6
$\begingroup$

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?

  • 3
    So you don't want to use the fact that any proper ideal is contained in a maximal ideal?2012-06-18
  • 0
    Doesn't seem to help much, but clearly you may assume that the ring is reduced.2012-06-18
  • 0
    There won't be any proof in ZF, only in ZF + Ultrafilter Principle.2012-06-18
  • 0
    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

3 Answers 3