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Unformatted text preview: Math 4124 Monday, April 25 April 25, 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 = for all d ∈ D , we see that φ ( d ) is invertible for all d ∈ D . By the universal property for D 1 R , we may extend φ to a map α : D 1 R → S 1 R . Specifically if θ : R → D 1 R is the natural homomorphism defined by θ ( r ) = r / 1, then φ = αθ . All that remains to prove is that α is a monomorphism, equivalently ker α = 0, so suppose r / d ∈ ker α where r ∈...
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 Spring '08
 Staff
 Algebra, Fractions, Homomorphism, Integral domain, Commutative ring, prime ideal

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