Finite Direct Product of Modules is Module

Theorem
Let $\left({R, +_R, \times_R}\right)$ be a ring.

Let $\left({G_1, +_1, \circ_1}\right)_R, \left({G_2, +_2, \circ_2}\right)_R, \ldots, \left({G_n, +_n, \circ_n}\right)_R$ be $R$-modules.

Let:
 * $\displaystyle G = \prod_{k \mathop = 1}^n G_k$

be the cartesian product of $G_1$ to $G_n$.

Then $\left({G, +, \circ}\right)_R$ is an $R$-module where:


 * $+$ is the operation induced on $G$ by the operations $+_1, +_2, \ldots, +_n$ on $G_1, G_2, \ldots, G_n$


 * $\circ$ is defined as $\lambda \circ \left({x_1, x_2, \ldots, x_n}\right) = \left({\lambda \circ_1 x_1, \lambda \circ_2 x_2, \ldots, \lambda \circ_n x_n}\right)$

If each $G_k$ is a unitary module, then so is $G$.

Proof
We need to show that:

$\forall x, y, \in G, \forall \lambda, \mu \in R$:


 * $(1): \quad \lambda \circ \left({x + y}\right) = \left({\lambda \circ x}\right) + \left({\lambda \circ y}\right)$


 * $(2): \quad \left({\lambda +_R \mu}\right) \circ x = \left({\lambda \circ x}\right) + \left({\mu \circ x}\right)$


 * $(3): \quad \left({\lambda \times_R \mu}\right) \circ x = \lambda \circ \left({\mu \circ x}\right)$

Checking the criteria in order:

Criterion 1

 * $(1): \quad \lambda \circ \left({x + y}\right) = \left({\lambda \circ x}\right) + \left({\lambda \circ y}\right)$

Let $x = \left({x_1, x_2, \ldots, x_n}\right), y = \left({y_1, y_2, \ldots, y_n}\right) \in G$.

So $(1)$ holds.

Criterion 2

 * $(2): \quad \left({\lambda +_R \mu}\right) \circ x = \left({\lambda \circ x}\right) + \left({\mu \circ x}\right)$

Let $x = \left({x_1, x_2, \ldots, x_n}\right) \in G$.

So $(2)$ holds.

Criterion 3

 * $(3): \quad \left({\lambda \times_R \mu}\right) \circ x = \lambda \circ \left({\mu \circ x}\right)$

Let $x = \left({x_1, x_2, \ldots, x_n}\right) \in G$.

So $(3)$ holds.

Criterion 4
Finally, we confirm that if each $G_k$ is unitary, then so is $G$:

That is, we need to show:
 * $(4): \quad \forall x \in G: 1_R \circ x = x$

Let $x = \left({x_1, x_2, \ldots, x_n}\right) \in G$.

So $(4)$ holds.

Hence the result.