Trivial Module is Module
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Theorem
Let $\struct {G, +_G}$ be an abelian group whose identity is $e_G$.
Let $\struct {R, +_R, \circ_R}$ be a ring.
Let $\struct {G, +_G, \circ}_R$ be the trivial $R$-module, such that:
- $\forall \lambda \in R: \forall x \in G: \lambda \circ x = e_G$
Then $\struct {G, +_G, \circ}_R$ is a module.
Proof
Checking the module axioms in turn:
- Module Axiom $\text M 1$: Distributivity over Module Addition: $\quad \lambda \circ \paren {x +_G y} = e_G = e_G +_G e_G = \paren {\lambda \circ x} +_G \paren {\lambda \circ y}$
- Module Axiom $\text M 2$: Distributivity over Scalar Addition: $\quad \paren {\lambda +_R \mu} \circ x = e_G = e_G +_G e_G = \paren {\lambda \circ x} +_G \paren {\mu \circ x}$
- Module Axiom $\text M 3$: Associativity: $\quad \paren {\lambda \times_R \mu} \circ x = e_G = \lambda \circ e_G = \lambda \circ \paren {\mu \circ x}$
Thus the trivial module is indeed a module.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {V}$: Vector Spaces: $\S 26$. Vector Spaces and Modules: Example $26.6$