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Let $R$ be a commutative ring with identity such that $R$ has exactly one prime ideal $P$.

Prove: all elements in $P$ are nilpotent.

While doing this problem, I used the fact that "the nilradical of $R$ is equal to the intersection of all prime ideals of $R$" (in this case, the intersection of all prime ideals $=P$), then I can solve this problem.

However, it seems to me that this fact overkills this problem because we have a strong condition that there is only one prime ideal.

I am here to ask if there is a much more simple and direct approach (which I have overlooked) to solving this problem.


Let me put it in another way: is there any proof without using Zorn's lemma?

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    Some intuition: There are models of ZF with some commutative non-trivial ring $R$, which has no prime ideal at all. So if we have some $R$ which has exactly one prime ideal $\mathfrak{p}$, this could just be since the existence of other prime ideals would need AC. I expect that some logician will show you how to produce a counterexample along these lines.2012-06-18

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When reading this post I began to wonder if $Nil(R)=\bigcap\{ P\mid P\text{ prime}\}$ was equivalent to choice, but that led me to this interesting post.

If I read it correctly, the nilradical equation is not equivalent to AC!

Hope this helps!

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    @MartinBrandenburg can you please give reference for fact that it is equivalent(i.e. implies) Ultrafilter Principle2015-09-23
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with this assumption one can show more: all element of $R$ is invertible or nilpotent .

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    How would you show an element not in the prime is invertible without using the fact that every proper ideal is contained in a prime ideal, whose proof (at least the one I have seen) uses Zorn's lemma?2012-06-18
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I would try something like the following. Let $a$ be in $P$ and suppose that $a$ is not nilpotent. Let $S$ be the multiplicative set of powers of $a$. Then $S$ does not contain $0$. By Zorn (I don't know if this is an acceptable strategy for you), there is a prime ideal $Q$, maximal with respect to not intersecting $S$. But then $Q$ cannot equal $P$, contradicting the fact that $P$ is the unique prime of $R$. Thus, $a$ is nilpotent.

This argument is basically a specialization of the result you quoted (see, e.g., p. 148 of Lang, Algebra, 1971 edition). Mohamed's result that if $a$ is not nilpotent then it must be a unit is pretty straightforward. Incidently, Theorem 1 of Kaplansky's "Commutative Rings" is the statement that an ideal $I$ that is maximal with respect to the property that it is disjoint from a multiiplicatively closed set $S$ is prime. Of course, one needs a way to get such a maximal element, e.g., Zorn. I don't know if Zorn can be avoided completely in this situation.

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    Sure, but not being an answer, this should probably have been a comment. Currently we have two comments masquerading as answers, when there really should be no answers (which helps to attract folks to the question).2012-06-18