Prove quotient ring of PID is PID too

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Discussion Overview

The discussion revolves around the problem of proving that the ring of quotients D^(-1)R of a principal ideal domain (PID) R, with respect to a multiplicatively closed subset D, is also a PID. Participants explore various approaches to the problem and clarify assumptions related to the inclusion of the element 1 in the set D.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Homework-related

Main Points Raised

  • One participant expresses difficulty in proving that the ideal I in D^(-1)R can be shown to be principal by relating it to an ideal in R.
  • Another participant suggests that the original problem is valid and that the first participant has effectively solved it by demonstrating that a certain set forms an ideal in R.
  • There is a discussion about the assumption that the multiplicative set D includes the element 1, which is considered important for the localization process.
  • Some participants question the necessity of including 1 in D, with one stating that it seems illogical to localize without it.
  • Clarifications are made regarding a subnote from the professor that specifies the inclusion of 1 in D, which some participants initially overlooked.

Areas of Agreement / Disagreement

Participants generally agree that the problem is valid, but there is disagreement regarding the necessity of the assumption that 1 is in D. Some participants believe it is essential, while others express uncertainty about the implications of localizing without it.

Contextual Notes

The discussion highlights the importance of assumptions in mathematical proofs, particularly regarding the multiplicative set used for localization. There is an acknowledgment of a potential oversight in the problem statement regarding the inclusion of 1 in D.

geor
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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!
 
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OK, there must be a mistake in this exercise so please ignore.
Excuse me for the inconvenience...
 
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.
 
morphism said:
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!
 
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).
 
morphism said:
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).

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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