# Is there an algebraic closure for the quaternions?

This post is a sequel of: Is the set of quaternions $\mathbb{H}$ algebraically closed?

This answer shows that:
1. $\mathbb{H}$ is algebraically closed for the polynomials of the form $\sum a_r x^r$
2. It is not for the polynomials freely generated by $\mathbb{H}$ and $x$, because $xi+ix-j$ has no root.

Question: Is there an algebraic closure (for the case 2)?
If so: What does it look like? What's its dimension over $\mathbb{H}$? What's its matrix representations?

• What is a matrix representation of an algebraic closure? You mean, like an algebra representation over a subdivision ring of finite index? Dec 17, 2014 at 14:22
• The "best" theorem I know for any notion of algebraic closure of $\Bbb H$ in terms of free-polynomials $\Bbb H\langle x\rangle$ is this: every polynomial in $\Bbb H\langle x\rangle$ whose highest degree term is a single monomial has a root in $\Bbb H$. The polynomial $xi+ix-j$ fails this because the part of degree $1$ has two pieces. But, for example, $xkx+xi+ix-j$ would have a root, since its highest degree is $2$ and it only has $xkx$ in that degree. Dec 17, 2014 at 14:30
• Sebastien, you want to read this paper by Lam on the quaternions for that theorem. It's an awesome paper for anyone interested in the quaternions :) Dec 17, 2014 at 14:43
• It's not clear to me what "an algebraic closure" for the case 2 would mean, exactly. I presume that it should contain all elements that satisfy "polynomials" in the free product $\mathbb{H} * \mathbb{Z}\langle x \rangle$. Does it need to be a division algebra? Does every element need to satisfy a "polynomial" over $\mathbb{H}$? Dec 18, 2014 at 1:36
• See Is the “algebraic closure” of the quaternions, finite dimensional? on mathoverflow. Dec 28, 2014 at 10:42

I don't think that there can be an associative $\Bbb{R}$-algebra $L$, containing $\Bbb{H}$ as a subring, such that the equation $$xi+ix=j\qquad(1)$$ has a solution $x\in L$.

Multiplying $(1)$ by $i$ from the left gives us $ixi+i^2x=ij$, or $ixi-ij=-i^2x$. As $i^2=-1$ and $ij=k$, this reads $$x=ixi-k.\qquad(2)$$ On the other hand multiplying $(1)$ by $i$ from the right gives us $xi^2+ixi=ji$, and using $i^2=-1, ji=-k$ this yields $$x=ixi+k.\qquad(3)$$ The equations $(2)$ and $(3)$ together imply $k=-k$. As $k$ is a unit of $L$ this implies that $2=0$ in $L$, so $L$ cannot be an extension of $\Bbb{H}$.

• Sorry about being dense. Clearly $-1$ is in the center. I don't need to assume that $L$ is an $\Bbb{R}$-algebra for the argument to work. Dec 31, 2014 at 20:06
• Thank you! Now the natural question is the following: Is there a (non-commutative) polynomial $p$ (called "allowed") having no root on $\mathbb{H}$ such that there is an associative $\mathbb{R}$-algebra $L$ containing $\mathbb{H}$ as a subring and such that $p$ has a root on $L$ ? If no, then $\mathbb{H}$ is "in some sense" algebraically closed, else, what is its "allowed" closure ? Dec 31, 2014 at 20:16
• I'm not sure. $\Bbb{H}$ is in a natural way a subring of $M_2(\Bbb{C})$ with $1\mapsto I_2$, $i\mapsto \pmatrix{i&0\cr0&-i\cr}$, $j\mapsto \pmatrix{0&1\cr-1&0\cr}$. I would try to find a polynomial that has a zero in the ring of 2x2 complex matrices, but does not have one in $\Bbb{H}$. Dec 31, 2014 at 20:23
• Did you find one? Jan 5, 2015 at 8:57

It looks like the answer is morally "no." Now, there is a formal closure for which we can solve free polynomials: As in the case of fields, you take an inductive limit. Let $\Bbb H=R$ be our normed division algebra. Then

$$\overline{R}=\varinjlim_{[L:R]<\infty} L$$

where the inductive system is taken relative to inclusions of algebra extensions $L/R$ of finite dimension over $R$, each of the form

$$L_p=R\{x\}/(p(x))$$

where $p(x)$ is irreducible over $R$ and $R\{x\}$ is the polynomials freely generated as in case ($2$). This certainly has the required property that all polynomials in $R$ have a root in $\overline{R}$, and any other such object has a copy of this inside of it for purely formal reasons.

I note that the directed system so-defined is indeed a directed system--in fact a lattice--so this should go through unless I'm missing something obvious.

rschweib has noted that the result is no longer a division algebra, so this is really not ideal, but the "algebraic closure" property holds, and necessarily it's a minimal ring where this property can hold, so it seems this is the best we can hope for. However we also cannot force algebraicness of the result since $R\{x\}/(xi+ix-j)$ doesn't make $x$ algebraic appropriately in the sense that you want to mimic the field case's excellent definition that algebraicness means $F(\alpha)/F$ is finite dimensional as an algebra over $F$, which doesn't hold in this setting.

• This would be the hope: that the same plan works for division rings. But there's still an awful lot up in the air. It isn't totally clear that $R\langle x\rangle/(p(x))$ does what we want it to do, or even if it's finite dimensional. $p(x)=xi+ix-j$ is a good example since $R\langle x\rangle/(p(x))$ looks like it might just be $R[x]$. Noncommutativity really throws a lot of wrenches into doing this whole scheme for $R\langle x\rangle$. Dec 17, 2014 at 14:13
• @AdamHughes Then maybe this really is the brick wall, since $R\langle x\rangle/(xi+ix-j)$ does not make $x$ algebraic, as was hoped. Dec 17, 2014 at 19:36
• You should be careful about how you define that colimit. For instance, if you take the diagram to include all injections between such extensions $L$ of $R$, the commutative analogue would be the zero ring for any $R$ that is not already algebraically closed (because it forces all Galois conjugates to be equal). As in the commutative case, you really want to construct a well-ordered sequence of extensions by induction (and for this reason existence of algebraic closures requires the axiom of choice in general). Dec 28, 2014 at 9:15
• Alternatively, you could take the colimit where the only maps you have are the inclusions $R\to L$ and you allow no maps between different $L$s. In this case, the commutative analogue will certainly not be a field (and you need to choose a maximal ideal of it to mod out to get an actual algebraic closure), and it is still not obvious to me that the noncommutative version gives a nonzero ring. Dec 28, 2014 at 9:31
• It is also not even obvious to me that $\mathbb{H}\{x\}/(p(x))$ is a nonzero ring for every nonconstant $p$. Dec 28, 2014 at 11:56