Localizations of $ \mathbb{Z}_{p^k}$

Question

Let $S \subseteq \mathbb{Z}_{p^k} $ be a multiplicative subset, where $p$ is a prime number, $k$ an integer.

Is it true that $$S^{-1} \mathbb{Z}_{p^k} \cong \mathbb{Z} /n\mathbb{Z} $$ for some integer $n$ ? Why ?

Answer 1

Yes.

The ring $\mathbb{Z}_{p^k}$ is local with unique maximal ideal $(p)$.

If $S\cap (p) = \emptyset$, you are inverting units and hence $$S^{-1} \mathbb{Z}_{p^k}\cong \mathbb{Z}_{p^k}=\mathbb{Z}/p^k\mathbb{Z}.$$

If $S\cap (p) \neq \emptyset$, then $ap\in S$ for some $a \in \mathbb{Z}_{p^k}$. But then you get $$\frac{1}{1}=\frac{a^{k-1}p^{k-1}}{a^{k-1}p^{k-1}}=\frac{0}{p}=\frac{0}{1}.$$ So you get $$S^{-1} \mathbb{Z}_{p^k}\cong 0=\mathbb{Z}/\mathbb{Z}.$$