Subring of a global field

Question

I have to prove the following implication:

Let $\mathbb{K}$ be a global field (i.e. finite extension of $\mathbb{Q}$ or $\mathbb{F}_q(x)$). And let $A$ be an arbitrary subring of $\mathbb{K}$ which is not a field than any non-zero ideal in $A$ has a finite index.

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It is one of the exercises in my course so maybe you can provide some hints, rather than exact answer...

Answer 1

For $A$ a subring of a number field and $I$ a non-zero ideal, take $a\in I-0$, the constant term of its minimal polynomial is a non-zero rational number $\in a\Bbb{Z}[a]\subset I$, thus its numerator is a non-zero integer $n\in I$.

$\Bbb{Q}A=Frac(A)$ being a $d$-dimensional $\Bbb{Q}$-vector space, $A/nA$ has at most $n^d$ elements.

This is because $A= \bigcup_j M_j$ where $M_j\subset M_{j+1}$ is an increasing sequence of finitely generated subgroups. $\Bbb{Z}$ being a PID, each $M_j$ is a free $\Bbb{Z}$-module, of rank $r_j=\dim(\Bbb{Q}M_j)\le d$, thus $M_j/nM_j$ has $n^{r_j}\le n^d$ elements, and $$|A/nA|=\lim |M_j/(M_j\cap nA)| \le \lim|M_j/nM_j| = n^d$$

Since $A/I$ is a quotient of $A/nA$, it has at most $n^d$ elements.

It works the same way when $Frac(A)$ is a finite extension of $\Bbb{F}_p(x)$, taking a non-constant element $y\in A$, replacing $n$ by $f\in \Bbb{F}_p[y]$ and finitely generated subgroup by finitely generated $\Bbb{F}_p[y]$-module.