Right Module Does Not Necessarily Induce Left Module over Ring

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Theorem

Let $\struct {R, +_R, \times_R}$ be a 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 $\struct{G, +_G, \circ’}$ is not necessarily a left module over $\struct {R, +_R, \times_R}$

Proof

Proof by Counterexample

Let $\struct {S, +, \times}$ be a ring with unity

Let $\struct {\map {\mathcal M_S} 2, +, \times}$ denote the ring of square matrices of order $2$ over $S$.

From Ring of Square Matrices over Ring is Ring, $\struct {\map {\mathcal M_S} 2, +, \times}$ is a ring.


Let $G = \set {\begin{bmatrix} x & y \\ 0 & 0 \end{bmatrix} : x, y \in S }$

Lemma

$G$ is a right ideal of $\struct {\map {\mathcal M_S} 2, +, \times}$.


From Right Ideal is Right Module over Ring, $\struct {G, +, \times}$ is a right module over $\struct {\map {\mathcal M_S} 2, +, \times}$.


Let $R = \map {\mathcal M_S} 2$.

Let $\mathbf A = \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}, \mathbf B = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \mathbf C = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$

Then:

\(\displaystyle \mathbf A \times \paren {\mathbf B \times \mathbf C}\) \(=\) \(\displaystyle \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix} \times \paren {\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \times \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} }\)
\(\displaystyle \) \(=\) \(\displaystyle \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix} \times \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\)
\(\displaystyle \) \(=\) \(\displaystyle \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\)

And:

\(\displaystyle \mathbf A \times \paren {\mathbf C \times \mathbf B}\) \(=\) \(\displaystyle \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix} \times \paren {\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \times \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} }\)
\(\displaystyle \) \(=\) \(\displaystyle \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix} \times \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)
\(\displaystyle \) \(=\) \(\displaystyle \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)


Hence for $\mathbf A, \mathbf B \in R, \mathbf C \in G$:

$ \mathbf A \times \paren{\mathbf B \times \mathbf C} \neq \mathbf A \times \paren {\mathbf C \times \mathbf B}$


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

$\forall \mathbf \Lambda \in R: \forall \mathbf X \in G: \mathbf \Lambda \circ \mathbf X = \mathbf X \times \mathbf \Lambda$

From Right Module induces Left Module over same Ring iff Actions are Commutative, $\struct {G, +, \circ}$ is not a left module over $\struct {R, +, \times}$

$\blacksquare$

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