Direct Product of Modules is Module

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

\(\ds \lambda \circ \paren {x + y}\) \(=\) \(\ds \lambda \circ \paren {\family {x_i}_{i \mathop \in I} + \family {y_i}_{i \mathop \in I} }\) by hypothesis
\(\ds \) \(=\) \(\ds \lambda \circ \family {x_i +_i y_i}_{i \mathop \in I}\) Definition of Module Direct Product/General Case
\(\ds \) \(=\) \(\ds \family {\paren {\lambda \circ_i \paren { x_i +_i y_i} } } _{i \mathop \in I}\) Definition of Module Direct Product/General Case
\(\ds \) \(=\) \(\ds \family {\paren {\lambda \circ_i x_i} +_i \paren {\lambda \circ_i y_i} }_{i \mathop \in I}\) Left Module Axiom $\text M 1$: (Left) Distributivity over Module Addition
\(\ds \) \(=\) \(\ds \family {\lambda \circ_i x_i}_{i \mathop \in I} + \family {\lambda \circ_i y_i}_{i \mathop \in I}\) Definition of Module Direct Product/General Case
\(\ds \) \(=\) \(\ds \paren {\lambda \circ \family {x_i}_{i \mathop \in I} } + \paren {\lambda \circ \family {y_i}_{i \mathop \in I} }\) Definition of Module Direct Product/General Case
\(\ds \) \(=\) \(\ds \paren {\lambda \circ x} + \paren {\lambda \circ y}\) by hypothesis

So $(1)$ holds.

$\Box$


Criterion 2

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

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

\(\ds \paren {\lambda +_R \mu} \circ x\) \(=\) \(\ds \paren {\lambda +_R \mu} \circ \family {x_i}_{i \mathop \in I}\) by hypothesis
\(\ds \) \(=\) \(\ds \family {\paren {\lambda +_R \mu} \circ_i x_i}_{i \mathop \in I}\) Definition of Module Direct Product/General Case
\(\ds \) \(=\) \(\ds \family {\paren {\lambda \circ_i x_i} +_i \paren {\mu \circ_i x_i} }_{i \mathop \in I}\) Left Module Axiom $\text M 2$: (Right) Distributivity over Scalar Addition
\(\ds \) \(=\) \(\ds \family {\lambda \circ_i x_i}_{i \mathop \in I} + \family {\mu \circ_i x_i}_{i \mathop \in I}\) Definition of Module Direct Product/General Case
\(\ds \) \(=\) \(\ds \paren {\lambda \circ \family {x_i}_{i \mathop \in I} } + \paren {\mu \circ \family {x_i}_{i \mathop \in I} }\) Definition of Module Direct Product/General Case
\(\ds \) \(=\) \(\ds \paren {\lambda \circ x} + \paren {\mu \circ x}\) by hypothesis

So $(2)$ holds.

$\Box$


Criterion 3

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

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

\(\ds \paren {\lambda \times_R \mu} \circ x\) \(=\) \(\ds \paren {\lambda \times_R \mu} \circ \family {x_i}_{i \mathop \in I}\) by hypothesis
\(\ds \) \(=\) \(\ds \family {\paren {\lambda \times_R \mu} \circ_i x_i}_{i \mathop \in I}\) Definition of Module Direct Product/General Case
\(\ds \) \(=\) \(\ds \family {\lambda \circ_i \paren {\mu \circ_i x_i} }_{i \mathop \in I}\) Left Module Axiom $\text M 3$: Associativity
\(\ds \) \(=\) \(\ds \lambda \circ \family {\mu \circ_i x_i}_{i \mathop \in I}\) Definition of Module Direct Product/General Case
\(\ds \) \(=\) \(\ds \lambda \circ \paren {\mu \circ \family {x_i}_{i \mathop \in I} }\) Definition of Module Direct Product/General Case
\(\ds \) \(=\) \(\ds \lambda \circ \paren {\mu \circ x}\) by hypothesis

So $(3)$ holds.

$\Box$


Thus all criteria are seen to hold.

The result follows.

$\blacksquare$


Also see

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