# Adv Alegbra HW Solutions 101 - 101 3.47 Let R S be a...

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101 3.47 Let ψ : R S be a homomorphism, where R and S are commutative rings, and let ker ψ = I . If J is an ideal in S , prove that ψ 1 ( J ) is an ideal in R which contains I . Solution. Absent. 3.48 If R is a commutative ring and c R , prove that the function ϕ : R [ x ] → R [ x ] , de fi ned by f ( x ) f ( x + c ) , is an isomorphism. In more detail, ϕ( i s i x i ) = i s i ( x + c ) i . Solution. It is clear that ϕ( 1 ) = 1. If f ( x ) = s j x j and g ( x ) = r j x j , then ϕ( f + g ) = ϕ( ( s i + r i ) x i ) = ( s i + r i )( x + c ) i = s i ( x + c ) i + r i ( x + c ) i = ϕ( f ) + ϕ( g ), as desired. To prove that ϕ( f g ) = f )(ϕ g ) , fi rst consider the special case f ( x ) = sx k , where s R and k 0. If g ( x ) = r i x i , then ϕ( sx k f ( x )) = ϕ( sr i x i + k ) = sr i ( x + c ) i + k = s ( x + c ) k r i ( x + c ) i = ϕ( sx k )ϕ( g ). We now prove that ϕ( f g ) = ϕ( f )ϕ( g ) by induction on deg ( f ) . The base step deg ( f ) = 0 is the special case just proved when
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