Lemma 60.5.5. In Situation 60.5.1. Let $P \to C$ be a surjection of $A$-algebras with kernel $J$. Write $D_{P, \gamma }(J) = (D, \bar J, \bar\gamma )$. Let $(D^\wedge , J^\wedge , \bar\gamma ^\wedge )$ be the $p$-adic completion of $D$, see Divided Power Algebra, Lemma 23.4.5. For every $e \geq 1$ set $P_ e = P/p^ eP$ and $J_ e \subset P_ e$ the image of $J$ and write $D_{P_ e, \gamma }(J_ e) = (D_ e, \bar J_ e, \bar\gamma )$. Then for all $e$ large enough we have

1. $p^ eD \subset \bar J$ and $p^ eD^\wedge \subset \bar J^\wedge$ are preserved by divided powers,

2. $D^\wedge /p^ eD^\wedge = D/p^ eD = D_ e$ as divided power rings,

3. $(D_ e, \bar J_ e, \bar\gamma )$ is an object of $\text{Cris}(C/A)$,

4. $(D^\wedge , \bar J^\wedge , \bar\gamma ^\wedge )$ is equal to $\mathop{\mathrm{lim}}\nolimits _ e (D_ e, \bar J_ e, \bar\gamma )$, and

5. $(D^\wedge , \bar J^\wedge , \bar\gamma ^\wedge )$ is an object of $\text{Cris}^\wedge (C/A)$.

Proof. Part (1) follows from Divided Power Algebra, Lemma 23.4.5. It is a general property of $p$-adic completion that $D/p^ eD = D^\wedge /p^ eD^\wedge$. Since $D/p^ eD$ is a divided power ring and since $P \to D/p^ eD$ factors through $P_ e$, the universal property of $D_ e$ produces a map $D_ e \to D/p^ eD$. Conversely, the universal property of $D$ produces a map $D \to D_ e$ which factors through $D/p^ eD$. We omit the verification that these maps are mutually inverse. This proves (2). If $e$ is large enough, then $p^ eC = 0$, hence we see (3) holds. Part (4) follows from Divided Power Algebra, Lemma 23.4.5. Part (5) is clear from the definitions. $\square$

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