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Let $S \subset \mathbb{Z}$ denote a multiplicative set, i.e. $1 \in S$ and if $a,b \in S$ then $ab \in S$.

Let $p_i \in \mathbb{Z}$ be a prime, and let $e_i \in \mathbb{Z}^+$. Furthermore, let $S$ be such that $\forall s \in S: \text{gcd}(s,p_i) = 1$.

I want to show that it follows that $S^{-1}(\mathbb{Z}/(p_i^{e_i})) \cong \mathbb{Z}/(p_i^{e_i})$, that is, the localization of $\mathbb{Z}/(p_i^{e_i})$ at the multiplicative set $S$ is isomorphic to $\mathbb{Z}/(p_i^{e_i})$ itself.

I am having a hard time showing that this is the case. I suppose I would have to show that every element on the form $\frac{x}{s}$ for $x \in \mathbb{Z}/(p_i^{e_i})$ is equivalent to some $x' \in \mathbb{Z}/(p_i^{e_i})$.

I´d be interested to se either an explicit isomorphism or an argument.

Ben123
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  • What does $\gcd(s,p)=1$ tell you about $s$? – Kolja Jan 19 '24 at 16:04
  • Things I can think of:

    $\exists a,b \in \mathbb{Z}$ so that $as+bp = 1$ ($\mathbb{Z}$ P.I.D. so Bezout Domain).

    It also tells me that $\frac{x}{s} \not \sim \frac{0}{1} = 0 \in S^{-1}(\mathbb{Z}/(p_i^{e_i}))$ since then there would have to exist $s' \in S$ so that $s'x = 0 \implies s'x \in (p_i^{e_i})$ but since $s' \neq (p_i^{e_i})$ we have that $x \in (p_i^{e_i})$ so that $x \equiv 0 \in \mathbb{Z}/(p_i^{e_i})$.

     – Ben123 Jan 19 '24 at 16:18
  • Edit: What I wrote is not correct, from $s’(x) = 0$ we can only say that either $x = 0$ or $s’ = 0$. If $0 \not \in S$ then $x = 0$. Am I correct that generally, when we say $$(s_1,t_1) \sim (s_2,t_2) \iff \exists s’ \in S: s’(s_1t_2 -s_2t_1) = 0$$ for a localization $S^{-1}R$, we mean that RHS above is $0 \in R$? – Ben123 Jan 19 '24 at 16:46

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