oct10 - D-1 R , we may extend to a map : D-1 R S-1 R ....

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Math 5125 Monday, October 10 October 10, Ungraded Homework Exercise 7.5.2 on page 264 Let R be an integral domain and let D be a multiplicatively closed subset of R which contains 1 but not 0. Prove that the ring of fractions D - 1 R is isomorphic to a subring of the quotient field of R (hence is also an integral domain). Let S = R - 0 and let φ : R S - 1 R denote the natural homomorphism defined by φ ( r ) = r / 1. Note that S - 1 R is the quotient field of the integral domain R , and φ is a monomorphism. In this situation we usually identify R with φ ( R ) , but we won’t do so here. Since φ ( d ) 6 = 0 for all d D , we see that φ ( d ) is invertible for all d D . By the universal property for
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Unformatted text preview: D-1 R , we may extend to a map : D-1 R S-1 R . Specically if : R D-1 R is the natural homomorphism dened by ( r ) = r / 1, then = . All that remains to prove is that is a monomorphism, equivalently ker = 0, so suppose r / d ker where r R and d D . Since ker D-1 R , we see that ( r / d )( d / 1 ) = r / 1 ker and hence ( r ) ker . Therefore ( r ) = ( r ) = 0 and since is a monomorphism, we deduce that r = 0. We conclude that r / d = 0 and the result follows....
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