Direct Product of Modules is Module

Theorem
Let $R$ be a ring.

Let $\left\{\left\langle {M_i,+_i,\circ_i} \right\rangle\right\}_{i \in I}$ be a family of $R$-modules.

Let $\left\langle{M, +, \circ}\right\rangle$ be their direct product.

Then $\left\langle{M, +, \circ}\right\rangle$ is a module.

Proof
From External Direct Product of Abelian Groups is Abelian Group it follows that $(M,+)$ is an abelian group.

We need to show that:

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


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


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


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

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 $x = \left({x_i}\right)_{i\in I}, y = \left({y_i}\right)_{i\in I} \in M$.

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 $x = \left({x_i}\right)_{i\in I} \in M$.

So $(2)$ holds.

Criterion 3

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

Let $x = \left({x_i}\right)_{i\in I} \in M$.

So $(3)$ holds.

Also see

 * Direct Product of Unitary Modules is Unitary Module