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:
\(\ds \lambda \circ' \paren {x \circ \mu}\) | \(=\) | \(\ds \paren {x \circ \mu} \circ \lambda\) | Definition of $\circ’$ | |||||||||||
\(\ds \) | \(=\) | \(\ds x \circ \paren {\mu \circ \lambda}\) | 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.
$\blacksquare$
Also see
Sources
- 2003: P.M. Cohn: Basic Algebra: Groups, Rings and Fields ... (previous) ... (next): Chapter $4$: Rings and Modules: $\S 4.1$: The Definitions Recalled