Let $R$ be a Noetherian ring, and $M$ a finitely generated $R$-module. If $P$ is prime minimal over $\operatorname{ann}M$, then $P=\operatorname{rad}(\operatorname{ann}M)$.
$\textit{Proof:}$ Since $\operatorname{ann}M \subset P$ is minimal, if we localize at $P$, then $(\operatorname{ann}M)R_P \subset PR_P$ and $PR_P$ is the unique prime ideal containing $(\operatorname{ann}M)R_P$. Hence $PR_P = \operatorname{rad}((\operatorname{ann}M)R_P)$. Since radical commutes with localization, we have $$PR_P = \operatorname{rad}((\operatorname{ann}M)R_P) = (\operatorname{rad}(\operatorname{ann}M))R_P.$$
So, $P = \operatorname{rad}(\operatorname{ann}M)$.
A counter example is $R=\mathbb Z$, $P=(2)$, $M = \mathbb Z/6\mathbb Z$ where we have $\operatorname{rad}(\operatorname{ann}M)=(6) \ne (2)$.
Best Answer
You obtain $$PR_P=(\operatorname{rad}(\operatorname{ann} M)R_P$$ and conclude from this that $P=\operatorname{rad}(\operatorname{ann} M)$. But this implication is not true; indeed, in your example where $R=\mathbb{Z}$, $P=(2)$ and $\operatorname{rad}(\operatorname{ann} M)=(6)$, you have $(2)R_P=(6)R_P$, as $3$ is a unit in $R_P$, but of course $(2) \ne (6)$ in $\mathbb{Z}$.