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?

  • $\begingroup$ I don't understand how to make sense of the definition. We can't iterate $f$ $n$ times since the domain is not equal to the range, and we also can't take an $n$th power since there is no ring structure on $N$. $\endgroup$
    – hunter
    Nov 22, 2013 at 13:52
  • $\begingroup$ Sorry, I make a stupid mistake, clearly it is $M$ and not $N$ $\endgroup$ Nov 22, 2013 at 13:53
  • $\begingroup$ A condition like "$M$ is a finitely generated $R$-module" will ensure that locally nilpotent implies nilpotent. I would be surprised but very much interested if there was any more to this. $\endgroup$ Nov 22, 2013 at 15:10

2 Answers 2


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 $n \in \mathbb N$.

  • $\begingroup$ @user As you wished here's the counterexample :) $\endgroup$ Nov 22, 2013 at 16:28
  • 1
    $\begingroup$ +1 for your answer. I have another proposal as an example of a locally (pointwise) nilpotent endomorphism which is not nilpotent: take $V$ a $K$-vector space with a countable basis $(e_n)_{n\ge 1}$ and $f:V\to V$ defined by $f(e_1)=0$ and $f(e_i)=e_{i-1}$ for $i\ge 2$. $\endgroup$
    – user89712
    Nov 23, 2013 at 20:53
  • $\begingroup$ Why "for finitely generated modules $f$ is pointwise nilpotent iff $f$ is nilpotent" ? I understand if $M$ is noetherian it holds. But finitely generated doesn't imply noetherian. $\endgroup$
    – user682141
    Nov 23, 2019 at 11:29
  • $\begingroup$ @user682141 That's because of the characterization of pointwise nilpotence, i.e. $M=\bigcup_n \ker f^n$. If $M$ is finitely generated then there must be a $n$ such that the finite set of generators are contained in $\ker f^n$, hence the sequence stabilizes, this proves that $f$ is nilpotent. Feel free to ask for additional details. $\endgroup$ Nov 26, 2019 at 17:10

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 $m$, $\mathrm{Ker}(f^m)=M$.

  • $\begingroup$ Though you say "another", noetherian implies finitely generated. $\endgroup$
    – user682141
    Nov 23, 2019 at 11:30

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