So, I StumbledUpon this really cool site and the last picture looked almost as if it had 3D structure. This reminded me of another website where I saw pictures of the order-8 Mandelbulb. I got to thinking about how to do a 3D version of the pictures in that first site and quaternions came to mind.
I know that $\mathbb{C}$ is closed, but is there a way to have quaternions as roots of some exotic polynomial or whatever?
Actually all the quaternions are zeros of quadratic polynomials with real coefficients. If $q=a+ib+jc+kd$, with $a,b,c,d$ real, then $$ (x-q)(x-\overline{q})=x^2-2ax+(a^2+b^2+c^2+d^2) $$ has $q$ as a zero, and also has real coefficients. Here $\overline{q}=a-ib-jc-kd$ is the 'conjugate' quaternion.
In a sense the quaternions are just a bunch of copies of $\mathbf{C}$ pointing in different directions. All the copies share the real axis, but we can freely rotate the `imaginary' axis in the 3D-space spanned by $\{i,j,k\}$. Whenever $b^2+c^2+d^2=1$ the quaternion $u=ib+jc+kd$ satisfies the equation $$ u^2=-1. $$ Thus $u$ can take the role of the complex number $i$ in the sense that the mapping $x+iy\mapsto x+uy$, ($x,y\in\mathbf{R}$) is a monomorphism of rings from the complex number into the quaternions, so its image is a copy of $\mathbf{C}$.
(Yes, there is more to quaternions than `just a bunch of copies of $\mathbf{C}$', but for the purposes of them being zeros of polynomials with real coefficients this point of view is enough.)
If you allow the coefficients of the polynomial to be quaternions, then some of the roots must be quaternions as well. However, since they do not commute, you need to represent them in a nice way, preferably as matrices.
Finding roots to matrix equations should not be too difficult, but you will need to solve a system of non-linear polynomial equations.
An analogue of the pictures might be to use matrix coefficients, where each entry is -1 or 1?
