Let $R$ be an artinian commutative ring and $A$ an algebra over $R$ with basis $\{a\_1, a\_2, \ldots , a\_n\}$ where each $a\_i$ is a unit and


$a\_ia\_j=u\_{ij}a\_ja\_i,$


where each $u\_{ij}\in R$ and is a unit. is true that for every element $x\in A$ the left annihilator of $x$ denoted by $ann\_l\{x\}$ equals the right annhilator $ann\_r\{x\}$?


I think that if exist an element $z$ with $ann\_l{z}\neq ann\_r{z}$, this lack of symmetry
must necessarily be reflected in the ring $R$ or in the elements of the base. Since, for the ring $R$ and for the elements of the base there is symmetry, this symmetry extends to all elements of $A$. This is my perception, but how to prove? 



Annihilators in algebras
It is false. The ring of two-by-two matrices has such a basis:


$$a\_1=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix},a\_2=\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix},a\_3=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},a\_4=\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$$


with $u\_{ij}=\pm 1$. (The $a\_i$ lie in a representation of the group D\_4, so $a\_ia\_ja\_i^{-1}a\_j^{-1}$ is central, thus is $\pm 1$.) But this ring obviously fails to satisfy the annihilator property.


