ahw10 - Math 4124 Wednesday, May 2 Tenth Homework Solutions...

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Math 4124 Wednesday, May 2 Tenth Homework Solutions 1. (a) Let s , t S and suppose st / S . Then stx = 0 for some nonzero x R . Then tx 6 = 0 because t is a nonzero divisor, and hence stx because s is a nonzero divisor and tx 6 = 0. It follows that S is a multiplicatively closed subset of R . Obviously it contains 1 but not 0. (b) Let r / s S - 1 R , where r R and s S . If r is a zero divisor, then rx = 0 for some nonzero x R , hence ( r / s )( x / 1 ) = 0 and x / 1 6 = 0. This shows that r / s is a zero divisor. On the other hand if r is not a zero divisor, then r S by definition of S and thus s / r S - 1 R . Since ( r / s )( s / r ) = 1, we see that r / s is a unit and we’re finished. 2. Let K = ker α and J = ker β . Then R / K = Z / 2 Z , a field, and we see that K is a maximal ideal of R . Similarly J is a maximal ideal of R . Since | R / J | 6 = | R / K | , we see that J 6 = K and hence J + K = R . By the Chinese Remainder theorem, we see that
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