# $(R,\mathcal m)$ be a Noetherian local ring and let $P$ be a prime ideal of $R$. If $P^2$ is a prime ideal of $R$, then $P=0$

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.

You do not need to further localize, since your ring is already local.

We have $P^2 \supset P \cdot P$, by the prime property we get $P^2 \supset P$. In particular $P \supset mP \supset P^2 \supset P$, hence $mP=P$. Nakayama finishes the proof.

Let $x\in p.$ Then $x^2\in p^2$ and since $p^2$ is prime we have $x\in p^2.$ So $p= p^2,$ and by Nakayama $p=0.$