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.



Krull dimension and elimination theory over the integers
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.


