[proofplan] We treat the noetherian case; the artinian case is identical with all ascending chains replaced by descending chains. We identify $N$ with its image in $M$ via the injective map, so that $N \leq M$ and $L \cong M/N$ with quotient map $\pi: M \to M/N$. For the forward direction, submodule chains in $N$ and $L$ lift to submodule chains in $M$, which stabilise by hypothesis. For the converse, given an ascending chain in $M$, we consider its intersection with $N$ and its image in $L$; both stabilise, and a short element-chase shows the original chain stabilises as well. [/proofplan] [step:Set up the identification $N \leq M$ and $L \cong M/N$] Write the short exact sequence as $0 \to N \xrightarrow{i} M \xrightarrow{\pi} L \to 0$, where $i$ is injective and $\pi$ is surjective with $\ker \pi = \operatorname{im} i$. Identify $N$ with its image $i(N) \leq M$ (this is justified since $i$ is injective, hence an isomorphism onto its image). Under this identification, $L \cong M / N$ via the map induced by $\pi$, and $\pi: M \to L$ is the quotient map. We prove the noetherian case. The artinian case follows by the same argument with all ascending chains replaced by descending chains and all inclusions reversed. [/step] [step:Prove the forward direction: $M$ noetherian implies $N$ and $L$ noetherian] Assume $M$ is noetherian. **$N$ is noetherian:** Let $N_0 \subseteq N_1 \subseteq \cdots$ be an ascending chain of submodules of $N$. Since $N \leq M$, each $N_k$ is also a submodule of $M$, so $N_0 \subseteq N_1 \subseteq \cdots$ is an ascending chain of submodules of $M$. By the ascending chain condition on $M$, this chain stabilises. Hence $N$ is noetherian. **$L$ is noetherian:** Let $L_0 \subseteq L_1 \subseteq \cdots$ be an ascending chain of submodules of $L$. Form the preimage chain $\pi^{-1}(L_0) \subseteq \pi^{-1}(L_1) \subseteq \cdots$ in $M$. Each $\pi^{-1}(L_k)$ is a submodule of $M$ (the preimage of a submodule under a module homomorphism is a submodule), and the chain is ascending because $L_k \subseteq L_{k+1}$ implies $\pi^{-1}(L_k) \subseteq \pi^{-1}(L_{k+1})$. By the ascending chain condition on $M$, there exists $k_0$ such that $\pi^{-1}(L_k) = \pi^{-1}(L_{k_0})$ for all $k \geq k_0$. Applying $\pi$ and using the surjectivity of $\pi$, we obtain $L_k = \pi(\pi^{-1}(L_k)) = \pi(\pi^{-1}(L_{k_0})) = L_{k_0}$ for all $k \geq k_0$. (The identity $\pi(\pi^{-1}(S)) = S$ holds for any submodule $S \leq L$ because $\pi$ is surjective.) Hence $L$ is noetherian. [guided] **$N$ is noetherian:** This is straightforward. A submodule of $N$ is also a submodule of $M$ (since $N \leq M$), so any ascending chain of submodules of $N$ is an ascending chain of submodules of $M$. The ascending chain condition on $M$ forces it to stabilise. **$L$ is noetherian:** The argument is slightly more involved because $L$ is a quotient of $M$, not a submodule. We cannot directly view chains in $L$ as chains in $M$. Instead, we lift: given an ascending chain $L_0 \subseteq L_1 \subseteq \cdots$ in $L$, we take preimages under $\pi$ to get the chain $\pi^{-1}(L_0) \subseteq \pi^{-1}(L_1) \subseteq \cdots$ in $M$. This chain stabilises by the ascending chain condition on $M$. To push back down: once $\pi^{-1}(L_k) = \pi^{-1}(L_{k_0})$ for $k \geq k_0$, applying $\pi$ gives $L_k = \pi(\pi^{-1}(L_k)) = \pi(\pi^{-1}(L_{k_0})) = L_{k_0}$. The key identity $\pi(\pi^{-1}(S)) = S$ uses the surjectivity of $\pi$: every element of $S \leq L$ has a preimage in $M$ under $\pi$, so $\pi(\pi^{-1}(S)) \supseteq S$; the reverse inclusion $\pi(\pi^{-1}(S)) \subseteq S$ holds for any map. [/guided] [/step] [step:Prove the converse: $N$ and $L$ noetherian implies $M$ noetherian] Assume $N$ and $L$ are both noetherian. Let $M_0 \subseteq M_1 \subseteq \cdots$ be an ascending chain of submodules of $M$. Consider the two induced chains: \begin{align*} &M_0 \cap N \subseteq M_1 \cap N \subseteq M_2 \cap N \subseteq \cdots &&\text{(submodules of } N\text{)}, \\ &\pi(M_0) \subseteq \pi(M_1) \subseteq \pi(M_2) \subseteq \cdots &&\text{(submodules of } L\text{)}. \end{align*} Since $N$ is noetherian, the first chain stabilises: there exists $k_1$ such that $M_i \cap N = M_{k_1} \cap N$ for all $i \geq k_1$. Since $L$ is noetherian, the second chain stabilises: there exists $k_2$ such that $\pi(M_i) = \pi(M_{k_2})$ for all $i \geq k_2$. Set $k = \max(k_1, k_2)$. We show $M_i = M_k$ for all $i \geq k$. Since $M_k \subseteq M_i$, it suffices to show $M_i \subseteq M_k$. Let $x \in M_i$. Since $\pi(M_i) = \pi(M_k)$ (as $i \geq k \geq k_2$), there exists $y \in M_k$ with $\pi(x) = \pi(y)$. Then $x - y \in \ker \pi = N$. Since $x \in M_i$ and $y \in M_k \subseteq M_i$, we have $x - y \in M_i$, so $x - y \in M_i \cap N = M_k \cap N$ (as $i \geq k \geq k_1$). In particular, $x - y \in M_k$. Since $y \in M_k$ as well, $x = (x - y) + y \in M_k$. Hence $M_i = M_k$ for all $i \geq k$, and the chain stabilises. Therefore $M$ is noetherian. [guided] The converse is the deeper direction. Given an ascending chain $M_0 \subseteq M_1 \subseteq \cdots$ in $M$, we cannot directly use the noetherian property of $N$ or $L$ — we need to translate the chain into chains living in $N$ and $L$. **The intersection chain** $M_i \cap N$ captures what the chain looks like "inside $N$." Each $M_i \cap N$ is a submodule of $N$, and the chain is ascending because $M_i \subseteq M_{i+1}$ implies $M_i \cap N \subseteq M_{i+1} \cap N$. Since $N$ is noetherian, this chain stabilises at some index $k_1$. **The image chain** $\pi(M_i)$ captures what the chain looks like "modulo $N$." Each $\pi(M_i)$ is a submodule of $L$, and the chain is ascending because $M_i \subseteq M_{i+1}$ implies $\pi(M_i) \subseteq \pi(M_{i+1})$. Since $L$ is noetherian, this chain stabilises at some index $k_2$. Now let $k = \max(k_1, k_2)$ and take any $x \in M_i$ with $i \geq k$. The element-chase proceeds in two stages: 1. **Match in the quotient:** Since $\pi(M_i) = \pi(M_k)$, the image $\pi(x) \in \pi(M_k)$, so there exists $y \in M_k$ with $\pi(y) = \pi(x)$. 2. **Match in the kernel:** Then $x - y \in \ker \pi = N$. But also $x - y \in M_i$ (since $x \in M_i$ and $y \in M_k \subseteq M_i$), so $x - y \in M_i \cap N = M_k \cap N \subseteq M_k$. Combining: $x = (x - y) + y \in M_k$. This shows $M_i \subseteq M_k$, and since $M_k \subseteq M_i$ by the ascending property, $M_i = M_k$. The chain stabilises at $k$. This argument is a prototype for "diagram chasing" in exact sequences: we decompose information about $M$ into information about $N$ (the kernel part) and $L$ (the quotient part), handle each piece separately, and reassemble. [/guided] [/step]