Disclaimer : I asked this question on Maths.StackExchange 20 days ago (and started a bounty) here but got no answer, so I'm asking it here now (with no modification).

$\newcommand{\l}{\ell} \newcommand{\Z}{\mathbb Z}$ I'm trying to understand a remark in Weil's conjectures for function fields I, by Gaitsgory-Lurie.

Specifically it's remark, which essentially states the following : an object $M$ in $\mathrm{Mod}_{\mathbb Z_\l}$ is perfect if and only if it's $\l$-complete and $M\otimes_{\Z_\l}\Z/\l$ is perfect in $\mathrm{Mod}_{\Z/\l}$ (this latter condition implies that this is also true for $M\otimes_{\Z_\l}\Z/\l^d$, for any $d\geq 0$)

They state it without proof, and it looks like there is no proof elsewhere in the book (as far as I can see, although I haven't checked the whole book).

What is clear to me is the forward direction : indeed $\Z_\l\otimes_{\Z_\l}\Z/\l$ is perfect over $\Z/\l$ and $\Z_\l$ is $\l$-complete, and these conditions are closed under retracts, so the sub-$\infty$-category of those $M$'s that satisfy the latter conditions is a stable subcategory, contains $\Z_\l$ and is closed under retracts so it contains all perfect $\Z_\l$-modules.

The reverse direction, however, does not seem so clear. I think one ought to use the fact that the $\infty$-category of $\l$-complete modules is equivalent to $\underset{d}{\varprojlim} \mathrm{Mod}_{\Z/\l^d}$ but that doesn't seem to be enough since we want $M$ to be compact in the whole $\mathrm{Mod}_{\Z_\l}$, not just in $\l$-complete modules.


Over a PID, you can check perfectness on homology. So the claim is that the homology of an $\ell$-complete $\mathbb{Z}_\ell$-module is finitely generated if and only if it is so mod $\ell$. The homology groups are $\ell$-complete, and so this follows from the long exact sequence of mod $\ell$-reduction.

  • $\begingroup$ Thank you for your answer, but I'm not sure I understand the end. What do you call the long exact sequence of mod $\ell$-reduction ? $\endgroup$ – Maxime Ramzi Jul 11 '20 at 9:54
  • $\begingroup$ Well, for any complex $X$, you have a cofiber sequence $X\xrightarrow{\ell} X\to X/\ell$, and this induces a long exact sequence in homology. $\endgroup$ – Achim Krause Jul 11 '20 at 14:09
  • $\begingroup$ This shows that the mod $\ell$-reduction of homology, $H_*(X)/\ell$, maps injectively into $H_*(X/\ell)$, so if the latter is finitely generated, so is the former. $\endgroup$ – Achim Krause Jul 11 '20 at 14:14
  • $\begingroup$ Right that makes sense - but it's not clear to me how to go from there to finite generation of $H_*(X)$. It seems like this reduces it to the same question about discrete $\mathbb Z_\ell$-modules, but I don't know how to prove that either $\endgroup$ – Maxime Ramzi Jul 12 '20 at 12:15
  • $\begingroup$ That's a completely elementary exercise, you just check that a map between discrete $\ell$-complete $\mathbb{Z}_\ell$-modules is surjective if it is surjective mod $\ell$. $\endgroup$ – Achim Krause Jul 13 '20 at 8:01

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