No extension to complex numbers? Complex numbers are 2D. It is a commonly cited result that there is no 3D or 4D analogue of the complex numbers.
I just want to be clear on exactly what this result says:


*

*It is impossible to construct a 4D field.

*It is impossible to construct a 4D field which the complex numbers are a subset of.

*It is impossible to construct a 4D field which the real numbers are a subset of.

*Something else?
(I'd also like to know why it's true - but given that I didn't understand the answer the first $2^7$ times, I doubt I'll understand it this time either. :-} Oh well...)
 A: The field of rational functions is infinite-dimensional. It includes the complex and real numbers. 
The same is correct about the field of analytic (holomorphic) functions.
A: 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 $K = \mathbb{C}$.
A: 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.
A: A question is, why one would need exactly a field. The bicomplex numbers aka tessarines are a commutative associative ring.
Yes, there are some elements that have no inverse, but that is not bad at all: dual numbers and split-complex numbers also have such elements.
Bicomplex numbers have subrings, isomorphic to complex numbers and hyperbolic (split-complex) numbers. They can be represented as commutative real matrices.
They are a really great thing, generalizing the various notions of complex number.
