If $\mathbb Z^n \subseteq V$ for an affine algebraic variety $V$ of $\mathbb C^n$ , then how to show that $V=\mathbb C^n$?

Let $f_1,...,f_k \in \mathbb C[X_1,...,X_n]$ and

$V:=V(f_1,...,f_k)=\{(a_1,...,a_n)\in \mathbb C^n : f_i(a_1,...,a_n)=0,\forall i=1,...,k\}$ .

If $\mathbb Z^n \subseteq V$, then how to show that $V=\mathbb C^n$ ?

Equivalently, due to $\mathbb C$ being algebraically closed, due to Hilbert's Nullstelensatz , how to show that $\sqrt {(f_1,...,f_k)}=\{0\}$

i.e. that all the polynomials $f_i$ s are $0$ ?

Hint: Your system of polynomials sure does have a lot of roots. How large must the degrees be?

• I don't understand. Could you please elaborate ?
– user495643
Nov 22, 2017 at 13:53
• @misao : Think about the proof of the Schwartz-Zippel lemma. Nov 22, 2017 at 14:05

NOTE (added after the comment below by misao:) I am unsure of my answer. Not able to see how to address the objection raised. But not deleting my answer in the hope that someone else may be able to salvage it!

Given the hypothesis all the points, for example, of the form $(k,1,1,\ldots,1)\in V$ for all integers $k$. That means the polynomials $g_j(X_1)=f_j(X_1,1,1,\ldots,1)$ (specializing all except the first variable to the value $1$) in the single variable $X_1$ have infinitely many roots ( namely all the integers). Hence $g_j$ is identically zero (as a polynomial in $X_1)$. This means in the polynomials $f_j$ all monomials involving $X_1$ are absent.

By symmetry you can show this for all the variables $X_i$.

• $g(X_1)$ identically zero need not imply $f$ has no monomial involving $X_1$. Take $f(X_1,X_2)=X_1- X_1X_2$ for instance
– user495643
Nov 22, 2017 at 13:53