Differences between nilpotent and pointwise nilpotent endomorphisms.

Question

Consider an endomorphism of a module $f:M\rightarrow M$. We have that $f$ is pointwise nilpotent if $\forall x\in M,\ \exists n,\ n\in \mathbb N$ such that $f^{n}(x)=0$. I already know that the definitions of nilpotent and pointwise nilpotent are not equivalent, but I do not know if it is possible to state when a pointwise nilpotent endomorphism is a nilpotent morphism. In particular are there some algebraic properties (for modules) that characterizes pointwise nilpotent morphisms?

Answer 1

For start it's easy to see that if the module $M$ is finitely generated then every pointwise nilpotent endomorphisms $f \colon M \to M$ is also a nilpotent endomorphism.

Another observation is that given a module $M$ and an endomorphism $f \colon M \to M$ we have the family $\langle \ker f^n \rangle_{n \in \mathbb N}$ which gives an ascending filtration of $M$ $$\{0\}= \ker f^0 \subseteq \ker f \subseteq \dots \subseteq \ker f^n \subseteq \cdots.$$ $f$ being pointwise nilpotent amounts saying that $M= \bigcup_{n \in \mathbb N} \ker f^n$, requiring that $f$ is also nilpotent is equivalent to require that $f$ is pointwise nilpotent and also the ascending filtration stabilizes: i.e. for a finite $n \in \mathbb N$ we have that $\ker f^n = \ker f^{n+k}$ for all $k \in \mathbb N$.

This also proves easily that for finitely generated modules $f$ is pointwise nilpotent iff $f$ is nilpotent, and so the equivalence between pointwise and global nilpotence holds for noetherian modules.

I'm not aware if there's a more useful or stronger characterization. Hope this helps.

Edit: since user asked for it here's an example of endomorphism of a module that is pointwise nilpotent but not nilpotent.

Consider the $\mathbb Z$ module $$M= \bigoplus_{n \in \mathbb N} \mathbb Z/\langle 2^n e_n \mid n \in \mathbb N\rangle$$ where the family $\langle e_n \rangle_{n \in \mathbb N}$ is the canonical basis of $\bigoplus_{n \in \mathbb N} \mathbb Z$, and the $\langle 2^n e_n \mid n \in \mathbb N\rangle$ is the submodule generated by the elements $2^ne_n$.

The module $M$ is generated by the images of the elements $e_n$ through the quotient projection, let's call $\bar e_n$ the image of $e_n$ through such projection.

There's the obvious homomorphism $f \colon M \to M$ given by $f(m)=2m$. For every $n \in \mathbb N$ we have that $f^n(m)=2^n m$ and so we have that for every $n \in \mathbb N$ and every $k \in \mathbb N$ such that $k \leq n$ the equality $$f^n(\bar e_k)= 2^n \bar e_k = 2^k \bar e_k=0$$ while for $k > n$ we have that $$f^n(\bar e_k)=2^n \bar e_k \ne 0$$ since $2^{n}e_k$ in not in the kernel of the canonical projection from $\bigoplus \mathbb Z$ to $M$.

Since for every $m \in M$ there must be an $n \in \mathbb N$ such that $m = a_1 \bar e_1 + \dots + a_n \bar e_n$ for some $a_1,\dots,a_n \in \mathbb Z$ it follows that $$f^n(m)=a_1 f^n(\bar e_1) + \dots + a_n f^n(\bar e_n) = 0$$ so $f$ is pointwise nilpotent, nonetheless for every $n \in \mathbb N$ we have seen that $f^n(\bar e_{n+1}) \ne 0$ so $f^n\ne 0$ for every $n \in \mathbb N$.

Answer 2

One condition that forces locally nilpotent endomorphisms to be nilpotent is finite generation of the underlying module. I can imagine that there are other conditions that force locally nilpotent endomorphisms to be nilpotent, but they will likely depend upon some further condition placed on the module or the base ring, or possibly the class of morphisms you allow in that category.

Here's another condition : if $M$ is a noetherian module, then locally nilpotent endomorphisms are nilpotent, as the kernels of $f^n$ form an increasing chain of submodules that eventually contains each and every element of $M$. Since that chain is stagnant by hypothesis, it has to be equal to $M$ after a finite amount of steps, that is to say : for some integer $m$, $\mathrm{Ker}(f^m)=M$.