[proofplan] We proceed by induction on $n$. The base case $n = 1$ is trivial. For the inductive step from $n - 1$ to $n$, we split the direct sum as $M_1 \oplus \cdots \oplus M_{n-1}$ and $M_n$ using the canonical short exact sequence, and apply the [Short Exact Sequence and Chain Conditions](/theorems/2906) theorem together with the inductive hypothesis. [/proofplan] [step:Establish the base case $n = 1$] When $n = 1$, the direct sum $M_1 \oplus \cdots \oplus M_n = M_1$ is noetherian (respectively artinian) by hypothesis. [/step] [step:Carry out the inductive step using the canonical short exact sequence] Assume that for some $n \geq 2$, the result holds for any collection of $n - 1$ noetherian (respectively artinian) $R$-modules. Let $M_1, \dots, M_n$ be noetherian $R$-modules. By the inductive hypothesis, $M_1 \oplus \cdots \oplus M_{n-1}$ is noetherian. Consider the canonical short exact sequence of $R$-modules: \begin{align*} 0 \to M_1 \oplus \cdots \oplus M_{n-1} \xrightarrow{\;\iota\;} M_1 \oplus \cdots \oplus M_n \xrightarrow{\;\pi\;} M_n \to 0, \end{align*} where $\iota$ is the inclusion into the first $n-1$ coordinates, defined by $\iota(m_1, \dots, m_{n-1}) = (m_1, \dots, m_{n-1}, 0)$, and $\pi$ is the projection onto the last coordinate, defined by $\pi(m_1, \dots, m_n) = m_n$. We verify this is exact: - **$\iota$ is injective:** If $\iota(m_1, \dots, m_{n-1}) = 0$, then $(m_1, \dots, m_{n-1}, 0) = (0, \dots, 0)$, so each $m_i = 0$. - **$\pi$ is surjective:** For any $m_n \in M_n$, we have $\pi(0, \dots, 0, m_n) = m_n$. - **$\ker \pi = \operatorname{im} \iota$:** An element $(m_1, \dots, m_n)$ lies in $\ker \pi$ if and only if $m_n = 0$, which is precisely the image of $\iota$. Both $M_1 \oplus \cdots \oplus M_{n-1}$ (noetherian by the inductive hypothesis) and $M_n$ (noetherian by assumption) are noetherian. By the [Short Exact Sequence and Chain Conditions](/theorems/2906) theorem, $M_1 \oplus \cdots \oplus M_n$ is noetherian. The artinian case is identical: the same short exact sequence and induction apply, with "noetherian" replaced by "artinian" throughout. [guided] The inductive step reduces the problem of $n$ modules to one about $n - 1$ modules plus a single module. The mechanism is the short exact sequence \begin{align*} 0 \to M_1 \oplus \cdots \oplus M_{n-1} \xrightarrow{\;\iota\;} M_1 \oplus \cdots \oplus M_n \xrightarrow{\;\pi\;} M_n \to 0. \end{align*} Why does this help? The [Short Exact Sequence and Chain Conditions](/theorems/2906) theorem tells us that the middle term $M_1 \oplus \cdots \oplus M_n$ is noetherian if and only if both the left term $M_1 \oplus \cdots \oplus M_{n-1}$ and the right term $M_n$ are noetherian. The right term $M_n$ is noetherian by hypothesis. The left term is noetherian by the inductive hypothesis (it is a direct sum of $n - 1$ noetherian modules). Hence the middle term is noetherian. One could also prove the base case $n = 2$ directly and then induct, but the argument is the same: the short exact sequence $0 \to M_1 \to M_1 \oplus M_2 \to M_2 \to 0$ and the theorem on short exact sequences give $M_1 \oplus M_2$ noetherian whenever $M_1$ and $M_2$ are. For the artinian case, the [Short Exact Sequence and Chain Conditions](/theorems/2906) theorem applies equally to the artinian property (the theorem statement includes "respectively artinian"). The same induction works. [/guided] [/step]