oct05 - a , b R-Q . Then a / P j and b / P k for some j , k...

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Math 5125 Wednesday, October 5 October 5, Ungraded Homework Exercise 7.4.36 on page 259 Assume that R 6 = 0 is commutative. Prove that the set of prime ideals in R has a minimal element with respect to inclusion. Let A denote the set of prime ideals of R . Since R 6 = 0, it has maximal ideals. Furthermore every maximal ideal is a prime ideal, consequently A 6 = /0. Partially order the prime ideals of A by reverse inclusion; that is P Q means Q P . Suppose { P j | j J } is a chain in A (where J is an indexing set). Let Q = T j P j . Then Q is certainly an ideal of R (the intersection of ideals is always an ideal), so we need to check that it is prime. Suppose
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Unformatted text preview: a , b R-Q . Then a / P j and b / P k for some j , k J . Since { P j } is a chain, without loss of generality we may assume that P j P k . Then a , b / P j and since P j is a prime ideal, we deduce that ab / P j and hence ab / Q . Therefore Q A and is an upper bound for the chain. We conclude by Zorns Lemma that A has maximal elements. This means that R has minimal prime ideals with respect to inclusion....
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