Non commutative \(R\), commutative \(R\,/\,I\)

Groups, Rings, Domains, Modules, etc, Galois theory
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Grigorios Kostakos
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Non commutative \(R\), commutative \(R\,/\,I\)

#1

Post by Grigorios Kostakos »

Find a non commutative ring $R$ which has an ideal $I$, such that the quotient ring $R\,/\,I$ is commutative.
Grigorios Kostakos
Papapetros Vaggelis
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Re: Non commutative \(R\), commutative \(R\,/\,I\)

#2

Post by Papapetros Vaggelis »

Hi Grigorios.

Let \(\displaystyle{\left(R,+,\cdot\right)}\) be any associative ring with unity and non-commutative.

(For example, \(\displaystyle{\left(R,+,\cdot\right)=\left(\mathbb{M}_{3}(\mathbb{R}),+,\cdot\right)}\) or (\(\displaystyle{(R,+,\cdot)=(\mathbb{H},+,\cdot)}\)) (quaternions)



Consider the left ideal \(\displaystyle{I=\langle{\left\{a\,b-b\,a\in R: a\,,b\in R\right\}\rangle}}\) i.e, the

"smallest" ideal generated by the set \(\displaystyle{X=\left\{a\,b-b\,a\in R:a\,,b\in R\right\}}\).

Let \(\displaystyle{r\in R\,\,,a\,b-b\,a\in X}\). Then,

\(\displaystyle{(a\,b-b\,a)\,r=a\,(b\,r)-a\,(r\,b)+(a\,r)\,b-b\,(a\,r)=a\,(b\,r-r\,b)+(a\,r)\,b-b\,(a\,r)\in I}\)

since \(\displaystyle{b\,r-r\,b\in X\subseteq I\,\,,(a\,r)\,b-b\,(a\,r)\in X\subseteq I\,,a\in R}\) and

\(\displaystyle{I}\) is a left ideal of \(\displaystyle{R}\).

Therefore, if \(\displaystyle{x=\sum_{i=1}^{n}x_i\,(a_i\,b_i-b_i\,a_i)\,y_i\in I}\) and \(\displaystyle{r\in R}\),

then, \(\displaystyle{x\,r=\sum_{i=1}^{n}x_i\,(a_i\,b_i\,y_i-b_i\,a_i\,y_i)\in I}\).

So, \(\displaystyle{I}\) is also a right ideal, that is a double ideal and we consider the quotient ring

\(\displaystyle{R/I}\).

If \(\displaystyle{r+I\,,s+I\in R/I}\), then,

\(\displaystyle{(r+I)\,(s+I)=(s+I)\,(r+I)\iff r\,s+I=s\,r+I\iff r\,s-s\,r\in I}\), where, the last relation

is true, so \(\displaystyle{R/I}\) is commutative.
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Grigorios Kostakos
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Re: Non commutative \(R\), commutative \(R\,/\,I\)

#3

Post by Grigorios Kostakos »

Thank you Vaggelis for the nice solution. Follows an answer based on finite elements principle.

For the subset
$$R=\bigg\{\bigg(\begin{smallmatrix}
[{\rm{k}}]_2 & [{\rm{m}}]_2\\
[0]_2 & [{\rm{n}}]_2
\end{smallmatrix}\bigg)\,\Big|\; {\rm{k,m,n}}\in\{0,1\}\bigg\}$$
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, $R$ 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 $R$ 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 $R\,/\,I$ has $\frac{8}{4}=2$ elements and is isomorphic with $\mathbb{Z}_2$, which is a commutative ring.
Grigorios Kostakos
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