Small clarification on D&F exercise prompt

Question

Consider the following problem:

Let $R$ be a P.I.D. and $a\in R$ with $a\ne 0$. Let $M=R/(a)$. For any prime $p$ of $R$ prove that $p^{k-1}M/p^kM \cong R/(p)$ if $k\leq n$ and $p^{k-1}M/p^kM \cong 0$ otherwise, where $n$ is the power of $p$ dividing $a$ in $R$.

I just solved this, but afterward I realized that we don't know $n\geq 1$. So isn't it possible that $p^{k-1}$ makes no sense when $k\leq n$? It's perfectly possible that $n=0$ for certain primes, right?

I'm just asking to see if anyone could clarify. Maybe I'm missing a conceptual detail, as there's nothing on Dummit and Foote's errata about this.

Answer 1

Nitpicking, the statement should be

$p^{k-1}M/p^kM \cong R/(p)$ for $1 \leq k \leq n$ and $p^{k-1}M/p^kM \cong 0$ for $k > n$.

In the case $n=0$, the first case never occurs and the second case is indeed always true for $k>0$, because $pM=M$ in the case where $p$ and $a$ are co-prime.