Perfect $\mathbb Z_\ell$-modules $\newcommand{\l}{\ell} \newcommand{\Z}{\mathbb Z}$
I'm trying to understand a remark in Weil's conjectures for function fields I, by Gaitsgory-Lurie.
Specifically it's remark 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.
 A: $\newcommand{\Z}{\mathbb Z_\ell}$
This question was answered here. Let me write a more detailed answer here (taking the comments into account).
Suppose $C$ is a discrete $\Z$-module which is $\ell$-complete (in $\mathcal D(\Z)$, so one might say "derived $\ell$-complete"), and such that $C\otimes \mathbb Z/\ell$ is perfect as a $\mathbb Z/\ell$-module.
In particular, $H_0(C\otimes \mathbb Z/\ell)$ is finitely generated, and therefore so is $C/\ell$. So we can find a map $\Z^n\to C$ which induces a surjection mod $\ell$, for some finite $n$.
The point is now :

A map $M\to N$ between (derived) $\ell$-complete discrete $\Z$-modules which is a surjection mod $\ell$ is a surjection.

This is very elementary if you consider discrete $\ell$-completion.
Note that $M\overset f\to N$ is surjective if and only if its fiber is connective, so let $K\to M\to N$ be the associated fiber sequence (note that $K$ is $\ell$-complete, as the $\infty$-category of $\ell$-complete modules is closed under limits). Then $K\otimes \mathbb Z/\ell \to M\otimes \mathbb Z/\ell \to N\otimes \mathbb Z/\ell$ is also a fiber sequence, and because of our assumption, $K\otimes \mathbb Z/\ell$ is connective.
But now let $K\to K\to K\otimes \mathbb Z/\ell$ be the obvious fiber sequence, and take its associated long exact sequence of homotopy groups : we get that $\ell : \pi_n(K)\to \pi_n(K)$ is surjective for all $n<0$.
$\pi_n(K)$ is also $\ell$-complete, and this implies that it is $0$ (indeed, if $X$ is a (derived) $\ell$-complete discrete $\Z$-module, $\hom(\Z[ 1/\ell ], X) = 0$).
Therefore $K$ is connective, and so $M\to N$ is indeed surjective.
Therefore $\Z^n\to C$ was surjective, so $C$ is finitely generated, and in particular it is a perfect $\Z$-module.
Now move on to the general case : assume $C\in \mathcal D(\Z)$ is $\ell$-complete and perfect mod $\ell$. Then because $C\simeq \bigoplus_n H_n(C)[n]$ ($\Z$ is a PID), the same is true for $H_n(C)$ for all $n$, in particular all $H_n(C)$ are finitely generated.
Finally, only finitely many $H_n(C)$ are nonzero because they are $\ell$-complete and thus can't be killed by $\otimes \mathbb Z/\ell$. This concludes the proof.
