The Stacks project

Lemma 23.4.5. Let $(A, I, \gamma )$ be a divided power ring. Let $p$ be a prime. If $p$ is nilpotent in $A/I$, then

  1. the $p$-adic completion $A^\wedge = \mathop{\mathrm{lim}}\nolimits _ e A/p^ eA$ surjects onto $A/I$,

  2. the kernel of this map is the $p$-adic completion $I^\wedge $ of $I$, and

  3. each $\gamma _ n$ is continuous for the $p$-adic topology and extends to $\gamma _ n^\wedge : I^\wedge \to I^\wedge $ defining a divided power structure on $I^\wedge $.

If moreover $A$ is a $\mathbf{Z}_{(p)}$-algebra, then

  1. for $e$ large enough the ideal $p^ eA \subset I$ is preserved by the divided power structure $\gamma $ and

    \[ (A^\wedge , I^\wedge , \gamma ^\wedge ) = \mathop{\mathrm{lim}}\nolimits _ e (A/p^ eA, I/p^ eA, \bar\gamma ) \]

    in the category of divided power rings.

Proof. Let $t \geq 1$ be an integer such that $p^ tA/I = 0$, i.e., $p^ tA \subset I$. The map $A^\wedge \to A/I$ is the composition $A^\wedge \to A/p^ tA \to A/I$ which is surjective (for example by Algebra, Lemma 10.96.1). As $p^ eI \subset p^ eA \cap I \subset p^{e - t}I$ for $e \geq t$ we see that the kernel of the composition $A^\wedge \to A/I$ is the $p$-adic completion of $I$. The map $\gamma _ n$ is continuous because

\[ \gamma _ n(x + p^ ey) = \sum \nolimits _{i + j = n} p^{je}\gamma _ i(x)\gamma _ j(y) = \gamma _ n(x) \bmod p^ eI \]

by the axioms of a divided power structure. It is clear that the axioms for divided power structures are inherited by the maps $\gamma _ n^\wedge $ from the maps $\gamma _ n$. Finally, to see the last statement say $e > t$. Then $p^ eA \subset I$ and $\gamma _1(p^ eA) \subset p^ eA$ and for $n > 1$ we have

\[ \gamma _ n(p^ ea) = p^ n \gamma _ n(p^{e - 1}a) = \frac{p^ n}{n!} p^{n(e - 1)}a^ n \in p^ e A \]

as $p^ n/n! \in \mathbf{Z}_{(p)}$ and as $n \geq 2$ and $e \geq 2$ so $n(e - 1) \geq e$. This proves that $\gamma $ extends to $A/p^ eA$, see Lemma 23.4.3. The statement on limits is clear from the construction of limits in the proof of Lemma 23.3.2. $\square$

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