[proofplan]
We prove both directions of the equivalence. For the forward direction, if $M$ is noetherian and some submodule $N$ were not finitely generated, we could construct a strictly ascending chain of finitely generated submodules inside $N$ that never stabilises, violating the ascending chain condition. For the converse, given an ascending chain $M_0 \subseteq M_1 \subseteq \cdots$, we form the union $N = \bigcup_{i \geq 0} M_i$, which is a submodule of $M$ and hence finitely generated by hypothesis; the finitely many generators all lie in some single $M_k$, forcing the chain to stabilise at $k$.
[/proofplan]
[step:Show that if $M$ is noetherian then every submodule is finitely generated]
Suppose $M$ is noetherian, and let $N \leq M$ be a submodule. Assume for contradiction that $N$ is not finitely generated. We construct a strictly ascending chain in $M$ as follows.
Choose any $x_1 \in N$; since $N$ is not finitely generated, $(x_1) \neq N$. Choose $x_2 \in N \setminus (x_1)$; since $N$ is not finitely generated, $(x_1, x_2) \neq N$. Inductively, having chosen $x_1, \dots, x_k \in N$ such that $(x_1, \dots, x_k) \subsetneq N$, choose $x_{k+1} \in N \setminus (x_1, \dots, x_k)$.
This construction produces a strictly ascending chain of submodules of $M$:
\begin{align*}
(x_1) \subsetneq (x_1, x_2) \subsetneq (x_1, x_2, x_3) \subsetneq \cdots
\end{align*}
Each inclusion is strict because $x_{k+1} \in (x_1, \dots, x_{k+1}) \setminus (x_1, \dots, x_k)$. This chain does not stabilise, contradicting the ascending chain condition on $M$. Hence $N$ is finitely generated.
[guided]
We want to show that in a noetherian module, no submodule can "escape" finite generation. The proof is by contradiction: if a submodule $N$ were not finitely generated, we could keep finding new elements outside any finite generating set built so far.
More precisely, start with any $x_1 \in N$. The submodule $(x_1)$ generated by $x_1$ alone cannot equal $N$ (since $N$ is not finitely generated). So we can pick $x_2 \in N \setminus (x_1)$. Now $(x_1, x_2)$ is a strictly larger finitely generated submodule of $N$, but it still cannot equal $N$. Continue inductively.
At each stage, $x_{k+1}$ is chosen from $N \setminus (x_1, \dots, x_k)$, guaranteeing that the inclusion $(x_1, \dots, x_k) \subsetneq (x_1, \dots, x_{k+1})$ is strict: the element $x_{k+1}$ witnesses the strictness. The resulting infinite strictly ascending chain
\begin{align*}
(x_1) \subsetneq (x_1, x_2) \subsetneq (x_1, x_2, x_3) \subsetneq \cdots
\end{align*}
lives inside $M$ (since $N \leq M$) and never stabilises. But $M$ is noetherian, so the ascending chain condition holds for all chains of submodules of $M$. This contradiction shows that $N$ must be finitely generated.
[/guided]
[/step]
[step:Show that if every submodule is finitely generated then $M$ is noetherian]
Suppose every submodule of $M$ is finitely generated. Let $M_0 \subseteq M_1 \subseteq M_2 \subseteq \cdots$ be an ascending chain of submodules of $M$. Define
\begin{align*}
N := \bigcup_{i \geq 0} M_i.
\end{align*}
[claim:$N$ is a submodule of $M$]
Let $x, y \in N$ and $r \in R$. Then $x \in M_j$ and $y \in M_k$ for some indices $j, k \geq 0$. Setting $\ell = \max(j, k)$, both $x$ and $y$ lie in $M_\ell$ (since $M_j \subseteq M_\ell$ and $M_k \subseteq M_\ell$). Because $M_\ell$ is a submodule, $x + y \in M_\ell \subseteq N$ and $rx \in M_j \subseteq N$. Also $0 \in M_0 \subseteq N$. Hence $N \leq M$.
[/claim]
[proof]
Contained in the claim statement above.
[/proof]
By hypothesis, $N$ is finitely generated: there exist $y_1, \dots, y_r \in N$ with $N = (y_1, \dots, y_r)$. Each generator $y_j$ lies in some $M_{k_j}$ (since $y_j \in N = \bigcup_i M_i$). Set $k = \max(k_1, \dots, k_r)$. Then all generators $y_1, \dots, y_r$ belong to $M_k$, so
\begin{align*}
N = (y_1, \dots, y_r) \subseteq M_k \subseteq N.
\end{align*}
Hence $M_k = N$. For all $i \geq k$, we have $M_k \subseteq M_i \subseteq N = M_k$, so $M_i = M_k$. The chain stabilises at index $k$, and $M$ is noetherian.
[guided]
For the converse direction, we are given an ascending chain and must show it stabilises. The key idea is to "take the limit" of the chain: the union $N = \bigcup_{i \geq 0} M_i$ is the smallest submodule containing every $M_i$.
Why is $N$ a submodule? Because any two elements $x, y \in N$ belong to some common $M_\ell$ (take $\ell$ large enough), and $M_\ell$ is closed under addition and scalar multiplication. The ascending property of the chain is essential here: it guarantees that for any finite collection of elements of $N$, there is a single $M_\ell$ containing all of them.
By hypothesis, $N$ is finitely generated, say $N = (y_1, \dots, y_r)$. Each $y_j$ entered the union at some finite stage $M_{k_j}$. Taking $k = \max(k_1, \dots, k_r)$, all generators land in $M_k$ (since the chain is ascending, $M_{k_j} \subseteq M_k$ for each $j$). But the generators of $N$ lie in $M_k$, so $N = (y_1, \dots, y_r) \subseteq M_k$. Since $M_k \subseteq N$ by construction, we get $M_k = N$. Every subsequent $M_i$ (for $i \geq k$) satisfies $M_k \subseteq M_i \subseteq N = M_k$, so the chain is constant from index $k$ onward.
[/guided]
[/step]
