Let $R$ be a PID, $\mathfrak{p}$ be a prime ideal in $R$, and $M$ an $R$-module. Then $$M(\mathfrak{p})=\{m\in M \mid \operatorname{Ann}_Rm=\mathfrak{p}^j, \, \text{ for some } j\geq 0 \}$$ is an $R$-module. $(\operatorname{Ann}_Rm=\{r\in R \mid rm=0\})$
Let $m,n\in M(\mathfrak{p})$ so say $\operatorname{Ann}_Rm=\mathfrak{p}^j$ and $\operatorname{Ann}_Rn=\mathfrak{p}^k$. I want to show $m-n\in M(\mathfrak{p})$. I am trying to figure out the exact form of $\operatorname{Ann}_R(m-n)$. As $R$ is a PID say $\mathfrak{p}=(p)$. I see $(p^{\max(j,k)})\subseteq \operatorname{Ann}_R(m-n)$ but I am not sure how to some some $r \in \operatorname{Ann}_R(m-n)$ is of the form $rp^t$ where $t=\max(j,k)$.
$\operatorname{Ann}_R(m-n)$ is a principal ideal generated by an element $a\in R$. Since $p^i\in (a)$, where $i=\max(k,l)$, then $a\mid p^i$, so $a$ is a power of $p$.