Trivial Module is Module

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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.

$\blacksquare$


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