# Thread: Ring of fractions & pid

1. ## Ring of fractions & pid

Let R is a principal ideal domain(PID) and D is a multiplicatively closed subset of R.Then prove that the RING OF FRACTIONS OF D WITH RESPECT TO R(D INVERSE R) is a PID.

How to do the sum?

I have proved it when D=R-0. When D is so, the ring of fractions has a field structure.And a field is a euclidean domain, & hence a PID.

But how to prove this for any subset D?

2. ## Re: Ring of fractions & pid

Since $\displaystyle R$ is an integral domain, we can view $\displaystyle R\subseteq D^{-1}R$.

Let $\displaystyle J$ be an ideal of $\displaystyle D^{-1}R$. Show that $\displaystyle I=J\cap R$ is an ideal of $\displaystyle R$, and then show that $\displaystyle J=D^{-1}I$. This is showing that every ideal in this ring of fractions corresponds to an ideal of $\displaystyle R$ in this particular way.

Once you have done this, it should be fairly easy to show that $\displaystyle J$ is principal.

3. ## Re: Ring of fractions & pid

What is 'D inverse I'. Is it the ring of fractions of D w.r.t I. For that we need D to be a subset of I which may not be true here. Just as when we talk about the ring of fractions of D w.r.t R , we need D to be a subset of R.

4. ## Re: Ring of fractions & pid

Originally Posted by jishnuray
What is 'D inverse I'. Is it the ring of fractions of D w.r.t I. For that we need D to be a subset of I which may not be true here. Just as when we talk about the ring of fractions of D w.r.t R , we need D to be a subset of R.
We don't need $\displaystyle D$ to be a subset of $\displaystyle I$. The definition is

$\displaystyle D^{-1}I=\{\frac{i}{d}\mid i\in I,d\in D\}$. This is certainly a subset of the ring $\displaystyle D^{-1}R$ and, if $\displaystyle I$ is an ideal in $\displaystyle R$, then $\displaystyle D^{-1}I$ is an ideal of $\displaystyle D^{-1}R$.

You can even extend this idea to discuss such things as modules of fractions -- modules where the numerators come from some $\displaystyle R$-module, and the denominators still come from $\displaystyle D$.

5. ## Re: Ring of fractions & pid

one of the shortcomings of rings, is that kernels are not sub-objects, they are just subgroups of the additive groups, with a certain multiplicative property.

convince yourself that the set of equivalence classes [i/d], where i is in I, is indeed a subgroup of $\displaystyle (D^{-1}R,+)$

and that for all $\displaystyle [r/d_1] \in D^{-1}R$, we have $\displaystyle [r/d_1][i/d] = [ri/(d_1d)] \in D^{-1}I$.

"technically" $\displaystyle R$ isn't actually in $\displaystyle D^{-1}R$, but $\displaystyle r \mapsto [rd/d]$ is a (ring) monomorphism (since a PID is an integral domain).