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

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.

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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zipirovich
zipirovich
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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}$$\mathscr{m}$ and $ \mathscr{n}$$\mathscr{n}$ be two ideals of $R.$ I$R$. I was able to prove that if there is an element $r\in R$ such that $\mathscr {m}= (\mathscr{n} :r)$$\mathscr{m}=(\mathscr{n}:r)$ and the $R-$$R$-module $R/\mathscr{n}$ is cyclic: $R/\mathscr{n} = <r+ \mathscr{n}>$$R/\mathscr{n}=\langle r+\mathscr{n}\rangle$, then $R/\mathscr{m}$ and $R/\mathscr{n}$ are isomorphic as $R-$module$R$-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.

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: $R/\mathscr{n} = <r+ \mathscr{n}>$, then $R/\mathscr{m}$ and $R/\mathscr{n}$ are isomorphic as $R-$module. 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.

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.

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Gordhob Brain
Gordhob Brain
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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: $R/\mathscr{n} = <r+ \mathscr{n}>$, then $R/\mathscr{m}$ and $R/\mathscr{n}$ are isomorphic as $R-$module. 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.