November 9th, 2007, 08:43 AM  #1 
Newbie Joined: Oct 2007 Posts: 3 Thanks: 0  field of fractions
Let R be an integral domain, F its field of fractions. Let S be a subset of R. Assume that 0 is not in S, that 1 is in S, and that every product st with s and t in S is itself in S. Prove that the fractions that can be written in the form b/s with s ∈ S is a subring of F containing R.

November 9th, 2007, 10:31 AM  #2 
Newbie Joined: Sep 2007 From: Nice, France Posts: 13 Thanks: 0 
Let call T this set of fractions. I assume that b is supposed to belongs to R. First, R C T because for all b in R, b=b/1. To see that T is a subring of F, you have to show that T is stable, ie : 1) 0 and 1 belongs to F 2) if b/s and c/t belongs to F, b/s*c/t=(bc)/(st) belongs to F 3) if b/s and c/t belongs to F, b/s+c/t=(bt+sc)/st belongs to F It seems very simple, no ? 1) 1 belongs to F by hypothesis, and 0=0/1 2) S is stable by product so st belongs to F (and obviously bc belongs to R) 3) idem (and... obviously too bt+sc belongs to R) 

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