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Exam3-Solns

# Exam3-Solns - Exam 3 Solutions 1 Since 0 A and is a...

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Exam 3 Solutions 1. Since 0 A and ϕ is a homomorphism, ϕ (0) = 0 ϕ ( A ) thus ϕ ( A ) 6 = Ø. Let ϕ ( a ) , ϕ ( b ) ϕ ( A ) and hence a, b A . ϕ ( a ) - ϕ ( b ) = ϕ ( a - b ) because ϕ is a homomor- phism. a - b A because A is an ideal, hence ϕ ( a ) - ϕ ( b ) ϕ ( A ) and ϕ ( A ) is an additive subgroup. Let s S and ϕ ( a ) ϕ ( A ). Since ϕ is onto, there exists r R such that ϕ ( r ) = s . A is an ideal, so ra , ar A . Thus ( a ) = ϕ ( r ) ϕ ( a ) = ϕ ( ra ) ϕ ( A ) and ϕ ( a ) s = ϕ ( a ) ϕ ( r ) = ϕ ( ar ) ϕ ( A ). Therefore ϕ ( A ) is an ideal of S . 2. A = { (3 x, y ) | x, y Z } is a maximal ideal. Ideal : note that for (3 a, b ) , (3 c, d ) A that (3 a, b ) - (3 c, d ) = (3( a - c ) , b - d ) A and for ( a, b ) Z × Z and (3 c, d ) A ( a, b )(3 c, d ) = (3 c, d )( a, b ) = (3 ac, bd ) A (and clearly A is nonempty since (0 , 0) is in A ). Thus A is an ideal. Maximal : Suppose A ( B and B is an ideal of Z × Z . Then there exists ( b 1 , b 2 ) B such that (3 , b 1 ) = 1. Therefore there exists α, β Z such that 3 α + b 1 β = 1. Thus (1 , 1) = (3 α + b 1 β, 1 - b 2 + b 2 ) = (3 α, 1 - b 2 )+( b 1 β, b 2 ) = (3 α, 1 - b 2 )+( β, 1)( b 1 , b 2 ) B .

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