Is the set of quaternions $\mathbb{H}$ algebraically closed?

Question

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?

Answer 1

If by $K[x]$ you mean the algebra freely generated by $K$ and an indeterminate $x$, then — as noted in my comment — $p(x)=ix+xi-j$ 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 finite-dimensional normed $\Bbb{R}$-algebra with $Z(K)=\Bbb{R}$, such that the subalgebra generated by each non-central element is isomorphic to $\Bbb{C}$. Then for any $k_0,\dots,k_{n-1} \in K$, there exists an $x \in K$ such that $x^n+\sum_{i=0}^{n-1} k_ix^i=0$.

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 non-central 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 orientation-preserving at its regular values; thus the restriction of $g$ to $S(K)$ has topological degree $n$.

Now, suppose for the sake of contradiction that $f(x)=x^n+\sum_{i=0}^{n-1} k_ix^i$ is never zero, and let $f_t(x)=t^nf(x/t)=x^n+\sum_{i=0}^{n-1} k_i t^{n-i} x^i$. 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 $\gamma_t=\dfrac{x^n+\sum_{i=0}^{n-1} k_i t^{n-i} x^i}{\left|x^n+\sum_{i=0}^{n-1} k_i t^{n-i} x^i\right|}$ 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 non-associativity. It's not too hard to adapt, though; for any $\omega \in \Bbb{O}$, left-multiplication 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 zero-divisors, so the statement is trivially false there.

Answer 2

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.

Answer 3

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.

Answer 4

I know the the question has been asked a long time ago but I found the following solution more satisfying. Actually, a stronger result holds: Let $R$ be a real-closed field and $D$ be the division ring over $R$.Then $D$ is right algebraically closed.

Proof. denote the conjugate of a quaternion $q$ on D by $\bar{q}$. It is easy to check that it is an anti-automorphism. Now suppose $P(X)=\sum q_i X^i $(I suppose the question is asked about right polynomials, otherwise first answer provide a genuine counter-example.) It is elementary to check that $\bar{\bar{P} P}=\bar{P}P$ therefore, $\bar{P}P \in R[X]$ so since $R$ is real closed field you can find a root in $R(i) \subset D$ (by $R(i)$ I mean the the algebraic closure of R) so now we have to case: First, $P$ has a root and we are done or $\bar{P}$ has a root. in this case denote the root of $\bar{P}$ by $\gamma$. therefore we have: $\bar{P}(\gamma)=\sum \bar{q_i} \gamma^i=0 \rightarrow \sum \bar{\gamma^i} q_i=0$ so $\bar{\gamma}$ is a left root of $P$

Now you can apply division algorithm (in division ring) to $P$ and find a polynomial $Q(X)$ such that $P(X)=(X-\bar{\gamma})Q(X)$ and $Q$ has lower degree than $P$. So by induction, $Q$ has a root which I denote it by $\beta$. Now, it is easy to see that $\beta$ is a root for $P$ and we are done!. (Note that valuation is not a ring homomorphism in non-commutative rings. Therefore, it is not totally trivial that $\beta$ is a root of $P$.)

Answer 5

I think that, with one important caveat, the case of the noncommutative algebraic closedness of the quaternions is not settled by the examples in the other answers. In my opinion the straight generalization of the definition for fields is not good enough for noncommutative rings (nor for commutative ones which are not integral domains) and we have to elaborate it a bit more. We can start by asking ourselves the question:

In the field case, why do we spare the nonzero constant polynomials from presenting a root?

The answer is because they "obviously" cannot have a root, nor in the base field nor in any extension. And why spare only these polynomials? Because

Lemma. Let $R$ be a unital ring and $f\in R[x]$ (resp. $R\langle x\rangle$). TFAE:

1. There is no extension of unital rings $R\to S$ in which $f$ has a root.
2. $(f)=R[x]$ (resp. $R\langle x\rangle$), where $(f)$ denotes the ideal generated by $f$.

In addition, if $R$ is an integral domain, then 2) iff 3) $f(x)=u$ with $u$ a unit of $R$.

Proof: We work with $R[x]$, the $R\langle x\rangle$ case is analogous. If $(f)=R[x]$ then $1\in (f)\subseteq R[x]$, so $1\in (f)\subseteq S[x]$ for any extension $S/R$, hence $f$ has no roots in $S$, as $1=gf$ for some $g\in S[x]$. Conversely, if $(f)\neq R[x]$ then $S=R[x]/(f)$ gives an extension of $R$ in which $f$ has a root. Lastly, if $R$ is an integral domain and $f(x)=u$ then $1\in (f)$, while if $\deg f>0$ then any polynomial $g$ in $(f)$ is of the form $g=hf$ with $\deg g>0$, so $1\not\in (f)$.

Thus in our search for algebraic closures in more general rings, we ought to remove those polynomials which are beyond repair, such as we do for fields:

Definition. Given a ring $R$ and a polynomial $f\in R[x]$ (resp. $f\in R\langle x\rangle$), we will call $f$ proper if $(f)\neq R[x]$ (resp. $(f)\neq R\langle x\rangle$), otherwise $f$ is improper.

Definition. We will say that a ring $R$ is algebraically closed (resp. noncommutatively algebraically closed) if every proper polynomial of $R[x]$ (resp. of $R\langle x\rangle$) has a root in $R$.

When there are zero divisors, or when the ring is not commutative, ideals of polynomials in one variable are more complicated than over integral domains, and "less obvious" improper polynomials can appear:

Examples:

• The polynomial $f(x)=2x+1\in\mathbb{Z}/4\mathbb{Z}[x]$ is improper: $1=-(2x-1)(2x+1)\in (f)$.
• The polynomial $f(x)=ix+xi -j\in\mathbb{H}\langle x\rangle$ from Micah's answer is improper: $if(x)i = -ix-xi-j$, so $f+ifi = -2j\in (f)$ and $1\in (f)$.

The polynomials suggested by Drike are also improper:

Proposition. Pick $0\neq a\in\mathbb{H}$ and define $\phi_a:\mathbb{H}\to \mathbb{H}, \phi_a(q):=aq-qa=:[a,q]$. Pick $z\in\mathbb{H}$ such that $z\not\in$im$\phi_a$; then the polynomial $f_{a,z}(x):= [a,x]-z$ is improper.

Proof: Note that $z\neq0$. Write $\mathbb{H}=$im$\phi_a\oplus V$. If $z=z_y+z_n$ with $z_y\in$im$\phi_a$ and $z_n\in V$, then there is $w\in\mathbb{H}$ such that $\phi_a(w)=z_y$ and we can translate our polynomial $f_{a,z}(x+w) = [a,x+w]-z = [a,x]+z_y-z = [a,x]-z_n = f_{a,z_n}(x)$, with $1\in (f_{a,z_n})$ iff $1\in (f_{a,z})$. Therefore we can assume that $z\in V$. In addition, if $a=a_r+a_i$ with $a_r\in\mathbb{R}, a_i\in K:=\mathbb Ri+\mathbb Rj+\mathbb Rk$, then $[a,x] = [a_r+a_i,x] = [a_i,x]$, so we can assume $a\in K$. Now, to determine $V$ from $a$, write the arbitrary but fixed $a=a_1i+b_1j+c_1k$ and the parameter $x=x_1+x_2i+x_3j+x_4k$ to get $\phi_a(x)= 2(b_1x_4 - c_1x_3)i + 2(-a_1x_4 + c_1x_2)j + 2(a_1x_3 - b_1x_2)k$, so by inspection we see that $V=\mathbb R1+V'$ with $V'=0$ unless $a\in \mathbb Rl$ with $l\in\{i,j,k\}$, in which case $V'=\mathbb Rl$. We study both cases separately:

1. If $V=\mathbb R1$ then $z=r\in\mathbb R\setminus\{0\}$ and $f_{a,z}(x) = [a,x]-r$, and since $(f_{a,z}) = (r^{-1}f_{a,z}) = (f_{r^{-1}a,1})$ we can assume $r=1$. Now by a theorem of Cohn (see Exercises 16.11,16.10 from A first course in noncommutative rings by Lam), when $D$ is a division ring with center $F$ and $a\in D$ is algebraic over $F$, the polynomial $ax-xa-1$ has a solution iff $a$ is not separable over $F$. In our case quaternions form a division ring over its center $\mathbb R$, which has characteristic $0$ (so is perfect), and every imaginary quaternion satisfies the quadratic polynomial $x^2+|x|^2$ over $\mathbb R$.
2. If $a = tl$ with $t\in\mathbb R$ and $l\in\{i,j,k\}$ we can assume $t=1$, and writing $z=r+sl$ with $r,s\in\mathbb R$ we have $f_{a,z}(x)=[l,x]-r-sl$. Now $lf_{a,z}(x)l = l^2xl-lxl^2-rl^2-sl^3 = -xl+lx + r+sl = [l,x]+r+sl$, and therefore $0\neq 2(r+sl) = lf_{a,z}(x)l - f_{a,z}(x)\in (f_{a,z})$ and we are done.

In conclusion: the examples exhibited above show only improper polynomials over the quaternions, hence we don't know yet if quaternions are noncommutatively algebraically closed.