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

 $\ds \lambda \circ \paren {x + y}$ $=$ $\ds \lambda \circ \paren {\family {x_i}_{i \mathop \in I} + \family {y_i}_{i \mathop \in I} }$ $\ds$ $=$ $\ds \lambda \circ \family {x_i +_i y_i}_{i \mathop \in I}$ $\ds$ $=$ $\ds \family {\paren {\lambda \circ_i x_i} +_i \paren {\lambda \circ_i y_i} }_{i \mathop \in I}$ $\ds$ $=$ $\ds \family {\lambda \circ_i x_i}_{i \mathop \in I} + \family {\lambda \circ_i y_i}_{i \mathop \in I}$ $\ds$ $=$ $\ds \paren {\lambda \circ \family {x_i}_{i \mathop \in I} } + \paren {\lambda \circ \family {y_i}_{i \mathop \in I} }$ $\ds$ $=$ $\ds \paren {\lambda \circ x} + \paren {\lambda \circ y}$

So $(1)$ holds.

$\Box$

### Criterion 2

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

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

 $\ds \paren {\lambda +_R \mu} \circ x$ $=$ $\ds \paren {\lambda +_R \mu} \circ \family {x_i}_{i \mathop \in I}$ $\ds$ $=$ $\ds \family {\paren {\lambda +_R \mu} \circ_i x_i}_{i \mathop \in I}$ $\ds$ $=$ $\ds \family {\paren {\lambda \circ_i x_i} +_i \paren {\mu \circ_i x_i} }_{i \mathop \in I}$ $\ds$ $=$ $\ds \family {\lambda \circ_i x_i}_{i \mathop \in I} + \family {\mu \circ_i x_i}_{i \mathop \in I}$ $\ds$ $=$ $\ds \paren {\lambda \circ \family {x_i}_{i \mathop \in I} } + \paren {\lambda \circ \family {y_i}_{i \mathop \in I} }$ $\ds$ $=$ $\ds \paren {\lambda \circ x} + \paren {\mu \circ x}$

So $(2)$ holds.

$\Box$

### Criterion 3

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

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

 $\ds \paren {\lambda \times_R \mu} \circ x$ $=$ $\ds \paren {\lambda \times_R \mu} \circ \family {x_i}_{i \mathop \in I}$ $\ds$ $=$ $\ds \family {\paren {\lambda \times_R \mu} \circ_i x_i}_{i \mathop \in I}$ $\ds$ $=$ $\ds \family {\lambda \circ_i \paren {\mu \circ_i x_i} }_{i \mathop \in I}$ $\ds$ $=$ $\ds \lambda \circ \family {\mu \circ_i x_i}_{i \mathop \in I}$ $\ds$ $=$ $\ds \lambda \circ \paren {\mu \circ \family {x_i}_{i \mathop \in I} }$ $\ds$ $=$ $\ds \lambda \circ \paren {\mu \circ x}$

So $(3)$ holds.

$\Box$

Thus all criteria are seen to hold.

The result follows.

$\blacksquare$