Let $R$ be a P.I.D., let $a$ be a nonzero element of $R$ and let $M = R/(a)$. For any prime $p$ of $R$ prove that $p^{k−1}M/p^kM \cong R/(p)$ if $k \le N$ and $p^{k−1}M/p^kM \cong 0$ if $k > n$.
I know I have to use the isomorphism theorems, but I don't know exactly how to go about it. I tried to use the proof of Lemma 12.8 in Dummit and Foote as a guide:
Let $R$ be a P.I.D. and let $p$ be a prime in $R$. Let $F$ denote the field $R/(p)$. Let $M = R/(a)$ where $a$ is a nonzero element of $R$. Then $M/p^kM \cong R/(p)$ if $p$ divides $a$ in $R$ and $p^{k−1}M/p^kM \cong 0$ if $p$ does not divide $a$ in $R$.
Proof: This follows from the Isomorphism Theorems: note first that $p(R/(a))$ is the image of the ideal $(p)$ in the quotient $R/(a)$, hence is $(p)+(a)/(a)$. The ideal $(p)+(a)$ is generated by a greatest common divisor of $p$ and $a$, hence is $(p)$ if $p$ divides $a$ and is $R = (1)$ otherwise. Hence $pM = (p)/(a)$ if $p$ divides $a$ and is $R/(a) = M$ otherwise. If $p$ divides $a$ then $M/pM = (R/(a))/((p)/(a)) = R/(p)$, and if $p$ does not divide $a$ then $M/pM = M/M = 0$.
Can anyone help with this proof and possibly explain the proof of the lemma? For example, I do not know how $p(R/(a)) = (p) + (a)/(a)$.