Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    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.
  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


    User Avatar
    Science Advisor
    Homework Helper

    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..
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?