Left Module Does Not Necessarily Induce Right Module over Ring/Lemma

Theorem
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 $S$.

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

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

Proof
From Test for Left Ideal, the following need to be proved:


 * $(1): \quad G \ne \varnothing$


 * $(2): \quad \forall \mathop {\mathbf X}, \mathop{\mathbf Y} \in G: \mathbf X + \paren {-\mathbf Y} \in G$


 * $(3): \quad \forall \mathop{\mathbf J} \in G, \mathop{\mathbf R} \in \map {\mathcal M_S} 2: \mathbf R \times \mathbf J \in G$

Condition $(1): \quad G \ne \varnothing$
By definition of $G$, $\quad \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \in G$

Condition $(2): \quad \forall \mathop {\mathbf X}, \mathop{\mathbf Y} \in G: \mathbf X + \paren {-\mathbf Y} \in G$
Let $\quad \mathbf X = \begin{bmatrix} x_1 & 0 \\ x_2 & 0 \end{bmatrix}, \quad \mathbf Y = \begin{bmatrix} y_1 & 0 \\ y_2 & 0 \end{bmatrix} \in G$

Then $\quad \mathbf X - \mathbf Y = \begin{bmatrix} x_1 - y_1 & 0 \\ x_2 - y_2 & 0 \end{bmatrix} \in G$

Condition $(3): \quad \forall \mathop{\mathbf J} \in G, \mathop{\mathbf R} \in \map {\mathcal M_S} 2: \mathbf R \times \mathbf J \in G$
Let $\quad \mathbf J = \begin{bmatrix} j_1 & 0 \\ j_2 & 0 \end{bmatrix} \in G, \quad \mathbf R = \begin{bmatrix} r_{1 1} & r_{2 1} \\ r_{1 2} & r_{2 2} \end{bmatrix} \in \map {\mathcal M_S} 2$

Then $\quad \mathbf R \times \mathbf J = \begin{bmatrix} r_{1 1} j_1 + r_{2 1} j_2 & 0 \\ r_{1 2} j_1 + r_{2 2} j_2 & 0 \end{bmatrix} \in G$