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.
$\blacksquare$