hcju/stacksqa huggingface dataset:
Set $D = D_{B, \gamma}(J)$. Choose elements $f_t \in J$ which generate $J/I$. Set $\mathcal{R} = \{(r_0, r_t) \in I \oplus \bigoplus\nolimits_{t \in T} B \mid \sum r_t f_t = r_0 \text{ in }B\}$ as in the proof of Lemma \ref{lemma-describe-divided-power-envelope}. This lemma shows that $$ D = B\langle x_t \rangle/ K $$ where $K$ is generated by the elements $x_t - f_t$ and $\delta_n(\sum r_t x_t - r_0)$ for $(r_0, r_t) \in \mathcal{R}$. Thus we see that \begin{equation} \label{equation-base-change} D \otimes_B B' = B'\langle x_t \rangle/K' \end{equation} where $K'$ is generated by the images in $B'\langle x_t \rangle$ of the generators of $K$ listed above. Let $f'_t \in B'$ be the image of $f_t$. By assumption (1) we see that the elements $f'_t \in J'$ generate $J'/I'$ and we see that $x_t - f'_t \in K'$. Set $$ \mathcal{R}' = \{(r'_0, r'_t) \in I' \oplus \bigoplus\nolimits_{t \in T} B' \mid \sum r'_t f'_t = r'_0 \text{ in }B'\} $$ To finish the proof we have to show that $\delta'_n(\sum r'_t x_t - r'_0) \in K'$ for $(r'_0, r'_t) \in \mathcal{R}'$, because then the presentation (\ref{equation-base-change}) of $D \otimes_B B'$ is identical to the presentation of $D_{B', \gamma'}(J')$ obtain in Lemma \ref{lemma-describe-divided-power-envelope} from the generators $f'_t$. Suppose that $(r'_0, r'_t) \in \mathcal{R}'$. Then $\sum r'_t f'_t = 0$ in $B'/I'$. As $B/I \to B'/I'$ is flat by assumption (1) we can apply the equational criterion of flatness (Algebra, Lemma \ref{algebra-lemma-flat-eq}) to see that there exist an $m > 0$ and $r_{jt} \in B$ and $c_j \in B'$, $j = 1, \ldots, m$ such that $$ r_{j0} = \sum\nolimits_t r_{jt} f_t  \in I \text{ for } j = 1, \ldots, m $$ and $$ i'_t  = r'_t - \sum\nolimits_j c_j r_{jt} \in I' \text{ for all }t $$ Note that this also implies that $r'_0 = \sum_t i'_t f_t + \sum_j c_j r_{j0}$. Then we have \begin{align*} \delta'_n(\sum\nolimits_t r'_t x_t  - r'_0) & = \delta'_n( \sum\nolimits_t i'_t x_t + \sum\nolimits_{t, j} c_j r_{jt} x_t - \sum\nolimits_t i'_t f_t - \sum\nolimits_j c_j r_{j0}) \\ & = \delta'_n( \sum\nolimits_t i'_t(x_t - f_t) + \sum\nolimits_j c_j (\sum\nolimits_t r_{jt} x_t  - r_{j0})) \end{align*} Since $\delta_n(a + b) = \sum_{m = 0, \ldots, n} \delta_m(a) \delta_{n - m}(b)$ and since $\delta_m(\sum i'_t(x_t - f_t))$ is in the ideal generated by $x_t - f_t \in K'$ for $m > 0$, it suffices to prove that $\delta_n(\sum c_j (\sum r_{jt} x_t  - r_{j0}))$ is in $K'$. For this we use $$ \delta_n(\sum\nolimits_j c_j (\sum\nolimits_t r_{jt} x_t  - r_{j0})) = \sum c_1^{n_1} \ldots c_m^{n_m} \delta_{n_1}(\sum r_{1t} x_t  - r_{10}) \ldots \delta_{n_m}(\sum r_{mt} x_t  - r_{m0}) $$ where the sum is over $n_1 + \ldots + n_m = n$. This proves what we want.
We have to check conditions (1) -- (5) of Divided Power Algebra, Definition \ref{dpa-definition-divided-powers}. We will prove this directly for this case, but please see the proof of the next lemma for a method which avoids calculations. Conditions (1) and (3) are clear. Condition (2) follows from \begin{align*} \delta_n(x + z)\delta_m(x + z) & = (\gamma_n(x) + \gamma_{n - 1}(x)z)(\gamma_m(x) + \gamma_{m - 1}(x)z) \\ & = \gamma_n(x)\gamma_m(x) + \gamma_n(x)\gamma_{m - 1}(x)z + \gamma_{n - 1}(x)\gamma_m(x)z \\ & = \frac{(n + m)!}{n!m!} \gamma_{n + m}(x) + \left(\frac{(n + m - 1)!}{n!(m - 1)!} + \frac{(n + m - 1)!}{(n - 1)!m!}\right) \gamma_{n + m - 1}(x) z \\ & = \frac{(n + m)!}{n!m!}\delta_{n + m}(x + z) \end{align*} Condition (5) follows from \begin{align*} \delta_n(\delta_m(x + z)) & = \delta_n(\gamma_m(x) + \gamma_{m - 1}(x)z) \\ & = \gamma_n(\gamma_m(x)) + \gamma_{n - 1}(\gamma_m(x))\gamma_{m - 1}(x)z \\ & = \frac{(nm)!}{n! (m!)^n} \gamma_{nm}(x) + \frac{((n - 1)m)!}{(n - 1)! (m!)^{n - 1}} \gamma_{(n - 1)m}(x) \gamma_{m - 1}(x) z \\ & = \frac{(nm)!}{n! (m!)^n}(\gamma_{nm}(x) + \gamma_{nm - 1}(x) z) \end{align*} by elementary number theory. To prove (4) we have to see that $$ \delta_n(x + x' + z + z') = \gamma_n(x + x') + \gamma_{n - 1}(x + x')(z + z') $$ is equal to $$ \sum\nolimits_{i = 0}^n (\gamma_i(x) + \gamma_{i - 1}(x)z) (\gamma_{n - i}(x') + \gamma_{n - i - 1}(x')z') $$ This follows easily on collecting the coefficients of $1$, $z$, and $z'$ and using condition (4) for $\gamma$.
Suppose we want to prove that property (4) of Divided Power Algebra, Definition \ref{dpa-definition-divided-powers} is satisfied. Pick $(x, z, w)$ and $(x', z', w')$ in $J$. Pick a map $$ A_0 = \mathbf{Z}\langle s, s'\rangle \longrightarrow A,\quad s \longmapsto x, s' \longmapsto x' $$ which is possible by the universal property of divided power polynomial rings. Set $M_0 = A_0 \oplus A_0$ and $N_0 = A_0 \oplus A_0 \oplus M_0 \otimes_{A_0} M_0$. Let $q_0 : M_0 \times M_0 \to N_0$ be the obvious map. Define $M_0 \to M$ as the $A_0$-linear map which sends the basis vectors of $M_0$ to $z$ and $z'$. Define $N_0 \to N$ as the $A_0$ linear map which sends the first two basis vectors of $N_0$ to $w$ and $w'$ and uses $M_0 \otimes_{A_0} M_0 \to M \otimes_A M \xrightarrow{q} N$ on the last summand. Then we see that it suffices to prove the identity (4) for the situation $(A_0, M_0, N_0, q_0)$. Similarly for the other identities. This reduces us to the case of a $\mathbf{Z}$-torsion free ring $A$ and $A$-torsion free modules. In this case all we have to do is show that $$ n! \delta_n(x, z, w) = (x, z, w)^n $$ in the ring $A$, see Divided Power Algebra, Lemma \ref{dpa-lemma-silly}. To see this note that $$ (x, z, w)^2 = (x^2, 2xz, 2xw + 2q(z, z)) $$ and by induction $$ (x, z, w)^n = (x^n, nx^{n - 1}z, nx^{n - 1}w + n(n - 1)x^{n - 2}q(z, z)) $$ On the other hand, $$ n! \delta_n(x, z, w) = (n!\gamma_n(x), n!\gamma_{n - 1}(x)z, n!\gamma_{n - 1}(x)w + n!\gamma_{n - 2}(x) q(z, z)) $$ which matches. This finishes the proof.
The empty product is $(C, 0, \emptyset)$. If $(B_t, J_t, \delta_t)$ is a family of objects of $\text{CRIS}(C/A)$ then we can form the product $(\prod B_t, \prod J_t, \prod \delta_t)$ as in Divided Power Algebra, Lemma \ref{dpa-lemma-colimits}. The map $C \to \prod B_t/\prod J_t = \prod B_t/J_t$ is clear. \medskip\noindent Given two objects $(B, J, \gamma)$ and $(B', J', \gamma')$ of $\text{CRIS}(C/A)$ we can form a cocartesian diagram $$ \xymatrix{ (B, J, \delta) \ar[r] & (B'', J'', \delta'') \\ (A, I, \gamma) \ar[r] \ar[u] & (B', J', \delta') \ar[u] } $$ in the category of divided power rings. Then we see that we have $$ B''/J'' = B/J \otimes_{A/I} B'/J' \longleftarrow C \otimes_{A/I} C $$ see Divided Power Algebra, Remark \ref{dpa-remark-forgetful}. Denote $J'' \subset K \subset B''$ the ideal such that $$ \xymatrix{ B''/J'' \ar[r] & B''/K \\ C \otimes_{A/I} C \ar[r] \ar[u] & C \ar[u] } $$ is a pushout, i.e., $B''/K \cong B/J \otimes_C B'/J'$. Let $D_{B''}(K) = (D, \bar K, \bar \delta)$ be the divided power envelope of $K$ in $B''$ relative to $(B'', J'', \delta'')$. Then it is easily verified that $(D, \bar K, \bar \delta)$ is a coproduct of $(B, J, \delta)$ and $(B', J', \delta')$ in $\text{CRIS}(C/A)$. \medskip\noindent Next, we come to coequalizers. Let $\alpha, \beta : (B, J, \delta) \to (B', J', \delta')$ be morphisms of $\text{CRIS}(C/A)$. Consider $B'' = B'/ (\alpha(b) - \beta(b))$. Let $J'' \subset B''$ be the image of $J'$. Let $D_{B''}(J'') = (D, \bar J, \bar\delta)$ be the divided power envelope of $J''$ in $B''$ relative to $(B', J', \delta')$. Then it is easily verified that $(D, \bar J, \bar \delta)$ is the coequalizer of $(B, J, \delta)$ and $(B', J', \delta')$ in $\text{CRIS}(C/A)$. \medskip\noindent By Categories, Lemma \ref{categories-lemma-almost-finite-colimits-exist} we have all finite nonempty colimits in $\text{CRIS}(C/A)$. The constructions above shows that (\ref{equation-forget-affine}) commutes with them. This formally implies part (3) as $\text{Cris}(C/A)$ is the fibre category of (\ref{equation-forget-affine}) over $C$.
Part (1) follows from Divided Power Algebra, Lemma \ref{dpa-lemma-extend-to-completion}. 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 \ref{dpa-lemma-extend-to-completion}. Part (5) is clear from the definitions.
The fact that $K \cap J(1) \subset J(1)$ is preserved by the divided power structure follows from the fact that $B(1) \to B$ is a homomorphism of divided power rings. \medskip\noindent Recall that $K/K^2$ has a canonical $B$-module structure. Denote $s_0, s_1 : B \to B(1)$ the two coprojections and consider the map $\text{d} : B \to K/K^2 +(K \cap J(1))^{[2]}$ given by $b \mapsto s_1(b) - s_0(b)$. It is clear that $\text{d}$ is additive, annihilates $A$, and satisfies the Leibniz rule. We claim that $\text{d}$ is a divided power $A$-derivation. Let $x \in J$. Set $y = s_1(x)$ and $z = s_0(x)$. Denote $\delta$ the divided power structure on $J(1)$. We have to show that $\delta_n(y) - \delta_n(z) = \delta_{n - 1}(y)(y - z)$ modulo $K^2 +(K \cap J(1))^{[2]}$ for $n \geq 1$. The equality holds for $n = 1$. Assume $n > 1$. Note that $\delta_i(y - z)$ lies in $(K \cap J(1))^{[2]}$ for $i > 1$. Calculating modulo $K^2 + (K \cap J(1))^{[2]}$ we have $$ \delta_n(z) = \delta_n(z - y + y) = \sum\nolimits_{i = 0}^n \delta_i(z - y)\delta_{n - i}(y) = \delta_{n - 1}(y) \delta_1(z - y) + \delta_n(y) $$ This proves the desired equality. \medskip\noindent Let $M$ be a $B$-module. Let $\theta : B \to M$ be a divided power $A$-derivation. Set $D = B \oplus M$ where $M$ is an ideal of square zero. Define a divided power structure on $J \oplus M \subset D$ by setting $\delta_n(x + m) = \delta_n(x) + \delta_{n - 1}(x)m$ for $n > 1$, see Lemma \ref{lemma-divided-power-first-order-thickening}. There are two divided power algebra homomorphisms $B \to D$: the first is given by the inclusion and the second by the map $b \mapsto b + \theta(b)$. Hence we get a canonical homomorphism $B(1) \to D$ of divided power algebras over $(A, I, \gamma)$. This induces a map $K \to M$ which annihilates $K^2$ (as $M$ is an ideal of square zero) and $(K \cap J(1))^{[2]}$ as $M^{[2]} = 0$. The composition $B \to K/K^2 + (K \cap J(1))^{[2]} \to M$ equals $\theta$ by construction. It follows that $\text{d}$ is a universal divided power $A$-derivation and we win.
Word for word the same as the proof of Lemma \ref{lemma-diagonal-and-differentials}. The only point that has to be checked is that the divided power ring $D = B \oplus M$ is an object of $\text{CRIS}(C/A)$ and that the two maps $B \to C$ are morphisms of $\text{CRIS}(C/A)$. Since $D/(J \oplus M) = B/J$ we can use $C \to B/J$ to view $D$ as an object of $\text{CRIS}(C/A)$ and the statement on morphisms is clear from the construction.
[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. \medskip\noindent 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 \ref{lemma-divided-power-first-order-thickening}. 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 $\vartheta$.
By Divided Power Algebra, Lemma \ref{dpa-lemma-extend-to-completion} we see that $\delta$ extends to $B_e = B/p^eB$ for all sufficiently large $e$. Hence the first limit make sense. The lemma also produces a divided power structure $\delta^\wedge$ on the completion $B^\wedge = \lim_e B_e$, hence the last limit makes sense. By Lemma \ref{lemma-omega} and the fact that $\text{d}p^e = 0$ (always) we see that the surjection $\Omega_{B/A, \delta} \to \Omega_{B_e/A, \bar\delta}$ has kernel $p^e\Omega_{B/A, \delta}$. Similarly for the kernel of  $\Omega_{B^\wedge/A, \delta^\wedge} \to \Omega_{B_e/A, \bar\delta}$. Hence the lemma is clear.
Omitted. Hints: If $T$ exists, then $T_0 = S_0 \times_{S'_0} U'$ (argue as in Divided Power Algebra, Remark \ref{dpa-remark-forgetful}). Since $T'$ is a divided power thickening, we see that $T$ (if it exists) will be a divided power thickening too. Hence we can define $T$ as the scheme with underlying topological space the underlying topological space of $T_0 = S_0 \times_{S'_0} U'$ and as structure sheaf on affine pieces the ring given by Lemma \ref{lemma-affine-thickenings-colimits}.
For the first assertion see Lemma \ref{lemma-divided-power-thickening-fibre-products}. This gives us a morphism of topoi $i : (X/S)_{\text{cris}} \to (X/S)_{\text{CRIS}}$ and a left adjoint $i_!$ such that $i^{-1}i_! = i^{-1}i_* = \text{id}$, see Sites, Lemmas \ref{sites-lemma-when-shriek}, \ref{sites-lemma-preserve-equalizers}, and \ref{sites-lemma-back-and-forth}. We claim that $i_!$ is exact. If this is true, then we can define $\pi$ by the rules $\pi^{-1} = i_!$ and $\pi_* = i^{-1}$ and everything is clear. To prove the claim, note that we already know that $i_!$ is right exact and preserves fibre products (see references given). Hence it suffices to show that $i_! * = *$ where $*$ indicates the final object in the category of sheaves of sets.  To see this it suffices to produce a set of objects $(U_i, T_i, \delta_i)$, $i \in I$ of $\text{Cris}(X/S)$ such that $$ \coprod\nolimits_{i \in I} h_{(U_i, T_i, \delta_i)} \to * $$ is surjective in $(X/S)_{\text{CRIS}}$ (details omitted; hint: use that $\text{Cris}(X/S)$ has products and that the functor $\text{Cris}(X/S) \to \text{CRIS}(X/S)$ commutes with them). In the affine case this follows from Lemma \ref{lemma-set-generators}. We omit the proof in general.
The fully faithful functor comes from thinking of objects of $\text{Cris}(X'/S')$ as divided power thickenings $(U, T, \delta)$ of $X$ where $U \to X$ factors through $X' \subset X$ (since then automatically $T \to S$ will factor through $S'$). This functor is clearly cocontinuous hence we obtain a morphism of topoi as indicated. Let $h_{X'} \in \Sh(X_{Zar})$ be the representable sheaf associated to $X'$ viewed as an object of $X_{Zar}$. It is clear that $\Sh(X'_{Zar})$ is the localization $\Sh(X_{Zar})/h_{X'}$. On the other hand, the category $\text{Cris}(X/S)/u_{X/S}^{-1}h_{X'}$ (see Sites, Lemma \ref{sites-lemma-localize-topos-site}) is canonically identified with $\text{Cris}(X'/S')$ by the functor above. This finishes the proof.
Note that $U = U(1)$ as $U \to X$ is an open immersion and as (\ref{equation-forget-small}) commutes with products. Hence we see that $\mathcal{K} \subset \mathcal{J}_{T(1)}$. Given this fact the lemma follows by working affine locally on $T$ and using Lemmas \ref{lemma-module-of-differentials-on-affine} and \ref{lemma-diagonal-and-differentials-affine-site}.
Part (1) follows from a combination of Lemmas \ref{lemma-module-differentials-divided-power-scheme} and \ref{lemma-module-of-differentials}. Part (2) follows from the fact that $(\Omega_{X/S})_T = \Omega_{T/S, \delta}$ is a quotient of $\Omega_{T/S}$ and that $f^*\Omega_{T'/S} \to \Omega_{T/S}$ is surjective.
As $U \to T$ is a thickening we see that $U$ is affine if and only if $T$ is affine, see Limits, Lemma \ref{limits-lemma-affine}. Having said this, let us apply Cohomology on Sites, Lemma \ref{sites-cohomology-lemma-cech-vanish-collection} to the collection $\mathcal{B}$ of affine objects $(U, T, \delta)$ and the collection $\text{Cov}$ of affine open coverings $\mathcal{U} = \{(U_i, T_i, \delta_i) \to (U, T, \delta)\}$. The {\v C}ech complex ${\check C}^*(\mathcal{U}, \mathcal{F})$ for such a covering is simply the {\v C}ech complex of the quasi-coherent $\mathcal{O}_T$-module $\mathcal{F}_T$ (here we are using the assumption that $\mathcal{F}$ is locally quasi-coherent) with respect to the affine open covering $\{T_i \to T\}$ of the affine scheme $T$. Hence the {\v C}ech cohomology is zero by Cohomology of Schemes, Lemma \ref{coherent-lemma-cech-cohomology-quasi-coherent} and \ref{coherent-lemma-quasi-coherent-affine-cohomology-zero}. Thus the hypothesis of Cohomology on Sites, Lemma \ref{sites-cohomology-lemma-cech-vanish-collection} are satisfied and we win.
We will use without further mention that $\mathcal{C}$ and $\text{Cris}(X/S)$ have products and fibre products, see Lemma \ref{lemma-divided-power-thickening-fibre-products}. Note that the inclusion functor $u : \mathcal{C} \to \text{Cris}(X/S)$ is fully faithful, continuous and commutes with products and fibre products. We claim it defines a morphism of ringed sites $$ f : (\text{Cris}(X/S), \mathcal{O}_{X/S}) \longrightarrow (\Sh(\mathcal{C}), \mathcal{O}_{X/S}|_\mathcal{C}) $$ To see this we will use Sites, Lemma \ref{sites-lemma-directed-morphism}. Note that $\mathcal{C}$ has fibre products and $u$ commutes with them so the categories $\mathcal{I}^u_{(U, T, \delta)}$ are disjoint unions of directed categories (by Sites, Lemma \ref{sites-lemma-almost-directed} and Categories, Lemma \ref{categories-lemma-split-into-directed}). Hence it suffices to show that $\mathcal{I}^u_{(U, T, \delta)}$ is connected. Nonempty follows from Lemma \ref{lemma-set-generators} and connectedness follows from the fact that $\mathcal{C}$ has products and that $u$ commutes with them (compare with the proof of Sites, Lemma \ref{sites-lemma-directed}). \medskip\noindent Note that $f_*\mathcal{F} = \mathcal{F}|_\mathcal{C}$. Hence the lemma follows if $R^pf_*\mathcal{F} = 0$ for $p > 0$, see Cohomology on Sites, Lemma \ref{sites-cohomology-lemma-apply-Leray}. By Cohomology on Sites, Lemma \ref{sites-cohomology-lemma-higher-direct-images} it suffices to show that $H^p((X, T, \delta), \mathcal{F}) = 0$ for all $(X, T, \delta)$. This follows from Lemma \ref{lemma-vanishing-lqc}.
Note that $H^q(X^p, \mathcal{F}) = 0$ for all $q > 0$ as any presheaf is a sheaf on $\mathcal{C}$. The assumption on $X$ is that $h_X \to *$ is surjective. Using that $H^q(X, \mathcal{F}) = H^q(h_X, \mathcal{F})$ and $H^q(\mathcal{C}, \mathcal{F}) = H^q(*, \mathcal{F})$ we see that our statement is a special case of Cohomology on Sites, Lemma \ref{sites-cohomology-lemma-cech-to-cohomology-sheaf-sets}.
We are going to use the principle of Simplicial, Lemma \ref{simplicial-lemma-functorial-homotopy} and more specifically Lemma \ref{lemma-homotopy-tensor} which tells us that homotopic maps between (co)simplicial objects are transformed by any functor into homotopic maps. The complex of the lemma is equal to the $p$-adic completion of the base change of the cosimplicial module $$ M_* = \left( \Omega_{P/A} \to \Omega_{P \otimes_A P/A} \to \Omega_{P \otimes_A P \otimes_A P/A} \to \ldots \right) $$ via the cosimplicial ring map $P\otimes_A \ldots \otimes_A P \to D(n)$. This follows from Lemma \ref{lemma-module-differentials-divided-power-envelope}, see comments following (\ref{equation-omega-D}). Hence it suffices to show that the cosimplicial module $M_*$ is homotopic to zero (uses base change and $p$-adic completion). We can even assume $A = \mathbf{Z}$ and $P = \mathbf{Z}[\{x_i\}_{i \in I}]$ as we can use base change with $\mathbf{Z} \to A$. In this case $P^{\otimes n + 1}$ is the polynomial algebra on the elements $$ x_i(e) = 1 \otimes \ldots \otimes x_i \otimes \ldots \otimes 1 $$ with $x_i$ in the $e$th slot. The modules of the complex are free on the generators $\text{d}x_i(e)$. Note that if $f : [n] \to [m]$ is a map then we see that $$ M_*(f)(\text{d}x_i(e)) = \text{d}x_i(f(e)) $$ Hence we see that $M_*$ is a direct sum over $I$ of copies of the module studied in Example \ref{example-cosimplicial-module} and we win.
By Lemma \ref{lemma-vanishing-omega-1} the endomorphisms $0$ and $1$ of $\Omega_{D(*)}$ are homotopic. If we apply the functor $\wedge^i$ we see that the same is true for the cosimplicial module $\wedge^i\Omega_{D(*)}$, see Lemma \ref{lemma-homotopy-tensor}. Another application of the same lemma shows the $p$-adic completion $\Omega^i_{D(*)}$ is homotopy equivalent to zero. Tensoring with $M_*$ we see that $M_* \otimes_{D(*)} \Omega^i_{D(*)}$ is homotopic to zero, see Lemma \ref{lemma-homotopy-tensor} again. A final application of the $p$-adic completion functor finishes the proof.
It suffices to show that the complex $$ E : (0 \to A \to P \to \Omega^1_{P/A, \delta} \to \Omega^2_{P/A, \delta} \to \ldots) $$ is homotopy equivalent to zero as a complex of $A$-modules. For every multi-index $K = (k_i)$ we can consider the subcomplex $E(K)$ which in degree $j$ consists of $$ \bigoplus\nolimits_{I = \{i_1, \ldots, i_j\} \subset \text{Supp}(K)} A \prod\nolimits_{i \not \in I} x_i^{[k_i]} \prod\nolimits_{i \in I} x_i^{[k_i - 1]} \text{d}x_{i_1} \wedge \ldots \wedge \text{d}x_{i_j} $$ Since $E = \bigoplus E(K)$ we see that it suffices to prove each of the complexes $E(K)$ is homotopic to zero. If $K = 0$, then $E(K) : (A \to A)$ is homotopic to zero. If $K$ has nonempty (finite) support $S$, then the complex $E(K)$ is isomorphic to the complex $$ 0 \to A \to \bigoplus\nolimits_{s \in S} A \to \wedge^2(\bigoplus\nolimits_{s \in S} A) \to \ldots \to \wedge^{\# S}(\bigoplus\nolimits_{s \in S} A) \to 0 $$ which is homotopic to zero, for example by More on Algebra, Lemma \ref{more-algebra-lemma-homotopy-koszul-abstract}.
Consider the decreasing filtration $F^*$ on $\Omega^*_{B/A, \delta}$ given by the subcomplexes $F^i(\Omega^*_{B/A, \delta}) = \sigma_{\geq i}\Omega^*_{B/A, \delta}$. See Homology, Section \ref{homology-section-truncations}. This induces a decreasing filtration $F^*$ on $\Omega^*_{P/A, \delta}$ by setting $$ F^i(\Omega^*_{P/A, \delta}) = F^i(\Omega^*_{B/A, \delta}) \wedge \Omega^*_{P/A, \delta}. $$ We have a split short exact sequence $$ 0 \to \Omega^1_{B/A, \delta} \otimes_B P \to \Omega^1_{P/A, \delta} \to \Omega^1_{P/B, \delta} \to 0 $$ and the last module is free on $\text{d}x_i$. It follows from this that $F^i(\Omega^*_{P/A, \delta}) \to \Omega^*_{P/A, \delta}$ is a termwise split injection and that $$ \text{gr}^i_F(\Omega^*_{P/A, \delta}) = \Omega^i_{B/A, \delta} \otimes_B \Omega^*_{P/B, \delta} $$ as complexes. Thus we can define a filtration $F^*$ on $M \otimes_B \Omega^*_{B/A, \delta}$ by setting $$ F^i(M \otimes_B \Omega^*_{P/A, \delta}) = M \otimes_B F^i(\Omega^*_{P/A, \delta}) $$ and we have $$ \text{gr}^i_F(M \otimes_B \Omega^*_{P/A, \delta}) = M \otimes_B \Omega^i_{B/A, \delta} \otimes_B \Omega^*_{P/B, \delta} $$ as complexes. By Lemma \ref{lemma-poincare} each of these complexes is quasi-isomorphic to $M \otimes_B \Omega^i_{B/A, \delta}$ placed in degree $0$. Hence we see that the first displayed map of the lemma is a morphism of filtered complexes which induces a quasi-isomorphism on graded pieces. This implies that it is a quasi-isomorphism, for example by the spectral sequence associated to a filtered complex, see Homology, Section \ref{homology-section-filtered-complex}. \medskip\noindent The proof of the second quasi-isomorphism is exactly the same.