Give an example of a non commutative ring $\mathcal{R}$ which has an ideal $I$ and $\mathcal{R}/I$ is commutative.
Solution
For the subset
$$\mathcal{R}=\bigg\{\bigg(\begin{smallmatrix}
[{\rm{k}}]_2 & [{\rm{m}}]_2\\
[0]_2 & [{\rm{n}}]_2
\end{smallmatrix}\bigg)\,\Big|\; {\rm{k,m,n}}\in\{0,1\}\bigg\}$$
of ${\cal{M}}_2(\mathbb{Z}_2)$, we have that
\begin{align*}
\bigg(\begin{smallmatrix}
[{\rm{k}}]_2 & [{\rm{m}}]_2\\
[0]_2 & [{\rm{n}}]_2
\end{smallmatrix}\bigg)-\bigg(\begin{smallmatrix}
[{\rm{a}}]_2 & [{\rm{b}}]_2\\
[0]_2 & [{\rm{c}}]_2
\end{smallmatrix}\bigg)&=\bigg(\begin{smallmatrix}
[{\rm{k}}]_2+[{\rm{a}}]_2 & [{\rm{m}}]_2+[{\rm{b}}]_2\\
[0]_2 & [{\rm{n}}]_2+[{\rm{c}}]_2
\end{smallmatrix}\bigg)\in R\,,\\
\bigg(\begin{smallmatrix}
[{\rm{k}}]_2 & [{\rm{m}}]_2\\
[0]_2 & [{\rm{n}}]_2
\end{smallmatrix}\bigg)\,\bigg(\begin{smallmatrix}
[{\rm{a}}]_2 & [{\rm{b}}]_2\\
[0]_2 & [{\rm{c}}]_2
\end{smallmatrix}\bigg)&=\bigg(\begin{smallmatrix}
[{\rm{ka}}]_2 & [{\rm{mc}}]_2+[{\rm{mc}}]_2\\
[0]_2 & [{\rm{nc}}]_2
\end{smallmatrix}\bigg)\in R\,,\\
\bigg(\begin{smallmatrix}
[1]_2 & [1]_2\\
[0]_2 & [1]_2
\end{smallmatrix}\bigg)\,\bigg(\begin{smallmatrix}
[0]_2 & [1]_2\\
[0]_2 & [1]_2
\end{smallmatrix}\bigg)=\bigg(\begin{smallmatrix}
[0]_2 & [0]_2\\
[0]_2 & [1]_2
\end{smallmatrix}\bigg)&\neq\bigg(\begin{smallmatrix}
[0]_2 & [1]_2\\
[0]_2 & [1]_2
\end{smallmatrix}\bigg)=\bigg(\begin{smallmatrix}
[0]_2 & [1]_2\\
[0]_2 & [1]_2
\end{smallmatrix}\bigg)\,\bigg(\begin{smallmatrix}
[1]_2 & [1]_2\\
[0]_2 & [1]_2
\end{smallmatrix}\bigg) \,.
\end{align*}
So, $\mathcal{R}$ ia a non-commutative ring with $2^3=8$ elements. For the subring
$$I=\bigg\{\bigg(\begin{smallmatrix}
[0]_2 & [{\rm{r}}]_2\\
[0]_2 & [{\rm{s}}]_2
\end{smallmatrix}\bigg)\,\Big|\; {\rm{r,s}}\in\{0,1\}\bigg\}$$
of $\mathcal{R}$ and for every
$\bigg(\begin{smallmatrix}
[{\rm{k}}]_2 & [{\rm{m}}]_2\\
[0]_2 & [{\rm{n}}]_2
\end{smallmatrix}\bigg)\in R\,,\; \bigg(\begin{smallmatrix}
[0]_2 & [{\rm{r}}]_2\\
[0]_2 & [{\rm{s}}]_2
\end{smallmatrix}\bigg)\in I$,
we have that
\begin{align*}
\bigg(\begin{smallmatrix}
[{\rm{k}}]_2 & [{\rm{m}}]_2\\
[0]_2 & [{\rm{n}}]_2
\end{smallmatrix}\bigg)\,\bigg(\begin{smallmatrix}
[0]_2 & [{\rm{r}}]_2\\
[0]_2 & [{\rm{s}}]_2
\end{smallmatrix}\bigg)&=\bigg(\begin{smallmatrix}
[0]_2 & [{\rm{kr}}]_2+[{\rm{ms}}]_2\\
[0]_2 & [{\rm{ns}}]_2
\end{smallmatrix}\bigg)\in I\,,\\
\bigg(\begin{smallmatrix}
[0]_2 & [{\rm{r}}]_2\\
[0]_2 & [{\rm{s}}]_2
\end{smallmatrix}\bigg)\,\bigg(\begin{smallmatrix}
[{\rm{k}}]_2 & [{\rm{m}}]_2\\
[0]_2 & [{\rm{n}}]_2
\end{smallmatrix}\bigg)&=\bigg(\begin{smallmatrix}
[0]_2 & [{\rm{rn}}]_2\\
[0]_2 & [{\rm{ns}}]_2
\end{smallmatrix}\bigg)\in I\,.
\end{align*}
Therefor $I$ is an ideal with $2^2=4$ elements. The quotient ring $\mathcal{R}\,/\,I$ has $\frac{8}{4}=2$ elements and is isomorphic with $\mathbb{Z}_2$, which is a commutative ring.
The exercise can also be found at mathimatikoi.org
Solution
For the subset
$$\mathcal{R}=\bigg\{\bigg(\begin{smallmatrix}
[{\rm{k}}]_2 & [{\rm{m}}]_2\\
[0]_2 & [{\rm{n}}]_2
\end{smallmatrix}\bigg)\,\Big|\; {\rm{k,m,n}}\in\{0,1\}\bigg\}$$
of ${\cal{M}}_2(\mathbb{Z}_2)$, we have that
\begin{align*}
\bigg(\begin{smallmatrix}
[{\rm{k}}]_2 & [{\rm{m}}]_2\\
[0]_2 & [{\rm{n}}]_2
\end{smallmatrix}\bigg)-\bigg(\begin{smallmatrix}
[{\rm{a}}]_2 & [{\rm{b}}]_2\\
[0]_2 & [{\rm{c}}]_2
\end{smallmatrix}\bigg)&=\bigg(\begin{smallmatrix}
[{\rm{k}}]_2+[{\rm{a}}]_2 & [{\rm{m}}]_2+[{\rm{b}}]_2\\
[0]_2 & [{\rm{n}}]_2+[{\rm{c}}]_2
\end{smallmatrix}\bigg)\in R\,,\\
\bigg(\begin{smallmatrix}
[{\rm{k}}]_2 & [{\rm{m}}]_2\\
[0]_2 & [{\rm{n}}]_2
\end{smallmatrix}\bigg)\,\bigg(\begin{smallmatrix}
[{\rm{a}}]_2 & [{\rm{b}}]_2\\
[0]_2 & [{\rm{c}}]_2
\end{smallmatrix}\bigg)&=\bigg(\begin{smallmatrix}
[{\rm{ka}}]_2 & [{\rm{mc}}]_2+[{\rm{mc}}]_2\\
[0]_2 & [{\rm{nc}}]_2
\end{smallmatrix}\bigg)\in R\,,\\
\bigg(\begin{smallmatrix}
[1]_2 & [1]_2\\
[0]_2 & [1]_2
\end{smallmatrix}\bigg)\,\bigg(\begin{smallmatrix}
[0]_2 & [1]_2\\
[0]_2 & [1]_2
\end{smallmatrix}\bigg)=\bigg(\begin{smallmatrix}
[0]_2 & [0]_2\\
[0]_2 & [1]_2
\end{smallmatrix}\bigg)&\neq\bigg(\begin{smallmatrix}
[0]_2 & [1]_2\\
[0]_2 & [1]_2
\end{smallmatrix}\bigg)=\bigg(\begin{smallmatrix}
[0]_2 & [1]_2\\
[0]_2 & [1]_2
\end{smallmatrix}\bigg)\,\bigg(\begin{smallmatrix}
[1]_2 & [1]_2\\
[0]_2 & [1]_2
\end{smallmatrix}\bigg) \,.
\end{align*}
So, $\mathcal{R}$ ia a non-commutative ring with $2^3=8$ elements. For the subring
$$I=\bigg\{\bigg(\begin{smallmatrix}
[0]_2 & [{\rm{r}}]_2\\
[0]_2 & [{\rm{s}}]_2
\end{smallmatrix}\bigg)\,\Big|\; {\rm{r,s}}\in\{0,1\}\bigg\}$$
of $\mathcal{R}$ and for every
$\bigg(\begin{smallmatrix}
[{\rm{k}}]_2 & [{\rm{m}}]_2\\
[0]_2 & [{\rm{n}}]_2
\end{smallmatrix}\bigg)\in R\,,\; \bigg(\begin{smallmatrix}
[0]_2 & [{\rm{r}}]_2\\
[0]_2 & [{\rm{s}}]_2
\end{smallmatrix}\bigg)\in I$,
we have that
\begin{align*}
\bigg(\begin{smallmatrix}
[{\rm{k}}]_2 & [{\rm{m}}]_2\\
[0]_2 & [{\rm{n}}]_2
\end{smallmatrix}\bigg)\,\bigg(\begin{smallmatrix}
[0]_2 & [{\rm{r}}]_2\\
[0]_2 & [{\rm{s}}]_2
\end{smallmatrix}\bigg)&=\bigg(\begin{smallmatrix}
[0]_2 & [{\rm{kr}}]_2+[{\rm{ms}}]_2\\
[0]_2 & [{\rm{ns}}]_2
\end{smallmatrix}\bigg)\in I\,,\\
\bigg(\begin{smallmatrix}
[0]_2 & [{\rm{r}}]_2\\
[0]_2 & [{\rm{s}}]_2
\end{smallmatrix}\bigg)\,\bigg(\begin{smallmatrix}
[{\rm{k}}]_2 & [{\rm{m}}]_2\\
[0]_2 & [{\rm{n}}]_2
\end{smallmatrix}\bigg)&=\bigg(\begin{smallmatrix}
[0]_2 & [{\rm{rn}}]_2\\
[0]_2 & [{\rm{ns}}]_2
\end{smallmatrix}\bigg)\in I\,.
\end{align*}
Therefor $I$ is an ideal with $2^2=4$ elements. The quotient ring $\mathcal{R}\,/\,I$ has $\frac{8}{4}=2$ elements and is isomorphic with $\mathbb{Z}_2$, which is a commutative ring.
The exercise can also be found at mathimatikoi.org
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