A Particular Quotient of Polynomials over Hyperbolic Numbers: is it Anything? The hyperbolic numbers are a two real dimensional algebra generated by $j$ such that $j^2=1$. In particular, denote $\mathcal{H} = \{ a+bj \ | \ a,b \in \mathbb{R} \}$. Multiplication is done in the natural manner,
$$ (a+bj)(x+jy) = ax+by+j(bx+ay) $$
and we may establish that $\mathcal{H}$ is an associative, unital, commutative algebra over $\mathbb{R}$. Consider $\mathcal{A}' = \mathcal{H}[x]/\langle x^2-(1+j)x+j \rangle$ where $\mathcal{H}[x]$ denotes the ring of polynomials in $x$ with hyperbolic number coefficients. The quotient has arithmetic defined in the usual fashion; add and multiply modulo $\mathcal{H}[x]$ multiples of $x^2-(1+j)x+j$. 
At first I had hoped $\mathcal{A}' \approxeq \mathcal{H} \times \mathcal{H}$ (as a direct product), but, after some thought I think my first thought is wrong. 
I don't think $\langle x-1 \rangle $ and $\langle x-j \rangle$ are comaximal, and, I suspect that suffices to deny my initial guess.

Question: is there a better way to look at $\mathcal{A}' = \mathcal{H}[x]/\langle x^2-(1+j)x+j \rangle$ ? 

In other words, can we see this as some sort of semi-direct product of copies of $\mathcal{H}$ etc. ?
Incidentally, this particular quotient is interesting since $x^2-(1+j)x+j = (x-1)(x-j)$ and $1-j$ is a zero divisor of $\mathcal{H}$. Thanks in advance for your insights into this problem!
 A: What you call the hyperbolic numbers, I call split complex numbers. In any case, CRT proffers an isomorphism (of rings / real vector spaces)
$$ \mathbb{R}[j] = \frac{\mathbb{R}[T]}{(T^2-1)} \cong \frac{\mathbb{R}[T]}{(T-1)}\oplus\frac{\mathbb{R}[T]}{(T+1)}\cong\mathbb{R}\oplus\mathbb{R}. $$
The elements corresponding to $(1,0)$ and $(0,1)$ must be idempotents that sum to unity; solving the equation $(a+bj)^2=a+bj$ yields $(1\pm j)/2$ as our idempotents. Thus we have
$$ \begin{array}{cc} \displaystyle \frac{1+j}{2}\leftrightarrow (1,0) & \quad & 1\leftrightarrow (1,1)~~~ \\[5pt] \displaystyle \frac{1-j}{2} \leftrightarrow (0,1) & \quad &  j\leftrightarrow (1,-1) \end{array} $$
There is also an isomorphism $(\mathbb{R}\oplus\mathbb{R})[x]\cong \mathbb{R}[x]\oplus \mathbb{R}[x]$. Consider the problem of quotienting this ring by the principal ideal generated by the tuple $(p(x),q(x))$. We can multiply the generator by the idempotents $(1,0)$ and $(0,1)$ to obtain $(p(x),0)$ and $(0,q(x))$ in the ideal, and conversely we can obtain the original tuple as the sum of these, so
$$ \frac{\mathbb{R}[x]\oplus\mathbb{R}[x]}{\langle \,(p(x),q(x))\,\rangle}=\frac{\mathbb{R}[x]\oplus\mathbb{R}[x]}{\langle\, (p(x),0), \, (0,q(x))\, \rangle} \cong \frac{\mathbb{R}[x]}{\langle p(x)\rangle} \oplus \frac{\mathbb{R}[x]}{\langle q(x)\rangle}.$$
In your case, the principal generator is $((x-1)^2,x^2-1)$, so the resulting ring is
$$ \frac{\mathbb{R}[x]}{\langle (x-1)^2\rangle} \oplus \frac{\mathbb{R}[x]}{\langle x^2-1\rangle}. $$
The first summand is an isomorphic copy of the dual numbers $\mathbb{R}[\varepsilon]/(\varepsilon^2)$ and the second summand is another copy of the hyperbolic numbers / split complex numbers.
We can interpret this decomposition as one of modules over the split complex numbers $\mathbb{R}\oplus\mathbb{R}$ by interpreting the first summand as a vector space over $\mathbb{R}\oplus 0$ and the second as a vector space over $0\oplus\mathbb{R}$ (and in which $0\oplus\mathbb{R}$ annihilates the first summand and $\mathbb{R}\oplus 0$ annihilates the second). In particular, the second summand is not a copy of the regular representation as a module, even though it is a copy of the split complex numbers as a ring and vector space.
