oct10 - D-1 R we may extend φ to a map α D-1 R → S-1 R...

This preview shows page 1. Sign up to view the full content.

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
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 ∈ R and d ∈ D . Since ker α ± D-1 R , we see that ( r / d )( d / 1 ) = r / 1 ∈ ker α and hence θ ( r ) ∈ ker α . Therefore φ ( r ) = αθ ( r ) = 0 and since φ is a monomorphism, we deduce that r = 0. We conclude that r / d = 0 and the result follows....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online