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I've been trying to solve the following problem.

Let $A$ be a commutative ring with identity and $a \in A$ a non-nilpotent element, i.e., $a^m \neq 0$ for all $m \in \mathbb{Z}^+$. Prove there exists a prime ideal that fails to contain $a$.

I've been trying to somehow define a partial order on the prime ideals of $A$ to apply Zorn's lemma and proves the existence of such a prime ideal, but I am not sure this approach would work. Any ideas?

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    Have you heard about the nilradical? – Jonas Lenz Mar 01 '19 at 07:09
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    Consider the set of all ideals that do not contain the set ${a^m\mid m\in\mathbb{Z}^+}$. Show it satisfies the hypotheses of Zorn’s Lemma, and then show that a maximal element in that set is in fact a prime ideal. – Arturo Magidin Mar 01 '19 at 07:12
  • I have not heard about nilradical. –  Mar 01 '19 at 07:14
  • @ArturoMagidin how do I know the set of all ideals that do not contain $a^m$ for all $m$ is not empty (not considering the $0$ ideal)? –  Mar 01 '19 at 07:15
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    @Peter: Because $(0)$ is one of them... Why do you not want to consider that ideal? It’s a perfectly fine ideal. It will even be prime in some rings. Are you a zeroist, discriminating against the zero ideal? ;-) – Arturo Magidin Mar 01 '19 at 07:17
  • @ArturoMagidin: I got it now. You are right: $(0)$ is an ideal ideal. –  Mar 01 '19 at 07:17
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    @Peter: I said “do not contain”, but it should be “that are disjoint from”... the ideals shouldn’t contain any of the elements in the set. – Arturo Magidin Mar 01 '19 at 07:22
  • Ok, anyway I understood the idea behind the words. –  Mar 01 '19 at 07:23
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    @Peter Hi: at the suggested duplicate, the last paragraphs outline a solution to exactly this. Happy to help if you have questions. – rschwieb Mar 01 '19 at 14:41

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