Lemma 60.2.3. Let $(A, I, \gamma )$ be a divided power ring. Let $\varphi : B' \to B$ be a surjection of $A$-algebras with kernel $K$. Let $IB \subset J \subset B$ be an ideal. Let $J' \subset B'$ be the inverse image of $J$. Write $D_{B', \gamma }(J') = (D', \bar J', \bar\gamma )$. Then $D_{B, \gamma }(J) = (D'/K', \bar J'/K', \bar\gamma )$ where $K'$ is the ideal generated by the elements $\bar\gamma _ n(k)$ for $n \geq 1$ and $k \in K$.

**Proof.**
Write $D_{B, \gamma }(J) = (D, \bar J, \bar\gamma )$. The universal property of $D'$ gives us a homomorphism $D' \to D$ of divided power algebras. As $B' \to B$ and $J' \to J$ are surjective, we see that $D' \to D$ is surjective (see remarks above). It is clear that $\bar\gamma _ n(k)$ is in the kernel for $n \geq 1$ and $k \in K$, i.e., we obtain a homomorphism $D'/K' \to D$. Conversely, there exists a divided power structure on $\bar J'/K' \subset D'/K'$, see Divided Power Algebra, Lemma 23.4.3. Hence the universal property of $D$ gives an inverse $D \to D'/K'$ and we win.
$\square$

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