# Perfect $\mathbb Z_\ell$-modules

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).


Specifically it's remark 2.3.4.3., 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.
• 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 ? – Maxime Ramzi Jul 11 '20 at 9:54
• 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. – Achim Krause Jul 11 '20 at 14:09
• 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. – Achim Krause Jul 11 '20 at 14:14
• 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 – Maxime Ramzi Jul 12 '20 at 12:15
• 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 $\ell$. – Achim Krause Jul 13 '20 at 8:01