# Colon ideal and Cyclic modules

I'm reading module theory as a beginner. The following problem might be super silly. I apologize for flooding SE with this kind of basic question.

Let $$R$$ be a ring with $$1$$, and $$\mathscr{m}$$ and $$\mathscr{n}$$ be two ideals of $$R$$. I was able to prove that if there is an element $$r\in R$$ such that $$\mathscr{m}=(\mathscr{n}:r)$$ and the $$R$$-module $$R/\mathscr{n}$$ is cyclic: $$R/\mathscr{n}=\langle r+\mathscr{n}\rangle$$, then $$R/\mathscr{m}$$ and $$R/\mathscr{n}$$ are isomorphic as $$R$$-modules. I believe the converse is also true. However, I don't know how to prove the converse of the proposition. Any help would be appreciated. Thank you very much.

Edit: $$(\mathscr{n}:r)$$ is defined as $$\{x \in R| xr \in \mathscr{n} \}$$

• Is $R$ commutative? Mar 28, 2020 at 19:21
• No. Do you think commutativity is required? Mar 28, 2020 at 19:22
• I'm not really sure what $(\mathscr{n}:r)$ is when $R$ is not commutative. Maybe you could add that to the question in case other people are not familiar with it. Mar 28, 2020 at 19:25
• I edited the question and defined the notation. Mar 28, 2020 at 19:31

An isomorphism of R-modules $$R/\mathfrak{m}\simeq R/\mathfrak{n}$$ yields a surjective $$R$$-linear map $$f: R\to R/\mathfrak{n}$$ with kernel $$\mathfrak{m}$$.
Now let $$r\in R$$ such that $$f(1)=\bar{r}$$. For all $$x\in R$$, we have $$f(x)=f(x\cdot 1)=x\cdot f(1)=x\cdot \bar{r}=\overline{xr}$$. Thus $$\mathfrak{m}=\ker(f)=\{ x\in R\mid xr \in\mathfrak{n}\}=(\mathfrak{n}:r)$$.
Surjectivity of $$f$$ and computations above show that $$R/\mathfrak{n}=\{ x\cdot \bar{r}, x\in R\}=R\cdot \bar{r}$$. Hence $$\bar{r}$$ is a generetor of $$R/\mathfrak{n}$$.