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# nov17 - Z 12 4(which in this case is the same as Z 12 h 4 i...

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Math 3124 Thursday, November 17 November 17, Ungraded Homework 1. Problem 39.1 on page 183. Deﬁne θ : Z 12 Z 12 by θ ([ a ] 12 ) = [ a ] 4 for each [ a ] 12 Z 12 . (a) Verify θ is well deﬁned. (b) Verify that θ is a homomorphism. (c) Use the Fundamental Homomorphism Theorem for rings to explain why Z 12 / ([ 4 ]) = Z 4 . (d) Construct the Cayley tables for the ring operations on Z 12 / ([ 4 ]) . (a) We need to show that if [ a ] 12 = [ b ] 12 , then θ [ a ] = θ [ b ] , that is [ a ] 4 = [ b ] 4 . However if [ a ] 12 = [ b ] 12 , then 12 | a - b , consequently 4 | a - b and we deduce that [ a ] 4 = [ b ] 4 as required. (b) θ ([ a ] 12 [ b ] 12 ) = θ ([ a + b ] 12 ) = [ a + b ] 4 = [ a ] 4 [ b ] 4 = θ ([ a ] 12 ) θ ([ b ] 12 ) , so θ respects addition. θ ([ a ] 12 ± [ b ] 12 ) = θ ([ ab ] 12 ) = [ ab ] 4 = [ a ] 4 ± [ b ] 4 = θ ([ a ] 12 ) ± θ ([ a ] 12 ) so θ re- spects multiplication. We have now established that θ is a homomorphism. (c) Clearly θ is onto. The fundamental homomorphism theorem for rings now tells us that Z 12 / ker θ = Z 4 , so we need to determine ker θ . But [ a ] 12 ker θ if and only if [ a ] 4 = [ 0 ] 4 , which is if an only if 4 | a , so ker θ = { [ 0 ] 12 , [ 4 ] 12 , [ 8 ] 12 } . This last set is by deﬁnition ([ 4 ] 12 ) , so the result is proven. (d) As in problem 22.3 on page 108, we may write the elements of
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Unformatted text preview: Z 12 / ([ 4 ]) (which in this case is the same as Z 12 / h [ 4 ] i ) as 0 = { [ ] , [ 4 ] , [ 8 ] } , 1 = { [ 1 ] , [ 5 ] , [ 9 ] } , 2 = { [ 2 ] , [ 6 ] , [ 10 ] } , 3 = { [ 3 ] , [ 7 ] , [ 11 ] } . Then the Cayley tables will be ⊕ 1 2 3 1 2 3 1 1 2 3 2 2 3 1 3 3 1 2 and ± 1 2 3 1 1 2 3 2 2 2 3 3 2 1 2. Problem 39.3 on page 183. Prove that if R is commutative and I is an ideal of R , then R / I is commutative. The general element of R / I is of the form I + a where a ∈ R . A ring is commutative means that the operation of multiplication is commutative. Therefore we need to prove that ( I + a )( I + b ) = ( I + b )( I + a ) for all a , b ∈ R , in other words I + ab = I + ba for all a , b ∈ R . Since R is commutative, we have ab = ba and the result follows....
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