[proofplan]
Let $M$ be a finitely generated $R$-module with generators $m_1, \dots, m_\ell$. The surjective $R$-module homomorphism $R^{\oplus \ell} \to M$ sending the $k$-th standard basis element to $m_k$ gives a short exact sequence $0 \to \ker \to R^{\oplus \ell} \to M \to 0$. Since $R$ is noetherian as an $R$-module, the direct sum $R^{\oplus \ell}$ is noetherian by the [Direct Sum Preserves Chain Conditions](/theorems/2818) theorem. The [Short Exact Sequence and Chain Conditions](/theorems/2906) theorem then implies the quotient $M$ is noetherian.
[/proofplan]
[step:Construct a surjection from $R^{\oplus \ell}$ onto $M$]
Let $M$ be a finitely generated $R$-module. Then there exist $m_1, \dots, m_\ell \in M$ such that $M = Rm_1 + \cdots + Rm_\ell$. Define the $R$-module homomorphism
\begin{align*}
\varphi: R^{\oplus \ell} &\to M \\
(r_1, \dots, r_\ell) &\mapsto r_1 m_1 + \cdots + r_\ell m_\ell.
\end{align*}
This map is $R$-linear (it sends $r \cdot (r_1, \dots, r_\ell) = (rr_1, \dots, rr_\ell)$ to $rr_1 m_1 + \cdots + rr_\ell m_\ell = r(r_1 m_1 + \cdots + r_\ell m_\ell) = r \varphi(r_1, \dots, r_\ell)$, and it preserves addition). It is surjective because every element of $M$ is an $R$-linear combination of $m_1, \dots, m_\ell$.
[/step]
[step:Show $R^{\oplus \ell}$ is noetherian using the direct sum theorem]
The ring $R$, viewed as a left $R$-module over itself, is noetherian by hypothesis (the noetherian condition on the ring $R$ means precisely that $R$ satisfies the ascending chain condition on ideals, which are the same as $R$-submodules of $R$).
Since each summand in $R^{\oplus \ell} = R \oplus \cdots \oplus R$ ($\ell$ copies) is a noetherian $R$-module, the [Direct Sum Preserves Chain Conditions](/theorems/2818) theorem gives that $R^{\oplus \ell}$ is noetherian.
[/step]
[step:Conclude that $M$ is noetherian via the short exact sequence]
The surjection $\varphi: R^{\oplus \ell} \to M$ yields the short exact sequence of $R$-modules
\begin{align*}
0 \to \ker \varphi \xrightarrow{\;\iota\;} R^{\oplus \ell} \xrightarrow{\;\varphi\;} M \to 0,
\end{align*}
where $\iota$ is the inclusion of $\ker \varphi$ into $R^{\oplus \ell}$.
Since $R^{\oplus \ell}$ is noetherian (established in the previous step), the [Short Exact Sequence and Chain Conditions](/theorems/2906) theorem (forward direction) implies that both $\ker \varphi$ and $M$ are noetherian. In particular, $M$ is noetherian, which is what we wanted to show.
The artinian case is identical: if $R$ is artinian as a ring, then $R$ is artinian as an $R$-module, the direct sum $R^{\oplus \ell}$ is artinian by the [Direct Sum Preserves Chain Conditions](/theorems/2818) theorem (artinian case), and the [Short Exact Sequence and Chain Conditions](/theorems/2906) theorem (artinian case) gives that $M$ is artinian.
[guided]
The strategy is to express $M$ as a quotient of something we already know is noetherian. Since $M$ is finitely generated with $\ell$ generators, there is a natural surjection from the free module $R^{\oplus \ell}$ onto $M$. This puts us in the framework of the short exact sequence $0 \to \ker \varphi \to R^{\oplus \ell} \to M \to 0$.
The [Short Exact Sequence and Chain Conditions](/theorems/2906) theorem states: $R^{\oplus \ell}$ is noetherian if and only if both $\ker \varphi$ and $M$ are noetherian. We know the middle term $R^{\oplus \ell}$ is noetherian (it is a finite direct sum of copies of $R$, each of which is noetherian as an $R$-module, and the [Direct Sum Preserves Chain Conditions](/theorems/2818) theorem handles the direct sum). By the forward direction of the short exact sequence theorem, both $\ker \varphi$ and $M$ are noetherian.
Why is $R$ noetherian as an $R$-module? The $R$-submodules of $R$ (viewed as a module over itself) are precisely the ideals of $R$. The statement "$R$ is a noetherian ring" means its ideals satisfy the ascending chain condition, which is exactly the noetherian condition on $R$ as an $R$-module.
Note that the converse is also true (though not needed here): if every finitely generated $R$-module is noetherian, then in particular $R$ itself (a finitely generated $R$-module with one generator $1_R$) is noetherian, so $R$ is a noetherian ring.
[/guided]
[/step]
