TITLE: Direct proof that, in a commutative ring with only one prime ideal $P$, every element of $P$ is nilpotent
QUESTION [6 upvotes]: 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?

REPLY [5 votes]: 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!