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 10.135.9. A composition of standard smooth ring maps is standard smooth.

Proof. Suppose that $R \to S$ and $S \to S'$ are standard smooth. We choose presentations $S = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ and $S' = S[y_1, \ldots , y_ m]/(g_1, \ldots , g_ d)$. Choose elements $g_ j' \in R[x_1, \ldots , x_ n, y_1, \ldots , y_ m]$ mapping to the $g_ j$. In this way we see $S' = R[x_1, \ldots , x_ n, y_1, \ldots , y_ m]/ (f_1, \ldots , f_ c, g'_1, \ldots , g'_ d)$. To show that $S'$ is standard smooth it suffices to verify that the determinant

\[ \det \left( \begin{matrix} \partial f_1/\partial x_1 & \ldots & \partial f_ c/\partial x_1 & \partial g_1/\partial x_1 & \ldots & \partial g_ d/\partial x_1 \\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\ \partial f_1/\partial x_ c & \ldots & \partial f_ c/\partial x_ c & \partial g_1/\partial x_ c & \ldots & \partial g_ d/\partial x_ c \\ 0 & \ldots & 0 & \partial g_1/\partial y_1 & \ldots & \partial g_ d/\partial y_1 \\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\ 0 & \ldots & 0 & \partial g_1/\partial y_ d & \ldots & \partial g_ d/\partial y_ d \end{matrix} \right) \]

is invertible in $S'$. This is clear since it is the product of the two determinants which were assumed to be invertible by hypothesis. $\square$


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