TITLE: Is the set of quaternions $\mathbb{H}$ algebraically closed?
QUESTION [30 upvotes]: A skew field $K$ is said to be algebraically closed if it contains a root for every non-constant polynomial in $K[x]$. I know that this is true for $\mathbb{C}$, which is the algebraic closure of $\mathbb{R}$ and a true field. I wonder if this is also true for the strictly skew field $\mathbb{H}$. I think it's not, but I can't find a counterexample. What about the set of octonions $\mathbb{O}$, which is no longer associative, or the set of sedenions $\mathbb{S}$, which is not even alternative? Is algebraic closedness even well-defined for non-alternative algebras?

REPLY [4 votes]: One can be a bit more general: if by $K[x]$ is meant the algebra of generalised polynomials,

Note. A division ring $K$ of finite dimension $[K:k]>1$ over its centre $k$ is never algebraically closed.

Proof. For every $a\in K\setminus k$, being $k$-linear of Kernel having dimension $\geqslant 1$, the polynomial $x\mapsto ax-xa$ is non-surjective by the Rank-nullity Theorem, which gives rise to degree $1$ rootless polynomials.