Theorem: Finite Direct Product of Modules is Module

Let $\struct {R, +_R, \times_R}$ be a ring.
Let $\struct {G_1, +_1, \circ_1}_R, \struct {G_2, +_2, \circ_2}_R, \ldots, \struct {G_n, +_n, \circ_n}_R$ be $R$-modules.
Let:
:$\ds G = \prod_{k \mathop = 1}^n G_k$
be their direct product.
Then $G$ is a module.

Proof:
=== {{Module-axiom|1|nolink}} ===
Let $x = \tuple {x_1, x_2, \ldots, x_n}, y = \tuple {y_1, y_2, \ldots, y_n} \in G$.
{{begin-eqn}}
{{eqn | l = \lambda \circ \paren {x + y}
      | r = \lambda \circ \paren {\tuple {x_1, x_2, \ldots, x_n} + \tuple {y_1, y_2, \ldots, y_n} }
      | c = 
}}
{{eqn | r = \lambda \circ \tuple {x_1 +_1 y_1, x_2 +_2 y_2, \ldots, x_n +_n y_n}
      | c = 
}}
{{eqn | r = \tuple {\lambda \circ_1 \paren {x_1 + y_1}, \lambda \circ_2 \paren {x_2 + y_2}, \ldots, \lambda \circ_n \paren {x_n + y_n} }
      | c = 
}}
{{eqn | r = \tuple {\lambda \circ_1 x_1, \lambda \circ_2 x_2, \ldots, \lambda \circ_n x_n} + \tuple {\lambda \circ_1 y_1, \lambda \circ_2 y_2, \ldots, \lambda \circ_n y_n}
      | c = 
}}
{{eqn | r = \paren {\lambda \circ \tuple {x_1, x_2, \ldots, x_n} } + \paren {\lambda \circ \tuple {y_1, y_2, \ldots, y_n} }
      | c = 
}}
{{eqn | r = \paren {\lambda \circ x} + \paren {\lambda \circ y}
      | c = 
}}
{{end-eqn}}
So $(1)$ holds.
{{qed|lemma}}
=== {{Module-axiom|2|nolink}} ===
Let $x = \tuple {x_1, x_2, \ldots, x_n} \in G$.
{{begin-eqn}}
{{eqn | l = \paren {\lambda +_R \mu} \circ x
      | r = \paren {\lambda +_R \mu} \circ \tuple {x_1, x_2, \ldots, x_n}
      | c = 
}}
{{eqn | r = \tuple {\paren {\lambda +_R \mu} \circ_1 x_1, \paren {\lambda +_R \mu} \circ_2 x_2, \ldots, \paren {\lambda +_R \mu} \circ_n x_n}
      | c = 
}}
{{eqn | r = \tuple {\paren {\lambda \circ_1 x_1} +_1 \paren {\mu \circ_1 x_1}, \paren {\lambda \circ_2 x_2} +_2 \paren {\mu \circ_2 x_2}, \ldots, \paren {\lambda \circ_n x_n} +_n \paren {\mu \circ_n x_n} }
      | c = 
}}
{{eqn | r = \tuple {\lambda \circ_1 x_1, \lambda \circ_2 x_2, \lambda \circ_n x_n} + \tuple {\mu \circ_1 x_1, \mu \circ_2 x_2, \ldots, \mu \circ_n x_n}
      | c = 
}}
{{eqn | r = \paren {\lambda \circ_1 \tuple {x_1, x_2, \ldots, x_n} } + \paren {\mu \circ_1 \tuple {x_1, x_2, \ldots, x_n} }
      | c = 
}}
{{eqn | r = \paren {\lambda \circ x} + \paren {\mu \circ x}
      | c = 
}}
{{end-eqn}}
So $(2)$ holds.
{{qed|lemma}}
=== {{Module-axiom|3|nolink}} ===
Let $x = \tuple {x_1, x_2, \ldots, x_n} \in G$.
{{begin-eqn}}
{{eqn | l = \paren {\lambda \times_R \mu} \circ x
      | r = \paren {\lambda \times_R \mu} \circ \tuple {x_1, x_2, \ldots, x_n}
      | c = 
}}
{{eqn | r = \tuple {\paren {\lambda \times_R \mu} \circ_1 x_1, \paren {\lambda \times_R \mu} \circ_2 x_2, \ldots, \paren {\lambda \times_R \mu} \circ_n x_n}
      | c = 
}}
{{eqn | r = \tuple {\lambda \circ_1 \paren {\mu \circ_1 x_1}, \lambda \circ_2 \paren {\mu \circ_2 x_2}, \ldots, \lambda \circ_n \paren {\mu \circ_n x_n} }
      | c = 
}}
{{eqn | r = \lambda \circ \tuple {\mu \circ_1 x_1, \mu \circ_2 x_2, \ldots, \mu \circ_n x_n}
      | c = 
}}
{{eqn | r = \lambda \circ \paren {\mu \circ \tuple {x_1, x_2, \ldots, x_n} }
      | c = 
}}
{{eqn | r = \lambda \circ \paren {\mu \circ x}
      | c = 
}}
{{end-eqn}}
So $(3)$ holds.
{{qed|lemma}}
Hence the result.
{{qed}}