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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 $\{x \in R| xr \in \mathscr{n} \}$

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    $\begingroup$ Is $R$ commutative? $\endgroup$ Mar 28, 2020 at 19:21
  • $\begingroup$ No. Do you think commutativity is required? $\endgroup$
    – Matha Mota
    Mar 28, 2020 at 19:22
  • $\begingroup$ 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. $\endgroup$ Mar 28, 2020 at 19:25
  • $\begingroup$ I edited the question and defined the notation. $\endgroup$
    – Matha Mota
    Mar 28, 2020 at 19:31

1 Answer 1

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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 $R/\mathfrak{n}$.

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  • $\begingroup$ Wow! A very natural and elegant proof. $\endgroup$
    – Matha Mota
    Mar 28, 2020 at 21:12

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