Proving R as a PID: Z ⊂ R ⊂ Q with R as an Integral Domain

[Mathsgirl]

Homework Statement

Prove that R is a PID (principal ideal domain) when R is a ring such that Z \subset R \subset Q (Z=integers, Q=rationals)

Homework Equations

The Attempt at a Solution

So I'm not really sure how to start this problem. I know that a principal ideal domain is an integral domain in which every ideal is principal. Z is euclidean domain and therefore a PID, Q is a field and therefore a PID. So R is 'between' 2 PIDs. Also if I could show it was a Euclidean Domain then it would be a PID.

Thanks!

 

[ystael]

Try to picture explicitly what a subring of \mathbb{Q} might look like. Take \mathbb{Z} and throw in a fraction or two; what does that generate?

 

[Hurkyl]

You could try attacking the problem directly. Let I be an ideal of R...

 

[Mathsgirl]

I can see that R must be all integers plus some rationals, an example of this could be {a*2^(-b): for an integer a and natural number b}, but I'm not sure how to write down a more general subring?

If I consider an ideal I of R, I need to show that it is principal, so of the form aR for some a in R. I know how to do this for integers, but not when there are rationals too.

I would also then need to show that R is an interal domain, so has no zero divisors. I think as Q has no zero divisors and R is a subring of Q I can deduce that R also has no zero divisors?