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Unformatted text preview: Homework #4 Solutions p 286, #8 Let : Z n Z n be a ring homomorphism. Let a = (1). Then for any 6 = r Z n = { 1 , 2 , . . . , n 1 } we have ( r ) = (1 + 1 + + 1  {z } r times ) = (1) + (1) + + (1)  {z } r times = r (1) = r a = ra mod n. Since a = (1) = (1 1) = (1) (1) = a 2 were finished. p 286, #10 Let I = h x 2 +1 i and let f ( x ) Z 3 [ x ]. By including zero coefficients if necessary we can write f ( x ) = n X i =0 a 2 i x 2 i + m X j =0 a 2 j +1 x 2 j +1 , for some a i Z 3 , i.e. we can write f ( x ) as the sum of its even degree and odd degree terms. In Z 3 [ x ] /I we have x 2 + I = 1 + I so that f ( x ) + I = n X i =0 ( a 2 i + I )( x 2 i + I ) + m X j =0 ( a 2 j +1 + I )( x 2 j +1 + I ) = n X i =0 ( a 2 i + I )( x 2 + I ) i + m X j =0 ( a 2 j +1 + I )( x + I )( x 2 + I ) j = n X i =0 ( a 2 i + I )( 1 + I ) i + m X j =0 ( a 2 j +1 + I )( x + I )( 1 + I ) j = n X i =0 (( 1) i a 2 i + I ) ! + ( x + I ) m X j =0 (( 1) j a 2 j +1 + I ) ! = n X i =0 ( 1) i a 2 i + x m X j =0 ( 1) j a 2 j +1 ! + I or, more succinctly, f ( x ) + I = a + bx + I for some a, b Z 3 . Moreover, if a + bx + I = c + dx + I for some a, b, c, d Z 3 then ( a c )+( b d ) x I , which means that x 2 +1 divides the linear polynomial ( a c )+( b d ) x , an obvious impossibility unless a c = b d = 0. That is, a + bx + I = c + dx + I implies that a + bx = c + dx . Hence, every element in Z 3 [ x ] /I can be expressed uniquely in the form a + bx + I , a, b Z 3 . We will use this fact below. Now define : Z 3 [ i ] Z 3 [ x ] /I by ( a + bi ) = a + bx + I . This is a homomorphism since for any a, b, c, d Z 3 we have (( a + bi )( c + di )) = (( ac bd ) + ( ad + bc ) i ) = ( ac bd ) + ( ad + bc ) x + I = ( a + bx + I...
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 Spring '07
 RyanDaileda

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