In his book *Introduction to arithmetic groups*, Dave Witte Morris implicitly gives a construction of central division algebras of degree 3 over $\mathbb{Q}$ in Proposition 6.7.4. More precisely, let $L/\mathbb{Q}$ be a cubic Galois extension and $\sigma$ a generator of its Galois group.If $p \in \mathbb{Z}^+$ and $p \neq t\sigma(t)\sigma^2(t)$ for all $t \in L$, then
$$ D=\left\{ \begin{pmatrix}
x & y & z\\
p\sigma(z) & \sigma(x) & \sigma(y)\\
p\sigma^2(y) & p\sigma^2(z) & \sigma^2(x)
\end{pmatrix} :(x,y,z)\in L^3 \right\}
$$
is a division algebra.


On page 145, just before Proposition 6.8.8, Morris claims that it is known that every division algebra of degree 3 arises in this manner. This should follow from the fact that every central division algebra of degree 3 is cyclic. I could not find this explicit construction in my references (e.g. Pierce - *Associative Algebras*, though maybe I missed something) and I would like to know if there is a reference or a quick way to see that this exhausts all central division algebras of degree 3 over $\mathbb{Q}$.



Explicit construction of division algebras of degree 3 over $\mathbb{Q}$
Looking more carefully in Pierce - *Associative algebras*, I found the answer I was looking for, which I'm going to describe here for future reference.


The algebra $D$ in the question is a realisation of a special type of crossed product algebra, which gives precisely the central division algebras of degree 3 over $\mathbb{Q}$.


The first step is to use the powerful result mentioned in the comments, which states that all central simple algebras over number fields are cyclic. For degree 3 algebras, this was proved by Wedderburn. Recall that an algebra $A$ over a field $F$ of degree $n$ is cyclic if there is a cyclic Galois extension $E/F$ of degree $n$ such that $E$ is a maximal subfield inside $A$.


Proposition **a** in Pierce, section 15.1, states that a cyclic algebra $A$ as above, where $\sigma$ generates the Galois group of $E/F$, contains an invertible element $u$ such that


1. $A=\bigoplus\_{0 \leq j <n} u^jE$
2. $u^{-1}du=\sigma(d)$ for all $d\in E$, and
3. $u^n=a \in F^\times$.


This is equivalent to saying that $A$ is isomorphic to the crossed product $(E,G,\Phi\_a)$, defined in Pierce. The only thing left to do is to check that the algebra $D$ in the question is a construction of this crossed product. You can check this by first embedding $E$ into $M\_3(E)$ by $$x \mapsto \text{diag}(x,\sigma(x),\sigma^2(x)).
$$ Then take a rational number $p\neq 0$, which will play the role of $a$ in Proposition **a**, and put $$ u = \begin{pmatrix}
 & & p\\
1 & & \\
 & 1 &
\end{pmatrix}.
$$
Then one easily checks that (at least after transposing to ensure the same order of the factors) the conditions in the proposition are satisfied.


The rest of the section in Pierce deals with the condition that $p$ is not a norm.


