The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Lemma 10.162.9. Let $A = \mathop{\mathrm{colim}}\nolimits _{i \in I} A_ i$ be a directed colimit of rings. Let $0 \in I$ and $\varphi _0 : B_0 \to C_0$ a map of $A_0$-algebras. Assume

  1. $A \otimes _{A_0} B_0 \to A \otimes _{A_0} C_0$ is syntomic (resp. a relative global complete intersection),

  2. $C_0$ is of finite presentation over $B_0$.

Then there exists an $i \geq 0$ such that the map $A_ i \otimes _{A_0} B_0 \to A_ i \otimes _{A_0} C_0$ is syntomic (resp. a relative global complete intersection).

Proof. Assume $A \otimes _{A_0} B_0 \to A \otimes _{A_0} C_0$ is a relative global complete intersection. By Lemma 10.134.12 there exists a finite type $\mathbf{Z}$-algebra $R$, a ring map $R \to A \otimes _{A_0} B_0$, a relative global complete intersection $R \to S$, and an isomorphism

\[ (A \otimes _{A_0} B_0) \otimes _ R S \longrightarrow A \otimes _{A_0} C_0 \]

Because $R$ is of finite type (and hence finite presentation) over $\mathbf{Z}$, there exists an $i$ and a map $R \to A_ i \otimes _{A_0} B_0$ lifting the map $R \to A \otimes _{A_0} B_0$, see Lemma 10.126.3. Using the same lemma, there exists an $i' \geq i$ such that $(A_ i \otimes _{A_0} B_0) \otimes _ R S \to A \otimes _{A_0} C_0$ comes from a map $(A_ i \otimes _{A_0} B_0) \otimes _ R S \to A_{i'} \otimes _{A_0} C_0$. Thus we may assume, after replacing $i$ by $i'$, that the displayed map comes from an $A_ i \otimes _{A_0} B_0$-algebra map

\[ (A_ i \otimes _{A_0} B_0) \otimes _ R S \longrightarrow A_ i \otimes _{A_0} C_0 \]

By Lemma 10.162.6 after increasing $i$ this map is an isomorphism. This finishes the proof in this case because the base change of a relative global complete intersection is a relative global complete intersection by Lemma 10.134.10.

Assume $A \otimes _{A_0} B_0 \to A \otimes _{A_0} C_0$ is syntomic. Then there exist elements $g_1, \ldots , g_ m$ in $A \otimes _{A_0} C_0$ generating the unit ideal such that $A \otimes _{A_0} B_0 \to (A \otimes _{A_0} C_0)_{g_ j}$ is a relative global complete intersection, see Lemma 10.134.15. We can find an $i$ and elements $g_{i, j} \in A_ i \otimes _{A_0} C_0$ mapping to $g_ j$. After increasing $i$ we may assume $g_{i, 1}, \ldots , g_{i, m}$ generate the unit ideal of $A_ i \otimes _{A_0} C_0$. The result of the previous paragraph implies that, after increasing $i$, we may assume the maps $A_ i \otimes _{A_0} B_0 \to (A_ i \otimes _{A_0} C_0)_{g_{i, j}}$ are relative global complete intersections. Then $A_ i \otimes _{A_0} B_0 \to A_ i \otimes _{A_0} C_0$ is syntomic by Lemma 10.134.4 (and the already used Lemma 10.134.15). $\square$


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