## Central commutators in rings

Posted: August 17, 2010 in Examples & Counter-Examples
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Theorem. Let $R$ be a ring and $a \in R.$ Let $Z(R)$ be the center of $R.$ If $Z(R)$ is reduced and $ar - ra \in Z(R)$ for all $r \in R,$ then $a \in Z(R)$.

Proof. Let $r \in R.$ Then

$ar(ar-ra)=a(ar-ra)r=(a(ar)-(ar)a)r=r(a(ar)-(ar)a)=ra(ar-ra).$

So $(ar-ra)^2=0.$ Hence $ar = ra$ because $Z(R)$ is reduced.  Thus $a \in Z(R). \ \Box$

One class of rings with reduced centers is the class of semiprime rings. If $Z(R)$ is not reduced, the result in the theorem need not hold. There is a nice example in Lam’s book, “A First Course in Noncommutative Rings”. Here it is:

Example. let $k$ be a ring with 1 and let

$R=\left \{ \begin{pmatrix} x & y & z \\ 0 & x & t \\ 0 & 0 & x \end{pmatrix}: \ x,y,z,t \in k \right \}.$

Let $a = e_{12}.$ Then $ar-ra=te_{13} \in Z(R)$ for every $r =\begin{pmatrix} x & y & z \\ 0 & x & t \\ 0 & 0 & x \end{pmatrix} \in R$ but obviously $a \notin Z(R)$ because, for example, $a$ does not commute with $e_{23}.$