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 10.131.6. Let $R \to S$ be a ring map. Let $M$ be an $S$-module. There is a canonical short exact sequence

\[ 0 \to \Omega _{S/R} \otimes _ S M \to P^1_{S/R}(M) \to M \to 0 \]

functorial in $M$ called the sequence of principal parts.

Proof. The map $P^1_{S/R}(M) \to M$ is given above. Let $N$ be an $S$-module and let $D : M \to N$ be a differential operator of order $1$. For $m \in M$ the map

\[ g \longmapsto D(gm) - gD(m) \]

is an $R$-derivation $S \to N$ by the axioms for differential operators of order $1$. Thus it corresponds to a linear map $D_ m : \Omega _{S/R} \to N$ determined by the rule $a\text{d}b \mapsto aD(bm) - abD(m)$ (see Lemma 10.130.3). The map

\[ \Omega _{S/R} \times M \longrightarrow N,\quad (\eta , m) \longmapsto D_ m(\eta ) \]

is $S$-bilinear (details omitted) and hence determines an $S$-linear map

\[ \sigma _ D : \Omega _{S/R} \otimes _ S M \to N \]

In this way we obtain a map $\text{Diff}^1(M, N) \to \mathop{\mathrm{Hom}}\nolimits _ S(\Omega _{S/R} \otimes _ S M, N)$, $D \mapsto \sigma _ D$ functorial in $N$. By the Yoneda lemma this corresponds a map $\Omega _{S/R} \otimes _ S M \to P^1_{S/R}(M)$. It is immediate from the construction that this map is functorial in $M$. The sequence

\[ \Omega _{S/R} \otimes _ S M \to P^1_{S/R}(M) \to M \to 0 \]

is exact because for every module $N$ the sequence

\[ 0 \to \mathop{\mathrm{Hom}}\nolimits _ S(M, N) \to \text{Diff}^1(M, N) \to \mathop{\mathrm{Hom}}\nolimits _ S(\Omega _{S/R} \otimes _ S M, N) \]

is exact by inspection.

To see that $\Omega _{S/R} \otimes _ S M \to P^1_{S/R}(M)$ is injective we argue as follows. Choose an exact sequence

\[ 0 \to M' \to F \to M \to 0 \]

with $F$ a free $S$-module. This induces an exact sequence

\[ 0 \to \text{Diff}^1(M, N) \to \text{Diff}^1(F, N) \to \text{Diff}^1(M', N) \]

for all $N$. This proves that in the commutative diagram

\[ \xymatrix{ 0 \ar[r] & \Omega _{S/R} \otimes _ S M' \ar[r] \ar[d] & P^1_{S/R}(M') \ar[r] \ar[d] & M' \ar[r] \ar[d] & 0 \\ 0 \ar[r] & \Omega _{S/R} \otimes _ S F \ar[r] \ar[d] & P^1_{S/R}(F) \ar[r] \ar[d] & F \ar[r] \ar[d] & 0 \\ 0 \ar[r] & \Omega _{S/R} \otimes _ S M \ar[r] \ar[d] & P^1_{S/R}(M) \ar[r] \ar[d] & M \ar[r] \ar[d] & 0 \\ & 0 & 0 & 0 } \]

the middle column is exact. The left column is exact by right exactness of $\Omega _{S/R} \otimes _ S -$. By the snake lemma (see Section 10.4) it suffices to prove exactness on the left for the free module $F$. Using that $P^1_{S/R}(-)$ commutes with direct sums we reduce to the case $M = S$. This case is a consequence of the discussion in Example 10.131.5. $\square$


Comments (3)

Comment #355 by Nuno on

I'm probably missing something here, but I don't see why it's enough "to prove the exactness on the left for the free modules Fi". I suspect that, in general, we can only guarantee that the sequence of principal parts is right exact, but could not find a counterexample.

Comment #356 by on

OK, the argument as it was given was not sufficient (but the result is correct). I've repaired it. Here is a link to the change. Please take another look and see if you agree.

Comment #358 by Nuno on

It's correct now, thanks. After consulting EGA IV I was able to show this using the natural isomorphism and the fact that the sequence splits for . But your way is better, since we don't need to talk about the other module structure for .


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