Proof. Pick any $\overline x \in M ^{\rm m}$. Since $\varphi$ on $M^{\rm m}$ is bijective, there exists a unique $\overline x_n \in M$ such that $\varphi ^n (\overline x_n ) = \overline x$. Select $x_n\in \mathfrak {M}$ a lift of $\overline x_n$ and define $[\overline x] : = \lim _{n \to \infty } \varphi ^n (x_n)$. We first check that $\varphi ^n (x_n)$ converges to an $x \in \mathfrak {M}$ such that $q(x) = \overline x$. Indeed, since $\varphi (\overline x_{n +1}) = \overline x_n$, $\varphi (x_{n +1}) -x_n = u y _n$ with $y _n \in \mathfrak {M}$. So $\varphi ^{n +1} (x_{n +1}) - \varphi ^n (x_n) = \varphi ^n (u) \varphi ^n (y )$ and hence $\varphi ^n (x_n)$ converges to an $x \in \mathfrak {M}$ and clearly $q(x)= \overline x$. Suppose that $x_n' \in \mathfrak {M}$ is another lift of $\overline x_n$. Then $x_n' - x_n = u z_n$ with $z_n \in \mathfrak {M}_n$. Then $\varphi ^n (x'_n)-\varphi ^n (x_n ) = u ^{p ^n} \varphi ^n (z_n)$. So $\{\varphi ^n (x_n')\}$ also converges to $x$. This implies that $x= [\overline x]$ does not depend on the choice of lift $x_n$ of $\overline x_n = \varphi ^{-n} (\overline x)$. Hence, the section $[\cdot ]: M^{\rm m}\to \mathfrak {M}$ is well defined and satisfies $q \circ [\cdot ] = \operatorname {id}_{M^{\rm m}}$. For any $a\in W(k)$, it is clear that $a [\overline x] = [a \overline x]$ from construction of $[\overline x]$. So $[\cdot ]$ is $W(k)$-linear. If $\overline y = \varphi (\overline x)$ then $\varphi (x_n)$ is a lift of $\varphi ^{-n} (y )$ and $[\overline y] = \lim _{n \to \infty } \varphi ^{n+1} (x_n) = \varphi (x)= \varphi ([\overline x])$. So $[\cdot ]$ is $\varphi$-equivariant. Finally, suppose there is another section $[\cdot ]': M^{\rm m} \to \mathfrak {M}$. Then $[ \overline x] - [\overline x]' \in u \mathfrak {M}$ for any $\overline x \in M ^{\rm m}$. Then $[ \overline x] - [\overline x]'= \varphi ^n ([\overline x_n]- [\overline x_n]')\in u ^{p ^n}\mathfrak {M}$. This forces $[\overline x]= [\overline x]'$.

Proof. Note that [Reference KisinKis09, Prop. (1.2.11)] has treated the situation that $\mathfrak {M}$ has no $u$-torsion, but our idea here is slightly different. By the above lemma, we can set $\mathfrak {M}^{\rm m}$ to be the ${\mathfrak {S}}$-submodule of $\mathfrak {M}$ generated by $[M^{\rm m}]$ and $\mathfrak {M}^{\rm n}: = \mathfrak {M} / \mathfrak {M} ^{\rm m}$. Clearly, $1 \otimes \varphi : {\mathfrak {S}} \otimes _{\varphi, {\mathfrak {S}}} \mathfrak {M}^{\rm m} \to \mathfrak {M}^{\rm m}$ is surjective. Consider the right exact sequence ${\mathfrak {S}} \otimes _{W(k)} [M^{\rm m}] \to \mathfrak {M} \to \mathfrak {M}^{\rm n}\to 0$. Modulo $u$, we have the right exact (indeed, exact) sequence $M^{\rm m}\to M \to \mathfrak {M}^{\rm n}/ u \mathfrak {M} ^{\rm n}\to 0$. So $\mathfrak {M}^{\rm n}/ u \mathfrak {M} ^{\rm n}\simeq M ^{\rm n}$ and also forces $\mathfrak {M}^{\rm m}/ u \mathfrak {M} ^{\rm m} = M^{\rm m}$. Hence, $\varphi$ on $\mathfrak {M}^{\rm n}$ is topologically nilpotent as $\varphi$ on $M^{\rm n}$ is nilpotent, thus $\mathfrak {M}^{\rm n}$ cannot have non-trivial multiplicative submodule. So we obtain exact sequence (2.4) which is functorial for $\mathfrak {M}$ because $[\cdot ]$ is clearly functorial for $\mathfrak {M}$ by the above lemma.

If $\mathfrak {M} \in \operatorname {Mod}_{{\mathfrak {S}} , \operatorname {tor}}^{\varphi, h, {\rm c}}$ then $\mathfrak {M}^{\rm m}$ has no $u$-torsion. Note that the exact sequence (2.4) modulo $u$ becomes the exact sequence $0 \to M^{\rm m} \to M \to M ^{\rm n}\to 0$. Then $\mathfrak {M}^{\rm n}$ cannot have $u$-torsion as $\mathfrak {M}$ has no $u$-torsion. Hence, both $\mathfrak {M} ^{\rm m}$ and $\mathfrak {M}^{\rm n}$ have no $u$-torsion. Then both $\mathfrak {M} ^{\rm m}$ and $\mathfrak {M}^{\rm n}$ have $E$-height $h$ by [Reference FontaineFon90, Prop. B 1.3.5] as required.

Proof. Suppose $x \in M$ is a lift of a $u$-torsion in $M/(M[p^n] + pM)$, hence satisfies ${u \cdot x = y + p \cdot z}$ for some $y \in M[p^n]$ and $z \in M$. Multiplying the equation by $p^n$, we get $u \cdot p^n \cdot x = p^{n+1} \cdot z$. As $M/p^{n+1}M$ also has no $u$-torsion by assumption, we see that $p^n \cdot x = p^{n+1} \cdot \widetilde {z}$ for some $\widetilde {z} \in M$. Writing $x = (x - p \cdot \widetilde {z}) + p \cdot \widetilde {z}$ shows that in fact $x \in M[p^n] + pM$, as required.

Proof. First let us treat the case when $M$ is torsion. In this case $M$ is killed by a power of $p$ (see [Reference Bhatt, Morrow and ScholzeBMS18, Proposition 4.3(i)]). Denote ${\rm Im}(M \xrightarrow {p} M) = pM =: M_1$. We claim $M_1/p^nM_1$ are also $u$-torsion-free for all $n > 0$. Granting this claim, by induction on the exponent of $p$ annihilating $M$, we know $M_1$ satisfies the conclusion. Here, for the starting point of induction, we used the fact that a finitely generated ${\mathfrak {S}}/p$-module is $u$-torsion-free if and only if it is free. Then by [Reference MinMin21, Lemma 5.9], we get the conclusion for $M$.

We now verify the claim. Applying the snake lemma to

yields an exact sequence

\[ 0 \to M/(M[p^n] + pM) \to M_1/p^nM_1 \to M/p^nM. \]

Here $M[p^n]$ denotes the $p^n$-torsion in $M$. Since $M/p^nM$ has no $u$-torsion by assumption, it suffices to show the same for $M/(M[p^n] + pM)$. Applying Lemma 2.5 gives the claim.

Next we turn to the general case. By [Reference Bhatt, Morrow and ScholzeBMS18, Proposition 4.3], we have two short exact sequences of generalized Breuil–Kisin modules

\[ 0 \to M_{\operatorname{tor}} \to M \to M_{\operatorname{tf}} \to 0 \]

and

\[ 0 \to M_{\operatorname{tf}} \to M_{\operatorname{fr}} \to M_0 \to 0. \]

Here $M_{\operatorname {tor}}$ is the torsion submodule, $M_{\operatorname {tf}}$ is the torsion-free quotient, $M_{\operatorname {fr}}$ is the reflexive hull of $M$ (which is free as ${\mathfrak {S}}$ is a two-dimensional regular Noetherian domain), and $M_0$ has finite length. The first sequence implies that $M_{\operatorname {tor}}/p^n$ injects into $M/p^n$, therefore $M_{\operatorname {tor}}$ satisfies the assumption. Since we have treated the torsion case, we see that $M_{\operatorname {tor}}$ satisfies the conclusion. Now we claim $M_0$ vanishes. This immediately implies that $M_{\operatorname {tf}} = M_{\operatorname {fr}}$ is free, hence the first sequence splits, and $M = M_{\operatorname {tor}} \oplus M_{\operatorname {tf}}$ has the shape of a $\mathbb {Z}_p$-module.

Finally, let us justify the claim that $M_0 = 0$. Taking the second sequence above, derived mod $p$, gives an inclusion $M_0[p] \subset M_{\operatorname {tf}}/p$. Since $M_0$ has finite length, we see that $M_0[p]$ must be $u^{\infty }$-torsion. If we can show that $M_{\operatorname {tf}}/p$ is $u$-torsion-free, then we get $M_0[p] = 0$ which implies $M_0 = 0$ as it must be $p^{\infty }$-torsion. We are now reduced to showing $M/(M_{\operatorname {tor}} + p \cdot M)$ is $u$-torsion-free. Since $M_{\operatorname {tor}} = M[p^n]$ for sufficiently large $n$, we finish the proof by appealing to Lemma 2.5.

Proof. Recall $I_+ = S \cap u K_0 [\![u]\!]$, $S/ I _+\simeq W(k)$ and $\varphi (x) = c_1^{-h} \varphi _h (E^h x)$. Write $S_n : = S/ p ^n S$ and assume that $\mathcal {M}$ is an $S_n$-module. We claim Lemma 2.2 still holds by replacing $\mathfrak {M}$ by $\mathcal {M}$, $M = \mathcal {M}/ I_+$ and $q : \mathcal {M} \to M = \mathcal {M} / I _+$. Indeed, the same proof goes through because $\varphi ^\ell (I_+)=0$ in $S_n$ for sufficient large $\ell$. Now we can set $\mathcal {M}^{\rm m}$ as the $S$-submodule of $\mathcal {M}$ generated by $[M^{\rm m}]$ and $\mathcal {M}^{\rm n} : = \mathcal {M} / \mathcal {M}^{\rm m}$. Using the same argument as in Lemma 2.3, the right exact sequence $S \otimes _{W(k)} [M^{\rm m}] \to \mathcal {M} \to \mathcal {M}^{\rm n} \to 0$ modulo $I_+$ becomes an exact sequence $0 \to M^{\rm m} \to M \to M^{\rm n} \to 0$. This forces $\mathcal {M}^{\rm m}/ I_+ = M ^{\rm m}$ and $\mathcal {M}^{\rm n} / I_+ = M ^{\rm n}$. Set $\operatorname {Fil} ^h \mathcal {M} ^m = \operatorname {Fil} ^h S \cdot \mathcal {M}^{\rm m}$ and $\operatorname {Fil} ^h \mathcal {M}^{\rm n} = \operatorname {Fil}^h \mathcal {M} / \operatorname {Fil}^h \mathcal {M}^{\rm m}$. It is clear that $\varphi _h: \operatorname {Fil} ^h \mathcal {M} ^{\rm m} \to \mathcal {M} ^{\rm m}$ and $\varphi _h : \operatorname {Fil} ^h \mathcal {M} ^{\rm n} \to \mathcal {M} ^{\rm n}$ are well defined. So we obtain an exact sequence $0 \to \mathcal {M} ^{\rm m} \to \mathcal {M} \to \mathcal {M}^{\rm n} \to 0$ in the category $^\sim \operatorname {Mod}^{\varphi, h}_S$.

To promote our exact sequence to the category $\operatorname {Mod}^{\varphi, h}_{S, \operatorname {tor}}$, we argue by induction on $n$ where $p ^n$ kills $\mathcal {M}$. The base case $n =1$ is most complicated and postponed to the end. For general $n$, by definition, $\mathcal {M}$ sits in the exact sequence in $\operatorname {Mod}^{\varphi, h}_{S, \operatorname {tor}}: \ 0 \to \mathcal {M}_1 \to \mathcal {M} \to \mathcal {M}_2 \to 0$ with $\mathcal {M}_1$, $\mathcal {M}_2$ killed by $p ^{n -1}$ and $p$, respectively. Consider the following commutative diagram:

(2.11)

We need to show that the first columns is short exact. Note that $\mathcal {M} _2$ is finite $S_1$-free, and the exact sequence in the second column yields the exact sequence $0 \to M _1 \to M \to M_2 \to 0$ where $M_i : = \mathcal {M}_i / I _+ \mathcal {M}_i$ for $i= 1, 2$. So the sequence $0 \to \mathcal {M}_1^{\rm m } / I _+ \to \mathcal {M} ^{\rm m}/I_+ \to \mathcal {M}_2^{\rm m}/ I_+ \to 0$ is also exact as it is the same as the exact sequence $0 \to M_1^{\rm m} \to M ^{\rm m} \to M_2^{\rm m} \to 0$. Note that $\mathcal {M}_i^{\rm m }$ is finite $S$-generated as they are generated by $[M_i^{\rm m}]$. Note that $S_n$ is a coherent ring (see [Reference Li and LiuLL20, Lemma 7.15]). By induction on $n$ and [Sta21, Tag 05CW], we see that $\mathcal {M}$ is coherent and then $\mathcal {M}^{\rm m}$ is coherent. Since $\mathcal {M}_1^{\rm m}$ is coherent by induction, $\mathcal {L} = \mathcal {M}^{\rm m}/ \mathcal {M}_1^{\rm m}$ is also coherent by [Sta21, Tag 05CW] again. Note that $f$ induces a map $f': \mathcal {L} \to \mathcal {M}_2 ^{\rm m}$. We need to show that $f'$ is an isomorphism. Let $L = \mathcal {L} / I_+$. Note that $\overline f' : = f' \ {\rm mod}\ I_+ : L \to M^{\rm m}_2$ is an isomorphism. Nakayama's lemma shows that $f'$ is surjective. Let $\mathcal {K}: = \ker (f')$, which is still coherent. Since $\mathcal {M}^{\rm m}_2$ is finite $S_1$-free by induction, $\operatorname {Tor}_1^S (\mathcal {M}^{\rm m }_2, S/I_+ )= 0$. So we obtain an exact sequence $0 \to \mathcal {K} / I_+ \to L \to M_2^{\rm m} \to 0.$ Hence $\mathcal {K} /I_+ = 0$ as $\overline f'$ is an isomorphism. By Nakayama's lemma, $\mathcal {K} = 0$ and the first column is exact as a finite $S$-module. Using that $\mathcal {M}_2^{\rm m}$ is finite $S_1$-free, we see that the sequence $0 \to \mathcal {M}_1^{\rm m}/ \operatorname {Fil} ^ h S \to \mathcal {M} ^{\rm m}/\operatorname {Fil}^h S\to \mathcal {M}_2^{\rm m}/ \operatorname {Fil} ^h S \to 0$ is exact. So the sequence $0 \to \operatorname {Fil} ^ h S \cdot \mathcal {M}_1^{\rm m} \to \operatorname {Fil} ^ h S \cdot \mathcal {M} ^{\rm m} \to \operatorname {Fil} ^ h S \cdot \mathcal {M}_2^{\rm m} \to 0$ is exact. Therefore, the first column of (

) is exact in $\operatorname {Mod}_{S, \operatorname {tor}}^{\varphi, h}$. Then it is standard to check that the last column is also exact sequence in $'\operatorname {Mod}_{S, \operatorname {tor}}^{\varphi, h}$. In particular, $\mathcal {M}^{\rm n}$ is an object in $\operatorname {Mod}_{S, \operatorname {tor}}^{\varphi, h}$ by induction on $n$. Once (2.10) is exact in $\operatorname {Mod}_{S, \operatorname {tor}}^{\varphi, h}$. Then $\varphi$ on $\mathcal {M}^{\rm m}$, $\mathcal {M}$ and $\mathcal {M}^{\rm n}$ defined from $\varphi (x) = c_1^{-h}\varphi _h (x)$ are compatible with maps in the sequence. Since $\mathcal {M}^{\rm m}$ is generated by $[M^{\rm m}]$ and $\mathcal {M}^{\rm n}/ I_+ = M^{\rm n}$, we see that $\mathcal {M} ^{\rm m}$ is multiplicative and $\mathcal {M}^{\rm n}$ is nilpotent.

Now we discuss the case $n =1$. We have already shown that $\mathcal {M}^{\rm m}$ is finite $S$-generated. Now the exact sequence $0 \to \mathcal {M}^{\rm m}/ I_+ \to \mathcal {M}/ I _+ \to \mathcal {M}^{\rm n}/ I_+ \to 0$ is an exact sequence of $k$-vector spaces. Pick $m_i \in \mathcal {M} ^{\rm m}$ and $n_j \in \mathcal {M}$ such that $m_i \ {\rm mod}\ I_+$ and $n_j \ {\rm mod}\ I_+$ are bases of $\mathcal {M}^{\rm m}/ I_+$ and $\mathcal {M}^{\rm n}/ I_+$ respectively. Using that $\mathcal {M}$ is finite $S_1$-free, it is easy to show that $m_i , n_j$ forms a basis of $\mathcal {M}$ and then both $\mathcal {M}^{\rm m}$ and $\mathcal {M}^{\rm n}$ are finite $S_1$-free. Now it remains to show that $\operatorname {Fil}^h \mathcal {M} \cap \mathcal {M} ^{\rm m} = \operatorname {Fil}^h S \mathcal {M}^{\rm m}$ such that $\operatorname {Fil}^h \mathcal {M}^{\rm n}= \operatorname {Fil}^h \mathcal {M} / \operatorname {Fil}^h \mathcal {M}^{\rm m}$ is a submodule of $\mathcal {M} ^{\rm n}$. Then it is easy to check that $(\mathcal {M}^{\rm n}, \operatorname {Fil}^h \mathcal {M}^{\rm n}, \varphi _h)$ is an object in $\operatorname {Mod}_{S, \operatorname {tor}}^{\varphi, h}$ and thus the sequence $0 \to \mathcal {M} ^{\rm m} \to \mathcal {M} \to \mathcal {M}^{\rm n} \to 0$ is in the category $\operatorname {Mod}^{\varphi, h}_{S, \operatorname {tor}}$. To show that $\operatorname {Fil}^h \mathcal {M}^{\rm n}= \operatorname {Fil}^h \mathcal {M} / \operatorname {Fil}^h \mathcal {M}^{\rm m}$, consider $\mathcal {F}: = \mathcal {M} ^{\rm m}/ \operatorname {Fil}^ p S_1\mathcal {M}^{\rm m}$. Write $\widetilde {\operatorname {Fil}}^h \mathcal {F} : = (\operatorname {Fil}^h \mathcal {M} \cap \mathcal {M} ^{\rm m}) / \operatorname {Fil} ^p S_1$ and $\operatorname {Fil}^h \mathcal {F} = \operatorname {Fil}^h S \mathcal {M}^{\rm m}/\operatorname {Fil}^p S_1$. Since $\operatorname {Fil}^h \mathcal {F}= u ^{eh}\mathcal {F} \subset \widetilde {\operatorname {Fil}}^h \mathcal {F} \subset \mathcal {F}$ which is a finite free ${k[\![u]\!]} /u^{pe}$-module, there exists a basis $e_1 , \ldots, e_d$ of $\mathcal {F}$ such that $\widetilde {\operatorname {Fil}}^h \mathcal {F}$ is generated by $u ^{a_i} e_i$ with $0 \leq a_i \leq eh$. Suppose one of the $a_i$ is less than $eh$, say $a_1< eh$. Let $\hat e_i \in \mathcal {M}^{\rm m}$ be a basis which lifts $e_i$. Then $u^{a_1} \hat e_1 \in \operatorname {Fil} ^h \mathcal {M} \cap \mathcal {M} ^{\rm m}$. So $\varphi _h (u ^{eh} \hat e _1) = \varphi _h (u^{eh -a_1 } u ^a_i \hat e_1)= \varphi (u^{eh -a_1}) \varphi _h (u ^{a_1} \hat e_1)\in I_+ \mathcal {M}$. This contradicts that $\varphi _h ( u ^{eh}\hat e_i) \ {\rm mod}\ I_+$ is a basis $M^m\subset M= \mathcal {M} / I_+$. So all the $a_i$ equal $eh$ and we have $\operatorname {Fil}^h \mathcal {M}^{\rm m} = \operatorname {Fil}^h S \cdot \mathcal {M} ^{\rm m}= \operatorname {Fil}^h \mathcal {M} \cap \mathcal {M} ^{\rm m}$ as required.

Proof. Since $\widetilde {\mathcal {M}}^{\rm n}$ is successive extension of finite free $S_1$-modules, $\operatorname {Tor}^1_{S} (\mathcal {M}^{\rm n}, S/I_+) = 0$. Hence, the sequence $0 \to \widetilde {\mathcal {M}}^{\rm m}/I_+ \to \mathcal {M}/I_+ \to \widetilde {\mathcal {M}}^{\rm n}/I_+ \to 0$ is exact. Since $\widetilde {\mathcal {M}}^{\rm m}$ is multiplicative, $\widetilde {\mathcal {M}}^{\rm m}/I_+ \subset M^{\rm m}$ and thus $\widetilde {\mathcal {M}}^{\rm m}/ I_+ = M^{\rm m}$, otherwise $\varphi$ on $\widetilde {\mathcal {M}}^{\rm n}/I_+$ cannot be nilpotent. So $[M^{\rm m}]\subset \widetilde {\mathcal {M}}^{\rm m}$. Hence, $\mathcal {M}^{\rm m}\subset \widetilde {\mathcal {M}}^{\rm m}$ as $\mathcal {M}^{\rm m}$ is constructed as $S$-submodule of $\mathcal {M}$ generated by $[M^{\rm m}]$. Since $\mathcal {M}^{\rm m}/I_+= M^{\rm m}$, we have $\widetilde {\mathcal {M}}^{\rm m}= \mathcal {M}^{\rm m}$ by Nakayama's lemma. By the definition of exact sequence in the category $\operatorname {Mod}_{S, \operatorname {tor}}^{\varphi, h}$, we see that

\[ \operatorname{Fil}^h \widetilde{\mathcal{M}}^{\rm m} = \widetilde{\mathcal{M}} ^{\rm m} \cap \operatorname{Fil}^h \mathcal{M} = \mathcal{M}^{\rm m} \cap \operatorname{Fil}^h \mathcal{M} = \operatorname{Fil} ^h \mathcal{M}^{\rm m}, \]

where the last equality was proved by the end of the proof of Lemma 2.9. Therefore, we have the desired equality $(\widetilde {\mathcal {M}}^{\rm m}, \operatorname {Fil}^h \widetilde {\mathcal {M}}^{\rm m}, \varphi _h )=( \mathcal {M}^{\rm m}, \operatorname {Fil}^h \mathcal {M}^{\rm m}, \varphi _h )$ as a subobject of $\mathcal {M}$.

Proof. (1) Condition (3) of being an object in $\operatorname {FM}_{W(k)}$ in the case of a multiplicative object translates to $\varphi _0$ being surjective, which is equivalent to being bijective due to length considerations. Conversely, if $\varphi _0$ is bijective, we let $M' \in \operatorname {FM}_{W(k)}$ be defined as follows: the underlying module is $M$ itself, with $\operatorname {Fil}^0 M' = M \supset \operatorname {Fil}^1 M' = 0$ and $\varphi _0$. Then there is an evident morphism $M' \to M$ in $\operatorname {FM}_{W(k)}$, which is necessarily strict with respect to filtrations (see [Reference Fontaine and LaffailleFL82, 1.10(b)]), hence $\operatorname {Fil}^1 M = \operatorname {Fil}^1 M' = 0$. The proof for nilpotent objects is at end of the proof of (2).

(2) By the Fitting lemma, we have $M = M^{\rm m} \oplus M^{\rm n}$ only as $\varphi$-modules, such that $\varphi _0$ on $M ^{\rm m}$ is bijective and $\varphi _0$ on $M ^{\rm n}$ is nilpotent. Let $\operatorname {Fil}^1 M^{\rm m} = 0$; we get the desired sequence. The fact that the quotient $M^{\rm n}$ with the induced filtration is nilpotent follows from (1). By the exact sequence (2.14), $M$ is nilpotent if and only if $M = M^{\rm n}$, whose $\varphi _0$ is nilpotent.

Proof. It is easy to check that if $M\in \operatorname {FM}_{W(k), \operatorname {tor}}$ (respectively, $\mathfrak {M} \in \operatorname {Mod}_{{\mathfrak {S}}, \operatorname {tor}}^{\varphi, h , {\rm c}}$) is multiplicative or nilpotent then so is $\underline { \mathcal {M}}_{\operatorname {FM}} (M)$ (respectively, $\underline { \mathcal {M}} (\mathfrak {M})$). Then the proposition follows from Corollary 2.12.

Proof. The ‘only if’ part follows from the standard direction of going from Fontaine–Laffaille modules to Breuil modules as discussed above. We prove the ‘if’ part, which is the only part that will be used later. To that end, we simply observe that $\mathcal {M}/(p,I_+) \cdot \mathcal {M} \cong M/p$. One checks that the induced map $\overline {\varphi _h} \colon \operatorname {Fil}^h\mathcal {M} \to M/p$ has image given by the image of $\sum _{i=0}^{h} \overline {\varphi _i} \colon \bigoplus _{i=0}^{h} \operatorname {Fil}^i M \to M/p$. Our condition now implies the reduction map is surjective. Since $M$ is $p$-adically complete, it follows that the map $\sum _{i=0}^{h} \varphi _i \colon \bigoplus _{i=0}^{h} \operatorname {Fil}^i M \to M$ before mod $p$ is also surjective, which is exactly what we need to show.

Proof. Write $\mathcal {M} : = \underline { \mathcal {M}} (\mathfrak {M})$. Then $\mathcal {M}$ is also nilpotent by Proposition 2.15. When $h \leq p-2$, $\iota$ is known to be an isomorphism (without assuming nilpotency of $\mathfrak {M}$) by [Reference Li and LiuLL20, Prop. 6.12]. So in the following, we assume $h = p-1$.

Since $T^h_{\mathfrak {S}}$ and $\underline { \mathcal {M}}$ are exact and $T_S$ is left exact, we can assume that $\mathfrak {M}$ is killed by $p$ such that $\mathfrak {M}$ is a finite free ${k[\![u]\!]}$-module with basis $e_1, \ldots, e_d$. Write $\varphi (e_1 , \ldots, e_d)= (e_1 , \ldots, e_d)A$ with $AB = B A =a_0^{-h} u ^{eh}I_d$. Let $\widetilde e_i : = 1 \otimes e_i$ be a basis of $\varphi ^*\mathfrak {M}$ and an $S_1$-basis of $\mathcal {M}$. Then $\operatorname {Fil} ^h \varphi ^* \mathfrak {M}$ is generated by $(\alpha _1 , \ldots, \alpha _d) = (\widetilde e_1 , \ldots, \widetilde e_d)B$ and $\operatorname {Fil}^h \mathcal {M}$ is generated by $(\alpha _1 , \ldots, \alpha _d)$ and $\operatorname {Fil}^p S_1 \mathcal {M}$. Note that $\iota (\widetilde e_1, \ldots, \widetilde e_d)= \frak c^h (\widetilde e_1, \ldots,\widetilde e_d)$, and any $x \in (\operatorname {Fil}^h \varphi ^*\mathfrak {M} \otimes _{\mathfrak {S}} A_{\inf })$ can be written as $x = (\alpha _1 ,\ldots, \alpha _d) X$ with $X\in (\mathcal {O}_\mathbf {C} ^\flat )^d$ and any $y \in \operatorname {Fil}^h \mathcal {M} \otimes _S A_{\operatorname {crys}}$ can be written as $y = \frak c^h (\alpha _1 , \ldots, \alpha _d) Y + \frak c^h (\widetilde e_1 , \ldots, \widetilde e_d) Z$ with $Y \in (\mathcal {O} _\mathbf {C}^\flat / u ^{ep})^d$ and $Z \in (\operatorname {Fil}^pA_{\operatorname {crys}, 1}) ^d$. Then $\iota$ is the same as

\begin{align*} \{X \in (\mathcal{O}_\mathbf{C}^\flat)^d \mid \varphi (X) = B X \}&\longrightarrow \{(Y, Z)\mid Y \in (\mathcal{O} _\mathbf{C}^\flat/ u ^{ep})^d ,Z \in (\operatorname{Fil}^pA_{\operatorname{crys}, 1}) ^d, \\ & \qquad \varphi (Y) + \varphi (A)\varphi _h (Z) = B Y +Z \} \end{align*}

by sending $X\mapsto (X, 0)$. We must show the above map is bijective. For injectivity, note that $X\in \ker (\iota )$ if and only if $B X \in (u^{pe}\mathcal {O}_\mathbf {C}^\flat )^d$. Then ${a_0 ^{-h }} u ^{eh} X=AB X \in (u^{pe}\mathcal {O}_\mathbf {C}^\flat )^d$. Hence, $Y=a_0 u^{-e} X\in (\mathcal {O}_C^\flat )^d$. Note that $\varphi (X) = B X$ implies that $A\varphi (X) = a_0^{-h}u^{eh} X$ and then $Y = A \varphi (Y)$. So $Y = A \varphi (A) \cdots \varphi ^m (A) \varphi ^{m +1} (Y)$. Since $\mathfrak {M}$ is nilpotent, $A \varphi (A) \cdots \varphi ^m (A)\to 0$ for $m \to \infty$, we see that $Y = 0$. This proves the injectivity of $\iota$.

To prove the surjectivity of $\iota$, consider the equation $\varphi (Y) + \varphi (A)\varphi _h (Z) - B Y =Z$. Note that $A_{\operatorname {crys}, 1} = (\mathcal {O}_\mathbf {C}^\flat / u ^{pe}) [\{z_i\}_{i \geq 1}]/ \{z_i ^p, i \geq 1\}$ with $z_i$ the image of $\gamma _{p ^i} (E)$ in $A_{\operatorname {crys},1}$. Since $\varphi _h (z_i) = a_0^{p ^i}$ or $0$, the left-hand side of the equation is in $(\mathcal {O}_\mathbf {C}^\flat /u ^{pe})^d$; this forces the right-hand side $Z= 0$ and we only have $\varphi (Y) = BY$ inside $(\mathcal {O}_\mathbf {C}^\flat /u ^{pe})^d$. So it suffices to show there exists $\widetilde Y\in (\mathcal {O}_C^\flat ) ^d$ such that $\varphi (\widetilde Y) = B \widetilde Y$ and $B \widetilde Y = B Y$ inside $(\mathcal {O} _\mathbf {C}^\flat / u ^{pe})^d$. To prove the existence of $\widetilde Y$, pick any lift $Y_0 \in (\mathcal {O}_\mathbf {C}^\flat )^d$ of $Y$. Then $\varphi (Y_0) = BY_0 + u ^{pe} W_0$. Since $u ^{pe} I_d = B A (a_0u ^e) ^h I _d$, we have $\varphi (Y_0)= B Y_1$ with $Y_1 = Y_0 + u^e A a^h_0W_0$. Then $\varphi (Y_1)= B Y_1 + u ^{pe}\varphi (A) W_1$ for some $W_1 \in (\mathcal {O}_\mathbf {C}^\flat )^d$. Continuing to construct $Y_n$ in this way, we have $\varphi (Y_{n}) = B Y_n + u ^{pe} A \varphi (A) \cdots \varphi ^n (A) W_n$ for some $W_n \in (\mathcal {O}_\mathbf {C}^\flat )^d$ and then $Y_{n+1} = Y_n + u^e A\varphi (A) \cdots \varphi ^n (A) W_n$. Then $Y_n$ converges to $\widetilde Y$ as $A\varphi (A) \cdots \varphi ^n (A)\to 0$. Since $\widetilde Y = Y_0 + u ^e A \widetilde W$ for some $\widetilde W \in (\mathcal {O}_C^\flat )^d$, we see that $B \widetilde Y = B Y_0 = B Y$ inside $(\mathcal {O}_\mathbf {C}^\flat / u ^{pe})^d$.

Proof. Note that $T^h_{\mathfrak {S}}$ is exact (see [Reference Li and LiuLL20, § 6.2]). Since $T_S$ is clearly left exact, by the exact sequence (2.4) and Lemma 2.18, it suffices to prove Corollary for $\mathfrak {M}$ being multiplicative. Clearly, we can assume $k = \overline k$ and then $\mathfrak {M}$ is direct sum of the ${\mathfrak {S}}_n \cdot e_1$ with $\varphi (e_1) =e_1$. Now our desired conclusion just follows from the above example.

Proof. For the first diagram, just apply [Reference Li and LiuLL20, Lemma 7.8(3)]. We see that $\varphi _{i-1}$ is injective in degree $i$ and $\varphi _i$ is an isomorphism in degree $i$. As for the second diagram, passage to $u^\infty$-torsion submodule commutes with $\varphi ^*$ as $\varphi _{{\mathfrak {S}}}$ is flat and sends $u$ to $u^p$, and the procedure of passing to a submodule clearly preserves injectivity.

Proof. The equality follows from the flatness of $\varphi _{{\mathfrak {S}}}$. The inclusion is because the multiplication by $E^{i-1}$ on $\varphi ^*\mathfrak {M}$ factors through $\mathfrak {N}$ (due to condition (1) of Definition 3.1) which is a submodule of $\mathfrak {M}$ (due to condition (2) of Definition 3.1).

Proof. Using Proposition 3.3, after modulo $(p)$, we get the inclusion

\[ E^{i-1} \cdot (u^\alpha) \subset \varphi^*(u^\alpha) = (u^{p\alpha}) \]

in ${\mathfrak {S}}/p = k[\![u]\!]$. Since $E \equiv u^e$ modulo $p$, the above inclusion translates to the inequality

\[ p \alpha \leq e(i-1) + \alpha, \]

which is exactly what we need to show.

Proof. Our assumption implies the existence of $\alpha$ in Corollary 3.4. In the situation of (1), the inequality in Corollary 3.4 gives $\alpha = 0$, hence ${\rm Ann}(\mathfrak {M}) + (p)$ is the unit ideal. Since $p$ is topologically nilpotent, this shows that ${\rm Ann}(\mathfrak {M})$ is already the unit ideal, hence $\mathfrak {M} = 0$.

In the situation of (2), the inequality in Corollary 3.4 gives $\alpha < 2$. Therefore, we have either $\mathfrak {M} = 0$ or ${\rm Ann}(\mathfrak {M}) + (p) = (u,p)$. Without loss of generality, we may assume $\mathfrak {M} \neq 0$, hence ${\rm Ann}(\mathfrak {M}) + (p) = (u,p)$. Let us pick an element $f = u + a \in {\rm Ann}(\mathfrak {M})$ with $a \in p \cdot W(k)$, and compute

\[ E^{i-1} \cdot f = (E(u))^{i-1} \cdot (u + a) = (u^{e \cdot (i-1)} + \cdots + a_1 \cdot u + a_0) \cdot (u+a) = \sum_{j = 0}^{p-1} \varphi_{{\mathfrak{S}}}(B_j) \cdot u^j. \]

Proposition 3.3 implies that all the $B_i$ are in ${\rm Ann}(\mathfrak {M})$. Let us consider $C_1 = B_1(0)$: the above equation says $\varphi (C_1) = a_1 \cdot a + a_0$. Since we know $v_p(a_1) \geq i-1$ and $v_p(a_0) = i-1$, we see that $v_p(C_1) = i-1$, which implies $(B_1) + (u) \supset (u, p^{i-1})$.

Finally, we turn to (3). Similarly arguing as above, we may assume $\mathfrak {M} \neq 0$ and ${\rm Ann}(\mathfrak {M}) + (p) = (u,p)$, and our first task is to show $u \in {\rm Ann} (\mathfrak {M})$. To that end, pick again an element $f = u + a \in {\rm Ann}(\mathfrak {M})$ with $a \in p \cdot W(k)$. Next we compute

\begin{align*} E^{i-1} \cdot f &= (u^e + p \cdot g_1)^{i-1} \cdot (u + a) = (u^{p-1} + p \cdot g_2) \cdot (u + a) \\ &= (u^p + p^{i-1} E(0)^{i-1} \cdot a) \cdot 1 + \sum_{j = 1}^{p-1} b_j \cdot u^j.\end{align*}

By Proposition 3.3, there is another element of the form $u + b \in {\rm Ann}(\mathfrak {M})$ with $b \in W(k)$ having a bigger $p$-adic valuation than that of $a$. Consequently, we have $u \in {\rm Ann} (\mathfrak {M})$, as $a-b$ and $a$ differ by a unit in $W(k)$. Now we do the trick again:

\[ E(u)^{i-1} \cdot u = (u ^e + p g(u)) ^ {i -1} \cdot u = u^p + \sum_{j = 1}^{i -1} {i-1 \choose j} u ^{1+ e (i -1 -j)} (p g(u)) ^{j} = u ^p + \sum_{j = 1}^{p-1} B_j u^j, \]

with $B_j\in W(k)$. Since $u ^p \in {\rm Ann} (\varphi ^* \mathfrak {M}_n ^i)$, we see that $\sum _{j = 1}^{p-1} B_j u^j \in {\rm Ann} (\varphi ^* \mathfrak {M}_n ^i)$ and hence each $\varphi ^{-1}(B_j)$ is in ${\rm Ann} (\mathfrak {M}_n ^i)$. From the above expansion, we see that

\[ E(u)^{i-1} \cdot u \equiv u^p + (i-1) u ^{1+ e (i - 2)} (p g(u))\quad {\rm mod}\ p^2. \]

Since $E(u)$ is an Eisenstein polynomial, we see that $g(0)$ is a $p$-adic unit. This implies that $v_p (B_{1+ e (i-2)}) =1$, so $p \in {\rm Ann} (\mathfrak {M}_n ^i)$.

Finally, we need to show the semi-linear Frobenius on $\mathfrak {M}$ is a bijection. The previous paragraph tells us that $\mathfrak {M} \simeq k^{\oplus r}$. Let us consider the diagram (⚀):

We claim the first arrow in the top row is surjective, the middle vertical arrow is injective with image $E^{i-1} \cdot \varphi ^*\mathfrak {M}$, and the map $h$ is an isomorphism. We know $\varphi ^*\mathfrak {M} \simeq (k[u]/u^p)^{\oplus r}$, hence $E^{i-1} \cdot \varphi ^*\mathfrak {M}$ is also abstractly isomorphic to $k^{\oplus r}$. Let $\ell (\cdot )$ denote the $k$-length. The above diagram gives a chain of inequalities of lengths

\[ r \leq \ell(\mathfrak{N}) \leq r = \ell(\mathfrak{M}), \]

where the first inequality follows from the previous sentence. So the above inequalities are both equalities, and the claim follows easily. The composition, which we have shown to be surjective, of

\[ \mathrm{Frob}_k^*\mathfrak{M} \cong \varphi_{{\mathfrak{S}}}^*\mathfrak{M}/u \xrightarrow{f} \mathfrak{N} \xrightarrow{h} \mathfrak{M} \]

is the linearization of the semi-linear Frobenius on $\mathfrak {M}$. This shows that the semi-linear Frobenius on $\mathfrak {M}$ is surjective, hence bijective by length/dimension considerations.

Proof. Combining Propositions 3.2 and 3.5 gives us what we want.

Proof. We always have inclusions ${\mathrm {H}}^i_{{{\mathbb {\Delta }}}}(\mathcal {X}/{\mathfrak {S}})/p^n \subset {\mathrm {H}}^i_{{\mathrm {qSyn}}}(\mathcal {X}, {{\mathbb {\Delta }}}_n)$. In the specified range, we know the latter has no $u$-torsion by Theorem 3.6(1). Applying Proposition 2.6 shows that there exists an isomorphism of ${\mathfrak {S}}$-modules

\[ {\mathrm{H}}^i_{{{\mathbb{\Delta}}}}(\mathcal{X}/{\mathfrak{S}}) \simeq N_i \otimes_{\mathbb{Z}_p} {\mathfrak{S}} \]

for some $\mathbb {Z}_p$-module $N_i$. To obtain $N_i \simeq {\mathrm {H}}^i_{\operatorname {\acute {e}t}}(\mathcal {X}_C, \mathbb {Z}_p)$, we simply use the étale comparison of Bhatt and Scholze (see [Reference Bhatt, Morrow and ScholzeBMS18, Theorem 1.8(iv)] and [Reference Bhatt and ScholzeBS22, Theorem 1.8(4)]). Here we are using the fact that the isomorphism class of a finitely generated $\mathbb {Z}_p$-module is determined by its base change to $W(C^{\flat })$.

Proof. Denote $\mathfrak {M} := {\mathrm {H}}^i_{{{\mathbb {\Delta }}}}(\mathcal {X}/{\mathfrak {S}})$. We shall use the two short exact sequences appearing in the proof of Proposition 2.6. Crystalline comparison implies $\mathfrak {M}/u \mathfrak {M} \hookrightarrow {\mathrm {H}}^i_{\operatorname {crys}}(\mathcal {X}_0/W)$. Let us derived modulo the sequence

\[ 0 \to \mathfrak{M}_{\operatorname{tor}} \to \mathfrak{M} \to \mathfrak{M}_{\operatorname{tf}} \to 0 \]

by $u$. Since $\mathfrak {M}_{\operatorname {tf}}$ is torsion-free, we have $\mathfrak {M}_{\operatorname {tor}}/u \hookrightarrow \mathfrak {M}/u$. The target is $p$-torsion-free by assumption whereas $\mathfrak {M}_{\operatorname {tor}}$ consists of $p$-power torsion, therefore $\mathfrak {M}_{\operatorname {tor}} = 0$ and $\mathfrak {M}_{\operatorname {tf}} = \mathfrak {M}$. Now we again derived modulo the sequence

\[ 0 \to \mathfrak{M} \to \mathfrak{M} _{\operatorname{fr}} \to \mathfrak{M}_0 \to 0 \]

by $u$ to get $\mathfrak {M}_0[u] \hookrightarrow \mathfrak {M} /u$; the same argument as above shows $\mathfrak {M}_0[u] = 0$, whereas $\mathfrak {M}_0$ is supported at the maximal ideal of ${\mathfrak {S}}$. Therefore, we again conclude $\mathfrak {M}_0 = 0$ and $\mathfrak {M} = \mathfrak {M}_{\operatorname {fr}}$.

Question 3.10 Recall $\mathfrak {M}^i_n := {\mathrm {H}}^i_{{\mathrm {qSyn}}}(\mathcal {X}, {{\mathbb {\Delta }}}_n)[u^\infty ]$.

1. (1) Let $\beta$ be the smallest exponent such that $p^\beta \in \mathrm {Ann}(\mathfrak {M}^i_n)$, and let $\gamma$ be the exponent such that ${\rm Ann}(\mathfrak {M}^i_n) + (u) = (u, p^{\gamma })$. Is there a bound on $\beta$ and $\gamma$ in terms of $e$ and $i$?

2. (2) In light of the example in § 6, is $\beta$ and/or $\gamma$ bounded above by a polynomial in $\log _p$ of a polynomial in $e \text { and } i$, perhaps simply bounded above by $\log _p(({e \cdot (i-1)})/({p-1})) + 1$ when $p$ is odd?Footnote ⁵