Note: this question is wrong – this is not a field, though it is not obvious why it wouldn't be.
So, I (first year undergraduate mathematics student) was looking around the internet and found an interesting enough piece of C++. It contained the definition of something they called a 'recomplex' number so I'm assuming it's some kind of hypercomplex construction, which it appears to be. It also obeys the field axioms and both the reals and the complex numbers are its subsets and closed under its operations. With fair warning that it is the work of a well-known man of many failures and delusions of grandeur, I went ahead and tested out the field axioms he was claiming it fulfilled.
In all simplicity for any numbers $x = (a,i,j,k), y = (b,l,m,n) \in \mathbb{R}^4$, their addition is normal vector addition $x+y = (a+b, i+l, j+m, k+n)$, but their multiplication is defined by $$xy = (a b - i l - j n - k m, a l + b i + j m - k n, a m + b j - i n - k l, a n + b k + i m + j l).$$
This can also be derived from the following multiplication table for basis vectors
| $\cdot$ | $1$ | $i$ | $j$ | $k$ |
|---|---|---|---|---|
| $1$ | $1$ | $i$ | $j$ | $k$ |
| $i$ | $i$ | $-1$ | $k$ | $-j$ |
| $j$ | $j$ | $k$ | $i$ | $-1$ |
| $k$ | $k$ | $-j$ | $-1$ | $-i$ |
Multiplication and addition are obviously commutative and associative, and the multiplication obviously simplifies to vector scalar multiplication if only the real component has a non-zero value. Similarly having only the first two components with a non-zero value in both numbers gives you the definition of complex multiplication.
Additive and multiplicative identity obviously exist as per the previous paragraph and are the same as for the reals. Additive inverses are similarly obvious.
Multiplicative inverses are tricky but can be derived. The components of a multiplicative inverse are in order $$\frac{a\left(a^2-i^2+2jk\right)+i\left(2ai-j^2+k^2\right)}{\left(a^2-i^2+2jk\right)^2+\left(2ai-j^2+k^2\right)^2},$$ $$\frac{-a\left(2ai-j^2+k^2\right)+i\left(a^2-i^2+2jk\right)}{\left(a^2-i^2+2jk\right)^2+\left(2ai-j^2+k^2\right)^2},$$ $$\frac{-j\left(a^2-i^2+2jk\right)-k\left(2ai-j^2+k^2\right)}{\left(a^2-i^2+2jk\right)^2+\left(2ai-j^2+k^2\right)^2}, \text{and}$$ $$\frac{j\left(2ai-j^2+k^2\right)-k\left(a^2-i^2+2jk\right)}{\left(a^2-i^2+2jk\right)^2+\left(2ai-j^2+k^2\right)^2}.$$ I have symbolically simplified multiplying by this and it does in fact give us $(1,0,0,0)$.
On top of all that, I've symbolically went and confirmed that distributivity is also obeyed. That may all be wrong in the end, as I did it quite tired and with rather simplistic tools. If anyone cares to check the work, it's in this gist
That kind of suggests that we have here a field which is also the 4-dimensional real vector space. My question is specifically what this (or this kind of, if this is one of many) field is called in mathematics? I was (and still am) under the impression that such a neat extension of the complex numbers should not be possible, as this doesn't seem to follow the pattern of dropping more and more properties as the dimension increases as seen in the Cayley-Dickson algebras. Quaternions for example have non-commutative multiplication.
EDIT: As suspected, it isn't a field. It's very nearly a field, but there are extra non-invertible numbers as found by John Palmieri. Crackpots are still crackpots and have not suddenly found something mathematically revolutionary. The complexity of inverting the numbers lured me into using a computational tool which did not check whether certain algebraic manipulations were exactly legal.
Here is another reason this algebra is not a field (aside from the Frobenius theorem and undefined inverses): according to CAS computations, it has two nontrivial idempotents, namely $(1/2,0,\pm\frac{1}{2\sqrt{2}},\mp\frac{1}{2\sqrt{2}})$, and so is isomorphic to $\Bbb C \times \Bbb C$.