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....
View
Full
Document
This note was uploaded on 01/02/2012 for the course MATH 4124 taught by Professor Staff during the Spring '08 term at Virginia Tech.
 Spring '08
 Staff
 Math, Algebra

Click to edit the document details