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.

  • 4
    $\begingroup$ Ehh, this is basically what most of algebraic number theory is about. There are many varieties that are conjectured to have finitely many rational or integral points, but in most cases we can't prove anything. Faltings got the Fields medal for proving that curves of genus $\geq 2$ have finitely many rational points. $\endgroup$ Dec 27, 2020 at 21:45
  • $\begingroup$ $n=2$, $r=1$, $p_1=x_2^2+1$ is a trivial counterexample. $\endgroup$
    – YCor
    Dec 28, 2020 at 7:18
  • 2
    $\begingroup$ I’m voting to close this question because it has been answered in a comment and no more satisfactory answer is likely. $\endgroup$ Jan 2, 2021 at 16:56
  • $\begingroup$ I agree and I also vote to close. $\endgroup$
    – Stein Chen
    Jan 12, 2021 at 22:06

1 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.

  • $\begingroup$ Ok, thanks. Do you happen to know, if there is a criterion which says sth. like There are finitely many solutions in the $K=\mathbb{Z}$ case $\Leftrightarrow \dots$ ? $\endgroup$
    – Stein Chen
    Dec 27, 2020 at 21:44
  • $\begingroup$ No, I'm sorry. But I think that you shouldn't expect a "nice" criterion. $\endgroup$
    – RumDiary
    Dec 27, 2020 at 21:46
  • $\begingroup$ Ok, thank you very much. Do you know a sufficient criterion in the $K=\mathbb{Z}$ case which guarantees that there are only finitely many solutions, if the criterion is fulfilled? $\endgroup$
    – Stein Chen
    Dec 27, 2020 at 21:49
  • 1
    $\begingroup$ Of course one sufficient criterion is that $R \otimes_{\mathbf Z} \mathbf Q$ has Krull dimension $0$, for then there are only finitely many rational solutions. $\endgroup$ Dec 27, 2020 at 21:52
  • 1
    $\begingroup$ There is provably no algorithm that determines whether a system of polynomials over $\mathbf Z$ has a solution (over $\mathbf Z$), so in particular no algorithm to find all solutions. However, in the baby case of my comment you may be able to do it (find the components of $R\otimes_{\mathbf Z}\mathbf Q$, determine which ones have length $1$ over $\mathbf Q$, and then determine which of these solutions have no denominators). Beyond that, already for smooth curves it is a very hard problem that seems out of reach with current technology. $\endgroup$ Dec 27, 2020 at 22:00

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