## Tag `07HQ`

## 54.6. Module of differentials

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

Definition 54.6.1. 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 $A$-derivationinto $M$ 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 54.6.2. Let $A$ be a ring. Let $(B, J, \delta)$ be a divided power ring and $A \to B$ a ring map.

- Consider $B[x]$ with divided power ideal $(JB[x], \delta')$ where $\delta'$ is the extension of $\delta$ to $B[x]$. Then $$ \Omega_{B[x]/A, \delta'} = \Omega_{B/A, \delta} \otimes_B B[x] \oplus B[x]\text{d}x. $$
- Consider $B\langle x \rangle$ with divided power ideal $(JB\langle x \rangle + B\langle x \rangle_{+}, \delta')$. Then $$ \Omega_{B\langle x\rangle/A, \delta'} = \Omega_{B/A, \delta} \otimes_B B\langle x \rangle \oplus B\langle x\rangle \text{d}x. $$
- 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

- $\text{d}(b + b') = \text{d}b + \text{d}b'$, $b, b' \in B$,
- $\text{d}a = 0$, $a \in A$,
- $\text{d}(bb') = b \text{d}b' + b' \text{d}b$, $b, b' \in B$,
- $\text{d}\delta_n(f) = \delta_{n - 1}(f)\text{d}f$, $f \in J$, $n > 1$.
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 54.6.3. 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 54.3.1. 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 54.6.4. 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 54.6.3 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 54.6.5. In Situation 54.5.1. 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 54.6.3. 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 54.6.6. 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 54.2.4 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 54.6.2. 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 54.6.2. 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 54.2.3. 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 54.6.2 and the description of $K'$ from Lemma 54.2.3. 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 54.6.7. Let $B$ be a ring. Write $\Omega_B = \Omega_{B/\mathbf{Z}}$ for the absolute

^{1}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 absolutede 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 54.6.8. 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 54.6.7. 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 54.6.9. 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 54.6.8 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 54.6.10. Let $B$ be a ring. Let $\Omega_B \to \Omega$ be a quotient satisfying the assumptions of Lemma 54.6.8. Let $M$ be a $B$-module. A

connectionis 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 isintegrableif 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 54.6.11. 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 54.6.8. 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 \begin{equation} \tag{54.6.11.1} 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 54.6.12. 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 54.6.2 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. $\square$

- 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 $\Omega_B$. ↑

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\section{Module of differentials}
\label{section-differentials}
\noindent
In this section we develop a theory of modules of differentials
for divided power rings.
\begin{definition}
\label{definition-derivation}
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 {\it 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 $x \in J$.
\end{definition}
\noindent
In the situation of the definition, just as in the case of usual
derivations, there exists a {\it 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.
\begin{lemma}
\label{lemma-omega}
Let $A$ be a ring. Let $(B, J, \delta)$ be a divided power ring and
$A \to B$ a ring map.
\begin{enumerate}
\item Consider $B[x]$ with divided power ideal $(JB[x], \delta')$
where $\delta'$ is the extension of $\delta$ to $B[x]$. Then
$$
\Omega_{B[x]/A, \delta'} =
\Omega_{B/A, \delta} \otimes_B B[x] \oplus B[x]\text{d}x.
$$
\item Consider $B\langle x \rangle$ with divided power ideal
$(JB\langle x \rangle + B\langle x \rangle_{+}, \delta')$. Then
$$
\Omega_{B\langle x\rangle/A, \delta'} =
\Omega_{B/A, \delta} \otimes_B B\langle x \rangle \oplus
B\langle x\rangle \text{d}x.
$$
\item 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 $k \in K$.
\end{enumerate}
\end{lemma}
\begin{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
\begin{enumerate}
\item $\text{d}(b + b') = \text{d}b + \text{d}b'$, $b, b' \in B$,
\item $\text{d}a = 0$, $a \in A$,
\item $\text{d}(bb') = b \text{d}b' + b' \text{d}b$, $b, b' \in B$,
\item $\text{d}\delta_n(f) = \delta_{n - 1}(f)\text{d}f$, $f \in J$, $n > 1$.
\end{enumerate}
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
$\text{d}x^{[n]} = x^{[n - 1]}\text{d}x$.
\end{proof}
\noindent
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 $I^{[0]} = A$.
\begin{lemma}
\label{lemma-diagonal-and-differentials}
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 $K = \Ker(B(1) \to B)$.
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.
\end{lemma}
\begin{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.
\medskip\noindent
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.
\medskip\noindent
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 \ref{lemma-divided-power-first-order-thickening}.
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.
\end{proof}
\begin{remark}
\label{remark-filtration-differentials}
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 \ref{lemma-diagonal-and-differentials}
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.
\end{remark}
\begin{lemma}
\label{lemma-diagonal-and-differentials-affine-site}
In Situation \ref{situation-affine}.
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
$K = \Ker(B(1) \to B)$. 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.
\end{lemma}
\begin{proof}
Word for word the same as the proof of
Lemma \ref{lemma-diagonal-and-differentials}.
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.
\end{proof}
\begin{lemma}
\label{lemma-module-differentials-divided-power-envelope}
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
$$
\Omega_{D/A, \bar\gamma} = \Omega_{B/A} \otimes_B D
$$
\end{lemma}
\begin{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 \ref{dpa-lemma-gamma-extends}.
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 \ref{lemma-describe-divided-power-envelope} 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 \ref{lemma-omega}. 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 \ref{lemma-omega}.
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 $A$.
\medskip\noindent
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 \ref{lemma-divided-power-envelop-quotient}.
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 \ref{lemma-omega} and the description of $K'$ from
Lemma \ref{lemma-divided-power-envelop-quotient}. 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 $k \in \Ker(P \to B)$ and since $\Omega_{B/A}$
is the quotient of $\Omega_{P/A} \otimes_A B$ by these elements
(Algebra, Lemma \ref{algebra-lemma-differential-seq}) we win.
\end{proof}
\begin{remark}
\label{remark-absolute-de-rham-complex}
Let $B$ be a ring. Write $\Omega_B = \Omega_{B/\mathbf{Z}}$
for the absolute\footnote{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 $\Omega_B$.} 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 \ref{algebra-section-tensor-algebra}.
The absolute {\it 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.
\end{remark}
\begin{lemma}
\label{lemma-de-rham-complex}
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
$$
\text{d} : \Omega^p \to \Omega^{p + 1},\quad
\text{d}\left(f_0\text{d}f_1 \wedge \ldots \wedge \text{d}f_p\right) =
\text{d}f_0 \wedge \text{d}f_1 \wedge \ldots \wedge \text{d}f_p
$$
\end{lemma}
\begin{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 \ref{remark-absolute-de-rham-complex}.
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 $\wedge^2\pi$.
\medskip\noindent
For $i > 1$ we note that $\wedge^i \pi$ is surjective with
kernel the image of $\Ker(\pi) \wedge \Omega^{i - 1}_B
\to \Omega_B^i$. For $\omega_1 \in \Ker(\pi)$ 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.
\end{proof}
\begin{remark}
\label{remark-divided-powers-de-rham-complex}
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 \ref{lemma-de-rham-complex} 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
{\it 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.
\end{remark}
\begin{remark}
\label{remark-connection}
Let $B$ be a ring. Let $\Omega_B \to \Omega$ be a quotient satisfying
the assumptions of Lemma \ref{lemma-de-rham-complex}.
Let $M$ be a $B$-module. A {\it 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 {\it 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.
\end{remark}
\begin{remark}
\label{remark-base-change-connection}
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 \ref{lemma-de-rham-complex}. 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 {\it 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
\begin{equation}
\label{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 $m \otimes \varphi(\eta)$.
\end{remark}
\begin{lemma}
\label{lemma-differentials-completion}
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
$$
\lim_e \Omega_{B_e/A, \bar\delta} =
\lim_e \Omega_{B/A, \delta}/p^e\Omega_{B/A, \delta} =
\lim_e \Omega_{B^\wedge/A, \delta^\wedge}/p^e \Omega_{B^\wedge/A, \delta^\wedge}
$$
see proof for notation and explanation.
\end{lemma}
\begin{proof}
By Divided Power Algebra, Lemma \ref{dpa-lemma-extend-to-completion}
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 $B^\wedge = \lim_e B_e$, hence the last limit makes
sense. By Lemma \ref{lemma-omega}
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.
\end{proof}
```

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