This is the crispest proof I’ve found that the complex numbers ℂ are the only finite field extension of the real numbers, ℝ.

In other words, if you want to continue the process which takes you from the number line ℝ to the number *plane* ℂ, you *can’t*. Once you’ve generalised from counting numbers to including zero, to negative numbers, to fractions, to irrational numbers, and then to complex numbers, you’re *done*. Number systems don’t keep on generalising. Complex numbers are *it*.

You can define an addition and multiplication on points in a *plane* which makes them obey all the rules of arithmetic which hold for the number *line* – that is, more or less, what it means to call that structure a *field* – but there’s no way of doing that with points in 3D (or 4D, or 5D… ) space.

You can define addition and multiplication on points in 4D space which obeys *most of* the rules of arithmetic which hold for the number line, and that creates a structure called *quaternions*, much used these days, so I’m told, in designing graphics for computer games. But in quaternions a.b=−b.a.

The proof is from *Abstract Algebra*, by Celine Carstensen, Benjamin Fine, and Gerhard Rosenberger (Walter de Gruyter, 2011), p.262 (and pp.186-7 for the Sylow theorems).

It proves the result together with the “Fundamental Theorem of Algebra”, namely, that in ℂ *every* polynomial has solutions, which also has a neat topological proof.

A field extension is a bigger field which differs from the smaller field in including solutions to polynomial equations insoluble in the smaller field.

If K is a field extension of ℝ, then there is another mathematical structure, K:ℝ, which is the *vector space* formed by K over ℝ. For example, ℂ forms a two-dimensional vector space over ℝ – each point in ℂ requires two numbers to describe it.

The dimension of K:ℝ can’t be odd, since polynomials of odd degree always have a solution in ℝ, and therefore the minimal polynomial of K, the polynomial of smallest degree which generates it, must be of even degree.

Suppose K is a field extension of ℂ. Then we look at another structure, the Galois group of K over ℝ, which is all the ways that the elements of K can be mapped onto each other while keeping the arithmetic structure of K and while mapping every element of ℝ into itself. (For example, mapping every complex number to its conjugate is a member of the Galois group of ℂ over ℝ).

Then the order (“size”) of the Galois group must be 2^{m}.q for some odd number q. A standard theorem of group theory, one of Sylow’s theorems, then tells us that this group has a subgroup of order 2^{m}. But that would be the Galois group of a smaller extension E with dimension of E:ℝ=q, which is impossible (from the above) unless q=1.

Another of Sylow’s theorems tells that the Galois group of K over ℂ which is of order 2^{m−1}, must have a subgroup of order 2^{m−2}, which corresponds to an intermediate field D which is a degree 2 extension of ℂ. But, no field can be a degree 2 extension of ℂ, because by the quadratic formula every quadratic in ℂ is soluble.

Therefore the finite field extension K is ℂ itself (or an identical copy of it). ∎

Mathematicians use lots of other number systems, and call them “number systems” – transfinite numbers, surreal numbers, p-adic numbers, and much more – and there are also *infinite* field extensions of ℝ – but in the direct sense ℂ is the *end* of the process of expanding our sense of number which we first embark on when we realise the counting numbers continue indefinitely.