Let R be an integral domain of characteristic 0 finitely generated as a ring over Z. Can the quotient group (R,+)/(Z,+) contain a divisible element? By a “divisible element” I mean an element e≠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∈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,+)/(Z,+), and Wilberd van der Kallen showed that all divisible elements arise in this way.

I wish I could accept both answers.

**Answer**

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

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

1p=b+amp=b+napr+1∈Z+nR.

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

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

**Final remark**: Q∩R=Z if and only if

Spec(R)→Spec(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∈R.

**Attribution***Source : Link , Question Author : Sidney Raffer , Answer Author : Qing Liu*