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Unformatted text preview: u 2 = 1 = u 21 = 0 = ( u1)( u + 1) = 0 = u1 = 0 or u + 1 = 0. (The last step uses the fact that R has no 0divisors.) So u = 1 or u =1. (c) Z 8 has 0divisors. 5. (a) Routine calculation. (b) R is closed under addition, multiplication, and negation, and it contains 0 and 1. All the ring axioms now follow from the fact that they hold for complex numbers. 6. No. If one existed, it would have to satisfy f ([0] 2 ) = 0 and f ([1] 2 ) = 1. But that function does not preserve sums: f ([1])+ f ([1]) = 1+1 = 2, but f ([1]+[1]) = f ([0]) = 0 6 = 2....
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 Spring '08
 BILLERA
 Algebra

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