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Proof. Set $D = D_ B(J)$ and denote $\bar J \subset D$ its divided power ideal with divided power structure $\bar\gamma $. The universal property of $D$ produces a $B$-algebra map $D \to D_{B'}(JB')$, whence a map as in the lemma. It suffices to show that the divided powers $\bar\gamma $ extend to $D \otimes _ B B'$ since then the universal property of $D_{B'}(JB')$ will produce a map $D_{B'}(JB') \to D \otimes _ B B'$ inverse to the one in the lemma.

Choose a surjection $P \to B'$ where $P$ is a polynomial algebra over $B$. In particular $B \to P$ is flat, hence $D \to D \otimes _ B P$ is flat by Algebra, Lemma 10.39.7. Then $\bar\gamma $ extends to $D \otimes _ B P$ by Divided Power Algebra, Lemma 23.4.2; we will denote this extension $\bar\gamma $ also. Set $\mathfrak a = \mathop{\mathrm{Ker}}(P \to B')$ so that we have the short exact sequence

Thus $\text{Tor}_1^ B(B', B/IB) = 0$ implies that $\mathfrak a \cap IP = I\mathfrak a$. Now we have the following commutative diagram

This diagram is exact even with $0$'s added at the top and the right. We have to show the divided powers on the ideal $\bar J \otimes _ B P$ preserve the ideal $\mathop{\mathrm{Im}}(\alpha ) \cap \bar J \otimes _ B P$, see Divided Power Algebra, Lemma 23.4.3. Consider the exact sequence

(which uses that $\mathfrak a \cap IP = I\mathfrak a$ as seen above). As $B'/IB'$ is flat over $B/IB$ this sequence remains exact after applying $B/J \otimes _{B/IB} -$, see Algebra, Lemma 10.39.12. Hence

is zero. Thus $\beta $ is injective. It follows that $\mathop{\mathrm{Im}}(\alpha ) \cap \bar J \otimes _ B P$ is the image of $\bar J \otimes \mathfrak a$. Now if $f \in \bar J$ and $a \in \mathfrak a$, then $\bar\gamma _ n(f \otimes a) = \bar\gamma _ n(f) \otimes a^ n$ hence the result is clear. $\square$