Actually, my headline pretty much sums up my question. I have done some exercises on nilpotent group and I'm thinking about an example of a division ring $(D,+,\cdot )$ where $(D,+)$ is an abelian group and $(D\setminus 0,\cdot)$ is a nilpotent group.
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If $x$ is a nilpotent element that is also a unit, then it is $0$.
This means that no nilpotent ring can have any nonzero invertible elements.
A division ring has at least its identity as a nonzero invertible element.
So no, there isn't any such ring.
Update
The user has since updated their question to indicate they are using a homemade definition of "nilpotent ring" that is not consistent with the standard notion I was using. The following is added after that came to light.
After doing a little research I came across this interesting survey:
In it, they mention that in 1950, Hua proved that when the multiplicative group of a division ring is solvable, the division ring is actually a field. Since nilpotent groups are solvable, this would say that all examples are actually fields.
Since finite fields are obvious examples of "nilpotent fields" in the sense you gave, I guess you were mainly interested in noncommutative examples, but the above account says that this is not possible.
rschwieb
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But, I've read the definition of nilpotent group as the central series is finite. – Achak0790 Sep 10 '19 at 10:16
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@AnindaChakraborty and that has no bearing on the definition of a nilpotent ring. Surely you already know this to the last question of yours we discussed this in? https://math.stackexchange.com/q/3329144/29335 – rschwieb Sep 10 '19 at 10:27
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I know, but I'm seeing nilpotent division ring (D,+,•) as (D,+) is an abelian group and (D{0},•) is a nilpotent group and I'm seeking for an example of this kind. – Achak0790 Sep 10 '19 at 10:30
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thanks for sharing with me this article. – Achak0790 Sep 10 '19 at 13:24