Isomorphism and cyclic modules

Question

Prove that $M$ is a cyclic $R$-module if and only if exists a left ideal $I\subset R$ such that $M \simeq R/I$.

I'm not sure how to even start this proof. I've been told that I could use an annhilator set to prove this, but I don't see the relation.

Answer 1

Suppose $M$ is cyclic, so there is some $x\in M$ such that $Rx=M$. You can consider the short exact sequence $0\to \operatorname{Ann}_R(x)\hookrightarrow R\xrightarrow{\varphi} M\to 0$, where the map $\varphi$ is defined by $\varphi(r)=rx$. Really, I am just saying that $\operatorname{Ann}_R(x)$ is the kernel of this $R$-module homomorphism $\varphi$. Then, by the first isomorphism theorem, $M\approx R/\operatorname{Ann}_R(x)$, and the annihilator is the left ideal you are looking for.

Conversely, if $M\approx R/I$, then for $\varphi$ defined the same way, $I$ is the kernel. For $[1]\in R/I$, $r[1]=0$ (i.e., $r1\in I$) if and only if $r\in I$, so $I$ is the annihilator of $[1]$. And, since $R[1]=R/I$, this shows $R/I$, and therefore $M$, is cyclic.