I'm messing around in the following setting: let $$\Bbb C' = \{a+bh \mid a,b \in \Bbb R, h \not\in \Bbb R, h^2=1\} \cong \frac{\Bbb R[x]}{(x^2-1)}$$be the ring of split-complex numbers. Define a linear operator $J: \Bbb R^2 \to \Bbb R^2$ by $J(e_1) = e_2$ and $J(e_2) =e_1$, where $(e_1,e_2)$ is the standard basis of $\Bbb R^2$. The split-complexification $\Bbb C' \otimes_{\Bbb R} \Bbb R^2 \cong (\Bbb C')^2$ has a $\Bbb C'$-module structure, and $(e_1,e_2)$ is a $\Bbb C'$-basis, great. We consider the unique $\Bbb C'$-linear extension of $J$ to $(\Bbb C')^2$, also denoted by $J$.
The characteristic polynomial of $J$ is $x^2-1$, which has four roots in $\Bbb C'$: $1$, $-1$, $h$ and $-h$. Since $\Bbb C'$ is commutative, each "eigenspace" $E_\lambda$ is s submodule of $(\Bbb C')^2$. I have computed
$$\begin{array}{|c|c|} \hline {\boldsymbol \lambda} & \textbf{basis} \\ \hline 1 & (1,1) \\ \hline -1 & (1,-1) \\ \hline h & (1,h) \\ \hline -h & (1,-h) \\ \hline \end{array}$$
Every basis of $(\Bbb C')^2$ over $\Bbb C'$ must have the same cardinality, so clearly $(\Bbb C')^2$ cannot be the direct sum of these $E_\lambda$. Also: $$(1+h)(1,1) + (-1-h)(1,h) = (0,0).$$This is very disturbing to me. I do realize that $1+h$ and $1-h$ are the only zero divisors (up to real scaling) in $\Bbb C'$. If $$E_1 = \{(a+bh,a+bh) \mid a,b \in \Bbb R \}\quad\mbox{and}\quad E_h = \{(a+bh, b+ah) \mid a,b \in \Bbb R\},$$the above says that these "eigenspaces" intersect along a "line" $a=b$ (corresponding to the zero divisor).
What is really going on here? Is there a "better" choice of decomposition for $(\Bbb C')^2$ here? Better yet, is there a way to develop "spectral" theory in this context of modules?
I feel like I'm walking on eggs here. Thanks for any help!