[proofplan] We treat the noetherian case; the artinian case follows by reversing all chain inclusions. Write the short exact sequence as $0 \to N \xrightarrow{i} M \xrightarrow{\pi} L \to 0$ with $i$ injective, $\pi$ surjective, and $\ker \pi = \operatorname{im} i$. For the forward direction, we show that ascending chains in $N$ embed into $M$ via $i$, and ascending chains in $L$ lift to $M$ via $\pi^{-1}$, so both stabilise. For the converse, given an ascending chain in $M$, we project it to $L$ and intersect it with $i(N)$; both stabilise by hypothesis, and an element-chase shows the original chain stabilises. [/proofplan] [step:Show that if $M$ is noetherian, then $N$ is noetherian] Write the sequence as $0 \to N \xrightarrow{i} M \xrightarrow{\pi} L \to 0$. Let $N_0 \subseteq N_1 \subseteq \cdots$ be an ascending chain of submodules of $N$. Since $i$ is an injective $R$-module homomorphism, the images $i(N_0) \subseteq i(N_1) \subseteq \cdots$ form an ascending chain of submodules of $M$. Since $M$ is noetherian, this chain stabilises: there exists $k_0$ such that $i(N_j) = i(N_{k_0})$ for all $j \geq k_0$. Since $i$ is injective, $i(N_j) = i(N_{k_0})$ implies $N_j = N_{k_0}$ for all $j \geq k_0$. The chain in $N$ stabilises, so $N$ is noetherian. [/step] [step:Show that if $M$ is noetherian, then $L$ is noetherian] Let $L_0 \subseteq L_1 \subseteq \cdots$ be an ascending chain of submodules of $L$. Since $\pi: M \to L$ is surjective, the preimages $\pi^{-1}(L_0) \subseteq \pi^{-1}(L_1) \subseteq \cdots$ form an ascending chain of submodules of $M$. Since $M$ is noetherian, this chain stabilises at some index $k_0$: $\pi^{-1}(L_j) = \pi^{-1}(L_{k_0})$ for all $j \geq k_0$. Applying $\pi$ to both sides and using the surjectivity of $\pi$ (which gives $\pi(\pi^{-1}(L_j)) = L_j$ for every submodule $L_j \subseteq L$), we obtain $L_j = L_{k_0}$ for all $j \geq k_0$. The chain in $L$ stabilises, so $L$ is noetherian. [/step] [step:Show that if both $N$ and $L$ are noetherian, then $M$ is noetherian] Let $M_0 \subseteq M_1 \subseteq \cdots$ be an ascending chain of submodules of $M$. Consider the two derived chains: \begin{align*} &i^{-1}(M_0) \subseteq i^{-1}(M_1) \subseteq \cdots \quad \text{(submodules of } N\text{)}, \\ &\pi(M_0) \subseteq \pi(M_1) \subseteq \cdots \quad \text{(submodules of } L\text{)}. \end{align*} Since $N$ is noetherian, the first chain stabilises at some index $k_1$. Since $L$ is noetherian, the second chain stabilises at some index $k_2$. Set $k = \max(k_1, k_2)$. We show $M_j = M_k$ for all $j \geq k$. Since $M_k \subseteq M_j$, it suffices to prove $M_j \subseteq M_k$. Let $x \in M_j$. Then $\pi(x) \in \pi(M_j) = \pi(M_k)$, so there exists $m \in M_k$ with $\pi(x) = \pi(m)$. Hence $\pi(x - m) = 0$, which means $x - m \in \ker \pi = \operatorname{im} i$. Write $x - m = i(n)$ for some $n \in N$. Since $x - m \in M_j$ (as $x \in M_j$ and $m \in M_k \subseteq M_j$), we have $i(n) \in M_j$, so $n \in i^{-1}(M_j) = i^{-1}(M_k)$. Therefore $i(n) \in M_k$, and $x = m + i(n) \in M_k$. This shows $M_j \subseteq M_k$, so the chain stabilises. [guided] For the converse, the key idea is to decompose an ascending chain in $M$ into two chains -- one living in the submodule $N$ (via intersection with $\operatorname{im} i$) and one living in the quotient $L$ (via the projection $\pi$) -- and then reconstruct stabilisation of the original chain from stabilisation of both derived chains. Let $M_0 \subseteq M_1 \subseteq \cdots$ be an ascending chain of submodules of $M$. Define the derived chains: \begin{align*} &i^{-1}(M_0) \subseteq i^{-1}(M_1) \subseteq \cdots \quad \text{in } N, \\ &\pi(M_0) \subseteq \pi(M_1) \subseteq \cdots \quad \text{in } L. \end{align*} Both are ascending chains. Since $N$ is noetherian, the first stabilises at some index $k_1$: $i^{-1}(M_j) = i^{-1}(M_{k_1})$ for all $j \geq k_1$. Since $L$ is noetherian, the second stabilises at some index $k_2$: $\pi(M_j) = \pi(M_{k_2})$ for all $j \geq k_2$. Set $k = \max(k_1, k_2)$. We claim $M_j = M_k$ for all $j \geq k$. The inclusion $M_k \subseteq M_j$ is automatic (the chain is ascending). For the reverse inclusion, take any $x \in M_j$. We need to show $x \in M_k$. Since $\pi(x) \in \pi(M_j) = \pi(M_k)$ (because $j \geq k \geq k_2$), there exists $m \in M_k$ with $\pi(x) = \pi(m)$. Then $x - m \in \ker \pi$. By exactness, $\ker \pi = \operatorname{im} i$, so $x - m = i(n)$ for some $n \in N$. Now, $x - m \in M_j$ because $x \in M_j$ and $m \in M_k \subseteq M_j$. So $i(n) \in M_j$, which means $n \in i^{-1}(M_j) = i^{-1}(M_k)$ (because $j \geq k \geq k_1$). Therefore $i(n) \in M_k$, and finally $x = m + i(n) \in M_k$. This shows $M_j \subseteq M_k$, so $M_j = M_k$ for all $j \geq k$. The chain stabilises, and $M$ is noetherian. Why does this argument work? It exploits the fact that exact sequences decompose information about $M$ into information about $N$ and $L$. Any element of $M$ can be analysed by first projecting to $L$ (to pin down its "quotient part") and then examining the remainder, which lives in $\operatorname{im} i \cong N$ (its "submodule part"). Stabilisation of both parts forces stabilisation of the whole. [/guided] [/step] [step:Deduce the artinian case by reversing chain directions] The artinian case is identical: replace "ascending" with "descending" and "ACC" with "DCC" throughout. Each argument transfers descending chains in $N$ and $L$ to descending chains in $M$ and vice versa by the same maps $i$, $\pi$, $i^{-1}$, $\pi^{-1}$. The element-chase in the converse direction is unchanged, since it uses only the inclusions $M_j \supseteq M_k$ (which hold for descending chains when $j \geq k$) and the exactness of the sequence. [/step]