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

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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

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 $R$.

$\blacksquare$


Sources

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