Krull dimension and elimination theory over the integers

Question

Let $K:=\mathbb{C}$, and let $R:=K[x_1,\dots , x_n]$.

Then, a system of polynomial equations $p_1=0, p_2=0, \dots , p_r = 0$, where the $p_i$ are polynomials in the $x_j$, has finitely many solutions $\Leftrightarrow$ the Krull dimension of $R/I$ is equal to $0$, where $I:=\langle p_1, p_2, \dots , p_r \rangle$.

My question is:

Is this also true, if we replace $\mathbb{C}$ by $\mathbb{Z}$ ?

If not, is there a criterion which says sth. like

There are finitely many solutions in the $K=\mathbb{Z}$ case $\Leftrightarrow \dots$ ?

Thank you.

Answer 1

This should be a comment, but my reputation is low, so I have to post this.

I assume that you want the solutions of $p_1=\dots=p_r=0$ to be in $\mathbb{Z}^n$. Then the answer is no. Take, for example, $r=1$, $n\geq 2$ and $p_1$ to be any polynomial in $\mathbb{Z}[X_1,\dots,X_n]$ that has no solutions in $\mathbb{Z}^n$ (say $p_1(x_1,\dots,x_n)=x_1^2+1$).

If the above seems pathological (as $p_1=\dots=p_r=0$ has 0 solutions), try to experiment further with similar choices. In particular, you might want to consider a polynomial on $n$ variables that has only finitely many solutions and see what happens in this case.