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

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.

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.

• Because this is the module I need to look at in order for my argument to work. I argue by induction, I first look at $R^2$ ($=R\oplus R$, the external direct sum of the two modules, i.e. the cartesian product as a set equipped with componentwise addition and scalar multiplication), then at $R^n$. In fact, it would do to just look at $R^n$ and decompose it as $R\oplus R^{n-1}$ since we already looked at $R$, but it's a little clearer this way. – Oliver Braun Mar 17 '14 at 12:40
• I thought so. Note that the external direct sum is a construction (of a new module) whereas the internal direct sum is a property of a ("pre-existing") module. The two concepts are related by the fact that the module $M_1\oplus_{\text{external}} M_2$ is the internal direct sum of the 2 submodules $M_1\oplus 0$ and $0\oplus M_2$. – Oliver Braun Mar 17 '14 at 12:45
• I meant what I wrote, the module $M:=M_1\oplus M_2$ (this is the external direct sum, it is common to denote that by $\oplus$, but you can use $\times$ as well) decomposes as the internal direct sum $M=(M_1\oplus 0)\oplus (0\oplus M_2)$, where $0$ abbreviates $\{0_{M_1}\}$ or $\{0_{M_2}\}$, respectively. Remember that the external direct sum is a construction, the internal one is a property of a module. So $\cong$ would rather denote an isomorphism of $M_1\times M_2$ with another external direct sum (i.e. a newly constructed module)... – Oliver Braun Mar 17 '14 at 12:57
• Consider $f ~:~ R\oplus R \to R, ~ (a,b) \mapsto a$. This is a surjective homomorphism of modules with kernel $N=0\oplus R$. Now the claim follows from the first isomorphism theorem. – Oliver Braun Mar 17 '14 at 13:13
• This is also a consequence of the isomorphism theorem. Let $M=\langle m_1,...,m_n\rangle$. Then the obvious morphism $R^n \to M$ is surjective and its kernel is the submodule I was referring to. I am getting the feeling you should do some revision on basic concepts of module theory... EDIT: The "obvious morphism" is given by $e_i \mapsto m_i$, i.e. the $i$-th standard basis vector is mapped to the $i$-th generator of $M$. – Oliver Braun Mar 17 '14 at 13:18

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

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