Definition 10.131.1. Let $R \to S$ be a ring map. Let $M$, $N$ be $S$-modules. Let $k \geq 0$ be an integer. We inductively define a *differential operator $D : M \to N$ of order $k$* to be an $R$-linear map such that for all $g \in S$ the map $m \mapsto D(gm) - gD(m)$ is a differential operator of order $k - 1$. For the base case $k = 0$ we define a differential operator of order $0$ to be an $S$-linear map.

## 10.131 Finite order differential operators

In this section we introduce differential operators of finite order.

If $D : M \to N$ is a differential operator of order $k$, then for all $g \in S$ the map $gD$ is a differential operator of order $k$. The sum of two differential operators of order $k$ is another. Hence the set of all these

is an $S$-module. We have

Lemma 10.131.2. Let $R \to S$ be a ring map. Let $L, M, N$ be $S$-modules. If $D : L \to M$ and $D' : M \to N$ are differential operators of order $k$ and $k'$, then $D' \circ D$ is a differential operator of order $k + k'$.

**Proof.**
Let $g \in S$. Then the map which sends $x \in L$ to

is a sum of two compositions of differential operators of lower order. Hence the lemma follows by induction on $k + k'$. $\square$

Lemma 10.131.3. Let $R \to S$ be a ring map. Let $M$ be an $S$-module. Let $k \geq 0$. There exists an $S$-module $P^ k_{S/R}(M)$ and a canonical isomorphism

functorial in the $S$-module $N$.

**Proof.**
The existence of $P^ k_{S/R}(M)$ follows from general category theoretic arguments (insert future reference here), but we will also give a construction. Set $F = \bigoplus _{m \in M} S[m]$ where $[m]$ is a symbol indicating the basis element in the summand corresponding to $m$. Given any differential operator $D : M \to N$ we obtain an $S$-linear map $L_ D : F \to N$ sending $[m]$ to $D(m)$. If $D$ has order $0$, then $L_ D$ annihilates the elements

where $g_0 \in S$ and $m, m' \in M$. If $D$ has order $1$, then $L_ D$ annihilates the elements

where $f \in R$, $g_0, g_1 \in S$, and $m \in M$. If $D$ has order $k$, then $L_ D$ annihilates the elements $[m + m'] - [m] - [m']$, $f[m] - [fm]$, and the elements

Conversely, if $L : F \to N$ is an $S$-linear map annihilating all the elements listed in the previous sentence, then $m \mapsto L([m])$ is a differential operator of order $k$. Thus we see that $P^ k_{S/R}(M)$ is the quotient of $F$ by the submodule generated by these elements. $\square$

Definition 10.131.4. Let $R \to S$ be a ring map. Let $M$ be an $S$-module. The module $P^ k_{S/R}(M)$ constructed in Lemma 10.131.3 is called the *module of principal parts of order $k$* of $M$.

Note that the inclusions

correspond via Yoneda's lemma (Categories, Lemma 4.3.5) to surjections

Example 10.131.5. Let $R \to S$ be a ring map and let $N$ be an $S$-module. Observe that $\text{Diff}^1(S, N) = \text{Der}_ R(S, N) \oplus N$. Namely, if $D : S \to N$ is a differential operator of order $1$ then $\sigma _ D : S \to N$ defined by $\sigma _ D(g) := D(g) - gD(1)$ is an $R$-derivation and $D = \sigma _ D + \lambda _{D(1)}$ where $\lambda _ x : S \to N$ is the linear map sending $g$ to $gx$. It follows that $P^1_{S/R} = \Omega _{S/R} \oplus S$ by the universal property of $\Omega _{S/R}$.

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

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

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

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

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

is exact because for every module $N$ the sequence

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

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

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

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. $\square$

Remark 10.131.7. Suppose given a commutative diagram of rings

a $B$-module $M$, a $B'$-module $M'$, and a $B$-linear map $M \to M'$. Then we get a compatible system of module maps

These maps are compatible with further composition of maps of this type. The easiest way to see this is to use the description of the modules $P^ k_{B/A}(M)$ in terms of generators and relations in the proof of Lemma 10.131.3 but it can also be seen directly from the universal property of these modules. Moreover, these maps are compatible with the short exact sequences of Lemma 10.131.6.

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