I used this "fact" when I hand in my take home exam. To my surprise this was returned with a remark "$M$ is not noetherian!". But I remember the criterion for noetherian is $M$'s submodules are all finitely generated. Then assume $M$ has submodule $M_{1}$, the map $$R^{n}\rightarrow M\rightarrow M_{1}$$ must be surjective since $M$ is finitely generated. So I do not know where I got wrong.
Commutative Algebra – Is Every Finitely Generated Module Noetherian?
commutative-algebramodules
Best Answer
What you claim is false; a finitely generated module need not be Noetherian. As a counterexample, consider the polynomial ring $\mathbb{Q}[x_0,x_1,\dots]$ in infinitely many variables as a module over itself. It is finitely generated (by 1), but its submodule $\langle x_0,x_1,\dots\rangle$ isn't, so the module itself isn' Noetherian.
What is true is that a module over a Noetherian ring is Noetherian iff it is finitely generated (this might require the ring to be commutative with unity).