Simple $R$-Modules are isomorphic to quotient of maximal ideal (proof explanation) I'm trying to understand the proof of the following proposition.

An $R$ -module is $M$ said to be cyclic if $M$ it is generated by one element, and simple if $M$ has no $R$ -submodules other than $0$ and $M$. Show that $M$ is a simple $R$ -module if, and only if, $M$ is isomorphic to $R / \mathfrak{m}$ for some maximal ideal $\mathfrak{m}$ of $R$.


Proof: Due to previous results, we know that:
(a) Any cyclic $R$ module is isomorphic to $R / I$ for some ideal $I$ of $R$, (b) Any simple $R$ -module is cyclic. Therefore, we must show that $R / I$ is simple if, and only if, $I$ is maximal. Let $f: R \rightarrow R / I$ be the natural quotient map. Then given any submodule $J$ of $R / I,$ its preimage $f^{-1}(J)$ is an ideal of $R$ containing $I .$ This gives a bijection between the ideals of $R$ containing $I$ and the submodules of $R / I .$ In particular we see that $R / I$ is simple if, and only if, the only ideals containing $I$ are $I$ itself and the unit ideal; that is, if and only if $I$ is maximal.

Here are my confusions.

*

*If $M$ is a simple $R$-module, then (b) implies it is cyclic, and (a) implies it is isomorphic to $R/I$ for some ideal $I \subset R$. How can we conclude from this that we must show $R / I$ is simple $\iff$ $I$ is maximal? I get that we want to show $M$ is a simple $R$-module $\implies$ $R / I$ is simple $\implies$ $I$ is maximal. But what about the converse direction?


*For such $f$ as above, I have managed to prove that for any $R$-submodule(!) $J$ of $R / I,$ its preimage $f^{-1}(J)$ is an ideal of $R$ containing $I .$ I have also proven that if $K \subset R$ is an ideal of $R$ and $I \subset K$, then $f(K)$ is an $R$-submodule of $R/I$ (i.e. $f(K) \subset R/I$). Is this enough to imply that $f$ is a bijection? I have seen this picture many times before, when $\phi: \text{set 1} \to \text{set 2}$ is a function, and $\phi^{-1}:\text{set 2} \to \text{set 1}$ is a function. But I don't know how to justify $\phi$ is a bijection from this.
Partial answers (answering only some of the questions) are very welcome! I will upvote them too.
Edit: I've moved some of my proofs in (2) to a different question.
 A: *

*Remember the original statement we want to prove:


$M$ is simple if and only if $M \cong R/\mathfrak{m}$ for some maximal ideal $\mathfrak{m}$.

We've already shown:

$M$ is simple if and only if $M \cong R/I$ for some ideal $I$ such that $R/I$ is simple.

Thus, we have left to prove:

$\color{green}{M \cong R/I ~\text{for some ideal}~ I ~\text{such that}~ R/I ~\text{is simple}}$ if and only if $\color{blue}{M \cong R/\mathfrak{m} ~\text{for some maximal ideal}~ \mathfrak{m}}$.

If we knew
$R/I$ simple $\iff$ $I$ maximal, the first (green) statement would be
equivalent to "$M \cong R/I$ for some ideal $I$ such that $I$ is
maximal", which is exactly the second (blue) statement. Thus, it suffices
to prove $R/I$ simple $\iff$ $I$ maximal.


*$f$ is never a bijection unless $I = 0$! I think you want to say that $J \mapsto f^{-1}(J)$ is a bijection from the set of submodules of $R/I$ to the set of ideals of $R$ which contain $I$. Anyway, your argument is not enough. You have correctly constructed the inverse map, but you never proved it's an inverse! You then need to show that $f^{-1}(f(K)) = K$ and $f(f^{-1}(J)) = J$ for all submodules $J$ of $R/I$ and all ideals $K$ of $R$ such that $I \subseteq K$. Then the two functions (which you correctly showed are well-defined) will be inverses, hence both bijective.

