Adv Alegbra HW Solutions 102

# Adv Alegbra HW Solutions 102 - 102 Therefore ϕ is a ring...

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Unformatted text preview: 102 Therefore, ϕ is a ring homomorphism. Finally, ϕ is an isomorphism, for it is easy to see that its inverse is the function ψ given by ψ( ∑ i s i x i ) = ∑ i s i ( x − c ) i . 3.49 If R is a fi eld, show that R ∼ = Frac ( R ) . More precisely, show that the homomorphism f : R → Frac ( R ) , given by r 7→ [ r , 1 ] , is an isomorphism. Solution. If R is a fi eld, de fi ne ψ : Frac ( R ) → R by [ r , s ] 7→ rs − 1 . Note that rs − 1 is de fi ned, because s = 0 and R is a fi eld. Also, ψ is single valued, for if [ r , s ] = [ a , b ] , then rb = sa , hence rs − 1 = ab − 1 , and so ψ( [ r , s ] ) = ψ( [ a , b ] ) . It is routine to check that ψ is a ring homomorphism inverse to f : R → Frac ( R ) . An alternative solution can be based on the observation that the homo- morphism f is surjective, for [ r , s ] = [ rs − 1 , 1 ] ∈ im f . 3.50 Let R be a domain and let F be a fi eld containing R as a subring....
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