Now we can apply Proposition 3.1.4 with $X=Z$ and $\partial=\delta_{i}/t$.∎

######  Lemma 3.3.7.

For $n>0$, $i=1,2$ and $\pi$ given by Proposition 3.3.6:

  1. (a)

$\mathcal{O}(Z)^{\circ}/(\pi^{n})$ is a free
$\mathcal{O}(X)^{\circ}/(\pi^{n})$-module on
$\\{y^{\ell}+(\pi^{n})\mid\ell\geqslant 0\\}$;

  2. (b)

$\mathcal{U}_{i}/(\pi^{n})$ is a free
$\mathcal{O}(X)^{\circ}/(\pi^{n})$-module on $\\{\partial_{i}^{\langle
k\rangle}y^{\ell}+(\pi^{n})\mid k,\ell\geqslant 0\\}$;

  3. (c)

$\mathcal{V}_{1}/(\pi^{n})$ is a free
$\mathcal{O}(X)^{\circ}/(\pi^{n})$-module on $\\{\partial_{x}^{\langle
k\rangle}T^{\ell}+(\pi^{n})\mid k,\ell\geqslant 0\\}$;

  4. (d)

$\mathcal{V}_{2}/(\pi^{n})$ is a free
$\mathcal{O}(X)^{\circ}/(\pi^{n})$-module on $\\{\partial_{x}^{\langle
k\rangle}T^{\ell}+(\pi^{n})\mid k\geqslant 0,\ell\leqslant 0\\}$.

###### Proof.

(a) $\mathcal{O}(Z)=\mathcal{O}(X)\langle y\rangle$, so
$\mathcal{O}(Z)^{\circ}=\mathcal{O}(X)^{\circ}\langle y\rangle$ and the result
follows easily.

(b) Since $\partial_{i}^{\langle k\rangle}y^{\ell}\equiv
y^{\ell}\partial_{i}^{\langle k\rangle}\operatorname{mod}\pi$ this follows
from (a) together with Proposition 3.1.4 and [32, I.2.3.17].

(c) and (d) follow from Proposition 3.1.4 together with the definitions of
$A[T;\delta]$ and $A[T^{-1};\delta]$ from §2.1. ∎

######  Lemma 3.3.8.

For each $k\geqslant 0$ we have the following equality in $\mathcal{U}_{2}$:

$y(\delta_{2}/t)^{\langle k\rangle}=\sum_{n\geqslant
0}(-\operatorname{ad}(x/s))^{n}\left((\delta_{2}/t)^{\langle
k\rangle}\right)y^{n+1}.$

###### Proof.

By Lemma 3.3.4, $\operatorname{ad}(x/s)$ acts locally nilpotently on
$\mathcal{D}$. Hence the sum on the right hand side of the formula is finite.
Since $\mathcal{U}_{2}$ is flat over $K^{\circ}$ we may invert $p$ and then
after multiplication by a scalar in $K$ reduce to proving that

$y\delta_{2}^{k}=\sum_{n\geqslant
0}(-\operatorname{ad}(x/s))^{n}(\delta_{2}^{k})y^{n+1}$

holds inside $\mathcal{O}(Z)[\delta_{2}]$. Now we recall that
$\ell_{\delta_{2}}=\operatorname{ad}(\delta_{2})+r_{\delta_{2}}$ so, using a
computation from the proof of Lemma 3.3.3, we have

$\displaystyle y\delta_{2}^{k}$ $\displaystyle=$
$\displaystyle(\ell_{\delta_{2}}-\operatorname{ad}(\delta_{2}))^{k}(y)$
$\displaystyle=$
$\displaystyle\sum_{n=0}^{k}\binom{k}{n}\delta_{2}^{k-n}(-\delta_{2})^{n}(y)$
$\displaystyle=$
$\displaystyle\sum_{n=0}^{k}\binom{k}{n}\delta_{2}^{k-n}\binom{n}{n}n!y^{n+1}/s^{n}.$

On the other hand, we saw in the proof of Lemma 3.3.4 that

$(-\operatorname{ad}(x/s))^{n}(\delta_{2}^{k})=1/s^{n}\binom{k}{n}n!\delta_{2}^{k-n}.$

The result now follows easily. ∎

######  Proposition 3.3.9.

There are $\mathcal{D}(X)$-linear isomorphisms of $K$-Banach algebras
$\phi_{i}\colon V_{i}\to U_{i}$ with $\phi_{1}(T)=y$ and $\phi_{2}(T^{-1})=y$.

###### Proof.

Since $\delta_{i}|_{\mathcal{O}(X)}=\partial_{x}$, the natural map
$\mathcal{O}(X)\to\mathcal{O}(Z)$ arising from the projection $Z\to X$ extends
to maps $\mathcal{D}(X)\to\mathcal{D}(Z)$ sending $\partial_{x}$ to
$\delta_{i}$. It is easy to verify that these in turn restrict to ring
homomorphisms $g_{i}\colon\mathcal{D}\to\mathcal{U}_{i}$.

Since $y-x/s$ is central in $\mathcal{U}_{1}$ it is easy to verify that for
each $P\in\mathcal{D}$,

$yg_{1}(P)-g_{1}(P)y=(x/s)g_{1}(P)-g_{1}(P)(x/s)=g_{1}(\operatorname{ad}(x/s)(P)).$

The universal property of $\mathcal{V}_{1}$ given by Proposition 2.1.1 thus
provides that $g_{1}$ extends to a unique ring homomorphism
$h_{1}\colon\mathcal{V}_{1}\to\mathcal{U}_{1}$ such that $h_{1}(T)=y$. This in
turn induces ring homomorphisms
$h_{1,n}\colon\mathcal{V}_{1}/(\pi^{n})\to\mathcal{U}_{1}/(\pi^{n})$ that we
claim are isomorphisms. To see this is suffices to observe that $h_{1,n}$ is
$\mathcal{O}(X)^{\circ}/(\pi^{n})$-linear and sends $\partial_{x}^{\langle
k\rangle}T^{\ell}+(\pi^{n})$ to $\delta_{1}^{\langle
k\rangle}y^{\ell}+(\pi^{n})$; now apply Lemma 3.3.7(b,c). It follows that
$\widehat{h_{1}}\colon\widehat{\mathcal{V}_{1}}\to\widehat{\mathcal{U}_{1}}$
is an isomorphism and we may take $\phi_{1}=(\widehat{h_{1}})_{K}$.

Similary, since every element of $\mathcal{D}$ is an
$\mathcal{O}(X)^{\circ}$-linear combination of the elements
$(\partial_{x}/t)^{\langle k\rangle}$ by Proposition 3.1.4(a), and since
$\mathcal{O}(Z)$ is commutative, it follows from Lemma 3.3.8 that inside
$\mathcal{U}_{2}$, for all $P\in\mathcal{D}$ we have

$yg_{2}(P)=\sum_{n\geqslant 0}g_{2}((-\operatorname{ad}(x/s))^{n}(P))y^{n+1}.$

Then the universal property of $\mathcal{V}_{2}$ given by Proposition 2.1.8
shows that $g_{2}$ extends to a unique ring homomorphism
$h_{2}\colon\mathcal{V}_{2}\to\mathcal{U}_{2}$ sending $T^{-1}$ to $y$. Again
this induces ring homomorphisms
$h_{2,n}\colon\mathcal{V}_{2}/(\pi^{n})\to\mathcal{U}_{2}/(\pi^{n})$ that we
claim are isomorphisms. To see this is suffices to observe that each map
$h_{2,n}$ is $\mathcal{O}(X)^{\circ}/(\pi^{n})$-linear and sends
$\partial^{\langle k\rangle}T^{-\ell}+(\pi^{n})$ to $\partial^{\langle
k\rangle}y^{\ell}+(\pi^{n})$ so we can appeal to Lemma 3.3.7(b,d) again and
then see that $\phi_{2}=(\widehat{h_{2}})_{K}$ is an isomorphism as before.∎

######  Proposition 3.3.10.

Let $i=1$ or $i=2$. There is an exact sequence of $V_{i}$-modules

$0\to V_{i}\stackrel{{\scriptstyle\cdot v_{i}}}{{\longrightarrow}}V_{i}\to
D_{i}\to 0,$

where the $v_{i}$ is the central element of $V_{i}$ given by

$v_{1}=sT-x\quad\mbox{and}\quad v_{2}=xT^{-1}-s.$

###### Proof.

Let $u_{1}=sy-x$ and $u_{2}=xy-s$. By Proposition 3.3.9 it suffices to show
that the sequences

$0\to U_{1}\stackrel{{\scriptstyle\cdot u_{1}}}{{\longrightarrow}}U_{1}\to
D_{1}\to 0\quad\mbox{and}\quad 0\to U_{2}\stackrel{{\scriptstyle\cdot
u_{2}}}{{\longrightarrow}}U_{2}\to D_{2}\to 0$

are exact. We may pick $n\geqslant 0$ such that
$\pi^{n}u_{1},\pi^{n}u_{2}\in\mathcal{O}(Z)^{\circ}$. Since

$0\to\mathcal{O}(Z)\stackrel{{\scriptstyle\cdot
u_{i}}}{{\to}}\mathcal{O}(Z)\stackrel{{\scriptstyle
p_{i}}}{{\to}}\mathcal{O}(X_{i})\to 0$

is exact, it follows that

$0\to\mathcal{O}(Z)^{\circ}\stackrel{{\scriptstyle\cdot\pi^{n}u_{i}}}{{\longrightarrow}}\mathcal{O}(Z)^{\circ}\to\mathcal{O}(X_{i})^{\circ}\to
0$

has $\pi$-torsion cohomology. Moreover the Banach Open Mapping Theorem, [41,
Proposition 8.6] tells us that the maps $\cdot u_{i}$ and $p_{i}$ are open
onto their image and so there is some $N>0$ such that $\pi^{N}$ annnihilates
the cohomology groups. We can then use Proposition 3.1.4 to deduce that

$0\to\mathcal{U}_{i}\stackrel{{\scriptstyle\cdot\pi^{n}u_{i}}}{{\to}}\mathcal{U}_{i}\to\mathcal{D}_{i}\to
0$

also has cohomology groups annihilated by $\pi^{N}$. Now apply [7, Lemma 3.6].
∎

######  Proposition 3.3.11.

Suppose that Hypothesis 3.3.2 holds. Then $D_{1}$ and $D_{2}$ are flat as
$D$-modules on both sides.

###### Proof.

Let $\pi$ be given by Proposition 3.3.6. Since $\widehat{\mathcal{D}}$ is a
$\pi$-adically complete and separated $K^{\circ}$-algebra, flat over
$K^{\circ}$ and $\widehat{\mathcal{D}}/(\pi)$ is a commutative
$K^{\circ}/(\pi)$-algebra of finite presentation we will apply the results
from §3.2 with $\mathcal{U}:=\widehat{\mathcal{D}}$.

By Proposition 3.3.10, $D_{1}\cong V_{1}/(sT-x)V_{1}$. By Lemma 3.3.4,
$[\frac{x}{s},\mathcal{D}]\subseteq\frac{1}{ts}\mathcal{D}$. By (16) we have
$|\pi ts|\geqslant|s|/\rho$. Since $\rho=\min(\rho(X),|s|)\leqslant|s|$, we
see that $|\frac{1}{ts}|\leqslant|\pi|$ which implies that
$\frac{1}{ts}\mathcal{D}\subseteq\pi\mathcal{D}$. We can now apply Theorem
3.2.7 with $y=x/s$ to deduce that $D_{1}\cong V_{1}/(T-\frac{x}{s})V_{1}$ is a
flat $D$-module.

Next, by Proposition 3.3.10, we have a presentation $D_{2}\cong
V_{2}/v_{2}V_{2}$ where $v_{2}=xT^{-1}-s\in V_{2}$. By [7, Proposition 4.4]
applied to the ring map $U\to V_{2}$ and the element $v_{2}\in V_{2}$, to
prove the flatness of $D_{2}$ as a $D$-module it suffices to prove that
$v_{2}$ is a regular element in $V_{2}$, $V_{2}$ is flat as a right $U$-module
and that $V_{2}\otimes_{D}M$ is $v_{2}$-torsion-free for all finitely
generated left $D$-modules $M$.

Note that in $\mathcal{V}_{2}$ we can use Lemma 2.1.6 to see that for all
$P\in\mathcal{D}$ we have

$\operatorname{ad}(T^{-1})(P)=\sum_{n=1}^{\infty}(-1)^{n}\operatorname{ad}(x/s)^{n}(P)T^{-n-1}.$

We saw above that Lemma 3.3.4 implies that
$\operatorname{ad}(x/s)^{n}(\mathcal{D})\subseteq\frac{1}{ts}\mathcal{D}\subseteq\pi\mathcal{D}$
for all $n\geqslant 1$. Hence
$\operatorname{ad}(T^{-1})(\mathcal{D})\subseteq\pi\mathcal{V}_{2}$, so the
hypotheses of Proposition 3.2.6 are satisfied with
$\mathcal{U}=\widehat{\mathcal{D}}$, $\mathcal{V}=\widehat{\mathcal{V}_{2}}$
and $y=T^{-1}$. The centraliser of $T^{-1}$ in $D$ certainly contains
$\mathcal{O}(X)$. Proposition 3.2.6 now tells us that $V_{2}$ is a flat
$U$-module on both sides, and that for every finitely generated $D$-module $M$
there is a natural isomorphism of $\mathcal{O}(X)\langle
T^{-1}\rangle$-modules $M\langle T^{-1}\rangle\cong V_{2}\otimes_{D}M$. We can
now apply [7, Lemma 4.1(a)] with $t=T^{-1}$ and $f=x/s$ to see that
$V_{2}\otimes_{D}M$ is $v_{2}$-torsion-free for every finitely generated
$D$-module $M$. In particular, setting $M=D$ shows that $v_{2}$ is a regular
element in $V_{2}$. ∎

###### Proof of Theorem 3.3.1.

We may certainly assume that $Y\neq\emptyset$ as the result is trivial
otherwise. By Lemma 3.2.3 and Lemma 2.3.11(a) we may replace $K$ by any finite
extension. In particular, by [8, Theorem 4.1.8] we may assume that $X$ and $Y$
are both split over $K$ and $r\in|K^{\times}|$.

If $Y=\bigcup_{i=1}^{n}Y_{i}$ is a disjoint union of cheeses then
$\mathcal{D}_{r}(Y)=\bigoplus_{i=1}^{n}\mathcal{D}_{r}(Y_{i})$ and so by [28,
Proposition 5.4.2] we may further assume that $Y$ is a single cheese.

Next we observe that as $r(Y)<r$, $\rho(Y)>\varpi/r$ by Corollary 2.3.17. That
is $Y=C_{K}(\alpha,\bf{s})$ for some $\alpha,\bf{s}$ satisfying the conditions
of [8, Definition 4.1.1] and moreover $|s_{i}|>\varpi/r$ for all
$i=0,\ldots,g$. Since $Y\subset X$ if we define
$X_{0}=X(\frac{x-\alpha_{0}}{s_{0}})$ and
$X_{i}=X_{i-1}(\frac{s_{i}}{x-\alpha_{i}})$ for $i=1,\ldots,g$ then we see
that $X_{g}=Y$. Then, since a composition of flat morphisms is flat, an
induction argument reduces us to the cases $Y=X(\frac{x-\alpha}{s})$ and
$Y=X(\frac{s}{x-\alpha})$ where $\alpha\in K$ and $s\in K^{\times}$. By a
change of coordinate $x\mapsto x-\alpha$, we may assume that $\alpha=0$.

Since $Y$ lies in $\mathbb{A}(\partial_{x}/r)$ by assumption, applying
Corollary 2.3.17 shows that

$|s|\geqslant\rho(Y)=\varpi/r(Y)>\varpi/r.$

Since $\varpi_{m}/r$ is a decreasing sequence converging to $\varpi/r$ from
above, we can find a sufficiently large $m$ such that
$\rho(Y)>|\varpi_{m}|/r$. Since $Y=X(x/s)$ or $X(s/x)$, it is clear from
Definition 2.3.14 that $\rho(X)\geqslant\rho(Y)$. Hence

$\rho=\min(\rho(X),|s|)\geqslant\rho(Y)>|\varpi_{m}|/r.$

By further enlarging $K$ if necessary, we may assume that $\varpi_{m}\in
K^{\times}$, and now all conditions of Hypothesis 3.3.2 are satisfied.

Choose any $t\in K^{\times}$ with $|t|=r/|\varpi_{m}|$. Theorem 3.1.9 tells us
that

$\mathcal{D}_{r}(X)\cong\widehat{\mathcal{D}_{t}^{(m)}(X)}=D\quad\mbox{and}\quad\mathcal{D}_{r}(X_{i})\cong
D_{i}\quad\mbox{for}\quad i=1,2.$

Hence $\mathcal{D}_{r}(X_{1})$ and $\mathcal{D}_{r}(X_{2})$ are flat
$\mathcal{D}_{r}(X)$-modules on both sides by Proposition 3.3.11. Since either
$Y=X_{1}=X(x/s)$ or $Y=X_{2}=X(s/x)$, we conclude that $\mathcal{D}_{r}(Y)$ is
a flat $\mathcal{D}_{r}(X)$-module on both sides. ∎

###  3.4. Flatness of divided power algebras in cotangent direction

Our goal in this section is to prove the following Theorem and then use it to
complete the proof of Theorem 3.2.1.

######  Theorem 3.4.1.

Let $s,r\in\sqrt{|K^{\times}|}$ be given with $s\geqslant r$, and suppose that
$X$ lies in $\mathbb{A}(\partial_{x}/r)$. Then $\mathcal{D}_{r}(X)$ is a flat
$\mathcal{D}_{s}(X)$-module on both sides.

As in $\S\ref{flatbasesec}$, we can enlarge $K$ to reduce to the case where
$X$ is split over $K$. We can also assume that $|K|^{\times}$ contains any
particular finite set of elements in $\sqrt{|K|^{\times}}$. We will first deal
with the special case where the following Hypothesis holds; we will see in the
proof of Theorem 3.4.1 given at the end of $\S\ref{flatcotdirsec}$ below that
the general case can be reduced to this one.

######  Hypothesis 3.4.2.

  * •

$X$ is an affinoid subdomain of $\mathbb{A}$ split over $K$;

  * •

$m\in\mathbb{N}$ is such that $p^{-1/p^{m}}\in K^{\times}$;

  * •

$s,r\in K^{\times}$ are such that $|s|>|r|>1/\rho(X)$ and
$|s/r|<|p|^{-1/p^{m}}$;

  * •

$\pi\in K^{\times}$ is such that
$\max\left\\{|p|,\frac{1}{|r|\rho(X)}\right\\}\leqslant|\pi|<1$.

We will assume that Hypothesis 3.4.2 holds until the end of the proof of
Corollary 3.4.9 below.

######  Definition 3.4.3.

Let $\mathcal{U}^{0}=\mathcal{D}_{s}^{(m)}(X)$. For each $i=0,\ldots,m$,
define inductively

$\mathcal{U}^{i+1}:=\mathcal{U}^{i}[(\partial_{x}/r)^{\langle
p^{i}\rangle}]\quad\subset\quad\mathcal{D}^{(m)}_{r}(X)$

so that
$\mathcal{U}^{0}\subset\mathcal{U}^{1}=\mathcal{D}_{s}^{(m)}(X)[\partial_{x}/r]\subset\mathcal{U}^{2}\subset\cdots\subset\mathcal{U}^{m+1}={\mathcal{D}}_{r}^{(m)}(X)$.

We will show that for each $i=0,\cdots,m$, $\widehat{\mathcal{U}^{i+1}_{K}}$
is flat as a $\widehat{\mathcal{U}^{i}_{K}}$-module on both sides. Then we
will appeal to transitivity of flatness to deduce that
$\widehat{\mathcal{D}}_{r}^{(m)}(X)_{K}$ is flat as a
$\widehat{\mathcal{D}}_{s}^{(m)}(X)_{K}$-module on both sides. Then Theorem
3.1.9 will imply that $\mathcal{D}_{|r\varpi_{m}|}(X)$ is flat over
$\mathcal{D}_{|s\varpi_{m}|}(X)$.

First we prove an important generalisation of Proposition 3.1.4.

######  Proposition 3.4.4.

Let $i=1,\ldots,m$.

  1. (a)

$\mathcal{U}^{i}$ is a free $\mathcal{O}(X)^{\circ}$-module with basis

$T_{i}:=\left\\{(\partial_{x}/r)^{\langle\alpha\rangle}(\partial_{x}/s)^{\langle
p^{i}\beta\rangle}:0\leqslant\alpha<p^{i},\beta\geqslant 0\right\\}.$

  2. (b)

$\mathcal{U}^{i}/(\pi)$ is a finitely presented commutative $K^{\circ}/\pi
K^{\circ}$-algebra.

###### Proof.

(a) Let $k\geqslant 0$. Recall from $(\ref{d<k>formula})$ that when we write
$k=\sum_{j=0}^{m-1}c_{j}p^{j}+cp^{m}$ for some
$c_{0},\ldots,c_{m-1}\in\\{0,\ldots,p-1\\}$ and $c\geqslant 0$, there is a
unit $u\in\mathbb{Z}_{p}^{\times}$ such that

(17) $\partial_{x}^{\langle
k\rangle}=u\left(\prod\limits_{j=0}^{m-1}(\partial_{x}^{\langle
p^{j}\rangle})^{c_{j}}\right)(\partial_{x}^{\langle p^{m}\rangle})^{c}.$

Using this equation together with Proposition 3.1.4, we see that $T_{i}$ is a
generating set for $\mathcal{U}^{i}$ as a $\mathcal{O}(X)^{\circ}$-algebra and
$T_{i}$ is linearly independent over $\mathcal{O}(X)^{\circ}$. To show that
$\mathcal{U}^{i}=\mathcal{O}(X)^{\circ}\cdot T_{i}$ it will suffice to prove
that $\mathcal{O}(X)^{\circ}\cdot T_{i}$ is closed under multiplication. Since
$\mathcal{O}(X)^{\circ}$ is closed under multiplication and
$K^{\circ}\subset\mathcal{O}(X)^{\circ}$, it will suffice to show that

  1. (1)

$T_{i}\cdot\mathcal{O}(X)^{\circ}\subseteq\mathcal{O}(X)^{\circ}\cdot T_{i}$,
and

  2. (2)

$T_{i}\cdot T_{i}\subseteq K^{\circ}\cdot T_{i}$.

(1) Let $0\leqslant\alpha<p^{i}$ and $\beta\geqslant 0$ be given, so that
$t_{\alpha,\beta}:=(\partial_{x}/r)^{\langle\alpha\rangle}(\partial_{x}/s)^{\langle
p^{i}\beta\rangle}$ is a typical element of $T_{i}$. We can of course rewrite
$t_{\alpha,\beta}=(r/s)^{p^{i}\beta}(\partial_{x}/r)^{\langle\alpha\rangle}(\partial_{x}/r)^{\langle
p^{i}\beta\rangle}$. By applying (17) to
$\partial_{x}^{\langle\alpha\rangle}$, $\partial_{x}^{\langle
p^{i}\beta\rangle}$ and $\partial_{x}^{\langle\alpha+p^{i}\beta\rangle}$ in
turn, and using the fact that $0\leqslant\alpha<p^{i}$, we find that

(18)
$t_{\alpha,\beta}=u_{\alpha,\beta}(r/s)^{p^{i}\beta}(\partial_{x}/r)^{\langle\alpha+p^{i}\beta\rangle}\quad\mbox{for
some}\quad u_{\alpha,\beta}\in\mathbb{Z}_{p}^{\times}.$

Let $f\in\mathcal{O}(X)^{\circ}$. Then Lemma 3.1.2b tells us that

(19)
$ft_{\alpha,\beta}=u_{\alpha,\beta}(r/s)^{p^{i}\beta}\sum\limits_{k=0}^{\alpha+p^{i}\beta}\genfrac{\\{}{\\}}{0.0pt}{}{\alpha+p^{i}\beta}{k}(\partial_{x}/r)^{\langle\alpha+p^{i}\beta-k\rangle}(f)(\partial_{x}/r)^{\langle
k\rangle}.$

For each $0\leqslant k\leqslant\alpha+p^{i}\beta$, write
$k=\alpha_{k}+p^{i}\beta_{k}$ with $0\leqslant\alpha_{k}<p^{i}$ and
$\beta_{k}\geqslant 0$, so that necessarily we have $\beta_{k}\leqslant\beta$.
Then using (18) we have

$(r/s)^{p^{i}\beta}(\partial_{x}/r)^{\langle
k\rangle}=\frac{(r/s)^{p^{i}(\beta-\beta_{k})}}{u_{\alpha_{k},\beta_{k}}}t_{\alpha_{k},\beta_{k}}\in
K^{\circ}\cdot T_{i}.$

This allows us to rewrite $(\ref{ftab})$ as follows:

(20)
$ft_{\alpha,\beta}=u_{\alpha,\beta}\sum\limits_{k=0}^{\alpha+p^{i}\beta}\genfrac{\\{}{\\}}{0.0pt}{}{\alpha+p^{i}\beta}{k}(\partial_{x}/r)^{\langle\alpha+p^{i}\beta-k\rangle}(f)\hskip
2.84544pt\frac{(r/s)^{p^{i}(\beta-\beta_{k})}}{u_{\alpha_{k},\beta_{k}}}t_{\alpha_{k},\beta_{k}}.$

We have chosen $\pi\in K^{\times}$ to satisfy
$\max\left\\{|p|,\frac{1}{|r|\rho(X)}\right\\}\leqslant|\pi|<1$, so
$\mathcal{D}^{(m)}_{r}(X)/(\pi)$ is a commutative algebra by Corollary
3.1.6(b). This implies that

(21) $(\partial_{x}/r)^{\langle
j\rangle}(f)\in\pi\mathcal{O}(X)^{\circ}\quad\mbox{for all}\quad j\geqslant
1.$

Using $(\ref{ftab2})$, we can now see that
$ft_{\alpha,\beta}\in\mathcal{O}(X)^{\circ}T_{i}$, and hence
$T_{i}\cdot\mathcal{O}(X)^{\circ}\subseteq\mathcal{O}(X)^{\circ}\cdot T_{i}$.

(2) Let $0\leqslant\alpha,\alpha^{\prime}<p^{i}$ be given. Using Lemma 3.1.2a
together with (17) we see that if $\alpha+\alpha^{\prime}<p^{i}$ then

$(\partial_{x}/r)^{\langle\alpha\rangle}(\partial_{x}/r)^{\langle\alpha^{\prime}\rangle}=\genfrac{\langle}{\rangle}{0.0pt}{}{\alpha+\alpha^{\prime}}{\alpha}(\partial_{x}/r)^{\langle\alpha+\alpha^{\prime}\rangle},$

whereas if $p^{i}\leqslant\alpha+\alpha^{\prime}<2p^{i}$ then

$(\partial_{x}/r)^{\langle\alpha\rangle}(\partial_{x}/r)^{\langle\alpha^{\prime}\rangle}=u(s/r)^{p^{i}}\genfrac{\langle}{\rangle}{0.0pt}{}{\alpha+\alpha^{\prime}}{\alpha}(\partial_{x}/r)^{\langle\alpha+\alpha^{\prime}-p^{i}\rangle}(\partial_{x}/s)^{\langle
p^{i}\rangle}$

for some $u\in\mathbb{Z}_{p}^{\times}$. Similarly for
$\beta,\beta^{\prime}\geqslant 0$, we have

$(\partial_{x}/s)^{\langle p^{i}\beta\rangle}(\partial_{x}/s)^{\langle
p^{i}\beta^{\prime}\rangle}=\genfrac{\langle}{\rangle}{0.0pt}{}{p^{i}(\beta+\beta^{\prime})}{p^{i}\beta}(\partial_{x}/s)^{\langle
p^{i}(\beta+\beta^{\prime})\rangle}.$

Thus to see that $K^{\circ}T_{i}$ is closed under multiplication it suffices
to show that
$(s/r)^{p^{i}}\genfrac{\langle}{\rangle}{0.0pt}{}{\alpha+\alpha^{\prime}}{\alpha}\in
K^{\circ}$ whenever $0\leqslant\alpha,\alpha^{\prime}<p^{i}$ but
$\alpha+\alpha^{\prime}\geqslant p^{i}$.

Now $0\leqslant\alpha,\alpha^{\prime}<p^{i}$ implies that
$\genfrac{\\{}{\\}}{0.0pt}{}{\alpha+\alpha^{\prime}}{\alpha}=1$. If in
addition we have $\alpha+\alpha^{\prime}\geqslant p^{i}$, there must be at
least one carry when adding $\alpha$ and $\alpha^{\prime}$ $p$-adically.
Therefore by Kummer’s Theorem (Theorem 6.4.3) we have

$v_{p}\left(\genfrac{\langle}{\rangle}{0.0pt}{}{\alpha+\alpha^{\prime}}{\alpha}\right)=v_{p}\left(\binom{\alpha+\alpha^{\prime}}{\alpha}\right)\geqslant
1.$

However we have assumed that $|s|$ and $|r|$ are sufficiently close, in the
sense that $|s/r|\leqslant|p|^{-1/p^{m}}$. This implies that
$|(s/r)^{p^{i}}|\leqslant|p|^{-p^{i}/p^{m}}\leqslant|p|^{-1}$ as $i\leqslant
m$, so $p(s/r)^{p^{i}}\in K^{\circ}$. Hence
$(s/r)^{p^{i}}\genfrac{\langle}{\rangle}{0.0pt}{}{\alpha+\alpha^{\prime}}{\alpha}\in
K^{\circ}$ as required.

(b) The commutativity of $\mathcal{U}^{i}/(\pi)$ is clear given equations
$(\ref{ftab2})$ and $(\ref{dxrjf})$.

It follows from (a) that $\mathcal{U}^{i}/(\pi)$ is a free
$\mathcal{O}(X)^{\circ}$-module on the image of $T_{i}$. Moreover writing
$\tau_{\alpha,\beta}$ for the image of $t_{\alpha,\beta}$ in
$\mathcal{U}^{i}/(\pi)$ we see, by (14), that the
$\mathcal{O}(X)^{\circ}/(\pi)$-submodule $U$ generated by
$\\{\tau_{0,p^{m-i}\gamma}\mid\gamma\geqslant 0\\}$ is a polynomial algebra
$U=\mathcal{O}(X)^{\circ}/(\pi)[\tau_{0,p^{m-i}}]$. Moreover $\mathcal{U}^{i}$
is a free $U$-module on the finite set $\\{\tau_{\alpha,\beta}\mid
0\leqslant\alpha<p^{i},0\leqslant\beta<p^{m-i}\\}$.

By the proof of Corollary 3.1.6(b), $\mathcal{O}(X)^{\circ}/(\pi)$ is a
finitely presented $K^{\circ}/(\pi)$-algebra. Thus, by [45, 00F4], $U$ is also
a finitely presented $K^{\circ}/(\pi)$-algebra. Finally
$\mathcal{U}^{i}/(\pi)$ is a finitely presented $K^{\circ}/(\pi)$-algebra by
[45, 0D46]. ∎

Recall that for each $u\in\mathcal{D}(X)$,
$\operatorname{ad}(u)\colon\mathcal{D}(X)\to\mathcal{D}(X)$ is a $K$-linear
derivation.

######  Lemma 3.4.5.

For any $i=0,\ldots,m$, we have
$\operatorname{ad}\left((\partial_{x}/{r})^{\langle
p^{i}\rangle}\right)(\mathcal{U}^{i})\subseteq\pi\mathcal{U}^{i}.$

###### Proof.

Let $f\in\mathcal{O}(X)^{\circ}$. By Lemma 3.1.2(b), we have

$\operatorname{ad}\left((\partial_{x}/r)^{\langle
p^{i}\rangle}\right)(f)=\sum_{k=0}^{p^{i}-1}\genfrac{\\{}{\\}}{0.0pt}{}{p^{i}}{k}(\partial_{x}/r)^{\langle{p^{i}-k}\rangle}(f)\cdot(\partial_{x}/r)^{\langle
k\rangle}.$

Now
$(\partial_{x}/r)^{\langle{p^{i}-k}\rangle}(f)\in\pi\mathcal{O}(X)^{\circ}$ by
$(\ref{dxrjf})$ and $(\partial_{x}/{r})^{\langle k\rangle}\in\mathcal{U}^{i}$
for all $k<p^{i}$ by (17), so the sum on the right hand side lies in
$\pi\mathcal{U}^{i}$. By Proposition 3.4.4(a), we know that $\mathcal{U}^{i}$
is a free left $\mathcal{O}(X)^{\circ}$-module with basis $T_{i}$. The result
follows because $[(\partial_{x}/r)^{\langle
p^{i}\rangle},ft]=[(\partial_{x}/r)^{\langle
p^{i}\rangle},f]t\in\pi\mathcal{U}^{i}$ for all $t\in T_{i}$. ∎

######  Lemma 3.4.6.

For each $i=0,\ldots,m-1$, $\left((\partial_{x}/r)^{\langle
p^{i}\rangle}\right)^{p}\in\mathcal{U}^{0}$.

###### Proof.

The assumption $i+1\leqslant m$ implies that $q_{p^{i}}=q_{p^{i+1}}=0$. Hence
$\partial^{\langle p^{i}\rangle}=\partial^{[p^{i}]}$ and $\partial^{\langle
p^{i+1}\rangle}=\partial^{[p^{i+1}]}$ by Definition 3.1.1(d), so

$\left((\partial_{x}/r)^{\langle
p^{i}\rangle}\right)^{p}=\frac{p^{i+1}!}{(p^{i}!)^{p}}(\partial_{x}/r)^{\langle
p^{i+1}\rangle}=\frac{p^{i+1}!}{(p^{i}!)^{p}}(s/r)^{p^{i+1}}(\partial_{x}/s)^{\langle
p^{i+1}\rangle}.$

Now as $v_{p}(p^{i}!)=\frac{p^{i}-1}{p-1}$, we can see that
$v_{p}\left(\frac{p^{i+1}!}{(p^{i}!)^{p}}\right)=1$. Since $i\leqslant m-1$
and $|s/r|\leqslant|p|^{-1/p^{m}}$, we see that
$|(s/r)^{p^{i+1}}|\leqslant|p|^{-1}$. Hence
$\frac{p^{i+1}!}{(p^{i}!)^{p}}(s/r)^{p^{i+1}}\in K^{\circ}$, so
$\left((\partial_{x}/r)^{\langle p^{i}\rangle}\right)^{p}\in
K^{\circ}\cdot(\partial_{x}/s)^{\langle
p^{i+1}\rangle}\subset\mathcal{D}_{s}^{(m)}(X)=\mathcal{U}^{0}$ as required.∎

######  Notation 3.4.7.

Let $i=0,\ldots,m$.

  * •

Let $y_{i}=(\partial_{x}/{r})^{\langle p^{i}\rangle}$.

  * •

Let $\delta_{i}$ be $\operatorname{ad}(y_{i})$ viewed as a derivation of
$\mathcal{U}^{i}$.

  * •

Let $\mathcal{V}^{i}$ be the skew-Ore extension
$\mathcal{U}^{i}[Y_{i};\delta_{i}]$.

  * •

Let $s_{i}=(r/s)^{p^{i}}\in K^{\circ}$.

Note that $\delta_{i}:\mathcal{U}^{i}\to\mathcal{U}^{i}$ is well-defined by
Lemma 3.4.5.

######  Proposition 3.4.8.

For each $0\leqslant i\leqslant m$ there is a surjective ring homomorphism
$\phi_{i}\colon\mathcal{V}^{i}\twoheadrightarrow\mathcal{U}^{i+1}$ extending
the inclusion map $\mathcal{U}^{i}\to\mathcal{U}^{i+1}$ and sending $Y_{i}$ to
$y_{i}$. The kernel contains the central element
$s_{i}Y_{i}-(\partial_{x}/s)^{\langle p^{i}\rangle}$ and

$s_{i}^{2p-1}\cdot\left(\ker\phi_{i}/\mathcal{V}^{i}(s_{i}Y_{i}-(\partial_{x}/s)^{\langle
p^{i}\rangle})\right)=0.$

###### Proof.

We write $z_{i}=(\partial_{x}/s)^{\langle
p^{i}\rangle}=s_{i}y_{i}\in\mathcal{U}^{i}$. Note that $z_{i}$ and $Y_{i}$
commute in $\mathcal{V}^{i}$. The existence of $\phi_{i}$ follows from the
universal property of the skew-Ore extension, Proposition 2.1.1. Note that
$\phi_{i}$ is surjective because $\mathcal{U}^{i+1}$ is generated by
$\mathcal{U}^{i}$ together with $y_{i}$, by Definition 3.4.3. The element
$s_{i}Y_{i}-z_{i}$ is central in the skew-Ore extension $\mathcal{V}^{i}$
because $s_{i}\delta_{i}$, $\operatorname{ad}(s_{i}y_{i})$ and
$\operatorname{ad}(z_{i})$ define the same derivation
$\mathcal{U}^{i}\to\mathcal{U}^{i}$. Moreover,
$\phi_{i}\left(s_{i}Y_{i}-z_{i}\right)=s_{i}y_{i}-z_{i}=0$, so
$\mathcal{V}^{i}(s_{i}Y_{i}-z_{i})\subseteq\ker\phi_{i}$.

We suppose first that $i\leqslant m-1$. Lemma 3.4.6 gives that
$y_{i}^{p}\in\mathcal{U}^{i}$, so that $(z_{i}/s_{i})^{p}\in\mathcal{U}^{i}$.
Then, writing $n=qp+r$ with $0\leqslant r<p$ and $q\geqslant 0$ we see that

(22)
$s_{i}^{p-1}(z_{i}/s_{i})^{n}=s_{i}^{p-1-r}z_{i}^{r}(z_{i}/s_{i})^{qp}\in\mathcal{U}^{i}\quad\mbox{for
all}\quad n\geqslant 0.$

It follows that for any $n\geqslant 0$ we have

(23)
$s_{i}^{p-1}(Y_{i}-z_{i}/s_{i})^{n}=s_{i}^{p-1}\sum\limits_{k=0}^{n}\binom{n}{k}(-z_{i}/s_{i})^{k}Y_{i}^{n-k}\in\mathcal{V}^{i}.$

Take an element $u=\sum\limits_{n=0}^{l}u_{n}Y_{i}^{n}\in\ker\phi_{i}$ with
$u_{0},\cdots,u_{l}\in\mathcal{U}^{i}$. Then

$\displaystyle s_{i}^{2p-1}u$ $\displaystyle=$
$\displaystyle\sum\limits_{n=0}^{l}s_{i}^{2p-1}u_{n}(Y_{i}-z_{i}/s_{i}+z_{i}/s_{i})^{n}$
$\displaystyle=$
$\displaystyle\sum\limits_{n=0}^{l}\sum\limits_{k=0}^{n}\binom{n}{k}u_{n}s_{i}^{p-1}(z_{i}/s_{i})^{n-k}s_{i}^{p}(Y_{i}-z_{i}/s_{i})^{k}.$

When $k\geqslant 1$, we can rewrite the summand on the right hand side as
follows:

$\left(\binom{n}{k}u_{n}\right)\cdot\left(s_{i}^{p-1}(z_{i}/s_{i})^{n-k}\right)\cdot\left(s_{i}^{p-1}(Y_{i}-z_{i}/s_{i})^{k-1}\right)\cdot(s_{i}Y_{i}-z_{i}).$

The first factor lies in $\mathcal{U}^{i}$, the second factor lies in
$\mathcal{U}^{i}$ by $(\ref{sip1})$ and the third factor lies in
$\mathcal{V}^{i}$ by $(\ref{sip2})$. Hence

$s_{i}^{2p-1}u\equiv\sum\limits_{n=0}^{l}s_{i}^{2p-1}u_{n}(z_{i}/s_{i})^{n}\quad\operatorname{mod}\mathcal{V}^{i}(s_{i}Y_{i}-z_{i}).$

But since $u\in\ker\phi_{i}$, we have

(24)
$0=\phi_{i}(s_{i}^{2p-1}u)=\sum\limits_{n=0}^{l}s_{i}^{2p-1}u_{n}y_{i}^{n}=\sum\limits_{n=0}^{l}u_{n}s_{i}^{2p-1}(z_{i}/s_{i})^{n}$

and we deduce that $s_{i}^{2p-1}u\in\mathcal{V}^{i}(s_{i}Y_{i}-z_{i})$ as
claimed.

Now we consider the case $i=m$. Again we suppose that
$u=\sum\limits_{n=0}^{l}u_{n}Y_{m}^{n}\in\ker\phi_{m}$ with
$u_{0},\cdots,u_{l}\in\mathcal{U}^{m}$. This time

$0=\phi_{m}(s_{m}^{l}u)=\sum\limits_{n=0}^{l}s_{m}^{l}u_{n}y_{m}^{n}=\sum\limits_{n=0}^{l}s_{m}^{l-n}u_{n}z_{m}^{n}.$

Thus

$\displaystyle s_{m}^{l}u$ $\displaystyle=$
$\displaystyle\sum\limits_{n=0}^{l}s_{m}^{l-n}u_{n}(s_{m}Y_{m}-z_{m}+z_{m})^{n}$
$\displaystyle=$
$\displaystyle\sum\limits_{n=0}^{l}\sum\limits_{k=0}^{n}s_{m}^{l-n}\binom{n}{k}u_{n}(s_{m}Y_{m}-z_{m})^{k}z_{m}^{n-k}$
$\displaystyle\in$
$\displaystyle\mathcal{V}^{m}(s_{m}Y_{m}-z_{m})+\sum_{n=0}^{l}s_{m}^{l-n}u_{n}z_{m}^{n}=\mathcal{V}^{m}(s_{m}Y_{m}-z_{m}).$

So for all $u\in\ker\phi_{m}$, we can find $t\in K^{\circ}$ and
$v=\sum\limits_{n\geqslant 0}v_{n}Y_{m}^{n}\in\mathcal{V}^{m}$ such that
$tu=v(s_{m}Y_{m}-z_{m})$. Expanding out this equation gives

$\left(-tu_{0}+\sum_{n\geqslant
1}(s_{m}v_{n-1}-tu_{n})Y_{m}^{n}\right)=\sum_{n\geqslant
0}v_{n}Y_{m}^{n}z_{m}$

and equating coefficients of $Y_{m}^{n}$ we obtain

(25) $\displaystyle v_{0}z_{m}=-tu_{0}\mbox{ and
}v_{n}z_{m}=(s_{m}v_{n-1}-tu_{n})\mbox{ for all }n\geqslant 1.$

Now we recall from Proposition 3.4.4(a) that $\mathcal{U}^{m}$ is a free
$\mathcal{O}(X)^{\circ}$-module on

$T_{m}=\\{(\partial_{x}/r)^{\langle\alpha\rangle}(\partial_{x}/s)^{\langle
p^{m}\beta\rangle}:0\leqslant\alpha<p^{m},\beta\geqslant 0\\}.$

Using $(\ref{d<k>formula})$, we see that $(\partial_{x}/s)^{\langle
p^{m}\beta\rangle}$ only differs from
$z_{m}^{\beta}=\left((\partial_{x}/s)^{\langle p^{m}\rangle}\right)^{\beta}$
by a $p$-adic unit, for any $\beta\geqslant 0$. Therefore

$T^{\prime}_{m}:=\\{(\partial_{x}/r)^{\langle\alpha\rangle}z_{m}^{\beta}:0\leqslant\alpha<p^{m},\beta\geqslant
0\\}$

is also a basis for $\mathcal{U}^{m}$ as a $\mathcal{O}(X)^{\circ}$-module.
Now right multiplication by $z_{m}$ in $\mathcal{U}^{m}$ maps $T^{\prime}_{m}$
to $T^{\prime}_{m}$ and so if $wz_{m}\in t\mathcal{U}^{m}$ for some
$w\in\mathcal{U}^{m}$ then $w\in t\mathcal{U}_{m}$. Using (25) we thus see
inductively that $v_{n}\in t\mathcal{U}_{m}$ for all $n\geqslant 0$. It
follows that $v/t\in\mathcal{V}^{m}$ and so
$u=(v/t)(s_{m}Y_{m}-z_{m})\in\mathcal{V}^{m}(s_{m}Y_{m}-z_{m})$. We have shown
that $\ker\phi_{m}=\mathcal{V}^{m}(s_{m}Y_{m}-z_{m})$, which is even stronger
than what we need. ∎

######  Corollary 3.4.9.

The homomorphism
$\widehat{\phi_{i,K}}\colon\widehat{\mathcal{V}^{i}_{K}}\to\widehat{\mathcal{U}^{i+1}_{K}}$
induces an isomorphism
$\widehat{\mathcal{V}^{i}_{K}}/(Y_{i}-y_{i})\stackrel{{\scriptstyle\cong}}{{\longrightarrow}}\widehat{\mathcal{U}^{i+1}_{K}}$.

###### Proof.

By Proposition 3.4.8, the complex
$0\to\mathcal{V}^{i}(s_{i}Y_{i}-z_{i})\to\mathcal{V}^{i}\stackrel{{\scriptstyle\phi_{i}}}{{\longrightarrow}}\mathcal{U}^{i+1}\to
0$ has cohomology killed by $s_{i}^{2p-1}$. In this situation, [7, Lemma 3.6]
tells us that the functor $\widehat{()}_{K}$ turns this complex into a short
exact sequence. Hence the result. ∎

###### Proof of Theorem 3.4.1.

We note that by Lemma 3.2.3 and Lemma 2.3.11(a) it will suffice to prove this
result after replacing $K$ by a finite extension. In particular by [8, Theorem
4.1.8] we may assume that $X$ splits over $K$ and $s,r\in|K^{\times}|$. Then
$r>r(X)=\varpi/\rho(X)$ by Corollary 2.3.17. We may also assume $s>r$.

Since the sequence $|\varpi_{m}|$ converges to $\varpi$ from above, we may
choose $m$ sufficiently large, so that $r>|\varpi_{m}|/\rho(X)$. We fix this
$m$ for the rest of this proof. Now, we enlarge $K$ if necessary, to ensure
that $\varpi_{m}\in K^{\times}$ and $p^{-1/p^{m}}\in K^{\times}$. We can then
choose $s^{\prime},r^{\prime}\in K$ such that $|\varpi_{m}s^{\prime}|=s$ and
$|\varpi_{m}r^{\prime}|=r$. By enlarging $K$ further, we may assume that
$K^{\times}$ contains an element $\pi$ such that
$\max\left\\{|p|,\frac{1}{|r^{\prime}|\rho(X)}\right\\}\leqslant|\pi|<1$.

Now since $r<s$ we see that $1/\rho(X)<|r^{\prime}|<|s^{\prime}|$. Then by
Theorem 3.1.9,

$\mathcal{D}_{s}(X)\cong\widehat{\mathcal{D}}_{s^{\prime}}^{(m)}(X)_{K}\quad\mbox{and}\quad\mathcal{D}_{r}(X)\cong\widehat{\mathcal{D}}_{r^{\prime}}^{(m)}(X)_{K}.$

We first consider the case when $s$ and $r$ are sufficiently close, in the
sense that

$1<s/r<|p|^{-1/p^{m}}.$

After this assumption, since $|s^{\prime}/r^{\prime}|=s/r$, we note that all
conditions of Hypothesis 3.4.2 are satisfied for the elements $s^{\prime}$ and
$r^{\prime}$ of $K^{\times}$ in place of $s$ and $r$, respectively.

We form $\mathcal{U}^{0}:=\mathcal{D}_{s^{\prime}}^{(m)}(X)$, and inductively,
$\mathcal{U}^{i+1}:=\mathcal{U}^{i}[(\partial_{x}/{r^{\prime}})^{\langle
p^{i}\rangle}]\subset\mathcal{D}_{r^{\prime}}^{(m)}(X)$ as in Definition
3.4.3, for each $i=0,\cdots,m$. We let $\mathcal{V}^{i}$ be the skew-Ore
extension $\mathcal{U}^{i}[Y_{i};\delta_{i}]$ where
$\delta_{i}:\mathcal{U}^{i}\to\mathcal{U}^{i}$ is the derivation
$\operatorname{ad}(y_{i})$ with $y_{i}:=(\partial_{x}/r^{\prime})^{\langle
p^{i}\rangle}$, as in Notation 3.4.7. Lemma 3.4.5 ensures that
$[y_{i},\mathcal{U}^{i}]\subseteq\pi\mathcal{U}^{i}$. Now Theorem 3.2.7
together with Corollary 3.4.9 imply that
$\widehat{\mathcal{U}^{i+1}_{K}}\cong\widehat{\mathcal{V}^{i}_{K}}/(Y_{i}-y_{i})$
is a flat $\widehat{\mathcal{U}^{i}_{K}}$-module on both sides. By the
transitivity of flatness, we see that
$\widehat{\mathcal{U}^{m+1}_{K}}=\widehat{\mathcal{D}}^{(m)}_{r^{\prime}}(X)_{K}=\mathcal{D}_{r}(X)$
is a flat
$\widehat{\mathcal{U}^{0}_{K}}=\widehat{\mathcal{D}}^{(m)}_{s^{\prime}}(X)_{K}=\mathcal{D}_{s}(X)$-module
on both sides, as claimed.

Returning to full generality, we can find a sequence
$r=r_{0}<r_{1}<\ldots<r_{N}=s$, all in $|K^{\times}|$, such that
$1<r_{i}/r_{i-1}\leqslant|p|^{-1/p^{m}}$ for each $1\leqslant i\leqslant N$.
The transitivity of flatness now reduces us to the special case considered
above. ∎

###### Proof of Theorem 3.2.1.

By Proposition 3.2.5 it suffices to show that if $s,r\in\sqrt{|K^{\times}|}$
with $s\geqslant r$ and $Y\subseteq X$ both lie in
$\mathbb{A}(\partial_{x}/r)$, the map
$\mathcal{D}_{s}(X)\to\mathcal{D}_{r}(Y)$ is flat on both sides. By Theorem
3.4.1 $\mathcal{D}_{s}(X)\to\mathcal{D}_{r}(X)$ is flat on both sides, and by
Theorem 3.3.1, $\mathcal{D}_{r}(X)\to\mathcal{D}_{r}(Y)$ is flat on both
sides. We’re now done by the transitivity of flatness. ∎

##  4\. Overconvergent $\mathcal{D}$-modules

###  4.1. The Huber space and overconvergent sheaves

Recall from [48, §5] that to every rigid analytic space $X$ we can associate a
topological space $\tilde{X}$ which we call the _Huber space_ of $X$. The
elements of $\tilde{X}$ are the prime filters on the admissible open subsets
of $X$. The sets of the form $\tilde{U}:=\\{p\in\tilde{X}:U\in p\\}$ as $U$
ranges over the admissible open subsets of $X$ form a basis for the topology
on $\tilde{X}$. If $X$ is quasi-compact and quasi-separated (qcqs), then
$\tilde{X}$ is quasi-compact but in general not Hausdorff. There is a set-
theoretic embedding $X\hookrightarrow\tilde{X}$ given by
$x\mapsto\mathfrak{{m}}_{x}$, where $\mathfrak{{m}}_{x}$ is the principal
maximal filter consisting of the admissible open subsets of $X$ containing
$x$. We will identify $X$ with its image in $\tilde{X}$ via this embedding. By
[48, §5] there is an equivalence of categories between the abelian sheaves on
$X$ and the abelian sheaves on $\tilde{X}$. Given an abelian sheaf
$\mathcal{F}$ on $X$, we will denote the corresponding sheaf on $\tilde{X}$ by
$\tilde{\mathcal{F}}$; it follows from the proof of [48, Theorem 1] that
$\tilde{\mathcal{F}}(\tilde{W})=\mathcal{F}(W)$ for any admissible open subset
$W$ of $X$.

######  Definition 4.1.1.

Let $X$ be a rigid analytic space, let $U$ be an admissible open subset of $X$
and let $\mathcal{F}$ be an abelian sheaf on $X$.

  1. (a)

Let $i:\overline{\tilde{U}}\to\tilde{X}$ denote the inclusion of the closure
of $\tilde{U}$ into $\tilde{X}$.

  2. (b)

Let $\mathcal{F}_{\overline{U}}$ be the abelian sheaf on $X$ defined by

$\widetilde{\mathcal{F}_{\overline{U}}}=i_{\ast}i^{-1}\tilde{\mathcal{F}}.$

In the setting of Definition 4.1.1, $i^{-1}$ is left adjoint to $i_{\ast}$,
and the counit of the adjunction gives a natural morphism
$\mathcal{F}\to\mathcal{F}_{\overline{U}}$ called restriction.

Note that $\widetilde{\mathcal{F}_{\overline{U}}}$ is denoted
$\tilde{\mathcal{F}}_{\overline{\tilde{U}}}$ in [29, §2.3]. This construction
is local, in the following sense.

######  Lemma 4.1.2.

Let $X$ be a rigid analytic space, let $U$ be an admissible open subset of $X$
and let $\mathcal{F}$ be an abelian sheaf on $X$. Let $Y$ be another
admissible open subset of $X$. Then there is a natural isomorphism of abelian
sheaves on $Y$

$(\mathcal{F}_{\overline{U}})_{|Y}\cong(\mathcal{F}_{|Y})_{\overline{U\cap
Y}}.$

###### Proof.

Because $\tilde{Y}$ is open in $\tilde{X}$, we deduce from [19, Chapter I,
§1.6, Proposition 5] that $\tilde{Y}\cap\overline{\tilde{U}}$ is the closure
in $\tilde{Y}$ of $\tilde{Y}\cap\tilde{U}=\widetilde{Y\cap U}$. Let
$f:\tilde{Y}\hookrightarrow\tilde{X}$ be the open inclusion. Applying the
formula [29, (2.3.19)] with $Z:=\overline{\tilde{U}}$ gives

$\widetilde{(\mathcal{F}_{\overline{U}})_{|Y}}=f^{-1}(\tilde{\mathcal{F}}_{\overline{\tilde{U}}})=(f^{-1}\tilde{\mathcal{F}})_{f^{-1}(\overline{\tilde{U}})}=(f^{-1}\tilde{\mathcal{F}})_{\overline{\widetilde{U\cap
Y}}^{\tilde{Y}}}=\widetilde{(\mathcal{F}_{|Y})_{\overline{U\cap Y}}}.$

Now use the equivalence of categories
$\operatorname{Ab}(Y)\cong\operatorname{Ab}(\tilde{Y})$ mentioned above. ∎

The following result will prove useful later.

######  Proposition 4.1.3.

Suppose that $\mathcal{U}$ is an admissible cover of $X$ that is totally
ordered by inclusion and $\mathcal{F}$ is an abelian sheaf on $X$. Then the
restriction maps $\mathcal{F}\to\mathcal{F}_{\overline{U}}$ for
$U\in\mathcal{U}$ induce an isomorphism

$\mathcal{F}\stackrel{{\scriptstyle\cong}}{{\to}}\lim\limits_{\longleftarrow}\mathcal{F}_{\overline{U}}.$

###### Proof.

If $U,V\in\mathcal{U}$ with $U\subset V$ then, by [29, Proposition 2.3.6(iii)]

$\mathcal{F}_{\overline{U}}=(\mathcal{F}_{\overline{V}})_{\overline{U}}$

and so it is easy to verify that there is a natural morphism
$\mathcal{F}\to\lim\limits_{\longleftarrow}\mathcal{F}_{\overline{U}}$ with
the connecting maps in the limit also given by restriction.

Moreover, by Lemma 4.1.2, for $U\subset V$ in $\mathcal{U}$ we can compute

$\mathcal{F}_{\overline{V}}|_{U}=(\mathcal{F}|_{U})_{\overline{U\cap
V}}=\mathcal{F}|_{U}.$

Since for all $W\in\mathcal{U}$ there is $V\in\mathcal{U}$ containing both $U$
and $W$, $\\{V\in\mathcal{U}\mid U\subset V\\}$ is cofinal in $\mathcal{U}$,
and it follows that
$\mathcal{F}|_{U}\cong\left(\lim\limits_{\longleftarrow}\mathcal{F}_{\overline{V}}\right)|_{U}$
for all $U\in\mathcal{U}$. Since $\mathcal{U}$ is a cover of $X$ the result
follows. ∎

We note that the proof above works whenever $\mathcal{U}$ is a directed set
under inclusion; that is whenever any two elements of $\mathcal{U}$ have a
common upper bound in $\mathcal{U}$ under inclusion.

The operation $\mathcal{F}\mapsto\mathcal{F}_{\overline{U}}$ on
$\mathcal{O}_{X}$-modules behaves well with respect to tensor products.

######  Lemma 4.1.4.

If $\mathcal{F}$ and $\mathcal{G}$ are two $\mathcal{O}_{X}$-modules and $U$
is an admissible open subset of $X$ then there is a natural isomorphism

$(\mathcal{F}\underset{\mathcal{O}_{X}}{\otimes}{}\mathcal{G})_{\overline{U}}\cong\mathcal{F}_{\overline{U}}\underset{(\mathcal{O}_{X})_{\overline{U}}}{\otimes}{}\mathcal{G}_{\overline{U}}.$

###### Proof.

By [29, Proposition 2.3.10] we have

$\mathcal{F}_{\overline{U}}\cong(\mathcal{O}_{X})_{\overline{U}}\underset{\mathcal{O}_{X}}{\otimes}\mathcal{F}\quad\mbox{and}\quad\mathcal{G}_{\overline{U}}\cong(\mathcal{O}_{X})_{\overline{U}}\underset{\mathcal{O}_{X}}{\otimes}\mathcal{G}.$

Now tensor these together over $(\mathcal{O}_{X})_{\overline{U}}$ and contract
tensors. ∎

We will now restrict to the case where $X$ is an affinoid variety, and explain
how $\mathcal{F}_{\overline{U}}$ is related to the more well-known operations
in rigid analytic geometry.

######  Definition 4.1.5.

Suppose that $X$ is affinoid, and
$U=X\left(\frac{g_{1}}{g_{0}},\frac{g_{2}}{g_{0}},\cdots,\frac{g_{n}}{g_{0}}\right)$
is a rational subdomain of $X$, where $g_{0},\ldots,g_{n}\in\mathcal{O}(X)$
generate the unit ideal in $\mathcal{O}(X)$. Following [22, Exercise
7.1.12(5)], for each $s\in\sqrt{|K^{\times}|}$ such that $s>1$, define

$U(s):=\\{a\in X:|g_{i}(a)|\leqslant s|g_{0}(a)|\quad\mbox{for all}\quad
i=1,\ldots,n\\}.$

These slightly larger rational subdomains of $X$ form a cofinal system of
_wide open neighbourhoods_ of $U$ in $X$, by [22, Exercise 7.1.12(5)(a)]. Note
that this notation is slightly misleading because the sets $U(s)$ depend in
general on the choice of presentation
$U=X\left(\frac{g_{1}}{g_{0}},\frac{g_{2}}{g_{0}},\cdots,\frac{g_{n}}{g_{0}}\right)$
of $U$ as a rational subdomain of $X$.

######  Lemma 4.1.6.

Let $V$ be an admissible open subset of $X$. Then $\tilde{V}$ contains the
closure $\overline{\tilde{U}}$ in $\tilde{X}$ if and only if $V$ contains
$U(s)$ for some $s\in\sqrt{|K^{\times}|}$ with $s>1$.

###### Proof.

Let $r:\tilde{X}\to\mathscr{M}(X)$ be the retraction map onto the Berkovich
space $\mathscr{M}(X)$ associated to $X$. We claim that
$\overline{\tilde{U}}=r^{-1}(\mathscr{M}(U))$. To see this, note that
$\mathscr{M}(U)=r(\tilde{U})$ is closed in $\mathscr{M}(X)$ because
$\mathscr{M}(U)$ is compact and $\mathscr{M}(X)$ is Hausdorff; since $r$ is
continuous this gives $\overline{\tilde{U}}\subseteq r^{-1}(\mathscr{M}(U))$.
For the other inclusion, let $x\in r^{-1}(\mathscr{M}(U))$ so that $r(x)=r(y)$
for some $y\in\tilde{U}$. Therefore $x\in r^{-1}(r(y))=\overline{\\{y\\}}$, by
[48, Lemma 3.2ii], and thus
$x\in\overline{\\{y\\}}\subseteq\overline{\tilde{U}}$ as claimed.

It follows that $\tilde{V}\supseteq\overline{\tilde{U}}$ if and only if
$r^{-1}(a)\subset\tilde{V}$ for all $a\in\mathscr{M}(U)$. Now,
$\tilde{V}\supseteq\overline{\tilde{U}}$ implies that $V$ is a wide open
neighbourhood of $U$ in the sense of [22, Exercise 7.1.12(4)]. Hence $V$
contains some $U(s)$ by [22, Exercise 7.1.12(5)(a)]. Conversely, if $V$
contains some $U(s)$ then $V$ is a wide open neighbourhood of $U$ so for all
$a\in\mathscr{M}(U)$ we can find an open neighbourhood $W_{a}$ of $r^{-1}(a)$
contained in $V$. But then
$\overline{\tilde{U}}=r^{-1}(\mathscr{M}(U))=\bigcup\limits_{a\in\mathscr{M}(U)}r^{-1}(a)\subseteq\bigcup\limits_{a\in\mathscr{M}(U)}\tilde{W_{a}}\subseteq\tilde{V}$
as required.∎

######  Lemma 4.1.7.

Let $\mathcal{F}$ be an abelian presheaf on an affinoid variety $X$, let $U$
be a rational subdomain of $X$, and let $\mathcal{F}^{\dagger}_{U}$ be the
presheaf

$\mathcal{F}^{\dagger}_{U}:Y\mapsto\underset{s\in\sqrt{|K^{\times}|},s>1}{\operatorname{colim}}{}\hskip
2.84544pt\mathcal{F}(Y\cap U(s))$

on the weak $G$-topology of $X$. Then $\mathcal{F}^{\dagger}_{U}$ is a sheaf
whenever $\mathcal{F}$ is a sheaf, and $\mathcal{F}^{\dagger}_{U}$ is acyclic
whenever $\mathcal{F}$ is acyclic.

###### Proof.

If $\mathcal{Y}:=\\{Y_{1},\ldots,Y_{n}\\}$ is a finite affinoid covering of an
affinoid subdomain $Y$ of $X$, and if $s\in\sqrt{|K^{\times}|}$ and $s>1$,
then $\mathcal{Y}_{s}:=\\{Y_{1}\cap U(s),\ldots,Y_{n}\cap U(s)\\}$ is a finite
affinoid covering of $Y\cap U(s)$. Hence there is an isomorphism of augmented
Čech complexes

$C^{\bullet}_{\operatorname{aug}}(\mathcal{Y},\mathcal{F}^{\dagger}_{U})=\underset{s\in\sqrt{|K^{\times}|},s>1}{\operatorname{colim}}{}\hskip
2.84544ptC^{\bullet}_{\operatorname{aug}}(\mathcal{Y}_{s},\mathcal{F}).$

The result follows, because cohomology commutes with filtered colimits. ∎

######  Theorem 4.1.8.

We have $\mathcal{F}_{\overline{U}}=\mathcal{F}^{\dagger}_{U}$ for every
abelian sheaf $\mathcal{F}$ on an affinoid variety $X$ and every rational
subdomain $U$ of $X$.

###### Proof.

For every open subset $V$ of $C:=\overline{\tilde{U}}$, let

$(i_{\operatorname{pre}}^{-1}\tilde{\mathcal{F}})(V)=\operatorname{colim}\tilde{\mathcal{F}}(W)$

where the colimit runs over all open subsets $W$ of $\tilde{X}$ containing
$i(V)=V$. This defines a presheaf
$i_{\operatorname{pre}}^{-1}\tilde{\mathcal{F}}$ on $C$, and
$i^{-1}\tilde{\mathcal{F}}$ is the sheafification of this presheaf. We will
first show that

(26) $(i^{-1}_{\operatorname{pre}}\tilde{\mathcal{F}})(\tilde{Z}\cap
C)=\underset{s\in\sqrt{|K^{\times}|},s>1}{\operatorname{colim}}{}\hskip
2.84544pt\mathcal{F}(Z\cap U(s))$

for every affinoid subdomain $Z$ of $X$. Consider the open subsets $W$ of
$\tilde{X}$ containing $\tilde{Z}\cap C$. Clearly those $W$ already contained
in $\tilde{Z}$ are cofinal; since $\tilde{Z}\cap C$ is the closure of
$\tilde{Z}\cap\tilde{U}=\widetilde{Z\cap U}$ in $\tilde{Z}$, Lemma 4.1.6 tells
us that the subsets of the form $(Z\cap U)(s)^{\tilde{}}$ are cofinal amongst
these. Now $(\ref{ipreForm})$ follows since $(Z\cap U)(s)=Z\cap U(s)$ for each
$s$.

Because the functor
$Z\mapsto\operatorname{colim}\limits_{s\in\sqrt{|K^{\times}|},s>1}\mathcal{F}(Z\cap
U(s))$ satisfies the sheaf condition with respect to finite coverings of $Z$
by affinoid subdomains of $Z$ by Lemma 4.1.7, and because the $\tilde{Z}\cap
C$ form a basis for the topology of $C$, $(\ref{ipreForm})$ now implies that
the values of $i^{-1}\tilde{\mathcal{F}}$ agree with those of
$i^{-1}_{\operatorname{pre}}\tilde{\mathcal{F}}$ on these basic open sets.

Hence we can calculate that

$\mathcal{F}_{\overline{U}}(Y)=(i_{\ast}i^{-1}\tilde{\mathcal{F}})(\tilde{Y})=(i^{-1}\tilde{\mathcal{F}})(\tilde{Y}\cap
C)=\underset{s\in\sqrt{|K^{\times}|},s>1}{\operatorname{colim}}{}\hskip
2.84544pt\mathcal{F}(Y\cap U(s))$

for every affinoid subdomain $Y$ of $X$ as claimed. ∎

Theorem 4.1.8 implies, in particular, that the sheaf
$\mathcal{F}^{\dagger}_{U}$ in fact does _not_ depend on the choice of
presentation of $U$ as a rational subdomain of $X$. We also see that we should
think of $\mathcal{F}_{\overline{U}}$ as being the sheaf of _sections of
$\mathcal{F}$ overconvergent into the complement of $U$_.

######  Corollary 4.1.9.

Let $U$ be a rational subdomain of the affinoid variety $X$ and let
$\mathcal{F}$ be a coherent $\mathcal{O}_{X}$-module. Then the higher Čech
cohomology of $\mathcal{F}_{\overline{U}}$ is zero.

###### Proof.

This follows from Theorem 4.1.8, Lemma 4.1.7 and Tate’s Acyclicity Theorem
[22, Theorem 4.5.2 and Theorem 4.2.2]. ∎

###  4.2. The $\mathcal{D}^{\dagger}_{\varpi/t}$-module
$\mathcal{M}(\mathcal{S},u,d)_{\overline{U_{t}}}$

In this section we study a Kummer-type finite étale covering of the rigid-
analytic affine line minus finitely many rational points. We recall that any
$K$-scheme $X$ of locally finite type has an analytification
$X^{\operatorname{an}}$ that is a rigid analytic space over $K$. The
construction $X\mapsto X^{\operatorname{an}}$ is functorial in $X$, and we
refer the reader to [35, §1.6] or [16, §5.4] for more details.

We fix $\mathcal{S}=\\{a_{1},\ldots,a_{h}\\}$, a finite subset of $K$.

######  Definition 4.2.1.

Given $u\in K(x)$ a rational function with factorisation in $\overline{K}(x)$

$u=\lambda\prod_{i=1}^{n}(x-b_{i})^{k_{i}}\in K(x)$

for some $\lambda\in K^{\times}$, pairwise distinct
$b_{1},\ldots,b_{n}\in\overline{K}$ and non-zero integers
$k_{1},\ldots,k_{n}$. The _set of finite zeroes/poles_ of $u$ is

$\mathcal{S}(u):=\\{b_{i}\mid i=1,\ldots n\\}.$

We write $v_{b_{i}}(u)=k_{i}\mbox{ for }i=1,\ldots n\mbox{ and
}v_{b}(u)=0\mbox{ for }b\in\overline{K}\backslash\\{b_{1},\ldots,b_{n}\\}$.

We fix $u$ such that $\mathcal{S}(u)$ is a subset of $\mathcal{S}$. Thus

$u=\lambda\prod_{a\in\mathcal{S}}(x-a)^{v_{a}(u)}$

with the $v_{a}(u)$ integers that may be zero and some $\lambda\in
K^{\times}$. We consider the open embedding of rigid $K$-analytic spaces

$j:=\mathbb{A}-\mathcal{S}\hookrightarrow\mathbb{A}:=(\operatorname{Spec}K\left[x\right])^{\operatorname{an}},$

and we introduce the finite étale covering

$f:Z(u,d)\to\mathbb{A}-\mathcal{S}$

where

(27)
$Z(u,d):=\left(\operatorname{Spec}K\left[x,\frac{1}{x-a_{1}},\cdots,\frac{1}{x-a_{h}}\right][z]/(z^{d}-u)\right)^{\operatorname{an}}.$

Note that $f_{\ast}\mathcal{O}_{Z(u,d)}$ is a free
$\mathcal{O}_{\mathbb{A}-\mathcal{S}}$-module of rank $d$:

$f_{\ast}\mathcal{O}_{Z(u,d)}=\bigoplus_{m=0}^{d-1}\mathcal{O}_{\mathbb{A}-\mathcal{S}}z^{m}.$

Since $f:Z(u,d)\to\mathbb{A}-\mathcal{S}$ is étale, the discussion in
$\S\ref{DivTwDer}$ shows that $f_{\ast}\mathcal{O}_{Z(u,d)}$ is in fact a
$\mathcal{D}_{\mathbb{A}-\mathcal{S}}$-module, and that this is a
decomposition of $f_{\ast}\mathcal{O}_{Z(u,d)}$ into a direct sum of $d$ line
bundles with flat connection on $\mathbb{A}-\mathcal{S}$. Applying (10) with
$u$ replaced by $u^{-m}$ shows that the
$\mathcal{D}_{\mathbb{A}-\mathcal{S}}$-action on each summand in the above
decomposition is given by

(28) $\partial_{x}\cdot
z^{m}=\frac{m}{d}\left(\sum_{a\in\mathcal{S}}\frac{v_{a}(u)}{x-a}\right)\hskip
2.84544ptz^{m}.$

Our main goal in $\S\ref{KummerSect}$ is to study the following
$\mathcal{D}_{\mathbb{A}}$-module.

######  Definition 4.2.2.

Define
$\mathcal{M}(\mathcal{S},u,d):=j_{\ast}(\mathcal{O}_{\mathbb{A}-\mathcal{S}}z)$.

The other line bundles with connection
$\mathcal{O}_{\mathbb{A}-\mathcal{S}}z^{m}$ can be dealt with by replacing $u$
by $u^{m}$ so no generality is lost by restricting to the case $m=1$.

######  Definition 4.2.3.

For each $t\in\sqrt{|K^{\times}|}$ we define

$U(\mathcal{S})_{t}:=U_{t}:=\\{z\in\mathbb{A}:|z-a|\geqslant t\quad\mbox{for
all}\quad a\in\mathcal{S}\\}.$

Clearly $U_{t}$ is a subset of $\mathbb{A}-\mathcal{S}$, we have
$U_{t}\subseteq U_{t^{\prime}}$ whenever $t^{\prime}<t$, and

$\mathbb{A}-\mathcal{S}=\bigcup\limits_{t>0}U_{t}.$

We will study the $\mathcal{D}$-module $\mathcal{M}(\mathcal{S},u,d)$ through
its approximations $\mathcal{M}(\mathcal{S},u,d)_{\overline{U}_{t}}$, as $t$
shrinks to zero: one can use Proposition 4.1.3 to see that

$\mathcal{M}(\mathcal{S},u,d)=\lim\limits_{\stackrel{{\scriptstyle\longleftarrow}}{{t>0}}}\mathcal{M}(\mathcal{S},u,d)_{\overline{U}_{t}}.$

We will give a proof of another form of this statement in Lemma 4.3.3.

We assume until the end of $\S\ref{KummerSect}$ that
$t\in\sqrt{|K^{\times}|}$.

Next let us see what the local sections of these objects
$\mathcal{M}(\mathcal{S},u,d)_{\overline{U}_{t}}$ look like.

######  Lemma 4.2.4.

Let $\mathcal{F}$ be an abelian sheaf on $\mathbb{A}$. Then

$(\mathcal{F}_{\overline{U_{t}}})(X)=\underset{s\in\sqrt{|K^{\times}|},s<t}{\operatorname{colim}}{}\hskip
2.84544pt\mathcal{F}(X\cap U_{s}).$

for every affinoid subdomain $X$ of $\mathbb{A}$.

###### Proof.

It is easy to verify by looking at the $\bf{C}$-points that

$X\cap U_{t}=X\left(\frac{t}{x-a_{1}},\ldots\frac{t}{x-a_{h}}\right).$

Recall now the wide open neighbourhoods $(X\cap U_{t})(r)$ of the affinoid
subdomain $X\cap U_{t}$ of $X$ from Definition 4.1.5. We can now apply Lemma
4.1.2 together with Theorem 4.1.8 to compute

$(\mathcal{F}_{\overline{U_{t}}})(X)=(\mathcal{F}_{|X})_{\overline{X\cap
U_{t}}}(X)=\underset{r\in\sqrt{|K^{\times}|},r>1}{\operatorname{colim}}{}\hskip
2.84544pt\mathcal{F}((X\cap U_{t})(r)).$

But for any $r\in\sqrt{|K^{\times}|}$ with $r>1$, we have

$(X\cap U_{t})(r)(\mathbf{C})=\bigcap\limits_{i=1}^{h}\\{y\in
X(\mathbf{C}):t\leqslant r|(y-a_{i})|\\}=X\cap U_{t/r},$

so
$(\mathcal{F}_{\overline{U_{t}}})(X)=\underset{s\in\sqrt{|K^{\times}|},s<t}{\operatorname{colim}}{}\hskip
2.84544pt\mathcal{F}(X\cap U_{s})$ as required.∎

######  Corollary 4.2.5.

We have
$\mathcal{M}(\mathcal{S},u,d)_{\overline{U_{t}}}=(\mathcal{O}_{\mathbb{A}})_{\overline{U_{t}}}\hskip
2.84544ptz$ as sheaves of $\mathcal{O}_{\mathbb{A}}$-modules.

###### Proof.

Let $X$ be an affinoid subdomain of $\mathbb{A}$ and let
$s\in\sqrt{|K^{\times}|}$. Since $X\cap
U_{s}\subset\mathbb{A}\backslash\mathcal{S}$, we see that
$\mathcal{M}(\mathcal{S},u,d)(X\cap U_{s})=\mathcal{O}(X\cap U_{s})z$. So by
Lemma 4.2.4,

(29)
$\mathcal{M}(\mathcal{S},u,d)_{\overline{U_{t}}}(X)=\underset{s\in\sqrt{|K^{\times}|},s<t}{\operatorname{colim}}{\hskip
2.84544pt}\mathcal{O}(X\cap U_{s})\hskip 2.84544ptz.\qed$

Recall from Definition 2.3.9(a) that
$\mathbb{A}(t\partial_{x}/\varpi)^{\dagger}$ is the $G$-topology on
$\mathbb{A}$ which consists of those affinoid subdomains such that
$r(X)\leqslant\varpi/t$. In view of Corollary 2.3.17, we see that
$X\in\mathbb{A}(t\partial_{x}/\varpi)^{\dagger}$ if and only if
$\rho(X)\geqslant t$. The following technical Lemma is necessary to prove
Lemma 4.2.7 and Proposition 4.2.9.

######  Lemma 4.2.6.

Let $X$ be a $(t\partial_{x}/\varpi)^{\dagger}$-admissible affinoid subdomain
of $\mathbb{A}$, and let $s\in\sqrt{|K^{\times}|}$ be such that $s<t$. Then
$r(X\cap U_{s})\leqslant\varpi/s$.

###### Proof.

In view of Lemma 2.3.13, we can replace $K$ by some finite field extension and
thereby assume that $s\in K^{\times}$. Choose a disc
$\mathbb{D}:=\operatorname{Sp}K\langle x/s_{0}\rangle$ with $s_{0}\in
K^{\times}$ large enough to contain $X$ and note that $\mathbb{D}\cap
U_{s}=C(\alpha,\mathbf{s})$ where $s_{1}:=s_{2}:=\cdots:=s_{h}:=s$,
$\alpha_{0}=0$ and $\alpha_{i}:=a_{i}$ for each $i=1,\ldots,h$. Then $X\cap
U_{s}=X\cap C(\alpha,\mathbf{s})$ and
$r(C(\alpha,\mathbf{s}))=\frac{\varpi}{\rho(C(\alpha,\mathbf{s}))}=\frac{\varpi}{s}$
by Corollary 2.3.17, so

$r(X\cap U_{s})=r(X\cap
C(\alpha,\mathbf{s}))\leqslant\max\\{r(X),r(C(\alpha,\mathbf{s}))\\}=\max\\{r(X),\varpi/s\\}$

by Proposition 2.3.2. Our assumptions on $X$, $s$ and $t$ imply that
$r(X)\leqslant\varpi/t<\varpi/s$, so
$\max\\{r(X),r(C(\alpha,\mathbf{s}))\\}=\varpi/s$. ∎

######  Lemma 4.2.7.

The $\mathcal{D}$-action on the restriction of
$(\mathcal{O}_{\mathbb{A}})_{\overline{U_{t}}}$ to
$\mathbb{A}(t\partial_{x}/\varpi)^{\dagger}$ extends to
$\mathcal{D}^{\dagger}_{\varpi/t}$.

###### Proof.

Let $X\in\mathbb{A}(t\partial_{x}/\varpi)^{\dagger}$; we must show that the
$\mathcal{D}(X)$-action on $(\mathcal{O}_{\mathbb{A}})_{\overline{U_{t}}}(X)$
extends to an action of $\mathcal{D}^{\dagger}_{\varpi/t}(X)$.

Let $f\in\mathcal{O}_{\overline{U_{t}}}(X)$ and
$Q\in\mathcal{D}^{\dagger}_{\varpi/t}(X)$. Then by Lemma 4.2.4 and Definition
2.3.9(d) we can find some $s^{\prime}<t$ and $r>\varpi/t$ such that
$f\in\mathcal{O}(X\cap U_{s^{\prime}})$ and $Q\in\mathcal{D}_{r}(X)$. Choose
$s\in\sqrt{|K^{\times}|}$ such that

$\max\\{\varpi/r,s^{\prime}\\}<s<t.$

Then $f\in\mathcal{O}(X\cap U_{s})$ and $r>\varpi/s$, so $Q\cdot
f\in\mathcal{O}(X\cap U_{s})\subset\mathcal{O}_{\overline{U_{t}}}(X)$ by Lemma
4.2.6 and Lemma 2.3.12(a). Thus the $\mathcal{D}(X)$-action on
$(\mathcal{O}_{\mathbb{A}})_{\overline{U_{t}}}(X)$ extends to an action of
$\mathcal{D}^{\dagger}_{\varpi/t}(X)$ as required. ∎

We assume for the remainder of $\S\ref{KummerSect}$ that $p\nmid d$.

######  Lemma 4.2.8.

Let $r>0$ and let $Y$ be a $\partial_{x}/r$-admissible affinoid subdomain of
$\mathbb{A}-\mathcal{S}$. Then the natural $\mathcal{D}(Y)$-action on
$\mathcal{O}(Y)z$ extends to a $\mathcal{D}_{r}(Y)$-action.

###### Proof.

Since $\partial_{x}(z)=\frac{1}{d}\frac{\partial_{x}(u)}{u}z$, the
$\mathcal{D}(Y)$-action on $\mathcal{O}(Y)z$ is given by

$(P,fz)\mapsto\sigma_{r}(\theta_{u^{-1},d}(P))(f)z,$

where $\theta_{u^{-1},d}$ was defined in Lemma 2.4.1 and $\sigma_{r}$ was
defined in Lemma 2.3.12. Thus to extend it to a $\mathcal{D}_{r}(Y)$-action it
suffices to extend the automorphism $\theta_{u^{-1},d}$ of $\mathcal{D}(Y)$ to
an automorphism of $\mathcal{D}_{r}(Y)$; that this can be done is guaranteed
by Theorem 2.4.9. ∎

This result enables us to extend Lemma 4.2.7 in the following manner.

######  Proposition 4.2.9.

The $\mathcal{D}$-action on the restriction of
$\mathcal{M}(\mathcal{S},u,d)_{\overline{U_{t}}}$ to
$\mathbb{A}(t\partial_{x}/\varpi)^{\dagger}$ extends to
$\mathcal{D}^{\dagger}_{\varpi/t}$.

###### Proof.

Let $X\in\mathbb{A}(t\partial_{x}/\varpi)^{\dagger}$ and suppose that
$f\in\mathcal{M}(\mathcal{S},u,d)_{\overline{U_{t}}}(X)$ and
$Q\in\mathcal{D}^{\dagger}_{\varpi/t}(X)$. Then by (29) and Definition
2.3.9(d) we can find some $s^{\prime}<t$ and $r>\varpi/t$ such that
$f\in\mathcal{O}(X\cap U_{s^{\prime}})z$ and $Q\in\mathcal{D}_{r}(X)$.

Choose $s\in\sqrt{|K^{\times}|}$ such that

$\max\\{\varpi/r,s^{\prime}\\}<s<t.$

Then $f\in\mathcal{O}(X\cap U_{s})z$ so if the $\mathcal{D}(X)$-action on
$\mathcal{O}(X\cap U_{s})z$ extends to $\mathcal{D}_{r}(X)$ then $Q\cdot
f\in\mathcal{M}(\mathcal{S},u,d)_{\overline{U_{t}}}(X)$.

Now $r(X\cap U_{s})\leqslant\varpi/s$ by Lemma 4.2.6 and $\varpi/s<r$, so
$X\cap U_{s}$ is $\partial_{x}/r$-admissible. Hence the $\mathcal{D}(X\cap
U_{s})$-action on $\mathcal{O}(X\cap U_{s})z$ extends to
$\mathcal{D}_{r}(X\cap U_{s})$ by Lemma 4.2.8. Since $X$ and $X\cap U_{s}$ are
both $\partial_{x}/r$-admissible, by Proposition 2.3.6(b) there is a ring
homomorphism $\mathcal{D}_{r}(X)\to\mathcal{D}_{r}(X\cap U_{s})$ extending
$\mathcal{D}(X)\to\mathcal{D}(X\cap U_{s})$, and we deduce that the
$\mathcal{D}(X)$-action on $\mathcal{O}(X\cap U_{s})z$ extends to
$\mathcal{D}_{r}(X)$ as required. ∎

In the remainder of this section we will establish an explicit presentation of
$\mathcal{M}(\mathcal{S},u,d)_{\overline{U_{t}}}$ as a
$\mathcal{D}^{\dagger}_{\varpi/t}$-module, assuming $t\in\sqrt{|K^{\times}|}$
is sufficiently small and a positivity condition on the exponents of $u$.

######  Definition 4.2.10.

We define

  1. (a)

$\Delta_{\mathcal{S}}:=\prod\limits_{a\in\mathcal{S}}(x-a)$, and

  2. (b)

$R_{\mathcal{S}}(u,d):=\Delta_{\mathcal{S}}\partial_{x}-\frac{1}{d}\sum\limits_{a\in\mathcal{S}}v_{a}(u)\prod\limits_{b\in\mathcal{S}\backslash\\{a\\}}(x-b)=\Delta_{\mathcal{S}}(\partial_{x}-\frac{1}{d}\operatorname{dlog}(u))$.

Note that $R_{\mathcal{S}}(u,d)\in\mathcal{D}(\mathbb{A})$. We will write
$R(u,d)=R_{\mathcal{S}(u)}(u,d)$ and $\Delta=\Delta_{\mathcal{S}(u)}$.

######  Lemma 4.2.11.

There is a complex of $\mathcal{D}^{\dagger}_{\varpi/t}$-modules on
$\mathbb{A}(t\partial_{x}/\varpi)^{\dagger}$

$0\to\mathcal{D}^{\dagger}_{\varpi/t}\stackrel{{\scriptstyle\cdot
R_{\mathcal{S}}(u,d)}}{{\longrightarrow}}\mathcal{D}^{\dagger}_{\varpi/t}\stackrel{{\scriptstyle\cdot
z}}{{\longrightarrow}}\mathcal{M}(\mathcal{S},u,d)_{\overline{U_{t}}}\to 0.$

###### Proof.

Note that the restriction of $\mathcal{M}(\mathcal{S},u,d)_{\overline{U_{t}}}$
to $\mathbb{A}(t\partial_{x}/\varpi)^{\dagger}$ is a
$\mathcal{D}^{\dagger}_{\varpi/t}$-module by Proposition 4.2.9. The second
non-zero arrow sends $Q\in\mathcal{D}^{\dagger}_{\varpi/t}$ to $Q\cdot
z\in\mathcal{M}(\mathcal{S},u,d)_{\overline{U_{t}}}$. By definition,
$R_{\mathcal{S}}(u,d)=\Delta_{\mathcal{S}}\partial_{x}-\frac{1}{d}\sum_{i=1}^{h}v_{a_{i}}(u)\prod_{j\neq
i}(x-a_{j})\in\mathcal{D}$, so
$R_{\mathcal{S}}(u,d)\in\mathcal{D}_{\varpi/t}^{\dagger}$ as well. A direct
calculation shows that $R_{\mathcal{S}}(u,d)\cdot z=0$. ∎

We will now carry out three local calculations that will be used to prove that
the complex appearing in Lemma 4.2.11 is exact, at least when the $a_{i}$ live
in different holes of $U_{t}$, and thus provides a finite presentation for
$\mathcal{M}(u,d)_{\overline{U_{t}}}$ as a
$\mathcal{D}^{\dagger}_{\varpi/t}$-module. The first shows that the
restriction of the complex to $U_{t}$ is exact.

######  Proposition 4.2.12.

Let $Y$ be an $(t\partial_{x}/\varpi)^{\dagger}$-admissible affinoid subdomain
of $U_{t}$. Then the following complex is exact:

$0\to\mathcal{D}^{\dagger}_{\varpi/t}(Y)\stackrel{{\scriptstyle\cdot
R_{\mathcal{S}}(u,d)}}{{\longrightarrow}}\mathcal{D}^{\dagger}_{\varpi/t}(Y)\stackrel{{\scriptstyle\cdot
z}}{{\longrightarrow}}\mathcal{M}(\mathcal{S},u,d)_{\overline{U_{t}}}(Y)\to
0.$

###### Proof.

Since $Y$ is contained in $U_{t}$, $u$ and $\Delta$ are both units in
$\mathcal{O}(Y)$ and it follows from Lemma 4.1.2 that
$\mathcal{O}(Y)z\to\mathcal{M}(\mathcal{S},u,d)_{\overline{U_{t}}}(Y)$ is an
isomorphism. By Corollary 2.4.11, $Q\mapsto z\hskip 2.84544ptQ\hskip
2.84544ptz^{-1}$ induces an automorphism $\theta_{u,d}$ of
$\mathcal{D}^{\dagger}_{\varpi/t}(Y)$, then by Definition 4.2.10(b) and Lemma
2.4.1

$\theta_{u,d}^{-1}(R_{\mathcal{S}}(u,d))=\Delta\partial_{x}.$

Since $\Delta$ is a unit in $\mathcal{O}(Y)$, we see that
$\mathcal{D}^{\dagger}_{\varpi/t}(Y)\cdot
R_{\mathcal{S}}(u,d)=\mathcal{D}^{\dagger}_{\varpi/t}(Y)\cdot\theta_{u,d}(\partial_{x})$.
By construction of the rings $\mathcal{D}_{r}(Y)$, there is a direct sum
decomposition of $\mathcal{O}(Y)$-modules:
$\mathcal{D}_{r}(Y)=\mathcal{O}(Y)\oplus\mathcal{D}_{r}(Y)\cdot\partial_{x}$
whenever $r>\varpi/t$. Therefore we also have the decomposition

$\mathcal{D}^{\dagger}_{\varpi/t}(Y)=\mathcal{O}(Y)\oplus\mathcal{D}^{\dagger}_{\varpi/t}(Y)\theta_{u,d}(\partial_{x})$

and the result now follows easily. ∎

We record an easy but important consequence of the proof of Proposition
4.2.12.

######  Corollary 4.2.13.

Let $Y$ be a $(t\partial_{x}/\varpi)^{\dagger}$-admissible subdomain of
$U_{t}$. Then the $\mathcal{O}(Y)$-module
$\mathcal{M}(\mathcal{S},u,d)_{\overline{U_{t}}}(Y)$ is free of rank $1$, and
it is simple as a $\mathcal{D}^{\dagger}_{\varpi/t}(Y)$-module whenever $Y$ is
connected as an affinoid domain.

###### Proof.

The first statement follows from the proof of Proposition 4.2.12.

For the second statement it suffices to show that
$\mathcal{M}(\mathcal{S},u,d)_{\overline{U_{t}}}(Y)$ is a simple
$\mathcal{D}(Y)$-module via restriction whenever $Y$ is connected. Given the
first part this follows from [8, Lemma 4.3.2], because $\mathcal{T}(Y)$ is
freely generated by $\partial_{x}$ as a $\mathcal{O}(Y)$-module. ∎

Our next step is to prove that the complex in Lemma 4.2.11 is exact in the
case when $|\mathcal{S}|=1$ under appropriate mild conditions on $u$. First we
need a couple of technical results to enable us to a make a certain estimate.
Together, they amount to a slightly more general version of [12, Lemme 4.2.1].

######  Lemma 4.2.14.

Suppose that $V$ is a a reduced $K$-affinoid variety and $f\in\mathcal{O}(V)$.
Then there is $s\in\sqrt{|K|^{\times}}$ with $s<t$ such that whenever
$g\in\mathcal{O}(V)$ vanishes on $V(t/f)$, $g$ also vanishes on $V(s/f)$.

###### Proof.

We first consider the case where $V$ is irreducible so that $\mathcal{O}(V)$
is an integral domain. Suppose first that $V(t/f)$ is non-empty. Then the
restriction map $\mathcal{O}(V)\to\mathcal{O}(V(t/f))$ is injective by the
proof of [10, Proposition 4.2], so any choice of $s$ will do. On the other
hand, if $V(t/f)=\emptyset$, then $|f|_{V}<t$ by the maximum modulus
principle, so choosing $|f|_{V}<s<t$ ensures that $V(s/f)=\emptyset$ and
yields the result.

In the general case, let $V_{1},\ldots,V_{n}$ be the irreducible components of
$V$ and let $f_{i}:=f_{|V_{i}}$ be the restriction of $f$ to $V_{i}$ for each
$i=1,\cdots,n$. Now, for each $s<t$ with $s\in\sqrt{|K^{\times}|}$ there is a
commutative diagram

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The vertical arrow on the left is injective because $V$ is reduced. Hence the
middle vertical arrow is injective, because the restriction map
$\mathcal{O}(V)\to\mathcal{O}(V(s/f))$ is flat.

Since each $V_{i}$ is irreducible, we may find $s_{i}<t$ in
$\sqrt{|K^{\times}|}$ such that every $g\in\mathcal{O}(V_{i})$ vanishing on
$V_{i}(t/f_{i})$ also vanishes on $V_{i}(s_{i}/f_{i})$. By enlarging each
$s_{i}$ if necessary, we may assume that $s_{1}=\cdots=s_{n}=:s$, say.

Suppose now that $g\in\mathcal{O}(V)$ vanishes on $V(t/f)$. Then by the
commutativity of the diagram, $g_{|V_{i}}$ vanishes on $V_{i}(t/f_{i})$ for
all $i$. Hence by the above paragraph, $g_{|V_{i}}$ also vanishes on
$V_{i}(s/f_{i})$ for all $i$. Using the fact that the middle arrow in the
diagram is injective, we deduce that $g$ also vanishes on $V(s/f)$ as
required. ∎

######  Lemma 4.2.15.

Let $V$ be a reduced $K$-affinoid variety and $f\in\mathcal{O}(V)$. Suppose
moreover that $(v_{n})$ is a sequence in $\mathcal{O}(V)$ such that

  1. (i)

$\sum\limits_{n=0}^{\infty}v_{n}T^{n}\in\mathcal{O}(V)[[T]]$ has radius of
convergence greater than $1/t$, and

  2. (ii)

the series $\sum\limits_{n=0}^{\infty}v_{n}f^{-n}$ converges to zero in
$\mathcal{O}(V(t/f))$.

Then, writing $w_{m}:=\sum\limits_{n=0}^{m}v_{n}f^{m-n}\in\mathcal{O}(V)$ for
each $m\geqslant 0$, the radius of convergence of
$\sum\limits_{m=0}^{\infty}w_{m}T^{m}\in\mathcal{O}(V)[[T]]$ is also greater
than $1/t$.

###### Proof.

By condition (i) we may choose $r\in\sqrt{|K|^{\times}}$ such that $r<t$ and

$\lim_{n\to\infty}|v_{n}|_{V}/r^{n}=0.$

Thus we may define $L:=\sup_{n\geqslant 0}|v_{n}|_{V}/r^{n}<\infty$ and
$W:=V(r/f)$. Because $|f^{-1}|_{W}\leqslant 1/r$, we have
$|v_{n}f^{-n}|_{W}\leqslant|v_{n}|_{V}/r^{n}\to 0$ as $n\to\infty$, so the
series $\sum\limits_{n=0}^{\infty}v_{n}f^{-n}$ converges in $\mathcal{O}(W)$
to an element $w$, say. Note that condition (ii) implies that $w$ vanishes on
$W(t/f)$.

Using Lemma 4.2.14 we may choose $s\in\sqrt{|K|^{\times}}$ with $s<t$ such
that whenever $g\in\mathcal{O}(W)$ vanishes on $W(t/f)$, $g$ also vanishes on
$W(s/f)$. By enlarging $s$ if necessary, we may also assume that $r<s$. Thus,
$w_{|W(s/f)}=0$.

We cover $V$ by its two affinoid subdomains $V(f/s)$ and $V(s/f)$. Noting that
$w_{|V(s/f)}=0$ because $V(s/f)=W(s/f)$, we can estimate $|w_{m}|_{V(s/f)}$ as
follows:

$|w_{m}|_{V(s/f)}=\left|-\sum_{n>m}v_{n}f^{m-n}\right|_{V(s/f)}\leqslant\sup_{n>m}Lr^{-n}s^{m-n}\leqslant
Ls^{m}.$

To estimate $|w_{m}|_{V(f/s)}$, we note that
$|v_{n}|_{V(f/s)}\leqslant|v_{n}|_{V}\leqslant Lr^{n}$ for all $n\geqslant 0$,
so

$|w_{m}|_{V(f/s)}\leqslant\max_{0\leqslant n\leqslant
m}|v_{n}f^{m-n}|_{V_{1}}\leqslant\max_{0\leqslant n\leqslant
m}Lr^{n}s^{m-n}\leqslant Ls^{m}$

since $r<s$. Therefore
$|w_{m}|_{V}=\max\\{|w_{m}|_{V(f/s)},|w_{m}|_{V(s/f)}\\}\leqslant Ls^{m}$ and
hence $\sum\limits_{m=0}^{\infty}w_{m}T^{m}$ has radius of convergence at
least $1/s$. ∎

We’re now ready for our next local calculation, dealing with the critical case
when $|\mathcal{S}|=1$. Note that, in this case, if
$X\in\mathbb{A}(t\partial_{x}/\varpi)^{\dagger}$ and $a_{1}\not\in X(K)$ then
$X\subseteq U_{t}$ since if $X$ splits over $K^{\prime}$ then
$\rho(X_{K^{\prime}})\geqslant t$. Thus, given Proposition 4.2.12, we might as
well assume $a_{1}\in X(K)$.

######  Theorem 4.2.16.

Suppose that $X\in\mathbb{A}(t\partial_{x}/\varpi)^{\dagger}$ and $a_{1}\in
X(K)$. Suppose also that $h=1$ and write $a:=a_{1}$ and $k=v_{a}(u)$ so that
$u=(x-a)^{k}$. Then the complex

$0\to\mathcal{D}^{\dagger}_{\varpi/t}(X)\stackrel{{\scriptstyle\cdot
R_{\mathcal{S}}(u,d)}}{{\longrightarrow}}\mathcal{D}^{\dagger}_{\varpi/t}(X)\stackrel{{\scriptstyle\cdot
z}}{{\longrightarrow}}\mathcal{M}(\mathcal{S},u,d)_{\overline{U_{t}}}(X)\to 0$

is exact whenever $\frac{k}{d}\notin\mathbb{N}$.

###### Proof.

This is based on [12, Proposition 5.1.2(ii)], which Berthelot attributes to
Laumon. Recall that the sequence is a complex follows from Lemma 4.2.11. Note
also that $R_{\mathcal{S}}(u,d)=R(u,d)$ since $k\neq 0$.

First we prove that the last non-zero map is surjective. To the end we
consider a typical element
$P=\sum\limits_{n=0}^{\infty}P_{n}\partial_{x}^{n}\in\mathcal{D}^{\dagger}_{\varpi/t}(X)$
with $P_{n}\in\mathcal{O}(X)$. The required convergence condition is that
there is some $r>\varpi/t$ such that $|P_{n}|r^{n}\to 0$ as $n\to\infty$. We
furthermore compute that

(30) $P\cdot
z=\sum\limits_{n=0}^{\infty}\binom{\frac{k}{d}}{n}n!P_{n}(x-a)^{-n}\hskip
2.84544ptz.$

Take an element $g\in\mathcal{M}(\mathcal{S},u,d)(X)_{\overline{U_{t}}}$.
Using $(\ref{OvConvMudSects}$), we can find some $s\in\sqrt{|K|^{\times}}$
with $s<t$ such that $g=fz$ for some $f\in\mathcal{O}(X\cap U_{s})$. Since
$\mathcal{O}(X\cap U_{s})=\mathcal{O}(X)\langle s/(x-a)\rangle$, we can find a
sequence $(b_{n})$ in $\mathcal{O}(X)$ such that

$f=\sum\limits_{n=0}^{\infty}b_{n}(x-a)^{-n}\quad\mbox{with}\quad\lim\limits_{n\to\infty}|b_{n}|s^{-n}=0.$

Thus to prove surjectivity, it suffices to show that if $|b_{n}|s^{-n}\to 0$
as $n\to\infty$ for some $s<t$, then
$\left|\frac{b_{n}}{n!}\binom{\frac{k}{d}}{n}^{-1}\right|r^{n}\to 0$ as
$n\to\infty$ for some $r>\varpi/t$.

Now as noted in [30, Proposition 13.1.5, Corollary 13.1.7], the rational
number $\frac{k}{d}$ is $p$-adic non-Liouville and so the radius of
convergence of the series

$\sum\limits_{n=0}^{\infty}\frac{T^{n}}{n!}\binom{\frac{k}{d}}{n}^{-1}$

is at least $\varpi$; that is $\sup_{n\geqslant
0}\left|\frac{1}{n!}\binom{\frac{k}{d}}{n}^{-1}\right|w^{n}<\infty$ for all
$w<\varpi$. Now, because $s<t$, we can choose $w$ such that $\varpi
s/t<w<\varpi$. Then $r:=w/s>\varpi/t$ works.

Next we prove the injectivity of the first non-zero map. From Definition
4.2.10(b) we can see that

$R(u,d)=(x-a)\partial_{x}-\frac{k}{d}.$

It follows from this that for every
$Q=\sum\limits_{n=0}^{\infty}Q_{n}\partial_{x}^{n}\in\mathcal{D}^{\dagger}_{\varpi/t}(X)$
we have

(31) $Q\cdot
R(u,d)=\sum_{n=0}^{\infty}\left((n-\frac{k}{d})Q_{n}+(x-a)Q_{n-1}\right)\partial_{x}^{n}$

and so if $Q\cdot R(u,d)=0$ then

$(n-\frac{k}{d})Q_{n}+(x-a)Q_{n-1}=0\quad\mbox{for all}\quad n\geqslant 0,$

where $a_{-1}:=0$. Since $\frac{k}{d}\notin\mathbb{N}$, an easy recurrence
gives that $a_{n}=0$ for all $n$.

It remains to show that the kernel of $\cdot z$ is contained in the image of
$\cdot R(u,d)$. So we suppose that
$P=\sum\limits_{n=0}^{\infty}P_{n}\partial_{x}^{n}\in\mathcal{D}^{\dagger}_{\varpi/t}(X)$
satisfies $P\cdot z=0$, and deduce using (30) that

$\sum_{n=0}^{\infty}\binom{\frac{k}{d}}{n}n!P_{n}(x-a)^{-n}=0\in\mathcal{O}(X)_{\overline{U_{t}}}\subset\mathcal{O}\left(X\left(\frac{t}{x-a}\right)\right).$

Let $b_{m}:=\sum\limits_{n=0}^{m}\binom{\frac{k}{d}}{n}n!P_{n}(x-a)^{m-n}$.
Because $\frac{k}{d}\notin\mathbb{N}$ by assumption, we can define

$Q:=\sum_{n=0}^{\infty}\frac{c_{n}}{n!}\partial_{x}^{n}\quad\mbox{where}\quad
c_{n}:=\left[\binom{\frac{k}{d}}{n}(n-\frac{k}{d})\right]^{-1}b_{n}\quad\mbox{for
all}\quad n\geqslant 0.$

We wish to show that $Q\in\mathcal{D}^{\dagger}_{\varpi/t}(X)$. Granted this,
it is an easy computation, using (31), to check that $Q\cdot R(u,d)=P$ giving
the result.

Since $P\in\mathcal{D}^{\dagger}_{\varpi/t}$, there is some $r>\varpi/t$ such
that $|P_{n}|r^{n}\to 0$ as $n\to\infty$. Moreover, for any $s>\varpi$,
$|n!|/s^{n}$ is bounded by Lemma 2.3.16 and
$\left|\binom{k/d}{n}\right|\leqslant 1$ for all $n$ since $p\nmid d$. Thus,
choosing $s\in(\varpi,tr)$, we see that
$\sum\limits_{n=0}^{\infty}\binom{k/d}{n}n!P_{n}T^{n}$ has radius of
convergence at least $r/s>1/t$ in $\mathcal{O}(X)[[T]]$. Now we may appeal to
Lemma 4.2.15, with $f=x-a$ and $v_{n}=\binom{k/d}{n}{n!}P_{n}$ to see that
$\sum\limits_{m=0}^{\infty}b_{m}T^{m}$ has radius of convergence greater than
$1/t$. Since $\frac{k}{d}$ is non-Liouville,
$\sum\limits_{n=0}^{\infty}\frac{T^{n}}{\frac{k}{d}-n}$ has radius of
convergence $1$ and as noted above
$\sum\limits_{n=0}^{\infty}\frac{T^{n}}{n!}\binom{\frac{k}{d}}{n}^{-1}$ has
radius of convergence $\varpi$. Combining all of these estimates, we see that
there is some $u>\varpi/t$ such that $|\frac{c_{n}}{n!}|u^{n}\to 0$ as
$n\to\infty$ as required. ∎

Next, we collect several basic facts about the linear differential operator
$R_{\mathcal{S}}(u,d)$.

######  Lemma 4.2.17.

  1. (a)

$R_{\mathcal{S}}(u,d)=\Delta_{\mathcal{S}}\theta_{u,d}(\partial_{x})$.

  2. (b)

$R_{\mathcal{S}}(u,d)=\theta_{u\Delta_{\mathcal{S}}^{d},d}(\partial_{x})\Delta_{\mathcal{S}}$.

  3. (c)

Let $v\in K(x)$ with $\mathcal{S}(v)\subset\mathcal{S}$. Then

$R_{\mathcal{S}}(uv,d)=\theta_{u,d}(R_{\mathcal{S}}(v,d)).$

###### Proof.

(a) This follows directly from Definition 4.2.10(b) and Lemma 2.4.1.

(b) For any $v\in\mathcal{O}(X)^{\times}$, $\theta_{v^{d},d}$ is conjugation
by $v$ in $\mathcal{D}(X)$ by Lemma 2.4.1. Applying Lemma 2.4.3 together with
part (a), we see that

$\theta_{u\Delta_{\mathcal{S}}^{d},d}(\partial_{x})\Delta_{\mathcal{S}}=\theta_{\Delta_{\mathcal{S}}^{d},d}\left(\theta_{u,d}(\partial_{x})\right)\Delta_{\mathcal{S}}=\Delta_{\mathcal{S}}\theta_{u,d}(\partial_{x})=R_{\mathcal{S}}(u,d).$

(c) We can compute

$R_{\mathcal{S}}(uv,d)=\Delta_{\mathcal{S}}\theta_{uv,d}(\partial_{x})=\Delta_{\mathcal{S}}\theta_{u,d}(\theta_{v,d}(\partial_{x}))=\theta_{u,d}(R_{\mathcal{S}}(v,d))$

using part (a), Lemma 2.4.3 and the fact that $\theta_{u,d}$ is
$\mathcal{O}(X)$-linear. ∎

Here is our final local calculation.

######  Proposition 4.2.18.

Let $Y\in\mathbb{A}(t\partial_{x}/\varpi)^{\dagger}$ containing at most one
$a_{i}$. Then

$0\to\mathcal{D}^{\dagger}_{\varpi/t}(Y)\stackrel{{\scriptstyle\cdot
R_{\mathcal{S}}(u,d)}}{{\longrightarrow}}\mathcal{D}^{\dagger}_{\varpi/t}(Y)\stackrel{{\scriptstyle\cdot
z}}{{\longrightarrow}}\mathcal{M}(\mathcal{S},u,d)_{\overline{U_{t}}}(Y)\to
0.$

is exact, provided that $v_{a_{i}}(u)/d\notin\mathbb{N}$.

###### Proof.

If none of the $a_{i}$ live in $Y$ then we’re done by Proposition 4.2.12. So
we may suppose $a_{i}\in Y(K)$ and define $a:=a_{i}$, $k:=v_{a_{i}}(u)$ and
$v=u/(x-a)^{k}$ so that $v$ is a unit in $\mathcal{O}(Y)$. In particular
$\theta_{v,d}$ induces an automorphism of
$\mathcal{D}^{\dagger}_{\varpi/t}(Y)$ by Corollary 2.4.11. Moreover by Lemma
4.2.17(c),

$R_{\mathcal{S}}(u,d)=\theta_{v,d}(R_{\mathcal{S}}((x-a)^{k},d)).$

Thus
$\mathcal{D}^{{\dagger}}_{\varpi/t}(Y)R_{\mathcal{S}}(u,d)=\mathcal{D}^{\dagger}_{\varpi/t}(Y)\theta_{v,d}(R_{\mathcal{S}}((x-a)^{k},d))$.
Under $\theta^{-1}_{v,d}$ the $\mathcal{D}^{\dagger}_{\varpi/t}(Y)$-module
$\mathcal{M}(\mathcal{S},u,d)_{\overline{U_{t}}}(Y)$ corresponds to the
$\mathcal{D}^{\dagger}_{\varpi/t}(Y)$-module
$\mathcal{M}(\mathcal{S},(x-a)^{k},d)_{\overline{U_{t}}}(Y)$. Now we can apply
Theorem 4.2.16. ∎

######  Definition 4.2.19.

For $i=1,\ldots,h$ let $W_{i}$ be the affinoid subdomain of $\mathbb{A}$ whose
$\mathbf{C}$-points are obtained by removing the open discs $\\{z\in{\bf
C}\mid|z-a_{j}|<t\\}$ for $j\neq i$ from the closed disc $\\{z\in{\bf
C}\mid|z-a_{i}|\leqslant t\\}$.

We note that the condition $t\leqslant\min\limits_{i\neq j}|a_{i}-a_{j}|$
means that each $W_{i}$ contains $a_{i}$ but does not contain $a_{j}$ for
$i\neq j$.

######  Lemma 4.2.20.

  1. (a)

Each $W_{i}$ is $\mathbb{A}(t\partial_{x}/\varpi)^{\dagger}$-admissible.

  2. (b)

Let $X\in\mathbb{A}(t\partial_{x}/\varpi)^{\dagger}$, and suppose that

$t\leqslant\min\\{|a_{i}-a_{j}|:a_{i},a_{j}\in X({\bf C})\quad\mbox{and}\quad
i\neq j\\}.$

Then $\\{X\cap W_{1},\ldots,X\cap W_{h},X\cap U_{t}\\}$ is an admissible
covering of $X$.

###### Proof.

(a) We choose a finite field extension $K^{\prime}$ of $K$ such that
$t\in|K^{\prime}|$; then $W_{i}\times_{K}K^{\prime}$ becomes a cheese such
that $\rho(W_{i}\times_{K}K^{\prime})=t$. Hence

$r(W_{i})=r(W_{i}\times_{K}K^{\prime})=\varpi/t$

by Lemma 2.3.13 and Corollary 2.3.17.

(b) Applying Lemma 2.3.13 and Corollary 2.3.17 again shows

$\rho(X\times_{K}{{\bf C}})=\varpi/r(X\times_{K}{\bf
C})\geqslant\frac{\varpi}{\varpi/t}=t.$

Suppose that $a\in X({\bf C})$ and $|a-a_{i}|<t$ for some $i$. If $a_{i}\notin
X({\bf C})$, then because $\rho(X\times_{K}{\bf{C}})\geqslant t$, the
intersection of $X({\bf C})$ with the open disc of radius $t$ around $a_{i}$
is empty. But $|a-a_{i}|<t$ and $a\in X({\bf C})$ — contradiction — so in fact
$a_{i}\in X({\bf C})$ whenever $a\in X({\bf C})$ and $|a-a_{i}|<t$. We will
now show that

$X({\bf C})=(X\cap U_{t})({\bf C})\cup(X\cap W_{1})({\bf
C})\cup\cdots\cup(X\cap W_{h}({\bf C}).$

Indeed, if $a\in X({\bf C})-(X\cap U_{t})({\bf C})$ then $a\in X(\bf C)$ and
$|a-a_{i}|<t$ for some $i$, so that $a_{i}\in X({\bf C})$ by the above. But if
also $|a-a_{j}|<t$ for some $j\neq i$, then $a_{j}\in X({\bf C})$ by the above
as well, and then

$|a_{i}-a_{j}|\leqslant\max\\{|a-a_{i}|,|a-a_{j}|\\}<t$

contradicts our hypothesis on $t$. So in fact $|a-a_{j}|\geqslant t$ for all
$j\neq i$ which forces $a\in(X\cap W_{i})(\bf{C})$ and establishes our claim.
Since $\\{X\cap U_{t},X\cap W_{1},\ldots,X\cap W_{h}\\}$ is a finite set of
affinoid subdomains of $X$, which covers $X$ on ${\bf C}$-points, we conclude
that it is an admissible covering of $X$. ∎

Here is the main result of $\S\ref{KummerSect}$.

######  Theorem 4.2.21.

Let $X\in\mathbb{A}(t\partial_{x}/\varpi)^{\dagger}$. Suppose that

  * •

$t\in\sqrt{|K^{\times}|}$ and $p\nmid d$,

  * •

$t\leqslant\min\\{|a_{i}-a_{j}|:a_{i},a_{j}\in X({\bf C})\quad\mbox{and}\quad
i\neq j\\}$, and

  * •

$k_{i}/d\not\in\mathbb{N}$ whenever $a_{i}\in X({\bf C})$.

Then the complex of $\mathcal{D}^{\dagger}_{\varpi/t}(X)$-modules

$0\to\mathcal{D}^{\dagger}_{\varpi/t}(X)\stackrel{{\scriptstyle\cdot
R_{\mathcal{S}}(u,d)}}{{\longrightarrow}}\mathcal{D}^{\dagger}_{\varpi/t}(X)\to\mathcal{M}(\mathcal{S},u,d)_{\overline{U_{t}}}(X)\to
0$

from Lemma 4.2.11 is exact.

###### Proof.

Lemma 4.2.20(b) tells us that $\\{X\cap U_{t},X\cap W_{1},\ldots,X\cap
W_{h}\\}$ is an $\mathbb{A}(t\partial_{x}/\varpi)^{\dagger}$-admissible
covering of $X$. By Theorem 2.3.10 and Theorem 4.1.9, we know that
$\mathcal{D}^{\dagger}_{\varpi/t}$ and
$\mathcal{M}(\mathcal{S},u,d)_{\overline{U_{t}}}$ are sheaves on
$\mathbb{A}(t\partial_{x}/\varpi)^{\dagger}$ with vanishing higher Čech
cohomology. Therefore it suffices to show that the sequence

$0\to\mathcal{D}^{\dagger}_{\varpi/t}(Y)\stackrel{{\scriptstyle\cdot
R_{\mathcal{S}}(u,d)}}{{\longrightarrow}}\mathcal{D}^{\dagger}_{\varpi/t}(Y)\to\mathcal{M}(\mathcal{S},u,d)_{\overline{U_{t}}}(Y)\to
0$

is exact whenever $Y$ is an intersection of any members of our covering; note
that this $Y$ is still $(t\partial_{x}/\varpi)^{\dagger}$-admissible. Since
each member of our covering contains at most one $a_{i}$ the exactness follows
from Proposition 4.2.18.∎

Finally, we record a purely algebraic calculation which tells us how the the
relators $R_{\mathcal{S}(u)}(u,d)$ from Definition 4.2.10 interact with the
maps $g^{\mathcal{D}}:\mathcal{D}\to g^{\ast}\mathcal{D}$ defining the
$G$-equivariant structure on $\mathcal{D}$. Recall from $\S\ref{AffTrans}$
that $\mathbb{B}$ denotes the subgroup scheme
$\\{g\in\mathbb{GL}_{2}:g_{21}=0\\}$ of upper-triangular matrices in
$\mathbb{GL}_{2}$. Recall also the isomorphism
$g^{\dagger}_{r}\colon\mathcal{D}_{r}^{\dagger}\stackrel{{\scriptstyle\cong}}{{\longrightarrow}}g^{\ast}\mathcal{D}_{\frac{r}{\varrho(g)}}^{\dagger}$
from Corollary 2.5.6(b).

######  Lemma 4.2.22.

Let $g\in\mathbb{B}(K)$, let $W$ be an affinoid subdomain of $\mathbb{A}$ and
let $u\in K(x)\cap\mathcal{O}(W)^{\times}$.

  1. (a)

We have $g\cdot R_{\mathcal{S}(u)}(u,d)=\varrho(g)^{1-|\mathcal{S}(u)|}\hskip
2.84544ptR_{\mathcal{S}(g\cdot u)}(g\cdot u,d)$.

  2. (b)

Suppose that $W\in\mathbb{A}(\partial_{x}/r)^{\dagger}$ for some
$r\in\mathbb{R}_{>0}$. Then

$g^{\dagger}_{r}(W)\left(R_{\mathcal{S}(u)}(u,d)\right)=\varrho(g)^{1-|\mathcal{S}(u)|}\
R_{\mathcal{S}(g\cdot u)}(g\cdot u,d).$

###### Proof.

(a) By Lemma 2.5.2(e), we have $g\cdot(x-z)=\varrho(g)^{-1}(x-g\cdot z)$ for
any $z\in\overline{K}$. Therefore $v_{g\cdot z}(g\cdot u)=v_{z}(u)$, so
$\mathcal{S}(g\cdot u)=g\cdot\mathcal{S}(u)$ and

$g\cdot\Delta_{u}=\prod_{z\in\mathcal{S}(u)}\varrho(g)(x-g\cdot
z)=\varrho(g)^{-|\mathcal{S}(u)|}\Delta_{g\cdot u}.$

Because we also have $g\cdot\partial_{x}=\varrho(g)\partial_{x}$ by Lemma
2.5.2(e), Lemma 2.5.7 gives

$\begin{array}[]{lll}g\cdot R_{\mathcal{S}(u)}(u,d)&=&g\cdot\Delta_{u}\hskip
2.84544ptg\cdot\theta_{u,d}(\partial_{x})=\varrho(g)^{-|\mathcal{S}(u)|}\Delta_{g\cdot
u}\hskip 2.84544pt\theta_{g\cdot u,d}(g\cdot\partial_{x})\\\
&=&\varrho(g)^{1-|\mathcal{S}(u)|}R_{\mathcal{S}(g\cdot u)}(g\cdot
u,d).\end{array}\qed$

(b) This follows from part (a) and the commutative square (13).

###  4.3. Line bundles with connection on the local Drinfeld space

In $\S\ref{UpperHP}$, we work with the following data.

######  Notation 4.3.1.

  1. (a)

$\Omega:=\mathbb{P}^{1,\operatorname{an}}-\mathbb{P}^{1}(F)$ is the the
Drinfeld upper half plane,

  2. (b)

$\Upsilon=\mathbb{D}\cap\Omega$,

  3. (c)

$j\colon\Upsilon\hookrightarrow\mathbb{D}$ is the open embedding,

  4. (d)

$[\mathscr{L}]\in\operatorname{PicCon}(\Upsilon)[p^{\prime}]$,

  5. (e)

$d$ is the order of $[\mathscr{L}]$ in $\operatorname{PicCon}(\Upsilon)$. Thus
$d$ is coprime to $p$.

We refer the reader to [8, Definition 3.2.1] for the terminology, and we fix a
$\mathcal{D}$-linear isomorphism $\psi\colon\mathscr{L}^{\otimes
d}\stackrel{{\scriptstyle\cong}}{{\longrightarrow}}\mathcal{O}_{\Upsilon}$.

######  Definition 4.3.2.

Let $n\geqslant 0$ be an integer.

  1. (a)

$V_{n}:=\mathbb{A}\backslash\bigcup\limits_{a\in\mathcal{O}_{F}}\\{|z-a|<|\pi_{F}|^{n}\\}$,

  2. (b)

$\Upsilon_{n}:=V_{n}\cap\mathbb{D}$,

  3. (c)

$\mathscr{L}_{n}:=j_{\ast}(\mathscr{L}_{\overline{\Upsilon_{n}}})$, and

  4. (d)

$h_{n}:=q^{n+1}$.

See Definition 4.1.1 for the meaning of the sheaf
$\mathscr{L}_{\overline{\Upsilon_{n}}}$ on $\Upsilon$.

We note that if
$\mathcal{S}:=\\{a_{1},\ldots,a_{h_{n}}\\}\subset\mathcal{O}_{F}$ is any set
of coset representatives for the group
$\mathcal{O}_{F}/\pi_{F}^{n+1}\mathcal{O}_{F}$, then $V_{n}$ was denoted
$U(\mathcal{S})_{|\pi_{F}|^{n}}$ in §4.2.

We also note that $\Upsilon_{n}$ is the cheese obtained by removing $h_{n}$
open balls of radius $|\pi_{F}|^{n}$ from $\mathbb{D}$ with centres at
$a_{1},\ldots,a_{h_{n}}$ and $\\{\Upsilon_{n}\\}_{n\geqslant 0}$ forms an
admissible cover of $\Upsilon$ totally ordered by inclusion.

######  Lemma 4.3.3.

The maps $j_{\ast}\mathscr{L}\to\mathscr{L}_{n}$ given by restriction induce
an isomorphism
$j_{\ast}\mathscr{L}\cong\mathop{\underset{\scriptstyle}{\underset{\raisebox{0.0pt}[0.75346pt][-1.50694pt]{$\longleftarrow$}}{\operatorname{lim}}}}\mathscr{L}_{n}$
of $\mathcal{D}$-modules on $\mathbb{D}$.

###### Proof.

Since $j_{\ast}$ is a right adjoint it commutes with limits, and so it
suffices to show that the restriction maps induce an isomorphism
$\mathscr{L}\cong\mathop{\underset{\scriptstyle}{\underset{\raisebox{0.0pt}[0.75346pt][-1.50694pt]{$\longleftarrow$}}{\operatorname{lim}}}}\mathscr{L}_{\overline{\Upsilon_{n}}}$
of $\mathcal{D}$-modules on $\Upsilon$. Since the $\Upsilon_{n}$ forms an
admissible cover of $\Upsilon$ which is totally ordered by inclusion, by
Proposition 4.1.3, the restriction maps induce such an isomorphism of abelian
sheaves on $\Upsilon$. It is easy to see that all relevant maps are
$\mathcal{D}$-linear. ∎

Note that there is a natural bijection

$h(\Upsilon_{n})\backslash\\{D_{\infty}\\}\to\mathcal{O}_{F}/\pi^{n+1}\mathcal{O}_{F}$

sending $D$ to $D\cap\mathcal{O}_{F}$ for each $D\in
h(\Upsilon_{n})\backslash\\{D_{\infty}\\}$.

######  Notation 4.3.4.

For each $a\in\mathcal{O}_{F}$ and $n\geqslant 0$, let $D_{a,n}$ be the
element of $h(\Upsilon_{n})\backslash\\{D_{\infty}\\}$ corresponding to
$a+\pi_{F}^{n+1}\mathcal{O}_{F}$ under this bijection.

We introduce the natural map abelian groups

$M_{n,d}\colon\operatorname{PicCon}(\Upsilon)[d]\to
M_{0}(h(\Upsilon_{n}),\mathbb{Z}/d\mathbb{Z})$

obtained by composing the restriction map

$\operatorname{PicCon}(\Upsilon)[d]\to\operatorname{Con}(\Upsilon_{n})[d];\hskip
14.22636pt\mathscr{L}\mapsto\mathscr{L}|_{\Upsilon_{n}}$

with the isomorphisms $\theta_{d}$ and $\mu_{\Upsilon_{n},d}$ from [8,
Definition 3.1.15 and Corollary 4.3.3]:

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For each $n\geqslant 0$, there is a unique $\mathbb{Z}/d\mathbb{Z}$-valued
measure on $h(\Upsilon_{n})$ of total value zero, whose value on each
$D_{a,n}$ is $1$:

$\nu_{n}(D_{a,n})=1+d\mathbb{Z}\quad\mbox{for all}\quad
a\in\mathcal{O}_{F},\quad\mbox{and}\quad\nu_{n}(D_{\infty})=-q^{n+1}+d\mathbb{Z}.$

######  Lemma 4.3.5.

Let $n\geqslant 0$.

  1. (a)

$M_{0}(h(\Upsilon_{n}),\mathbb{Z}/d\mathbb{Z})^{I}$ is freely generated as a
$\mathbb{Z}/d\mathbb{Z}$-module by $\nu_{n}$.

  2. (b)

The restriction map
$r_{n+1}:M_{0}(h(\Upsilon_{n+1}),\mathbb{Z}/d\mathbb{Z})^{I}\to
M_{0}(h(\Upsilon_{n}),\mathbb{Z}/d\mathbb{Z})^{I}$ sends $\nu_{n+1}$ to
$q\nu_{n}$.

  3. (c)

$r_{n+1}$ is an isomorphism of $\mathbb{Z}/d\mathbb{Z}$-modules.

  4. (d)

The restriction map
$\operatorname{PicCon}(\Upsilon)^{I}[d]\to\operatorname{Con}(\Upsilon_{n})^{I}[d]$
is an isomorphism.

  5. (e)

$\operatorname{PicCon}(\Upsilon)^{I}[d]$ is a free
$\mathbb{Z}/d\mathbb{Z}$-module of rank $1$.

###### Proof.

(a) The natural $I$-action on
$\mathcal{O}_{F}/\pi_{F}^{n+1}\mathcal{O}_{F}\cup\\{\infty\\}$ has two orbits:
$\mathcal{O}_{F}/\pi_{F}^{n+1}\mathcal{O}_{F}$ and $\\{\infty\\}$. Therefore
the natural $I$-action on $h(\Upsilon_{n})$ also has two orbits, namely
$\\{D_{b,n}:b\in\mathcal{O}_{F}\\}$ and $\\{D_{\infty}\\}$. If $\nu$ is an
$I$-invariant measure on $h(\Upsilon_{n})$, then it must be constant on these
orbits. If, in addition, it has total value zero, then we easily see that
$\nu=\nu(D_{0,n})\nu_{n}$ as required.

(b) Fix $b\in\mathcal{O}_{F}$. In view of part (a), the map $r_{n+1}$ must
send $\nu_{n+1}$ to some integer multiple of $\nu_{n}$. However the
restriction map $h(\Upsilon_{n+1})\to h(\Upsilon_{n})$ sends
$D_{b+\pi_{F}^{n+2}c,n+1}$ to $D_{b,n}$ for each $c\in\mathcal{O}_{F}$, and
the preimage of $D_{b,n}$ under this map consists of precisely $q$ such holes.
Hence $r_{n+1}(\nu_{n+1})=q\nu_{n}$.

(c) This follows immediately from parts (a) and (b) since $d$ and $q$ are
coprime.

(d) For each $m\geqslant n$ there is a natural commutative square

$\textstyle{\operatorname{Con}(\Upsilon_{m+1})^{I}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\cong}$$\scriptstyle{\mu_{\Upsilon_{m+1},d}\circ\theta_{d}}$$\textstyle{\operatorname{Con}(\Upsilon_{m})^{I}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\cong}$$\scriptstyle{\mu_{\Upsilon_{m},d}\circ\theta_{d}}$$\textstyle{M_{0}(h(\Upsilon_{m+1}),\mathbb{Z}/d\mathbb{Z})^{I}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{r_{m+1}}$$\textstyle{M_{0}(h(\Upsilon_{m}),\mathbb{Z}/d\mathbb{Z})^{I}}$

whose vertical arrows are isomorphisms by [8, Corollary 4.3.4]. Using part (b)
we see that the top horizontal arrow is bijective. Hence the connecting maps
in the inverse system $(\operatorname{Con}(\Upsilon_{m})^{I})_{m\geqslant n}$
are all isomorphisms, so the projection map

$\mathop{\underset{\scriptstyle}{\underset{\raisebox{0.0pt}[0.75346pt][-1.50694pt]{$\longleftarrow$}}{\operatorname{lim}}}}\operatorname{Con}(\Upsilon_{m})^{I}\to\operatorname{Con}(\Upsilon_{n})^{I}$

is an isomorphism as well. Next, we have
$\operatorname{PicCon}(\Upsilon_{m})=\operatorname{Con}(\Upsilon_{m})$ by [8,
Proposition 4.1.11]. The family $(\Upsilon_{m})_{m\geqslant n}$ forms an
increasing admissible covering of $\Upsilon$ by geometrically connected
affinoid subdomains, so [8, Proposition 3.1.9] implies that the restriction
map

$\operatorname{PicCon}(\Upsilon)^{I}\longrightarrow\mathop{\underset{\scriptstyle}{\underset{\raisebox{0.0pt}[0.75346pt][-1.50694pt]{$\longleftarrow$}}{\operatorname{lim}}}}\operatorname{PicCon}(\Upsilon_{m})^{I}$

is an isomorphism. The composition of this restriction map with the projection
$\operatorname{PicCon}(\Upsilon_{m})^{I}\to\operatorname{PicCon}(\Upsilon_{n})^{I}$
is the restriction map
$\operatorname{PicCon}(\Upsilon)^{I}\to\operatorname{PicCon}(\Upsilon_{n})^{I}$
in the statement of (d), and it is therefore an isomorphism as required.

(e) Use part (d), [8, Corollary 4.3.4] and part (a). ∎

######  Definition 4.3.6.

Let $X$ be an affinoid subdomain of $\mathbb{D}$. A section
$\dot{z}\in\mathscr{L}_{n}(X)$ is said to be an _algebraic generator_ if there
exists $u\in K(x)$ such that

  1. (a)

$\mathscr{L}_{n|X}=(\mathcal{O}_{X})_{\overline{X\cap V_{n}}}\dot{z}$,

  2. (b)

$\psi(\dot{z}^{\otimes d})=u$,

  3. (c)

$\mathcal{S}(u)\subset F$ (see Definition 4.2.1 for this notation),

  4. (d)

$|a-b|\geqslant|\pi_{F}|^{n}$ for all $a\neq b\in\mathcal{S}(u)\cap X$, and

  5. (e)

$\frac{v_{a}(u)}{d}\not\in\mathbb{N}$ for all $a\in\mathcal{S}(u)\cap X$.

We call $u$ the _associated rational function_.

We will now prove that algebraic generators exist.

######  Theorem 4.3.7.

Let $X$ be an affinoid subdomain of $\mathbb{D}$.

  1. (a)

Let $\mathcal{S}\subset\mathcal{O}_{F}$ be any set of coset representatives
for the group $\mathcal{O}_{F}/\pi^{n+1}\mathcal{O}_{F}$ and let
$(k_{a})_{a\in\mathcal{S}}$ be a family of integers such that for all
$a\in\mathcal{S}$ we have

(32) $\frac{k_{a}}{d}\notin\mathbb{N}\quad\mbox{and}\quad
M_{n,d}([\mathscr{L}])(D_{a,n})={k_{a}}+d\mathbb{Z}.$

Then there exists an algebraic generator $\dot{z}\in\mathscr{L}_{n}(X)$ with
associated rational function of the form

$u=\lambda\prod_{a\in\mathcal{S}}(x-a)^{k_{a}}\quad\mbox{for
some}\quad\lambda\in K^{\times}.$

  2. (b)

Algebraic generators always exist.

###### Proof.

(a) Using Lemma 4.1.2 we can see that if
$\dot{z}\in\mathscr{L}_{n}(\mathbb{D})$ is algebraic generator then
$\dot{z}|_{X}$ is an algebraic generator with the same associated rational
function and so we only need to consider the case $X=\mathbb{D}$ itself.

Let
$u_{0}:=\prod\limits_{a\in\mathcal{S}}(x-a)^{k_{a}}\in\mathcal{O}(\Upsilon_{n})^{\times}$.
Then the assumption (32) tells us that

$\mu_{\Upsilon_{n},d}(u_{0}K^{\times}\mathcal{O}(\Upsilon_{n})^{\times
d})=M_{n,d}([\mathscr{L}|)=\mu_{\Upsilon_{n},d}(\theta_{d}(\mathscr{L}|_{\Upsilon_{n}})).$

Since $\mu_{\Upsilon_{n},d}$ is an isomorphism, we see that

(33)
$\theta_{d}(\mathscr{L}|_{\Upsilon_{n}})=u_{0}K^{\times}\mathcal{O}(\Upsilon_{n})^{\times
d}.$

Write $\mathcal{S}=\\{a_{1},\ldots,a_{h_{n}}\\}$, choose an integer $\ell$
coprime to $p$ and define

$V:=\operatorname{Sp}K\left\langle x,\frac{\pi_{F}^{\ell
n+1}}{(x-a_{1})^{\ell}},\cdots,\frac{\pi_{F}^{\ell
n+1}}{(x-a_{h_{n}})^{\ell}}\right\rangle.$

This is a $K$-form of the $\mathbf{C}$-cheese $V_{\mathbf{C}}$ obtained from
$\mathbb{D}_{\mathbf{C}}$ by removing the open balls of radius
$|\pi_{F}|^{n+\frac{1}{\ell}}$ around the $a_{i}$. Observe that $V$ is a wide
open neighbourhood of $\Upsilon_{n}$ contained in $\Upsilon_{n+1}$. Since $V$
and $\Upsilon_{n}$ are geometrically connected, we have the following
commutative square whose vertical arrows are given by restriction:

$\textstyle{\operatorname{Con}(V)[d]\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\theta_{d}}$$\scriptstyle{\cong}$$\textstyle{\mathcal{O}(V)^{\times}/K^{\times}\mathcal{O}(V)^{\times
d}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{r_{n}}$$\textstyle{\operatorname{Con}(\Upsilon_{n})[d]\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\theta_{d}}$$\scriptstyle{\cong}$$\textstyle{\mathcal{O}(\Upsilon_{n})^{\times}/K^{\times}\mathcal{O}(\Upsilon_{n})^{\times
d}.}$

We claim that $r_{n}$ is injective. Assuming this, and noting that $u_{0}$
also lies in $\mathcal{O}(V)^{\times}$, it follows from (33) that
$\theta_{d}([\mathscr{L}|_{V}])=u_{0}K^{\times}\mathcal{O}(V)^{\times d}$
inside $\mathcal{O}(V)^{\times}/K^{\times}\mathcal{O}(V)^{\times d}$. Using
the construction of $\theta_{d}$ given at [8, Proposition 3.1.14], we can find
a section $\dot{z}\in\mathscr{L}(V)$ and some $\lambda\in K^{\times}$ such
that $\mathscr{L}(V)=\mathcal{O}(V)\dot{z}$ and $\psi(\dot{z}^{\otimes
d})=u:=\lambda u_{0}$. Since $V$ is affinoid and $\mathscr{L}$ is a coherent
$\mathcal{O}$-module, in fact we also have
$\mathscr{L}_{|V}=\mathcal{O}_{|V}\dot{z}$. Since $V$ is a wide open
neighbourhood of $\Upsilon_{n}$, i.e. $\widetilde{V}$ contains the closure of
$\widetilde{\Upsilon_{n}}$, we may view $\dot{z}$ as an element of
$\mathscr{L}_{n}(\mathbb{D})$ and deduce that
$\mathscr{L}_{n|\mathbb{D}}=(\mathcal{O}_{\mathbb{D}})_{\overline{\Upsilon_{n}}}\dot{z}$.

Finally, $v_{a}(u)=k_{a}$ for all $a\in\mathcal{S}$. As none of the $k_{a}$
can be zero, given the assumption that $\frac{k_{a}}{d}\notin\mathbb{N}$ for
all $a\in\mathcal{S}$, we have $\mathcal{S}(u)=\mathcal{S}$ and this is
contained in $\mathcal{O}_{F}$. We now see that Definition 4.3.6(c,d,e) is
satisfied and $\dot{z}\in\mathscr{L}_{n}(\mathbb{D})$ is an algebraic
generator, once we have checked the injectivity of $r_{n}$ claimed above.

Let $L=K(\pi_{F}^{\frac{1}{\ell}})$. Then in the notation of [8, Definition
4.1.1], $V_{L}$ is isomorphic to the $L$-cheese
$C_{L}(0,a_{1},\cdots,a_{h_{n}},1,\pi_{F}^{n+\frac{1}{\ell}},\cdots,\pi_{F}^{n+\frac{1}{\ell}})$
as an $L$-affinoid variety. The map $r_{n}$ appears in the following
commutative diagram:

$\textstyle{\mathcal{O}(V)^{\times}/K^{\times}\mathcal{O}(V)^{\times
d}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{r_{n}}$$\textstyle{\mathcal{O}(V_{L})^{\times}/L^{\times}\mathcal{O}(V_{L})^{\times
d}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\mu_{V_{L},d}}$$\textstyle{M_{0}(h(V_{L}),\frac{\mathbb{Z}}{d\mathbb{Z}})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathcal{O}(\Upsilon_{n})^{\times}/K^{\times}\mathcal{O}(\Upsilon_{n})^{\times
d}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathcal{O}(\Upsilon_{n,L})^{\times}/L^{\times}\mathcal{O}(\Upsilon_{n,L})^{\times
d}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\mu_{\Upsilon_{n,L},d}}$$\textstyle{M_{0}(h(\Upsilon_{n,L}),\frac{\mathbb{Z}}{d\mathbb{Z}}).}$

Since $|h(V_{L})|=|h(\Upsilon_{n,L})|=h_{n}$, the vertical arrow on the right
is bijective. Since $\mu_{V_{L},d}$ is an isomorphism by [8, Corollary
4.3.3(c)], it remains to show that the first horizontal arrow in the top row
is injective, or equivalently, that

$L^{\times}\mathcal{O}(V_{L})^{\times d}\hskip 2.84544pt\cap\hskip
2.84544pt\mathcal{O}(V)^{\times}\subseteq K^{\times}\mathcal{O}(V)^{\times
d}.$

Let $S$ denote the subgroup of $\mathcal{O}(V)^{\times}$ generated by
$x-a_{1},\cdots,x-a_{h_{n}}$. View $S$ also as a subgroup of
$\mathcal{O}(V_{L})^{\times}$. Applying [8, Proposition 4.1.10] to $V_{L}$, we
have

$\mathcal{O}(V_{L})^{\times}=\mathcal{O}(V_{L})^{\times\times}\cdot
L^{\times}\cdot S.$

Now $L^{\times}=L^{\times\times}\cdot
K^{\times}\cdot\pi_{F}^{\frac{1}{\ell}\mathbb{Z}}$ because
$L=K(\pi_{F}^{\frac{1}{\ell}})$ is a totally ramified extension of $K$. Using
[8, Lemma 4.3.2(a)] we then see that

$L^{\times}\mathcal{O}(V_{L})^{\times
d}=\pi^{\frac{d}{\ell}\mathbb{Z}}\cdot\mathcal{O}(V_{L})^{\times\times}\cdot
K^{\times d}\cdot S^{d}.$

Since $K^{\times d}S^{d}\subseteq K^{\times}\mathcal{O}(V)^{\times d}$, by the
modular law it remains to show that

$\pi_{F}^{\frac{d}{\ell}\mathbb{Z}}\mathcal{O}(V_{L})^{\times\times}\cap\mathcal{O}(V)^{\times}\subseteq
K^{\times}\mathcal{O}(V)^{\times d}.$

Let $\zeta\in\overline{K}$ be a primitive $\ell^{th}$ root of unity and let
$M:=L(\zeta)$, a Galois extension of $K$ with Galois group $\mathcal{G}$. Let
$u\in\pi_{F}^{\frac{d}{\ell}\mathbb{Z}}\mathcal{O}(V_{L})^{\times\times}\cap\mathcal{O}(V)^{\times}$.
Then $u$ is fixed by the natural action of $\mathcal{G}$ on
$\mathcal{O}(V_{M})$ and we can write $u=\pi_{F}^{\frac{m}{\ell}}(1+\epsilon)$
for some $m\in d\mathbb{Z}$ and $\epsilon\in\mathcal{O}(V_{L})^{\circ\circ}$.
Choose $\sigma\in\mathcal{G}$ which sends $\pi_{F}^{\frac{1}{\ell}}$ to
$\zeta\pi_{F}^{\frac{1}{\ell}}$; then

$\sigma(\pi_{F}^{\frac{m}{\ell}})(1+\sigma(\epsilon))=\pi_{F}^{\frac{m}{\ell}}(1+\epsilon),\quad\mbox{so}\quad\zeta^{m}=\frac{\sigma(\pi_{F}^{\frac{m}{\ell}})}{\pi_{F}^{\frac{m}{\ell}}}=\frac{1+\epsilon}{1+\sigma(\epsilon)}\in\mathcal{O}(V_{M})^{\times\times}.$

Since $p\neq\ell$ and $(\zeta^{m})^{\ell}=1$, it follows from [8, Lemma
4.3.2(a)] that $\zeta^{m}=1$. We conclude that $\ell\mid m$. Therefore
$\pi_{F}^{\frac{m}{\ell}}\in K^{\times}$ and
$\epsilon=\pi_{F}^{-\frac{m}{\ell}}u-1\in\mathcal{O}(V)$. Since
$\epsilon\in\mathcal{O}(V_{L})^{\circ\circ}$, we see that
$\epsilon\in\mathcal{O}(V)^{\circ\circ}$ and $u\in
K^{\times}\mathcal{O}(V)^{\times\times}$. Applying [8, Lemma 4.3.2(a)] again
shows that $u\in K^{\times}\mathcal{O}(V)^{\times d}$ and completes the proof.

(b) Choose any set $\mathcal{S}$ of coset representatives for
$\pi_{F}^{n+1}\mathcal{O}_{F}$ in $\mathcal{O}_{F}$, and choose any family
$(\ell_{a})_{a\in\mathcal{S}}$ of integers such that
$M_{n,d}([\mathscr{L}])(a+\pi_{F}^{n+1}\mathcal{O}_{F})=\frac{\ell_{a}}{d}+\mathbb{Z}$
for all $a\in\mathcal{S}$. Let $M$ be any positive integer such that
$M>\max_{a\in\mathcal{S}}\frac{\ell_{a}}{d}$ and let $k_{a}:=\ell_{a}-Md<0$
for all $a\in\mathcal{S}$. Then $\frac{k_{a}}{d}\notin\mathbb{N}$ and
$\frac{\ell_{a}}{d}+\mathbb{Z}=\frac{k_{a}}{d}+\mathbb{Z}$ for all
$a\in\mathcal{S}$. We can now apply part (a) to the family of integers
$(k_{a})_{a\in\mathcal{S}}$ to conclude.∎

We recall the Definition 4.2.2 of $\mathcal{M}(\mathcal{S},u,d)$ for any $u\in
K(x)$ with $\mathcal{S}(u)\subset\mathcal{S}$.

######  Proposition 4.3.8.

Let $X$ be an affinoid subdomain of $\mathbb{D}$ and $\mathcal{S}_{n}$ be a
set of coset representatives for the group
$\mathcal{O}_{F}/\pi_{F}^{n+1}\mathcal{O}_{F}$. Moreover, let
$\dot{z}\in\mathscr{L}_{n}(X)$ be an algebraic generator, with associated
rational function $u_{n}$ such that
$\mathcal{S}(u_{n})\subset\mathcal{S}_{n}$. There is a
$\mathcal{D}_{X}$-linear isomorphism

$\left(\mathcal{M}(\mathcal{S}_{n},u_{n},d)_{\overline{V_{n}}}\right)_{|X}\stackrel{{\scriptstyle\cong}}{{\longrightarrow}}\mathscr{L}_{n|X}$

which sends $z$ to $\dot{z}$.

###### Proof.

By Corollary 4.2.5 applied with $t=|\pi_{F}|^{n}$, and by Lemma 4.1.2, we have

$\left(\mathcal{M}(\mathcal{S}_{n},u_{n},d)_{\overline{V_{n}}}\right)|_{X}=\left((\mathcal{O}_{\mathbb{A}})_{\overline{V_{n}}}\right)|_{X}z=(\mathcal{O}_{X})_{\overline{V_{n}\cap
X}}z.$

By Definition 4.3.6(a) we know that
$\mathscr{L}_{n|X}=(\mathcal{O}_{X})_{\overline{X\cap V_{n}}}\dot{z}$. Hence
sending $z$ to $\dot{z}$ defines an $\mathcal{O}_{X}$-linear isomorphism

$\left(\mathcal{M}(\mathcal{S}_{n},u_{n},d)_{\overline{V_{n}}}\right)_{|X}\stackrel{{\scriptstyle\cong}}{{\longrightarrow}}\mathscr{L}_{n|X}.$

The fact that $\psi(\dot{z}^{\otimes d})=u_{n}$ implies that this isomorphism
is in fact $\mathcal{D}$-linear. ∎

We are now in the situation of $\S\ref{KummerSect}$ and can begin to reap the
rewards.

######  Definition 4.3.9.

Let $n\geqslant 0$.

  1. (a)

$\mathbb{D}_{n}$ will denote the $G$-topology
$\mathbb{D}_{n}:=\mathbb{D}(|\pi_{F}|^{n}\partial_{x}/\varpi)^{\dagger}$ on
$\mathbb{D}$.

  2. (b)

$\mathscr{D}_{n}$ will denote the sheaf
$\mathcal{D}^{\dagger}_{\varpi/|\pi_{F}|^{n}}$ on $\mathbb{D}_{n}$.

In view of Corollary 2.3.17 and Lemma 2.3.13, we see that an affinoid
subdomain $X$ of $\mathbb{D}$ lies in $\mathbb{D}_{n}$ if and only if
$\rho(X_{\mathbf{C}})\geqslant|\pi_{F}|^{n}$. We note the following flatness
properties that follow from work done in §3.

######  Lemma 4.3.10.

  1. (a)

$\mathscr{D}_{n}(X)$ is a flat right $\mathscr{D}_{n+1}(X)$-module whenever
$X\in\mathbb{D}_{n+1}$.

  2. (b)

$\mathscr{D}_{n}(Y)$ is a flat right $\mathscr{D}_{n}(X)$-module whenever
$Y\subseteq X$ in $\mathbb{D}_{n}$.

###### Proof.

This is a consequence of Theorem 3.2.1. ∎

######  Corollary 4.3.11.

The natural action of the sheaf of finite order differential operators
$\mathcal{D}$ on the restriction of $\mathscr{L}_{n}$ to $\mathbb{D}_{n}$
extends to $\mathscr{D}_{n}$.

###### Proof.

Let $\mathcal{S}_{n}$ be any set of coset representatives for the group
$\mathcal{O}_{F}/\pi^{n+1}\mathcal{O}_{F}$. Theorem 4.3.7 and Proposition
4.3.8 together give us a $\mathcal{D}_{\mathbb{D}}$-linear isomorphism

$\mathcal{M}(\mathcal{S}_{n},u_{n},d)_{\overline{V_{n}}}\stackrel{{\scriptstyle\cong}}{{\longrightarrow}}\mathscr{L}_{n}.$

We can now apply Proposition 4.2.9 with $t=|\pi_{F}|^{n}$ to transport the
$\mathscr{D}_{n}$-action on
$\mathcal{M}(\mathcal{S}_{n},u_{n},d)_{\overline{V_{n}}}$ to $\mathscr{L}_{n}$
along this isomorphism. ∎

Recall the relator $R_{\mathcal{S}(u)}(u,d)$ from Definition 4.2.10(b).

######  Corollary 4.3.12.

Let $X\in\mathbb{D}_{n}$ and let $\dot{z}\in\mathscr{L}_{n}(X)$ be an
algebraic generator, with associated rational function $u_{n}$. Then
$\mathscr{L}_{n}(X)$ is a finitely presented $\mathscr{D}_{n}(X)$-module:

$\mathscr{L}_{n}(X)=\mathscr{D}_{n}(X)\cdot\dot{z}\cong\mathscr{D}_{n}(X)/\mathscr{D}_{n}(X)R_{\mathcal{S}(u_{n})}(u_{n},d).$

###### Proof.

By Proposition 4.3.8, there is a $\mathcal{D}_{X}$-linear isomorphism

$\left(\mathcal{M}(\mathcal{S}(u_{n}),u_{n},d)_{\overline{V_{n}}}\right)_{|X}\stackrel{{\scriptstyle\cong}}{{\longrightarrow}}(\mathscr{L}_{n})_{|X}$

sending $z$ to $\dot{z}$. Note that $|a-b|\geqslant|\pi_{F}|^{n}$ for all
$a\neq b\in\mathcal{S}(u_{n})\cap X$ by Definition 4.3.6(d) and
$\frac{v_{a}(u_{n})}{d}\notin\mathbb{N}$ for all $a\in\mathcal{S}(u_{n})\cap
X$ by Definition 4.3.6(e). Now we may apply Theorem 4.2.21 with
$t=|\pi_{F}|^{n}$. ∎

###  4.4. Local irreducibility

We assume throughout $\S\ref{LocIrred}$ that our ground field $K$ is
discretely valued, with uniformiser $\pi_{K}$ and residue field $k$. Let $X$
be an affinoid subdomain of the closed unit disc $\mathbb{D}$ obtained by
removing finitely many open discs of radius $1$ _not_ containing the point
$x=0$. Thus

$X=\operatorname{Sp}K\left\langle
x,\frac{1}{x-\alpha_{1}},\cdots,\frac{1}{x-\alpha_{v}}\right\rangle$

for some $\alpha_{1},\ldots,\alpha_{v}\in K$ with
$|\alpha_{1}|=\cdots=|\alpha_{v}|=1$. We also consider the case where $v=0$
when $X=\mathbb{D}$. The main result of $\S\ref{LocIrred}$ is the following

######  Theorem 4.4.1.

Let $\lambda\in K\backslash\mathbb{Z}$ and let $X$ be as above. Then the
$\mathcal{D}^{\dagger}_{\varpi}(X)$-module
$\mathcal{D}^{\dagger}_{\varpi}(X)/\mathcal{D}^{\dagger}_{\varpi}(X)(x\partial-\lambda)$
is simple if $\lambda\in\mathbb{Z}_{p}$, and zero otherwise.

Let $\mathscr{X}:=\operatorname{Spf}\mathcal{O}(X)^{\circ}$ and note that
$\mathscr{X}$ is the formal completion of the punctured affine line
$\operatorname{Spec}K^{\circ}\left[x,\frac{1}{x-\alpha_{1}},\cdots,\frac{1}{x-\alpha_{d}}\right]$
along its special fibre. Recall Berthelot’s ring
$\widehat{\mathscr{D}}^{(m)}_{\mathscr{X},\mathbb{Q}}(\mathscr{X})$ of
level-$m$ arithmetic differential operators on $\mathscr{X}$ for some fixed
integer $m\geqslant 0$ from [13, p. 46, (2.4.1.4)].

######  Definition 4.4.2.

For each $\lambda\in K$, define

$\mathscr{M}^{(m)}(\lambda):=\frac{\widehat{\mathscr{D}}^{(m)}_{\mathscr{X},\mathbb{Q}}}{\widehat{\mathscr{D}}^{(m)}_{\mathscr{X},\mathbb{Q}}(x\partial-\lambda)}\in\operatorname{coh}(\widehat{\mathscr{D}}^{(m)}_{\mathscr{X},\mathbb{Q}}).$

We will actually prove the following stronger statement.

######  Theorem 4.4.3.

Let $\lambda\in K\backslash\mathbb{Z}$, let $\mathscr{X}$ be as above and let
$m\geqslant 0$. Then the
$\widehat{\mathscr{D}}^{(m)}_{\mathscr{X},\mathbb{Q}}(\mathscr{X})$-module
$\mathscr{M}^{(m)}(\lambda)(\mathscr{X})$ is simple if
$\lambda\in\bigcup\limits_{i=0}^{p^{m}-1}i+p^{m}K^{\circ}$, and zero
otherwise.

The following elementary statement may help to explain the sets appearing in
Theorem 4.4.3.

######  Lemma 4.4.4.

We have
$\mathbb{Z}_{p}=\bigcap\limits_{m=0}^{\infty}\bigcup\limits_{i=0}^{p^{m}-1}i+p^{m}K^{\circ}$.

###### Proof.

Certainly the forward inclusion holds. On the other hand, if
$\lambda\notin\mathbb{Z}_{p}$ then because $\mathbb{Z}_{p}$ is compact and $K$
is Hausdorff, we can find $m\geqslant 0$ sufficiently large so that
$|\lambda-\alpha|>|p^{m}|$ for all $\alpha\in\mathbb{Z}_{p}$. Then
$\lambda\notin\bigcup\limits_{i=0}^{p^{m}-1}i+p^{m}K^{\circ}$. ∎

###### Proof of Theorem 4.4.1.

Write $D:=\mathcal{D}^{{\dagger}}_{\varpi}(X)$ and
$D_{m}:=\widehat{\mathscr{D}}^{(m)}_{\mathscr{X},\mathbb{Q}}(\mathscr{X})$ for
each $m\geqslant 0$. Let $r_{m}:=|p^{m}!|^{\frac{1}{p^{m}}}$; then by Theorem
3.1.9 there is a (not necessarily isometric) isomorphism of $K$-Banach
algebras $D_{m}\cong\mathcal{D}_{r_{m}}(X)$. Because the sequence of real
numbers $r_{m}$ approaches $\varpi$ as $m\to\infty$, it follows from
Definition 2.3.9(d) that $D=\operatorname{colim}\limits_{m\geqslant 0}D_{m}$.

Suppose that $\lambda\notin\mathbb{Z}_{p}$. Then by Lemma 4.4.4,
$\lambda\notin\bigcup\limits_{i=0}^{p^{m}-1}i+p^{m}K^{\circ}$ for some
$m\geqslant 0$, and then $M_{m}:=\mathscr{M}^{(m)}(\lambda)(\mathscr{X})$ is
zero by Theorem 4.4.3. Therefore
$D/D(x\partial-\lambda)=D\otimes_{D_{m}}M_{m}$ is zero as well.

Now suppose that $\lambda\in\mathbb{Z}_{p}$. Since colimits are exact, the
$D$-module $M:=D/D(x\partial-\lambda)$ is the colimit of the $D_{m}$-modules
$M_{m}=D_{m}/D_{m}(x\partial-\lambda)$. Note that the image of the canonical
generator $v_{m}$ of $M_{m}$ inside $M$ equals the canonical generator $v$ of
$M$. Now given any non-zero $w\in M$, because
$M=\operatorname{colim}\limits_{m\geqslant 0}M_{m}$ we can find $m\geqslant 0$
such that $w$ is the image in $M$ of some non-zero $w_{m}\in M_{m}$. Since
$\lambda\in\mathbb{Z}_{p}$, $M_{m}$ is a simple $D_{m}$-module by Theorem
4.4.3 and Lemma 4.4.4, we can find $Q_{m}\in D$ such that $Q_{m}\cdot
w_{m}=v_{m}$ in $M_{m}$. Hence $Q_{m}\cdot w=v$ in $M$. So any non-zero
$D$-submodule of $M$ contains $v$ and is hence equal to $M$. ∎

We will now use Berthelot’s theory of _Frobenius descent_ to reduce Theorem
4.4.3 to the case $m=0$. Let $q:=p^{m}$, let $x^{\prime}$ be a new local
coordinate and let $\mathscr{X}^{\prime}$ be the formal completion of the
punctured affine line
$\operatorname{Spec}K^{\circ}\left[x^{\prime},\frac{1}{x^{\prime}-\alpha_{1}^{q}},\cdots,\frac{1}{x^{\prime}-\alpha_{d}^{q}}\right]$
along its special fibre. There is a natural lift of the relative Frobenius
morphism

$F:\mathscr{X}\to\mathscr{X}^{\prime}$

which sends $x^{\prime}\in\mathcal{O}_{\mathscr{X}^{\prime}}$ to
$x^{q}\in\mathcal{O}_{\mathscr{X}}$ and which is completely determined by this
property as a morphism of formal $K^{\circ}$-schemes. In this situation,
Berthelot’s Frobenius descent theorem, [15, Théorème 2.3.6], tells us that the
functors of _Frobenius pullback_ and _Frobenius descent_

$\begin{array}[]{llllll}F^{\ast}&:&\operatorname{coh}(\mathscr{D}^{(0)}_{\mathscr{X}^{\prime}})&\longrightarrow&\operatorname{coh}(\mathscr{D}^{(m)}_{\mathscr{X}}),&\mathscr{N}\mapsto\mathcal{O}_{\mathscr{X}}\underset{\mathcal{O}_{\mathscr{X}^{\prime}}}{\otimes}{}\mathscr{N},\quad\mbox{and}\\\
F^{\natural}&:&\operatorname{coh}(\mathscr{D}^{(m)}_{\mathscr{X}})&\longrightarrow&\operatorname{coh}(\mathscr{D}^{(0)}_{\mathscr{X}^{\prime}}),&\mathscr{M}\mapsto\left(\mathscr{D}^{(0)}_{\mathscr{X}^{\prime}}\underset{\mathcal{O}_{\mathscr{X}^{\prime}}}{\otimes}{}\mathcal{O}_{\mathscr{X}}^{\vee}\right)\underset{\mathscr{D}^{(m)}_{\mathscr{X}}}{\otimes}{}\mathscr{M}\end{array}$

are mutually inverse equivalences of categories, where
$\mathcal{O}_{\mathscr{X}}^{\vee}:=\mathpzc{Hom}_{\mathcal{O}_{\mathscr{X}^{\prime}}}(\mathcal{O}_{\mathscr{X}},\mathcal{O}_{\mathscr{X}^{\prime}})$.
These equivalences extend in a natural way to $p$-adic completions

$F^{\natural}:\operatorname{coh}(\widehat{\mathscr{D}}^{(m)}_{\mathscr{X}})\stackrel{{\scriptstyle\cong}}{{\longrightarrow}}\operatorname{coh}(\widehat{\mathscr{D}}^{(0)}_{\mathscr{X}^{\prime}})\quad\mbox{and}\quad
F^{\natural}:\operatorname{coh}(\widehat{\mathscr{D}}^{(m)}_{\mathscr{X},\mathbb{Q}})\stackrel{{\scriptstyle\cong}}{{\longrightarrow}}\operatorname{coh}(\widehat{\mathscr{D}}^{(0)}_{\mathscr{X}^{\prime},\mathbb{Q}})$

which, abusing notation, we will denote by the same symbols. We are now in a
position to be able to compute the Frobenius descent of
$\mathscr{M}^{(m)}(\lambda)$.

######  Proposition 4.4.5.

Let $\lambda\in K$ and set $\lambda_{i}:=\frac{\lambda-i}{q}$ for each
$i=0,\ldots,q-1$. Then

$F^{\natural}(\mathscr{M}^{(m)}(\lambda))\cong\bigoplus_{i=0}^{q-1}\frac{\widehat{\mathscr{D}}^{(0)}_{\mathscr{X}^{\prime},\mathbb{Q}}}{\widehat{\mathscr{D}}^{(0)}_{\mathscr{X}^{\prime},\mathbb{Q}}(x^{\prime}\partial^{\prime}-\lambda_{i})}$

where $\partial^{\prime}\in\mathcal{T}_{\mathscr{X}^{\prime}}$ is the unique
derivation such that $\partial^{\prime}(x^{\prime})=1$.

###### Proof.

In [15, Proposition 4.1.2(ii)], Berthelot established a canonical isomorphism
of $\widehat{\mathscr{D}}^{(m)}_{\mathscr{X}}$-bimodules

$\Phi:F^{\ast}F^{\flat}\widehat{\mathscr{D}}^{(0)}_{\mathscr{X}^{\prime}}:=\mathcal{O}_{\mathscr{X}}\underset{\mathcal{O}_{\mathscr{X}^{\prime}}}{\otimes}{}\widehat{\mathscr{D}}^{(0)}_{\mathscr{X}^{\prime}}\underset{\mathcal{O}_{\mathscr{X}^{\prime}}}{\otimes}{}\mathcal{O}_{\mathscr{X}}^{\vee}\stackrel{{\scriptstyle\cong}}{{\longrightarrow}}\widehat{\mathscr{D}}^{(m)}_{\mathscr{X}}.$

We will use the following explicit description of $\Phi$ due to Garnier [24];
we will follow the clear exposition of Garnier’s results found at [4, §2.2].
Note that $\mathcal{O}_{\mathscr{X}}$ is a free
$\mathcal{O}_{\mathscr{X}^{\prime}}$-module with basis
$\\{1,x,\cdots,x^{q-1}\\}$; let
$\\{\theta_{0},\theta_{1},\cdots,\theta_{q-1}\\}$ be the dual basis for
$\mathcal{O}_{\mathscr{X}}^{\vee}$ so that

$\mathcal{O}_{\mathscr{X}}\underset{\mathcal{O}_{\mathscr{X}^{\prime}}}{\otimes}{}\widehat{\mathscr{D}}^{(0)}_{\mathscr{X}^{\prime}}\underset{\mathcal{O}_{\mathscr{X}^{\prime}}}{\otimes}{}\mathcal{O}_{\mathscr{X}}^{\vee}=\bigoplus_{i,j=0}^{q-1}x^{i}\otimes\widehat{\mathscr{D}}^{(0)}_{\mathscr{X}^{\prime}}\otimes\theta_{j}.$

Then the isomorphism $\Phi$ is defined as follows:

  * •

$\Phi(x^{i}\otimes Q\otimes\theta_{j})=x^{i}Q^{\circ}Hx^{-j}$ for any
$Q\in\widehat{\mathscr{D}}^{(0)}_{\mathscr{X}^{\prime}}$ and $0\leqslant
i,j<q$, where

  * •

$H:=\frac{1}{q}\sum\limits_{\zeta^{q}=1}\sum\limits_{k=0}^{\infty}(\zeta-1)^{k}x^{k}\partial^{[k]}\in\widehat{\mathscr{D}}^{(m)}_{\mathscr{X}}x^{q-1}$
is the _Dwork operator_ , and

  * •

$Q\mapsto Q^{\circ}$ is the non-unital ring homomorphism
$\widehat{\mathscr{D}}^{(0)}_{\mathscr{X}^{\prime}}\hookrightarrow\widehat{\mathscr{D}}^{(m)}_{\mathscr{X}}$
determined by the rules $1^{\circ}=H,(x^{\prime})^{\circ}=x^{q}H$ and
$(\partial^{\prime})^{\circ}=(qx^{q-1})^{-1}\partial H$.

Note that $(x^{\prime}\partial^{\prime})^{\circ}=x^{q}(qx^{q-1})^{-1}\partial
H=\frac{1}{q}x\partial H$. Therefore

$\Phi(x^{i}\otimes(x^{\prime}\partial^{\prime}-\lambda_{i})\otimes\theta_{i})=x^{i}\left(\frac{1}{q}x\partial
H-\frac{\lambda-i}{q}H\right)Hx^{-i}=\frac{1}{q}(x\partial-\lambda)x^{i}Hx^{-i}$

for any $i=0,\ldots,q-1$, where we used the relations
$x^{i}(x\partial)=(x\partial-i)x^{i}$ and $H^{2}=H$. Because
$\sum\limits_{i=0}^{q-1}x^{i}Hx^{-i}=1$ in
$\widehat{\mathscr{D}}^{(m)}_{\mathscr{X}}$ by [24, Proposition 2.5.1], we see
that

(34)
$\Phi^{-1}\left(\frac{x\partial-\lambda}{q}\right)=\sum\limits_{i=0}^{q-1}x^{i}\otimes(x^{\prime}\partial^{\prime}-\lambda_{i})\otimes\theta_{i}.$

By the Frobenius descent theorem, [15, Théorème 2.3.6],

$F^{\natural}(\widehat{\mathscr{D}}^{(m)}_{\mathscr{X},\mathbb{Q}})\quad\underset{\cong}{\overset{F^{\natural}(\Phi^{-1})}{{\longrightarrow}}}\quad
F^{\natural}F^{\ast}F^{\flat}\widehat{\mathscr{D}}^{(0)}_{\mathscr{X}^{\prime}}\quad\cong\quad
F^{\flat}(\widehat{\mathscr{D}}^{(0)}_{\mathscr{X}^{\prime},\mathbb{Q}})=\bigoplus\limits_{j=0}^{q-1}\widehat{\mathscr{D}}^{(0)}_{\mathscr{X}^{\prime},\mathbb{Q}}\otimes\theta_{j}$

which is a free left
$\widehat{\mathscr{D}}^{(0)}_{\mathscr{X}^{\prime},\mathbb{Q}}$-module of rank
$q$. By definition, $\mathscr{M}^{(m)}(\lambda)$ is the cokernel of right-
multiplication by $\frac{x\partial-\lambda}{q}$ on
$\widehat{\mathscr{D}}^{(m)}_{\mathscr{X},\mathbb{Q}}$. Using (34), we
conclude that $F^{\natural}(\mathscr{M}^{(m)}(\lambda))$ is the cokernel of
the ‘diagonal’ left
$\widehat{\mathscr{D}}^{(0)}_{\mathscr{X}^{\prime},\mathbb{Q}}$-linear
endomorphism

$\bigoplus_{i=0}^{q-1}\cdot(x^{\prime}\partial^{\prime}-\lambda_{i})\otimes
1:\bigoplus\limits_{j=0}^{q-1}\widehat{\mathscr{D}}^{(0)}_{\mathscr{X}^{\prime},\mathbb{Q}}\otimes\theta_{j}\to\bigoplus\limits_{j=0}^{q-1}\widehat{\mathscr{D}}^{(0)}_{\mathscr{X}^{\prime},\mathbb{Q}}\otimes\theta_{j}.$

The result follows.∎

######  Corollary 4.4.6.

Let $\lambda\in K$.

  1. (a)

If $\lambda\notin\bigcup\limits_{i=0}^{q-1}i+qK^{\circ}$, then
$x\partial-\lambda$ is a unit in
$\widehat{\mathscr{D}}^{(m)}_{\mathscr{X},\mathbb{Q}}$ and
$\mathscr{M}^{(m)}(\lambda)=0$.

  2. (b)

Otherwise there exists a unique integer $i$ such that $0\leqslant i<q$ and
$\lambda\in i+qK^{\circ}$, and then

$F^{\natural}(\mathscr{M}^{(m)}(\lambda))\quad\cong\quad\frac{\widehat{\mathscr{D}}^{(0)}_{\mathscr{X}^{\prime},\mathbb{Q}}}{\widehat{\mathscr{D}}^{(0)}_{\mathscr{X}^{\prime},\mathbb{Q}}(x^{\prime}\partial^{\prime}-\frac{\lambda-i}{q})}.$

###### Proof.

(a) In this case, $\lambda_{i}=\frac{\lambda-i}{q}\notin K^{\circ}$ for all
$i=0,\ldots,q-1$, so $|\lambda_{i}|>1$ for all $i$. But then
$x^{\prime}\partial^{\prime}-\lambda_{i}$ is a unit in
$\widehat{\mathscr{D}}^{(0)}_{\mathscr{X}^{\prime},\mathbb{Q}}$, so
$F^{\natural}(\mathscr{M}^{(m)}(\lambda))=0$ by Proposition 4.4.5. So
$\mathscr{M}^{(m)}(\lambda)=0$ and $x\partial-\lambda$ is a unit in
$\widehat{\mathscr{D}}^{(m)}_{\mathscr{X},\mathbb{Q}}$ because $F^{\natural}$
is an equivalence of categories.

(b) Note that if $0\leqslant i<j<q$, then $q$ does not divide $j-i$, so
$|j-i|>|q|$. Hence $(i+qK^{\circ})\cap(j+qK^{\circ})=\emptyset$, and if
$\lambda\in i+qK^{\circ}$ then $\lambda\notin j+qK^{\circ}$ for any other $j$.
This means that $|\lambda_{j}|>1$ whenever $j\neq i$, and then
$x^{\prime}\partial^{\prime}-\lambda_{j}$ is a unit in
$\widehat{\mathscr{D}}^{(0)}_{\mathscr{X}^{\prime},\mathbb{Q}}$ as above. Now
apply Proposition 4.4.5. ∎

We now assume that $m=0$ and focus on the algebra
$D:=\widehat{\mathscr{D}}^{(0)}_{\mathscr{X}}(\mathscr{X})$, which is
$\pi_{K}$-adically complete and separated. It is a lattice inside the Tate-
Weyl algebra
$D_{K}=\widehat{\mathscr{D}}^{(0)}_{\mathscr{X},\mathbb{Q}}(\mathscr{X})$. If
$I$ is a left ideal in $D$, then we say that $I$ is _$\pi_{K}$ -closed_ if the
$K^{\circ}$-module $D/I$ is torsionfree, or equivalently, if $I=I_{K}\cap D$.

######  Lemma 4.4.7.

  1. (a)

The algebra $D$ is Noetherian.

  2. (b)

Let $J\leqslant I$ be two left ideals of $D$ and suppose that $I$ is
$\pi_{K}$-closed. Then $I+\pi_{K}D=J+\pi_{K}D$ implies that $I=J$.

###### Proof.

(a) $\overline{D}:=D/\pi_{K}D$ is the skew-polynomial ring
$B[\overline{\partial};\frac{d}{dx}]$ where

$B:=k\left[x,\frac{1}{x-\overline{\alpha_{1}}},\cdots,\frac{1}{x-\overline{\alpha_{d}}}\right]=\mathcal{O}(\mathscr{X}\times_{K^{\circ}}k)$

and $\overline{\alpha_{i}}$ denotes the image of $\alpha_{i}$ in the residue
field $k$ of $K^{\circ}$. Since $k$ is a field, both $B$ and $\overline{D}$
are Noetherian rings 111This uses the assumption that $K^{\circ}$ is a
discrete valuation ring.. The associated graded ring of $D$ with respect to
the $\pi_{K}$-adic filtration is isomorphic to a polynomial ring
$\overline{D}[s]$ and is therefore is Noetherian. Hence $D$ is Noetherian by
[33, Chapter II, §1.2, Proposition 3(1)].

(b) Note that $I\cap\pi_{K}D=\pi_{K}I$ since $I$ is $\pi_{K}$-closed. Hence,
by the modular law

$I=I\cap(I+\pi_{K}D)=I\cap(J+\pi_{K}D)=J+(I\cap\pi_{K}D)=J+\pi_{K}I$

Now $I$ is finitely generated by part (a), and $\pi_{K}$ lies in the Jacobson
radical of $D$ because $D$ is $\pi_{K}$-adically complete, so $I=J$ by
Nakayama’s Lemma. ∎

We will assume until the end of $\S\ref{LocIrred}$ that $\lambda\in
K^{\circ}$.

######  Lemma 4.4.8.

Let $I$ be a $\pi_{K}$-closed left ideal of $D$ containing
$t:=x\partial-\lambda$. Then $I$ is stable under the operators
$E_{n}:=\binom{\operatorname{ad}(x\partial)}{n}:D_{K}\to D_{K}$ for all
$n\geqslant 0$.

###### Proof.

We first consider the case where $I=D$ itself. For any
$f\in\mathcal{O}(\mathscr{X})$ and any $m\geqslant 0$ we calculate
$\operatorname{ad}(x\partial)(f\partial^{m})=[x\partial,f\partial^{m}]=(x\partial-m)(f)\partial^{m}$.
Therefore for any $n\geqslant 0$ we see that

(35)
$\binom{\operatorname{ad}(x\partial)}{n}(f\partial^{m})=\binom{x\partial-m}{n}(f)\partial^{m}=\sum\limits_{i=0}^{n}\binom{-m}{n-i}x^{i}\partial^{[i]}(f)\hskip
2.84544pt\partial^{m}$

which implies that $\binom{\operatorname{ad}(x\partial)}{n}$ preserves $D$ as
required. Now, $\operatorname{ad}(x\partial)=\operatorname{ad}(t)$. Letting

$\operatorname{ad}(t)_{[n]}:=\operatorname{ad}(t)(\operatorname{ad}(t)-1)\cdots(\operatorname{ad}(t)-n+1)=n!\binom{\operatorname{ad}(t)}{n}$

we see that $\operatorname{ad}(t)_{[n]}(D)\subseteq n!D$. Returning to the
general case where $I$ is not necessarily equal to $D$, note that
$\operatorname{ad}(t)(I)=[t,I]\subseteq I$ because $t\in I$ by assumption.
Hence

$\operatorname{ad}(t)_{[n]}(I)\subseteq I\cap n!D=n!I$

because $D/I$ is torsionfree. Dividing through by $n!$ we conclude that $I$ is
stable under
$\binom{\operatorname{ad}(t)}{n}=\binom{\operatorname{ad}(x\partial)}{n}$ as
required. ∎

######  Proposition 4.4.9.

Let $I$ be a $\pi_{K}$-closed left ideal of $D$ properly containing $Dt$. Then
for some $n\geqslant 0$, either $x^{n}\in I+\pi_{K}D$ or $\partial^{n}\in
I+\pi_{K}D$.

###### Proof.

Let $\overline{I}:=(I+\pi_{K}D)/\pi_{K}D$. Since $I$ is $\pi_{K}$-closed, we
see that $\overline{I}$ strictly contains
$\overline{Dt}=(Dt+\pi_{K}D)/\pi_{K}D$ by Lemma 4.4.7(b). Now $\overline{I}$
is a left ideal in $\overline{D}=B[\overline{\partial};\frac{d}{dx}]$ which is
an Ore localisation of the Weyl algebra
$W:=k[x][\overline{\partial};\frac{d}{dx}]$ at the Ore set consisting of
powers of the polynomial $\prod_{i=1}^{d}(x-\overline{\alpha_{i}})$.
Therefore, by [36, Proposition 2.1.16(iii)], $\overline{I}=\overline{D}\cdot
J$ where $J:=\overline{I}\cap W$; note that $J$ strictly contains
$W\overline{t}$ because $\overline{I}$ strictly contains
$\overline{D}\overline{t}$.

Now, $W/W\overline{t}$ is a $k$-vector space with basis
$\\{v_{m}:m\in\mathbb{Z}\\}$ where $v_{m}:=\overline{x^{m}}+W\overline{t}$ if
$m\geqslant 0$ and $v_{m}:=\overline{\partial^{-m}}+W\overline{t}$ if $m<0$.
It follows from (35) that

$E_{n}(v_{m})=\binom{m}{n}v_{m}\quad\mbox{for all}\quad n\geqslant
0,m\in\mathbb{Z}.$

Pick a non-zero element $u\in J/W\overline{t}$ ; then we can write $u$ as a
finite sum

$u=\sum_{m\in\mathbb{Z}}u_{m}v_{m}$

for some $u_{m}\in k$, not all zero. Suppose that in fact $u_{r}\neq 0$ for
some $r\in\mathbb{Z}$. The binomial coefficients $\binom{z}{n}$ form a basis
for the ring of locally constant $k$-valued functions on $\mathbb{Z}_{p}$. We
can find such a function, $\epsilon$ say, such that $\epsilon(r)=1$ and
$\epsilon(m)=0$ for all $m\in\mathbb{Z}\backslash\\{r\\}$ such that $u_{m}\neq
0$. Write $\epsilon=\sum\limits_{n=0}^{N}\epsilon_{n}\binom{z}{n}$; then

$\left(\sum\limits_{n=0}^{N}\epsilon_{n}E_{n}\right)(u)=\sum_{n=0}^{N}\sum_{m\in\mathbb{Z}}\epsilon_{n}u_{m}\binom{m}{n}v_{m}=\sum_{m\in\mathbb{Z}}u_{m}v_{m}\epsilon(m)=u_{r}v_{r}.$

Now $E_{n}(J)\subseteq J$ for all $n\geqslant 0$ by Lemma 4.4.8, so
$u_{r}v_{r}\in J/W\overline{t}$. Since $k$ is a field, we conclude that
$v_{r}\in J/W\overline{t}$ and the result follows. ∎

######  Corollary 4.4.10.

Let $I$ be a $\pi_{K}$-closed left ideal of $D$ properly containing $Dt$. Then
either

  1. (a)

there exist $a_{n},a_{n-1},\cdots,a_{0}\in K^{\circ}\langle\partial\rangle$
with $a_{n}\neq 0$ and

$a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}\in I,\quad\mbox{or}\quad$

  2. (b)

there exist $a_{n},a_{n-1},\cdots,a_{0}\in\mathcal{O}(\mathscr{X})$ with
$a_{n}\neq 0$ and

$a_{n}\partial^{n}+a_{n-1}\partial^{n-1}+\cdots+a_{1}\partial+a_{0}\in I.$

###### Proof.

By Proposition 4.4.9, we know that either $\overline{x}^{n}\in\overline{I}$ or
$\overline{\partial}^{n}\in\overline{I}$ for some $n\geqslant 0$. Suppose
first that $\overline{x}^{n}\in\overline{I}$, and let
$S:=\\{1,x,\cdots,x^{n-1}\\}$. Because the image of $S$ in
$B/B\overline{x}^{n}$ spans it as a $k$-vector space, its image also generates
$\overline{D}/\overline{D}\overline{x}^{n}$ as a $k[\partial]$-module; hence
its image in $\overline{D}/\overline{I}=\overline{D/I}$ generates it as a
$k[\partial]$-module. Since $D/I$ is $\pi_{K}$-adically separated by Lemma
4.4.7(a) and [33, Chapter II, §1.2, Theorem 10(1)], we conclude using [33,
Chapter I, Theorem 5.7] that the image of $S$ in $D/I$ generates it as a
$K^{\circ}\langle\partial\rangle$-module. Hence we may write $x^{n}$ as a
$K^{\circ}\langle\partial\rangle$-linear combination vectors in this image,
and hence (a) holds. Alternatively, if
$\overline{\partial}^{n}\in\overline{I}$ then the same argument using
$T:=\\{1,\partial,\cdots,\partial^{n-1}\\}$ instead of $S$ and
$\mathcal{O}(\mathscr{X})$ instead of $K^{\circ}\langle\partial\rangle$ shows
that (b) holds. ∎

######  Proposition 4.4.11.

Let $I$ be a $\pi_{K}$-closed left ideal of $D$ properly containing $Dt$ and
suppose that $\lambda\in K^{\circ}\backslash\mathbb{Z}$. Then either $I\cap
K^{\circ}\langle\partial\rangle\neq\\{0\\}$ or
$I\cap\mathcal{O}(\mathscr{X})\neq\\{0\\}$.

###### Proof.

Let $v_{\lambda}$ be the canonical generator of $D/Dt$ and let $m\geqslant 1$.
Then

$\begin{array}[]{lll}\partial\cdot x^{m}v_{\lambda}&=&\partial x\cdot
x^{m-1}v_{\lambda}=(x^{m-1}\partial x+[\partial x,x^{m-1}])\cdot
v_{\lambda}=\\\
&=&x^{m-1}(x\partial+1+(m-1))v_{\lambda}=(\lambda+m)x^{m-1}v_{\lambda}.\end{array}$

It follows that for any $a\in K^{\circ}\langle\partial\rangle$ and any
$m\geqslant 1$, we have

$\partial\cdot ax^{m}\equiv a(\lambda+m)x^{m-1}\operatorname{mod}Dt.$

Suppose now that $a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}\equiv
0\operatorname{mod}I$, for some $a_{i}\in K^{\circ}\langle\partial\rangle$
with $a_{n}\neq 0$ and $n\geqslant 0$ _least possible_. Suppose for a
contradiction that $n\geqslant 1$. Then

$0=\partial\cdot
0\equiv\partial\cdot\sum\limits_{m=0}^{n}a_{m}x^{m}\equiv\sum\limits_{m=1}^{n}a_{m}(\lambda+m)x^{m-1}+\partial
a_{0}\quad\operatorname{mod}I.$

Since $\lambda\notin\mathbb{Z}$ and since $K^{\circ}\langle\partial\rangle$ is
a domain, we see that $a_{m}(\lambda+m)\neq 0$, so this congruence contradicts
the minimality of $n$. Therefore $n=0$ and $I\cap
K^{\circ}\langle\partial\rangle\neq\\{0\\}$ as claimed. Otherwise, by
Corollary 4.4.10, we have the congruence
$a_{n}\partial^{n}+a_{n-1}\partial^{n-1}+\cdots+a_{1}\partial+a_{0}\equiv
0\operatorname{mod}I$ for some $a_{i}\in\mathcal{O}(\mathscr{X})$ with
$a_{n}\neq 0$. A similar calculation shows that

(36) $x\cdot a\partial^{m}\equiv
a(\lambda-m+1)\partial^{m-1}\quad\operatorname{mod}Dt\quad\mbox{for all}\quad
m\geqslant 1,a\in\mathcal{O}(\mathscr{X})$

and using this congruence we conclude using a similar argument that in fact we
can choose $n=0$ and therefore $I\cap\mathcal{O}(\mathscr{X})\neq\\{0\\}$. ∎

######  Corollary 4.4.12.

If $\lambda\in K^{\circ}\backslash\mathbb{Z}$, then the only $\pi_{K}$-closed
left ideal $I$ of $D$ properly containing $D(x\partial-\lambda)$ is $D$
itself.

###### Proof.

Every non-zero ideal in the Tate algebra $K\langle\partial\rangle$ is
generated by a non-zero element of $K[\partial]$. So, if $I\cap
K^{\circ}\langle\partial\rangle$ is non-zero, then $I_{K}\cap K[\partial]$ is
non-zero. Clearing denominators, we may even assume that $I\cap
K^{\circ}[\partial]$ is non-zero. By applying an appropriate linear