I proved by routine check that: let $R$ be a commutative ring with unit, $f$ be $R$-regular, $m$ be a maximal ideal s.t. $f\in m$. Then $(R/fR)_m\cong (R/fR)_{m/fR}$ as rings with the isomorphism $\overline{r}/{s}\mapsto \overline{r}/\overline{s} $.
The routine check is so obvious that it hardly takes any effort. Even before I know the problem, I believe it is true.
My question is, is the isomorphism correct? And is there any kind of intuition to acknowledge isomorphisms like this?
I have two questions here. Hope you can answer
You obtain an isomorphism $(R/fR)_{\mathfrak m}\simeq R_{\mathfrak m}\!\Bigm/\!\frac f1R_{\mathfrak m}$ considering the commutative diagram of exact sequences $$\begin{array}{*{10}{c}} 0&\longrightarrow &R&\!\xrightarrow{\enspace\times f\enspace}&R&\xrightarrow{\quad\enspace}&R/fR&\longrightarrow 0 \\ &&\downarrow &&\downarrow &&\downarrow \\ 0&\longrightarrow &R_{\mathfrak m}&\!\xrightarrow{\enspace\smash{\times \frac f1}\enspace}&R_{\mathfrak m}&\xrightarrow{\quad\enspace}& (R/fR)_{\mathfrak m}&\longrightarrow 0 \end{array}$$