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 10.130.14. If $S = R[x_1, \ldots , x_ n]$, then $\Omega _{S/R}$ is a finite free $S$-module with basis $\text{d}x_1, \ldots , \text{d}x_ n$.

Proof. We first show that $\text{d}x_1, \ldots , \text{d}x_ n$ generate $\Omega _{S/R}$ as an $S$-module. To prove this we show that $\text{d}g$ can be expressed as a sum $\sum g_ i \text{d}x_ i$ for any $g \in R[x_1, \ldots , x_ n]$. We do this by induction on the (total) degree of $g$. It is clear if the degree of $g$ is $0$, because then $\text{d}g = 0$. If the degree of $g$ is $>0$, then we may write $g$ as $c + \sum g_ i x_ i$ with $c\in R$ and $\deg (g_ i) < \deg (g)$. By the Leibniz rule we have $\text{d}g = \sum g_ i \text{d} x_ i + \sum x_ i \text{d}g_ i$, and hence we win by induction.

Consider the $R$-derivation $\partial / \partial x_ i : R[x_1, \ldots , x_ n] \to R[x_1, \ldots , x_ n]$. (We leave it to the reader to define this; the defining property being that $\partial / \partial x_ i (x_ j) = \delta _{ij}$.) By the universal property this corresponds to an $S$-module map $l_ i : \Omega _{S/R} \to R[x_1, \ldots , x_ n]$ which maps $\text{d}x_ i$ to $1$ and $\text{d}x_ j$ to $0$ for $j \not= i$. Thus it is clear that there are no $S$-linear relations among the elements $\text{d}x_1, \ldots , \text{d}x_ n$. $\square$


Comments (1)

Comment #1087 by Nuno Cardoso on

Leibniz is spelled wrong as "Leibnize". There are also two more instances of this wrong spelling in Tags 55.12 and 82.4.6.

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  • 6 comment(s) on Section 10.130: Differentials

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