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Math Help - Ring of fractions & pid

  1. #1
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    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?
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  2. #2
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    Re: Ring of fractions & pid

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

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

    Once you have done this, it should be fairly easy to show that J is principal.
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  3. #3
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    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.
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  4. #4
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    Re: Ring of fractions & pid

    Quote Originally Posted by jishnuray View Post
    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 D to be a subset of I. The definition is

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

    You can even extend this idea to discuss such things as modules of fractions -- modules where the numerators come from some R-module, and the denominators still come from D.
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  5. #5
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    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 (D^{-1}R,+)

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

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