aware of this lemma:

Lemma: Let $k$ be a field (or, more generally, an integral domain) and let $S_1,\cdots, S_n\subseteq k$ be infinite subsets. The only polynomial $p\in k[X_1,\cdots, X_n]$ such that $p(x)=0$ for all $x\in \prod_{j=1}^n S_j$ is the zero polynomial.

This proves easily that $\Bbb A^n(\Bbb Z)$ is Zariski dense in $\Bbb A^n(\Bbb C)$, which is in turn Zariski dense in $\Bbb P^n(\Bbb C)$. We may also substitute $\Bbb C$ with any field of characteristic $0$.
The proof of the lemma may be found in many places (I think I saw it in Lang's Algebra). However, it's an easy induction. Assume the lemma works for all $1\le n<m$, let $S_1,\cdots,S_m\subseteq k$ infinite sets and let $p\in k[X_1,\cdots, X_n]\setminus\{0\}$. Then, $$p(X_1,\cdots,X_n)=\sum_{j=0}^h p_j(X_1,\cdots,X_{m-1})X_m^j$$
with $p_h\in k[X_1,\cdots, X_{m-1}]\setminus\{ 0\}$. It is understood that if $m=1$, then $k[X_1,\cdots,X_0]:=k$. For $m>1$, by hypothesis there is some $(\xi_1,\cdots,\xi_{m-1})\in \prod_{j=1}^{m-1}S_j$ such that $p_j(\xi_1,\cdots, \xi_{m-1})\ne 0$. Therefore, $p(\xi_1,\cdots, \xi_{m-1},X_m)$ is an univariate polynomial of degree $h$, and we know that it can have at most $h$ roots. This implies the existence of some $\xi_m\in S_m$ such that $p(\xi_1,\cdots,\xi_{m-1},\xi_m)\ne 0$. $\xi=(\xi_1,\cdots,\xi_m)$ is thus an element of $\prod_{j=1}^m S_j$ such that $p(\xi)\ne 0$, as desired.
