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 10.162.5. 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 unramified,

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

Then for some $i \geq 0$ the map $A_ i \otimes _{A_0} B_0 \to A_ i \otimes _{A_0} C_0$ is unramified.

Proof. Set $B_ i = A_ i \otimes _{A_0} B_0$, $C_ i = A_ i \otimes _{A_0} C_0$, $B = A \otimes _{A_0} B_0$, and $C = A \otimes _{A_0} C_0$. Let $x_1, \ldots , x_ m$ be generators for $C_0$ over $B_0$. Then $\text{d}x_1, \ldots , \text{d}x_ m$ generate $\Omega _{C_0/B_0}$ over $C_0$ and their images generate $\Omega _{C_ i/B_ i}$ over $C_ i$ (Lemmas 10.130.14 and 10.130.9). Observe that $0 = \Omega _{C/B} = \mathop{\mathrm{colim}}\nolimits \Omega _{C_ i/B_ i}$ (Lemma 10.130.4). Thus there is an $i$ such that $\text{d}x_1, \ldots , \text{d}x_ m$ map to zero and hence $\Omega _{C_ i/B_ i} = 0$ as desired. $\square$

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