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.

(c) Show that H is a division ring (defined in Exercise 42). [ Hint: If M = a1 + bi + cj + dk,, then verify that the solution of the equation Mx= 1 is the matrix ta1 - tbi - tcj - tdk, where t = 1/(a² + b² + c² + d².]

Short answer

It is proved that H is a division ring.

Step-by-step solution

Check for axiom 11

Prove that H satisfies Axiom 11, that is, whenever $a,b\in R$ and ab = 0_R then a = 0_R or b = 0_R .

Prove Axiom 11 by contraposition, assume that role="math" localid="1659350895139" $A=a_{1}1+b_{1}i+c_{1}j+d_{1}k$, and $B=a_{2}1+b_{2}i+c_{2}j+d_{2}k$ are any elements of H such that role="math" localid="1659351047009" $A\ne 0$and role="math" localid="1659351065258" $B\ne 0$. Now, we have to show that $AB\ne 0$.

Since $A\ne 0$and $B\ne 0$, this implies that there exists $x_1\in \left\{a_1,b_1,c_1,d_1\right\}$and $y_2\in \left\{a_2,b_2,c_2,d_2\right\}$, such that $x_1\ne 0$and $y_2\ne 0$.

From part (b), H is a noncommutative ring with identity. As x₁y₂ appear $\left(a_1{a}_2,a_1{b}_2...d_1{d}_2\right)$ as summands, in some elements in AB, this implies that $AB\ne 0$.

Therefore, H satisfies Axiom 11.

Prove the result

Prove that H satisfies Axiom 12, that is, for each $a\ne 0_R$ in R, the equation $ax=1_R$ has a solution in R.

Prove Axiom 12 as:

Assume that $A=a_{1}1+b_{1}i+c_{1}j+d_{1}k$ is some arbitrary element; now prove that AB = 1 where $B=ta1-tbi-tcj-tdk$and $t\frac{1}{a^2+b^2+c^2+d^2}$.

Then:

$\begin{array}{rcl}AB& =& \left(a_{1}1+b_{1}i+c_{1}j+d_{1}k\right)\left(ta1-tbi-tcj-tdk\right)\\ & =& t\left(a^{2}1-abi-acj-adk+abi+b^{2}1-bck+bdj+acj+bck+c^{2}1-cdi+adk-bdj+cdi+d^{2}1\right)\\ & =& t\left(a^{2}1+b^{2}1+c^{2}1+d^{2}1\right)\\ & =& t\left(a^2+b^2+c^2+d^2\right)1\\ & =& \frac{1}{a^2+b^2+c^2+d^2}\left(a^2+b^2+c^2+d^2\right)1\\ & =& 1\end{array}$

This implies that matrix B is the solution of the equation Ax = 1. It implies H satisfies Axiom 12.

Hence it is proved that H is a division ring.