Explanation Given information: R ' = { [ 0 ] , [ 2 ] , [ 4 ] , [ 6 ] } is a subring of ℤ 8 . The addition and multiplication tables for ring R = { a , b , c , d } . + a b c d a a b c d b b c d a c c d a b d d a b c ⋅ a b c d a a a a a b a c a c c a a a a d a c a c Formula used: Isomorphism: Let θ be a homomorphism from ring R to ring R ' . If θ is a one-to-one correspondence (both onto and one-one), then θ is an isomorphism. Explanation: Let R = { a , b , c , d } and R ' = { [ 0 ] , [ 2 ] , [ 4 ] , [ 6 ] } be rings with respect to addition and multiplication. Addition and multiplication tables for ring R = { a , b , c } : + a b c d a a b c d b b c d a c c d a b d d a b c ⋅ a b c d a a a a a b a c a c c a a a a d a c a c Addition and multiplication tables for ring R ' = { [ 0 ] , [ 2 ] , [ 4 ] , [ 6 ] } : + [ 0 ] [ 2 ] [ 4 ] [ 6 ] [ 0 ] [ 0 ] [ 2 ] [ 4 ] [ 6 ] [ 2 ] [ 2 ] [ 4 ] [ 6 ] [ 0 ] [ 4 ] [ 4 ] [ 6 ] [ 0 ] [ 2 ] [ 6 ] [ 6 ] [ 0 ] [ 2 ] [ 4 ] ⋅ [ 0 ] [ 2 ] [ 4 ] [ 6 ] [ 0 ] [ 0 ] [ 0 ] [ 0 ] [ 0 ] [ 2 ] [ 0 ] [ 4 ] [ 0 ] [ 4 ] [ 4 ] [ 0 ] [ 0 ] [ 0 ] [ 0 ] [ 6 ] [ 0 ] [ 4 ] [ 0 ] [ 4 ] In order to define a mapping θ : R → R ' that is an isomorphism From the addition and multiplication for R and R ' , Ring R and R ' have no unity element. Now, an additive identity element a of R must map an additive identity element [ 0 ] of R ' . ⇒ θ ( a ) = [ 0 ] . θ must map an additive inverse element of R to the additive inverse element of R ' . In R , b + d = a = d + b . ⇒ b and d are additive inverses of each other. In R ' , [ 2 ] + [ 6 ] = [ 8 ] = [ 0 ] = [ 6 ] + [ 2 ] . That is, [ 2 ] and [ 6 ] are additive inverses of each other. So, if we choose θ ( b ) = [ 2 ] , then θ ( b − 1 ) = ( θ ( b ) ) − 1 = ( [ 2 ] ) − 1 = [ 6 ] . ∴ θ ( d ) = [ 6 ] . The remaining element c is its own inverse in R and also [ 4 ] is its own inverse in R ' . ⇒ θ ( c ) = [ 4 ] . Thus, the mapping θ : R → R ' is defined by θ ( a ) = [ 0 ] , θ ( b ) = [ 2 ] , θ ( c ) = [ 4 ] , θ ( d ) = [ 6 ] . Now check the isomorphism of θ : For every element in R , there is a different image in R ' . So, θ : R → R ' is one-one. That is, x ≠ y ⇒ θ ( x ) ≠ θ ( y ) . For any element [ z ] ∈ R ' there exists an element x ∈ R such that θ ( x ) = [ z ]

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ISBN: 9781285463230

Author: Gilbert, Linda, Jimmie

Publisher: Cengage Learning,

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Chapter 6.2, Problem 26E

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An isomorphism from R to R'.

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