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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.

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1 Answer 1

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).

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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

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