Finitely generated modules over a Noetherian ring are Noetherian

Solution 1:

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.

Will be waiting if you have more questions... good luck!