Trivial Module is Module

Theorem
Let $\left({G, +_G}\right)$ be an abelian group whose identity is $e_G$.

Let $\left({R, +_R, \circ_R}\right)$ be a ring.

Let $\left({G, +_G, \circ}\right)_R$ be the trivial $R$-module, such that:


 * $\forall \lambda \in R: \forall x \in G: \lambda \circ x = e_G$

Then $\left({G, +_G, \circ}\right)_R$ is a module.

Proof
Checking the criteria for module in turn:


 * $(1): \quad \lambda \circ \left({x +_G y}\right) = e_G = e_G +_G e_G = \left({\lambda \circ x}\right) +_G \left({\lambda \circ y}\right)$


 * $(2): \quad \left({\lambda +_R \mu}\right) \circ x = e_G = e_G +_G e_G = \left({\lambda \circ x}\right) +_G \left({\mu \circ x}\right)$


 * $(3): \quad \left({\lambda \times_R \mu}\right) \circ x = e_G = \lambda \circ e_G = \lambda \circ \left({\mu \circ x}\right)$

Thus the trivial module is indeed a module.