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 10.135.4. Let $R \to S$ be a smooth ring map. Let $R \to R'$ be any ring map. Then the base change $R' \to S' = R' \otimes _ R S$ is smooth.

Proof. Let $\alpha : R[x_1, \ldots , x_ n] \to S$ be a presentation with kernel $I$. Let $\alpha ' : R'[x_1, \ldots , x_ n] \to R' \otimes _ R S$ be the induced presentation. Let $I' = \mathop{\mathrm{Ker}}(\alpha ')$. Since $0 \to I \to R[x_1, \ldots , x_ n] \to S \to 0$ is exact, the sequence $R' \otimes _ R I \to R'[x_1, \ldots , x_ n] \to R' \otimes _ R S \to 0$ is exact. Thus $R' \otimes _ R I \to I'$ is surjective. By Definition 10.135.1 there is a short exact sequence

\[ 0 \to I/I^2 \to \Omega _{R[x_1, \ldots , x_ n]/R} \otimes _{R[x_1, \ldots , x_ n]} S \to \Omega _{S/R} \to 0 \]

and the $S$-module $\Omega _{S/R}$ is finite projective. In particular $I/I^2$ is a direct summand of $\Omega _{R[x_1, \ldots , x_ n]/R} \otimes _{R[x_1, \ldots , x_ n]} S$. Consider the commutative diagram

\[ \xymatrix{ R' \otimes _ R (I/I^2) \ar[r] \ar[d] & R' \otimes _ R (\Omega _{R[x_1, \ldots , x_ n]/R} \otimes _{R[x_1, \ldots , x_ n]} S) \ar[d] \\ I'/(I')^2 \ar[r] & \Omega _{R'[x_1, \ldots , x_ n]/R'} \otimes _{R'[x_1, \ldots , x_ n]} (R' \otimes _ R S) } \]

Since the right vertical map is an isomorphism we see that the left vertical map is injective and surjective by what was said above. Thus we conclude that $\mathop{N\! L}\nolimits (\alpha ')$ is quasi-isomorphic to $\Omega _{S'/R'} \cong S' \otimes _ S \Omega _{S/R}$. And this is finite projective since it is the base change of a finite projective module. $\square$


Comments (1)

Comment #720 by Keenan Kidwell on

In the fourth line of the proof, should be .


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