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

## Theorem

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

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

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

Then:

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

## Proof

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

$(1): \quad G \ne \O$
$(2): \quad \forall \mathop {\mathbf X}, \mathop{\mathbf Y} \in G: \mathbf X + \paren {-\mathbf Y} \in G$
$(3): \quad \forall \mathop{\mathbf J} \in G, \mathop{\mathbf R} \in \map {\MM_S} 2: \mathbf J \times \mathbf R \in G$

### Condition $(1): \quad G \ne \O$

By definition of $G$:

$\quad \begin {bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \in G$

$\Box$

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

Let:

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

Then:

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

$\Box$

### Condition $(3): \quad \forall \mathop{\mathbf J} \in G, \mathop{\mathbf R} \in \map {\MM_S} 2: \mathbf J \times \mathbf R \in G$

Let:

$\quad \mathbf J = \begin{bmatrix} j_1 & j_2 \\ 0 & 0 \end{bmatrix} \in G, \quad \mathbf R = \begin{bmatrix} r_{1 1} & r_{2 1} \\ r_{1 2} & r_{2 2} \end{bmatrix} \in \map {\MM_S} 2$

Then:

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

$\blacksquare$