Question: 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.

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.
I will also add a more "topological" proof but requires some familiarity with manifolds:

For a multiplication to "expand" the complex multiplication we would require that $|u\_1\times u\_2|=|u\_1||u\_2|$ which is saying that $S^2$ would take the structure of a group (as is the case in $\mathbb{C}$ and $S^1$). But since $S^n$ are manifolds , this would $S^2$ a *Lie group* and we know that for every *Lie Group* $TM=M\times M$ , namely is is parallelizable.

Using the (very deep) fact that only $S^1,S^3,S^7$ are parallelizable we get that the only multiplications *like* $\mathbb{C}$ exist in $\mathbb{R^2},\mathbb{R^4},\mathbb{R^8}$ which ofcourse correspond to the complex numbers, quartenions and octonions. 

Notice that for tha case of $S^2$ and in general $S^{2k}$ we can just use the hairy-ball theorem to show that they aren't parallelizable!