The following lemma in commutative algebra is important for the foundations of algebraic geometry:
If $A$ is a commutative ring, $U \subseteq A$ is a finite subset generating the unit ideal, then for every $A$-module $M$ the sequence $M \to \prod_{u \in U} M_u \rightrightarrows \prod_{u,v \in U} M_{u \cdot v}$ is an equalizer.
Namely it implies $M \cong \Gamma(\mathrm{Spec}(A),\widetilde{M})$. The usual proofs I know of are fiddly computations with fractions (e.g. Theorem 1.3.7 in EGA I, or Theorem 2.33 in Görtz-Wedhorn). I have found two more conceptual proofs on my own (which actually work in a more general setting), which probably are not new at all. Therefore I would like to ask if they appear in the literature. Any comments are appreciated.
Both proofs use that $M_u$ is the colimit of the sequence $M \xrightarrow{u} M \xrightarrow{u} \dotsc$.
A. First observe that if $M_w=0$ for all $w \in U$, then $M=0$ (if $m \in M$, then $w^k m = 0$ for some $k$, hence $m=0$ since the $w^k$ generate the unit ideal). By exactness of localizations, it suffices to prove that each $M_w \to \prod_{u \in U} M_{uw} \rightrightarrows \prod_{u,v \in U} M_{uvw}$ is an equalizer. But in fact it is a split equalizer via the maps $\mathrm{pr}_w : \prod_{u \in U} M_{uw} \to M_w$ and $(\mathrm{pr}_{u,w})_u : \prod_{u,v} M_{uvw} \to \prod_{u} M_{uw}$. $\square$
B. Consider the category $I$ with objects $U \sqcup U^2$ and morphisms $u \to (u,v) \leftarrow v$. Define a diagram $D : I \times \mathbb{N} \to \mathsf{Mod}(A)$ as follows: It is constant $M$ on objects, and on morphisms $D(u,k \to k+1)=(u : M \to M)$, $D(u \to (u,v),k) = (v^k : M \to M)$ (similarily with $v \to (u,v)$). Now we use that finite limits commute with filtered colimits (Mac Lane, IX.2). $\mathrm{lim}_{i \in I} \mathrm{colim}_{k \in \mathbb{N}} D(u,v)$ is the equalizer of the two maps $\prod_{u \in U} M_u \rightrightarrows \prod_{u,v \in U} M_{uv}$, whereas $\mathrm{colim}_{k \in \mathbb{N}} \mathrm{lim}_{i \in I} D(u,v)$ is the colimit of the modules $M[k] := \{a \in M^U : u^k a(v) = v^k a(u)\}$. The diagonal is an isomorphism $M \cong M[0]$ and the composite $M \to M[k]$ is given by $m \mapsto (u^k m)_u$, which is injective, but also surjective: If $a \in M[k]$, choose $\lambda \in A^U$ with $\sum_u \lambda(u) u^k=1$, then $a(v)=\sum_u \lambda(u) u^k a(v) = \sum_u \lambda(u) v^k a(u)$ shows that $\sum_u \lambda(u) a(u)$ is a preimage. Hence the colimit is just $M$. $\square$
