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