Let $R \rightarrow R'$ be a faithfully flat ring map, let $M$ be an $R$-module, and let $M\_n$ be the base change of $M$ to the tensor product of $n + 1$ copies of $R'$ over $R$. One way to formulate the statement of faithfully flat descent is as the claim that the complex 
$$ 0 \rightarrow M \rightarrow M\_0 \rightarrow M\_1 \rightarrow \dotsc$$
is exact, where the maps are the Cech style alternating sums of pullbacks. This is a fairly common way that is discussed further here <https://stacks.math.columbia.edu/tag/023M>.


Another, probably less common, way to formulate faithfully flat descent is to say that 
$$ M \xrightarrow{\sim} Rlim\_{\Delta} M\_n$$
where $\Delta$ is the usual simplex category that is used to define simplicial objects. 


**My question is**: why are these two formulations equivalent? How does one see that one implies the other?


I suspect that one probably uses the Dold-Kan equivalence and then some "standard" but possibly difficult to locate in the literature simplicial calculus. I am guessing that the core of my question is not really about descent for modules, but rather about simplicial or bisimplicial objects and Rlim. I would very much appreciate if someone could explain the equivalence between the two points of view in detail (or give a precise reference to the literature where this is treated).



Faithfully flat descent for modules from the simplicial point of view
Essentially, your question is: for any co-simplicial abelian group $M\_{\bullet}$, why is ${\rm R lim}\_{\Delta} M\_{\bullet}$ given by the complex $$C:= M\_0 \to M\_1 \to M\_2 \to \dots ?$$ (If $A$ is an abelian group $A \to C$ is a quasi-isomorphism if and only if $M \to M\_0 \to M\_1 \to M\_2 \to \dots $ is acyclic).


There is a direct lowbrow proof, given by resolving the constant cosimplicial abelian group $\mathbb Z \*$ given by $\mathbb Z \*\_n = \mathbb Z$ and all maps the identity. Since we have a natural isomorphism $${\rm Hom}\_{\Delta}(\mathbb Z \*, M\_{\bullet}) \cong {\rm lim}\_{\Delta} M\_{\bullet}$$ there is a quasi-isomorphism ${\rm RHom}\_{\Delta}(\mathbb Z \*, M\_{\bullet}) \simeq {\rm R lim}\_{\Delta} M\_{\bullet}$. To compute it, we resolve $\mathbb Z \*$ by free (or if you prefer, representable) co-simplicial abelian groups. 


The resolution looks like $$\mathbb Z \Delta(0,-) \leftarrow \mathbb Z \Delta (1, -) \leftarrow \mathbb Z \Delta(2, -) \leftarrow \mathbb Z \Delta(3,-) \leftarrow \dots $$


Here, the number $n$ represents the totally ordered set $\{0, \dots, n\}$ and the cosimplicial abelian group $\mathbb Z \Delta(n, -)$ is given by $\mathbb Z \Delta(n, -)\_m = \mathbb Z \Delta(n, m)$. The differential $\mathbb Z \Delta(n, -) \to \mathbb Z \Delta(n-1, -)$ is given by the alternating sum of precomposition by the $n+1$ face maps in $\Delta(n-1, n)$. 


The complex resolves $\mathbb Z \*$, because if we look at level $n$, we see that we get exactly the simplicial homology complex for the $n$-simplex! $$\mathbb Z \Delta(0,n) \leftarrow \mathbb Z \Delta(1,n) \leftarrow \dots$$ Since we know the homology of the simplex, we know that the complex is a resolution. 


Now by the Yoneda lemma, we have that ${\rm Hom}\_{\Delta}( \mathbb Z \Delta(n,-), M) = M\_n$. In particular, this resolution is projective. Using it to compute ${\rm RHom}(\mathbb Z \*, M)$ we get the complex $${\rm Hom}\_{\Delta}( \mathbb Z \Delta(0,-), M) \to {\rm Hom}\_{\Delta}( \mathbb Z \Delta(1,-), M) \to {\rm Hom}\_{\Delta}( \mathbb Z \Delta(2,-), M) \to \dots, $$ and applying Yoneda, this is exactly the complex $C$ we wanted.


