Unique factorization consequence Let R be principal ideal domain and $p$ prime element and $b \in R$, $b\neq 0$, $E=R/(pb)$ module over $R$ and let $E_p$ be submodule consisting of elements with exponent $p$ (there exist positive intiger $r$ so that $p^rx=0$). I need to prove  $E_p=bR/(pb)$. One inclusion is trivial, and for other solution say it follows at once from unique factorization in $R$. 
My try:
Let $x \in E_p$ so $x=a+(pb)$ and $p^ra+(pb)=(pb)$ which implies $p^ra=cpb$ for some $c \in R$. How do I use unique factorization here to prove that $a=by$ for some $y \in R$? What I can conclude from unique factorization in general? In some way I think i can "divide by $p$"?
 A: The correct statement should be that $E_p=b'R/pbR$, where $b'$ is $b/p^v$, where $p^v$ is the maximum power of $p$ appearing in the factorization of $b$.
The inclusion $b'R/pbR\subseteq E_p$ is obvious, because $p^{v+1}b'z=pbz\in pbR$ for any $z\in R$.
The submodule $E_p$ can be described as the set of cosets $x+(pb)$ such that, for some $k>0$, $p^kx\in(pb)$ or, equivalently, the set of cosets $x+(pb)$ such that, for some $k\ge0$, $p^kx\in(b)$.
If $x+(pb)\in E_p$, then $p^kx=by$. Let $r$ be the maximum exponent such that $p^r$ divides $y$. If $r\ge k$, we have
$$
x=b\frac{y}{p^k}
$$
so $x\in bR$. Otherwise, we get
$$
p^{k-r}x=b\frac{y}{p^r}
$$
and so we can assume $p^kx=pby$ where $p$ does not divide $y$ and $k\ge0$.
Now, consider the factorizations of $x$, $b$ and $y$, writing them as
$$
x=x_0p^up_1^{u_1}\dots p_s^{u_s},\quad
b=b_0p^vp_1^{v_1}\dots p_s^{v_s},\quad
y=y_0p_1^{w_1}\dots p_s^{w_s}
$$
where the exponent can be zero, the primes are all distinct and $x_0$, $b_0$, $y_0$ are units.
Thus we have
$$
x_0p^{u+k}p_1^{u_1}\dots p_s^{u_s}=
b_0y_0p^vp_1^{v_1}\dots p_s^{v_s}p_1^{w_1}\dots p_s^{w_s}
$$
This implies
$$
\begin{cases}
u+k=v\\
u_1=v_1+w_1\\
\dots\\
u_s=v_2+w_s
\end{cases}
$$
In particular $v_i\le u_i$ (for $i=1,2,\dots,s$) and so $b'$ divides $x$.
Note that when $b=p^v$ we have $b'=1$ and indeed $E_p=R/pbR$ in this case. On the other hand, if $p$ does not divide $b$, then $b'=b$ and $E_p=bR/pbR$.
