Ring is Module over Itself
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Theorem
Let $\struct {R, +, \circ}$ be a ring.
Then $\struct {R, +, \circ}_R$ is an $R$-module.
Ring with Unity
Let $\struct {R, +, \circ}$ be a ring with unity $1_R$.
Then $\struct {R, +, \circ}_R$ is a unitary $R$-module.
Proof 1
Note that:
$\struct {R, +, \circ}$ is a ring by assumption.
$\struct {R, +}$ is an abelian group by the definition of a ring.
Let us verify the module axioms:
\((1)\) | $:$ | \(\ds \forall x, y, z \in R:\) | \(\ds x \circ \paren {y + z} = \paren {x \circ y} + \paren {x \circ z} \) | ||||||
\((2)\) | $:$ | \(\ds \forall x, y, z \in R:\) | \(\ds \paren {x + y} \circ z = \paren {x \circ z} + \paren {y \circ z} \) | ||||||
\((3)\) | $:$ | \(\ds \forall x, y, z \in R:\) | \(\ds \paren {x \circ y} \circ z = x \circ \paren {y \circ z} \) |
Axiom $(1)$ and $(2)$ follow from distributivity of $\circ$.
Axiom $(3)$ follows from associativity of $\circ$.
$\blacksquare$
Proof 2
This is a special case of Module on Cartesian Product is Module:
- $\struct {R^n, +, \circ}_R$ is an $R$-module
where $n = 1$.
$\blacksquare$