Exact principal-open Cech resolution

Write $C^{\bullet }$ for the augmented alternating Cech complex of the statement: $C^{-1}=M$, $C^p={⨁}_{i_0<\cdots <i_p}{M}_{f_{i_0}\cdots f_{i_p}}$ for $p\ge 0$, with the alternating differentials ${\delta }^p$ and $\delta \circ \delta =0$ [F1]; the augmentation is $M\to C^0$, $m\mapsto (m\mathrm{/}1{)}_i$.

Let $\mathfrak{p}$ be a prime ideal of $A$. Since ${\sum }_i{a}_i{f}_i=1\mathrm{\notin }\mathfrak{p}$ we have $f_i\mathrm{\notin }\mathfrak{p}$ for at least one $i$; fix such an index $i$ for this $\mathfrak{p}$.

Localising the complex $C^{\bullet }$ at $\mathfrak{p}$ and using that localisation commutes with finite direct sums [F3] gives a complex $C_{\mathfrak{p}}^{\bullet }$ whose degree-$p$ term is ${⨁}_{i_0<\cdots <i_p}{M}_{f_{i_0}\cdots f_{i_p},\mathfrak{p}}$, the localisation of each principal localisation at $\mathfrak{p}$; by [F4] any factor $f_{i_j}$ with $i_j=i$ may be dropped, and [F5] shows that the remaining terms are computed in the ring $A_{\mathfrak{p}}$ in which $f_i$ is a unit.

Extend each cochain's components from increasing tuples to arbitrary tuples by alternating signs under permutations, and set the component to zero when an index is repeated. These are the same cochains, expressed with redundant indices; all restrictions to intersection localisations respect the sign rule. For $s\in C_{\mathfrak{p}}^q$ with $q\ge 0$, define $h(s{)}_{i_0\cdots i_{q-1}}=s_{i,i_0,\dots ,i_{q-1}}$, using this alternating convention, and for $q=0$ define $h(s)=s_i\in M_{\mathfrak{p}}=C_{\mathfrak{p}}^{-1}$. Since $f_i$ is a unit in $A_{\mathfrak{p}}$, [F4] identifies the component on the intersection with $D(f_i)$ with the required target component, making $h$ well defined even when $i$ lies among the other indices. For an increasing tuple $J=(i_0,\dots ,i_p)$ and $p\ge 0$, the alternating differential gives $((\delta h+h\delta )s{)}_J={\sum }_{j=0}^p(-1{)}^j{s}_{i,J\setminus i_j}+(\delta s{)}_{i,J}={\sum }_{j=0}^p(-1{)}^j{s}_{i,J\setminus i_j}+s_J+{\sum }_{j=0}^p(-1{)}^{j+1}{s}_{i,J\setminus i_j}=s_J$. In degree $-1$, $h\delta (m)$ is the $i$-component of the augmentation of $m$, hence equals $m$. Thus $\delta h+h\delta =1$ in every degree, the localised augmented complex is contractible, and it is exact.

Since for every prime ideal $\mathfrak{p}$ the localisation $C_{\mathfrak{p}}^{\bullet }$ is exact, the local criterion [F2] applied to each consecutive pair of differentials of the complex $C^{\bullet }$ shows that $C^{\bullet }$ is exact at every term, that is, the augmented Cech complex of the statement is exact.

Now fix $g\in A$ and localise at $g$. The images of $f_1,\dots ,f_r$ in $A_g$ still generate the unit ideal (apply the localisation ring map to $1={\sum }_i{a}_i{f}_i$), and by [F3] and [F4] the localisation at $g$ of each $M_{f_{i_0}\cdots f_{i_p}}$ is canonically the module $(M_g{)}_{f_{i_0}\cdots f_{i_p}}$ computed in $A_g$; hence the $g$-localisation of the displayed complex is the corresponding augmented Cech complex of $M_g$ with respect to those generators, and step 4.1 applied in the ring $A_g$ proves it is exact.

The Axiom of Choice is used exactly through the local criterion [F2]; the index $i$ is chosen for one fixed prime $\mathfrak{p}$ at a time in the pointwise argument of step 1.2, so no simultaneous selection over primes is made, and no other step uses a choice principle.