Associative division subalgebras of split Cayley-Dickson algebra 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.
 A: 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).
