Direct Product of Modules is Module

Theorem
Let $R$ be a ring.

Let $\family {\struct {M_i, +_i, \circ_i} }_{i \mathop \in I}$ be a family of $R$-modules.

Let $\struct {M, +, \circ}$ be their direct product.

Then $\struct {M, +, \circ}$ 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:

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


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


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


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

Checking the criteria in order:

Criterion 1

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

Let $x = \family {x_i}_{i \mathop \in I}, y = \family {y_i}_{i \mathop \in I} \in M$.

So $(1)$ holds.

Criterion 2

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

Let $x = \family {x_i}_{i \mathop \in I} \in M$.

So $(2)$ holds.

Criterion 3

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

Let $x = \family {x_i}_{i \mathop \in I} \in M$.

So $(3)$ holds.

Thus all criteria are seen to hold.

The result follows.

Also see

 * Direct Product of Unitary Modules is Unitary Module