21
$\begingroup$

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?

$\endgroup$
8
  • $\begingroup$ What is a matrix representation of an algebraic closure? You mean, like an algebra representation over a subdivision ring of finite index? $\endgroup$
    – rschwieb
    Dec 17, 2014 at 14:22
  • 6
    $\begingroup$ 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. $\endgroup$
    – rschwieb
    Dec 17, 2014 at 14:30
  • 3
    $\begingroup$ 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 :) $\endgroup$
    – rschwieb
    Dec 17, 2014 at 14:43
  • 1
    $\begingroup$ 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 $\mathbb{H}$? $\endgroup$
    – Manny Reyes
    Dec 18, 2014 at 1:36
  • 1
    $\begingroup$ See Is the “algebraic closure” of the quaternions, finite dimensional? on mathoverflow. $\endgroup$
    – Sebastien Palcoux
    Dec 28, 2014 at 10:42

2 Answers 2

Reset to default
10
$\begingroup$

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 $\Bbb{H}$.

$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$
    – Jyrki Lahtonen
    Dec 31, 2014 at 20:06
  • $\begingroup$ 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 ? $\endgroup$
    – Sebastien Palcoux
    Dec 31, 2014 at 20:16
  • 2
    $\begingroup$ 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 $\Bbb{H}$. $\endgroup$
    – Jyrki Lahtonen
    Dec 31, 2014 at 20:23
  • $\begingroup$ Did you find one? $\endgroup$
    – Sebastien Palcoux
    Jan 5, 2015 at 8:57
2
$\begingroup$

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.

$\endgroup$
18
  • 3
    $\begingroup$ 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 $R\langle x\rangle$. $\endgroup$
    – rschwieb
    Dec 17, 2014 at 14:13
  • 3
    $\begingroup$ @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. $\endgroup$
    – rschwieb
    Dec 17, 2014 at 19:36
  • 3
    $\begingroup$ 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). $\endgroup$
    – Eric Wofsey
    Dec 28, 2014 at 9:15
  • 2
    $\begingroup$ 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. $\endgroup$
    – Eric Wofsey
    Dec 28, 2014 at 9:31
  • 3
    $\begingroup$ It is also not even obvious to me that $\mathbb{H}\{x\}/(p(x))$ is a nonzero ring for every nonconstant $p$. $\endgroup$
    – Eric Wofsey
    Dec 28, 2014 at 11:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.