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?


Note: I previously posted this to Math StackExchange here: <https://math.stackexchange.com/questions/71031/a-question-about-the-additive-group-of-a-finitely-generated-integral-domain>


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.



A question about the additive group of a finitely generated integral domain
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 
$$\frac{1}{p}=b+ \frac{am}{p}=b+n\frac{a}{p^{r+1}}\in \mathbb Z + nR.$$ 


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 $1/p\in R$. 


