# Prove quotient ring of PID is PID too

Hello everybody,
I'm a bit stuck here. I have a problem tha goes like this:

Let R be a principal ideal domain (PID). Let D a subset of R
which is multiplicatively closed. Show that the ring of quotients D^(-1)R is
a PID too.

I've tried several different ways but I couldn't get to the result.
For example, take an ideal I of D^(-1)R. I was able to show that
all r \in R such as r/d \in I for some d \in D, form an ideal of R,
(Thus a principal ideal). If I could show that I=(a)/d for some d..

Anyway, I don't know..
Any hints for that?

OK, there must be a mistake in this exercise so please ignore.
Excuse me for the inconvenience...

morphism
Homework Helper
There's nothing wrong with the problem -- and you've actually solved it. You said that you've shown that the set of all r in R such that r/d is in I for some d in D is an ideal of R, hence is generated by some a in R. This means that I itself is generated by a in D-1R.

There's nothing wrong with the problem -- and you've actually solved it. You said that you've shown that the set of all r in R such that r/d is in I for some d in D is an ideal of R, hence is generated by some a in R. This means that I itself is generated by a in D-1R.
Actually, I assumed D includes 1, which was not in the question.
Though I just realized that there was a subnote of the prof (this was in
an assignment) stating we should use the extra assumption that 1 is in D.
So yes, I was right but I missed his extra note and wasted a couple of
hours!!

You assumed that 1 is in D, too, right?

Thanks a lot!

morphism