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\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 10.130.13. Let $R \to S$ be a ring map. Let $J = \mathop{\mathrm{Ker}}(S \otimes _ R S \to S)$ be the kernel of the multiplication map. There is a canonical isomorphism of $S$-modules $\Omega _{S/R} \to J/J^2$, $a \text{d} b \mapsto a \otimes b - ab \otimes 1$.

First proof. Apply Lemma 10.130.10 to the commutative diagram

\[ \xymatrix{ S \otimes _ R S \ar[r] & S \\ S \ar[r] \ar[u] & S \ar[u] } \]

where the left vertical arrow is $a \mapsto a \otimes 1$. We get the exact sequence $0 \to J/J^2 \to \Omega _{S \otimes _ R S/S} \otimes _{S \otimes _ R S} S \to \Omega _{S/S} \to 0$. By Lemma 10.130.5 the term $\Omega _{S/S}$ is $0$, and we obtain an isomorphism between the other two terms. We have $\Omega _{S \otimes _ R S/S} = \Omega _{S/R} \otimes _ S (S \otimes _ R S)$ by Lemma 10.130.12 as $S \to S \otimes _ R S$ is the base change of $R \to S$ and hence

\[ \Omega _{S \otimes _ R S/S} \otimes _{S \otimes _ R S} S = \Omega _{S/R} \otimes _ S (S \otimes _ R S) \otimes _{S \otimes _ R S} S = \Omega _{S/R} \]

We omit the verification that the map is given by the rule of the lemma. $\square$

Second proof. First we show that the rule $a \text{d} b \mapsto a \otimes b - ab \otimes 1$ is well defined. In order to do this we have to show that $\text{d}r$ and $a\text{d}b + b \text{d}a - d(ab)$ map to zero. The first because $r \otimes 1 - 1 \otimes r = 0$ by definition of the tensor product. The second because

\[ (a \otimes b - ab \otimes 1) + (b \otimes a - ba \otimes 1) - (1 \otimes ab - ab \otimes 1) = (a \otimes 1 - 1\otimes a)(1\otimes b - b \otimes 1) \]

is in $J^2$.

We construct a map in the other direction. We may think of $S \to S \otimes _ R S$, $a \mapsto a \otimes 1$ as the base change of $R \to S$. Hence we have $\Omega _{S \otimes _ R S/S} = \Omega _{S/R} \otimes _ S (S \otimes _ R S)$, by Lemma 10.130.12. At this point the sequence of Lemma 10.130.9 gives a map

\[ J/J^2 \to \Omega _{S \otimes _ R S/ S} \otimes _{S \otimes _ R S} S = (\Omega _{S/R} \otimes _ S (S \otimes _ R S))\otimes _{S \otimes _ R S} S = \Omega _{S/R}. \]

We leave it to the reader to see it is the inverse of the map above. $\square$


Comments (2)

Comment #1694 by Jeroen Sijsling on

A shorter proof seems available by using 02HP: there one takes to be and to be . This is possible because as mentioned the maps and are in fact -algebra maps, with moreover the latter a right inverse of the former.

We get the exact sequence . By 00RP the term is trivial, and one obtains an isomorphism between the other two terms. One then rewrites as as in the current proof.

Comment #1742 by on

Dear Jeroen, I have added your proof as a first proof and left in the original as well. Thanks! For changes see here.

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  • 6 comment(s) on Section 10.130: Differentials

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