$\color{Green}{Background:}$
$\textbf{Assumed facts:}$
$\textbf{Theorem 1:}$ Let $R$ be a ring. Then the following conditions are equivalent:
$(1)$ Every ideal of $R$ is finitely generated
$(2)$ Every ascending sequences $I_1\subset I_2\subset I_3\subset\cdots$ of $R$ becomes stationary
$(3)$ Every set $M$ of ideals contiains a maximal element, i.e., an element $J\in M$ which is not properly contained in any $I\in M$
$\textbf{Definition:}$ A ring satisfying the equivalent conditions of theroem 1 is called a $\textbf{Noetherian ring.}$
$\textbf{Theorem 2:}$ Every principal ideal ring is Noetherian.
$\textbf{Exercise 4.1.}$ 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 Exercise 4.1 above, before I can use the correspondence theorem, I am having a lot of trouble figuring out how to show that the only submodules are the ones in the chain $0=p^k M\subset p^{k-1}M\subset\ldots\subset pM\subset p^0 M=M.$ Is the reason for it being that $(p^k)\subset (p)\subset M=Rx,$ and $(p^k)\subset (p).$ Since $R$ is a principal ideal domain and $R=Rx,$ Can I say there exists a $p \in R$ such that $R=(p)$ and since $\text{Ann}_R(x)=p^kR,$ then $p^kR=R$
Thank you in advance
Let $R$ be a principal ideal domain and let $M$ be an $R$-module. Assume that there exists $x \in M$ such that $M=Rx$ and $\text{Ann}_R(x) = p^kR$.
Consider a submodule of $M$. Since $M \cong R/\text{Ann}_R(x)$, we can say that this submodule is the image of some submodule of $R/\text{Ann}_R(x)$ under the isomorphism $R/\text{Ann}_R(x) \to M$.
By the correspondence theorem, every submodule of $R/\text{Ann}_R(x)$ is of the form $I/\text{Ann}_R(x)$, where $I$ is an ideal of $R$ containing $\text{Ann}_R(x)$.
Since $\text{Ann}_R(x) = p^kR$ and $R$ is a principal ideal domain, the ideals of $R$ containing $\text{Ann}_R(x)$ are of the form $p^lR$, where $l \in \{0,\dots,k\}$. (Check this!)
Finally, prove that the image of $p^lR/\text{Ann}_R(x)$ under the isomorphism $R/\text{Ann}_R(x) \to M$ is $p^lM$.