Prove quotient ring of PID is PID too

In summary, the problem is that the author is trying to solve an equation but he doesn't seem to know how to get started. He has an idea of what he needs to do but he isn't sure how to get there.
  • #1
geor
35
0
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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  • #2
OK, there must be a mistake in this exercise so please ignore.
Excuse me for the inconvenience...
 
  • #3
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.
 
  • #4
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!
 
  • #5
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).
 
  • #6
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..
 

What is a PID?

A PID, or principal ideal domain, is a type of ring in abstract algebra that has a unique factorization property. This means that every nonzero element in the ring can be written as a product of irreducible elements, and this factorization is unique up to rearrangement and multiplication by units.

What is a quotient ring?

A quotient ring is a ring obtained by dividing a larger ring by one of its ideals. The elements of the quotient ring are the cosets of the ideal, and the operations are defined by the corresponding operations in the larger ring.

How do you prove that the quotient ring of a PID is also a PID?

To prove that the quotient ring of a PID is also a PID, we need to show that every ideal in the quotient ring is principal. This can be done by considering the ideal generated by any element in the ideal. If this ideal is equal to the original ideal, then the ideal is principal. Additionally, we need to show that every nonzero element in the quotient ring can be written as a product of irreducible elements, which follows from the unique factorization property of the PID.

Why is it important to prove that the quotient ring of a PID is also a PID?

Proving that the quotient ring of a PID is also a PID is important because it allows us to apply the unique factorization property to elements in the quotient ring. This makes it easier to study the quotient ring and understand its structure, as well as solving equations and problems involving the quotient ring.

What are some examples of quotient rings of PIDs?

Some examples of quotient rings of PIDs include the quotient ring of integers modulo a prime number, the quotient ring of polynomials over a field, and the quotient ring of Gaussian integers (complex numbers of the form a + bi, where a and b are integers) modulo a prime number.

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