Finite dimensional associative division algebras over Q

Question

I know the following two facts of division algebras:

• The finite dimensional associative division algebras over $\mathbb{R}$ are $\mathbb{R}, \mathbb{C}$ and $\mathbb{H}$.
• The finite dimensional associative division algebras over $\mathbb{F}_q$ are just the finite field extensions.

So I was wondering, what do we know about finite dimensional field extensions over $\mathbb{Q}$? Or more precisely: Can they all be embedded into $\mathbb{H}$?

I know that division algebras in the form of $\mathbb{Q}(\alpha)$ are just number fields. Is it possible to do something with that fact?

Answer 1

There are division algebras with centre $\Bbb Q$ of dimension $n^2$ for any positive integer $n$. For $n\ge3$ they do not embed in $\Bbb H$. These division algebras can be constructed as cyclic algebras.

Answer 2

A finite-dimensional field extension $F/\mathbb{Q}$ is algebraic. Moreover, since char$(\mathbb{Q})=0$, the extension is separable, so by the primitive element theorem there exists some $\alpha\in F$ such that $F=\mathbb{Q}(\alpha)$.

As for the second question: See $\mathbb{Q}(\alpha)$ as a subfield of $\mathbb{C}$ and see $\mathbb{C}$ as the subring $\mathbb{R}[1,i]\subseteq\mathbb{H}$. Then you have an injective ring homomorphism from $\mathbb{Q}(\alpha)$ to $\mathbb{H}$.

ADDED: The noncommutative case

The noncommutative case is harder, with proofs based in ramification of ideals and $p$-adic fields (proofs were recopiled by Albert in his Structure of Algebras of 1939, Chapter IX).

If $D$ is a noncommutative algebra over $\mathbb{Q}$, then $D$ is a cyclic algebra over its center $Z(D)$, i.e., there are a finite-dimensional field extension $F/Z(D)$ of dimension $n$ with cyclic Galois group generated by $\sigma$, and a nonzero element $\alpha\in Z(D)$ such that for every $x\in D$ we have $x=x_0+x_1y+\ldots+x_{n-1}y^{n-1}$, with $x_i\in F$, $y\in D$ such that $y^n=\alpha$ and noncommutative multiplication given by $yz:=\sigma(z)y$ for every $z\in D$.