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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

T C
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1 Answers1

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You obtain an isomorphism $(R/fR)_{\mathfrak m}\simeq R_{\mathfrak m}\!\Bigm/\!\frac f1R_{\mathfrak m}$ considering the commutative diagram of exact sequences $$\begin{array}{*{10}{c}} 0&\longrightarrow &R&\!\xrightarrow{\enspace\times f\enspace}&R&\xrightarrow{\quad\enspace}&R/fR&\longrightarrow 0 \\ &&\downarrow &&\downarrow &&\downarrow \\ 0&\longrightarrow &R_{\mathfrak m}&\!\xrightarrow{\enspace\smash{\times \frac f1}\enspace}&R_{\mathfrak m}&\xrightarrow{\quad\enspace}& (R/fR)_{\mathfrak m}&\longrightarrow 0 \end{array}$$

Bernard
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  • I think this is not an answer. My isomorphism is different, or do you mean something here? – T C Jun 28 '18 at 19:18
  • I'm not sure if the vertical arrows are of any use here or if this verifies your claimed isomorphism, but the things you wanted to be isomorphic are isomorphic via SES. The second sequence above is exact by exactness of localization, and clearly $R_m/fR_m$ could also fill the last spot, so $(R/fR)_m \cong (R/fR)_m$. – forget this Jun 28 '18 at 23:29
  • Of course, then you must explicit that what you just did for modules also says something about rings. – forget this Jun 28 '18 at 23:30