Problem IV.6.4 If $R$ is a principal ideal domain and $A$ is a cyclic $R$-module of order $r$, then
(i) every submodule of $A$ is cyclic, with order dividing $r$;
(ii) for every ideal $(s)$ containing $(r)$, $A$ has exactly one submodule, which is cyclic of order $s$.
Here is my attempt about problem IV.6.4
(i)
Since $A$ is cycle, say $A=(a)$, and take any submodule $B$ of $A$. It's obvious that $I:=\{t\in R:ta\in B\}$ is an ideal of $R$. Since $R$ is a principal ideal domain, there exists $m\in R$ such that $I=(m)$.
To prove $B=(ma)$, on one hand, suppose $b\in B\subset A=(a)=Ra$. There exists $u\in R$ such that $b=ua$, whence $u\in I=(m)$. Hence $u=sm$ for some $s\in R$ which implies $b=sma$, then $B\subset(ma)$. On the other hand, $ma\in B$ since $m\in I=(m)$, then $(ma)\subset B$. Therefore $B=(ma)$.
Let $d$ be the order of $B=(ma)$, then $$(r)=\mathcal{O}_a=\{t\in R:ta=0\}\subset\{t\in R:tma=0\}=\mathcal{O}_{ma}=(d)$$ which implies $d|r$.
(ii)
$r=ns$ for some non-zero element $n\in R$ since $(r)\subset(s)$. Let $B=(na)$, then $B$ is a submodule of $A$. To prove $\mathcal{O}_{na}=(s)$, on one hand, $sna=nsa=ra=0$ since $A$ has the order of $r$, which implies $(s)\subset\mathcal{O}_{na}$. On the other hand, suppose $t\in\mathcal{O}_{na}$, then $tna=0$. Therefore $tn\in(r)$, whence $tn=ur=uns$ for some $u\in R$. Since $R$ is a principal ideal domain and $n\neq0$, we have $t=us\in(s)$, which implies $\mathcal{O}_{na}\subset(s)$. Therefore $B$ is cyclic of order $s$.
My question is how to prove the uniqueness of $B$ in part (ii).
If $C$ is another submodule, which is cyclic of order $s$. We have $$B=(na)=Rna\cong R/\mathcal{O}_{na}=R/(s)\cong C$$ by using Theorem IV.6.4. But I think it's still not sufficient to imply $B=C$.
Theorem IV.6.4 Let $A$ be a left module over an integral domain $R$ and for each $a\in A$ and let $\mathcal{O}_a=\{r\in R:ra=0\}$.
(i) $\mathcal{O}_a$ is an ideal of $R$ for each $a\in A$.
(ii) $A_t=\{a\in A:\mathcal{O}_a\neq0\}$ is a submodule of $A$.
(iii) For each $a\in A$ there is an isomorphism of left modules $$R/\mathcal{O}_a\cong Ra=\{ra:r\in R\}.$$
Any help would be appreciated!
You can just use the fact that $A\cong R/(r)$ as $R$-modules and then do everything in there, which might be a bit clearer. Let me stick to your notation though:
Let $C=(c)$ be of order $s$. Let $(A:C)$ be the ideal $(A:C):=\{t\in R\mid ta\in C\}$, and let $n'$ be a generator. In part (i) you showed that then $C=(n'a)$. Note also that $n'$ divides $r$ as $r\in(A:C)$.
Thus we have \begin{align} &\forall t\in R: \quad& t\in\mathcal{O}_c&\iff tn'\in\mathcal{O}_a\\ \implies &\forall t\in R: \quad& t\in(s)&\iff tn'\in(r)\\ \implies &\forall t\in R: \quad& tn'\in(n's)&\iff tn'\in(r)\\ \implies &\forall t\in (n'): \quad& t\in(n's)&\iff t\in(r). \end{align} As we have $(n's),(r)\subset(n')$, we thus obtain $(n's)=(r)$. Hence if $B$ is another submodule of order $s$, with $(A:B)=(n)$ and thus $B=(na)$, we obtain $(n's)=(r)=(ns)$ and hence $n's=u_1ns$ and $ns=u_2n's$ for some $u_1,u_2\in R$. That is, there is a unit $u\in R$ such that $n'=un$. ($n$ and $n'$ are associates.) Therefore $B=C$.