Question

In the ring M(C), let

$\mathbf{1}\mathbf{=}\mathbf{\left(}\begin{array}{cc}\mathbf{1}& \mathbf{0}\\ \mathbf{0}& \mathbf{1}\end{array}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{i}\mathbf{=}\mathbf{\left(}\begin{array}{cc}\mathbf{i}& \mathbf{0}\\ \mathbf{0}& \mathbf{-}\mathbf{i}\end{array}\mathbf{\right)}\mathbf{}\mathbf{}\mathit{j}\mathbf{=}\mathbf{\left(}\begin{array}{cc}\mathbf{0}& \mathbf{1}\\ \mathbf{-}\mathbf{1}& \mathbf{0}\end{array}\mathbf{\right)}\mathbf{}\mathbf{}\mathit{k}\mathbf{=}\mathbf{\left(}\begin{array}{cc}\mathbf{0}& \mathbf{i}\\ \mathbf{i}& \mathbf{0}\end{array}\mathbf{\right)}$

The Product of a real number and a matrix is the matrix given by this rule:

$\mathit{r}\mathbf{\left(}\begin{array}{cc}\mathbf{t}& \mathbf{u}\\ \mathbf{v}& \mathbf{w}\end{array}\mathbf{\right)}\mathbf{=}\mathbf{\left(}\begin{array}{cc}\mathbf{rt}& \mathbf{ru}\\ \mathbf{rv}& \mathbf{rw}\end{array}\mathbf{\right)}$

The set H of real quaternions consists of all matrices of the form

$\begin{array}{rcl}\mathit{a}\mathbf{1}\mathbf{+}\mathit{b}\mathit{i}\mathbf{+}\mathit{c}\mathit{j}\mathbf{+}\mathit{d}\mathit{k}& \mathbf{=}& \mathit{a}\mathbf{\left(}\begin{array}{cc}\mathbf{1}& \mathbf{0}\\ \mathbf{0}& \mathbf{1}\end{array}\mathbf{\right)}\mathbf{+}\mathit{b}\mathbf{\left(}\begin{array}{cc}\mathbf{i}& \mathbf{0}\\ \mathbf{0}& \mathbf{-}\mathbf{i}\end{array}\mathbf{\right)}\mathbf{+}\mathit{c}\mathbf{\left(}\begin{array}{cc}\mathbf{0}& \mathbf{1}\\ \mathbf{-}\mathbf{1}& \mathbf{0}\end{array}\mathbf{\right)}\mathbf{+}\mathit{d}\mathbf{\left(}\begin{array}{cc}\mathbf{0}& \mathbf{i}\\ \mathbf{i}& \mathbf{0}\end{array}\mathbf{\right)}\\ & \mathbf{=}& \mathbf{\left(}\begin{array}{cc}\mathbf{a}& \mathbf{0}\\ \mathbf{0}& \mathbf{a}\end{array}\mathbf{\right)}\mathbf{+}\mathbf{\left(}\begin{array}{cc}\mathbf{bi}& \mathbf{0}\\ \mathbf{0}& \mathbf{-}\mathit{b}\mathbf{i}\end{array}\mathbf{\right)}\mathbf{+}\mathbf{\left(}\begin{array}{cc}\mathbf{0}& c\\ \mathbf{-}\mathit{c}& \mathbf{0}\end{array}\mathbf{\right)}\mathbf{+}\mathbf{\left(}\begin{array}{cc}\mathbf{0}& \mathbf{di}\\ \mathbf{di}& \mathbf{0}\end{array}\mathbf{\right)}\\ & \mathbf{=}& \mathbf{\left(}\begin{array}{cc}\mathbf{a}\mathbf{+}\mathbf{bi}& \mathbf{c}\mathbf{+}\mathbf{di}\\ \mathbf{-}\mathbf{c}\mathbf{+}\mathbf{di}& \mathbf{a}\mathbf{-}\mathbf{bi}\end{array}\mathbf{\right)}\end{array}$

Where a,b,c are d real numbers.

(b) Show that H is a noncommutative ring with identity.

Short answer

It is proved that H is a noncommutative ring with identity.

Step-by-step solution

Define

The set H is defined as:

$\begin{array}{rcl}H& =& \left\{a1+bi+cj+dk\overline{)a,b,c,d\in \mathrm{ℝ}}\right\}\\ & =& \left\{\left(\begin{array}{cc}a+bi& c+di\\ -c+di& a-bi\end{array}\right)\overline{)a,b,c,d\in \mathrm{ℝ}}\right\}\end{array}$

As H is the subset of a ring M(C), we have to show that H is the subring of a ring M(C) .

Prove that H is a subring of M(C) 

Verify the conditions as:

1. Show that$A+B\in H$ as:
   $\begin{array}{rcl}A& =& \left(\begin{array}{cc}{a}_1+b_{1}i& \\ -c_1+b_{1}i& \end{array}\begin{array}{c}{c}_1+d_{1}i\\ a_1+b_{1}i\end{array}\right),B=\left(\begin{array}{ccc}{a}_2+b_{2}i& & c_2+d_{2}i\\ -c_2+d_{2}i& & a_2-b_{2}i\end{array}\right)\in H\\ A+B& =& \left(\begin{array}{cc}{a}_1+b_{1}i& \\ -c_1+b_{1}i& \end{array}\begin{array}{c}{c}_1+d_{1}i\\ a_1+b_{1}i\end{array}\right)+\left(\begin{array}{ccc}{a}_2+b_{2}i& & c_2+d_{2}i\\ -c_2+d_{2}i& & a_2-b_{2}i\end{array}\right)\\ & =& \left(\begin{array}{cc}\left(a_1+a_2\right)i& +\\ -\left(c_1+c_2\right)& +\end{array}\begin{array}{c}\left(b_1+b_2\right)i\\ \left(d+d_2\right)i\end{array}\begin{array}{cc}\begin{array}{cc}\left(c_1+c_2\right)& +\end{array}& \left(d_1+d_2\right)\\ \begin{array}{cc}\left(a_1+a_2\right)i& +\end{array}& \left(b_1+b_2\right)\end{array}\begin{array}{c}i\\ i\end{array}\right)\end{array}$
   

2. Show that$AB\in H$ as:
   role="math" localid="1659349242071" $\begin{array}{rcl}A& =& \left(\begin{array}{ccc}{a}_1+b_{1}i& & c_1+d_{1}i\\ -c_1+d_{1}i& & a_1-b_{1}i\end{array}\right),B=\left(\begin{array}{ccc}{a}_2+b_{2}i& & c_2+d_{2}i\\ -c_2+d_{2}i& & a_2-b_{2}i\end{array}\right)\in H\\ AB& =& \left(\begin{array}{ccc}{a}_1+b_{1}i& & c_1+d_{1}i\\ -c_1+d_{1}i& & a_1-b_{1}i\end{array}\right)\left(\begin{array}{ccc}{a}_2+b_{2}i& & c_2+d_{2}i\\ -c_2+d_{2}i& & a_2-b_{2}i\end{array}\right)\\ & =& \left(\begin{array}{cc}\left(a_1{a}_2-b_1{b}_2-c_1{c}_2-d_1{d}_2\right)+\left(a_1{b}_2+a_2{b}_1+c_1{d}_2-c_2{d}_1\right)i& \\ -\left(a_1{c}_2-b_1{d}_2+a_2{c}_1+b_2{d}_1\right)+\left(a_1{b}_2+b_1{c}_2-b_2{c}_1+a_2{d}_1\right)i& \end{array}\begin{array}{c}\left(a_1{c}_2-b_1{d}_2+a_2{c}_1+b_2{d}_1\right)+\left(a_1{b}_2+b_1{c}_2-b_2{c}_1+a_2{d}_1\right)i\\ \left(a_1{a}_2-b_1{b}_2-c_1{c}_2-d_1{d}_2\right)+\left(a_1{b}_2+a_2{b}_1+c_1{d}_2-c_2{d}_1\right)i\end{array}\right)\in H\end{array}$
   

3. Assume that 0 denotes 0_M(C) then:
   $\begin{array}{rcl}0& =& \left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)\\ & =& \left(\begin{array}{cc}0+0i& 0+0i\\ -0+0i& 0-0i\end{array}\right)\\ & \in & H\end{array}$
   

4. Consider $\begin{array}{rcl}A& =& \left(\begin{array}{ccc}{a}_1+b_{1}i& & c_1+d_{1}i\\ -c_1+d_{1}i& & a_1-b_{1}i\end{array}\right)\in H\end{array}$and$B\in M\left(C\right)$ , such that A + B = 0 . Then:
   $\begin{array}{rcl}B& =& \left(\begin{array}{ccc}-a+\left(-b\right)i& & -c+\left(-d\right)i\\ -\left(-c\right)+{\left(-d\right)}_{1}i& & \left(-a\right)-\left(-b\right)i\end{array}\right)\\ & \in & H\end{array}$
   

This implies that H is a subring of M(C) . Hence H is a ring.