Proof. Let $\bar{M}=M\otimes _{A}R$ . There is an exact sequence of finite projective $R$ -modules

(3.4) $$\begin{eqnarray}\bar{M}^{\unicode[STIX]{x1D70E}}\xrightarrow[{}]{\bar{\unicode[STIX]{x1D711}}}\bar{M}\xrightarrow[{}]{\bar{\unicode[STIX]{x1D713}}}\bar{M}^{\unicode[STIX]{x1D70E}}\xrightarrow[{}]{\bar{\unicode[STIX]{x1D711}}}\bar{M},\end{eqnarray}$$

and we have to show that $\operatorname{Im}(\bar{\unicode[STIX]{x1D713}})$ is a direct summand of $\bar{M}^{\unicode[STIX]{x1D70E}}$ . This holds if and only if for each maximal ideal $\mathfrak{m}\subset R$ the base change of (3.4) to $k=R/\mathfrak{m}$ is exact, or equivalently if the base change to $k^{\operatorname{per}}$ is exact.

Let $\unicode[STIX]{x1D6E5}:A\rightarrow W(A)$ be the homomorphism with $w_{n}\circ \unicode[STIX]{x1D6E5}=\unicode[STIX]{x1D70E}^{n}$ , where $w_{n}$ is the $n$ th Witt polynomial; see [Reference BourbakiBou83, IX, § 1.2, Proposition 2]. The composition of $\unicode[STIX]{x1D6E5}$ with the homomorphism $W(A)\rightarrow W(R)\rightarrow W(k^{\operatorname{per}})$ is a homomorphism $A\rightarrow W(k^{\operatorname{per}})$ that commutes with $\unicode[STIX]{x1D70E}$ . Then $M\otimes _{A}W(k^{\operatorname{per}})$ is a Dieudonné module whose reduction mod $p$ is (3.4) $\otimes _{R}\,k^{\operatorname{per}}$ , which is therefore exact as required. ☐

Proof. Since the ideal $J_{0}$ is finitely generated, the $J_{0}$ -adic topology of $R_{0}$ coincides with the linear topology defined by the ideals $\unicode[STIX]{x1D719}^{n}(J_{0})$ for $n\geqslant 0$ . Thus $R^{\flat }$ is the $J_{0}$ -adic completion of $R_{0}$ . Then the assertion is clear.☐

Proof. Clearly $\unicode[STIX]{x1D719}(\operatorname{Ker}(\unicode[STIX]{x1D6FC}))\subseteq \operatorname{Ker}(\unicode[STIX]{x1D6FC})^{p}$ . To prove the opposite inclusion, let $x_{1},\ldots ,x_{p}\in \operatorname{Ker}(\unicode[STIX]{x1D6FC})$ be given, and choose $y_{i}\in R^{\prime }$ with $\unicode[STIX]{x1D719}(y_{i})=x_{i}$ . Then $\unicode[STIX]{x1D6FC}(y_{i})\in \bar{J}=\operatorname{Ker}(\unicode[STIX]{x1D719}:R\rightarrow R)$ . Since $R$ is balanced we have $\unicode[STIX]{x1D6FC}(\prod y_{i})=0$ , thus $\prod x_{i}=\unicode[STIX]{x1D719}(\prod y_{i})\in \unicode[STIX]{x1D719}(\operatorname{Ker}(\unicode[STIX]{x1D6FC}))$ as required.☐

Proof. If $R$ is balanced then $J^{p}=\unicode[STIX]{x1D719}(J)$ by Lemma 4.5. The rest is clear.☐

Proof. The composition $\unicode[STIX]{x1D6FC}:R^{\prime }\xrightarrow[{}]{\unicode[STIX]{x1D70B}}R\rightarrow R^{\operatorname{bal}}$ is an isogeny since $R$ is iso-balanced. Lemma 4.5 implies that $\operatorname{Ker}(\unicode[STIX]{x1D6FC})^{p^{n}}=\unicode[STIX]{x1D719}^{n}(\operatorname{Ker}(\unicode[STIX]{x1D6FC}))$ , which is zero for large $n$ .☐

Proof. If $R$ is a complete intersection, $J$ is generated by a regular sequence $(u_{1},\ldots ,u_{r})$ ; see Lemma 4.2. Let $J^{\prime }=([u_{1}],\ldots ,[u_{r}])$ in $W(R^{\flat })$ . The ring $A=W(R^{\flat })/J^{\prime }$ is $p$ -adically complete and $p$ -torsion free with $A/pA=R$ . The ideal $J^{\prime }$ is stable under $\unicode[STIX]{x1D70E}$ since $\unicode[STIX]{x1D70E}([u_{i}])=[u_{i}]^{p}$ . Thus $A$ is a straight lift of $R$ .

Assume that $R$ is balanced. Let $J^{\prime }\subseteq W(R^{\flat })$ be the set of all Witt vectors $a=(a_{0},a_{1},\ldots )$ with $a_{i}\in \unicode[STIX]{x1D719}^{i}(J)$ . We claim that $J^{\prime }$ is an ideal. Indeed, the ring structure of $W(R^{\flat })$ is given by $(x_{0},x_{1},\ldots )\ast (y_{0},y_{1},\ldots )=(g_{0}^{\ast }(x,y),g_{1}^{\ast }(x,y),\ldots )$ where $\ast$ is $+$ or $\times$ , with certain polynomials $g_{n}^{\ast }$ . If the variables $x_{i},y_{i}$ have degree $p^{i}$ , then $p_{n}^{+}$ is homogeneous of degree $p^{n}$ , and $p_{n}^{\times }$ is bihomogeneous of bidegree $(p^{n},p^{n})$ . Since $R$ is balanced we have $\unicode[STIX]{x1D719}(J)=J^{p}$ ; see Lemma 4.6. It follows that $J^{\prime }$ is an ideal. We have $J^{\prime }\cap pW(R^{\flat })=pJ^{\prime }$ , and $J^{\prime }$ is the closure of the ideal generated by the elements $[a]$ for all $a\in J$ . Clearly $J^{\prime }$ is stable under $\unicode[STIX]{x1D70E}$ . Thus $A=W(R^{\flat })/J^{\prime }$ is a straight lift of $R$ .☐

Proof. The first assertion holds because the natural $\unicode[STIX]{x1D70E}$ -equivariant homomorphism $W(R^{\flat })\rightarrow A$ is surjective, and $\unicode[STIX]{x1D70E}$ is bijective on $W(R^{\flat })$ ; see Remark 4.11. Let $B=\varprojlim (A,\unicode[STIX]{x1D70E})$ . Since $A$ is $p$ -torsion free the same holds for $B$ . We take the limit over $\unicode[STIX]{x1D70E}$ of the exact sequence $0\rightarrow A\xrightarrow[{}]{}A\rightarrow A_{n}\rightarrow 0$ , where the first map is $p^{n}$ . It follows that $B/p^{n}B=\varprojlim (A_{n},\unicode[STIX]{x1D70E})$ , which implies that $\mathop{\varprojlim }\nolimits_{n}(B/p^{n}B)\,=\,B$ ; moreover $B/pB\,=\,R^{\flat }$ . Therefore $B=W(R^{\flat })$ .☐

Proof. Since $\unicode[STIX]{x1D70E}$ is a lift of $\unicode[STIX]{x1D719}$ , for $a\in A$ we have $a\in \tilde{\text{Fil}}\,A$ if and only if $\unicode[STIX]{x1D70E}(a)\in pA$ if and only if $a^{p}\in pA$ . For $a\in \tilde{\text{Fil}}\,A$ let $b=a^{p}/p\in A$ . We have to show that $b\in \tilde{\text{Fil}}\,A$ , or equivalently that $\unicode[STIX]{x1D70E}(b)\in pA$ . But $\unicode[STIX]{x1D70E}(b)=\unicode[STIX]{x1D70E}(a)^{p}/p=p^{p-1}(\unicode[STIX]{x1D70E}(a)/p)^{p}$ .☐

Proof. Let $I$ be the kernel of the surjective homomorphism $\unicode[STIX]{x1D70E}:A\rightarrow A$ . If we write $A=W(R^{\flat })/J^{\prime }$ (see Remark 4.11) then $I=J^{\prime }/\unicode[STIX]{x1D70E}(J^{\prime })$ . Thus $\unicode[STIX]{x1D70E}=p\unicode[STIX]{x1D70E}_{1}$ is zero on $I$ . Since $A$ is $p$ -torsion free it follows that $\unicode[STIX]{x1D70E}_{1}:I\rightarrow I$ is zero, and the lemma follows from the general deformation lemma [Reference LauLau10, Theorem 3.2].☐

Proof. Let $J=\operatorname{Ker}(R^{\flat }\rightarrow R)$ and $A=W(R^{\flat })/J^{\prime }$ . Since $A$ is straight, there are generators $a_{i}$ of $J$ with $[a_{i}]\in J^{\prime }$ . The elements $b_{i}=\unicode[STIX]{x1D719}^{-1}(a_{i})+J$ of $R$ generate the ideal $\operatorname{Ker}(\unicode[STIX]{x1D719})$ . We claim that $b_{i}^{[p]}=0$ , which proves the lemma. The element $c_{i}=[\unicode[STIX]{x1D719}^{-1}(a_{i})]+J^{\prime }$ of $A$ is an inverse image of $b_{i}$ . We have $c_{i}^{p}=[a_{i}]+J^{\prime }=0$ in $A$ , thus $c_{i}^{[p]}=0$ in $A$ , and thus $b_{i}^{[p]}=0$ in $R$ .☐

Proof. The diagram (5.3) commutes by the functoriality of $\unicode[STIX]{x1D6F7}_{A}$ with respect to the frame endomorphism $\unicode[STIX]{x1D70E}:\text{}\underline{A}\rightarrow \text{}\underline{A}$ . The factorization (5.1) of $\unicode[STIX]{x1D70E}$ induces the following extension of (5.3).

(5.4)

Here $\unicode[STIX]{x1D70B}^{\ast }$ is an equivalence by Lemma 5.2. Thus (5.3) is equivalent to the left-hand square of (5.4). For a $p$ -divisible group $G$ over $R$ let $\text{}\underline{M}=\unicode[STIX]{x1D6F7}_{\text{}\underline{A}/\unicode[STIX]{x1D719}}(G)$ in $\operatorname{Win}(\text{}\underline{A}/\unicode[STIX]{x1D719})$ . Then $M\,\otimes _{A}\,R$ is the value of $\mathbb{D}(G)$ at the PD extension $\unicode[STIX]{x1D719}:R\rightarrow R$ . Thus lifts of the Hodge filtration of $G$ under $\unicode[STIX]{x1D719}$ correspond to lifts of the Hodge filtration of $\text{}\underline{M}$ under $\unicode[STIX]{x1D704}^{\ast }$ . The latter correspond to lifts of $\text{}\underline{M}$ under $\unicode[STIX]{x1D704}^{\ast }$ by [Reference LauLau10, Lemma 4.2], and the former correspond to lifts of $G$ under $\unicode[STIX]{x1D719}$ by the Grothendieck–Messing Theorem [Reference MessingMes72] since the divided powers on $\operatorname{Ker}(\unicode[STIX]{x1D719})$ are pointwise nilpotent.☐

Proof. Proposition 5.5 gives the following cartesian diagram.

The upper limit category is equivalent to $\operatorname{BT}(\operatorname{Spec}R^{\flat })$ by the obvious analogue of [Reference MessingMes72, ch. II, Lemma 4.16]; see also [Reference de JongdeJ95, Lemma 2.4.4]. Since we have $\varprojlim (\text{}\underline{A},\unicode[STIX]{x1D70E})=\text{}\underline{W}(R^{\flat })$ by Lemma 4.14, the lower limit category is equivalent to $\operatorname{Win}(\text{}\underline{W}(R^{\flat }))$ by [Reference LauLau10, Lemma 2.12].☐

Proof of Theorem 5.7.

Since $R^{\flat }$ is a perfect ring, the functor $\unicode[STIX]{x1D6F7}_{W(R^{\flat })}$ is an equivalence by a theorem of Gabber; see [Reference LauLau13, Theorem 6.4]. Every window over $\text{}\underline{A}$ can be lifted to a window over $\text{}\underline{W}(R^{\flat })$ . Indeed, the projections $A\rightarrow R$ and $W(R^{\flat })\rightarrow R^{\flat }\rightarrow R$ induce bijective maps of the sets of isomorphisms classes of finite projective modules, and thus the same holds for $W(R^{\flat })\rightarrow A$ . Hence a normal representation of an $\text{}\underline{A}$ -window in the sense of [Reference LauLau10, Lemma 2.6] can be lifted to $\text{}\underline{W}(R^{\flat })$ . Now Lemma 5.9 below applied to the diagram of Corollary 5.6 gives the result.◻

Proof. The case of additive categories is reduced to the case of groupoids using that a homomorphism $u:X\rightarrow Y$ can be encoded by the automorphism $(\!\begin{smallmatrix}1 & 0\\ u & 1\end{smallmatrix}\!)$ of $X\oplus Y$ . Consider the groupoid case. We may assume that ${\mathcal{A}}$ is equal to the fibered product of ${\mathcal{C}}$ and ${\mathcal{B}}$ over ${\mathcal{D}}$ , which is the category of triples $(C,\unicode[STIX]{x1D6FF},B)$ with $C\in {\mathcal{C}}$ , $B\in {\mathcal{B}}$ , and $\unicode[STIX]{x1D6FF}:g(C)\cong \unicode[STIX]{x1D70B}(B)$ .

1. (1) The functor $g$ is surjective on isomorphism classes: This holds for $\unicode[STIX]{x1D70B}$ and $f$ .

2. (2) The functor $\unicode[STIX]{x1D713}$ is surjective on isomorphism classes: Let $C\in {\mathcal{C}}$ . Find $B\in {\mathcal{B}}$ and $\unicode[STIX]{x1D6FF}:g(C)\cong \unicode[STIX]{x1D70B}(B)$ . Then $A=(C,\unicode[STIX]{x1D6FF},B)$ satisfies $\unicode[STIX]{x1D713}(A)=C$ .

3. (3) The functor $g$ is faithful: We have to show that if $C\in {\mathcal{C}}$ and $\unicode[STIX]{x1D6FE}\in \operatorname{Aut}(C)$ with $g(\unicode[STIX]{x1D6FE})=\operatorname{id}$ then $\unicode[STIX]{x1D6FE}=\operatorname{id}$ . Extend $C$ to $A=(C,\unicode[STIX]{x1D6FF},B)\in {\mathcal{A}}$ . Then $\unicode[STIX]{x1D6FC}=(\unicode[STIX]{x1D6FE},\operatorname{id}_{B})$ lies in $\operatorname{Aut}(A)$ with $f(\unicode[STIX]{x1D6FC})=\operatorname{id}$ . Thus $\unicode[STIX]{x1D6FC}=\operatorname{id}$ and $\unicode[STIX]{x1D6FE}=\operatorname{id}$ .

4. (4) The functor $g$ is full: Let $C,C^{\prime }\in {\mathcal{C}}$ and $\unicode[STIX]{x1D6FF}:g(C)\cong g(C^{\prime })$ . Extend $C^{\prime }$ to $A^{\prime }=(C^{\prime },\unicode[STIX]{x1D6FF}^{\prime },B^{\prime })\in {\mathcal{A}}$ . Let $A=(C,\unicode[STIX]{x1D6FF}^{\prime }\unicode[STIX]{x1D6FF},B^{\prime })\in {\mathcal{A}}$ . Then $f(A)=B^{\prime }=f(A^{\prime })$ , and $\operatorname{id}_{B^{\prime }}$ lifts to a unique $\unicode[STIX]{x1D6FC}:A\cong A^{\prime }$ , which consists of $(\unicode[STIX]{x1D6FE},\operatorname{id}_{B^{\prime }})$ with $\unicode[STIX]{x1D6FE}:C\cong C^{\prime }$ such that $\unicode[STIX]{x1D6FF}^{\prime }\circ g(\unicode[STIX]{x1D6FE})=\unicode[STIX]{x1D6FF}^{\prime }\unicode[STIX]{x1D6FF}$ , thus $g(\unicode[STIX]{x1D6FE})=\unicode[STIX]{x1D6FF}$ . ☐

Proof. Let $N\subseteq A_{\operatorname{cris}}(R)$ be the kernel of $\unicode[STIX]{x1D718}$ . Since $A$ is $p$ -torsion free we have $N\cap p^{n}A_{\operatorname{cris}}(R)=p^{n}N$ . Since $\unicode[STIX]{x1D718}$ is continuous for the $p$ -adic topology and $A$ is $p$ -adically complete, $N$ is closed in the $p$ -adic topology of $A_{\operatorname{cris}}(R)$ , and it follows that $N$ is $p$ -adically complete. We have an exact sequence $0\rightarrow N/p\rightarrow A_{\operatorname{cris}}(R)/p\rightarrow R\rightarrow 0$ , and this is the PD envelope over $\mathbb{F}_{p}$ of the ideal $J=\operatorname{Ker}(R^{\flat }\rightarrow R)$ .

Clearly $N$ is stable under $\unicode[STIX]{x1D70E}_{1}$ . We claim that $\unicode[STIX]{x1D70E}_{1}$ is nilpotent on $N/p$ ; cf. [Reference Scholze and WeinsteinSW13, Lemma 4.2.4]. Let $A=W(R^{\flat })/J^{\prime }$ . The hypothesis means that there are generators $a_{i}$ of $J$ such that $[a_{i}]\in J^{\prime }$ . The ideal $N/pN$ of $A_{\operatorname{cris}}(R)/p$ is generated by the elements $a_{i}^{[n]}$ for $n\geqslant 1$ . The elements $[a_{i}]^{[n]}\in N$ satisfy

(5.7) $$\begin{eqnarray}\unicode[STIX]{x1D70E}_{1}([a_{i}]^{[n]})=\frac{(pn)!}{p\cdot n!}[a_{i}]^{[pn]};\end{eqnarray}$$

see [Reference Scholze and WeinsteinSW13, Lemma 4.1.8]. Since the integer $(pn)!/(p\cdot n!)$ is divisible by $p$ when $n\geqslant p$ it follows that $\unicode[STIX]{x1D70E}_{1}\circ \unicode[STIX]{x1D70E}_{1}=0$ on $N/pN$ .

We consider the frames $\text{}\underline{B}_{n}=(A_{\operatorname{cris}}(R)/p^{n}N,\operatorname{Fil}A_{\operatorname{cris}}(R)/p^{n}N,R,\unicode[STIX]{x1D70E},\unicode[STIX]{x1D70E}_{1})$ for $n\geqslant 0$ . Since $\unicode[STIX]{x1D70E}_{1}$ is nilpotent on $N/p^{n}N$ , the projection $\text{}\underline{B}_{n}\rightarrow \text{}\underline{B}_{0}=\text{}\underline{A}$ is crystalline by the general deformation lemma [Reference LauLau10, Theorem 3.2]. We have $\varprojlim \text{}\underline{B}_{n}=\text{}\underline{A}_{\operatorname{cris}}(R)$ , and the proposition follows; see [Reference LauLau10, Lemma 2.12].☐

Proof. By Theorem 5.7 and Proposition 5.10 we have equivalences

$$\begin{eqnarray}\operatorname{BT}(\operatorname{Spec}R)\xrightarrow[{}]{\unicode[STIX]{x1D6F7}_{A}\,}\operatorname{Win}(\text{}\underline{A})\xleftarrow[{}]{\unicode[STIX]{x1D718}^{\ast }}\operatorname{Win}(\text{}\underline{A}_{\operatorname{ cris}}(R)).\Box\end{eqnarray}$$

Proof. If $R$ is a complete intersection, the ring $A_{\operatorname{cris}}(R)$ is $p$ -torsion free; see Remark 4.3. Therefore we have a sequence of functors

$$\begin{eqnarray}\operatorname{BT}(\operatorname{Spec}R)\xrightarrow[{}]{\operatorname{DF}_{R}}\operatorname{DF}(\operatorname{Spec}R)\xrightarrow[{}]{e}\operatorname{Win}(A_{\operatorname{cris}}(R))\xrightarrow[{}]{\unicode[STIX]{x1D718}^{\ast }}\operatorname{Win}(\text{}\underline{A}),\end{eqnarray}$$

where $e$ is the evaluation functor, and the composition is $\unicode[STIX]{x1D6F7}_{A}$ . The functor $e$ is an equivalence; see [Reference Cais and LauCL14, Proposition 2.6.4]. Here no connection appears because $R^{\flat }$ is perfect, and thus $\unicode[STIX]{x1D6FA}_{R^{\flat }}=0$ . The functors $\unicode[STIX]{x1D718}^{\ast }$ and $\unicode[STIX]{x1D6F7}_{A}$ are equivalences by Theorem 5.7 and Proposition 5.10. Thus $\operatorname{DF}_{R}$ is an equivalence as well.☐

Proof. For a PD extension $S\xrightarrow[{}]{\unicode[STIX]{x1D70B}}R$ of $\mathbb{F}_{p}$ -algebras the Frobenius $\unicode[STIX]{x1D719}_{S}$ factors through a homomorphism $\unicode[STIX]{x1D719}_{S/R}:R\rightarrow S$ , i.e.  $\unicode[STIX]{x1D719}_{S/R}\circ \unicode[STIX]{x1D70B}=\unicode[STIX]{x1D719}_{S}$ . An object of $\operatorname{DF}(\operatorname{Spec}R)$ is a triple $({\mathcal{M}},F,V)\in \operatorname{D}(\operatorname{Spec}R)$ together with a direct summand of ${\mathcal{M}}_{R}$ whose base change under each $\unicode[STIX]{x1D719}_{S/R}$ is determined by $({\mathcal{M}},F)$ ; see [Reference Cais and LauCL14, Definition 2.4.1]. The lift $A$ of $R$ makes $\unicode[STIX]{x1D719}:R\rightarrow R$ into a PD extension, which we write as $S\rightarrow R$ ; see Remark 5.3. The corresponding $\unicode[STIX]{x1D719}_{S/R}$ is the identity of $R$ , and the lemma follows.☐

Proof. For $A\in \operatorname{Nil}_{\unicode[STIX]{x1D6EC}/R}$ the fiber product $A^{\prime }=A\times _{R}\unicode[STIX]{x1D6EC}$ lies in $\operatorname{Nil}_{\unicode[STIX]{x1D6EC}/\unicode[STIX]{x1D6EC}}$ . Let $\operatorname{LF}(A)$ denote the category of finite projective $A$ -modules. Then the obvious functor $\operatorname{LF}(A^{\prime })\rightarrow \operatorname{LF}(A)\times _{\operatorname{LF}(R)}\operatorname{LF}(\unicode[STIX]{x1D6EC})$ is an equivalence. It follows that the natural map $\operatorname{Def}_{G^{\prime }}(A^{\prime })\rightarrow \operatorname{Def}_{G}(A)$ is bijective, which proves the lemma.☐

Proof. Let $A\in \tilde{\operatorname{Nil}}_{\unicode[STIX]{x1D6EC}/R}$ . We have to show that the natural map $\operatorname{Hom}(B,A)\rightarrow \operatorname{Def}_{G}(A)$ is bijective. For each pair of homomorphisms $f_{1},f_{2}:B\rightarrow A$ in $\operatorname{Aug}_{\unicode[STIX]{x1D6EC}/R}$ there is a finitely generated ideal $\mathfrak{b}\subseteq J_{A}$ such that the projection $A\rightarrow \bar{A}=A/\mathfrak{b}$ equalizes $f$ and $g$ . For each pair of deformations $G_{1},G_{2}$ of $G$ over $A$ the reduction map $\operatorname{Hom}_{A}(G_{1},G_{2})\rightarrow \operatorname{End}_{R}(G)$ is injective with cokernel annihilated by $p^{r}$ for some $r$ ; see [Reference LauLau14, Lemma 3.4]. Thus there is a unique isogeny $\unicode[STIX]{x1D713}:G_{1}\rightarrow G_{2}$ which lifts $p^{r}\operatorname{id}_{G}$ . Its kernel is finitely presented; see [Reference LauLau14, Lemma 3.6]. Thus there is a finitely generated ideal $\mathfrak{b}\subseteq A$ such that $\operatorname{Ker}(\unicode[STIX]{x1D713})$ and $G_{1}[p^{r}]$ coincide over $A/\mathfrak{b}$ , which means that $G_{1}$ and $G_{2}$ map to the same element of $\operatorname{Def}_{G}(A/\mathfrak{b})$ . Moreover $G_{1}$ and $G_{2}$ are equal as deformations of $G$ if and only if they are equal as deformations of $G_{1}\otimes _{A}A/\mathfrak{b}$ . In view of these remarks it suffices to show that $B(A)\rightarrow \operatorname{Def}_{G}(A)$ is bijective when $R$ is replaced by $R^{\prime }=A/\mathfrak{b}$ for varying finitely generated ideals $\mathfrak{b}$ . Then $A$ lies in $\operatorname{Nil}_{\unicode[STIX]{x1D6EC}/R^{\prime }}$ , and the lemma follows from Lemma 6.1.☐

Proof. This is similar to [Reference LauLau14, Theorem 3.19].

Let $G\mapsto (M(G),\operatorname{Fil}M(G),F)$ be as defined in the theorem. We have to find a functorial map $F_{1}:\operatorname{Fil}M(G)\rightarrow M(G)$ which gives a window $\text{}\underline{M}(G)$ , and verify that $F_{1}$ is unique. If $A_{\operatorname{cris}}(R)$ is $p$ -torsion free then $F_{1}$ and thus $\text{}\underline{M}(G)$ are well defined; see [Reference LauLau14, Proposition 3.17]. This applies in particular when $R$ is perfect since then $A_{\operatorname{cris}}(R)=W(R)$ .

In general let $\unicode[STIX]{x1D70B}:R^{\flat }\rightarrow R$ be the projection. We write $\unicode[STIX]{x1D70B}^{\ast }$ for the base change functor of modules or windows from $W(R^{\flat })$ to $A_{\operatorname{cris}}(R)$ . Note that $p$ -divisible groups can be lifted under $\unicode[STIX]{x1D719}:R\rightarrow R$ by [Reference IllusieIll85, Theorem 4.4], and thus $p$ -divisible groups can be lifted under $\unicode[STIX]{x1D70B}$ . Let $G\in \operatorname{BT}(\operatorname{Spec}R)$ be given. We choose a lift $G_{1}\in \operatorname{BT}(\operatorname{Spec}R^{\flat })$ of $G$ . Then $M(G)=\unicode[STIX]{x1D70B}^{\ast }M(G_{1})$ as modules with $\operatorname{Fil}$ and $F$ , and necessarily we have to define $\text{}\underline{M}(G)=\unicode[STIX]{x1D70B}^{\ast }\text{}\underline{M}(G_{1})$ as windows. We have to show that this construction of $F_{1}$ does not depend on the choice of $G_{1}$ , i.e. if $G_{2}\in \operatorname{BT}(\operatorname{Spec}R^{\flat })$ is another lift of $G$ , then the composite isomorphism of modules

$$\begin{eqnarray}\unicode[STIX]{x1D70B}^{\ast }M(G_{1})\cong M(G)\cong \unicode[STIX]{x1D70B}^{\ast }M(G_{2})\end{eqnarray}$$

preserves the homomorphisms $F_{1}$ defined on the outer terms by the windows $\text{}\underline{M}(G_{i})$ .

We want to lift the situation to perfect rings. More precisely, we claim that one can find a commutative diagram of rings

where $S$ and $S^{\prime }$ are perfect, and $p$ -divisible groups $H_{1},H_{2}\in \operatorname{BT}(\operatorname{Spec}S^{\prime })$ together with an isomorphism $\unicode[STIX]{x1D6FC}:u^{\ast }H_{1}\cong u^{\ast }H_{2}$ over $S$ and isomorphisms $f^{\ast }H_{i}\cong G_{i}$ over $R^{\flat }$ for $i=1,2$ such that $\unicode[STIX]{x1D6FC}$ induces the given isomorphism $\unicode[STIX]{x1D70B}^{\ast }G_{1}\cong \unicode[STIX]{x1D70B}^{\ast }G_{2}$ over $R$ , i.e. the composition

$$\begin{eqnarray}\unicode[STIX]{x1D70B}^{\ast }G_{1}\cong \unicode[STIX]{x1D70B}^{\ast }f^{\ast }H_{1}\cong g^{\ast }u^{\ast }H_{1}\xrightarrow[{}]{g^{\ast }\unicode[STIX]{x1D6FC}}g^{\ast }u^{\ast }H_{2}\cong \unicode[STIX]{x1D70B}^{\ast }f^{\ast }H_{2}\cong \unicode[STIX]{x1D70B}^{\ast }G_{2}\end{eqnarray}$$

is the given isomorphism. Then the homomorphisms $F_{1}$ of $H_{1}$ and of $H_{2}$ coincide over $S$ since $S$ is perfect, and by base change under $g$ it follows that the homomorphisms $F_{1}$ of $G_{1}$ and of $G_{2}$ coincide over $R$ as required.

Let us prove the claim. Let $G^{\prime }$ be a lift of $G$ to $R^{\flat }$ , for example $G^{\prime }=G_{1}$ . Let $B=R^{\flat }[[Q]]$ be the universal deformation ring of $G^{\prime }$ as in § 6.1 and let ${\mathcal{G}}$ over $B$ be the universal deformation. By Lemma 6.2, $B$ represents the deformation functor $\operatorname{Def}_{G}$ on the category $\tilde{\operatorname{Nil}}_{R^{\flat }/R}$ of augmented algebras $R^{\flat }\rightarrow A\rightarrow R$ such that the kernel of $A\rightarrow R$ is bounded nilpotent. The system $(\unicode[STIX]{x1D719}^{n}:R\rightarrow R)_{n}$ is a pro-object of $\tilde{\operatorname{Nil}}_{R^{\flat }/R}$ with limit $R^{\flat }\rightarrow R$ in $\operatorname{Aug}_{R^{\flat }/R}$ . Thus there are homomorphisms $\unicode[STIX]{x1D6FD}_{i}:B\rightarrow R^{\flat }$ in $\operatorname{Aug}_{R^{\flat }/R}$ with $\unicode[STIX]{x1D6FD}_{i}^{\ast }{\mathcal{G}}\cong G_{i}$ as deformations of $G$ over $R^{\flat }$ .

We put $S=R^{\flat }$ with $g=\unicode[STIX]{x1D70B}$ and $S^{\prime }=B^{\operatorname{per}}=\varinjlim (B,\unicode[STIX]{x1D719})$ with $u=\unicode[STIX]{x1D6FD}_{1}^{\operatorname{per}}$ and $f=\unicode[STIX]{x1D6FD}_{2}^{\operatorname{per}}$ . Let $H_{1}$ be the base change of $G_{1}$ under $R^{\flat }\rightarrow B\rightarrow S^{\prime }$ and let $H_{2}$ be the base change of ${\mathcal{G}}$ under $B\rightarrow S^{\prime }$ . Then $u^{\ast }H_{1}\cong G_{1}\cong u^{\ast }H_{2}$ and $f^{\ast }H_{1}\cong G_{1}$ and $f^{\ast }H_{2}\cong G_{2}$ as deformations of $G$ . This proves the claim; the required equality of isomorphisms $\unicode[STIX]{x1D70B}^{\ast }G_{1}\cong \unicode[STIX]{x1D70B}^{\ast }G_{2}$ is automatic because the reduction map $\operatorname{Hom}(G_{1},G_{2})\rightarrow \operatorname{End}(G)$ is injective.☐

Proof. The functor $\unicode[STIX]{x1D6F7}_{R}^{\operatorname{cris}}$ without $F_{1}$ is given by the Dieudonné crystal evaluated at $A_{\operatorname{cris}}(R)$ . Thus the functor $\unicode[STIX]{x1D718}^{\ast }\circ \unicode[STIX]{x1D6F7}_{R}^{\operatorname{cris}}$ without $F_{1}$ is given by the Dieudonné crystal evaluated at $A$ . Since $A$ is $p$ -torsion free, for the frame $\text{}\underline{A}$ the functor of forgetting $F_{1}$ is fully faithful, and the lemma follows.☐

Proof. If $A$ is a straight lift of $R$ , the functor $\unicode[STIX]{x1D6F7}_{A}$ is an equivalence by Theorem 5.7 together with Lemma 5.4, and the functor $\unicode[STIX]{x1D718}^{\ast }$ is an equivalence by Proposition 5.10. By Lemma 6.4 it follows that $\unicode[STIX]{x1D6F7}_{R}^{\operatorname{cris}}$ is an equivalence. The final assertion is clear.☐

Proof. To prove the assertion we may replace $R$ by an isogenous ring; see [Reference Scholze and WeinsteinSW13, Proposition 4.1.5]. Thus we can assume that $R$ is balanced, so $R$ has a straight lift. For $G\in \operatorname{BT}(\operatorname{Spec}R)$ and $\unicode[STIX]{x1D6F7}_{R}^{\operatorname{cris}}(G)=\text{}\underline{M}=(M,\operatorname{Fil}M,F,F_{1})$ , the Dieudenné crystal $\mathbb{D}(G)$ is given by the pair $(M,F)$ . For $G,G^{\prime }\in \operatorname{BT}(\operatorname{Spec}R)$ we have to show that the composition

$$\begin{eqnarray}\operatorname{Hom}(G,G^{\prime })\otimes \mathbb{Q}\rightarrow \operatorname{Hom}(\text{}\underline{M},\text{}\underline{M}^{\prime })\otimes \mathbb{Q}\rightarrow \operatorname{Hom}((M,F),(M^{\prime },F))\otimes \mathbb{Q}\end{eqnarray}$$

is bijective. The first map is bijective without $\otimes \mathbb{Q}$ by Corollary 6.5, the second map is bijective because the $A_{\operatorname{cris}}(R)$ -module $M$ is of finite type.☐

Proof. The universal property of $A_{\operatorname{cris}}(R)$ gives a PD homomorphism $\unicode[STIX]{x1D718}:A_{\operatorname{cris}}\rightarrow A$ over the identity of $R$ , and $\unicode[STIX]{x1D718}$ commutes with $\unicode[STIX]{x1D70E}$ . To show that $\unicode[STIX]{x1D718}$ is a frame homomorphism it suffices to verify that $\unicode[STIX]{x1D718}(\unicode[STIX]{x1D70E}_{1}(y))=\unicode[STIX]{x1D70E}_{1}(\unicode[STIX]{x1D718}(y))$ for generators $y$ of the ideal $\operatorname{Fil}A_{\operatorname{cris}}(R)$ . A set of generators of this ideal is formed by $p$ and the elements $[x]^{[n]}$ for generators $x\in J$ and $n\geqslant 1$ . We have $\unicode[STIX]{x1D718}(\unicode[STIX]{x1D70E}_{1}(p))=1=\unicode[STIX]{x1D70E}_{1}(\unicode[STIX]{x1D718}(p))$ , moreover (5.7) gives

(7.1) $$\begin{eqnarray}\unicode[STIX]{x1D718}(\unicode[STIX]{x1D70E}_{1}([x]^{[n]}))=\frac{(np)!}{p\cdot n!}\unicode[STIX]{x1D718}([x]^{[np]})=\frac{(np)!}{p\cdot n!}\unicode[STIX]{x1D718}([x])^{[np]}\end{eqnarray}$$

and

(7.2) $$\begin{eqnarray}\unicode[STIX]{x1D70E}_{1}(\unicode[STIX]{x1D718}([x]^{[n]}))=\unicode[STIX]{x1D70E}_{1}(\unicode[STIX]{x1D718}([x])^{[n]}).\end{eqnarray}$$

Since the weak lift $A$ is straight, the generators $x$ of $J$ can be chosen such that $[x]\in J^{\prime }$ . Then $\unicode[STIX]{x1D718}([x])=0$ , and (7.1) and (7.2) are both zero.☐

Proof. As in the proof of Lemma 4.13 we see that $W(J_{\ast })$ is an ideal. This ideal is closed in $W(R^{\flat })$ , and thus $A=A(J_{\ast })$ is $p$ -adically complete. Since $J_{0}=J$ we have $A/pA=R$ . Clearly $W(J_{\ast })$ is stable under the endomorphism $\unicode[STIX]{x1D70E}$ of $W(R^{\flat })$ , so $\unicode[STIX]{x1D70E}$ induces $\unicode[STIX]{x1D70E}:A\rightarrow A$ . We have $pA=pW(R^{\flat })/(W(J_{\ast })\cap pW(R^{\flat }))$ , and an element $a\in W(J_{\ast })$ lies in $pW(R^{\flat })$ if and only if $a_{0}=0$ . Since the sequence $J_{\ast }$ is descending we have $\unicode[STIX]{x1D70E}_{1}(W(J_{\ast })\cap pW(R^{\flat }))\subseteq W(J_{\ast })$ , so $\unicode[STIX]{x1D70E}_{1}$ induces $\unicode[STIX]{x1D70E}_{1}:pA\rightarrow A$ . It follows that $A$ is a weak lift, which is straight because for every $a\in J$ we have $[a]\in W(J_{\ast })$ .☐

Proof. The homomorphism $\unicode[STIX]{x1D70B}$ exists because $J_{i}\subseteq J_{i}^{\prime }$ . Let $\mathfrak{a}=\operatorname{Ker}(\unicode[STIX]{x1D70B})=W(J_{\ast }^{\prime })/W(J_{\ast })$ . Then $\unicode[STIX]{x1D70E}_{1}$ induces an endomorphism of $\mathfrak{a}$ , and $(\unicode[STIX]{x1D70E}_{1})^{n}$ is zero on $\mathfrak{a}$ because $J_{i+n}^{\prime }\subseteq J_{i}$ . The result follows from the deformation lemma [Reference LauLau10, Theorem 3.2] if we find a sequence of ideals $\mathfrak{a}=\mathfrak{a}_{0}\supseteq \cdots \supseteq \mathfrak{a}_{n}=0$ which are stable under $\unicode[STIX]{x1D70E}_{1}$ such that $\unicode[STIX]{x1D70E}(\mathfrak{a}_{m})\subseteq \mathfrak{a}_{m+1}$ for $m<n$ .

This sequence can be constructed as follows. We have $\unicode[STIX]{x1D719}^{n}(J_{i}^{\prime })\subseteq J_{i}^{\prime p^{n}}\subseteq J_{i+n}^{\prime }\subseteq J_{i}$ and thus $J_{i}\subseteq J_{i}^{\prime }\subseteq \unicode[STIX]{x1D719}^{-n}(J_{i})$ . For each $m$ with $0\leqslant m\leqslant n$ let $K_{m,i}=J_{i}^{\prime }\cap \unicode[STIX]{x1D719}^{m-n}(J_{i})$ . Then $K_{m,\ast }$ is an admissible sequence, moreover $J_{i}^{\prime }=K_{0,i}\supseteq K_{1,i}\supseteq \cdots \supseteq K_{n,i}=J_{i}$ and thus $W(J_{\ast }^{\prime })=W(K_{0,\ast })\supseteq \cdots \supseteq W(K_{n,\ast })=W(J_{\ast })$ . Let $\mathfrak{a}_{m}=W(K_{m,\ast })/W(J_{\ast })$ . Then $\mathfrak{a}_{m}$ is stable under $\unicode[STIX]{x1D70E}_{1}$ because $K_{m,\ast }$ is a decreasing sequence; see the proof of Lemma 7.6. We have $\unicode[STIX]{x1D70E}(\mathfrak{a}_{m})\subseteq \mathfrak{a}_{m+1}$ because $\unicode[STIX]{x1D719}(K_{m,i})\subseteq K_{m+1,i}$ .☐

Proof. Let $R\rightarrow R^{\prime }$ be an isogeny with balanced $R^{\prime }$ whose kernel is annihilated by $\unicode[STIX]{x1D719}^{n}$ . Then $R^{\prime }=R^{\flat }/J^{\prime }$ with $\unicode[STIX]{x1D719}(J^{\prime })={J^{\prime }}^{p}$ and $\unicode[STIX]{x1D719}^{n}(J^{\prime })\subseteq J\subseteq J^{\prime }$ . Let $K_{i}=J\cap \unicode[STIX]{x1D719}^{i}(J^{\prime })$ for $i\geqslant 0$ . The sequence $K_{\ast }$ is admissible. For $i\geqslant 0$ we have $K_{n+i}=\unicode[STIX]{x1D719}^{n+i}(J^{\prime })\subseteq J_{i}\subseteq K_{i}$ . Thus the natural frame homomorphism $\unicode[STIX]{x1D70B}:\text{}\underline{A}(J_{\ast })\rightarrow \text{}\underline{A}(K_{\ast })$ is crystalline by Lemma 7.7, and it suffices to show that the composition

$$\begin{eqnarray}\unicode[STIX]{x1D718}^{\prime }=\unicode[STIX]{x1D70B}\circ \unicode[STIX]{x1D718}:\text{}\underline{A}_{\operatorname{ cris}}(R)\rightarrow \text{}\underline{A}(K_{\ast })\end{eqnarray}$$

is crystalline. Let $N=\operatorname{Ker}(\unicode[STIX]{x1D718}^{\prime })$ .

Proof. Let $J_{i}^{\prime }=\unicode[STIX]{x1D719}^{i}(J^{\prime })$ . Then $J_{\ast }^{\prime }$ is an admissible sequence of ideals of $W(R^{\flat })$ with respect to $R^{\prime }=R^{\flat }/J^{\prime }$ ; note that $R^{\flat }={R^{\prime }}^{\flat }$ . We have $K_{i}\subseteq J_{i}^{\prime }$ with equality for $i\geqslant n$ , so there is a projection $A(K_{\ast })\rightarrow A(J_{\ast }^{\prime })$ whose kernel is annihilated by $p^{n}$ . The ring $A(J_{\ast }^{\prime })$ is the straight lift of $R^{\prime }$ constructed in Lemma 4.13, which is $p$ -torsion free. This proves (i), and (ii) follows.

Let us prove (iii). The ring $A_{\operatorname{cris}}(R)$ is the $p$ -adic completion of a $W(R^{\flat })$ -algebra generated by the elements $[x]^{[i]}$ for $x\in J$ and $i\geqslant 1$ , and these elements map to zero in $A(K_{\ast })$ . Thus for each $m\geqslant 1$ the image of $N$ in $A_{\operatorname{cris}}(R)/p^{m}$ is generated as an ideal by $W(K_{\ast })$ and the elements $[x]^{[i]}$ . By (ii) it follows that $N/pN$ is generated as an $A_{\operatorname{cris}}(R)$ -module by $W(K_{\ast })$ and the elements $[x]^{[i]}$ . We check these elements separately.

First, the explicit formula (5.7) for $\unicode[STIX]{x1D70E}_{1}$ implies that for $x\in J$ the element $(\unicode[STIX]{x1D70E}_{1})^{2}([x]^{[i]})$ lies in $pN$ ; see the proof of Proposition 5.10. Second, since $\unicode[STIX]{x1D719}^{i}(K_{n})=K_{n+i}$ for $i\geqslant 0$ , each element of $W(K_{\ast })/pW(K_{\ast })$ is represented by an element $a=(a_{0},a_{1},a_{2},\ldots )\in W(K_{\ast })$ with $a_{i}=0$ for $i>n$ . Then

$$\begin{eqnarray}(\unicode[STIX]{x1D70E}_{1})^{n+2}(a)=(\unicode[STIX]{x1D70E}_{1})^{n+2}([a_{0}])+\cdots +(\unicode[STIX]{x1D70E}_{1})^{2}([a_{n}])\end{eqnarray}$$

lies in $pN$ , using that $a_{i}\in K_{i}\subseteq J$ . Thus $(\unicode[STIX]{x1D70E}_{1})^{n+2}$ is zero on $N/pN$ , and Lemma 7.9 is proved. ☐

We continue the proof of Proposition 7.8. Since $A(K_{\ast })$ is $p$ -adically complete, the ideal $N$ is closed in $A_{\operatorname{cris}}(R)$ , and thus $N$ is $p$ -adically complete by Lemma 7.9(ii). Since $\unicode[STIX]{x1D70E}_{1}:N\rightarrow N$ stabilizes $p^{m}N$ , the ring $A_{\operatorname{cris}}(R)/p^{m}N$ carries a natural frame structure, denoted by $\text{}\underline{A}_{\operatorname{cris}}(R)/p^{m}N$ . We have $\text{}\underline{A}_{\operatorname{cris}}(R)/N=\text{}\underline{A}(K_{\ast })$ and $\text{}\underline{A}_{\operatorname{cris}}(R)=\varprojlim \text{}\underline{A}_{\operatorname{cris}}(R)/p^{m}N$ . This limit preserves the window categories by [Reference LauLau10, Lemma 2.12]. Thus it suffices to show that the frame homomorphism $\text{}\underline{A}_{\operatorname{cris}}(R)/p^{m}N\rightarrow \text{}\underline{A}(K_{\ast })$ is crystalline for each $m$ . Since $\unicode[STIX]{x1D70E}_{1}:N/p^{m}N\rightarrow N/p^{m}N$ is nilpotent by Lemma 7.9(iii), this follows from [Reference LauLau10, Theorem 3.2].☐

Proof. By Corollary 6.5 the theorem holds for balanced rings. An isogeny from $R$ to a balanced ring has nilpotent kernel by Lemma 4.5. Therefore it suffices to show the following. Let $\unicode[STIX]{x1D70B}:R^{\prime }\rightarrow R$ be an isogeny of iso-balanced rings such that $\operatorname{Ker}(\unicode[STIX]{x1D70B})^{p}=0$ . If $\unicode[STIX]{x1D6F7}_{R}^{\operatorname{cris}}$ is an equivalence then so is $\unicode[STIX]{x1D6F7}_{R^{\prime }}^{\operatorname{cris}}$ .

To prove this we use some auxiliary frames. Let $J=\operatorname{Ker}(R^{\flat }\rightarrow R)$ and $J^{\prime }=\operatorname{Ker}(R^{\flat }\rightarrow R^{\prime })$ , thus $J^{p}\subseteq J^{\prime }\subseteq J$ . We define $J_{i}=J^{p^{i}}$ and $J_{i}^{\prime }={J^{\prime }}^{p^{i}}$ for $i\geqslant 0$ , and we define $K_{0}=J^{\prime }$ and $K_{i}=J_{i}$ for $i\geqslant 1$ . Then $J_{\ast }$ is an admissible sequence with respect to $R$ , while $K_{\ast }$ and $J_{\ast }^{\prime }$ are admissible sequences with respect to $R^{\prime }$ . There are obvious frame homomorphisms

$$\begin{eqnarray}\text{}\underline{A}(J_{\ast }^{\prime })\xrightarrow[{}]{a}\text{}\underline{A}(K_{\ast })\xrightarrow[{}]{q}\text{}\underline{A}(J_{\ast }),\end{eqnarray}$$

where $a$ lies over $\operatorname{id}_{R^{\prime }}$ and $q$ lies over $\unicode[STIX]{x1D70B}$ . Here $a$ is crystalline by Lemma 7.7, using that $K_{i+1}\subseteq J_{i}^{\prime }\subseteq K_{i}$ . We want to factor $q$ over another frame ${\mathcal{F}}=(A,I,R,\unicode[STIX]{x1D70E},\unicode[STIX]{x1D70E}_{1})$ with $A=A(K_{\ast })$ , thus $I$ is the kernel of $A(K_{\ast })\rightarrow A(J_{\ast })\rightarrow R$ . We only have to define $\unicode[STIX]{x1D70E}_{1}:I\rightarrow A$ . It is easy to see that the natural map $\operatorname{Ker}(q)\rightarrow \operatorname{Ker}(\unicode[STIX]{x1D70B})$ is bijective and that

$$\begin{eqnarray}I=\operatorname{Ker}(q)\oplus pA\end{eqnarray}$$

as a direct sum of ideals. We have $\operatorname{Ker}(q)^{p}=0$ , and $\unicode[STIX]{x1D70E}(x)=0$ for $x\in \operatorname{Ker}(q)$ . We extend the homomorphism $\unicode[STIX]{x1D70E}_{1}$ and the divided powers defined on $pA$ to $I$ by $\unicode[STIX]{x1D70E}_{1}(x)=0$ and $x^{[p]}=0$ for $x\in \operatorname{Ker}(q)$ . This defines a PD frame ${\mathcal{F}}$ as above. Together we have homomorphisms of PD frames

$$\begin{eqnarray}\text{}\underline{A}(J_{\ast }^{\prime })\xrightarrow[{}]{a}\text{}\underline{A}(K_{\ast })\xrightarrow[{}]{b}{\mathcal{F}}\xrightarrow[{}]{c}\text{}\underline{A}(J_{\ast })\end{eqnarray}$$

over $R^{\prime }\xrightarrow[{}]{\operatorname{id}}R^{\prime }\xrightarrow[{}]{\unicode[STIX]{x1D70B}}R\xrightarrow[{}]{\operatorname{id}}R$ , where $c$ is given by $q$ and $b$ is given by $\operatorname{id}_{A}$ . Since $\unicode[STIX]{x1D70E}_{1}$ is zero on $\operatorname{Ker}(c)=\operatorname{Ker}(q)$ , $c$ is crystalline by [Reference LauLau10, Theorem 3.2].

Since $A\rightarrow R$ is a $p$ -adic PD extension, the universal property of $A_{\operatorname{cris}}(R)$ gives a unique homomorphism $\tilde{\unicode[STIX]{x1D718}}:A_{\operatorname{cris}}(R)\rightarrow A$ of PD extensions of $R$ , and $\tilde{\unicode[STIX]{x1D718}}$ commutes with $\unicode[STIX]{x1D70E}$ . We claim that $\tilde{\unicode[STIX]{x1D718}}$ is a frame homomorphism $\text{}\underline{A}_{\operatorname{cris}}(R)\rightarrow {\mathcal{F}}$ , i.e. that $\tilde{\unicode[STIX]{x1D718}}$ commutes with $\unicode[STIX]{x1D70E}_{1}$ . As in the proof of Lemma 7.4 it suffices to show that $\tilde{\unicode[STIX]{x1D718}}(\unicode[STIX]{x1D70E}_{1}(y))=\unicode[STIX]{x1D70E}_{1}(\tilde{\unicode[STIX]{x1D718}}(y))$ when $y=[x]^{[n]}$ with $x\in J$ and $n\geqslant 1$ . Let $z=\tilde{\unicode[STIX]{x1D718}}([x])$ in $A$ . Then $z\in \operatorname{Ker}(q)$ and thus $z^{[np]}=0$ , moreover $z^{[n]}\in \operatorname{Ker}(q)$ as well and thus $\unicode[STIX]{x1D70E}_{1}(z^{[n]})=0$ . Therefore (7.1) and (7.2) with $\tilde{\unicode[STIX]{x1D718}}$ in place of $\unicode[STIX]{x1D718}$ show that $\tilde{\unicode[STIX]{x1D718}}(\unicode[STIX]{x1D70E}_{1}(y))=0$ and $\unicode[STIX]{x1D70E}_{1}(\tilde{\unicode[STIX]{x1D718}}(y))=0$ . Thus $\tilde{\unicode[STIX]{x1D718}}$ is a frame homomorphism. Since $a,b,c$ are PD homomorphisms, we obtain a commutative diagram of frames

where the homomorphisms $\unicode[STIX]{x1D718}$ are given by Lemma 7.4, and $\unicode[STIX]{x1D718}^{\prime }=a\circ \unicode[STIX]{x1D718}$ . The two homomorphisms $\unicode[STIX]{x1D718}$ are crystalline by Proposition 7.8. Since $a$ and $c$ are crystalline, the same holds for $\unicode[STIX]{x1D718}^{\prime }$ and $\tilde{\unicode[STIX]{x1D718}}$ . We have the following commutative diagram of categories.

The functors $\operatorname{BT}(\operatorname{Spec}R^{\prime })\rightarrow \operatorname{Win}(\text{}\underline{A}(K_{\ast }))$ and $\operatorname{BT}(\operatorname{Spec}R)\rightarrow \operatorname{Win}({\mathcal{F}})$ are given by the Dieudonné crystal with an additional $F_{1}$ . Since $\operatorname{Ker}(\unicode[STIX]{x1D70B})^{p}=0$ , the ideal $\operatorname{Ker}(\unicode[STIX]{x1D70B})$ can be equipped with the trivial divided powers. Then the projection $A(K_{\ast })=A\rightarrow R^{\prime }$ is a homomorphism of PD extensions of $R$ . It follows that for $G\in \operatorname{BT}(\operatorname{Spec}R)$ with associated $\text{}\underline{M}\in \operatorname{Win}({\mathcal{F}})$ there is a natural isomorphism $M\otimes _{A}R^{\prime }\cong \mathbb{D}(G)_{R^{\prime }}$ . By the Grothendieck–Messing theorem [Reference MessingMes72] and its trivial counterpart for the frame homomorphism $b$ in [Reference LauLau10, Lemma 4.2] it follows that the lifts of $G$ under $\unicode[STIX]{x1D70B}$ and the lifts of $\text{}\underline{M}$ under $b$ coincide; cf. the proof of Proposition 5.5. Therefore if $\unicode[STIX]{x1D6F7}_{R}^{\operatorname{cris}}$ is an equivalence, the same holds for $\unicode[STIX]{x1D6F7}_{R^{\prime }}^{\operatorname{cris}}$ . This finishes the proof of Theorem 7.10.☐