# Ideal of Ring/Examples/Order 2 Matrices with some Zero Entries

## Example of Ideal of Ring

Let $R$ be the set of all order $2$ square matrices of the form $\begin{pmatrix} x & y \\ 0 & z \end{pmatrix}$ with $x, y, z \in \R$.

Let $S$ be the set of all order $2$ square matrices of the form $\begin{pmatrix} x & y \\ 0 & 0 \end{pmatrix}$ with $x, y \in \R$.

Then $R$ is a ring and $S$ is an ideal of $R$.

### Corollary

$R / S \cong \R$

where:

$R / S$ is the quotient ring of $R$ by $S$
$\cong$ denotes ring isomorphism,

## Proof 1

Let $\begin{pmatrix} x_1 & y_1 \\ 0 & z_1 \end{pmatrix}, \begin{pmatrix} x_2 & y_2 \\ 0 & z_2 \end{pmatrix} \in R$.

Then:

 $\ds \begin{pmatrix} x_1 & y_1 \\ 0 & z_1 \end{pmatrix} + \begin{pmatrix} -x_2 & -y_2 \\ 0 & -z_2 \end{pmatrix}$ $=$ $\ds \begin{pmatrix} x_1 - x_2 & y_1 - y_2 \\ 0 & z_1 - z_2 \end{pmatrix}$ $\ds$ $\in$ $\ds R$ $\ds \begin{pmatrix} x_1 & y_1 \\ 0 & z_1 \end{pmatrix} \begin{pmatrix} x_2 & y_2 \\ 0 & z_2 \end{pmatrix}$ $=$ $\ds \begin{pmatrix} x_1 \times x_2 + y_1 \times 0 & x_1 \times y_2 + y_1 \times z_2 \\ 0 \times x_2 + z_1 \times 0 & 0 \times y_2 + z_1 \times z_2 \end{pmatrix}$ $\ds$ $=$ $\ds \begin{pmatrix} x_1 x_2 & x_1 y_2 + y_1 z_2 \\ 0 & z_1 z_2 \end{pmatrix}$ $\ds$ $\in$ $\ds R$

Thus by the Subring Test $R$ is a subring of the ring of order $2$ matrices over $\R$.

We have that, for example, $\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \in S$

Hence $S \ne \O$.

Let $\begin{pmatrix} x_1 & y_1 \\ 0 & z_1 \end{pmatrix}, \begin{pmatrix} x_2 & y_2 \\ 0 & z_2 \end{pmatrix} \in S$.

 $\ds \begin{pmatrix} x_1 & y_1 \\ 0 & 0 \end{pmatrix} + \begin{pmatrix} -x_2 & -y_2 \\ 0 & 0 \end{pmatrix}$ $=$ $\ds \begin{pmatrix} x_1 - x_2 & y_1 - y_2 \\ 0 & 0 \end{pmatrix}$ $\ds$ $\in$ $\ds S$

and so $S$ is closed under matrix subtraction.

Now let $\begin{pmatrix} a & b \\ 0 & 0 \end{pmatrix} \in S$ for real $a, b$.

Let $\begin{pmatrix} x & y \\ 0 & z \end{pmatrix} \in R$.

Then we have:

 $\ds \begin{pmatrix} x & y \\ 0 & z \end{pmatrix} \begin{pmatrix} a & b \\ 0 & 0 \end{pmatrix}$ $=$ $\ds \begin{pmatrix} x \times a + y \times 0 & x \times b + y \times 0 \\ 0 \times a + z \times 0 & 0 \times b + z \times 0 \end{pmatrix}$ $\ds$ $=$ $\ds \begin{pmatrix} x a & x b \\ 0 & 0 \end{pmatrix}$ $\ds$ $\in$ $\ds S$

and:

 $\ds \begin{pmatrix} a & b \\ 0 & 0 \end{pmatrix} \begin{pmatrix} x & y \\ 0 & z \end{pmatrix}$ $=$ $\ds \begin{pmatrix} a \times x + b \times 0 & a \times y + b \times z \\ 0 \times x + 0 \times 0 & 0 \times y + 0 \times z \end{pmatrix}$ $\ds$ $=$ $\ds \begin{pmatrix} a x & a y + b z \\ 0 & 0 \end{pmatrix}$ $\ds$ $\in$ $\ds S$

Thus, by the Test for Ideal, $S$ is an ideal of $R$.

$\blacksquare$

## Proof 2

Consider the mapping $\phi: R \to \R$ defined as:

$\forall \mathbf A \in R: \map \phi {\begin {pmatrix} x & y \\ 0 & z \end {pmatrix} } = z$

It is to be demonstrated that $\phi$ is a ring homomorphism whose kernel is $S$.

Thus:

 $\ds \map \phi {\begin {pmatrix} x_1 & y_1 \\ 0 & z_1 \end {pmatrix} + \begin {pmatrix} x_2 & y_2 \\ 0 & z_2 \end {pmatrix} }$ $=$ $\ds \map \phi {\begin {pmatrix} x_1 + x_2 & y_1 + y_2 \\ 0 & z_1 + z_2 \end {pmatrix} }$ Definition of Matrix Entrywise Addition $\ds$ $=$ $\ds z_1 + z_2$ Definition of $\phi$ $\ds$ $=$ $\ds \map \phi {\begin {pmatrix} x_1 & y_1 \\ 0 & z_1 \end {pmatrix} } + \map \phi {\begin {pmatrix} x_2 & y_2 \\ 0 & z_2 \end {pmatrix} }$ Definition of $\phi$

and:

 $\ds \map \phi {\begin {pmatrix} x_1 & y_1 \\ 0 & z_1 \end {pmatrix} \begin {pmatrix} x_2 & y_2 \\ 0 & z_2 \end {pmatrix} }$ $=$ $\ds \map \phi {\begin {pmatrix} x_1 x_2 & x_1 y_2 + y_1 x_2 \\ 0 & z_1 z_2 \end {pmatrix} }$ Definition of Matrix Product (Conventional) $\ds$ $=$ $\ds z_1 z_2$ Definition of $\phi$ $\ds$ $=$ $\ds \map \phi {\begin {pmatrix} x_1 & y_1 \\ 0 & z_1 \end {pmatrix} } \times \map \phi {\begin {pmatrix} x_2 & y_2 \\ 0 & z_2 \end {pmatrix} }$ Definition of $\phi$

Thus by definition $\phi$ is a ring homomorphism.

By definition of $S$ itself, we have that:

$S \subseteq \map \ker \phi$

Then we have that:

 $\ds \mathbf A$ $\in$ $\ds \map \ker \phi$ $\ds \leadsto \ \$ $\ds \map \phi {\mathbf A}$ $=$ $\ds 0$ Definition of Kernel of Ring Homomorphism $\ds \leadsto \ \$ $\ds \map \phi {\begin {pmatrix} x & y \\ 0 & z \end {pmatrix} }$ $=$ $\ds 0$ where $\mathbf A = \begin {pmatrix} x & y \\ 0 & z \end {pmatrix}$ $\ds \leadsto \ \$ $\ds z$ $=$ $\ds 0$ Definition of $\phi$ $\ds \leadsto \ \$ $\ds \mathbf A$ $\in$ $\ds S$ Definition of $S$ $\ds \leadsto \ \$ $\ds \map \ker \phi$ $\subseteq$ $\ds S$ Definition of $S$

Hence:

$\map \ker \phi = S$
$S$ is an ideal of $R$.

$\blacksquare$