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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


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?

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up vote 2 down vote accepted

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.

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Thank you. It is a nice example to prove that we have not $ann_l x=ann_rs$. But, probably, their cardinality is the same. I'm trying to find a proof of this. – Miguel Jun 13 '12 at 18:35

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