I've been trying to understand the Adams spectral sequence, and I've gotten myself confused about how derived descent is supposed to work, so I would like to understand a simple example.


Given a faithfully flat map of commutative rings $A \to B$, the usual rules of faithfully flat descent let us identify $A$-modules as $B$-modules equipped with descent data (one nice presentation of this is that a $B$-module with descent data is a comodule over the $B$-coalgebra $B \otimes\_A B$). If I understand correctly, derived descent is supposed to allow us to generalize this to more general maps of rings, and so in particular I'd like to understand derived descent along the morphism $\mathbb{Z} \to \mathbb{F}\_2$.


I've made a bit of progress, working in the derived category of abelian groups. I believe a cofibrant replacement for $\mathbb{F}\_2$ is
$$ \widetilde{\mathbb{F}}\_2 = \mathbb{Z}[\eta; d \eta = 2], $$
and I can compute the coalgebra structure on
$$ \widetilde{\mathbb{F}}\_2 \otimes\_\mathbb{Z} \widetilde{\mathbb{F}}\_2 = \mathbb{Z} [\eta\_1, \eta\_2; d \eta\_i = 2]. $$
But now I get stuck. Somehow, a (cofibrant?) comodule over $\widetilde{\mathbb{F}}\_2 \otimes\_\mathbb{Z} \widetilde{\mathbb{F}}\_2$ is supposed to be the same thing as a 2-completed abelian group, but I don't even see how we would distinguish, say, $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Z}/4\mathbb{Z}$ from different comodule structures on $\widetilde{\mathbb{F}}\_2$. Maybe the answer is in a choice of coassociator for the comodule structure?


In any case, I would greatly appreciate help doing this calculation, or advice on a different way to think about derived descent along $\mathbb{Z} \to \mathbb{F}\_2$ that makes manifest the equivalence between 2-completed abelian groups and $\mathbb{F}\_2$-modules with derived descent data.



Basic example of derived descent
$\newcommand{\Z}{\mathbf{Z}}\newcommand{\FF}{\mathbf{F}}\newcommand{\H}{\mathrm{H}}$Let me do the universal case of your situation, which is understanding descent along the map $\Z[t] \to \Z$ sending $t\mapsto 0$. [To get your case, note that $\FF\_2 = \Z \otimes\_{\Z[t]} \Z$, where one of the maps $\Z[t] \to \Z$ sends $t\mapsto 0$, and the other map sends $t\mapsto 2$. In other words, replace $t$ everywhere below by $2$.]
How does one do descent along $\Z[t] \to \Z$? In other words, suppose I have a (derivedly) $t$-complete $\Z[t]$-module $M$; how do I encode in terms of a $\Z$-module and descent data?


Let $R\_1$ denote the complex $\Z \otimes\_{\Z[t]}^L \Z$, so that $\H\_\ast(R)$ is given by $\Z[\epsilon]/\epsilon^2$, where $\epsilon$ lives in $\H\_1$. If $M$ is a $\Z[t]$-module, then the associated $\Z$-module is the complex $M \otimes\_{\Z[t]}^L \Z$. I'll write this as $M\_0$.


What is the $R\_1 = \Z \otimes\_{\Z[t]}^L \Z$-comodule structure on $M\_0$? This is a map $M\_0 \to M\_0 \otimes\_\Z R\_1$, which on homology gives a map $M\_0 \to M\_0 \otimes\_\Z \Z[\epsilon]/\epsilon^2$. Projecting onto the second factor of $\Z[\epsilon]/\epsilon^2 \cong \Z \oplus \Z \cdot \epsilon$, we obtain a map $\delta\_1: M\_0 \to M\_0 [1]$. (The shift by $1$ is because $\epsilon$ lives in homological degree $1$.) Explicitly, $\delta\_1$ is the composite
$$M\_0 \to M\_0 \otimes\_\Z \Z[\epsilon]/\epsilon^2 \to M\_0 [1].$$


What is $\delta\_1$? Well, there is a cofiber sequence/distinguished triangle in chain complexes of $\Z[t]$-modules given by
$$\Z \cong (t)/(t^2) \to \Z[t]/t^2 \to \Z[t]/t = \Z.$$
Rotating this gives a cofiber sequence
$$\Z[t]/t^2 \to \Z \to \Z[1].$$
Applying $M \otimes\_{\Z[t]}^L -$, we get a cofiber sequence
$$M \otimes\_{\Z[t]}^L \Z[t]/t^2 \to M \otimes\_{\Z[t]}^L \Z = M\_0 \to M \otimes\_{\Z[t]}^L \Z[1] = M\_0 [1].$$
This second map $M\_0 \to M\_0[1]$ is precisely $\delta\_1$. If you want, this map $\Z \to \Z[1]$ is the "universal example" of a $\delta\_1$.


Summarizing, the $R\_1$-comodule structure on $M\_0$ encodes $\delta\_1$, and taking the fiber of $\delta\_1$ produces $M \otimes\_{\Z[t]}^L \Z[t]/t^2$. In other words, knowing the $R\_1$-comodule structure is the same as knowing $M \otimes\_{\Z[t]}^L \Z[t]/t^2$.


In the derived setting, one has not just the $R\_1$-comodule structure, but a whole bunch of other structure coming from $R\_n = \Z^{\otimes\_{\Z[t]}^L n+1}$. Note that $R\_n = R\_{n-1} \otimes\_{\Z[t]}^L \Z$. When one keeps track of the structure arising from $R\_n$, the same argument as above shows that one has a new $\delta$ showing up. This new $\delta$ will correspond to the cofiber sequence
$$\Z[t]/t^{n-1} \cong (t)/(t^{n-1}) \to \Z[t]/t^n \to \Z[t]/t = \Z,$$
which rotates to a cofiber sequence
$$\Z[t]/t^n \to \Z \to \Z[t]/t^{n-1} [1].$$
If you want, this map $\Z \to \Z[t]/t^{n-1} [1]$ is the "universal example" of a $\delta\_{n-1}$.


If one knows all this structure in a coherent way (which is exactly what the derived comodule structure is encoding), then the above discussion shows that you can recover $M \otimes\_{\Z[t]}^L \Z[t]/t^n$ for every $n\geq 1$. If $M$ is derived $t$-complete, then knowing $M \otimes\_{\Z[t]}^L \Z[t]/t^n$ for every $n\geq 1$ is the same as knowing $M$ itself. In other words, the derived comodule structure in all its coherent glory is the descent data for $\Z[t] \to \Z$. (One could think of the $\delta\_n$'s as being the differentials in the associated graded spectral sequence for the $t$-adic filtration on $M$, i.e., the $t$-Bockstein spectral sequence. This whole story can then be viewed as a version of Koszul duality between $\Z[t]^\wedge\_t = \Z[\![t]\!]$ and $\Z \otimes\_{\Z[t]}^L \Z = \Z[\epsilon]/\epsilon^2$.)


