# Right Module over Commutative Ring induces Bimodule

## Theorem

Let $\struct {R, +_R, \times_R}$ be a commutative ring.

Let $\struct {G, +_G, \circ}$ be a right module over $\struct {R, +_R, \times_R}$.

Let $\circ' : R \times G \to G$ be the binary operation defined by:

$\forall \lambda \in R: \forall x \in G: \lambda \circ' x = x \circ \lambda$

Then $\struct {G, +_G, \circ', \circ}$ is a bimodule over $\struct {R, +_R, \times_R}$.

## Proof

From Right Module over Commutative Ring induces Left Module, $\struct {G, +_G, \circ'}$ is a left module.

Let $\lambda, \mu \in R$ and $x \in G$.

Then:

 $\ds \lambda \circ' \paren {x \circ \mu}$ $=$ $\ds \paren {x \circ \mu} \circ \lambda$ Definition of $\circâ€™$ $\ds$ $=$ $\ds x \circ \paren {\mu \circ \lambda}$ Right module axiom $(\text {RM} 3)$ : Associativity of Scalar Multiplication $\ds$ $=$ $\ds x \circ \paren {\lambda \circ \mu}$ Ring product $\circ$ is commutative $\ds$ $=$ $\ds \paren {x \circ \lambda} \circ \mu$ Right module axiom $(\text {RM} 3)$ : Associativity of Scalar Multiplication $\ds$ $=$ $\ds \paren {\lambda \circ' x} \circ \mu$ Definition of $\circ'$

Hence $\struct{G, +_G, \circ', \circ}$ is a bimodule over $\struct {R, +_R, \times_R}$ by definition.

$\blacksquare$