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im asked to prove that the set consisting of all zero divisors in a commutative ring with unity contains at least one prime ideal.

i cant even start in the proof , ive just defined my set but cant move on construction the ideal !

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Nice Question.. Integral domain has $(??)$ as prime ideal... Can you see how and can you think of what would be the case of general commutative ring? – Praphulla Koushik Jan 4 '14 at 16:59
@PraphullaKoushik for integral domains (p) is a prime ideal iff p is a prime number, then? – Enas Jan 4 '14 at 17:10
Check to see that your set is an ideal, then use the fact that $R/I$ is an integral domain if and only if $I$ is prime. By the way I think there are integral domains with prime ideals not generated by a single prime. It holds for principal ideal domains though. – user111013 Jan 4 '14 at 17:19
up vote 4 down vote accepted

One way to approach this is to combine these two lemmas:

  1. In a commutative ring, there exist minimal prime ideals.

  2. In a commutative ring with identity, minimal primes consist entirely of zero divisors (I use the convention that $0$ is a zero divisor here.)

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If you don't mind me asking a follow-up question: is it true or false that an ideal which is maximal with respect to being contained in the set of zero divisors is necessarily prime? That's what I started trying to prove... I can post a new question if that's preferred, but I figured I might as well ask you first. – Dustan Levenstein Jan 4 '14 at 17:33
@DustanLevenstein : I keep thinking about this extended question, but I don't have a conclusion either way. A counterexample would have to lie outside of reduced Noetherian rings. In those rings, the zero divisors are a union of prime ideals, and then prime avoidance tells you that an ideal maximal w.r.t. being all zero divisors is exactly one of these ideals. – rschwieb Jan 5 '14 at 0:21
Posted a question. – Dustan Levenstein Jan 5 '14 at 4:47

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