Examples of division algebras

Question

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.

Answer 1

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.

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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.