Adv Alegbra HW Solutions 88

Adv Alegbra HW Solutions 88 - 88 On the other hand, a (b c)...

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88 On the other hand, a ( b c ) = a ( b + c bc ) = a + b + c bc a ( b + c bc ) = a + b + c bc ab ac + a ( bc ). (ii) Prove that a commutative ring R is a f eld if and only if the set R # ={ r R : r ±= 1 } is an abelian group under the circle operation. Solution. Assume that R is a f eld. To show that R # is a group, we must f rst show that is an operation on R # ; that is, if a ±= 1 and c ±= 1, then a c ±= 1. If, on the contrary, 1 = a c = a + c = a + c ( 1 a ), then 1 a = c ( 1 a ) . Since a ±= 1, we may cancel 1 a , and this gives the contradiction c = 1. Part (i) shows that is associative and that 0 acts as the identity. It remains to f nd the inverse of an element a ±= 1; that is, to f nd b ±= 1 such that 0 = a b = a + b .De f ne b = a ( a 1 ) 1 . Conversely, assume that R # is a group. It suf f ces to f nd a (multi- plicative) inverse for every a ±= 0. Now a ±= 0 implies a + 1 ±= 1, and so there exists b R # with ( a + 1 ) b = 0; that is, 0 = a + 1 + b ( a + 1 ) b = a + 1 + b
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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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