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Unformatted text preview: Exam 3 Solutions 1. Since 0 A and is a homomorphism, (0) = 0 ( A ) thus ( A ) 6 = . Let ( a ) , ( b ) ( A ) and hence a,b A . ( a ) ( b ) = ( a b ) because is a homomor phism. a b A because A is an ideal, hence ( a ) ( b ) ( A ) and ( A ) is an additive subgroup. Let s S and ( a ) ( A ). Since is onto, there exists r R such that ( r ) = s . A is an ideal, so ra , ar A . Thus s ( a ) = ( r ) ( a ) = ( ra ) ( A ) and ( a ) s = ( a ) ( r ) = ( ar ) ( A ). Therefore ( 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 , Z such that 3 + b 1 = 1. Thus (1 , 1) = (3 + b 1 , 1 b 2 + b 2 ) = (3 , 1 b 2 )+( b 1 ,b 2 ) = (3 , 1 b...
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This note was uploaded on 04/19/2011 for the course MATH 21373 taught by Professor Johnson during the Spring '11 term at Carnegie Mellon.
 Spring '11
 Johnson
 Algebra

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