Let $R$ be a ring (not necessarily commutative or unital) that is generated by idempotents. I'd like to know if $\text{Ann}(R)=0$ must hold. Here I use $\text{Ann}(R)$ to denote the set of all elements $r\in R$ such that $rR=Rr=0$. All I knew is that it holds when $R$ is commutative.
$\begingroup$
$\endgroup$
2
-
$\begingroup$ generated by idempotents in which sense? as an additive group? $\endgroup$– YCorCommented Mar 31, 2015 at 8:04
-
$\begingroup$ @YCor As a ring. $\endgroup$– Censi LICommented Mar 31, 2015 at 8:17
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
No, $\mathrm{Ann}(R)$ does not necessarily hold when $R$ is generated as a ring by idempotents.
Let $K$ be a field, or more generally any commutative (associative) ring with 1. Let $R$ be the (associative, non-unital) $K$-algebra of matrices $m(e,a,b,c)=\begin{pmatrix}0 & a & c\\0 & e & b\\ 0 & 0 & 0\end{pmatrix}$ with $a,b,c,e\in A$. Then $m(0,0,0,1)$ belongs to $\mathrm{Ann}(R)$. On the other hand, all elements of the form $m(1,a,b,ab)$ are idempotents, and the idempotents $m(1,0,0,0)$, $m(1,1,0,0)$, $m(1,0,1,0)$, $m(1,1,1,1)$ form a basis of $R$ as a $K$-module; in particular they generate $R$ as a ring (and even as an additive group) when $K=\mathbf{Z}$ or $\mathbf{Z}/n\mathbf{Z}$.
