For a homomorphism of commutative rings $f:R\to S$, there are at least two notions of a descent datum for this map. One of these is to be an $S$-module $M$, with an isomorphism $M\otimes_R S\cong S\otimes_R M$ satisfying the cocycle condition. This can be thought of as saying something about "agreeing on intersections" since it's demanding that the two ways of tensoring $M$ up to an $S\otimes_R S$ module, i.e. either along the left unit or the right unit $S\to S\otimes_R S$ are equivalent. This is basically, I think, the dual of saying that it agrees on projections.

Another common way of phrasing descent data is to say that $M$ is an $S$-module which is also an $S\otimes_R S$-comodule, where $S\otimes_R S$ is a coring with structure map $\Delta:S\otimes_R S\to S\otimes_R S\otimes_R S\cong S\otimes_R S\otimes_S S \otimes_R S$ using the unit map of $S$.

I have written down some vague things about how these two are the same, but is there a functorial equivalence between the two categories of descent data? How does one obtain a comodule structure from the isomorphism $M\otimes_R S\cong S\otimes_R M$? And vice versa? And if you're feeling pedagogical, you might even mention how these two notions are equivalent to the notion of being a coalgebra for a certain comonad!


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    $\begingroup$ They are indeed equivalent notions, and in fact, this generalises to any symmetric monoidal category. You can prove this by manipulating string diagrams. Here is a proof. $\endgroup$ – Zhen Lin Dec 2 '13 at 11:24
  • $\begingroup$ Dear @Zhen Lin, for future reference, what is the source of the pdf file in your link? Is it part of a book or some notes? $\endgroup$ – Ricardo Andrade Dec 3 '13 at 15:23
  • $\begingroup$ It's part of an essay I wrote about a year ago. $\endgroup$ – Zhen Lin Dec 3 '13 at 17:04

I find it easier to use geometric notation, so let $X = \mathrm{Spec}(S)$, $Y=\mathrm{Spec}(R)$, and $\phi: X \to Y$ be the morphism corresponding to $f$. We have adjoint functors $$ \phi_\ast : S\text{-}\mathrm{mod} \to R\text{-}\mathrm{mod}: \phi^\ast. $$

Consider the (beginning of the) Cech simplicial set correponding to the map $\phi$.

$$ \ldots X \times_Y X \rightrightarrows ^{\widehat{\phi}_1} _{\widehat{\phi}_2} X \stackrel{\phi}{\rightarrow} Y $$

(sorry for the bad latex). It might be helpful to have in mind the example where $X$ is an open cover of $Y$, so that the space $X\times _Y X$ is consists of double intersections.

First of all, note that the structure of being a comodule for the coring $S \otimes_R S$ is a map

$$M \stackrel{a}{\to} M \otimes _S (S \otimes _R S) \simeq M \otimes _R S \simeq \phi ^\ast \phi _\ast M$$

satisfying the usual conditions. You may notice that the functor $\phi^\ast \phi_\ast$ has the structure of a comonad, and indeed, being a comodule for $S \otimes_R S$ in the category of $S$-modules is the same thing as being a comodule (aka coalgebra) for the comonad $\phi^\ast \phi_\ast$.

By base change, we have $\phi ^\ast \phi_\ast M \simeq \widehat{\phi}_{1\ast} \widehat{\phi}_2 ^\ast M$, so by adjunction, a map $a$ as above corresponds to a gluing map

$$\widetilde{a}: \widehat{\phi}^\ast _1 M \to \widehat{\phi}^\ast_2 M$$

i.e. a map

$$ M\otimes_R S \to S \otimes_R M$$.

This is the basic mechanism of the correspondence between comodules for $S \otimes _R S$ and the more down-to-earth notion of descent data. One can check that the conditions for $a$ to define a comodule correspond to the cocycle condition for the gluing map $\widetilde{a}$.

If the map $\phi$ is actually nice enough for descent (e.g. faithfully flat), then descent is a concequence of the Barr-Beck theorem: an $S$-module $M$ which is a comodule for $\phi^\ast \phi_\ast$ descends to an $R$-module $N$.

As you are a homotopy theorist, I should note that if you want to do this with complexes of modules (or more fancy things), then you will need the entire Cech simplicial set $\mathcal C(\phi)_\bullet$. Then, the data of being a comodule for $\phi^\ast \phi_\ast$ will be equivalent to being a simplicial module on $\mathcal C(\phi)_\bullet$.

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    $\begingroup$ Wow, fantastic. This is super helpful. Thanks @Sam! $\endgroup$ – Jonathan Beardsley Dec 2 '13 at 18:26
  • $\begingroup$ Hey @Sam for the above equivalence, does one need faithful flatness? $\endgroup$ – Jonathan Beardsley Feb 11 '14 at 4:09
  • $\begingroup$ I find it tricky to verify that co-multiplication (given that the cocycle condition holds) is co-unital if and only if the flip morphism in the definition of the descent data is an isomorphism (which is not just verifying an identity). Do you have some explanations for this? $\endgroup$ – Yai0Phah Nov 5 '19 at 21:07

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