
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. $\square$

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 $P^1_{S/R}(M)\cong P^1_{S/R} \otimes_S M$ and the fact that the sequence splits for $M=S$. But your way is better, since we don't need to talk about the other module structure for $P^1_{S/R}(M)$.

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