Definition:Trivial 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 $$\circ$$ be defined as $$\forall \lambda \in R: \forall x \in G: \lambda \circ x = e_G$$.

Then $$\left({G, +_G: \circ}\right)_R$$ is an $R$-module.

Such a module is called a trivial module.

Unless $$R$$ is a ring with unity and $$G$$ contains only one element, this is not a unitary module.

Proof
Checking the criteria for module in turn:


 * VS1: $$\lambda \circ \left({x +_G y}\right) = e_G = e_G +_G e_G = \left({\lambda \circ x}\right) +_G \left({\lambda \circ y}\right)$$.


 * VS2: $$\left({\lambda +_R \mu}\right) \circ x = e_G = e_G +_G e_G = \left({\lambda \circ x}\right) +_G \left({\mu \circ x}\right)$$.


 * VS3: $$\left({\lambda \times_R \mu}\right) \circ x = e_G = \lambda \circ e_G = \lambda \circ \left({\mu \circ x}\right)$$.

Thus the trivial module is indeed a module.


 * By definition, for the trivial module to be unitary, then $$R$$ needs to be a ring with unity.

For VS4 to apply, we require $$\forall x \in G: 1_R \circ x = x$$.

But for the trivial module, $$\forall x \in G: 1_R \circ x = e_G$$.

So VS4 can apply only when $$\forall x \in G: x = e_G$$.

Thus for the trivial module to be unitary, it is necessary that $$G$$ is the Trivial Group, and thus contains one element.