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If $R/\langle a \rangle$ is the quotient module over the commutative ring $R$ with identity 1, then, $R/\langle a \rangle=\{r+\langle a \rangle: r \in R\}=\langle1+\langle a \rangle \rangle$
Is it correct? If it is ,then it means that every quotient module with identity is cyclic R-module.

Am I correct?

Esha
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    Use \langle and \rangle, not < and >. The latter are relation symbols, the former are delimiters. – Arturo Magidin Jun 09 '22 at 18:24
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    $R$ itself, as a $R$-module, is cyclic, generated by $1$, so all its homomorphic images/quotients of $R$ are of course cyclic (generated by the image of $1$). Note that ideals of $R$ may not be principal, so I would not write $\langle a \rangle$ where you in fact need just an arbitrary ideal $I$ of $R$. –  Jun 09 '22 at 18:25

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Yes, and in fact, every cyclic $R$-module is isomorphic to one of the form $R/I$ with $I$ an ideal of $R$: if $M$ is a cyclic $R$-module generated by an element $x \in M$, then $M \cong R/\operatorname{Ann}_R x$, where $\operatorname{Ann}_R x = \{r \in R : rx = 0\}$ is the annihilator of $x$. The isomorphism is induced by the map $R \to M$ given by $r \mapsto rx$, which is surjective because $x$ generates $M$, and whose kernel by construction is the annihilator of $x$. Note that this does not require $I$ to be a principal ideal.

Also, conversely, every ideal $I$ is the annihilator of some cyclic module, because $\operatorname{Ann}_R(R/I) = I$.

Daniel Hast
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  • But if R is not commutative then the annihilator of x will not be an ideal of R, instead it will be only a one sided ideal of R. So R/kernel cannot be said to be a quotient ring, right? But in any case we can say that it is an R-module. Am I right? – Esha Jun 09 '22 at 18:37
  • Also is the converse of your stated result also true? That is if R/I is any arbitrary R-module then is I the annihilator of some x(since R/I is cyclic R-module then I guess the converse also holds)? – Esha Jun 09 '22 at 18:43
  • For my last doubt asked in the comments, I figured it out just now that I=$Ann_R(1+I)$. That means the converse is also true. Am I correct? – Esha Jun 09 '22 at 18:52
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    I just edited my answer to include the converse. Also, I think you're right about it not quite working out for noncommutative rings, so I've removed that assertion. – Daniel Hast Jun 09 '22 at 22:10