Finite presentation data descend to Noetherian algebra and module stages

Choose presentations $S\cong R[X_1,\dots ,X_n]\mathrm{/}(f_1,\dots ,f_u)$ and $S^a\stackrel{D}{\to }{S}^b\to M\to 0$ as in [F1]. The coefficients of the $f_j$ and of representatives in $R[X_1,\dots ,X_n]$ for the finitely many entries of $D$ form a finite subset $E\subseteq R$. Let $\mathrm{\Lambda }$ be the directed set of finite subsets $i\subseteq R$ containing $E$, ordered by inclusion, and put $R_i=\mathbb{Z}[i]\subseteq R$. The images of the same polynomial equations and matrix coefficients over $R_i$ define $S_i=R_i[X_1,\dots ,X_n]\mathrm{/}(f_{1,i},\dots ,f_{u,i})$ and $M_i=\mathrm{coker}(S_i^a\stackrel{D_i}{\to }{S}_i^b)$.

For $i\subseteq j$, the equations and matrix entries at stage $j$ are the images of those at stage $i$. By [F3], tensoring their finite presentations gives canonical isomorphisms $S_i{\otimes }_{R_i}{R}_j\cong S_j$ and $M_i{\otimes }_{S_i}{S}_j\cong M_j$. Every element of $R$ lies in some $R_i$, so ; applying the same finite presentations gives and . Each $R_i$ is finitely generated over $\mathbb{Z}$, and [F2] makes $R_i$ and $S_i$ Noetherian. This proves clauses 1 and 2.

The contractions ${\mathfrak{p}}_i$ and ${\mathfrak{q}}_i$ are compatible primes, with ${\mathfrak{q}}_i\cap R_i={\mathfrak{p}}_i$. Localize the system at their complements. Any numerator of $R_{\mathfrak{p}}$ or $S_{\mathfrak{q}}$ occurs at some stage, and any denominator outside the selected prime already occurs outside its contracted stage prime; therefore and . The same argument for representatives of module elements, using exact localization [F3], gives .

For $i\le j$, the source of the asserted local transition is the localization of $S_i{\otimes }_{R_i}{R}_j=S_j$ obtained by inverting $S_i\setminus {\mathfrak{q}}_i$ and $R_j\setminus {\mathfrak{p}}_j$. Both sets avoid ${\mathfrak{q}}_j$. Localizing this ring once more at the prime induced by ${\mathfrak{q}}_j$ yields exactly $(S_j{)}_{{\mathfrak{q}}_j}$; hence the transition is a localization. Localizing the matrix presentation and using [F3] yields $(M_i{)}_{{\mathfrak{q}}_i}{\otimes }_{(S_i{)}_{{\mathfrak{q}}_i}}(S_j{)}_{{\mathfrak{q}}_j}\cong (M_j{)}_{{\mathfrak{q}}_j}$. This proves clause 3.

If $R$ or $S$ is the zero ring, the global construction still uses the same finite equations and matrices; there are no primes for clause 3. The Axiom of Choice only selects the finite presentations and their finite lifts; no infinite stage selection is needed. The next flatness-descent result must be proved separately before this approximation can be used to transfer flatness hypotheses. [F1, step 2.1, step 4.1] $\mathrm{\square }$