Adv Alegbra HW Solutions 102

Adv Alegbra HW Solutions 102 - 102 Therefore, is a ring...

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Unformatted text preview: 102 Therefore, is a ring homomorphism. Finally, is an isomorphism, for it is easy to see that its inverse is the function given by ( 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 : 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, is single valued, for if [ r , s ] = [ a , b ] , then rb = sa , hence rs 1 = ab 1 , and so ( [ r , s ] ) = ( [ a , b ] ) . It is routine to check that 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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This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.

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