No extension to complex numbers?

Solution 1:

It is impossible to have a field which is $n$ dimensional over $\mathbb{R}$ for any $n\geq 3$. The reason why this is true boils down to the following two statements.

Any field $K\supseteq \mathbb{R}$ which is finite dimensional over $\mathbb{R}$ is algebraic over $\mathbb{R}$.
The complex numbers are the algebraic closure of $\mathbb{R}$.

Thus is $K\supseteq \mathbb{R}$ is a field which is finite dimensional over $\mathbb{R}$, then it is algebraic over $\mathbb{R}$, and hence is contained in the algebraic closure of $\mathbb{R}$, i.e., $K\subseteq \mathbb{C}$. Since $\mathbb{C}$ has dimension $2$ over $\mathbb{R}$, this implies that $K$ has dimension either $1$ or $2$ over $\mathbb{R}$. In the first case, $K = \mathbb{R}$, and in the second $K = \mathbb{C}$.

Solution 2:

Since $\mathbb{C}$ is algebraically closed, every finite extension of $\mathbb{C}$ is $\mathbb{C}$ itself. The quaternions are not a field because they are not commutative; they are what is called a normed division algebra. Hurwitz's theorem gives a complete description of the possible normed division algebras (over $\mathbb{R}$) - the are either the real numbers, the complex numbers, the quaternions or the octonions. Only the first two are fields, and only the first three are associative.

The nonexistence of a three-dimensional real normed algebra can be viewed as a consequence of the hairy ball theorem. If $\mathbb{R}^3$ could be given a normed division algebra structure, the unit sphere $S^2$ could be endowed with a smooth group structure (the elements of norm $1$ form a group). But the hairy ball theorem implies that there is no such thing, because otherwise, by letting an infinitesimal element of the group act on the left on the sphere, we would obtain a nowhere vanishing, continuous vector field on the sphere, which is impossible by the hairy ball theorem.