I am trying to solve the following exercise:
Prove that complex multiplication does not extend to a multiplication
on $\mathbb R^3$ so as to make $\mathbb R^3$ into a real division
algebra.
I am aware of Frobenius' theorem that there are only three finite dimensional real associative division algebras. But the exercise would be pointless if the theorem was assumed.
Let $x,y \in \mathbb R^3$. Since $x y$ has to extend the complex multiplication:
$$xy = (x_1y_1-x_2y_2, x_1y_2 + x_2 y_1, ?)$$
I have no idea how I can proceed from here. Could someone tell me how to do this? It should be easy as the other exercises I did so far were also easy.
Best Answer
Assume that $D$ is a 3-dimensional division algebra over $\Bbb{R}$. Let $a\in D\setminus\Bbb{R}$. Consider the linear mapping $\rho_a:z\mapsto az, z\in D,$ from $D$ to itself. Let $M$ be the matrix representing $\rho_a$ (with respect to some basis). The eigenvalue polynomial $$ \chi_a(x)=\det(xI_3-M)\in\Bbb{R}[x] $$ is monic of degree three. Thus $\lim_{x\to\pm\infty}\chi_a(x)=\pm\infty$. Therefore, by continuity of $\chi_a(x)$, there exists a real number $r$ such that $\chi_a(r)=0$. This means that the mapping $$ L:D\to D, z\mapsto az-rz=(a-r)z $$ has a non-trivial kernel. Therefore the element $a-r$ cannot be invertible. Because $a\notin\Bbb{R}$ we have $a-r\neq0_D$. This contradicts the assumption that $D$ is a division algebra.
Edit: The above was a bit of overkill for the task at hand (=proving that complex multiplication cannot be extended to a 3D-space). A simpler argument follows.
If the multiplication of $D$ is an extension of the multiplication of $\Bbb{C}$, then $D$ has a subalgebra isomorphic to $\Bbb{C}$. Therefore $D$ has a structure of a (left) vector space over $\Bbb{C}$. Thus $D$ is a finite dimensional vector space over $\Bbb{C}$. But this implies that the dimension of $D$ as a vector space over $\Bbb{R}$ is an even number.