Ideal of Ring/Examples/Order 2 Matrices with some Zero Entries/Proof 1

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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 $R$.


Proof

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 $R$.

$\blacksquare$


Sources

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