This is the first part of exercise 11 on page 168 of Advanced linear algebra third edition of Steven Roman
Let $M$ an $R$-module, then prove that $M$ is cyclic if and only if there is some ideal $I\subset R$ such that $R/I\cong M$ as $R$-modules.
(Note: here I assumed that $R$ doesn't have necessarily an identity or it is commutative.)
I had proven that if $M$ is cyclic then such ideal exists but I dont know how to prove the other direction. Suppose that
$$ \tau: R/I\to M,\quad r+I\mapsto w $$
is a module isomorphism, then I want to show that there is some $v\in M$ such that $\tau(r+I)=r v$, but I dont know how to continue from here.
Trying by contradiction: suppose that $M$ is not cyclic, then there are $v,w\in M$ such that $\mathrm{span}(v,w)\neq\mathrm{span}(x)$ for $x\in\{v,w\}$, then I must show that doesn't exists any ideal $I$ of $R$ such that $R/I\cong M$ as $R$-modules. Then I tried to proceed as follows:
Set $H:=\mathrm{span}(v)$ and $N:=\mathrm{span}(v,w)$, then $H$ is cyclic and so there is some ideal $I\subset R$ such that $R/I\cong H$ as $R$-modules. Then we have two possibilities here: or $H\cong N$ or there is some ideal $J\neq I$ such that $N\cong R/J$. Well, it is easy to check that cannot be the case that $H\cong N$, because if suck isomorphism exists then we will have that there is some $r,s\in R$ such that $f(rv)=w$ and $f(sv)=v$, but this would imply that $(r+s) f(v)=v+w$ so $N$ would be cyclic, a contradiction.
Then we must shows that there is not such $J$. However Im stuck here again. I tried other different ways to get a proof by contradiction but I get stuck in any of them. Some help will be appreciated.
I believe that this is false unless $R$ itself is a cyclic $R$-module...
Just so we're on the same page, I'm assuming that $M$ is a left $R$-module. Firstly, if $R$ is a cyclic $R$-module, then let $r$ be a generator, i.e. $Rr=R$. In this case, if $I$ is an ideal of $R$ and $\tau:R/I\to M$ is an $R$-module isomorphism, then $\tau(r+I)$ will generate $M$.
On the other hand, consider $R=2\mathbb{Z}$. This is not a cyclic module over itself, since for any $x\in R$, $Rx\subseteq 4\mathbb{Z}\neq R$. However, we can take the ideal $I=(0)$, so $R/I\cong R$, which isn't cyclic.