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 non-triviality condition? For example, if $R$ is commutative non-reduced, and $S$ is non-commutative 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.

  1. Use Jeremy Rickard's direct product construction.

  2. 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 "nilpotent-central" 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 similar-looking 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.