Challenging problem in Hungerford's $\textit{Algebra}$ Here is a problem from the text $\textit{Algebra}$ by Hungerford, which I seem to be stuck on for quite some time now:  
Let $A$ be a cyclic $R$-module ($R$ is assumed to be a principal ideal domain) of order $r \in R$.
(a) If $\gcd(s,r)=1$ for some $s \in R$, then $sA=A$ and $A[s]=0$.
(b) If $s$ divides $r$, say $sk=r$, then $sA \cong R/(k)$ and $A[s] \cong R/(s)$.  
For (a), since $s,r$ are relatively prime, there exist $x,y \in R$ such that $sx + ry=1$ by Bézout's Identity. Now by definition,  $sA= \{sa: a\in A\}$ and $A[s]= \{a \in A: sa=0\}$. Isn't one inclusion clear, i.e. $sA \subset A$? If so, how should I approach the reverse inclusion just on the basis of relative primality, $A$ being cyclic, and $R$ being a PID of order $r$? For (b), a hint was given: use the First Isomorphism Theorem, but so far I'm not sure what $R$-module homomorphism(s) to use, but I'm sure that they will have to involve cosets and must be surjective in order to apply the First Isomorphism Theorem (that is quite clear to me). I'd appreciate any useful feedback.
 A: Let's first deal with (a). Let $b$ be a generator for $A$. Then $rb = 0$. Since $A$ is an $R$-module, $sA \subset A$. On the other hand, using a relation $sx + ry = 1$ ($x,y\in R$) and the equation $rb = 0$, we find that
\begin{equation}
b = 1b = (sx + ry)b = s(xb) + y(rb) = x(sb) + y(0) = s(xb) \in sA.
\end{equation}
Hence $A \subset sA$ and therefore $A = sA$. 
Suppose $a\in A[s]$. Then $sa = 0 = ra$. Therefore
\begin{equation}
a = 1a = (sx + ry)a = x(sa) + y(ra) = x(0) + y(0) = 0.
\end{equation}
Thus $A[s] = 0$.
To answer (b), consider the mapping $\phi : R \to A[s]$ defined by the equation $\phi(t) = ktb$. This is well-defined, since for any $t \in R$, 
\begin{equation}
s(ktb) = (sk)tb = rtb = t(rb) = t(0) = 0.
\end{equation}
Given a triple $t, u, v\in R$,
\begin{equation}
\phi(tu + v) = k(tu + v)b = t(kub) + kvb = t\phi(u) + \phi(v)
\end{equation}
So, $\phi$ is an $R$-homomorphism. 
Now we show that $\text{Ker}(\phi) = (s)$. Given $t \in \text{Ker}(\phi)$, $ktb = 0$ and so $kt\in (r)$. Write $kt = ru$ for some $u \in R$. Since $r = sk$, $kt = ksu$, whence $t = su$ (since $R$ is an integral domain and $k$ is nonzero). Thus $t \in (s)$, showing that $\text{Ker}(\phi) \subset (s)$. On the other hand, $\phi(s) = ksb = rb = 0$. Therefore $(s) \subset \text{Ker}(\phi)$ and consequently $\text{Ker}(\phi) = (s)$. 
The mapping $\phi$ is also onto. For given $a\in A[s]$, $a = tb$ for some $t\in R$ and $sa = 0$, i.e., $stb = 0$. Thus $st \in (r)$, and we can write $st = ru$ for some $u\in R$. Then $st = sku$, implying that $t = ku$. Therefore $\phi(u) = kub = tb = a$.
Since $\phi$ is an onto $R$-homomorphism with kernel $(s)$, the first isomorphism theorem gives $A[s] \cong R/(s)$.
By a similar argument, the mapping $\psi : R \to sA$ defined by the equation $\psi(t) = stb$ is an onto $R$-homomorphism with kernel $(k)$. So $sA \cong R/(k)$ by the first isomorphism theorem. 
