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