I am currently learning for an exam in commutative algebra due in a few weeks. After a year of introductory algebra courses it was the only possible lecture I could attend to extend my knowledge in Algebra. This is to say: I don't have any background in algebraic geometry and would prefer to leave it aside, since I only have quite limited time left.
One of the topics treated was Grothendiek descent on modules framed like:
Given a faithfully flat algebra $A$ over a commutative, unital ring $R$, under which conditions is an $A$-module $N$ of the form $A \otimes_R M$ where $M$ is an $R$-module?
We answered it by Grothendiek's theorem saying that there is an equivalence of categories between $R$-$Mod$ and $A$-$Mod$-$DD$ where the latter is the category of descent dates of $A|R$. For us a descent date is just a tuple $(N, \phi)$ where $N$ is an $A$-module and $\phi: A \otimes_R N \longrightarrow N \otimes_R A$ is an $A \otimes_R A$-module-isomorphism satisfying the cocycle condition $$(\phi \otimes 1_A)\circ(1_A \otimes \phi) \overset{!}{=} (\tau \otimes 1_A) \circ (1_A \otimes \phi) \circ (\tau \otimes 1_N): A \otimes_R A \otimes_R N \rightarrow N \otimes_R A \otimes_R A$$ as $A \otimes_R A \otimes_R A$-module homomorphims where $\tau: X \otimes_R Y \rightarrow Y \otimes_R X: x \otimes y \mapsto y \otimes x$ on the corresponding modules.
Now (finally!) to my question:
Is there a purely algebraic reason for the descent data to be defined like they are?
So far I understand that any $R$-module $M$ gives rise to the exact sequence
$$0 \longrightarrow M \overset{\partial^0}{\longrightarrow} A \otimes_R M \overset{\partial^1}{\longrightarrow} A \otimes_R A \otimes_R M \overset{\partial^2}{\longrightarrow} A \otimes_R A \otimes_R A \otimes_R M$$
defined via
$\partial^0(m) = 1 \otimes m$
$\partial^1(a \otimes m) = 1 \otimes a \otimes m - a \otimes 1 \otimes m$
$\partial^2(a \otimes b \otimes m) = 1 \otimes a \otimes b \otimes m - a \otimes 1 \otimes b \otimes m + a \otimes b \otimes 1 \otimes m$
called Amitsur's Complex. It might be motivated as follows:
Given $M$ and its scalar extension $A \otimes_R M$ there are two canonical $A$-Module structures on $A \otimes_R A \otimes_R M$, that is multiplication into the first and multiplication into the second $A$. These structures seem to overlap exactly on the $R$-submodule $1 \otimes_R 1 \otimes_R M$, so we can identify $M$ with the $R$-kernel of the difference of both structures: $\partial^1$. The motivation of $\partial^2$, besides making the sequence one exact step longer, is yet to be clarified.
Clearly we can recover $M$ from the complex via $M = \operatorname{im} \partial^0 = \ker \partial^1$. So we would like to define this complex for an arbitrary $A$-module $N$. Now the problem is that the definition above heavyly relies on the structure of $N$ being of the form $A \otimes_R M$. We must find another way of defining it via morphisms we can define without knowing of this structure. Therefore we analyse the case of $N$ being $M_A = A \otimes_R M$:
There is the $A \otimes_R A$-linear braiding isomorphism $$\phi: A \otimes_R M_A \rightarrow M_A \otimes_R A, \;\; a \otimes (b \otimes m) \mapsto (a \otimes m) \otimes b$$ With $\iota^1: M_A \rightarrow A \otimes M_A, \;\; a \otimes m \mapsto 1 \otimes (a \otimes m)$ we can write $$\partial^1 = (1_{A \otimes M_A} - \tau\phi)\iota^1$$
Given $\iota^2: A \otimes_R M_A \rightarrow A \otimes_R A \otimes_R M_A, \;\, a \otimes (b \otimes m) \mapsto 1 \otimes a \otimes b \otimes m$, we can redefine $\partial^2$ in two different ways though:
$$\partial^2 = (1_{A \otimes A \otimes M_A} - \tau \otimes 1_{M_A} + (1_A \otimes \tau)(1_A \otimes \phi)(\tau \otimes 1_{M_A}))\iota^2$$
and
$$\partial^2 = (1_{A \otimes A \otimes M_A} - \tau \otimes 1_{M_A} + (1_A \otimes \tau)(\tau \otimes 1_A)(\phi \otimes 1_A)(1_A \otimes \phi))\iota^2$$
By construction they match for $\phi$ being the braiding isomorphism. Moreover the two representations of $\partial^2$ match if and only if they match in the third part of the sum. By elementary calculations we see that this is equivalent to $\phi$ satisfying the cocycle condition mentioned above, where we could omit the $\iota^2$ due to the $A\otimes_R A \otimes_R A$-linearity.
Now we have seen all of the descent data (which seem to be needed) naturally arising from Amitsur's complex if we are in the desirable case of our $A$-module $N$ being of the form $M_A = A \otimes_R M$.
Having a descent datum $(N,\phi)$ we can imitate Amitsur's complex with it by replacing the braiding isomorphism with the given $\phi$. We denote its morphisms by $\delta^i$ instead of $\partial^i$ to highlight this replacement. We then can define the $R$-module corresponding to $N$ via $N_R := \ker \delta^1$.
I thus can reformulate my question(s) in a more precise manner:
- Why are the two representations of $\partial^2$ via $\phi$ and $\tau$ the only ones possible?
I'd bet there is a combinatorical reason or one might check all possibilities by computation, so I don't think it is that important. - Why is $\partial^2$ important in the first place?
We could simply define the $R$-kernel of $\delta^1$ as soon as we have $\delta^1$. So I guess it has something to do with the requirement of $A \otimes_R N_R \cong N$ as $A$-Modules, but I didn't find a way to relate it yet. - Given a descent datum $(N,\phi)$, is the imitated Amitsur complex always exact? If so, is it important?
I gloriously failed to show its exactness, so I'd really like to know if its due to my incompetence or due to it simply being false... - Given a descent datum $(N,\phi)$ and $N_R = \ker \delta^1$. Why does the proof of Grothendiek's theorem compare $$0 \longrightarrow A \otimes_R N_R \overset{1 \otimes i}{\longrightarrow} A \otimes_R N \overset{1 \otimes \delta^1}{\longrightarrow} A \otimes_R A \otimes_R N$$ (ie. the imitated Amitsur complex tensored with $A$) and $$0 \longrightarrow N \overset{\partial^0}{\longrightarrow} A \otimes_R N \overset{\partial^1}{\longrightarrow} A \otimes_R A \otimes_R N$$ (ie. the Amitsur complex of $N$ considered as $R$-module)
instead of comparing $$0 \longrightarrow N_R \overset{\partial^0}{\longrightarrow} A \otimes_R N_R \overset{\partial^1}{\longrightarrow} A \otimes_R (A \otimes_R N_R) \overset{\partial^2}{\longrightarrow} A \otimes_R A \otimes_R (A \otimes_R N_R)$$ (ie. the Amitsur complex of $N_R$) to $$0 \longrightarrow N_R \overset{\ker \delta^1}{\longrightarrow} N \overset{\delta^1}{\longrightarrow} A \otimes_R N \overset{\delta^2}{\longrightarrow} A \otimes_R A \otimes_R N$$ (ie. the imitated Amitsur complex)?
I must admit that my questions (especially the last one) are quite vague, and am absolutely grateful if anyone read up to this point. It is clear to me that one could define the descent data as they are just because they work, but it's clear that this is not really a satisfying answer to the question why they are defined in this way. Everyone I asked for intuition referred me to algebraic geometry but quickly denied my request for explaining it thoroughly since it seems to be a rather deep result. I found the only source explaining descent from an algebraic point of view in the lecture notes of Marco Garuti found here, but I got lost in notation and it does not seem to tackle the questions I asked.
I hope anybody can help me out. Big thanks to everyone, who read this question to the end...