Suppose $P$ is a nontrivial prime ideal in an arbitrary ring. Can $P^2 = P?$ I suppose this might be true for some ring but I cannot think of an example...
Context: In a Dedekind domain, $P^2 \neq P$ as otherwise this would violate unique prime factorization of ideals. This is why I was wondering if equality could actually hold for some ring.
Here's a commutative ring with a non-zero maximal ideal $M$ with $M^2=M$. Let $k$ be a field and $R$ consist of "polynomials with rational exponents" over $k$, that is each element of $R$ is a formal $k$-linear combination of $x^r$ where $r\in\Bbb Q$ and $r\ge0$. The map $R\to k$ taking $a_0+\sum_{r>0}a_r x^r$ to $a_0$ is a surjective homomorphism with kernel $M$ generated by the $x^r$ for $r>0$. Each of these is a square of an element of $M$, so $M^2=M$.
Of course, this $R$ is non-Noetherian.