Let rings be associative, but not necessarily unital or commutative. We say that a ring is locally nilpotent if every finitely generated subring is a nilpotent ring.
I am interested in finding some examples (or ideally, families) of nilpotent rings and some examples (or ideally, families) of locally nilpotent rings which are not nilpotent. I am interested in both “generic” families and weird pathological examples.
Best Answer
Let $p \in \mathbb{Z}$ be prime and $n \geq 2$ be an integer. Let $R_n$ be the subset of $\mathbb{Z}/p^n\mathbb{Z}$ containing ${0}, {p}, {2p}, \ldots, {(p^n-1)p}$. In other words, $R_n$ is the $\mathbb{Z}$-module $p\mathbb{Z}/p^n\mathbb{Z}$.
Since $p\mathbb{Z}$ is a (commutative, but not unital) subring of $\mathbb{Z}$, it follows that $R_n$ is in fact a ring. Furthermore, it is clear that $R_n^n = 0$, since each element of $R_n$ is a multiple of $p$, so any element of $R_n^n$ is a multiple of $p^n$.
Now we can use this construction to obtain a ring that is locally nilpotent, but not nilpotent. Define the ring
$$ A = \bigoplus_{n = 2}^\infty R_n $$ where the $\bigoplus$ notation means that $A$ consists of sequences of elements of the $R_n$ with $\textbf{finitely}$ many nonzero terms, with elementwise addition and multiplication.
Then for any finitely generated subring $B$ of $A$, we have $$ B \subseteq \bigoplus_{n = 2}^N R_n $$ for some $N \in \mathbb{N}$ (since each generator of the subring is in such a subset, and hence any combination of those generators will be as well).
For $n \leq N$, we have $R_n^N = 0$, so $B^N = 0$, and hence $B$ is nilpotent.
However, $A$ is not a nilpotent ring, because for every $n \in \mathbb{N}$, for $k > n$, for $p \in R_k$, we have $p^n \neq 0$ in $R_k$. Thus we have constructed a ring that is locally nilpotent, but not nilpotent.
This argument doesn't seem to rely on the fact that our original ring is $\mathbb{Z}$, or even that $p$ is prime, so you should be able to generalise it to a pretty broad class of examples. Off the top of my head, I think it would work if you started with any nonzero ideal $I$ in a Noetherian ring and defined $R_n = I/I^n$. I haven't worked through the details myself, but it seems like the argument should carry over (although you need to show that $I^n \neq I^k$ for $n < k$, which you can do with Nakayama's lemma).