Left Module Does Not Necessarily Induce Right Module over Ring

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Theorem

Let $\struct {R, +_R, \times_R}$ be a 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 $\struct{G, +_G, \circ’}$ is not necessarily a right 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 $G = \set {\begin{bmatrix} x & 0 \\ y & 0 \end{bmatrix} : x, y \in S }$

Lemma

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


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


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

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

Then:

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

And:

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


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

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


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

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

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

$\blacksquare$

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