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? $\endgroup$ – darij grinberg Apr 18 '19 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 '19 at 19:51
  • $\begingroup$ Ah! But what about A = Z/4, P = A, Q = 0 and f = multiplication by 2? $\endgroup$ – darij grinberg Apr 18 '19 at 20:08
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    $\begingroup$ Actually, traces of nilpotent maps are nilpotent, not 0. So your conjecture should be modified accordingly. $\endgroup$ – darij grinberg Apr 18 '19 at 20:09
  • $\begingroup$ You're so right @darijgrinberg, I've changed the conjecture. Thank you very much. $\endgroup$ – Stabilo Apr 18 '19 at 21:42

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. $\square$

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