In what follows, $x$ is always taken to commute with the coefficient ring. This means that for any given polynomial, you can put the coefficients to the right or the left of $x$ as you please. This doesn't make a difference to the ring itself, but it will make a difference for the roots: a right root is an element that yields zero upon evaluation when you write $x$ to the right of the coefficients. (Similarly for left root. I suppose a two-sided root would be an element that is both a right and a left root.) For example, $\mathrm{j}$ is a right root but not a left root of the following polynomial (working in the quaternions). $$ \mathrm{i}x-\mathrm{k}. $$
Definition: Let $\mathbb{K}$ be a ring. Then, $\mathbb{K}$ is algebraically-closed iff every nonconstant polynomial with coefficients in $\mathbb{K}$ can be written as a product of polynomials with degree $1$.
Definition: A $\mathbb{K}$-algebra $\mathbb{K}\rightarrow \mathbb{L}$ is right algebraic over $\mathbb{K}$ iff every $\alpha \in \mathbb{L}$ is a right root of some $p\in \mathbb{K}[x]$.
Definition: A right algebraic-closure of a ring $\mathbb{K}$ is an algebraically-closed $\mathbb{K}$-algebra $\mathbb{K}\rightarrow \mathbb{L}$ that is right algebraic over $\mathbb{K}$.
(There are of course left and two-sided versions of the above definitions.)
It is well-known that every field possesses an algebraic-closure, which is unique up to (nonunique) isomorphism. What about division rings?
Does every division ring $\mathbb{K}$ possess a right/left/two-sided algebraic-closure? If so, is it unique up to isomorphism? If not, is it because I'm using the "wrong" definitions, or is it really not fixable?
