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
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $9$: Rings: Exercise $10$