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.
Best Answer
We've already shown:
Thus, we have left to prove:
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.