Prove quotient ring of PID is PID too

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SUMMARY

The discussion centers on proving that the quotient ring D^(-1)R of a principal ideal domain (PID) R, where D is a multiplicatively closed subset, is also a PID. Participants confirm that if an ideal I of D^(-1)R can be shown to be generated by an element a in R, then I is also generated by a in D^(-1)R, establishing that D^(-1)R retains the PID property. A critical assumption is that the set D includes the element 1, which is necessary for the localization process to be valid. The oversight regarding this assumption led to initial confusion among participants.

PREREQUISITES
  • Understanding of principal ideal domains (PIDs)
  • Knowledge of localization of rings, specifically D^(-1)R
  • Familiarity with ideals in ring theory
  • Concept of multiplicatively closed sets
NEXT STEPS
  • Study the properties of principal ideal domains (PIDs) in depth
  • Learn about the localization of rings and its implications
  • Explore examples of multiplicatively closed sets in ring theory
  • Investigate the role of the element 1 in localization processes
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Mathematicians, particularly those studying abstract algebra, graduate students in mathematics, and anyone interested in the properties of rings and ideals in algebraic structures.

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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