TITLE: Is there an algebraic closure for the quaternions?
QUESTION [21 upvotes]: This post is a sequel of: 
Is the set of quaternions $\mathbb{H}$ algebraically closed? 
This answer shows that:
1.  $\mathbb{H}$ is algebraically closed for the polynomials of the form $\sum a_r x^r$
2.  It is not for the polynomials freely generated by $\mathbb{H}$ and $x$, because $xi+ix-j$ has no root.
Question: Is there an algebraic closure (for the case 2)?
If so:   What does it look like?   What's its dimension over $\mathbb{H}$? What's its matrix representations?

REPLY [10 votes]: I don't think that there can be an associative $\Bbb{R}$-algebra $L$, containing $\Bbb{H}$ as a subring, such that the equation
$$xi+ix=j\qquad(1)$$
has a solution $x\in L$.
Multiplying $(1)$ by $i$ from the left gives us
$ixi+i^2x=ij$, or $ixi-ij=-i^2x$. As $i^2=-1$ and $ij=k$, this reads
$$
x=ixi-k.\qquad(2)
$$
On the other hand multiplying $(1)$ by $i$ from the right gives us
$xi^2+ixi=ji$, and using $i^2=-1, ji=-k$ this yields
$$
x=ixi+k.\qquad(3)
$$
The equations $(2)$ and $(3)$ together imply $k=-k$. As $k$ is a unit of $L$ this implies that $2=0$ in $L$, so $L$ cannot be an extension of $\Bbb{H}$.