A skew field $K$ is said to be algebraically closed if it contains a root for every nonconstant 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 welldefined for nonalternative algebras?

11$p(x)=ix+xij$ lacks a root (since $ix+xi$ lies in the plane generated by $\{1,i\}$ for all $x \in \Bbb{H}$). Does that satisfy your definition of "polynomial", or do you want to enforce that $x$ is always multiplied on the right? – Micah Jun 15 '13 at 01:24

4What does $K[x]$ mean if $K$ is noncommutative? Does it consist of expressions of the form $\sum k_i x^i$ where $k_i \in K$? If so, it isn't closed under multiplication... – Qiaochu Yuan Jun 15 '13 at 03:38

3@Qiaochu: This is not correct. $R[x]$ is defined for every ring. The multiplication is defined as usual, in such a way that $x$ commutes with the image of $R$. – Martin Brandenburg Jun 15 '13 at 07:09

1@Martin: yes, but then you can't evaluate $x$ at an element of $R$ not in the center. Presumably the OP doesn't want this. (I should have said, "does it consist of *functions* of the form...") – Qiaochu Yuan Jun 15 '13 at 07:51

1@QiaochuYuan: Sure you can. The problem is that evaluation isn't a homomorphism anymore. – Chris Eagle Jun 15 '13 at 14:46

5@Chris: okay, sure. But my point is that something desirable breaks. – Qiaochu Yuan Jun 15 '13 at 18:44
3 Answers
If by $K[x]$ you mean the algebra freely generated by $K$ and an indeterminate $x$, then — as noted in my comment — $p(x)=ix+xij$ has no root in $\Bbb{H}$, because $ix+xi$ always lies in the plane spanned by $\{1, i\}$.
On the other hand, if by $K[x]$ you mean the subset of that free algebra consisting of expressions of the form $\sum k_i x^i$, we can generalize a topological proof of the fundamental theorem of algebra, as follows.
Theorem: Let $K$ be a finitedimensional normed $\Bbb{R}$algebra with $Z(K)=\Bbb{R}$, such that the subalgebra generated by each noncentral element is isomorphic to $\Bbb{C}$. Then for any $k_0,\dots,k_{n1} \in K$, there exists an $x \in K$ such that $x^n+\sum_{i=0}^{n1} k_ix^i=0$.
Proof: Let $g(x)=x^n$; let $S(K)$ and $B(K)$ be the unit sphere and unit ball in $K$. Since the subalgebra generated by each noncentral element is isomorphic to $\Bbb{C}$, every element of $S(K)$ except $\pm 1$ has $n$ preimages under $g$. A lengthy but straightforward Jacobian computation shows that $g$ is orientationpreserving at its regular values; thus the restriction of $g$ to $S(K)$ has topological degree $n$.
Now, suppose for the sake of contradiction that $f(x)=x^n+\sum_{i=0}^{n1} k_ix^i$ is never zero, and let $f_t(x)=t^nf(x/t)=x^n+\sum_{i=0}^{n1} k_i t^{ni} x^i$. Then $f_t$ is also nonvanishing. Define a map $\gamma_t:B(K) \to S(K)$ by $\gamma_t(x)=\dfrac{f_t(x)}{f_t(x)}$. Since $B(K)$ is contractible, the restriction of $\gamma_t$ to $S(K)$ has topological degree $0$.
But $\gamma_t=\dfrac{x^n+\sum_{i=0}^{n1} k_i t^{ni} x^i}{\leftx^n+\sum_{i=0}^{n1} k_i t^{ni} x^i\right}$ is homotopic to $g$; since topological degree is a homotopy invariant, we have a contradiction.
This proves exactly the statement you were looking for when $K=\Bbb{H}$, and for monic polynomial functions of this form in general.
When $K=\Bbb{O}$, this doesn't quite prove the statement you were looking for. Given an arbitrary polynomial function in the form you want, you can't necessarily divide through by the leading coefficient to get a monic polynomial function of the form you want, because of nonassociativity. It's not too hard to adapt, though; for any $\omega \in \Bbb{O}$, leftmultiplication by $\omega$ is a nonsingular linear map, so $g(x)=\omega x^n$ has nonzero degree and the rest of the proof still goes through.
$K=\Bbb{S}$ has zerodivisors, so the statement is trivially false there.
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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 $x\mapsto axxa$ is nonsurjective by the Ranknullity Theorem, which gives rise to degree $1$ rootless polynomials.
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Eilenberg and Steenrod, in Foundations of Algebraic Topology, prove a form of algebraic closure for both the quarternions and the octononions (ch. 11, sec. 5). The proof uses topological degree theory. The polynomial is required to be of the form $m+g$, where $m$ is a monomial of degree $n$ and $g$ is a sum of monomials all of degree less than $n$. However, monomials like $axbxc$ (of degree 2) are allowed.
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