Let $A$ be a commutative ring with unit and let $P$ be a projective $A$-module finitely generated. By definition, there exists an $A$-module $P'$ such that $P\oplus P'$ is free of finite rank $r$. If $f$ is an endomorphism of $P$, its trace $\operatorname{Tr}(f|P)$ is defined to be $\operatorname{Tr}(f\oplus 0|P\oplus P')$. We can show that this definition does not depend on the choice of $P'$.

Let $Q$ be a submodule of $P$ stable by $f$ which is projective and finitely generated. Assume that there is a positive integer $n$ such that $f^n(P)\subset Q$. If $P/Q$ would have been projective and finitely generated, I would have conclude that $\operatorname{Tr}(f|P)=\operatorname{Tr}(f|Q)+s$, where $s$ is a nilpotent element of $A$, as $f$ acts nilpotently on $P/Q$. But, does this still holds in general?

Any idea is welcome. Many thanks!

(EDIT: As Darij Grinberg pointed me out in comments, the trace of a nilpotent endomorphism of a projective finitely generated module is a sum of nilpotents elements of $A$. As $A$ is commutative, the trace is again nilpotent and not necessarily zero)

  • $\begingroup$ Do you have examples where $P/Q$ is not projective? $\endgroup$ Apr 18, 2019 at 19:44
  • $\begingroup$ Take the inclusion 2Z into Z as Z-modules, and f is multiplication by 2. Z/2Z is torsion, so not projective $\endgroup$
    – Stabilo
    Apr 18, 2019 at 19:51
  • $\begingroup$ Ah! But what about A = Z/4, P = A, Q = 0 and f = multiplication by 2? $\endgroup$ Apr 18, 2019 at 20:08
  • 1
    $\begingroup$ Actually, traces of nilpotent maps are nilpotent, not 0. So your conjecture should be modified accordingly. $\endgroup$ Apr 18, 2019 at 20:09
  • $\begingroup$ You're so right @darijgrinberg, I've changed the conjecture. Thank you very much. $\endgroup$
    – Stabilo
    Apr 18, 2019 at 21:42

1 Answer 1


This is true. It's more natural to prove a generalisation:

Lemma. Let $A$ be a commutative ring, let $F_\bullet$ be a bounded complex of finite projective $A$-modules (say in degrees $[a,b]$), and let $f_\bullet \colon F_\bullet \to F_\bullet$ be an endomorphism such that $f_\bullet^n$ is homotopic to $0$ for some $n \in \mathbf Z_{>0}$. Then $\operatorname{tr}(f_\bullet) := \sum_i (-1)^i\operatorname{tr}(f_i)$ is nilpotent.

The original question is the case $[a,b] = [0,1]$ and $d \colon F_1 \to F_0$ is injective. The advantage of the version above is that it's stable under base change (whereas injectivity of $d$ is not). We use homological indexing to avoid confusion when writing $f_\bullet^n$. I'm not sure if there is a fancy homological algebra explanation for this lemma, but here is a hands-on proof:

Proof of Lemma. When $A = k$ is a field, this follows by additivity of trace and nilpotence of $H_i(f)$ for all $i$. In general, note that the question is stable under base change: for any ring map $\phi \colon A \to B$, we have $$\operatorname{tr}\left(f_\bullet \underset A\otimes \mathbf 1_B\right) = \phi\big(\operatorname{tr}(f_\bullet)\big).$$ Applying this to $A \to \kappa(\mathfrak p)$ for any prime $\mathfrak p \subseteq A$, we conclude from the field case that $\operatorname{tr}(f_\bullet) \in \mathfrak p$. Applying this to all $\mathfrak p$, we see that $\operatorname{tr}(f_\bullet)$ is nilpotent. $\square$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.