- #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!
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!