Finite Direct Sum of Noetherian Module is Noetherian

Theorem

Let $A$ be a commutative ring with unity.

Let $n \in \N_{>0}$.

Let $M_1, \ldots, M_n$ be $A$-modules.

Then the direct sum:

$\ds \bigoplus_{i \mathop = 1}^n M_i$

is an $A$-Noetherian module.

Proof

By Direct Sum of Modules is Module, it is an $A$-module.

Thus we only need to show that it is Noetherian.

We prove it by induction.

For $n = 1$, there is nothing to prove.

Let $n \ge 2$.

Suppose that the claim is true for $n-1$, i.e.:

$\ds \bigoplus_{i \mathop = 1}^{n-1} M_i$

is Noetherian.

Then, consider the short exact sequence:

$\ds 0 \longrightarrow \bigoplus_{i \mathop = 1}^{n-1} M_i \stackrel {\alpha} {\longrightarrow} \bigoplus_{i \mathop = 1}^n M_i \stackrel {\beta} {\longrightarrow} M_n \longrightarrow 0$

where:

$\map \alpha {\tuple {x_1, \ldots, x_{n-1} } } := \tuple {x_1, \ldots, x_{n-1}, 0}$

$\map \beta { \tuple {x_1, \ldots, x_n} } := x_n$

Recall that $M_n$ is also Noetherian by hypothesis.

Thus by Short Exact Sequence Condition of Noetherian Modules:

$\ds \bigoplus_{i \mathop = 1}^n M_i$

is Noetherian.

$\blacksquare$