Are there any central simple algebras admitting a standard basis?

Question

Are there any central simple algebras admitting a standard basis?

By a standard basis I mean a normal basis that has a cyclic property generalizing that of the familiar basis $1, i, j, k$ for Quaternion algebras $(a, b \mid F)$, satisfying relations $i^2=a, j^2=b, k=ij=-ji$ for $a, b \in F^\times$.

I saw Cayley–Dickson construction which yields $F$-algebras of dimension $2^n$, but these algebras are not associative so not central simple in general)

One way to construct quaternion algebras is to use abelian varieties: Like in the case of elliptic curves, by using $\ell$-adic Tate modules we have the following:

the endomorphism ring $\text{End}(A)$ is finitely generated $\mathbb{Z}$-module of rank at most $4g^2$.

In general, the endomorphism ring $\text{End(A)}$ of an abelian vairety $A$ is an order in a semi-simple algebra over $\mathbb{Q}$. When $B$ is a semi-simple algebra over $\mathbb{Q}$ admitting a positive definite anti-involution, such algebras have been classified by A. Albert and G. Scorza.

Type I: Totally real number field
Type II/III: Definite or indefinite quaternion algebra over totally real field
Type IV: Central simple algebra over a CM-field.

Central simple algebras of dimension $n^2$ are all quaternion algebras when $n=2$. I was wondering if other form of basis with cyclic property appears when $n$ is larger, but at least in the construction above only quaterinon algebras appear.

Answer 1

As suggested by @Kimball I develop my comment. An important class of central simple algebras consists of the cyclic algebras: assume that the field $k$ contains a primitive $n$-th root of unity $\zeta $, with $n$ prime to $\operatorname{char}(k) $. Let $a,b\in k^*$; the cyclic algebra $R_{a,b}$ is the quotient of $k\langle x,y \rangle$ by the relations $x^n=a, y^n=b, yx=\zeta xy$. The monomials $x^{i}y^{j}$ for $0\leq i,j<n$ form a natural basis of $R_{a,b}$ over $k$, generalizing the standard basis of the quaternions. Over a number field, every central simple algebra is cyclic — this is a deep result of Albert–Hasse–Brauer–Noether. It is a classical conjecture that for $n$ prime this should hold over any field.

Answer 2

According to the Wedderburn–Artin Theorem:

Any finite-dimensional simple algebra is isomorphic to the matrix algebra $M_n(D)$ for some division ring $D$.

So it is enough to look for examples in this setting. In the particular case when $D=K$ is the base field, it is a central-simple algebra. If such field $K$ contains a primitive $n$-th root of 1 (let us denote it by $\varepsilon$) one can define the following matrices:

$$A=\begin{pmatrix} \varepsilon^{n-1} & 0 & \cdots & 0 & 0\\ 0& \varepsilon^{n-2} & \cdots & 0 & 0\\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & \varepsilon^{n-1} & 0\\ 0 & 0 & \cdots & 0 & 1\\ \end{pmatrix} \quad\text{and}\quad B=\begin{pmatrix} 0 & 1 & 0 & \cdots & 0\\ 0 & 0 & 1 & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1\\ 1 & 0 & 0 & \cdots & 0\\ % \end{pmatrix}.$$

The above matrices satisfy $A^n=B^n=I_n$ and $AB=\varepsilon BA$.

By straightforward computations and using the Vandermonde determinant one can prove that the set $\{A^kB^l\,|\,0\leq k, l \leq n-1\}$ is a linear basis of $M_n(K)$.

The above basis elements satisfy $A^kB^l A^rB^s=\varepsilon^{-lr}A^{k+r}B^{l+s}$.

This may be seen as a generalization of such properties of basis elements of quaternions mentioned in the question.

Remark: this algebra is isomorphic to the cyclic algebras mentioned in @abx's answer. The isomorphism can be defined by sending $A$ to $y$ and $B$ to $x$. Nevertheless, it seems more natural to me to consider an example on matrix algebras, in light of the Wedderburn-Artin Theorem.