Is there an algebraic closure for the quaternions?

Question

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?

Answer 1

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 $\Bbb{H}$.

Answer 2

It looks like the answer is morally "no." Now, there is a formal closure for which we can solve free polynomials: As in the case of fields, you take an inductive limit. Let $\Bbb H=R$ be our normed division algebra. Then

$$\overline{R}=\varinjlim_{[L:R]<\infty} L$$

where the inductive system is taken relative to inclusions of algebra extensions $L/R$ of finite dimension over $R$, each of the form

$$L_p=R\{x\}/(p(x))$$

where $p(x)$ is irreducible over $R$ and $R\{x\}$ is the polynomials freely generated as in case ($2$). This certainly has the required property that all polynomials in $R$ have a root in $\overline{R}$, and any other such object has a copy of this inside of it for purely formal reasons.

I note that the directed system so-defined is indeed a directed system--in fact a lattice--so this should go through unless I'm missing something obvious.

rschweib has noted that the result is no longer a division algebra, so this is really not ideal, but the "algebraic closure" property holds, and necessarily it's a minimal ring where this property can hold, so it seems this is the best we can hope for. However we also cannot force algebraicness of the result since $R\{x\}/(xi+ix-j)$ doesn't make $x$ algebraic appropriately in the sense that you want to mimic the field case's excellent definition that algebraicness means $F(\alpha)/F$ is finite dimensional as an algebra over $F$, which doesn't hold in this setting.