Suppose $R$ is a commutative ring, $P$ is a prime ideal of $R$, $P$ is finitely generated, and $\operatorname{Ann}(P)=0$. Show that $$\operatorname{Ann}(P/P^2)=P.$$
These are my efforts:
$\operatorname{Ann}(P)=\{r \in R$ (perhaps)$\mid rP=0 \}=0 $, $\operatorname{Ann}(\frac{P}{P^2})=\{r \in R$ (perhaps)$\mid r\frac{P}{P^2}=P^2 \}=P$ $$r\cdot\frac{P}{P^2}=r\cdot\{p+P^2\mid p\in P\}=P^2 \Rightarrow r=p, p\in P$$ $$r\cdot(p+P^2)=P^2 \Rightarrow $$
$P$ is finitely generated.
Lets make a remark, if $\{p_i\}$ generates $P$ then any element of $P^2$ is represented as $\sum q_ip_i$ with $q_i \in P$. This is quite trivial: an element $p$ of $P$ is $p=\sum r_ip_i$, so if $q \in P$ then
$$qp=\sum (qr_i)p_i$$ and $qr_i \in P$
Now let $a \in ann (P/P^2)$
then $$ap_k=\sum q_{ki} p_i$$ with $q_{ki} \in P$. So by looking at the characteristic polynomial, like in linear algebra, we have that
$$a^n+b_1a^{n-1}+\cdots +b_n=0$$ and each $b_i \in P$ since they are sums of products of $q_{ki}$. This implies that $a^n \in P$ and since $P$ is prime $a \in P$.