TITLE: Examples of division algebras
QUESTION [6 upvotes]: Together with the Grunwald–Wang theorem, the Albert–Brauer–Hasse–Noether theorem implies that every central simple algebra over an algebraic number field is cyclic, i.e. can be obtained by an explicit construction from a cyclic field extension $L/K$ .
Are there any kind of division algebras not arising from quaternions or cyclic algebras? I have never seen this and it would be nice to know.

REPLY [4 votes]: This is a summary of the construction of a non-cyclic division algebra of degree four from Nathan Jacobson's book Finite-Dimensional Division Algebras over Fields. Jacobson tells that Albert was the first to construct such division algebras, and the presented construction may (?) be a modification of his method.
The construction:

Let $D(F,\alpha,\beta)$ be the quaternion algebra with center $F$,
and basis $\{1,u,v,uv\}$, where $u^2=\alpha$, $v^2=\beta$ and
$uv=-vu$.
Let $F_0$ be a subfield of reals, and $F=F_0(\xi,\eta)$ a purely transcendental extension with $\xi,\eta$ algebraically independent over $F_0$.
Let $D_1=D(F,\eta,\eta)$ and $D_2=D(F,\xi,\xi\eta)$.
Then $D=D_1\otimes_FD_2$ is a non-cyclic division algebra of degree four with center $F$ (variations are possible, and we may use as $\beta$-elements polynomials on $\xi$ and $\eta$ such that the parities of the exponents of the leading term are as above).

Why is it a division algebra?
Here the key step is that the tensor product of two quaternion division algebras over a field $F$ is NOT a division algebra if and only if $D_1$ and $D_2$ contain isomorphic quadratic extensions of $F$ as subfields.
This condition can be re-expressed in terms of the reduced norms as follows. Let $D_i', i=1,2$ be the kernels of the reduced trace maps of $D_1,D_2$ respectively that is, the $F$-spans of the respective sets $\{u,v,uv\}$. Then on $D_1'\oplus D_2'$ we can define a quadratic form $n$ by the recipe $n(x_1,x_2)=n_1(x_1)-n_2(x_2)$. The reformulation says that $D_1\otimes_F D_2$ is a division algebra, iff $n$ is anisotropic. The equivalence of these two conditions is easy to believe. For if $n_1(x_1)=n_2(x_2)$ for some $x_i\in D_i', i=1,2,$ then the quadratic fields $F(x_1)$ and $F(x_2)$ are isomorphic. The other direction is not too difficult either.
Why is it not cyclic?
This depends on a Lemma due to Albert: If $F$ is a field, $\sqrt{-1}\notin F$, and $E/F$ is a cyclic quartic extension, then the unique quadratic intermediate field is of the form $F(\sqrt{u^2+v^2})$, where $u,v\in F$ and (obviously) $u^2+v^2$ is a non-square of $F$.
This leads to the idea. If $D\otimes_F K$ remains a division algebra in 
every extension of scalars from $F$ to $K=F(\sqrt{u^2+v^2})$, then $D$ can't contain a copy of $K$, and hence, by Albert's Lemma, won't contain a cyclic quartic extension field either. Jacobson then proceeds to prove that this holds with the above $D$. The parity constraint mentioned above saves the day, as using it allows to show that the quadratic form $n$ remains anisotropic under quadratic extensions of scalars of the prescribed type.

I'm afraid this is about as far as I have ever made in Jacobson's book. I'm not very conversant with the details here. Anyway, I hope this gives you an idea of what tools and tricks the construction requires. All this takes a bit over four pages in the book.