## 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.

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 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.
$\square$

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.
$\square$

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.
$\square$

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.
$\square$

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