# Proving a submodule is Noetherian

I'm doing some independent study with a professor in Ring/Field theory (I'm an undergraduate) and I have been having a hell of a time wrapping my head around problems involving Noetherian rings and modules. Here's an example of a problem I can't figure out:

Let $R$ be a left Noetherian ring, $_{R}M$ an $R$-module, and $m\in M$. Show $Rm$ is a left Noetherian $R$-Module.

I went about showing that $Rm$ was a submodule of $_RM$ and I feel decent about it. I'm really having trouble showing that $R$ left Noetherian tells us anything about whether $Rm$ is as well.

In general, is it standard practice in any problem involving Noetherian rings (modules) to apply Zorn's Lemma to say that it has maximal ideals (submodules)? Any intuition or help in this arena would be greatly appreciated.

• You can translate the Noetherian property on submodules of $Rm$, to the Noetherian property on ideals of $R$. $M$ here is more or less irrelevant. You could replace $M$ with $Rm$ and the problem wouldn't change. – Prometheus Sep 18 '14 at 7:13

The left $R$-module $R$ is noetherian. The left $R$-module $Rm$ is an image of the noetherian left $R$-module $R$. Now apply a result about images of noetherian modules.
• I appreciate your solution, and by combining some earlier results I was able to get to a solution. As for the general theory, I suppose you answered me by showing me that this solution required "none of the above." :) For the future, is there a kind of correspondence between the ideals of $R$ and the submodules of the $R$-module $R$? It seems that there are complimentary theorems for these two sets but I can't think of a way to connect them more concretely. – Nico Sep 18 '14 at 23:14
• You're welcome. There is a correspondence between modules generated by $1$ element ( cyclic modules) and ideals. Let $M$ be cyclic module. Choose $m$ a generator of $M$. The map $a \mapsto a m$ is a surjective morphism. The kernel of this map is the annihilator of $m$, a left ideal $I$ of $A$. We have $A/I$ isomorphic to $M$ by $\bar a \mapsto a m$. In general the ideal $A$ themselves are not generated by one element. – Orest Bucicovschi Sep 19 '14 at 0:15