Central commutators in rings

Posted: August 17, 2010 in Examples & Counter-Examples
Tags: , , ,

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}.

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