Let R be the real numbers and C be the complex numbers. Can CxC be made into a field as RxR can?

Question

There is no obvious way to do so and after several attempts I suspect the answer is no, but I can't see how you would show it. Is the reason because CxC is four dimensional ?

Answer 1

You can go to a higher dimension, but you lose some properties of a field. You get the Quaternions(denoted $\mathbb{H}$), but it is no longer a field but a skew-field--it is non-commutiative.

Perhaps the following theorem of Frobenius is insightful here:

Let $\mathcal{A}\neq 0$ be an associative quadratic real algebra without zero divisors, then there are three and only three possibilities: 1. $\mathcal{A}\cong \mathbb{R}$ 2. $\mathcal{A}\cong \mathbb{C}$ 3. $\mathcal{A}\cong \mathbb{H}$

You can go further to the octonians(8 dimensional $\mathbb{R}$ algebra), but you lose associatvity. You only get an alternative division algebra.

I believe this is the farthest you can go if you want a division algebra.(or are there any other restrictions? Sorry I'm no expert here so if I claimed something wrong pls comment)

If you're interested in this topic I recommend the book ${\it Numbers}$ by Ebbinghaus(and others)

Answer 2

$\mathbb C\times\mathbb C$ is called bicomplex numbers (or tessarines). It is commutative and associative and has all properties of a field except it has zero divisors.

You have an error in the question, $\mathbb R\times\mathbb R$ is isomorphic to the set of split-complex numbers and also has zero divisors. It is thus also not a field.