An Algebraic Construction of the Split-Complex Numbers

In a previous post, I constructed the split-complex numbers in a somewhat handwavy fashion. It went as so: suppose $j$ is a solution to $x^2=1$ but $j\neq\pm 1$, then the split-complex numbers are defined by $a+jb$. In the last post, we examined a new characterization for the complex numbers via quotients. Particularly, we wanted to find a solution to $x^2+1$. However no such solution exists in $\mathbb{R}$, so by quotienting $\mathbb{R}[x]$ by $x^2+1$, we were able to get solutions.

We know that for a polynomial $p(x)$ over a field $\mathbb{F}$, $\mathbb{F}[x]/(p(x))$ is a field if and only if $p(x)$ is irreducible over $\mathbb{F}$. Let us consider the case of $p(x) = x^2-1$. We know that this polynomial has solutions in $\mathbb{R}$: $\pm 1$. We, however, can still quotient $\mathbb{R}[x]$ by $x^2-1$. Let us see what happens if we do so. Repeating similar logic as in the case of $x^2+1$, we have that

$$\mathbb{R}[x]/(x^2-1) = \{[a+bx]:a,b\in\mathbb{R}\}$$

where $[a+bx] = \{q(x):p(x)\mid (q(x)-a-bx)\}$. Again, we have an algebra structure on $\mathbb{R}[x]/(x^2-1)$ since we can multiply polynomials. Multiplying two elements of the $\mathbb{R}[x]/(x^2-1)$, we get

$$(a + bx + (p(x)))(c + dx + (p(x)))= ac + ad x + bc x + bd x^2 + (p(x)).$$

However since $x^2 \equiv 1 \pmod{x^2+1}$, it follos that $bd x^2 \equiv bd$ and so

$$(a + bx + (p(x)))(c + dx + (p(x))) = (ac + bd) + (ad + bc) x + (p(x)).$$

If you recall the structure of the split-complex numbers, $(a+jb)(c+jd) = (ac+bd)+(ad+bc)j$. Thus there is a natural isomorphism between $\mathbb{R}[x]/(x^2-1)$ and $\mathbb{H}$ as they have the same additive and multiplicative structures. Before, it was noted that not every element in $\mathbb{H}$ has an inverse; particularly, those elements living on the lines $y = \pm x$.

The ring $\mathbb{R}[x]/(x^2-1)$ is commutative but as we noted, it cannot be a field since $x^2-1$ is reducible over $\mathbb{R}$. This is in exact agreement with what we showed before. Thus the general algebraic theory gave to us quite naturally the fact that $\mathbb{H}$ is not a field. Moreover, we have a matrix representation for the split-complex numbers by the same logic as in the previous post:

$$a+jb \Longleftrightarrow \left(\begin{array}{cc} a & b \\ b & a\end{array}\right).$$

The matix corresponding to complex numbers is

$$a+ib \Longleftrightarrow \left(\begin{array}{rr} a & -b \\ b & a\end{array}\right)$$

which is invertible if $a,b\neq 0$ since its determinant is nonzero. The former matrix is not invertible since whenever $a^2 = b^2$, the determinant is zero. This again agrees with the previous analysis.