# Constructing noncommutative nilpotent rings of given index

When I read about algebra I often see a certain disregard for examples or perhaps a disregard for a reader whose knowledge of examples is limited. When I'm interested in a property $p$ of an algebraic structure $X$, I often find theorems (and answers on forums like this one) saying that "any structure with property $p'$ also has the property $p.$" Which leaves me with nothing if I know no examples of $p'$-structures. In this particular case I would like to understand the nilpotency of a noncommutative ring better.

A ring (necessarily without unity, unless perhaps the ring is trivial -- I'm not sure about this one) is called nilpotent when there exists a natural number $n$ such that the product of any $n$ elements is $0.$ The nilpotency index of a nilpotent ring is the smallest such $n$.

It would be interesting for me to see examples of sequences $$R_N,R_{N+1},R_{N+2},\ldots$$

of noncommutative rings such that $R_{N+i}$ is nilpotent with nilpotency index $N+i$. I would like the examples to be concrete, that is not of the form "take any $\phi$-ring and a $\psi$-group and make a group ring out of them." Please take concrete examples of $\phi$-rings and $\psi$-groups.

Also, I would be very happy to see an abundancy of examples, which is why I'm adding the big-list tag and asking the mods to make it CW.

• $R_N = X\mathbb{Z}[X]/(X^N)$ should be the canonical example. – Slade Oct 12 '14 at 8:06
• For the entire class of finitely generated examples, look at quotients of $(S)\mathbb{Z}[S]/(S)^N$, where $S$ is any finite collection of independent variables. For that matter, let $S$ be a set of arbitrary cardinality and you have the entire class of examples, period. – Slade Oct 12 '14 at 8:11
• Concrete: $R_3 = (X,Y)\mathbb{Z}[X,Y]/(X^3, Y^3, XY)$, etc. etc. I don't mean to show "disregard" for examples, but when you ask for something so general, and no indication of what other properties you're interested in, it's hard to avoid responding in a general fashion. – Slade Oct 12 '14 at 8:15

## 1 Answer

Here are some examples coming from linear algebra.

Let $R$ be a ring ( commutative or not). Let $n$ and $1\le k \le n-1$ natural numbers. Consider the ring $N_{n,k}(R)$ of $n\times n$ matrices with elements in $R$ that have zero entries below the diagonal $\{ (i,j)\ | \ j-i =k\}$. Then $N_{n,k}(R)$ has nilpotency index $\le (n-k+1)$, and if $R$ has a unit the index is exactly $(n-k+1)$.

• The only nilpotent ring with identity is the zero ring, so maybe the first line is not necessary :) Triangular matrix rings are an awesome noncommutative example +1 – rschwieb Oct 12 '14 at 11:59
• @rschwieb: I removed the redundant note. Thanks! – Orest Bucicovschi Oct 12 '14 at 14:51