A new $4$-dimensional number system?

Solution 1:

Such a ring $S$ is automatically isomorphic to the 4-fold Cartesian product $\Bbb{R}\times\Bbb{R}\times\Bbb{R}\times\Bbb{R}$ with addition and multiplication done componentwise. The reason is the following. Multiplication in any ring $\Bbb{R}$-algebra that is also a 4-dimensional vector space can be represented with $4\times4$-matrices.

• Because all the basis elements $1,i,j,k$ have finite order, the corresponding matrices are diagonalizable.
• Because the order is one or two, the eigenvalues are $\pm1$, and they are diagonalizable using a change of bases matrix with real coefficients.
• Because the basis elements commute, those matrices are automatically simultaneously diagonalizable.
• We have just embedded $S$ into the ring of diagonal $4\times4$ real matrices. Because $S$ is 4-dimensional, the embedding is an isomorphism, and we know what $S$ looks like.

An explicit isomorphism is given by $$ \begin{aligned} 1&\mapsto I_4,\\ i&\mapsto diag(1,-1,1,-1),\\ j&\mapsto diag(1,1,-1,-1),\\ k&\mapsto diag(1,-1,-1,1). \end{aligned} $$ The reader who has taken a course on representation theory recognizes these as coming from the characters of the Klein Viergruppe $V_4$. This explains the connection between my answer and that of Chris Culter.

Solution 2:

This number system does have a name: $\mathbb R[V_4]$. In other words, it is the group ring of the Klein four-group over the real numbers.

Solution 3:

If you want your new system to have "nice" properties then the possibilities are severely limited by the Frobenius Theorem.

Being more precise about "nice", if you want a field so that familiar properties such as division and commutative multiplication are retained then the only finite dimension extension of the the reals is the complex numbers.

If you want to retain division but are willing to sacrifice commutativity then you have one more finite dimension extension: the quaternions.

Yours is not either of those so it must have some "not nice" property. As others have said, this is the presence of zero divisors. In your system, you could not deduce $x = y$ from $ax = ay$ even if you knew that $a \neq 0$.

You can go further but you need to give up more nice properties. E.g. the Octonians and Sedenions but you have now lost associativity.