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Prove that the set \[\displaystyle{\mathbb{H}=\left\{\begin{pmatrix}

z & w\\

-\bar{w} & \bar{z}

\end{pmatrix} : z,w\in\mathbb{C}\right\}\subseteq \mathbb{M}_{2}\,\left(\mathbb{C}\right)}\,\] equipped with the usual operations of addition and multiplication of matrices, is a division ring, known as Tetranion division ring of \(\displaystyle{\rm{Hamilton}}\) .

1st part : \(\displaystyle{\mathbb{H}=\left\{\begin{pmatrix}
\,\,\,\,z && w\\
-\overline{w} && \overline{z}
\end{pmatrix} : z\,,w\in\mathbb{C}\right\}\subseteq \mathbb{M_2}\,(\mathbb{C})}\)

Obviously, \(\displaystyle{\mathbb{H}\neq \varnothing}\) cause

\(\displaystyle{\mathbb{O}= \begin{pmatrix}
0 && 0\\
0 && 0
\end{pmatrix}\in \mathbb{H}\,\,,I_2=\begin{pmatrix}
1 && 0\\
0 && 1
\end{pmatrix}=1_{\mathbb{M_2}\,(\mathbb{C})}\in\mathbb{H}}\)

Consider now

\(\displaystyle{A=\begin{pmatrix}
\,\,\,\,z_1 && w_1\\
-\overline{w_1} && \overline{z_1}
\end{pmatrix}\in\mathbb{H}\,\,,B=\begin{pmatrix}
\,\,\,\,z_2 && w_2\\
-\overline{w_2} && \overline{z_2}
\end{pmatrix}\in \mathbb{H}}\)

where \(\displaystyle{z_i\,,w_i\in\mathbb{C}\,,i\in\left\{1,2\right\}}\) .

\(\displaystyle{A+B=\begin{pmatrix}
z_1+z_2 && w_1+w_2\\
-\overline{w_1}-\overline{w_2} && \overline{z_1}+\overline{z_2}
\end{pmatrix}=\begin{pmatrix}
\,\,z_1+z_2 && w_1+w_2\\
-(\overline{w_1+w_2}) && \overline{z_1+z_2}
\end{pmatrix}\in\mathbb{H}}\)

and \(\displaystyle{-A=\begin{pmatrix}
-z_1 && -w_1\\
-(-\overline{w_1}) && -\overline{z_1}
\end{pmatrix}=\begin{pmatrix}
-z_1 && -w_1\\
-(\overline{-w_1}) && \overline{-z_1}
\end{pmatrix}\in\mathbb{H}}\)

Also,

\(\displaystyle{A\cdot B=\begin{pmatrix}
\,\,z_1\,z_2-w_1\,\overline{w_2} && z_1\,w_2+w_1\,\overline{z_2}\\
-z_2\,\overline{w_1}-\overline{z_1\,w_2} && -w_2\,\overline{w_1}+\overline{z_1\,z_2}\end{pmatrix}}\)

where :

\(\displaystyle{
-z_2\,\overline{w_1}-\overline{z_1\,w_2}=-(z_2\,\overline{w_1}+\overline{z_1\,w_2})=-\overline{z_1\,w_2+w_1\,z_2}}\)

and

\(\displaystyle{-w_2\,\overline{w_1}+\overline{z_1\,z_2}=\overline{z_1\,z_2-w_1\,\overline{w_2}}}\)

so, \(\displaystyle{A\cdot B\in\mathbb{H}}\) .

Therefore, the triplet \(\displaystyle{\left(\mathbb{H},+,\cdot\right)}\) is a ring with \(\displaystyle{\mathbb{O}= \begin{pmatrix}
0 && 0\\
0 && 0
\end{pmatrix}}\)

as the zero-element and

\(\displaystyle{I_2=\begin{pmatrix}
1 && 0\\
0 && 1
\end{pmatrix}}\)

as its unity.

Let \(\displaystyle{A=\begin{pmatrix}
\,\,\,\,z && w\\
-\overline{w} && \overline{z}
\end{pmatrix}\in\mathbb{H}-\left\{\mathbb{O}\right\}}\). We have that :

\(\displaystyle{z\neq 0}\) or \(\displaystyle{w\neq 0}\) and

\(\displaystyle{\det\,(A)=z\,\overline{z}+w\,\overline{w}=\left|z\right|^2+\left|w\right|^2>0}\), so the matrix \(\displaystyle{A}\)

is invertible. There is \(\displaystyle{A^{-1}\in\mathbb{M}_{2}\,(\mathbb{C})}\) such that \(\displaystyle{A\cdot A^{-1}=I_{2}=A^{-1}\cdot A}\) .

We''ll prove that \(\displaystyle{A^{-1}\in\mathbb{H}}\) and then we get the desired. It's known that

\(\displaystyle{A\cdot \mathrm{adj}(A)=\det\,(A)\,I_{2}=\mathrm{adj}(A)\cdot A}\) , so

\(\displaystyle{A^{-1}=\dfrac{1}{\det\,(A)}\,\mathrm{adj}(A)}\)


where:

\(\displaystyle{\mathrm{adj}(A)=\begin{pmatrix}
\overline{z} && -w\\
\overline{w} && z
\end{pmatrix}}\)

so, \(\displaystyle{A^{-1}=\begin{pmatrix}
\overline{z/a} && -w/a\\
-(\overline{-w/a}) && \overline{\overline{z/a}}
\end{pmatrix}\in\mathbb{H}\,\,,a=\det\,(A)}\)

Finally, \(\displaystyle{\left(\mathbb{H},+,\cdot\right)}\) is a division ring but not a field because

\(\displaystyle{\begin{pmatrix}
i && 0\\
0 && -i
\end{pmatrix}\in\mathbb{H}\,\,,\begin{pmatrix}
0 && 1\\
-1 && 0
\end{pmatrix}\in\mathbb{H}}\) and

\(\displaystyle{\begin{pmatrix}
i && 0\\
0 && -i
\end{pmatrix}\cdot \begin{pmatrix}
0 && 1\\
-1 && 0
\end{pmatrix}=\begin{pmatrix}
0 && i\\
i && 0
\end{pmatrix} }\)

but

\(\displaystyle{
\begin{pmatrix}
0 && 1\\
-1 && 0
\end{pmatrix}\cdot \begin{pmatrix}
i && 0\\
0 && -i
\end{pmatrix}= \begin{pmatrix}
0 && -i\\
-i && 0
\end{pmatrix}\neq \begin{pmatrix}
0 && i\\
i && 0
\end{pmatrix}}\)

After requesting permission from Vaggelis, i am adding a similar question to the above (for those interested), instead of opening a new topic. Here it is:

Show that the set
\[ \displaystyle S = \left\{ m + ni\sqrt{3} \in \mathbb{C} | \text{ either both m,n } \in \mathbb{Z} \text{ or both } m,n \in \mathbb{Q} \text{ such that } 2m,2m \text{ are odd integers } \right\} \]
is a (commutative) ring ( where by ring we mean an associative ring with multiplicative identity ). Find, if possible, \( \displaystyle U(R) \), that is the group of invertible elements in \( \displaystyle R\).

Some thoughts :

We define \(\displaystyle{f:\mathbb{H}\longrightarrow \mathbb{R}^{4}\,\,,f\,\left(\begin{pmatrix}
\,\,\,\,z && w\\
-\overline{w} && \overline{z}
\end{pmatrix}\right)=\left(Re(z),Im(z),Re(w),Im(w)\right)}\)

The function \(\displaystyle{f}\) is well defined . Let

\(\displaystyle{A=\begin{pmatrix}
\,\,\,\,z_1 && w_1\\
-\overline{w_1} && \overline{z_1}
\end{pmatrix}\,\,,B=\begin{pmatrix}
\,\,\,\,z_2 && w_2\\
-\overline{w_2} && \overline{z_2}
\end{pmatrix}\in\mathbb{H}}\) .

Then,

\(\displaystyle{\begin{aligned}f\,(A+B)&=\left(Re(z_1+z_2),Im(z_1+z_2),Re(w_1+w_2),Im(w_1+w_2)\right)\\&=\left(Re(z_1)+Re(z_2),Im(z_1)+Im(z_2),Re(w_1)+Re(w_2),Im(w_1)+Im(w_2)\right)\\&=\left(Re(z_1),Im(z_1),Re(z_2),Im(z_2)\right)+\left(Re(z_2),Im(z_2),Re(w_2),Im(w_2)\right)\\&=f\,(A)+f\,(B)\end{aligned}}\)

Now, let \(\displaystyle{A=\begin{pmatrix}
\,\,\,\,z_1 && w_1\\
-\overline{w_1} && \overline{z_1}
\end{pmatrix}\,\,,B=\begin{pmatrix}
\,\,\,\,z_2 && w_2\\
-\overline{w_2} && \overline{z_2}
\end{pmatrix}\in\mathbb{H}}\)

with \(\displaystyle{f\,(A)=f\,(B)}\). Then,

\(\displaystyle{\left(Re(z_1),Im(z_1),Re(w_1),Im(w_1)\right)=\left(Re(z_2),Im(z_2),Re(w_2),Im(w_2)\right)}\),

so :

\(\displaystyle{Re(z_1)=Re(z_2)\,\land Im(z_1)=Im(z_2)\,\land Re(w_1)=Re(w_2)\,\land Im(w_1)=Im(w_2)}\)

and thus :

\(\displaystyle{z_1=z_2\,\land w_1=w_2\implies A=B}\) , which means that \(\displaystyle{f}\) is \(\displaystyle{1-1}\) at \(\displaystyle{\mathbb{H}}\)

Consider \(\displaystyle{\left(a,b,x,y\right)\in\mathbb{R}^4}\).

Setting \(\displaystyle{z=a+i\,b\,,w=x+i\,y\in\mathbb{C}}\), we have that


\(\displaystyle{\begin{pmatrix}
\,\,\,\,z && w\\
-\overline{w} && \overline{z}
\end{pmatrix}\in\mathbb{H}}\) and

\(\displaystyle{f\,\left(\begin{pmatrix}
\,\,\,\,z && w\\
-\overline{w} && \overline{z}
\end{pmatrix}\right)=\left(Re(z),Im(z),Re(w),Im(w)\right)=\left(a,b,x,y\right)}\)

which means that \(\displaystyle{f}\) is onto.


Therefore, \(\displaystyle{\left(\mathbb{H},+\right)\simeq \left(\mathbb{R}^4,+\right)}\) .

Also,

\(\displaystyle{f\,(I_2)=f\,(1_{\mathbb{M_{2}}\,(\mathbb{C}}))=\left(Re(1),Im(1),Re(0),Im(0)\right)=\left(1,0,0,0\right)}\)



If \(\displaystyle{A=\begin{pmatrix}
\,\,\,\,z_1 && w_1\\
-\overline{w_1} && \overline{z_1}
\end{pmatrix}\,\,,B=\begin{pmatrix}
\,\,\,\,z_2 && w_2\\
-\overline{w_2} && \overline{z_2}
\end{pmatrix}\in\mathbb{H}}\) ,

then :

\(\displaystyle{A\,B=A\cdot B=\begin{pmatrix}
\,\,z_1\,z_2-w_1\,\overline{w_2} && z_1\,w_2+w_1\,\overline{z_2}\\
-z_2\,\overline{w_1}-\overline{z_1\,w_2} && -w_2\,\overline{w_1}+\overline{z_1\,z_2}\end{pmatrix}}\)

and if \(\displaystyle{z_1=a_1+i\,b_1\,,w_1=c_1+i\,d_1\,,z_2=a_2+i\,b_2\,,w_2=c_2+i\,d_2}\), then :

\(\displaystyle{\begin{aligned}z_1\,z_2-w_1\,\overline{w_2}&=\left(a_1+i\,b_1 \right )\,\left(a_2+i\,b_2 \right )-\left(c_1+i\,d_1 \right )\,\left(c_2-i\,d_2 \right )\\&=\left[(a_1\,a_2-b_1\,b_2)+i\,(a_2\,b_1+a_1\,b_2) \right ]-\left[(c_1\,c_2+d_1\,d_2)+i\,(d_1\,c_2-c_1\,d_2) \right ]\\&=\left(a_1\,a_2-b_1\,b_2-c_1\,c_2-d_1\,d_2 \right )+i\,(a_2\,b_1+a_1\,b_2-d_1\,c_2+c_1\,d_2) \end{aligned}}\)

and

\(\displaystyle{\begin{aligned}z_1\,w_2+w_1\,\overline{z_2}&=\left(a_1+i\,b_1 \right )\,\left(c_2+i\,d_2 \right )+\left(c_1+i\,d_1 \right )\,\left(a_2-i\,b_2 \right )\\&=\left[(a_1\,c_2-b_1\,d_2)+i\,(d_2\,a_1+b_1\,c_2) \right ]+\left[(c_1\,a_2+d_1\,b_2)+i\,(d_1\,a_2-c_1\,b_2) \right ]\\&=\left(a_1\,c_2-b_1\,d_2+c_1\,a_2+d_1\,b_2 \right )+i\,(d_2\,a_1+b_1\,c_2+d_1\,a_2-c_1\,b_2) \end{aligned}}\)

Therefore,
\begin{align*}
A\,B\stackrel{f}{\mapsto} &(a_1\,a_2-b_1\,b_2-c_1\,c_2-d_1\,d_2,a_2\,b_1+a_1\,b_2-d_1\,c_2+c_1\,d_2,a_1\,\\
&\quad\quad\quad c_2-b_1\,d_2+c_1\,a_2+d_1\,b_2,d_2\,a_1+b_1\,c_2+d_1\,a_2-c_1\,b_2)
\end{align*}
If we define multiplication on \(\displaystyle{\mathbb{R}^4}\) such that :

\(\displaystyle{a=\left(a_1,b_1,c_1,d_1\right)\,,b=\left(a_2,b_2,c_2,d_2\right)\in\mathbb{R}^4}\), then :
\begin{align*}
a\cdot b&=\big(a_1\,a_2-b_1\,b_2-c_1\,c_2-d_1\,d_2,a_2\,b_1+a_1\,b_2-d_1\,c_2+c_1\,d_2,a_1\,\\
&\quad\quad\quad\quad\quad c_2-b_1\,d_2+c_1\,a_2+d_1\,b_2,d_2\,a_1+b_1\,c_2+d_1\,a_2-c_1\,b_2\big)
\end{align*}
we have that (easy but painful) the triplet \(\displaystyle{\left(\mathbb{R}^4,+,\cdot\right)}\) is a ring with \(\displaystyle{1=\left(1,0,0,0\right)}\) as its unity,

(\(\displaystyle{\left(1,0,0,0\right)=f\,(I_2)=f\,(1_{\mathbb{M_{2}}\,(\mathbb{C}))}}\) ) ,

isomorphic to \(\displaystyle{\left(\mathbb{H},+,\cdot\right)}\)

because the function \(\displaystyle{f}\) maintains the operations of addition and multiplication ( painful) and additionally, is one to one and onto.

Therefore, \(\displaystyle{\left(\mathbb{R}^4,+,\cdot\right)}\) is a division ring but not a field.

Let \(\displaystyle{\left(a,b,c,d\right)\in\mathbb{R}^4-\left\{0\right\}}\) .

Then, \(\displaystyle{\left(a,b,c,d\right)=f\,\left(\begin{pmatrix}
a+i\,b && c+i\,d\\
-c+i\,d && a-i\,b
\end{pmatrix}\right)}\) .

\(\displaystyle{\left(\begin{pmatrix}
a+i\,b && c+i\,d\\
-c+i\,d && a-i\,b
\end{pmatrix}\right)^{-1}=\begin{pmatrix}
\dfrac{a-i\,b}{a^2+b^2+c^2+d^2} && -\dfrac{c+i\,d}{a^2+b^2+c^2+d^2}\\
\dfrac{c-i\,d}{a^2+b^2+c^2+d^2} && \dfrac{a-i\,b}{a^2+b^2+c^2+d^2}
\end{pmatrix}}\)

so, \(\displaystyle{\left(a,b,c,d\right)^{-1}=\dfrac{1}{a^2+b^2+c^2+d^2}\,\left(a,-b,-c,-d\right)}\) .

Additonal question :

Find, the subset \(\displaystyle{Z\,(\mathbb{H})=\left\{r\in\mathbb{H}: r\cdot x=x\cdot r\,,\forall\,x\in\mathbb{H}\right\}}\),

that is the center of \(\displaystyle{\left(\mathbb{H},+,\cdot\right)}\) and prove that it is a field.

Also, prove that \(\displaystyle{\mathbb{H}}\) is a \(\displaystyle{Z\,(\mathbb{H})}\) - vector space and find its dimension.

Additional question

Let \(\displaystyle{x=\left(x_0,x_1,x_2,x_3,x_4\right)\in Z\,(\mathbb{R}^4)}\). Then,

\(\displaystyle{x\cdot \left(0,1,0,0\right)=\left(0,1,0,1\right)\cdot x}\) and we have that :

\(\displaystyle{\left(-x_1,x_0,x_3,-x_2\right)=\left(-x_1,x_0,-x_3,x_2\right)\implies x_3=x_2=0}\)

so : \(\displaystyle{x=\left(x_0,x_1,0,0\right)}\) . Also,

\(\displaystyle{x\cdot \left(0,0,1,0\right)=\left(0,0,1,0\right)\cdot x}\) ad we get:

\(\displaystyle{\left(0,0,x_0,x_1\right)=\left(0,0,x_0,-x_1\right)\implies x_0\in\mathbb{R}\,\land x_1=0}\) .

So, \(\displaystyle{x=\left(x_0,0,0,0\right)}\). On the other hand, if \(\displaystyle{\left(x,0,0,0\right)\in\mathbb{R}^4}\), then :

for each \(\displaystyle{\left(a,b,c,d\right)\in\mathbb{R}^4}\) holds :

\(\displaystyle{\left(x,0,0,0\right)\cdot \left(a,b,c,d\right)=\left(a,b,c,d\right)\cdot \left(x,0,0,0\right)=\left(x\,a,0,0,0\right)}\) .

Therefore,

\(\displaystyle{Z\,(\mathbb{R}^4)=Z\,(\mathbb{H})=\left\{\left(x,0,0,0\right)\in\mathbb{R}^4: x\in\mathbb{R}\right\}\simeq \left\{\begin{pmatrix}
z&0 \\
0&z
\end{pmatrix} : z\in\mathbb{R}\right\}}\)

We define \(\displaystyle{g:\mathbb{R}\longrightarrow Z\,(\mathbb{R}^4)}\) by \(\displaystyle{g(x)=\left(x,0,0,0\right)}\)

and we see that \(\displaystyle{g}\) is an isomorphism

and thus \(\displaystyle{\left(Z\,(\mathbb{R}^4),+,\cdot\right)\simeq \left(\mathbb{R},+,\cdot\right)}\) and it is a field.

Besides, since \(\displaystyle{\left(\mathbb{R}^4,+,\cdot\right)}\) is a division ring, we have that \(\displaystyle{Z\,(\mathbb{R}^4)}\)

is a field. Check here: Vector space over the center of a division ring

The commutative additive group \(\displaystyle{\left(\mathbb{R}^4,+\right)}\)

equuiped with the scalar multiplication \(\displaystyle{\star:Z\,(\mathbb{R}^4)\times \mathbb{R}^4\longrightarrow \mathbb{R}^4}\)

\(\displaystyle{\left(\left(x,0,0,0\right),\left(a,b,c,d\right)\right)\mapsto \left(x,0,0,0\right)\star \left(a,b,c,d\right)\mapsto \left(x\,a,x\,b,x,c,x,d\right)}\)

is a left(right) \(\displaystyle{S=Z\,(\mathbb{R}^4)}\) -module, or else a \(\displaystyle{S=Z\,(\mathbb{R}^4)}\) - vector space, because :

if

\(\displaystyle{\left(x,0,0,0\right)\,,\left(y,0,0,0\right)\in S\,\,,\left(a,b,c,d\right)\,,\left(e,f,g,h\right)\in\mathbb{R}^4}\), then :

\(\displaystyle{\begin{aligned} \left[(x,0,0,0)+(y,0,0,0)\right]\star (a,b,c,d)&=(x+y,0,0,0)\star(a,b,c,d)\\&=((x+y)\,a,(x+y)\,b,(x+y)\,c,(x+y)\,d)\\&=(x\,a,x\,b,x\,c,x\,d)+(y\,a,y\,b,y\,c,y\,d)\\&=(x,0,0,0)\star (a,b,c,d)+(y,0,0,0)\star (a,b,c,d)\end{aligned}}\)

\(\displaystyle{\begin{aligned} (x,0,0,0)\star \left[(a,b,c,d)+(e,f,g,h)\right]&=(x,0,0,0)\star (a+e,b+f,c+g,d+h)\\&=(x(a+e),x(b+f),x(c+g),x(d+h))\\&=(x\,a,x\,b,x\,c,x\,d)+(x\,e,x\,f,x\,g,x\,h)\\&=(x,0,0,0)\star (a,b,c,d)+(x,0,0,0)\star (e,f,g,h)\end{aligned}}\)

\(\displaystyle{\begin{aligned} \left[(x,0,0,0)\cdot (y,0,0,0)\right]\star (a,b,c,d)&=(xy,0,0,0)\star (a,b,c,d)\\&=((x\,y)\,a,(x\,y)\,b,(x\,y)\,c,(x\,y)\,d)\\&=(x,0,0,0)\star (y\,a,y\,b,y\,c,y\,d)\\&=(x,0,0,0)\star\left[(y,0,0,0)\star (a,b,c,d)\right]\end{aligned}}\)

\(\displaystyle{1_{S}\star (a,b,c,d)=(1,0,0,0)\star (a,b,c,d)=(a,b,c,d)}\) .

Let now \(\displaystyle{x=\left(x_0,x_1,x_2,x_3\right)\in\mathbb{R}^4}\). Then,

\(\displaystyle{x=(x_0,0,0,0)+(0,x_1,0,0)+(0,0,x_2,0)+(0,0,0,x_3)=\sum_{i=1}^{4}(x_i,0,0,0)\star e_{i}}\), where :

\(\displaystyle{e_1=(1,0,0,0)\,,e_2=(0,1,0,0)\,,e_3=(0,0,1,0)\,,e_4=(0,0,0,1)}\) .

Futhermore, if \(\displaystyle{\left(a_{i},0,0,0\right)\in S\,,1\leq i\leq 4}\) such that \(\displaystyle{\sum_{i=1}^{4}a_{i}\star e_{i}=\left(0,0,0,0\right)}\), then :

\(\displaystyle{(a_1,0,0,0)\star (1,0,0,0)+(a_2,0,0,0)\star (0,1,0,0)+(a_3,0,0,0)\star (0,0,1,0)+(a_4,0,0,0)\star (0,0,0,1)=\left(0,0,0,0\right)}\)

\(\displaystyle{\implies (a_1,0,0,0)+(0,a_2,0,0)+(0,0,a_3,0)+(0,0,0,a_4)=\left(0,0,0,0\right)}\)

\(\displaystyle{\implies \left(a_1,a_2,a_3,a_4\right)=\left(0,0,0,0\right)}\)

\(\displaystyle{\implies \forall\,i\in\left\{1,2,3,4\right\}: a_{i}=0}\)

\(\displaystyle{\implies \forall\,i\in\left\{1,2,3,4\right\}: \left(a_{i},0,0,0\right)=\left(0,0,0,0\right)=0_{S}}\)

In conclusion, the vectors \(\displaystyle{e_{i}\,,1\leq i\leq 4}\) are linear-indepedent and thus the set

\(\displaystyle{A=\left\{e_{i}\in\mathbb{R}^4: 1\leq i\leq 4\right\}}\) is a basis of \(\displaystyle{\left(\mathbb{R}^4,+,\star\right)}\)

and \(\displaystyle{\dim_{S}\,\mathbb{R}^4=4\neq 2}\) .

The basis in \(\displaystyle{\mathbb{H}=\left\{\begin{pmatrix}
z&w \\
-\overline{w}&\overline{z}
\end{pmatrix}: z\,,w\in\mathbb{C}\right\}}\)

is (according to the isomorphism \(\displaystyle{f}\) above) the set

\(\displaystyle{A'=\left\{I_{2},\begin{pmatrix}
i&0 \\
0&-i
\end{pmatrix}\,,\begin{pmatrix}
0&1 \\
-1 & 0
\end{pmatrix}\,,\begin{pmatrix}
0&i \\
i& 0
\end{pmatrix}\right\}}\)