# Property of the trace on finitely generated projective modules

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)

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

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. $$\square$$