Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let's consider the split Cayley-Dickson algebra $C$ over an arbitrary field $F$ (It is well known that all split composition algebras having the same dimension over base field are isomorphic, e.g., all split Cayley-Dickosn algebras are isomorphic to Zorn's vector-matrix algebra).
The question is: what are the conditions for existence of associative division subalgebra $A$ of $C$ with $dim_F A = 4?$
Of course, it is necessary for $F$ not to be algebraically closed. It also seems to me that it's not possible for $A$ to be a field because of its dimension and $A$ being a composition algebra, so $F$ must be infinite by Wedderburn's theorem.
I'm also wondering if $A$ is unique, because there may be more than one division composition algebra of dimension $4$ over $F$, but I'm not sure that all of them can be proper subalgebras of $C$. Let's consider this subspace of $C$: $ V = \left\{\begin{pmatrix} 0 & 0 \\ \alpha & v \\ \end{pmatrix}|\ \alpha \in F, v \in F^3\right\}. $
It's easy to see that $dim_F V=4$ and $V \cap A =0$ because every element of $V$ is not invertible, so by dimension counting we have $V \oplus A = C$.
We also have $F \subset A$, so i think $A$ must look like $ A = \left\{\begin{pmatrix} \beta & u \\ \varphi(u) & \beta \\ \end{pmatrix}|\ \beta \in F, u \in F^3\right\}, $
where $\varphi$ is an invertible linear map of $F^3$ satisfying certain identities (for example, $\varphi(v) \cdot u = v \cdot \varphi(u) $ , where $\cdot$ is the ordinary dot product — it can be easily obtained by multiplying elements from $A$ and comparing results).
For example, quaternions are the only division subalgebra of split octonions, and they can be represented (with multiplication mentioned above) as $ \mathbb{H} = \left\{\begin{pmatrix} \alpha & v \\ -v & \alpha \\ \end{pmatrix}|\ \alpha \in \mathbb{R}, v \in \mathbb{R}^3\right\} $ , where $\varphi(v)=-v$.
So, that's another question: am I correct about the construction of these division subalgebras? If yes, what else can we say about $\varphi?$
Sorry for my poor English and LaTeX skills.
Thank you in advance.

share|cite|improve this question

It seems likely to me that an exhaustive answer to your question would fill in a couple volumes. I will write a summary of a point of view I know about. Others can hopefully shed more light.

Let $D$ be a 4-dimensional associative division algebra over its center $F$. Assume that $char F\neq 2$. If $a\in D\setminus F$, then $E=F(a)$ is a Galois extension of $F$ (this was the reason, why I excluded characteristic two). As the minimal polynomial $m(x)$ of $a$ over $F$ is quadratic, both its roots are in the field $F(\theta)$, where $\theta=\sqrt d$ and $d$ is the discriminant of $m(x)$. Therefore $E=F(\sqrt d)$ is Galois, and $\sigma:\theta\mapsto -\theta$ gives the non-trivial element $\sigma$ of the Galois group $G=Gal(E/F)$.

Because $D$ is a central simple $F$-algebra, it follows from the Skolem-Noether theorem that $\sigma$ can be realized as a conjugation by an element $u\in D^*$, i.e. for all $z\in E$ we have $$ \sigma z = u^{-1}zu, $$ or $u\sigma z=zu$ for all $z\in E$. Consequently we also have $$ u^{-2}zu^2=u^{-1}\sigma(z) u=\sigma^2(z)=z $$ for all $z\in E$, so $\gamma=u^2$ commutes with all of $E$. The element $\gamma$ thus commutes with $E$ and it obviously commutes with $u$ also. As $u\notin E$, we see that $\dim C_D(F(\gamma))\ge3.$

The double centralizer theorem states that $$ \dim F(\gamma) \cdot \dim C_D(F(\gamma))=\dim D=4, $$ so the only possibility is that $\gamma\in F$.

The element $u\theta$ cannot be in the $F$-span of $1,\theta,u$ for then we would have $u\in E$. Therefore $\{1,\theta,u,u\theta\}$ forms an $F$-basis of $D$, and $\{1,u\}$ is a basis of $D$, when viewed as an $E$-space with $E$ acting from the right. The left regular representation of $D$ thus gives rise to the following homomorphism of $F$-algebras $\rho:D\rightarrow M_{2\times 2}(E)$ $$ \theta\mapsto\pmatrix{\theta&0\cr0&-\theta\cr},\qquad u\mapsto\pmatrix{0&\gamma\cr1&0\cr}. $$ The image of this homomorphism consists of the matrices of the form $$ A=\pmatrix{z_1&\gamma\sigma(z_2)\cr z_2&\sigma(z_1)\cr} $$ where $z_1,z_2\in E$ are arbitrary. As $D$ is simple, the homomorphism must be injective.

When do we get a division algebra from the datum $(E/F,\sigma,\gamma)$? The only thing we need to check is that all those matrices are invertible, i.e. the determinants are non-zero. Here $\det A= z_1\sigma(z_)-\gamma z_2\sigma(z_2)=N(z_1)-\gamma N(z_2)$, where $N:E\rightarrow F, z\mapsto z\cdot\sigma(z)$ is the norm map. We see that (assuming that $A\neq0$) $$\det A=0\Leftrightarrow N(z_1/z_2)=\gamma.$$ We have proven.

Theorem. The above construction gives a 4-dimensional division algebra with center $F$, iff $\gamma\notin N(E^*)$.

Thus the answer I offer reads. There exists a 4-dimensional associative division algebra with center $F$, if the field $F$ has a quadratic Galois extension $E$ such that the norm map $N:E^*\to F^*$ is not surjective.

This condition is clearly sufficient also, when $char F =2$. We showed that it is necessary, when $char F\neq2$.

The task of classifying those 4-dimensional associative division algebras is a formidable one. For number fields it is known that there will be infinitely many non-isomorphic associative division algebras like htis. A more precise answer in that case is given by global class field theory. I am the wrong person to say more about that. I am also the wrong person to say whether the above condition is necessary also in characteristic two. I would think that we can always find a separable quadratic extension of $F$ inside $D$, when the above argument would go through??

Note that the determinant of the representation of $D$ as 2x2 matrices over $E$ always gives you a multiplicative norm $N:D\to F$ (as required in the definition of a composition algebra).

share|cite|improve this answer
But a big question that I cannot answer is, whether all the algebras I described can be described as subalgebras of a Cayley-Dickson algebra? I think so, because this looks a lot like a (generalized) quaternion algebra, but I'm off to bed, now. – Jyrki Lahtonen May 1 '12 at 21:48
Yes, it seems to be that all 4-dimensional division algebras must be generalised quaternions; I found some information about that in Cohn's Algebra, Vol.3. Theorem there states that every division algebra possessing a two-dimensional split field must be quaternion algebra. It's also interesting to notice that your proof mirrors the proof of Frobenius theorem once given to me, except for Galois Theory — I'm still not familiar with that. – Yury Popov May 2 '12 at 3:05
But still, it would be interesting to find a way to represent these generalised quaternion algebras (in this case it's easier to identify them, because if restriction of norm to this subalgebra is nondegenerate, then it must be a generalized quaternion algebra) as proper subalgebras of $C$. Construction suggested above seems to be right, but I think there must be a way to describe $\varphi$ explicitly. Since the expression for $C$ norm remains the same in $A$, $\varphi$ must have something to do with «native» norm in $A$. – Yury Popov May 2 '12 at 3:24
And yet another comment, I realized that I still can't vote up your answer. So… Thank you, @Jyrki :) – Yury Popov May 2 '12 at 3:33

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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