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*}

55.6 Module of differentials

In this section we develop a theory of modules of differentials for divided power rings.

Definition 55.6.1. Let $A$ be a ring. Let $(B, J, \delta )$ be a divided power ring. Let $A \to B$ be a ring map. Let $M$ be an $B$-module. A divided power $A$-derivation into $M$ is a map $\theta : B \to M$ which is additive, annihilates the elements of $A$, satisfies the Leibniz rule $\theta (bb') = b\theta (b') + b'\theta (b)$ and satisfies

\[ \theta (\delta _ n(x)) = \delta _{n - 1}(x)\theta (x) \]

for all $n \geq 1$ and all $x \in J$.

In the situation of the definition, just as in the case of usual derivations, there exists a universal divided power $A$-derivation

\[ \text{d}_{B/A, \delta } : B \to \Omega _{B/A, \delta } \]

such that any divided power $A$-derivation $\theta : B \to M$ is equal to $\theta = \xi \circ d_{B/A, \delta }$ for some $B$-linear map $\Omega _{B/A, \delta } \to M$. If $(A, I, \gamma ) \to (B, J, \delta )$ is a homomorphism of divided power rings, then we can forget the divided powers on $A$ and consider the divided power derivations of $B$ over $A$. Here are some basic properties of the divided power module of differentials.

Lemma 55.6.2. Let $A$ be a ring. Let $(B, J, \delta )$ be a divided power ring and $A \to B$ a ring map.

  1. Consider $B[x]$ with divided power ideal $(JB[x], \delta ')$ where $\delta '$ is the extension of $\delta $ to $B[x]$. Then

    \[ \Omega _{B[x]/A, \delta '} = \Omega _{B/A, \delta } \otimes _ B B[x] \oplus B[x]\text{d}x. \]
  2. Consider $B\langle x \rangle $ with divided power ideal $(JB\langle x \rangle + B\langle x \rangle _{+}, \delta ')$. Then

    \[ \Omega _{B\langle x\rangle /A, \delta '} = \Omega _{B/A, \delta } \otimes _ B B\langle x \rangle \oplus B\langle x\rangle \text{d}x. \]
  3. Let $K \subset J$ be an ideal preserved by $\delta _ n$ for all $n > 0$. Set $B' = B/K$ and denote $\delta '$ the induced divided power on $J/K$. Then $\Omega _{B'/A, \delta '}$ is the quotient of $\Omega _{B/A, \delta } \otimes _ B B'$ by the $B'$-submodule generated by $\text{d}k$ for $k \in K$.

Proof. These are proved directly from the construction of $\Omega _{B/A, \delta }$ as the free $B$-module on the elements $\text{d}b$ modulo the relations

  1. $\text{d}(b + b') = \text{d}b + \text{d}b'$, $b, b' \in B$,

  2. $\text{d}a = 0$, $a \in A$,

  3. $\text{d}(bb') = b \text{d}b' + b' \text{d}b$, $b, b' \in B$,

  4. $\text{d}\delta _ n(f) = \delta _{n - 1}(f)\text{d}f$, $f \in J$, $n > 1$.

Note that the last relation explains why we get “the same” answer for the divided power polynomial algebra and the usual polynomial algebra: in the first case $x$ is an element of the divided power ideal and hence $\text{d}x^{[n]} = x^{[n - 1]}\text{d}x$. $\square$

Let $(A, I, \gamma )$ be a divided power ring. In this setting the correct version of the powers of $I$ is given by the divided powers

\[ I^{[n]} = \text{ideal generated by } \gamma _{e_1}(x_1) \ldots \gamma _{e_ t}(x_ t) \text{ with }\sum e_ j \geq n\text{ and }x_ j \in I. \]

Of course we have $I^ n \subset I^{[n]}$. Note that $I^{[1]} = I$. Sometimes we also set $I^{[0]} = A$.

Lemma 55.6.3. Let $(A, I, \gamma ) \to (B, J, \delta )$ be a homomorphism of divided power rings. Let $(B(1), J(1), \delta (1))$ be the coproduct of $(B, J, \delta )$ with itself over $(A, I, \gamma )$, i.e., such that

\[ \xymatrix{ (B, J, \delta ) \ar[r] & (B(1), J(1), \delta (1)) \\ (A, I, \gamma ) \ar[r] \ar[u] & (B, J, \delta ) \ar[u] } \]

is cocartesian. Denote $K = \mathop{\mathrm{Ker}}(B(1) \to B)$. Then $K \cap J(1) \subset J(1)$ is preserved by the divided power structure and

\[ \Omega _{B/A, \delta } = K/ \left(K^2 + (K \cap J(1))^{[2]}\right) \]

canonically.

Proof. The fact that $K \cap J(1) \subset J(1)$ is preserved by the divided power structure follows from the fact that $B(1) \to B$ is a homomorphism of divided power rings.

Recall that $K/K^2$ has a canonical $B$-module structure. Denote $s_0, s_1 : B \to B(1)$ the two coprojections and consider the map $\text{d} : B \to K/K^2 +(K \cap J(1))^{[2]}$ given by $b \mapsto s_1(b) - s_0(b)$. It is clear that $\text{d}$ is additive, annihilates $A$, and satisfies the Leibniz rule. We claim that $\text{d}$ is an $A$-derivation. Let $x \in J$. Set $y = s_1(x)$ and $z = s_0(x)$. Denote $\delta $ the divided power structure on $J(1)$. We have to show that $\delta _ n(y) - \delta _ n(z) = \delta _{n - 1}(y)(y - z)$ modulo $K^2 +(K \cap J(1))^{[2]}$ for $n \geq 1$. We will show this by induction on $n$. It is true for $n = 1$. Let $n > 1$ and that it holds for all smaller values. Note that

\[ \delta _ n(z - y) = \sum \nolimits _{i = 0}^ n (-1)^{n - i}\delta _ i(z)\delta _{n - i}(y) \]

is an element of $K^2 +(K \cap J(1))^{[2]}$. From this and induction we see that working modulo $K^2 +(K \cap J(1))^{[2]}$ we have

\begin{align*} & \delta _ n(y) - \delta _ n(z) \\ & = \delta _ n(y) + \sum \nolimits _{i = 0}^{n - 1} (-1)^{n - i}\delta _ i(z)\delta _{n - i}(y) \\ & = \delta _ n(y) + (-1)^ n\delta _ n(y) + \sum \nolimits _{i = 1}^{n - 1} (-1)^{n - i}(\delta _ i(y) - \delta _{i - 1}(y)(y - z))\delta _{n - i}(y) \end{align*}

Using that $\delta _ i(y)\delta _{n - i}(y) = \binom {n}{i} \delta _ n(y)$ and that $\delta _{i - 1}(y)\delta _{n - i}(y) = \binom {n - 1}{i} \delta _{n - 1}(y)$ the reader easily verifies that this expression comes out to give $\delta _{n - 1}(y)(y - z)$ as desired.

Let $M$ be a $B$-module. Let $\theta : B \to M$ be a divided power $A$-derivation. Set $D = B \oplus M$ where $M$ is an ideal of square zero. Define a divided power structure on $J \oplus M \subset D$ by setting $\delta _ n(x + m) = \delta _ n(x) + \delta _{n - 1}(x)m$ for $n > 1$, see Lemma 55.3.1. There are two divided power algebra homomorphisms $B \to D$: the first is given by the inclusion and the second by the map $b \mapsto b + \theta (b)$. Hence we get a canonical homomorphism $B(1) \to D$ of divided power algebras over $(A, I, \gamma )$. This induces a map $K \to M$ which annihilates $K^2$ (as $M$ is an ideal of square zero) and $(K \cap J(1))^{[2]}$ as $M^{[2]} = 0$. The composition $B \to K/K^2 + (K \cap J(1))^{[2]} \to M$ equals $\theta $ by construction. It follows that $\text{d}$ is a universal divided power $A$-derivation and we win. $\square$

Remark 55.6.4. Let $A \to B$ be a ring map and let $(J, \delta )$ be a divided power structure on $B$. The universal module $\Omega _{B/A, \delta }$ comes with a little bit of extra structure, namely the $B$-submodule $N$ of $\Omega _{B/A, \delta }$ generated by $\text{d}_{B/A, \delta }(J)$. In terms of the isomorphism given in Lemma 55.6.3 this corresponds to the image of $K \cap J(1)$ in $\Omega _{B/A, \delta }$. Consider the $A$-algebra $D = B \oplus \Omega ^1_{B/A, \delta }$ with ideal $\bar J = J \oplus N$ and divided powers $\bar\delta $ as in the proof of the lemma. Then $(D, \bar J, \bar\delta )$ is a divided power ring and the two maps $B \to D$ given by $b \mapsto b$ and $b \mapsto b + \text{d}_{B/A, \delta }(b)$ are homomorphisms of divided power rings over $A$. Moreover, $N$ is the smallest submodule of $\Omega _{B/A, \delta }$ such that this is true.

Lemma 55.6.5. In Situation 55.5.1. Let $(B, J, \delta )$ be an object of $\text{CRIS}(C/A)$. Let $(B(1), J(1), \delta (1))$ be the coproduct of $(B, J, \delta )$ with itself in $\text{CRIS}(C/A)$. Denote $K = \mathop{\mathrm{Ker}}(B(1) \to B)$. Then $K \cap J(1) \subset J(1)$ is preserved by the divided power structure and

\[ \Omega _{B/A, \delta } = K/ \left(K^2 + (K \cap J(1))^{[2]}\right) \]

canonically.

Proof. Word for word the same as the proof of Lemma 55.6.3. The only point that has to be checked is that the divided power ring $D = B \oplus M$ is an object of $\text{CRIS}(C/A)$ and that the two maps $B \to C$ are morphisms of $\text{CRIS}(C/A)$. Since $D/(J \oplus M) = B/J$ we can use $C \to B/J$ to view $D$ as an object of $\text{CRIS}(C/A)$ and the statement on morphisms is clear from the construction. $\square$

Lemma 55.6.6. Let $(A, I, \gamma )$ be a divided power ring. Let $A \to B$ be a ring map and let $IB \subset J \subset B$ be an ideal. Let $D_{B, \gamma }(J) = (D, \bar J, \bar\gamma )$ be the divided power envelope. Then we have

\[ \Omega _{D/A, \bar\gamma } = \Omega _{B/A} \otimes _ B D \]

Proof. We will prove this first when $B$ is flat over $A$. In this case $\gamma $ extends to a divided power structure $\gamma '$ on $IB$, see Divided Power Algebra, Lemma 23.4.2. Hence $D = D_{B, \gamma '}(J)$ is equal to a quotient of the divided power ring $(D', J', \delta )$ where $D' = B\langle x_ t \rangle $ and $J' = IB\langle x_ t \rangle + B\langle x_ t \rangle _{+}$ by the elements $x_ t - f_ t$ and $\delta _ n(\sum r_ t x_ t - r_0)$, see Lemma 55.2.4 for notation and explanation. Write $\text{d} : D' \to \Omega _{D'/A, \delta }$ for the universal derivation. Note that

\[ \Omega _{D'/A, \delta } = \Omega _{B/A} \otimes _ B D' \oplus \bigoplus D' \text{d}x_ t, \]

see Lemma 55.6.2. We conclude that $\Omega _{D/A, \bar\gamma }$ is the quotient of $\Omega _{D'/A, \delta } \otimes _{D'} D$ by the submodule generated by $\text{d}$ applied to the generators of the kernel of $D' \to D$ listed above, see Lemma 55.6.2. Since $\text{d}(x_ t - f_ t) = - \text{d}f_ t + \text{d}x_ t$ we see that we have $\text{d}x_ t = \text{d}f_ t$ in the quotient. In particular we see that $\Omega _{B/A} \otimes _ B D \to \Omega _{D/A, \gamma }$ is surjective with kernel given by the images of $\text{d}$ applied to the elements $\delta _ n(\sum r_ t x_ t - r_0)$. However, given a relation $\sum r_ tf_ t - r_0 = 0$ in $B$ with $r_ t \in B$ and $r_0 \in IB$ we see that

\begin{align*} \text{d}\delta _ n(\sum r_ t x_ t - r_0) & = \delta _{n - 1}(\sum r_ t x_ t - r_0)\text{d}(\sum r_ t x_ t - r_0) \\ & = \delta _{n - 1}(\sum r_ t x_ t - r_0) \left( \sum r_ t\text{d}(x_ t - f_ t) + \sum (x_ t - f_ t)\text{d}r_ t \right) \end{align*}

because $\sum r_ tf_ t - r_0 = 0$ in $B$. Hence this is already zero in $\Omega _{B/A} \otimes _ A D$ and we win in the case that $B$ is flat over $A$.

In the general case we write $B$ as a quotient of a polynomial ring $P \to B$ and let $J' \subset P$ be the inverse image of $J$. Then $D = D'/K'$ with notation as in Lemma 55.2.3. By the case handled in the first paragraph of the proof we have $\Omega _{D'/A, \bar\gamma '} = \Omega _{P/A} \otimes _ P D'$. Then $\Omega _{D/A, \bar\gamma }$ is the quotient of $\Omega _{P/A} \otimes _ P D$ by the submodule generated by $\text{d}\bar\gamma _ n'(k)$ where $k$ is an element of the kernel of $P \to B$, see Lemma 55.6.2 and the description of $K'$ from Lemma 55.2.3. Since $\text{d}\bar\gamma _ n'(k) = \bar\gamma '_{n - 1}(k)\text{d}k$ we see again that it suffices to divided by the submodule generated by $\text{d}k$ with $k \in \mathop{\mathrm{Ker}}(P \to B)$ and since $\Omega _{B/A}$ is the quotient of $\Omega _{P/A} \otimes _ A B$ by these elements (Algebra, Lemma 10.130.9) we win. $\square$

Remark 55.6.7. Let $B$ be a ring. Write $\Omega _ B = \Omega _{B/\mathbf{Z}}$ for the absolute1 module of differentials of $B$. Let $\text{d} : B \to \Omega _ B$ denote the universal derivation. Set $\Omega _ B^ i = \wedge ^ i_ B(\Omega _ B)$ as in Algebra, Section 10.12. The absolute de Rham complex

\[ \Omega _ B^0 \to \Omega _ B^1 \to \Omega _ B^2 \to \ldots \]

Here $\text{d} : \Omega _ B^ p \to \Omega _ B^{p + 1}$ is defined by the rule

\[ \text{d}\left(b_0\text{d}b_1 \wedge \ldots \wedge \text{d}b_ p\right) = \text{d}b_0 \wedge \text{d}b_1 \wedge \ldots \wedge \text{d}b_ p \]

which we will show is well defined; note that $\text{d} \circ \text{d} = 0$ so we get a complex. Recall that $\Omega _ B$ is the $B$-module generated by elements $\text{d}b$ subject to the relations $\text{d}(a + b) = \text{d}a + \text{d}b$ and $\text{d}(ab) = b\text{d}a + a\text{d}b$ for $a, b \in B$. To prove that our map is well defined for $p = 1$ we have to show that the elements

\[ a\text{d}(b + c) - a\text{d}b - a\text{d}c \quad \text{and}\quad a\text{d}(bc) - ac\text{d}b - ab\text{d}c,\quad a,b,c \in B \]

are mapped to zero by our rule. This is clear by direct computation (using the Leibniz rule). Thus we get a map

\[ \Omega _ B \otimes _\mathbf {Z} \ldots \otimes _\mathbf {Z} \Omega _ B \longrightarrow \Omega _ B^{p + 1} \]

defined by the formula

\[ \omega _1 \otimes \ldots \otimes \omega _ p \longmapsto \sum (-1)^{i + 1} \omega _1 \wedge \ldots \wedge \text{d}(\omega _ i) \wedge \ldots \wedge \omega _ p \]

which matches our rule above on elements of the form $b_0\text{d}b_1 \otimes \text{d}b_2 \otimes \ldots \otimes \text{d}b_ p$. It is clear that this map is alternating. To finish we have to show that

\[ \omega _1 \otimes \ldots \otimes f\omega _ i \otimes \ldots \otimes \omega _ p \quad \text{and}\quad \omega _1 \otimes \ldots \otimes f\omega _ j \otimes \ldots \otimes \omega _ p \]

are mapped to the same element. By $\mathbf{Z}$-linearity and the alternating property, it is enough to show this for $p = 2$, $i = 1$, $j = 2$, $\omega _1 = a_1 \text{d}b_1$ and $\omega _2 = a_2 \text{d}b_2$. Thus we need to show that

\begin{align*} & \text{d}fa_1 \wedge \text{d}b_1 \wedge a_2\text{d}b_2 - fa_1 \text{d}b_1 \wedge \text{d}a_2 \wedge \text{d}b_2 \\ & = \text{d}a_1 \wedge \text{d}b_1 \wedge fa_2\text{d}b_2 - a_1 \text{d}b_1 \wedge \text{d}fa_2 \wedge \text{d}b_2 \end{align*}

in other words that

\[ (a_2 \text{d}fa_1 + fa_1 \text{d}a_2 - fa_2 \text{d}a_1 - a_1 \text{d}fa_2) \wedge \text{d}b_1 \wedge \text{d}b_2 = 0. \]

This follows from the Leibniz rule.

Lemma 55.6.8. Let $B$ be a ring. Let $\pi : \Omega _ B \to \Omega $ be a surjective $B$-module map. Denote $\text{d} : B \to \Omega $ the composition of $\pi $ with $\text{d}_ B : B \to \Omega _ B$. Set $\Omega ^ i = \wedge _ B^ i(\Omega )$. Assume that the kernel of $\pi $ is generated, as a $B$-module, by elements $\omega \in \Omega _ B$ such that $\text{d}_ B(\omega ) \in \Omega _ B^2$ maps to zero in $\Omega ^2$. Then there is a de Rham complex

\[ \Omega ^0 \to \Omega ^1 \to \Omega ^2 \to \ldots \]

whose differential is defined by the rule

\[ \text{d} : \Omega ^ p \to \Omega ^{p + 1},\quad \text{d}\left(f_0\text{d}f_1 \wedge \ldots \wedge \text{d}f_ p\right) = \text{d}f_0 \wedge \text{d}f_1 \wedge \ldots \wedge \text{d}f_ p \]

Proof. We will show that there exists a commutative diagram

\[ \xymatrix{ \Omega _ B^0 \ar[d] \ar[r]_{\text{d}_ B} & \Omega _ B^1 \ar[d]_\pi \ar[r]_{\text{d}_ B} & \Omega _ B^2 \ar[d]_{\wedge ^2\pi } \ar[r]_{\text{d}_ B} & \ldots \\ \Omega ^0 \ar[r]^{\text{d}} & \Omega ^1 \ar[r]^{\text{d}} & \Omega ^2 \ar[r]^{\text{d}} & \ldots } \]

the description of the map $\text{d}$ will follow from the construction of $\text{d}_ B$ in Remark 55.6.7. Since the left most vertical arrow is an isomorphism we have the first square. Because $\pi $ is surjective, to get the second square it suffices to show that $\text{d}_ B$ maps the kernel of $\pi $ into the kernel of $\wedge ^2\pi $. We are given that any element of the kernel of $\pi $ is of the form $\sum b_ i\omega _ i$ with $\pi (\omega _ i) = 0$ and $\wedge ^2\pi (\text{d}_ B(\omega _ i)) = 0$. By the Leibniz rule for $\text{d}_ B$ we have $\text{d}_ B(\sum b_ i\omega _ i) = \sum b_ i \text{d}_ B(\omega _ i) + \sum \text{d}_ B(b_ i) \wedge \omega _ i$. Hence this maps to zero under $\wedge ^2\pi $.

For $i > 1$ we note that $\wedge ^ i \pi $ is surjective with kernel the image of $\mathop{\mathrm{Ker}}(\pi ) \wedge \Omega ^{i - 1}_ B \to \Omega _ B^ i$. For $\omega _1 \in \mathop{\mathrm{Ker}}(\pi )$ and $\omega _2 \in \Omega ^{i - 1}_ B$ we have

\[ \text{d}_ B(\omega _1 \wedge \omega _2) = \text{d}_ B(\omega _1) \wedge \omega _2 - \omega _1 \wedge \text{d}_ B(\omega _2) \]

which is in the kernel of $\wedge ^{i + 1}\pi $ by what we just proved above. Hence we get the $(i + 1)$st square in the diagram above. This concludes the proof. $\square$

Remark 55.6.9. Let $A \to B$ be a ring map and let $(J, \delta )$ be a divided power structure on $B$. Set $\Omega _{B/A, \delta }^ i = \wedge ^ i_ B \Omega _{B/A, \delta }$ where $\Omega _{B/A, \delta }$ is the target of the universal divided power $A$-derivation $\text{d} = \text{d}_{B/A} : B \to \Omega _{B/A, \delta }$. Note that $\Omega _{B/A, \delta }$ is the quotient of $\Omega _ B$ by the $B$-submodule generated by the elements $\text{d}a = 0$ for $a \in A$ and $\text{d}\delta _ n(x) - \delta _{n - 1}(x)\text{d}x$ for $x \in J$. We claim Lemma 55.6.8 applies. To see this it suffices to verify the elements $\text{d}a$ and $\text{d}\delta _ n(x) - \delta _{n - 1}(x)\text{d}x$ of $\Omega _ B$ are mapped to zero in $\Omega ^2_{B/A, \delta }$. This is clear for the first, and for the last we observe that

\[ \text{d}(\delta _{n - 1}(x)) \wedge \text{d}x = \delta _{n - 2}(x) \text{d}x \wedge \text{d}x = 0 \]

in $\Omega ^2_{B/A, \delta }$ as desired. Hence we obtain a divided power de Rham complex

\[ \Omega ^0_{B/A, \delta } \to \Omega ^1_{B/A, \delta } \to \Omega ^2_{B/A, \delta } \to \ldots \]

which will play an important role in the sequel.

Remark 55.6.10. Let $B$ be a ring. Let $\Omega _ B \to \Omega $ be a quotient satisfying the assumptions of Lemma 55.6.8. Let $M$ be a $B$-module. A connection is an additive map

\[ \nabla : M \longrightarrow M \otimes _ B \Omega \]

such that $\nabla (bm) = b \nabla (m) + m \otimes \text{d}b$ for $b \in B$ and $m \in M$. In this situation we can define maps

\[ \nabla : M \otimes _ B \Omega ^ i \longrightarrow M \otimes _ B \Omega ^{i + 1} \]

by the rule $\nabla (m \otimes \omega ) = \nabla (m) \wedge \omega + m \otimes \text{d}\omega $. This works because if $b \in B$, then

\begin{align*} \nabla (bm \otimes \omega ) - \nabla (m \otimes b\omega ) & = \nabla (bm) \otimes \omega + bm \otimes \text{d}\omega - \nabla (m) \otimes b\omega - m \otimes \text{d}(b\omega ) \\ & = b\nabla (m) \otimes \omega + m \otimes \text{d}b \wedge \omega + bm \otimes \text{d}\omega \\ & \ \ \ \ \ \ - b\nabla (m) \otimes \omega - bm \otimes \text{d}(\omega ) - m \otimes \text{d}b \wedge \omega = 0 \end{align*}

As is customary we say the connection is integrable if and only if the composition

\[ M \xrightarrow {\nabla } M \otimes _ B \Omega ^1 \xrightarrow {\nabla } M \otimes _ B \Omega ^2 \]

is zero. In this case we obtain a complex

\[ M \xrightarrow {\nabla } M \otimes _ B \Omega ^1 \xrightarrow {\nabla } M \otimes _ B \Omega ^2 \xrightarrow {\nabla } M \otimes _ B \Omega ^3 \xrightarrow {\nabla } M \otimes _ B \Omega ^4 \to \ldots \]

which is called the de Rham complex of the connection.

Remark 55.6.11. Let $\varphi : B \to B'$ be a ring map. Let $\Omega _ B \to \Omega $ and $\Omega _{B'} \to \Omega '$ be quotients satisfying the assumptions of Lemma 55.6.8. Assume that the map $\Omega _ B \to \Omega _{B'}$, $b_1\text{d}b_2 \mapsto \varphi (b_1)\text{d}\varphi (b_2)$ fits into a commutative diagram

\[ \xymatrix{ B \ar[r] \ar[d] & \Omega _ B \ar[r] \ar[d] & \Omega \ar[d]^{\varphi } \\ B' \ar[r] & \Omega _{B'} \ar[r] & \Omega ' } \]

In this situation, given any pair $(M, \nabla )$ where $M$ is a $B$-module and $\nabla : M \to M \otimes _ B \Omega $ is a connection we obtain a base change $(M \otimes _ B B', \nabla ')$ where

\[ \nabla ' : M \otimes _ B B' \longrightarrow (M \otimes _ B B') \otimes _{B'} \Omega ' = M \otimes _ B \Omega ' \]

is defined by the rule

\[ \nabla '(m \otimes b') = \sum m_ i \otimes b'\text{d}\varphi (b_ i) + m \otimes \text{d}b' \]

if $\nabla (m) = \sum m_ i \otimes \text{d}b_ i$. If $\nabla $ is integrable, then so is $\nabla '$, and in this case there is a canonical map of de Rham complexes

55.6.11.1
\begin{equation} \label{crystalline-equation-base-change-map-complexes} M \otimes _ B \Omega ^\bullet \longrightarrow (M \otimes _ B B') \otimes _{B'} (\Omega ')^\bullet = M \otimes _ B (\Omega ')^\bullet \end{equation}

which maps $m \otimes \eta $ to $m \otimes \varphi (\eta )$.

Lemma 55.6.12. Let $A \to B$ be a ring map and let $(J, \delta )$ be a divided power structure on $B$. Let $p$ be a prime number. Assume that $A$ is a $\mathbf{Z}_{(p)}$-algebra and that $p$ is nilpotent in $B/J$. Then we have

\[ \mathop{\mathrm{lim}}\nolimits _ e \Omega _{B_ e/A, \bar\delta } = \mathop{\mathrm{lim}}\nolimits _ e \Omega _{B/A, \delta }/p^ e\Omega _{B/A, \delta } = \mathop{\mathrm{lim}}\nolimits _ e \Omega _{B^\wedge /A, \delta ^\wedge }/p^ e \Omega _{B^\wedge /A, \delta ^\wedge } \]

see proof for notation and explanation.

Proof. By Divided Power Algebra, Lemma 23.4.5 we see that $\delta $ extends to $B_ e = B/p^ eB$ for all sufficiently large $e$. Hence the first limit make sense. The lemma also produces a divided power structure $\delta ^\wedge $ on the completion $B^\wedge = \mathop{\mathrm{lim}}\nolimits _ e B_ e$, hence the last limit makes sense. By Lemma 55.6.2 and the fact that $\text{d}p^ e = 0$ (always) we see that the surjection $\Omega _{B/A, \delta } \to \Omega _{B_ e/A, \bar\delta }$ has kernel $p^ e\Omega _{B/A, \delta }$. Similarly for the kernel of $\Omega _{B^\wedge /A, \delta ^\wedge } \to \Omega _{B_ e/A, \bar\delta }$. Hence the lemma is clear. $\square$

[1] This actually makes sense: if $\Omega _ B$ is the module of differentials where we only assume the Leibniz rule and not the vanishing of $\text{d}1$, then the Leibniz rule gives $\text{d}1 = \text{d}(1 \cdot 1) = 1 \text{d}1 + 1 \text{d}1 = 2 \text{d}1$ and hence $\text{d}1 = 0$ in $\Omega _ B$.

Comments (5)

Comment #1467 by sdf on

line 966: ideal generate -> ideal generated

Comment #1603 by Rakesh Pawar on

In the definition 48.6.1 the compatibilty relation between gamma and theta should be between delta and theta as (B, J, delta) is the divided power ring in your notation.

Comment #3054 by Dingxin Zhang on

There are some paragraphs of Stacks Project containing (sometimes quite useful) comments, conventions, and notations that are not tagged.

The definition of "divided power differential" (between tags 07HR and 07HS) and the definition "powers of divided power ideals" (between 07HS and 07HT) are examples. (There are a lot of other instances in other chapters, I am sure.)

If one reads Stacks Project in a tag-by-tag fashion (e.g., using the "next-tag" feature to skip the long sections), these paragraphs could be missed. When these contain crucial information such as a definition, it could affect user experience; not to mention that one (you?) may at some point want to have references to them.

I don't know the infrastructure of Stacks Project website, but if it is based on reading "being-end" patterns in source .tex files to generate pages, one solution could be to enclose these paragraphs in some theorem environment without names and numbers, and then giving tags to these environments. It should not be hard to locate them in source files using suitable search method.

Comment #3055 by on

@3054: Dingxin, this is indeed a problem that needs to be addressed at some point. See comment #2460 for another instance and a bit of a discussion, and see also issue #67 on the Stacks Project infrastructure.

Your guess as to how the pages are generated is roughly correct. One issue I possibly have with the proposed solution is in decided which paragraphs intermediate to Lemmas, Propositions, etc. should be labelled and tagged. Fixing this now would require an enormous effort in tweaking the source code, and anyway, feels rather unnatural. In any case, if you have further ideas, please let's discuss!


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