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:
- $\map \ker \phi = S$
From Kernel of Ring Homomorphism is Ideal:
- $S$ is an ideal of $R$.
$\blacksquare$