apr20 - is a prime ideal of R and P contains no zero...

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Math 4124 Wednesday, April 20 April 20, Ungraded Homework Exercise 7.4.4 on page 256 Let R be a commutative ring with identity 1 = 0. Prove that R is a field if and only if 0 is a maximal ideal. First suppose R is a field. To prove that 0 is a maximal ideal, we must show that there are no ideals strictly between 0 and R , equivalently 0 and R are the only ideals of R . Let I be a nonzero ideal of R . Let 0 = x I . Then xx - 1 = 1 I and we deduce that 1 r I for all r R . This shows that I = R and it follows that 0 is a maximal ideal of R . Conversely suppose 0 is a maximal ideal of R . To show that R is a field, we need to prove that every nonzero element of R has a multiplicative inverse, so let 0 = x R . Since xR is a nonzero ideal of R (remember that R is commutative), we see that xR = R and hence there exists y R such that xy = 1. This proves that x is invertible as required. Exercise 7.4.10 on page 257 Let R be a commutative ring with identity 1 = 0. Prove that
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Unformatted text preview: is a prime ideal of R and P contains no zero divisors, then R is an integral domain. Let 0 6 = a , b ∈ R . We want to prove that ab 6 = 0. If a or b ∈ P , this is true because P contains no (nonzero) zero divisors. Therefore we may assume that a , b / ∈ P . But then ab / ∈ P because P is a prime ideal and the result follows. Let R be a commutative ring identity 1 6 = 0 and let M be a maximal ideal of R . Suppose M is nilpotent, that is there exists n ∈ N such that M n = 0. Prove that M is the only prime ideal of R . Let P be a prime ideal of R . If M is not contained in P , we may choose x ∈ M \ P . Then x n = 0, hence x n ∈ P and since P is prime, we deduce that x ∈ P . This is a contradiction. Therefore M ⊆ P and since M is maximal and P 6 = R , we deduce that P = M and the result follows....
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