Fundamental theorem of algebra for Quaternions and Octonions -- and the completeness

Question

The fundamental theorem of algebra states that every non-constant single-variable polynomial with complex $\mathbb{C}$ coefficients has at least one complex root. This includes polynomials with real coefficients, since every real number is a complex number with its imaginary part equal to zero.

In this sense, the $\mathbb{C}$ is complete -- more complete than $\mathbb{R}$, both are fields.

1. Do we have the fundamental theorem of algebra for Quaternions $\mathbb{Q}$ and Octonions $\mathbb{O}$? How do we (could you) sketch the ideas of the proof?

2. Are Quaternions $\mathbb{Q}$ and Octonions $\mathbb{O}$ complete, or not? in the sense of they are NOT fields? Thanks to @Qiaochu Yuan.

3. Are there better or other ways to measure the completeness of $\mathbb{R}$, $\mathbb{C}$, $\mathbb{Q}$ and $\mathbb{O}$?

Merci beaucoup.

Answer 1

There is a proof of the fundamental theorem of algebra for $\mathbb{H}$ which has not been cited in the links provided in the comments, and which does not use any topological argument. This concerns polynomials with a central indeterminate $X$ (that is $aX=Xa$ for any quaternion $a$).

The source is Lam's First course in Noncommutative rings.

I sketch the proof here.

Let $D$ be a division ring and let $f=\sum_n d_n X^n\in D[X]$.

We say that $\alpha\in D$ is a right root of $f $ if $f(\alpha):=\sum_n d_n\alpha^n=0$, and a left root if $\sum_n \alpha^n d_n=0$ (note that the latter sum is NOT denoted by $f(\alpha)$.)

First step. Using left/right long division of polynomials, we see that $\alpha $is a left/right root if $f=(X-\alpha)g / f=g (X-\alpha)$, for some $g\in D[X]$. This result generalizes to arbitrary rings (not only division ones).

Second step. Assume that $f=gh$, and let $d\in D$ such that $a:=h(d)\neq 0$. Then $f(d)=g(ada^{-1}))h(d)$ (easy computations). In particular, if $\alpha$ is a right root of $f$, then $\alpha$ is a right root of $h$ or there exists a conjugate of $\alpha$ which is a right root of $g$.

Third step (Main theorem). Let $D=\mathbb{H}$. Then every nonzero polynomial of $D[X]$ has a left/right root in $D$.

For $d=x+yi+zj+tk$, set $d^*=x-yi-zj-tk$.

If $f=\sum_n d_n X^n$ set $f^*=\sum_n d_n^*X^n$. For all $f,g\in D[X]$, one may check that $(fg)^*=g^* f^*$. In particular $(ff^*)^*=f^{**}f^*=ff^*$, and $ff^*\in\mathbb{R}[X]$.

We now prove the theorem by induction on the degree of $f$. For degree $1$, this is obvious.

If $f$ has at least degree $2$, then $ff^*\in\mathbb{R}[X]$ has a root of $\alpha\in \mathbb{R}(i)$ (which is a copy of $\mathbb{C}$ inside $D$) by the classical fundamental theorem of algebra. By step two, either $\alpha$ is a right root of $f^*$ or some conjugate of $\alpha$ is a right root of $f$. In the second case, we are done. In the first case, $\alpha^*$ is then a left root of $f$, so $f=(X-\alpha^*)g$ for some $g$. Since $1\leq \deg(g)<\deg(f)$, by induction, $g$ has a right root $\beta$. But $\beta$ is also a right root of $f.$ Indeed, we have $g=h(X-\beta)$, and so $f=[(X-\alpha^*)h](X-\beta)$ (note that we cannot use evaluation at $\beta$ to conclude, since it is not a ring morphism).

By induction, we are done. Note that you can prove similarly that any $f$ has a left root.

Concerning octonions : it seems that the partial results are known for polynomials with a noncentral indeterminates. Thus it is possible that the fundamental theorem of algebra is true for octonions if you consider (as I did for quaternions) the indeterminate $X$ to be central.

However, i'm not sure that the proof above maybe easily generalized to octonions because of the lack of associativity.

Answer 2

Let me address 2 and 3 by offering one perspective on what you might mean by "complete" (not in the topological sense, and not quite in the sense of "algebraic closure" which may also be of interest to you), or more precisely one way in which the quaternions and octonions can do more for you.

As you observed, one way to think about what more $\mathbb C$ does for you than $\mathbb R$ is that now you can factor any 1-variable polynomial $f(x)$ (say with coefficients in $\mathbb R$) into linear factors. Alternatively, all real polynomials $f(x)$ have roots in $\mathbb C$ but not in $\mathbb R$.

Over $\mathbb H$, there are real polynomials you can factor into linear polynomials that you can't over $\mathbb C$, but this only shows up in higher variables. E.g. $$ x^2+y^2+z^2+w^2 = (x+iy+jz + kw)(x-iy-jz - kw)$$

Similarly, over $\mathbb O$ you can factor a sum of 8 squares into 2 linear factors. You still can't factor most polynomials, but these are especially interesting cases related to quadratic forms and composition laws. And of course there is a trade-off that you lose properties like commutativity and associativity if you want to work in these more general number systems.

You also see more geometry, as described in Conway and Smith's book On quaternions and octonions, but I think this is not along the lines you were asking about.