nov10 - Math 3124 Thursday, November 10 November 10,...

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Math 3124 Thursday, November 10 November 10, Ungraded Homework Problem 25.3 on page 127 Verify that ([ 2 ] , [ 0 ]) is a zero divisor in Z 3 × Z 3 . Obviously ([ 2 ] , [ 0 ]) 6 = ([ 0 ] , [ 0 ]) . Also ([ 2 ] , [ 0 ])([ 0 ] , [ 2 ]) = ([ 0 ] , [ 0 ]) . Since ([ 0 ] , [ 2 ]) 6 = 0, this shows that ([ 2 ] , [ 0 ]) is a zero divisor in Z 3 × Z 3 . Problem 25.6 on page 127 What is the smallest subring of R containing 1/2? Let R be the smallest subring of R containing 1/2. Then 1 / 2 + 1 / 2 = 1 R . Since R is an abelian group under +, we deduce that Z R . Also R is closed under multiplication, consequently a / 2 b R for all a Z and all non-negative integers b . Now write S for this subset, that is S = { a / 2 b | a , b Z , b 0 } . Then S R . All that remains to show is that S is a subring of R . First note that 0 S . Suppose a / 2 b and c / 2 d S . Then a / 2 b + c / 2 d = ( 2 d a + 2 b c ) / 2 b + d and - ( a / 2 b ) = ( - a ) / 2 b , which shows that S is closed under + and taking
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nov10 - Math 3124 Thursday, November 10 November 10,...

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