Let $R$ be a Noetherian ring. Then all finitely generated $R$-modules are Noetherian

Question

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.

Answer 1

Every finitely generated $R$-module is a quotient of a free $R$-module, and since quotienting preserves Noetherian-ness, it suffices to show that every free $R$-module is Noetherian. This can be proved by induction. The simplest free $R$-module is $R^{1} = R$, which we already know is Noetherian. The next one to consider is $R^{2}$ (recall that all free $R$-modules look like $R^n$ for some $n$). Consider the exact sequence

$$ 0 \to R \to R^{2} \to R \to 0$$

Where $R \to R^2$ is given by inclusion (in the first coordinate), and $R^2 \to R$ is projection onto the second coordinate. By Claim I, $R^2$ is Noetherian. Proceed inductively using the exact sequence

$$0 \to R^{n-1} \to R^{n} \to R \to 0$$

Answer 2

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.