Let $R$ be a commutative ring with one and $S$ commutative faithfully flat $R$-algebra (that is there is a faithfully flat ring map let $\phi: R \to S$). Then the so called Amitsur complex $R \to S^{\otimes \bullet +1}$ is exact and determines a faithfully flat resolution of $R$.
Applying the multiplicative group scheme functor $\mathbb{G}_m$ to this complex and taking cohomology gives exactly the fppf Cech cohomology $H^i_{Cech, fppf}(S^{\otimes \bullet +1}, \mathbb{G}_m)$ with respect the fppf cover $Spec(S) \to Spec(R)$.
We want to compare the fppf Cech cohomology with fppf cohomology of $H^i_{fppf}(R, \mathbb{G}_m)$. In general there exist only a morphism
$$ H^i_{Cech, fppf}(S^{\otimes \bullet +1}, \mathbb{G}_m) \to H^i_{fppf}(R, \mathbb{G}_m) $$
since often Cech cannot resolve 'fine' enough. For which class of rings $R$ there exist a fppf resolution $R \to S \to ...$ of $R$ which induces an isomorphism between associated Cech and fppf cohomology? What about the same problem with etale cohomology instead of fppf? Where (if that's true) a complete proof can be found? For which rings do moreover the fppf cohomology $H^i_{fppf}(R, \mathbb{G}_m)$ and etale cohomology $H^i_{etale}(R, \mathbb{G}_m)$ coinside?
Note that this question is identical to the question I recently asked in mse.