Definition 10.133.1. 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.
10.133 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
\[ \text{Diff}^ k(M, N) = \text{Diff}^ k_{S/R}(M, N) \]
is an $S$-module. We have
\[ \text{Diff}^0(M, N) \subset \text{Diff}^1(M, N) \subset \text{Diff}^2(M, N) \subset \ldots \]
Lemma 10.133.2. 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
\[ D'(D(gx)) - gD'(D(x)) = D'(D(gx)) - D'(gD(x)) + D'(gD(x)) - gD'(D(x)) \]
is a sum of two compositions of differential operators of lower order. Hence the lemma follows by induction on $k + k'$. $\square$
Lemma 10.133.3. 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 $N$.
\[ \text{Diff}^ k_{S/R}(M, N) = \mathop{\mathrm{Hom}}\nolimits _ S(P^ k_{S/R}(M), N) \]
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
\[ [m + m'] - [m] - [m'],\quad g_0[m] - [g_0m] \]
where $g_0 \in S$ and $m, m' \in M$. If $D$ has order $1$, then $L_ D$ annihilates the elements
\[ [m + m'] - [m] - [m'],\quad f[m] - [fm], \quad g_0g_1[m] - g_0[g_1m] - g_1[g_0m] + [g_1g_0m] \]
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
\[ g_0g_1\ldots g_ k[m] - \sum g_0 \ldots \hat g_ i \ldots g_ k[g_ im] + \ldots +(-1)^{k + 1}[g_0\ldots g_ km] \]
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 10.133.4. Let $R \to S$ be a ring map. Let $M$ be an $S$-module. The module $P^ k_{S/R}(M)$ constructed in Lemma 10.133.3 is called the module of principal parts of order $k$ of $M$.
Note that the inclusions
\[ \text{Diff}^0(M, N) \subset \text{Diff}^1(M, N) \subset \text{Diff}^2(M, N) \subset \ldots \]
correspond via Yoneda's lemma (Categories, Lemma 4.3.5) to surjections
\[ \ldots \to P^2_{S/R}(M) \to P^1_{S/R}(M) \to P^0_{S/R}(M) = M \]
Example 10.133.5. 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 10.133.6. 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.
\[ 0 \to \Omega _{S/R} \otimes _ S M \to P^1_{S/R}(M) \to M \to 0 \]
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 10.131.3). 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 10.133.5. $\square$
Remark 10.133.7. 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 10.133.3 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 10.133.6.
\[ \xymatrix{ B \ar[r] & B' \\ A \ar[u] \ar[r] & A' \ar[u] } \]
\[ \xymatrix{ \ldots \ar[r] & P^2_{B'/A'}(M') \ar[r] & P^1_{B'/A'}(M') \ar[r] & P^0_{B'/A'}(M') \\ \ldots \ar[r] & P^2_{B/A}(M) \ar[r] \ar[u] & P^1_{B/A}(M) \ar[r] \ar[u] & P^0_{B/A}(M) \ar[u] } \]
Lemma 10.133.8. Let $A \to B$ be a ring map. The differentials $\text{d} : \Omega ^ i_{B/A} \to \Omega ^{i + 1}_{B/A}$ are differential operators of order $1$.
Proof. Given $b \in B$ we have to show that $\text{d} \circ b - b \circ \text{d}$ is a linear operator. Thus we have to show that
\[ \text{d} \circ b \circ b' - b \circ \text{d} \circ b' - b' \circ \text{d} \circ b + b' \circ b \circ \text{d} = 0 \]
To see this it suffices to check this on additive generators for $\Omega ^ i_{B/A}$. Thus it suffices to show that
\[ \text{d}(bb'b_0\text{d}b_1 \wedge \ldots \wedge \text{d}b_ i) - b\text{d}(b'b_0\text{d}b_1 \wedge \ldots \wedge \text{d}b_ i) - b'\text{d}(bb_0\text{d}b_1 \wedge \ldots \wedge \text{d}b_ i) + bb'\text{d}(b_0\text{d}b_1 \wedge \ldots \wedge \text{d}b_ i) \]
is zero. This is a pleasant calculation using the Leibniz rule which is left to the reader. $\square$
Lemma 10.133.9. Let $A \to B$ be a ring map. Let $g_ i \in B$, $i \in I$ be a set of generators for $B$ as an $A$-algebra. Let $M, N$ be $B$-modules. Let $D : M \to N$ be an $A$-linear map. In order to show that $D$ is a differential operator of order $k$ it suffices to show that $D \circ g_ i - g_ i \circ D$ is a differential operator of order $k - 1$ for $i \in I$.
Proof. Namely, we claim that the set of elements $g \in B$ such that $D \circ g - g \circ D$ is a differential operator of order $k - 1$ is an $A$-subalgebra of $B$. This follows from the relations
\[ D \circ (g + g') - (g + g') \circ D = (D \circ g - g \circ D) + (D \circ g' - g' \circ D) \]
and
\[ D \circ gg' - gg' \circ D = (D \circ g - g \circ D) \circ g' + g \circ (D \circ g' - g' \circ D) \]
Strictly speaking, to conclude for products we also use Lemma 10.133.2. $\square$
Lemma 10.133.10. Let $A \to B$ be a ring map. Let $M, N$ be $B$-modules. Let $S \subset B$ be a multiplicative subset. Any differential operator $D : M \to N$ of order $k$ extends uniquely to a differential operator $E : S^{-1}M \to S^{-1}N$ of order $k$.
Proof. By induction on $k$. If $k = 0$, then $D$ is $B$-linear and hence we get the extension by the functoriality of localization. Given $b \in B$ the operator $L_ b : m \mapsto D(bm) - bD(m)$ has order $k - 1$. Hence it has a unique extension to a differential operator $E_ b : S^{-1}M \to S^{-1}N$ of order $k - 1$ by induction. Moreover, a computation shows that $L_{b'b} = L_{b'} \circ b + b' \circ L_ b$ hence by uniqueness we obtain $E_{b'b} = E_{b'} \circ b + b' \circ E_ b$. Similarly, we obtain $E_{b'} \circ b - b \circ E_{b'} = E_ b \circ b' - b' \circ E_ b$. Now for $m \in M$ and $g \in S$ we set
\[ E(m/g) = (1/g)(D(m) - E_ g(m/g)) \]
To show that this is well defined it suffices to show that for $g' \in S$ if we use the representative $g'm/g'g$ we get the same result. We compute
\begin{align*} (1/g'g)(D(g'm) - E_{g'g}(g'm/gg')) & = (1/gg')(g'D(m) + E_{g'}(m) - E_{g'g}(g'm/gg')) \\ & = (1/g'g)(g'D(m) - g' E_ g(m/g)) \end{align*}
which is the same as before. It is clear that $E$ is $R$-linear as $D$ and $E_ g$ are $R$-linear. Taking $g = 1$ and using that $E_1 = 0$ we see that $E$ extends $D$. By Lemma 10.133.9 it now suffices to show that $E \circ b - b \circ E$ for $b \in B$ and $E \circ 1/g' - 1/g' \circ E$ for $g' \in S$ are differential operators of order $k - 1$ in order to show that $E$ is a differential operator of order $k$. For the first, choose an element $m/g$ in $S^{-1}M$ and observe that
\begin{align*} E(b m/g) - bE(m/g) & = (1/g)(D(bm) - bD(m) - E_ g(bm/g) + bE_ g(m/g)) \\ & = (1/g)(L_ b(m) - E_ b(m) + gE_ b(m/g)) \\ & = E_ b(m/g) \end{align*}
which is a differential operator of order $k - 1$. Finally, we have
\begin{align*} E(m/g'g) - (1/g')E(m/g) & = (1/g'g)(D(m) - E_{g'g}(m/g'g)) - (1/g'g)(D(m) - E_ g(m/g)) \\ & = -(1/g')E_{g'}(m/g'g) \end{align*}
which also is a differential operator of order $k - 1$ as the composition of linear maps (multiplication by $1/g'$ and signs) and $E_{g'}$. We omit the proof of uniqueness. $\square$
Lemma 10.133.11. Let $R \to A$ and $R \to B$ be ring maps. Let $M$ and $M'$ be $A$-modules. Let $D : M \to M'$ be a differential operator of order $k$ with respect to $R \to A$. Let $N$ be any $B$-module. Then the map is a differential operator of order $k$ with respect to $B \to A \otimes _ R B$.
\[ D \otimes \text{id}_ N : M \otimes _ R N \to M' \otimes _ R N \]
Proof. It is clear that $D' = D \otimes \text{id}_ N$ is $B$-linear. By Lemma 10.133.9 it suffices to show that
\[ D' \circ a \otimes 1 - a \otimes 1 \circ D' = (D \circ a - a \circ D) \otimes \text{id}_ N \]
is a differential operator of order $k - 1$ which follows by induction on $k$. $\square$
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