Right Module over Commutative Ring induces Bimodule

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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:

\(\displaystyle \lambda \circ’ \paren{x \circ \mu}\) \(=\) \(\displaystyle \paren{x \circ \mu} \circ \lambda\) Definition of $\circ’$
\(\displaystyle \) \(=\) \(\displaystyle x \circ \paren {\mu \circ \lambda}\) Right module axiom $(RM \,3)$ : Associativity of Scalar Multiplication
\(\displaystyle \) \(=\) \(\displaystyle x \circ \paren {\lambda \circ \mu}\) Ring product $\circ$ is commutative
\(\displaystyle \) \(=\) \(\displaystyle \paren{x \circ \lambda} \circ \mu\) Right module axiom $(RM \,3)$ : Associativity of Scalar Multiplication
\(\displaystyle \) \(=\) \(\displaystyle \paren{\lambda \circ’ x} \circ \mu\) Definition of $\circ’$

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

$\blacksquare$

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