Here is an excerpt of my lecture notes:
" Claim I: Let $M$ be $R$- module and $N$ be submodule of $M.$ Then $M$ is Noetherian iff $N, \ M/N$ are Noetherian.
Def: The ring $R$ is Noetherian iff the regular $R$-module is Noetherian.
Claim II: Let $R$ be a Noetherian ring. Then all finitely generated $R$-modules are Noetherian
Proof: By Claim I, every free $R$-module of finite rank is Noetherian and hence every finitely generated $R$-module. "
How do we deduce from Claim I that every free $R$-module of finite rank is Noetherian and hence every finitely generated $R$-module? Please advise/instruct me. Thank you.
Best Answer
If $R$ is Noetherian, then the regular module ($R$ as a module over itself, say as a left module) is Noetherian. Denote the regular module by $R$ as well. Then $R\oplus R$ is Noetherian as it has $N=R\oplus 0$ as a Noetherian submodule with Noetherian quotient (isomorphic to $R$). By induction, $R^n$ is Noetherian.
If $M$ is finitely generated by $n$ elements it is a quotient of $R^n$ modulo the submodule consisting of the relations between those generators. Hence $M$ is Noetherian.