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 10.130.6. In diagram (10.130.5.1), suppose that $S \to S'$ is surjective with kernel $I \subset S$. Then $\Omega _{S/R} \to \Omega _{S'/R'}$ is surjective with kernel generated as an $S$-module by the elements $\text{d}a$, where $a \in S$ is such that $\varphi (a) \in \beta (R')$. (This includes in particular the elements $\text{d}(i)$, $i \in I$.)

Proof. Consider the map of presentations above. Clearly the right vertical map of free modules is surjective. Thus the map is surjective. A diagram chase shows that the following elements generate the kernel as an $S$-module for sure: $i\text{d}a, i\in I, a \in S$, and $\text{d}a$, with $a \in S$ such that $\varphi (a) = \beta (r')$ for some $r' \in R'$. Note that $\varphi (i) = \varphi (ia) = 0 = \beta (0)$, and that $\text{d}(ia) = i\text{d}a + a \text{d}i$. Hence $i\text{d}a = \text{d}(ia) - a \text{d}i$ is an $S$-linear combination of elements of the second kind. $\square$


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