Property of a Noetherian ring: How come $P \setminus P^2$ is non-empty? ($P$ is a prime ideal)

Question

Let $A$ be a Noetherian ring, and let $P$ be a prime ideal. How come we know that $P \setminus P^2$ is non-empty? Thank you!

Answer 1

You can't, without the stronger hypothesis that $A$ is also a connected ring, i.e. a ring which has no nontrivial idempotents. For any Noetherian ring $A$ and ideal $I \subseteq A$, $I^{2} = I$ if and only if $I$ is a principal ideal generated by an idempotent. This is a consequence of Nakayama's lemma, and has been covered before on this site: see Georges Elencwajg's excellent answer here, for instance. If you assume $A$ is connected, this proves your claim, provided you add the extra hypothesis that $P$ is a nonzero prime ideal.

For an explicit counterexample when $A$ is not connected, take $A = \mathbb{Z}/6\mathbb{Z}, P = \langle [3]_{6} \rangle$. $A$ is obviously Noetherian, and it is not hard to see that $P$ is indeed a prime ideal of $A$, as $A/P \cong \mathbb{Z}/3\mathbb{Z}$. But $P$ is generated by an idempotent element of $A$, so it is clear that $P^{2} = P$.