Left Module over Commutative Ring induces Bimodule

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:

\(\ds \lambda \circ \paren{x \circ' \mu}\)

\(=\)

\(\ds \lambda \circ \paren{\mu \circ x}\)

Definition of $\circ’$

\(\ds \)

\(=\)

\(\ds \paren {\lambda \circ \mu} \circ x\)

Module Axiom $\text M 3$: Associativity

\(\ds \)

\(=\)

\(\ds \paren {\mu \circ \lambda} \circ x\)

Ring product $\circ$ is commutative

\(\ds \)

\(=\)

\(\ds \mu \circ \paren{\lambda \circ x}\)

Module Axiom $\text M 3$: Associativity

\(\ds \)

\(=\)

\(\ds \paren{\lambda \circ x} \circ' \mu\)

Definition of $\circ'$

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

$\blacksquare$

Also see

• Right Module over Commutative Ring induces Bimodule

Sources

• 2003: P.M. Cohn: Basic Algebra: Groups, Rings and Fields ... (previous): Chapter $4$: Rings and Modules: $\S 4.1$: The Definitions Recalled