Left 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 left module over $\struct {R, +_R, \times_R}$.

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

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


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

Proof

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

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

Then:

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