The Stacks project

First proof. Let $M$ be a $D$-module. We claim that an $A$-derivation $\vartheta : B \to M$ is the same thing as a divided power $A$-derivation $\theta : D \to M$. The claim implies the statement by the Yoneda lemma.

Consider the square zero thickening $D \oplus M$ of $D$. There is a divided power structure $\delta $ on $\bar J \oplus M$ if we set the higher divided power operations zero on $M$. In other words, we set $\delta _ n(x + m) = \bar\gamma _ n(x) + \bar\gamma _{n - 1}(x)m$ for any $x \in \bar J$ and $m \in M$, see Lemma 60.3.1. Consider the $A$-algebra map $B \to D \oplus M$ whose first component is given by the map $B \to D$ and whose second component is $\vartheta $. By the universal property we get a corresponding homomorphism $D \to D \oplus M$ of divided power algebras whose second component is the divided power $A$-derivation $\theta $ corresponding to $\vartheta $. $\square$

Second 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 60.2.4 for notation and explanation. Write $\text{d} : D' \to \Omega _{D'/A, \delta }$ for the universal derivation. Note that

see Lemma 60.6.2. 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 60.6.2. 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

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 60.2.3. 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 60.6.2 and the description of $K'$ from Lemma 60.2.3. 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 10.131.9) we win. $\square$