November 9th, 2007, 07:43 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, 09:31 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) 

Tags 
field, fractions 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
does cyclic field implies Galois field  rayman  Abstract Algebra  1  March 11th, 2014 03:27 PM 
Fin. gen field extension > intermediate field f.g. also?  watson  Abstract Algebra  1  September 20th, 2012 06:53 PM 
Show R (comm. domain) over a field k is a field if dimR<inft  watson  Abstract Algebra  1  September 14th, 2012 09:07 PM 
Simplifying fractions with variables in fractions  nova3421  Algebra  1  September 9th, 2009 10:03 AM 
Every quadratic field is contained in a cyclotomic field  brunojo  Abstract Algebra  0  June 5th, 2009 06:25 PM 