The following is taken from $\textit{Module Theory An Approach to Linear Algebra}$ By: T.S.Blyth
$\color{Green}{Background:}$
$\textbf{Exercise 4.1.}$An $R-$module $M$ is said to be $\textit{cyclic}$ if it is generated by a singleton subset. Let $M=Rx$ be a cyclic $R-$module. Prove that the $R-$module $M$ and $R/\textit{Ann}_R(x)$ are $R-$isomorphic.
[Recall that $\textit{Ann}_R(x)=\{\lambda; \lambda x=0\}.$] Dededuce that if $R$ is a principal ideal domain (i.e., a commutative integral domain in which every ideal is generated by a single subset) and if $x$ is such that $\textit{Ann}_R(x)=p^k R$ for some $p\in R$ then the only submodules of $M$ are those in the chain
$$0=p^k M\subset p^{k-1}M\subset\ldots\subset pM\subset p^0 M=M.$$
[Hint. Use the correspondence theorem.]
Also here is the screenshot of the exercise:
$\color{Red}{Questions:}$
In the above exercise, the notation: $\textit{Ann}_R(x)=p^k R.$ Specifically, is the $k$ in $p^k$ fixed? What i mean is if $p^k r\in p^kR,$ then $p^k r\in \textit{Ann}_R(x),$ and $p^krx=0.$ So say if we have $k=k_0\geq 1,$ then $p^{k_0} rx=0,$ $0=p^{k_0} M\subset p^{{k_0}-1}M\subset\ldots\subset pM\subset p^0 M=M.$ Each time I create the finite chain of subsets, I have to first settle on a particular value of $k$ where $p^kr\in p^kR?$
Thank you in advance
It is saying that if $p$ is a prime element of a PID, and $I=(p^k)=p^kR$, then the only $R$ submodules of $R/I$ are just $(p^iR)/I$ for $0\leq i\leq k$.