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 $\{x \in R| xr \in \mathscr{n} \}$
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 $R/\mathfrak{n}$.