# A question about the additive group of a finitely generated integral domain

Let $R$ be an integral domain of characteristic 0 finitely generated as a ring over $\mathbb{Z}$. Can the quotient group $(R,+)/(\mathbb{Z},+)$ contain a divisible element? By a “divisible element” I mean an element $e\ne 0$ such that for every positive integer $n$ there is an element f such that $e=nf$.

As Darji points out, another way to ask the question is this: Suppose $e\in R$ has the property that for all positive integers $n$, $e$ is congruent to an integer mod $nR$. Must $e$ be an integer?

TO SUMMARIZE:
Qing Liu showed that in fact any non-integer rational in $R$ determines a divisible element of $(R,+)/(\mathbb{Z},+)$, and Wilberd van der Kallen showed that all divisible elements arise in this way.
I wish I could accept both answers.

The answer is no in general ($e$ needs not to be in $\mathbb Z$), but one can show that $e$ is divisible in $R/\mathbb Z$ if and only if $e\in \mathbb Q\cap R$.

First let $R=\mathbb Z[1/p]$ for some prime number $p$. Then I claim that $1/p$ is divisible in $R/\mathbb Z$. Indeed for any $n\ge 1$, write $n=p^rm$ with $m$ prime to $p$. Let $a,b\in \mathbb Z$ such that $am+bp=1$. Then

For general $R$, denote by $D$ the elements $e\in R$ which are divisible in $R/\mathbb Z$. One can check directly that $D$ is a subring of $R$. Let us show $\mathbb Q\cap R\subseteq D$. If $e=k/q\in \mathbb Q\cap R$ with coprime $k, q$, then again using Bézout, $1/q\in R$. Then it is enough to show that $1/p\in D$ for all prime divisors $p$ of $q$. But this is done just above.

The converse is proved in Wilberd’s answer ($e\in \mathbb Z[1/a]$).

Final remark: $\mathbb Q\cap R=\mathbb Z$ if and only if
$\mathrm{Spec}(R)\to \mathrm{Spec}(\mathbb Z)$ is surjective. This is
because the fiber of this morphism above $p$ is the spectrum of
$R/pR$, and this spectrum is empty if and only if $1/p\in R$.