# Finitely generated modules over a Noetherian ring are Noetherian

I'm trying to prove that if the ring $R$ is Noetherian then every finitely generated $R$-module is Noetherian.

First of all, it is known that every module is a homomorphic image of a free module, so we can take $R$ as a module over itself, and then we will have a homomorphism $f:R \rightarrow M$, where $M$ is some arbitrary module. Next, according to the well-known theorem, $$\operatorname{Im}(f) \cong \frac{R}{\ker(f)},$$ and $\frac{R}{\ker(f)}$ is Noetherian because $R$ is Noetherian (we can take finite set of generators for some submodule in $R$ and then its canonical image in $\frac{R}{\ker(f)}$ generates submodule there). Thus, $\operatorname{Im}(f)$ is Noetherian as well, but $\operatorname{Im}(f) \subseteq M$. How then one could say the same about the whole module $M$?

• I'm going to add the proof-verification tag because that's what this question seems like. If it isn't, then it probably should be closed as a duplicate. – rschwieb Oct 21 '13 at 12:59

The answer is that $f$, for certain rings, cannot be onto $M$, so this line of reasoning isn't going to work. If, for example $M=R^n$ $n>1$, and $R$ is finite, then it's going to be impossible for $R$ to map onto $R^n$.
It is commendable that you had this idea that "$R$ is Noetherian, so let me find a map to $M$ from $R$..."! If we revise your idea slightly, you'll be on the right track.
The thing to notice is that $R^n$ is a Noetherian module if $R$ is Noetherian. After you know that, you can cover $M$ more completely than you could with simply $R$. Then proceed with your "covering $M$" idea.