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.

  • 4
    $\begingroup$ Cyclic algebras. $\endgroup$
    – abx
    Jan 20, 2022 at 6:56
  • $\begingroup$ @abx That's a good example. Can you add more number of generators while being central simple? $\endgroup$
    – Andy
    Jan 20, 2022 at 14:02
  • 1
    $\begingroup$ There aren't canonical bases, even for quaternion algebras. But as abx says, you can write down nice bases in cyclic algebras, which include division algebras over number fields. Does this answer your question? $\endgroup$
    – Kimball
    Jan 20, 2022 at 21:28
  • $\begingroup$ @Kimbell I was wondering if there are non-canonical bases, but yeah I'm good with abx's answer. $\endgroup$
    – Andy
    Jan 22, 2022 at 6:37
  • $\begingroup$ In that case, @abx are you willing to write up a brief answer, and then Andy can accept it so that this question gets marked as resolved? $\endgroup$
    – Kimball
    Jan 24, 2022 at 15:51

2 Answers 2


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.


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.

  • 1
    $\begingroup$ Actually, It is not a direct consequence od the above equations. But it is enough to prove that the set is linearly independent, and this follows from straightforward computations (and using the vandermonde determinant). I will edit the answer. Thank you for pointing this out. $\endgroup$
    – Thiago
    Jan 26, 2022 at 18:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.