Let $(R,\mathcal m)$ be a Noetherian local ring and let $P$ be a prime ideal of $R$. If $P^2$ is a prime ideal of $R$, then $P=0$.
I was thinking to use Nakayama lemma as:
$R_P$ is local with $PR_P$ as its maximal ideal. Also as $P^2$ is prime so $P^2R_P$ will also be the same maximal ideal. Then $PR_P=P^2R_P= P.PR_P$. Using Nakayama lemma here, we get $PR_p=0$. This implies $P=0$, as $PR_P$ is finitely generated.
Is it correct? Otherwise please provide some hints.
Thanks in advance.