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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 $A$-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 54.6.2. 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 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 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 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 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 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. $\square$

  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 $\Omega_B$.

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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}

Comments (5)

Comment #1467 by sdf on May 20, 2015 a 2:58 pm UTC

line 966: ideal generate -> ideal generated

Comment #1487 by Johan (site) on June 6, 2015 a 3:32 pm UTC

Thanks! Fixed here.

Comment #1603 by Rakesh Pawar on August 17, 2015 a 12:12 pm UTC

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 January 10, 2018 a 12:32 am UTC

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 Raymond Cheng (site) on January 10, 2018 a 4:35 pm UTC

@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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