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Prove quotient ring of PID is PID too

  1. Nov 25, 2008 #1
    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?

    Thank you in advance!
  2. jcsd
  3. Nov 25, 2008 #2
    OK, there must be a mistake in this exercise so please ignore.
    Excuse me for the inconvenience...
  4. Nov 25, 2008 #3


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    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.
  5. Nov 26, 2008 #4
    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

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

    Thanks a lot!
  6. Nov 26, 2008 #5


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    Yes I did. I don't think it makes a lot of sense to localize a domain at a multiplicative set that doesn't have 1 in it (but I may be wrong).
  7. Nov 26, 2008 #6
    Yes, I understand, you want to have R inside D^-1R..
    Anyway, I think the book was ommited this "extra" assumption
    by mistake anyway..
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