Uniqueness submodule of cyclic module

Question

I'm having some problems with an exercise from Hungerford's Book of Algebra.

It states:

Let $R$ be a PID, and $A$ a unitary $R$-module such that $A$ is cyclic of order $r$.

a) Prove that every submodule of $A$ is cyclic of order dividing $r$.

b) Prove that for each $s\mid r$, there exists exactly one submodule of $A$ of order $s$.

I was able to prove a), and the existence part of b) but I can't manage to prove the uniqueness.

In a) what I did was basically noting that by the structure theorem, $A\cong R/(r)$, and every submodule of $R/(r)$ has the form $(s)/(r)$ where $s\mid r$, and is indeed cyclic.

In b) the idea for the existence was noting that if $r=sx$, then the submodule $Rb$ with $b=ax$ ($a$ is the generator of $A$), then $Rb$ is a cyclic submodule whose order is $\mathcal{O}_b = (s)$.

Answer 1

The submodules of $R/(r)$ correspond by second isomorphism theorem to submodules of $R$ containing $(r)$. But the submodules of $R$ are just left ideals, or just ideals since $ R$ is commutative. Since $R$ is a PID, these are generated by a single element $t$, say, and $(r) \subset (t) $, so $r \in (t)$ and $r=ts$ for some $s$. Thus any submodule of $ R/(r)$ is cyclic (generated by $t+(r)$), of the form $(t)/(r)$. Map $R \rightarrow (t)/(r) $, by $\lambda \mapsto \lambda (t+(r))= \lambda t + (r)$. Now, $st \in (r)$ and if $ \lambda t \in (r)=(st) $ then $st |\lambda t$. This implies $s| \lambda$ since $R$ is an ID. Hence the kernel of this map is $ (s)$, so the module $(t)/(r) \cong R/(s) $ and is cyclic of order $s$, where $s|r$ since $r=ts$ Conversly if $s|r$ then $r=ts$ for some $t$. The submodule generated by $t+(r) $ is cyclic of order $s$ as required.