# Example of noncommutative central reduced rings which is not reduced

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?

• 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. – Jeremy Rickard Apr 17 '16 at 10:01
• I have deleted my answer; I totally missed the words "not reduced". Sorry! – LSpice Apr 17 '16 at 14:16
• What do you mean by "has given an example"? Is not every commutative ring central reduced by this definition?? – მამუკა ჯიბლაძე Apr 17 '16 at 18:58
• @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. – LSpice Apr 18 '16 at 0:19
• @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... – მამუკა ჯიბლაძე Apr 18 '16 at 4:59

## 1 Answer

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).

• 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 $Z(R)$." – LSpice Apr 18 '16 at 0:21
• @LSpice I didn't communicate what I meant very clearly, so I've modified my answer slightly to more clearly communicate what I meant. – Pace Nielsen Apr 18 '16 at 2:07