Trivial Module is Module

From ProofWiki
Jump to navigation Jump to search

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