A ring $R$ is called central reduced if every nilpotent element is central. Ungor et al. math.RA 14 Dec 2013 has given an example of a commutative ring which is central reduced but not reduced. Can we find an example of a ring which is noncommutative central reduced but not reduced?

5$\begingroup$ Do you want some kind of nontriviality condition? For example, if $R$ is commutative nonreduced, and $S$ is noncommutative reduced, then $R\times S$ is an example. $\endgroup$ – Jeremy Rickard Apr 17 '16 at 10:01

$\begingroup$ I have deleted my answer; I totally missed the words "not reduced". Sorry! $\endgroup$ – LSpice Apr 17 '16 at 14:16

$\begingroup$ What do you mean by "has given an example"? Is not every commutative ring central reduced by this definition?? $\endgroup$ – მამუკა ჯიბლაძე Apr 17 '16 at 18:58

$\begingroup$ @მამუკაჯიბლაძე, I think that you and I misread the poster's question in the same way. The crucial point, in bold, is that the example is commutative, and so central reduced, but not reduced. $\endgroup$ – LSpice Apr 18 '16 at 0:19

$\begingroup$ @LSpice No no my question was not about the question but about the first sentence. Specifically about the words "has given an example". I mean, if any commutative ring is an example... $\endgroup$ – მამუკა ჯიბლაძე Apr 18 '16 at 4:59
The answer is yes, and there are many ways to do it.
Use Jeremy Rickard's direct product construction.
Let $F$ be a field, let $R=F[x\ :\ x^2=0]$, and let $S=R\langle y,z\rangle$ be the extension of $R$ in the noncommuting variables $y,z$.
By the way, I'd personally use "nilpotentcentral" to describe this "central reduced" condition, since I think the first phrase is more descriptive of what is happening (we are forcing all nilpotent elements to be central).

$\begingroup$ I think the idea behind the name is just that one is replacing the reducedness condition "every nilpotent lies in $\{0\}$" by the similarlooking condition "every nilpotent lies in $Z(R)$." $\endgroup$ – LSpice Apr 18 '16 at 0:21

$\begingroup$ @LSpice I didn't communicate what I meant very clearly, so I've modified my answer slightly to more clearly communicate what I meant. $\endgroup$ – Pace Nielsen Apr 18 '16 at 2:07