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# apr18 - a = b for some elements a b ∈ R if and only if a...

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Math 4124 Monday, April 18 April 18, Ungraded Homework Exercise 7.4.7 on page 256 Let R be a commutative ring with identity 1 = 0. Prove that the principal ideal generated by x in the polynomial ring R [ x ] is a prime ideal if and only if R is an integral domain. Prove that ( x ) is a maximal ideal if and only if R is a field. Define a ring homomorphism θ : R [ x ] R by θ ( r ) = r for r R and θ ( x ) = 0; another way to describe this homomorphism is θ ( f ) = f ( 0 ) . Then θ is onto and ker θ = ( x ) . Therefore R [ x ] / ( x ) = R by the fundamental isomorphism theorem. Since ( x ) is a prime ideal if and only if R [ x ] / ( x ) is an integral domain, and ( x ) is a maximal ideal if and only if R [ x ] / ( x ) is a field, the result is proven. Exercise 7.4.8 on page 256 Let R be an integral domain. Prove that
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Unformatted text preview: ( a ) = ( b ) for some elements a , b ∈ R if and only if a = ub for some unit u of R . Suppose a = ub for some unit u of R . Then a ∈ bR , so ( a ) ⊆ ( b ) . Also if uv = 1, then b = va and we deduce that ( b ) ⊆ ( a ) . Therefore ( a ) = ( b ) . Conversely suppose ( a ) = ( b ) . The result is obvious if a = 0, because then b = 0 and we can take u = 1. Therefore we may assume that a 6 = 0. Then we may write a = br and b = as for some r , s ∈ R and we have a = ars . Since a 6 = 0 and R is an integral domain, we see that 1 = rs , so r is a unit and the result follows....
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