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
    $\begingroup$ So you don't want to use the fact that any proper ideal is contained in a maximal ideal? $\endgroup$ – Arturo Magidin Jun 18 '12 at 16:44
  • $\begingroup$ Doesn't seem to help much, but clearly you may assume that the ring is reduced. $\endgroup$ – Arturo Magidin Jun 18 '12 at 18:54
  • $\begingroup$ There won't be any proof in ZF, only in ZF + Ultrafilter Principle. $\endgroup$ – Martin Brandenburg Jun 18 '12 at 18:57
  • $\begingroup$ 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. $\endgroup$ – Martin Brandenburg Jun 18 '12 at 19:10
  • $\begingroup$ 8 years late but just in case someone wanders in and is confused by it, it's not true that in a ring with only one prime all elements are nilpotent (indeed, in a nonzero ring we always have a non-nilpotent 1). Instead, every element is either a unit or a nilpotent. $\endgroup$ – Wojowu Sep 20 '20 at 8:25

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!

  • 1
    $\begingroup$ Yes; it is equivalent to the Ultrafilter Principle, which is (strictly) weaker than AC. $\endgroup$ – Martin Brandenburg Jun 18 '12 at 18:58
  • $\begingroup$ @MartinBrandenburg Then, following that post, BenLi should have a proof without choice, yes? $\endgroup$ – rschwieb Jun 18 '12 at 19:12
  • $\begingroup$ @Benli See above comment! (I think we should be able to @ two people in each comment....) $\endgroup$ – rschwieb Jun 18 '12 at 19:12
  • $\begingroup$ At the beginning, I thought I overlooked a really direct and simple proof without using that characterization of nilradical but now I would say that characterization of nilradical seems to be the most direct approach.Anyway, the link in your post really give me a proof without Zorn lemma so you have answered my question :) $\endgroup$ – Ben Jun 19 '12 at 1:30
  • $\begingroup$ @MartinBrandenburg can you please give reference for fact that it is equivalent(i.e. implies) Ultrafilter Principle $\endgroup$ – Sushil Sep 23 '15 at 3:02

with this assumption one can show more: all element of $R$ is invertible or nilpotent .

  • 2
    $\begingroup$ 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? $\endgroup$ – Rankeya Jun 18 '12 at 18:36

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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    $\begingroup$ But this is precisely a well-known proof of the (nontrivial direction of the) nilradical characterization, which the OP knows and is trying to avoid. $\endgroup$ – Bill Dubuque Jun 18 '12 at 18:31
  • $\begingroup$ @BillDubuque - Exactly. My main point is that I don't know if Zorn can be avoided. I would thoroughly enjoy seeing a proof that doesn't use Zorn. As with a lot of things, a Zorn-free proof might be insightful, but I always worry that such a proof might be somewhat messy. I don't see how to do it "from scratch," but I'm a bit preoccupied to try it right now. $\endgroup$ – Chris Leary Jun 18 '12 at 19:55
  • $\begingroup$ 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). $\endgroup$ – Bill Dubuque Jun 18 '12 at 19:58

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