Proof. Cf. [Reference Bhatt and Scholze13, Lemma 2.31].

Proof. Note that for the second case, one can always assume $\min (r,s)=0$ , up to replacing d by $\varphi ^{\min (r,s)}(d)$ . Then the statement is proven in [Reference Anschütz and Le Bras1, Lemma 3.3] and [Reference Anschütz and Le Bras1, Lemma 3.6].

Proof. Cf. [Reference Bhatt and Scholze13, Theorem 3.9].

Proof. We can lift $R^{\prime }$ to a $(p,\ker (\theta ))$ -completely étale $A_{\mathrm {inf}}(R)$ -algebra B. By [Reference Bhatt and Scholze13, Lemma 2.18], the $\delta $ -structure on $A_{\mathrm {inf}}(R)$ extends uniquely to B. Reducing modulo p, we see that B is a perfect $\delta $ -ring as it is $(p,\ker (\theta ))$ -completely étale over $A_{\mathrm {inf}}(R)$ . Using Proposition 2.8, $R^{\prime }\cong B/\ker (\theta )B$ is therefore perfectoid. The statement on henselisations follows from this as henselisations are colimits along étale maps (cf. the proof of [Reference Project52, Tag 0A02]). (Note that since R has bounded $p^{\infty }$ -torsion, the p-completion of an étale R-algebra is p-completely étale.)

Proof. Cf. [Reference Bhatt and Scholze13, Lemma 4.7].

Proof. By Lemma 2.4 (which applies to B as B is bounded and, thus, classically $(p,[p]_q)$ -complete, by [Reference Bhatt and Scholze13, Lemma 3.7 (1)]), the element $[x^{1/p}]_{\tilde {\theta }}$ is of rank $1$ as it admits arbitrary $p^n$ -roots. Moreover, $[x^{1/p}]_{\tilde {\theta }}\in 1+\mathcal {N}^{\geq 1}B$ as $\varphi ([x^{1/p}]_{\tilde {\theta }})=[x]_{\tilde {\theta }}\equiv 1$ modulo J. By Lemma 2.13, we can therefore conclude.

Proof. We can assume that $X=\mathrm {Spf}(R)$ is affine. Write $\bar {A}=A/I$ . We want to prove that there is a canonical isomorphism

First, let us note that by the transitivity triangle for $A\to {\bar {A}} \to R$ , the cotangent complex $L_{R/A}\{-1\}[-1]^{\wedge _p}$ sits inside a triangle

$$ \begin{align*} R\cong R\otimes_{{\bar{A}}}L_{{\bar{A}}/A}\{-1\}[-1]^{\wedge_p}\to L_{R/A}\{-1\}[-1]^{\wedge_p}\to L_{R/{\bar{A}}}\{-1\}[-1]^{\wedge_p}, \end{align*} $$

and the outer terms are isomorphic to and

To construct the isomorphism $\alpha _R$ , it suffices to restrict to ${\bar {A}}\to R p$ -completely smooth first, and then Kan extend. Thus, assume from now on that R is p-completely smooth over ${\bar {A}}$ .

Let

, that is, $(B,J)$ is a prism over $(A,I)$ with a morphism $\iota \colon R\to B/J$ . Pulling back the extension of A-algebras

$$ \begin{align*} 0\to J/J^2\to B/J^2\to B/J\to 0 \end{align*} $$

along $\iota \colon R\to B/J$ defines an extension of R by $J/J^2\cong B/J\{1\}$ , and as such, is thus classified by a morphism

$$ \begin{align*} \alpha_R^{\prime}\colon L_{R/A}^{\wedge_p}\to B/J\{1\}[1]. \end{align*} $$

Passing to the (homotopy) limit over all

then defines (after shifting and twisting) the morphism

Concretely, if $R={\bar {A}}\langle x\rangle $ , then

$$ \begin{align*} L_{R/A}^{\wedge_p}\cong R\otimes_{{\bar{A}}} I/I^2[1]\oplus Rdx. \end{align*} $$

On the summand $R\otimes _{{\bar {A}}}I/I^2[1]$ , the morphism $\alpha _R^{\prime }$ is simply the base extension of $I/I^2\to J/J^2$ as follows by considering the case ${\bar {A}}=R$ . On the summand $Rdx$ , the morphism $\alpha _R^{\prime }$ is (canonically) represented by the $J/J^2$ -torsor of preimages of $\iota (x)$ in $B/J^2$ and factors as $R\xrightarrow {\iota }B/J\to B/J\{1\}[1]$ with the second morphism the connecting morphism for $0\to B/J\{1\}\to B/J^2\to B/J\to 0$ . Thus, after passing to the limit, we get a diagram

and on $H^0$ , the horizontal morphism induces the Bockstein differential

Thus, the image of $dx\in H^0(L_{R/A}^{\wedge _p})$ under $\alpha _R$ is $\beta (\iota (x))$ . Therefore, we see that on $H^0$ , the morphism $\alpha _R$ induces the identity under the identifications

$$ \begin{align*}(\Omega^{1}_{R/{\bar{A}}})^{\wedge_p}\cong H^0(L_{R/A}^{\wedge_p}) \end{align*} $$

and

(the second is the Hodge-Tate comparison). Moreover, the morphism

is the canonical one obtained by tensoring

with $I/I^2$ . By functoriality (and as $\Omega ^1_{R/A}$ is generated by $dr$ for $r\in R$ ), we can conclude that for every p-completely smooth algebra R over A

induces the canonical morphism, and thus, that $\alpha _R$ is an isomorphism in general.

Proof. See [Reference Illusie24, Chapter III.2.1.2.3], respectively, [Reference Illusie24, Chapter III.2.1.3.3].

Proof. Indeed, splits if and only if the class in

defined by vanishes. Proposition 3.12 shows that this class is the same as the class defined by the p-completed cotangent complex $L_{X/\mathrm {Spf}(A)}^{\wedge _p}\{-1\}$ . Lifting X to a p-completely flat formal scheme over $A/I^2$ is the same as lifting $X\otimes _{A/I}A/(I,p^n)$ to a flat scheme over $A/(I^2,p^n)$ for all $n\geq 1$ (here, we use that $(A,I)$ is bounded in order to know that $A/I$ is classically p-complete). One concludes by applying Proposition 3.13, together with the fact that the p-completion of the cotangent complex does not affect the (derived) reduction modulo $p^n$ .

Proof. See [Reference Bhatt, Morrow and Scholze12, Lemma 4.27] and [Reference Bhatt, Morrow and Scholze12, Lemma 4.29].

Proof. Since the proof is short, we recall it. Choose a surjection

$$ \begin{align*} A/I\langle x_j, j\in J \rangle \to R, \end{align*} $$

for some index set J. Set

$$\begin{align*}S = A/I\langle x_j^{1/p^{\infty}} \rangle \hat{\otimes}_{A/I\langle x_j, j\in J \rangle}^{\mathbb{L}} R. \end{align*}$$

Then $R \to S$ is a quasisyntomic cover, and by assumption, $A/I \to R$ is quasisyntomic: hence, the map $A/I \to S$ is quasisyntomic. Moreover the p-completion of $\Omega _{S/(A/I)}^1$ is zero. We deduce that the map $A/I \to S$ is such that $(L_{S/(A/I)})^{\wedge _p}$ has p-complete Tor-amplitude in degree $[-1,-1]$ . Therefore, by the Hodge-Tate comparison, the derived prismatic cohomology is concentrated in degree $0$ and the map is p-completely faithfully flat. One can thus just take .

Proof. This is [Reference Bhatt and Scholze13, Theorem 7.12]. Since the proof is also short, we recall it. Write $R=A/I$ , for a perfect prism $(A,I)$ (Proposition 2.8). The p-complete R-algebra $\tilde {R}$ obtained by adding roots of all possible monic polynomials over R is a quasisyntomic cover, so by Proposition 3.22, we can find a prism $(B,J)$ over $(A,I)$ with a p-completely faithfully flat morphism $\tilde {R} \to R_1:=B/J$ . Moreover, we can (and do) assume that $(B,J)$ is a perfect prism. Indeed, as $(A,I)$ is perfect, the underlying A-algebra of the perfectionFootnote ¹³ of $(B,J)$ is the $(p,I)$ -completion of a filtered colimit of $(p,I)$ -completely faithfully flat A-algebras, hence is $(p,I)$ -completely faithfully flat. Transfinitely iterating the construction $R\mapsto R_1$ produces the desired ring S.

Proof. Immediate from the definition (cf. [Reference Project52, Tag 00XJ]) and the previous proposition.

Proof. The first assertion is just saying that $u^{-1}$ is exact, as u is a cocontinuous functor ([Reference Project52, Tag 00XL]). The second assertion follows, as any finite locally free group scheme is syntomic (cf. [Reference Breuil16, Proposition 2.2.2]).

Proof. This is [Reference Bhatt and Scholze13, Proposition 7.10] (the second isomorphism, i.e. the fact that is bounded, follows from the last assertion of loc. cit.).

Proof. As $A_{\mathrm {crys}}(R)$ is p-torsion free (cf. [Reference Bhatt, Morrow and Scholze12, Theorem 8.14]) and carries a canonical Frobenius lift, there we get a natural morphism

Conversely, the kernel of the natural morphism (cf. Theorem 3.29, which does not depend on this lemma)

has divided powers (as one checks similarly to [Reference Bhatt, Morrow and Scholze12, Proposition 8.12], using that the proof of Theorem 3.7 goes through in the syntomic case, cf. Remark 3.8). This provides a canonical morphism

in the other direction. Similarly, to [Reference Bhatt, Morrow and Scholze12, Theorem 8.14], one checks that both are inverse to each other.

Proof. See [Reference Bhatt and Scholze13, Theorem 12.2].

Proof. Using [Reference Bhatt, Morrow and Scholze12, Construction 2.1] (respectively, [Reference Lurie40, Proposition 5.5.8.15]) the functor , which is the left Kan extension from p-completely smooth algebras to all derived p-complete simplicial $A/I$ -algebras, commutes with colimits if it preserves finite coproducts. Clearly, , that is preserves the final object. Moreover, for $R,S p$ -completely smooth over $A/I$ , the canonical morphism

is an isomorphisms because this I-completeness may be checked for where it follows from the Hodge-Tate comparison.

Proof. It is enough to produce such an isomorphism functorially on a basis of $(R)_{\mathrm {qsyn}}$ . By Proposition 3.21, we can thus assume that R is quasiregular semiperfectoid. In this case, we conclude by Theorem 3.29.

Proof. Because , it is clear that for all finite locally free $\mathcal {O}^{\mathrm {pris}}$ -modules $\mathcal {M}$ , the canonical morphism

$$ \begin{align*} \mathcal{M}\to v_{\ast}(v^{\ast}(\mathcal{M})) \end{align*} $$

is an isomorphism as this can be checked locally on $(R)_{\mathrm {qsyn}}$ . Conversely, let $\mathcal {N}$ be a finite locally free -module. We have to show that the counit

$$ \begin{align*} v^{\ast}v_{\ast}(\mathcal{N})\to \mathcal{N} \end{align*} $$

is an isomorphism. For any morphism $R\to R^{\prime }$ with $R^{\prime }$ quasisyntomic, there are equivalences

of slice topoi, where $h_{R^{\prime }}(B,J):=\mathrm {Hom}_R(R^{\prime },B/J)$ . By passing to a quasisyntomic cover $R\to R^{\prime }$ , we can therefore assume that R is quasiregular semiperfectoid, in particular, that the site has a final object given by . By $(p,I)$ -completely faithfully flat descent of finitely generated projective modules over $(p,I)$ -complete rings of bounded $(p,I)$ -torsion (cf. Proposition A.3), the category of finite locally free -modules on is equivalent to finitely generated projective -modulesFootnote ¹⁶ . As the morphism (the ‘ $\theta $ ’-map) is henselian along its kernel, cf. Lemma 4.28, finite locally free -modules split on the pullback of an open cover of $\mathrm {Spf}(R)$ . Thus, after passing to a quasisyntomic cover of $\mathrm {Spf}(R)$ , we may assume that $\mathcal {N}$ is finite free. Then the isomorphism

$$ \begin{align*} v^{\ast} v_{\ast}(\mathcal{N})\cong \mathcal{N} \end{align*} $$

is clear.

Proof. This follows from the definition, because by general properties of topoi, modules under $\mathcal {O}^{\mathrm {pris}}$ and $\mathcal {O}$ form a stack for the quasisyntomic topology on $(R)_{\mathrm {qsyn}}$ .

Proof. Let us call $G_R$ , respectively, $F_R$ , the first, respectively, the second, functor displayed in the statement of the proposition. Using Proposition 4.4 and the equivalence between finite locally free -modules and finite locally free -modules, one immediately gets that $F_R$ is an equivalence between the category of prismatic Dieudonné crystals over R and the category of prismatic Dieudonné modules over R, with quasi-inverse given by $G_R$ . Hence, we only need to check that the admissibility conditions on both sides agree.

Let $(M,\varphi _M)$ be an admissible Dieudonné module over R.

Lemma 4.14. Let $R \to R^{\prime }$ be a quasisyntomic morphism, with $R^{\prime }$ being also quasiregular semiperfectoid. Let be the base change of $(M,\varphi _M)$ . Then

The lemma follows from Proposition 4.29 (and Remark 4.21), which will be proved below; let us take it for granted and finish the proof. For any quasiregular semiperfectoid ring $R^{\prime }$ quasisyntomic over R, note that, using the notations from the lemma,

The lemma tells us that, in particular

This being true for any quasiregular semiperfectoid ring $R^{\prime }$ quasisyntomic over R, we deduce that we have a short exact sequence of sheaves on $(R)_{\mathrm {qsyn}}$

$$ \begin{align*}0 \to \varphi_{F_R(M)}^{-1}(\mathcal{I}^{\mathrm{pris}}.F_R(M)) \to F_R(M) \to \mathcal{O} \otimes_R M/\varphi_M^{-1}(I.M) \to 0. \end{align*} $$

By admissibility of $(M,\varphi _M)$ , the rightmost term is a finite locally free $\mathcal {O}$ -module, and thus, $F_R(M)$ is admissible.

Conversely, let $(\mathcal {M},\varphi _{\mathcal {M}})$ be an admissible Dieudonné crystal. Consider the exact sequence of sheaves

$$ \begin{align*}0 \to \varphi_{\mathcal{M}}^{-1}(\mathcal{I}^{\mathrm{pris}}.\mathcal{M}) \to \mathcal{M} \to \mathcal{M}/\varphi_{\mathcal{M}}^{-1}(\mathcal{I}^{\mathrm{pris}}.\mathcal{M}) \to 0, \end{align*} $$

and apply to it the functor $\Gamma (R,-)$ . We get an exact sequence

$$ \begin{align*}0 \to \Gamma(R, \varphi_{\mathcal{M}}^{-1}(\mathcal{I}^{\mathrm{pris}}.\mathcal{M})) = \varphi_{G_R(\mathcal{M})}^{-1}(I.G_R(\mathcal{M})) \to G_R(\mathcal{M}) \to \Gamma(R,\mathcal{M}/\varphi_{\mathcal{M}}^{-1}(\mathcal{I}^{\mathrm{pris}}.\mathcal{M})). \end{align*} $$

Since $\mathcal {M}/\varphi _{\mathcal {M}}^{-1}(\mathcal {I}^{\mathrm {pris}}.\mathcal {M})$ is a finite locally free $\mathcal {O}$ -module by admissibility of $(\mathcal {M},\varphi _{\mathcal {M}})$ , the rightmost term is a finite projective R-module, and it, therefore, suffices to show that the above sequence is also right exact. The map

$$ \begin{align*}G_R(\mathcal{M}) \to \Gamma(R,\mathcal{M}/\varphi_{\mathcal{M}}^{-1}(\mathcal{I}^{\mathrm{pris}}.\mathcal{M})) \end{align*} $$

factors through

which is a map of R-modules, and it suffices to show that this map is surjective. Since the target is a finitely generated R-module and R is p-complete, it suffices by Nakayama’s lemma to prove surjectivity after base change along any surjection $R \to k$ , with k a perfect field of characteristic p. After base change along such a morphism $R \to k$ , the above map factors through

Since $G_R(\mathcal {M})$ , respectively, $\mathcal {M}/\varphi _{\mathcal {M}}^{-1}(\mathcal {I}^{\mathrm {pris}}.\mathcal {M})$ , is a finite locally free $\mathcal {O}^{\mathrm {pris}}$ -module, respectively, a finite locally free $\mathcal {O}$ -module, this identifies with the map

$$ \begin{align*}G_k(\mathcal{M}_k) \to \Gamma(k,\mathcal{M}_k/\varphi_{\mathcal{M}_k}^{-1}(\mathcal{I}^{\mathrm{pris}}.\mathcal{M}_k)), \end{align*} $$

that is the same map as the one we originally wanted to prove is surjective, but now with R replaced by k (we denoted with an index k the restrictions of the various objects involved to the quasisyntomic site of k). But since k is perfect, $(G_k(\mathcal {M}_k),\varphi _{G_k(\mathcal {M}_k)})$ is automatically admissible, by definition of admissibility using that every k-module is free. Hence, as proved above, we have an exact sequence (using that $F_k \circ G_k \cong \mathrm {Id}$ )

$$ \begin{align*}0 \to \varphi_{\mathcal{M}_k}^{-1}(\mathcal{I}^{\mathrm{pris}}.\mathcal{M}_k) \to \mathcal{M}_k \to \mathcal{O} \otimes_R G_k(\mathcal{M}_k)/\varphi_{G_k(\mathcal{M}_k)}^{-1}(I.G_k(\mathcal{M}_k)) \to 0, \end{align*} $$

that is

$$ \begin{align*}\mathcal{M}_k/\varphi_{\mathcal{M}_k}^{-1}(\mathcal{I}^{\mathrm{pris}}.\mathcal{M}_k) \cong \mathcal{O} \otimes_R G_k(\mathcal{M}_k)/\varphi_{G_k(\mathcal{M}_k)}^{-1}(I.G_k(\mathcal{M}_k)), \end{align*} $$

hence

$$ \begin{align*}\Gamma(k,\mathcal{M}_k/\varphi_{\mathcal{M}_k}^{-1}(\mathcal{I}^{\mathrm{pris}}.\mathcal{M}_k)) \cong G_k(\mathcal{M}_k)/\varphi_{G_k(\mathcal{M}_k)}^{-1}(I.G_k(\mathcal{M}_k)). \end{align*} $$

This shows that the map

$$ \begin{align*}G_k(\mathcal{M}_k) \to \Gamma(k,\mathcal{M}_k/\varphi_{\mathcal{M}_k}^{-1}(\mathcal{I}^{\mathrm{pris}}.\mathcal{M}_k)) \end{align*} $$

is surjective, as desired.

Proof. This is a combination of [Reference Lau33, Remark 2.4] and [Reference Lau33, Lemma 2.5]. Let us give some details, and set $S:=A/\mathrm {Fil}~A$ . The module $S\otimes _{A}M$ decomposes, as $M/\mathrm {Fil}~M$ is finite projective, into a direct sum $S\otimes _{A}M\cong M/\mathrm {Fil}~M\oplus Q$ for some finite projective S-module Q. Let $L,T$ be finite projective A-modules, such that L is a lift of Q and T a lift of $M/\mathrm {Fil}~M$ . We can then lift the decomposition $S\otimes _{A}M$ to a decomposition $M=L\oplus T$ by projectivity. The property $\mathrm {Fil}~M=L\oplus \mathrm {Fil}~A T$ follows. Given $\varphi _{M}$ , we define $\Psi (l+t):=\varphi _{M,1}(l)+\varphi _M(t)$ for $l\in L, t\in T$ on $M=L\oplus T$ , and conversely, given $\Psi $ , we set $\varphi _M(l+t):=\varpi \Psi (l)+\Psi (t)$ and $\varphi _{M,1}(l+at):=\Psi (l)+\varphi _1(a)\Psi (t)$ for $l\in L,t\in T,a\in \mathrm {Fil}~A$ .

Proof. Let

$$ \begin{align*}M=L \oplus T \end{align*} $$

be a normal decomposition of M as in 4.22, and

$$ \begin{align*}\Psi= (\varphi_{M,1})_{|_L} + (\varphi_{M})_{|_T}, \end{align*} $$

so that $\mathrm {Fil} ~ M= L \oplus \mathrm {Fil}~ A.T$ . Let $x=l+t \in M$ , such that $\varphi _M(x) \in \varpi M$ . We have

$$ \begin{align*}\varphi_M(x) = \varpi \Psi(l) + \Psi(t) \end{align*} $$

so the condition is equivalent to requiring that $\Psi (t) \in \varpi. M$ . For simplicity, we assume that $L, T$ are free A-modules in the following. The general case follows by localisation. Fix a basis $t_1,\dots ,t_r$ of T and a basis $l_1,\dots ,l_s$ of L, as A-modules. Since $\Psi $ is a $\varphi $ -linear isomorphism, the family $(\Psi (t_1),\dots ,\Psi (t_r),\Psi (l_1),\dots ,\Psi (l_s))$ is a basis of M, and so the reduction of the family $(\Psi (t_1),\dots ,\Psi (t_r))$ modulo $\varpi $ is linearly independent. Write $t=\sum _{i=1}^r a_i t_i$ , with $a_i \in A$ for all $i=1,\dots , r$ . By assumption, we have that

$$ \begin{align*}\Psi(t) = \sum_{i=1}^r \varphi(a_i) \Psi(t_i) \in \varpi. M, \end{align*} $$

and therefore we must have $\varphi (a_i) \in \varpi A$ for all $i=1,\dots ,r$ , that is $a_i \in \mathrm {Fil} ~A$ for all $i=1,\dots ,r$ , by the condition on $\mathrm {Fil}~A$ . Hence, $t \in \mathrm {Fil}~A.T$ and thus $x \in \mathrm {Fil}~M$ , as desired. For the last statements note that the map $\varphi _M\colon M/\mathrm {Fil} M\cong T/\mathrm {Fil} A.T\to M/\varpi $ identifies with the map induced by $\Psi $ . As $\varphi _M(t_i)=\Psi (t_i), i=1,\ldots , r,$ are linearly independent (over $A/\varpi $ ), this map extends to an inclusion

$$\begin{align*}A/\varpi\otimes_{A/\mathrm{Fil} A} M/\mathrm{Fil} {M} \to M/\varpi \end{align*}$$

of a direct summand. This finishes the proof.

Proof. See [Reference Cais and Lau17, Lemma 2.1.16] (taking Remark 4.25 into account).

Proof. If $M,N$ are finite projective A-modules, then any isomorphism $M/IM\cong N/IN$ can be lifted to a morphism $M\to N$ by projectivity of M. As $I\subseteq A$ lies in the Jacobson radical of A, this lifted homomorphism is then automatically an isomorphism. Moreover, any finite projective $A/I$ -module can be lifted to a finite projective A-module by [Reference Project52, Tag 0D4A].

Proof. Because is $(p,\xi )$ -adically complete, it suffices to prove that the pair

is henselian (cf. [Reference Project52, Tag 0DYD]). We know . Hence, for every element $x\in \ker (\theta )$ , $x^p\in (p,\tilde {\xi })$ . As locally nilpotent ideals are henselian, the claim follows.

Proof. Thanks to Lemma 4.28, we can apply Lemma 4.23 to the frame . This yields fully faithfulness, and that for a window $(M,\mathrm {Fil}~M,\varphi _M,\varphi _{M,1})$ , the image of

$$\begin{align*}M\xrightarrow{\varphi_M}M\to M/I\cdot M \end{align*}$$

identifies with $M/\mathrm {Fil}~M$ . By Lemma 4.23, we can deduce admissibility. Assume conversely that $(M,\varphi _M)$ is an admissible prismatic Dieudonné module. Then the datum $(M,\varphi _M^{-1}(I\cdot M), \varphi _M, \frac {1}{\tilde {\xi }}\varphi _M)$ is a window over . Indeed, the condition that $\varphi _{M,1}(\mathrm {Fil} M)$ generates M follows from the definition of admissibility and Remark 4.7. This finishes the proof.

Proof of Proposition 4.12.

We know by Proposition 4.29 that the functor

$$ \begin{align*}(M, \varphi_M) \mapsto \left(M,\varphi_M^{-1}(\tilde{\xi}. M), \varphi_M, \frac{1}{\tilde{\xi}} \varphi_M\right) \end{align*} $$

is an equivalence between $\mathrm {DM}^{\mathrm {adm}}(R)$ and . Since R is perfectoid, and so

By Proposition 4.26, the functor

$$\begin{align*}(N, \mathrm{Fil}~N, \varphi_N) \mapsto \left(\mathrm{Fil}~N, \frac{\xi}{\tilde{\xi}} \varphi_{N}\right) \end{align*}$$

induces an equivalence between and $\mathrm {BK}_{\mathrm {min}}(A_{\mathrm {inf}}(R))$ (the category of minuscule Breuil-Kisin modules over $A_{\mathrm { inf}}(R)$ ). The latter category is, however, obviously equivalent to $\mathrm {DM}(R')$ , with $R'=A_{\mathrm {inf}}(R) / \xi $ . As $\varphi $ is bijective on , base change along $\varphi $ is also an equivalence between $\mathrm {DM}(R')$ and $\mathrm {DM}(R)$ . Composing these equivalences, we obtain an equivalence

$$ \begin{align*}\mathrm{DM}^{\mathrm{adm}}(R) \to \mathrm{DM}(R). \end{align*} $$

But this composite functor is nothing but the identity functor.

Proof. To prove fully faithfulness, it suffices to show (by passing to internal homs) that for every $\varphi $ -module $(M,\varphi _M)$ over A, the map

$$ \begin{align*}M^{\varphi_M=1}\to (M/JM)^{\varphi_M=1} \end{align*} $$

is bijective. Let $m\in M^{\varphi _M=1}\cap JM$ , and write $m=\sum \limits _{i=1}^na_im_i$ with $a_i\in J$ and $m_i\in M$ . Then

$$ \begin{align*}m=\varphi^j_M(m)=\sum\limits_{i=1}^n\varphi^j(a_i)\varphi_M^j(m_i), \end{align*} $$

where the $\varphi ^j(a_i)$ converge to $0$ if $j\to \infty $ by our assumption on $\varphi $ . Thus, $m=\varphi ^j_M(m)\to 0$ if $j\to \infty $ and therefore $m=0$ , which proves injectivity. Conversely, let $m\in M$ and assume that $\varphi _M(m)\equiv m$ modulo $JM$ . Write

$$ \begin{align*}z:=\varphi_M(m)-m\in JM. \end{align*} $$

As above, the sequence $\varphi _M^j(z)$ converges to $0$ if $j\to \infty $ . Set

$$ \begin{align*}\tilde{m}:=m+\sum\limits_{j=0}^{\infty} \varphi_M^j(z). \end{align*} $$

Then $\tilde {m}\equiv m$ modulo $JM$ and $\varphi _M(\tilde {m})=\tilde {m}$ . Thus, we showed that

$$ \begin{align*}M^{\varphi_M=1}\cong (M/JM)^{\varphi_M=1} \end{align*} $$

and the functor $\varphi -\mathrm {Mod}_A^{\mathrm {unit}}\to \varphi -\mathrm {Mod}_B^{\mathrm {unit}}$ is fully faithful, and we are left with essential surjectivity. For this, let $(N,\varphi _N)\in \varphi -\mathrm {Mod}_B^{\mathrm {unit}}$ . By assumption, A is henselian along J and, thus, we can write $N\cong M\otimes _A B$ for some finite projective A-module M. Using projectiviy of $\varphi ^{\ast } M$ over A, we can lift $\varphi _N\colon \varphi ^{\ast } N\to N$ to some homomorphism $\varphi _M\colon \varphi ^{\ast } M\to M$ . As J lies in the radical of A, the homomorphism $\varphi _M$ will automatically be an isomorphism as $\varphi _N$ is. Thus, we have lifted $(N,\varphi _N)$ to $(M,\varphi _M)$ , which finishes the proof.

Proof. In this proof, we will use the following convenient notation: if $\sigma : S\to S$ is a ring endomorphism and $f: M \to N$ is a $\sigma $ -linear map between two S-modules, we will denote by $f^{\sharp }: \sigma ^{\ast } M \to N$ its linearisation. We will also abbreviate $\underline {A}_{\mathrm {Nyg}}$ as $\underline {A}$ and $(B,\mathcal {N}^{\geq 1} A/J,A/\mathcal {N}^{\geq 1}A, \varphi ,\varphi _1)$ as $\underline {B}$ .

By the existence of normal decompositions (cf. Proposition 4.22: we can apply it since the proof of Lemma 4.28 shows that A is henselian along $\mathcal {N}^{\geq 1} A$ , and this implies that finite projective $B/\mathrm {im}(\mathcal {N}^{\geq 1} A)$ -modules can be lifted to finite locally free B-modules — even to finite projective A-modules) and the fact that A is henselian along J, the base change functor

$$ \begin{align*}\mathrm{Win}(\underline{A})\to \mathrm{Win}(\underline{B}) \end{align*} $$

is essentially surjective. Let $\underline {M}, \underline {N}$ be two windows over $\underline {A}$ . We want to prove that

$$ \begin{align*}\mathrm{Hom}_{\underline{A}}(\underline{M},\underline{N})\cong \mathrm{Hom}_{\underline{B}}(\underline{M}/J,\underline{N}/J), \end{align*} $$

where $\underline {M}/J, \underline {N}/J$ denote the base change of $\underline {M},\underline {N}$ to $\underline {B}$ . The idea of the proof is similar to Lemma 4.31 (and [Reference Lau33, Theorem 3.2]). Let

$$ \begin{align*}\beta\colon M\to JN \end{align*} $$

be an arbitrary homomorphism of A-modules. Then the A-module homomorphism

$$ \begin{align*}U(\beta)\colon M\to JN,\ m\mapsto \varphi_{N,1}^{\sharp} (\mathrm{Id}\otimes \beta)(\varphi_{M,1}^{\sharp})^{-1}(m) \end{align*} $$

is well defined. Indeed, $\varphi _M^{\sharp } \colon \varphi ^{\ast } M\to M$ is injective with cokernel killed by $\tilde {\xi }$ (which follows from the fact that $\varphi _{M,1}(\mathrm {Fil} M)$ generates M and that $M,\varphi ^{\ast }(M)$ are $\tilde {\xi }$ -torsion free), and thus, on $\tilde {\xi } M$ , there exists a partial inverse ${(\varphi _M^{\sharp })}^{-1}\colon \tilde {\xi } M\to \varphi ^{\ast }M$ of ${\varphi _M^{\sharp }}$ . Moreover, as $\beta $ has image in $JN$ , the composition $\varphi _N^{\sharp } (\mathrm {Id}\otimes \beta )$ has image in $\tilde {\xi } N$ . The module M is finitely generated: choose generators $x_1, \dots , x_r$ . For each $n\geq 1$ , and each $x \in M$ , we can write

$$ \begin{align*}{((\varphi^{n-1})^{\ast} (\varphi_{M,1}^{\sharp})^{-1} \circ \dots \circ \varphi^{\ast} (\varphi_{M,1}^{\sharp})^{-1} \circ (\varphi_{M,1}^{\sharp})^{-1} (x) } = \sum_{i=1}^r b_{i,n}(x) \otimes x_i \in (\varphi^n)^{\ast} M, \end{align*} $$

with $b_{i,n}(x) \in A$ . Hence, we get

$$ \begin{align*}U^n(\beta)(x)= {(((\varphi_{N,1})^{\sharp} \circ \varphi^{\ast} (\varphi_{N,1})^{\sharp} \circ \dots \circ (\varphi^{n-1})^{\ast} (\varphi_{N,1})^{\sharp}) }\circ (\varphi^n)^{\ast} \beta) \left( \sum_{i=1}^r b_{i,n}(x) \otimes x_i \right), \end{align*} $$

whence

$$ \begin{align*}U^n(\beta)(x)= \sum_{i=1}^r \varphi^n(b_{i,n}(x)) \varphi_{N,1}^n(\beta(x_i)). \end{align*} $$

Write for each $i=1,\dots , r$ ,

$$ \begin{align*}\beta(x_i)=\sum_{k=1}^{s_r} j_{i,k} y_{i,k}, \end{align*} $$

with $j_{i,k} \in J$ , $y_{i,k} \in N$ . We have, for each $i=1,\dots , r$ ,

$$ \begin{align*}\varphi_{N,1}^n(\beta(x_i)) = \varphi_{N,1}^n \left(\sum_{k=1}^{s_r} j_{i,k} y_{i,k}\right) = \sum_{k=1}^{s_r} \varphi_1^n(j_{i,k}) \varphi_{N}^n(y_{i,k}). \end{align*} $$

By our assumption, $\varphi _1$ on J is pointwise topologically nilpotent, and so, in particular, for each $m_0 \geq 0$ , we can find $m\geq 0$ , such that $\varphi _1^m(j_{i,k}) \in (p,\tilde {\xi })^{m_0}$ , for all $i=1,\dots , r$ , $j=1,\dots , s_r$ . The above equalities show that for all $n\geq m$ and for all $x\in M$ ,

$$ \begin{align*}U^n(\beta)(x) \in (p,\tilde{\xi})^{m_0} N. \end{align*} $$

Hence, we deduce from the above that for every $\beta \colon M\to JN$ , the sequence

$$ \begin{align*}\beta, U(\beta),U(U(\beta)),\ldots, U^n(\beta),\ldots \end{align*} $$

converges to $0$ (because A is $(p,\tilde {\xi })$ -adically complete as $(A,\tilde {\xi })$ is bounded). Now let $\alpha \colon M\to N$ be a homomorphism of windows, such that $\alpha \equiv 0$ modulo J. Then $U^n(\alpha )=\alpha $ for all n because $\alpha \circ \varphi _M=\varphi _N\circ \alpha $ , which implies $\alpha =0$ as the sequence $U^n(\alpha )$ converges to $0$ as we saw above. Conversely, assume that $\alpha \colon M\to N$ is an A-module homomorphism, such that $\alpha $ modulo J is an homomorphism of windows over $\underline {B}$ . Then $\alpha $ maps $\mathrm {Fil} M$ to $\mathrm {Fil} N$ because this can be checked modulo J. Note that $(\varphi _M^{\sharp })^{-1}(\tilde {\xi }.M)= \varphi ^{\ast }(\mathrm {Fil} M)$ as follows from Lemma 4.23. Hence, $U(\alpha )$ sends M to N. Set