Finite Direct Product of Modules is Module

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

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

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

Then $$\left({G, +: \circ}\right)_R$$ is an $R$-module where:
 * $$+$$ is the operation induced on $$G$$ by the operations 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:

$$\forall x, y, \in G, \forall \lambda, \mu \in R$$:
 * VS 1: $$\lambda \circ \left({x + y}\right) = \left({\lambda \circ x}\right) + \left({\lambda \circ y}\right)$$;
 * VS 2: $$\left({\lambda +_R \mu}\right) \circ x = \left({\lambda \circ x}\right) + \left({\mu \circ x}\right)$$;
 * VS 3: $$\left({\lambda \times_R \mu}\right) \circ x = \lambda \circ \left({\mu \circ x}\right)$$.

Checking the criteria in order:


 * VS 1:

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

$$ $$ $$ $$ $$ $$

VS 2:

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

$$ $$ $$ $$ $$ $$

VS 3:

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

$$ $$ $$ $$ $$ $$

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

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

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

$$ $$ $$ $$