TITLE: Differences between nilpotent and pointwise nilpotent endomorphisms.
QUESTION [5 upvotes]: 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?

REPLY [4 votes]: 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 $n \in \mathbb N$.