# Deformed Laurent series rings and completions of the Weyl division ring

Gang Han
School of Mathematics, Zhejiang University
<EMAIL_ADDRESS>
Yulin Chen
School of Mathematics, Zhejiang University
<EMAIL_ADDRESS>
Zhennan Pan
School of Mathematics, Zhejiang University
<EMAIL_ADDRESS>
(Jan. 26, 2024)

###### Abstract

Let $L((T^{-1}))$ be the space of (inverse) Laurent series with coefficients
in some field $L$. It has a standard degree map and the induced topology. With
its usual addition and a new product on this space which is continuous and
preserves the standard degree map, it will be a complete topological division
ring, and called a deformed Laurent series ring. Under mild restrictions, we
give the necessary and sufficient conditions for a product on $L((T^{-1}))$ to
make it a deformed Laurent series ring. Then we apply the above theory to
construct the completions of the Weyl division ring $D_{1}$, over some field
of characteristic 0, with respect to a class of discrete valuations on it.
Such completions are topological division rings with nice properties. For
instance, their valuation rings are non-commutative Henselian rings; the
centralizer of each element not in the center is commutative.

MSC2020: 16K40, 16L30

Key words: deformed Laurent series ring, Weyl division ring, complete
topological division ring, non-commutative Henselian ring, commutative
centralizer condition

###### Contents

  1. 1 Introduction
  2. 2 A class of degree maps on $D_{1}$
  3. 3 Deformed Laurent series rings with the standard degree map
  4. 4 Construction of $\widetilde{D}_{r,s}$ and $\check{D}_{r,s}$ and embedding of $\widehat{D}_{r,s}$
  5. 5 Topological generators of $\widehat{D}_{r,s}$
  6. 6 $\widehat{D}_{r,s}$ satisfies the commutative centralizer condition

##  1 Introduction

Let $K$ be a field of Characteristic 0. Let $A_{1}(K)$ be the first Weyl
algebra over $K$ generated by 2 generators $\textbf{p},\textbf{{q}}$
satisfying the relation

$[\textbf{q},\textbf{p}]=\textbf{qp}-\textbf{pq}=1.$ (1.1)

We adopt the convention (1.1) following Joseph [J1973], from which we benefit
a lot. The algebra $A_{1}(K)$ is a simple Noetherian domain of GK dimension 2.
The quotient division algebra $D_{1}(K)$ of $A_{1}(K)$ is usually called the
Weyl division ring or Weyl skew field.

People have been interested in $A_{1}(K)$ and $D_{1}(K)$ for quite some time,
and have conducted extensive research in this area. For example in [ML1983]
Makar-Limanov showed that $D_{1}(K)$ contains a two-generator free subalgebra.

There were also some research about the valuations on $A_{1}(K)$ and
$D_{1}(K)$. A class of discrete valuations of $D_{1}(K)$ compatible with the
Bernstein filtration were classified in [W1997]. All the valuations on the
real Weyl algebra $A_{1}(\mathbb{R})$ whose residue field is $\mathbb{R}$ are
classified in [V2024].

A map $u:D_{1}(K)\rightarrow\mathbb{R}_{-\infty}$ such that
$-u:D_{1}(K)\rightarrow\mathbb{R}_{\infty},f\mapsto-u(f)$ is a rank 1
valuation will be called a degree map on $D_{1}(K)$. The degree maps $u$ on
$A_{1}(K)$ over $K$ with the usual basis
$\\{\textbf{p}^{i}\textbf{q}^{j}|i,j\geq 0\\}$ being an evaluation basis for
$u$ are classified in Proposition 2.4. It is interesting that they are in 1-1
correspondence with the set $\\{(\rho,\sigma)\in\mathbb{R}^{2}|\rho+\sigma\geq
0\\}$. Up to equivalence, those degree maps $\mathsf{v}_{\rho,\sigma}$ with
$(\rho,\sigma)\in\mathbb{Z}^{2}$, $\rho$ and $\sigma$ are coprime, and
$\rho+\sigma\geq 0$, correspond to discrete valuations. See Section 2.

In the influential paper [D1968], Dixmier studied the structure of $A_{1}(K)$
in great detail, and in the end he posed six problems. The first problem asks
if every endomorphism of the Weyl algebra $A_{1}(K)$ is an automorphism. The
conjecture that its answer is positive was later called Dixmier Conjecture.
One approach to study this conjecture is to find all the solutions (up to
automorphisms of $A_{1}(K)$) of the equation $[x,y]=1$ for $x,y$ in
$A_{1}(K)$. We believe that it is better to study this equation in $D_{1}(K)$
and its completions. One purpose of this note is to construct the completions
of $D_{1}(K)$ with respect to such discrete valuations. In fact, certain skew
Laurent series algebras and the algebra of formal pseudo-differential
operators are completions of $D_{1}(K)$ with respect to different valuations.
Since there are infinitely many nonequivalent degree maps on $D_{1}(K)$, it
will be interesting to find the relationship between different completions.

To construct such completions, we did the following. Let $L((T^{-1}))$ be the
space of (inverse) Laurent series with coefficients in some field $L$, which
has a standard degree map and the corresponding topology. We define a new
product on this space (while the addition is as usual) which is continuous and
preserves the standard degree map. We call it a deformed Laurent series ring.
See Def. 3.1. Assume that it satisfies

$T\cdot a=aT+\sum_{i=0}^{-\infty}\delta_{i}(a)T^{i},a\in L,$

where $\delta_{i}:L\rightarrow L,i\in\mathbb{Z}_{\leq 0},$ are additive maps.
Then the necessary and sufficient condition for it to make $L((T^{-1}))$ a
deformed Laurent series ring is (See Theorem 3.10)

$\delta_{i}(ab)=a\delta_{i}(b)+\delta_{i}(a)b+\sum_{k>0}\sum_{(l,j_{1},\cdots,j_{k})}^{j_{1}+\cdots+j_{k}-k+l=i}\binom{l}{k}\delta_{l}(a)\delta_{j_{1}}\cdots\delta_{j_{k}}(b),a,b\in
L,$

where the summation is over all $l,j_{1},\cdots,j_{k}\leq 0$ and
$j_{1}+\cdots+j_{k}-k+l=i$. The proof uses a lot of combinatorial identities.

This ring, denoted by $L((T^{-1};\tilde{\delta}))$, is a complete topological
skew field. It is a division ring of formal Laurent series with a deformed
multiplication. Its center is the field $F=\cap_{i\leq
0}\text{Ker}(\delta_{i})$, and such $\delta_{i}$ are $F$-linear differential
operators on $L$.

The proof of above results is in Section 3, in which a symmetric version of
such results is also given.

Let $\widehat{D}_{r,s}$ be the completion of $D_{1}(K)$ with respect to
$\mathsf{v}_{r,s}$, where $(r,s)\in\mathbb{Z}^{2}$ with $r+s>0$. To construct
$\widehat{D}_{r,s}$, we need to first construct $\widetilde{D}_{r,s}$ and
$\check{D}_{r,s}$ using the theory of deformed Laurent series ring, which are
also complete skew fields. Then we embed $D_{1}(K)$ into it. See Proposition
4.1 and Corollary 4.9. Non-commutative Henselian rings are defined and studied
in [A2009]. The respective valuation rings of $\widetilde{D}_{r,s}$,
$\check{D}_{r,s}$, and $\widehat{D}_{r,s}$ are in fact non-commutative
Henselian rings. See Proposition 4.5.

In Section 5, a pair of topological generators of $\widehat{D}_{r,s}$ is
given.

In [B06], a noncommutative algebra $A$ is said to satisfy the commutative
centralizer condition (the ccc, for short) if the centralizer of each element
not in the center of $A$ is a commutative algebra. Bavula found that many
algebras, including $D_{1}(K)$, satisfy the ccc [B06]. As in [B06] one can
show that $\widetilde{D}_{r,s}$ and $\check{D}_{r,s}$, thus
$\widehat{D}_{r,s}$, also satisfy the ccc. See Section 6.

Here are some of the conventions of the paper.

All the rings and algebras are associative and unital, with the multiplicative
identity $1$. All the ring (and algebra) homomorphisms are unital.

For a ring $R$, $R^{\times}$ is the set of nonzero elements in $R$.

For $a,b\in\mathbb{R}$, let

$[a,b]_{k}=\left\\{\begin{array}[]{ll}\prod_{i=0}^{k-1}(a-ib),&k>0;\\\
1,&k=0.\end{array}\right.$ (1.2)

For $k\geq 0$, let $[a]_{k}=[a,1]_{k}$. Then $[a]_{k}=a(a-1)\cdots(a-k+1)$ for
$k\geq 1$, and $[a]_{0}=1$.

For $a\in\mathbb{R},k\geq 0$, $\binom{a}{k}=[a]_{k}/k!$.

$x,y,z,w$ are elements in $A_{1}(K)$, where $K$ is a field.

For $f\in K(X),i\geq 0$, $f^{(i)}$ denotes the $i$-th derivative of $f$.

$X,Y,T$ are indeterminants.

$char\ L$ denotes the characteristic of the field $L$.

$\mathbb{R}_{\infty}=\mathbb{R}\cup\\{\infty\\}$; $\mathbb{R}_{\geq
0}=\\{a\in\mathbb{R}|a\geq 0\\}$; and similar notations will be used.

For integers $m$ and $n$, $gcd\\{m,n\\}$ is the non-negative greatest common
divisor of $m$ and $n$.

”iff” stands for ”if and only if”.

All the indices under the sign $\sum$ will be assumed to be integers in
$\mathbb{Z}_{\leq 0}$, unless otherwise stated. For example,
$\sum_{(j_{1},\cdots,j_{k})}f(j_{1},\cdots,j_{k})$ means to sum
$f(j_{1},\cdots,j_{k})$ over all $(j_{1},\cdots,j_{k})$ with
$j_{i}\in\mathbb{Z}_{\leq 0}$ for $i=1,\cdots,k$. Here is another example,

$\sum_{(j_{1},\cdots,j_{k})}^{j_{1}+\cdots+j_{k}-k=1}f(j_{1},\cdots,j_{k})$
(1.3)

means to make summations of $f(j_{1},\cdots,j_{k})$ over all
$(j_{1},\cdots,j_{k})\in(\mathbb{Z}_{\leq 0})^{k}$ satisfying
$j_{1}+\cdots+j_{k}-k=1$.

Acknowledgements

We would like to heartily thank Binyong Sun for his advice to improve the
results. We are very thankful to Dongwen Liu, Fangyang Tian, Chengbo Wang,
Shilin Yu, Tao Feng and Yeshun Sun for helpful discussions during the
research. We are grateful to Qinghu Hou at Nankai University, who told us
Lemma 4.3 and its proof.

##  2 A class of degree maps on $D_{1}$

Let $R$ be a ring.

######  Def. 2.1.

A map $v:R\rightarrow\mathbb{R}_{\infty}$ is called a rank 1 valuation on $R$
if it satisfies

  1. 1.

$v(fg)=v(f)+v(g),\forall f,g\in R$;

  2. 2.

$v(f+g)\geq min\\{v(f),v(g)\\},\forall f,g\in R$;

  3. 3.

$v(f)=\infty$ iff $f=0$.

If $v(R^{\times})=0$ then the valuation $v$ is called trivial; If
$v(R^{\times})\cong\mathbb{Z}$ then the valuation $v$ is called discrete.

Two rank 1 valuations $v_{1},v_{2}$ on $R$ are called equivalent if there
exists some $c\in\mathbb{R}_{>0}$, $v_{2}(f)=cv_{1}(f)$ for all $f\in
R^{\times}$.

All the valuations and degree maps considered in this paper will be assumed to
be rank 1.

A map $u:R\rightarrow\mathbb{R}_{-\infty}$ such that
$-u:R\rightarrow\mathbb{R}_{\infty},f\mapsto-u(f)$ is a rank 1 valuation will
be called a degree map on $R$, i.e., $u$ is a degree map on $R$ iff $u$
satisfies

$\begin{array}[]{lll}(1)\ u(fg)=u(f)+u(g),\forall f,g\in R;\\\ (2)\ u(f+g)\leq
max\\{u(f),u(g)\\},\forall f,g\in R;\\\ (3)\ u(f)=-\infty\ \text{iff}\
f=0.\end{array}$ (2.1)

A degree map $u$ is called trivial (resp. discrete) if so is the valuation
$-u$. Two degree maps $u_{1},u_{2}$ on $R$ are called equivalent if so are
$-u_{1},-u_{2}$.

For a degree map $u$ on $R$, there exists a metric on $R$ defined by
$d(f,g)=e^{u(f-g)},f,g\in R.$ Given a degree map $u$ on $R$, we will always
equip $R$ with the topology induced by this metric.

Let $K$ be a field of characteristic 0. Assume that $A$ is an algebra over
$K$.

######  Def. 2.2.

A valuation $v:A\rightarrow\mathbb{R}_{\infty}$ is said to be over $K$, or $v$
is a $K$-valuation, if $v|_{K}$ is the trivial valuation on $K$. Similarly, A
degree map $u:A\rightarrow\mathbb{R}_{-\infty}$ is said to be over $K$, or $u$
is a $K$-degree map, if $-u$ is a valuation of $A$ over $K$.

Let $\mathcal{C}=\\{y_{i}|i\in I\\}$ be a $K$-basis of $A$. $\mathcal{C}$ is
called an evaluation basis for the $K$-degree map $u$ on $A$, if

$u(\sum_{i\in I}\lambda_{i}y_{i})=sup\\{u(y_{i})|i\in I,\lambda_{i}\neq 0\\},\
\sum_{i\in I}\lambda_{i}y_{i}\in A^{\times}.$

See Definition 2 of [KKL2007].

Assume $u:A\rightarrow\mathbb{Z}_{-\infty}$ is a discrete degree map over $K$.
Let $A^{\leq i}=u^{-1}(\mathbb{Z}_{\leq i}\cup\\{-\infty\\})$. Then by (2.1)
one has $A^{\leq i}$ is a $K$-vector space for each $i$, and $A^{\leq i}\cdot
A^{\leq j}\subseteq A^{\leq i+j}$. So one has a graded $K$-algebra

$gr(A)=\oplus_{i\in\mathbb{Z}}\ gr_{i}(A),gr_{i}(A)=A^{\leq i}/A^{\leq i-1},$

called the associated graded algebra of $A$ with respect to $u$.

For simplicity, denote the Weyl algebra $A_{1}(K)$ and the Weyl division ring
$D_{1}(K)$ by $A_{1}$ and $D_{1}$ respectively.

One knows that
$\mathcal{B}=\\{\textbf{p}^{i}\textbf{q}^{j}|i,j\in\mathbb{Z}_{\geq 0}\\}$ is
a basis of $A_{1}$. For $z=\sum{\lambda_{i,j}}\textbf{p}^{i}\textbf{q}^{j}\in
A_{1}\setminus\\{0\\}$, let $supp\\{z\\}=\\{(i,j)\in(\mathbb{Z}_{\geq
0})^{2}|\lambda_{i,j}\neq 0\\},supp\\{0\\}=\o$.

The formula for the product of the basis elements is

$\textbf{p}^{i}\textbf{q}^{j}\cdot\textbf{p}^{s}\textbf{q}^{t}=\textbf{p}^{i+s}\textbf{q}^{j+t}+\sum_{k\geq
1}\frac{[j]_{k}[s]_{k}}{k!}\textbf{p}^{i+s-k}\textbf{q}^{j+t-k}.$ (2.2)

Assume $(\rho,\sigma)\in\mathbb{R}^{2}$. The map

$\mathbf{v}_{\rho,\sigma}:A_{1}\rightarrow\mathbb{R}_{-\infty},\ \
0\mapsto-\infty;\textbf{p}^{i}\textbf{q}^{j}\mapsto\rho i+\sigma
j,0\neq\sum\lambda_{i,j}\textbf{p}^{i}\textbf{q}^{j}\mapsto
sup\\{\mathsf{v}_{\rho,\sigma}(\textbf{p}^{i}\textbf{q}^{j})|\lambda_{i,j}\neq
0\\}$

is originally defined in [D1968] in the case
$(\rho,\sigma)\in\mathbb{R}^{2}\setminus(0,0)$ and $\rho+\sigma>0$.

######  Lemma 2.3.

For $(\rho,\sigma)\in\mathbb{R}^{2}\setminus(0,0)$ and $\rho+\sigma\geq 0$,
$\mathsf{v}_{\rho,\sigma}:A_{1}\rightarrow\mathbb{R}_{-\infty}$ is a degree
map.

###### Proof.

It follows from (2.2) that
$\mathsf{v}_{\rho,\sigma}(\textbf{p}^{i}\textbf{q}^{j}\cdot\textbf{p}^{s}\textbf{q}^{t})=\mathsf{v}_{\rho,\sigma}(\textbf{p}^{i}\textbf{q}^{j})+\mathsf{v}_{\rho,\sigma}(\textbf{p}^{s}\textbf{q}^{t}),\
\textbf{p}^{i}\textbf{q}^{j},\textbf{p}^{s}\textbf{q}^{t}\in\mathcal{B}$. In
fact in Lemma 2.4 of [D1968], it is shown that
$\mathsf{v}_{\rho,\sigma}(zw)=\mathsf{v}_{\rho,\sigma}(z)+\mathsf{v}_{\rho,\sigma}(w)\
(\forall z,w\in A_{1})$ if $\rho+\sigma>0$, which can be proven similarly in
the case $\rho+\sigma=0$. Thus (2.1) (1) holds.

As $supp\\{z+w\\}\subseteq supp\\{z\\}\cup supp\\{w\\}$, one has
$\mathsf{v}_{\rho,\sigma}(z+w)\leq\max\\{\mathsf{v}_{\rho,\sigma}(z),\mathsf{v}_{\rho,\sigma}(w)\\}$,
which is (2.1) (2). It is clear that $\mathsf{v}_{\rho,\sigma}$ also satisfies
(2.1) (3). Therefore $\mathsf{v}_{\rho,\sigma}$ is a degree map on $A_{1}$.

∎

######  Proposition 2.4.

(1) The degree maps on $A_{1}$ over $K$ with $\mathcal{B}$ being an evaluation
basis are exactly those $\mathsf{v}_{\rho,\sigma}$ with $\rho+\sigma\geq 0$.
See figure 1.

(2) Two degree maps $\mathsf{v}_{\rho_{i},\sigma_{i}}(i=1,2)$ on $A_{1}$ are
equivalent iff there exists some $r\in\mathbb{R}_{>0}$,
$(\rho_{2},\sigma_{2})=r(\rho_{1},\sigma_{1})$.

(3) $\mathsf{v}_{\rho,\sigma}$ is the trivial degree map on $A_{1}$ iff
$(\rho,\sigma)=(0,0)$.

The degree map $\mathsf{v}_{\rho,\sigma}$, where $\rho+\sigma\geq 0$ and
$(\rho,\sigma)\neq(0,0)$, is discrete iff $\rho,\sigma$ are
$\mathbb{Q}$-dependent. In this case, up to equivalence one can choose
$(\rho,\sigma)$ to be coprime integers.

$\rho+\sigma\geq 0$$\rho$$\sigma$

figure 1

###### Proof.

(1) Let $u:A_{1}\rightarrow\mathbb{R}_{-\infty}$ be a degree map. Assume
$u(\textbf{p})=\rho,u(\textbf{q})=\sigma$. Then by (2.1) (1),
$u(\textbf{p}^{i}\textbf{q}^{j})=\rho i+\sigma j,i,j\geq 0$; by (2.1) (2), for
$0\neq z=\sum\lambda_{i,j}\textbf{p}^{i}\textbf{q}^{j}\in A_{1}$,
$u(z)=sup\\{\mathsf{v}_{\rho,\sigma}(\textbf{p}^{i}\textbf{q}^{j})|\lambda_{i,j}\neq
0\\}$. Thus $u=\mathsf{v}_{\rho,\sigma}$.

As $1=\textbf{qp}-\textbf{pq}$, by (2.1) (2),

$0=\mathsf{v}_{\rho,\sigma}(1)\leq
max\\{\mathsf{v}_{\rho,\sigma}(\textbf{qp}),\mathsf{v}_{\rho,\sigma}(\textbf{pq})\\}=\rho+\sigma.$

Thus one has $\rho+\sigma\geq 0$.

Conversely, for any $(\rho,\sigma)$ with $\rho+\sigma\geq 0$,
$\mathsf{v}_{\rho,\sigma}$ is a degree map on $A_{1}$ by last lemma.

(2) The ’if’ part is clear. Now we prove ’only if’. Let
$\mathsf{v}_{i}=\mathsf{v}_{\rho_{i},\sigma_{i}},i=1,2$. As
$\mathsf{v}_{1},\mathsf{v}_{2}$ are equivalent, there exists some $r>0$,
$\mathsf{v}_{2}=r\mathsf{v}_{1}$. Thus

$(\rho_{2},\sigma_{2})=(\mathsf{v}_{2}(\textbf{p}),\mathsf{v}_{2}(\textbf{q}))=(r\mathsf{v}_{1}(\textbf{p}),r\mathsf{v}_{1}(\textbf{q}))=r(\rho_{1},\sigma_{1}).$

(3) This follow from
$\mathsf{v}_{\rho,\sigma}(A_{1}^{\times})=\rho\mathbb{Z}_{\geq
0}+\sigma\mathbb{Z}_{\geq 0}.$

∎

As $A_{1}$ is an Ore domain, a degree map
$u:A_{1}\rightarrow\mathbb{R}_{-\infty}$ can be extended to

$u:D_{1}\rightarrow\mathbb{R}_{-\infty},zw^{-1}\mapsto u(z)-u(w)$

(cf. Proposition 9.1.1 of [Co1995]), and a degree map on $D_{1}$ restricts to
a degree map on $A_{1}$. Thus there is a 1-1 correspondence between degree
maps on $A_{1}$ and degree maps on $D_{1}$. We will regard
$\mathsf{v}_{\rho,\sigma}$ as degree maps on $D_{1}$.

######  Example 2.5.

1\. $(\rho,\sigma)=({1,-1})$. Consider the skew Laurent series algebra
$L((T^{-1};\gamma)),L=K(\alpha)$ with
$\gamma\in\text{Aut}_{K}(L),\gamma(\alpha)=\alpha-1,Tf=\gamma(f)T\ for\ f\in
L$ (cf. [B06]). There is an embedding

$D_{1}\rightarrow L((T^{-1};\gamma)),\textbf{p}\mapsto T,\textbf{q}\mapsto
T^{-1}\alpha\ (\text{thus\ }\textbf{pq}\mapsto\alpha),$

and $L((T^{-1};\gamma))$ is the completion of $D_{1}$ with respect to the
topology induced by the degree map $\mathsf{v}_{1,-1}$.

2\. $(\rho,\sigma)=({0,1})$. Consider the algebra of formal pseudo-
differential operators $L((T^{-1};\delta)),L=K(\alpha)$, with
$\delta\in\text{Der}_{K}(L),\delta(f)=f^{\prime}$ for $f\in L$. One has
$Tf=fT+\delta(f)$. See [G1983]. There is an embedding

$D_{1}\rightarrow L((T^{-1};\delta)),\textbf{p}\mapsto\alpha,\textbf{q}\mapsto
T,$

and $L((T^{-1};\delta))$ is the completion of $D_{1}$ with respect to the
topology induced by the degree map $\mathsf{v}_{0,1}$.

In this paper we will construct the completions of $D_{1}$ with respect to the
topology induced by the degree map $\mathsf{v}_{\rho,\sigma}$, where
$(\rho,\sigma)\in\mathbb{Z}^{2}$ with $\rho+\sigma>0$.

##  3 Deformed Laurent series rings with the standard degree map

Let $L$ be a field, and $L((T^{-1}))=\\{\sum_{i\leq k}a_{i}T^{i}|a_{i}\in
L,k\in\mathbb{Z}\\}$ be the left $L$-vector space of (inverse) Laurent series.
We temporarily forget its usual multiplication.

The map

$deg:L((T^{-1}))\rightarrow\mathbb{\mathbb{missing}}Z_{-\infty},\ \
0\mapsto-\infty,L\setminus\\{0\\}\mapsto 0,\sum_{i\leq k}a_{i}T^{i}\mapsto k\
(a_{k}\neq 0),$ (3.1)

satisfies (2) and (3) of (2.1), and we call it the standard degree map on
$L((T^{-1}))$. Equip $L((T^{-1}))$ with the topology induced by the metric

$d(f,g)=e^{deg(f-g)},f,g\in L((T^{-1})).$

######  Def. 3.1.

Let $L((T^{-1}))$ have the usual addition, the standard degree map and the
corresponding topology. Assume $L((T^{-1}))$ has a multiplication

$\tilde{\delta}:L((T^{-1}))\times L((T^{-1}))\rightarrow
L((T^{-1})),(f,g)\mapsto f\cdot g$

that satisfies

  1. 1.

the multiplication $\tilde{\delta}$ is associative, and distributive over
addition ; $\tilde{\delta}$ is continuous;

  2. 2.

$\tilde{\delta}$ agrees with the left multiplication of $L$ on
$L((T^{-1}))$,i.e., $b\cdot\sum_{i\leq k}a_{i}T^{i}=\sum_{i\leq
k}(ba_{i})T^{i}$;

  3. 3.

$T^{i}\cdot T^{j}=T^{i+j},i,j\in\mathbb{Z};$ ($T^{0}=1$);

  4. 4.

$deg(f\cdot g)=deg(f)+deg(g),f,g\in L((T^{-1}))$.

Then $L((T^{-1}))$ is a ring, and $deg$ satisfies all the 3 axioms in (2.1)
thus is a degree map on it. We denote it by $L((T^{-1};\tilde{\delta}))$.

As $1\cdot aT^{i}=aT^{i}$ and $aT^{i}\cdot 1=a(T^{i}\cdot 1)=aT^{i}$ by the
axioms, the multiplicative identity 1 of $L$ is also the multiplicative
identity of $L((T^{-1};\tilde{\delta}))$, and $L((T^{-1}))$ contains $L$ as a
subring.

######  Lemma 3.2.

The ring $L((T^{-1};\tilde{\delta}))$ is a topological division ring which is
complete with respect to the topology induced by the degree map $deg$.

###### Proof.

It is clear that the addition and multiplication of
$L((T^{-1};\tilde{\delta}))$ is continuous with respect to its topology. So we
only need to show that any nonzero element of $L((T^{-1};\tilde{\delta}))$ has
a multiplicative inverse and the inverse map is continuous.

Assume $0\neq z\in L((T^{-1};\tilde{\delta}))$. One has $z=\sum_{i\leq
n}a_{i}T^{i}$. Then

$z=a_{n}T^{n}(1+w),w=\sum_{k\geq
1}w_{k},w_{k}=T^{-n}a_{n}^{-1}a_{n-k}T^{n-k}.$

Note that $deg(w_{k})=-k$ for $k\geq 1$, so $\sum_{k\geq 1}w_{k}$ is
convergent, and $deg(w)\leq-1$. Thus

$z^{-1}=(1+w)^{-1}T^{-n}a_{n}^{-1},\ (1+w)^{-1}=\sum_{j\geq 0}(-1)^{j}w^{j}.$

So any nonzero element of $L((T^{-1};\tilde{\delta}))$ has a multiplicative
inverse and $L((T^{-1};\tilde{\delta}))$ is a division ring. It follows from
the above formula that the inverse map is continuous.

∎

We will call $L((T^{-1};\tilde{\delta}))$ a deformed Laurent series (division)
ring with the standard degree map. Note that $L((T^{-1};\tilde{\delta}))$
should not be confused with the deformation of algebras in [F1993].

The purpose of this section is to find the necessary and sufficient conditions
for $\tilde{\delta}$ to make $L((T^{-1};\tilde{\delta}))$ a deformed Laurent
series ring with the standard degree map.

As the multiplication of $L((T^{-1};\tilde{\delta}))$ is distributive over
addition, one need to know the formula for $T^{i}\cdot a$ for $a\in
L,i\in\mathbb{Z}$, and $aT^{i}\cdot bT^{j},i,j\in\mathbb{Z}$. One first
consider $T\cdot a$.

######  Lemma 3.3.

Assume that $L((T^{-1};\tilde{\delta}))$ is a deformed Laurent series ring
with the standard degree map. Then there exist maps $\delta_{i}:L\rightarrow
L,i\in\mathbb{Z}_{\leq 1}$ such that

$T\cdot a=\delta_{1}(a)T+\sum_{i=0}^{-\infty}\delta_{i}(a)T^{i},a\in L.$ (3.2)

One has $\delta_{i}$ preserve addition for all $i$; and
$\delta_{1}:L\rightarrow L$ is an injective ring homomorphism.

###### Proof.

by $T(a+b)=Ta+Tb$, one has $\delta_{i}$ preserve addition for all $i$. $T1=T$
implies $\delta_{1}(1)=1$; $(Ta)b=Tab$ implies
$\delta_{1}(ab)=\delta_{1}(a)\delta_{1}(b)$ for all $a,b\in L$, so
$\delta_{1}:L\rightarrow L$ is a ring homomorphism.

Then $\text{Ker}(\delta_{1})$ is a proper ideal of $L$, which must be 0, thus
$\delta_{1}$ is injective.

∎

######  Lemma 3.4.

Let $L((T^{-1};\tilde{\delta}))$ be a deformed Laurent series ring with the
standard degree map satisfying (3.2). Assume that $\delta_{1}=1$. Then

$T\cdot a=aT+\sum_{i=0}^{-\infty}\delta_{i}(a)T^{i},a\in L,$ (3.3)

and for $n,m\in\mathbb{Z}$,

$T^{n}\cdot a=\sum_{k\geq
0}\sum_{(j_{1},\cdots,j_{k})}\binom{n}{k}\delta_{j_{1}}\cdots\delta_{j_{k}}(a)T^{j_{1}+\cdots+j_{k}-k+n}$
(3.4)
$=aT^{n}+\sum_{k>0}\sum_{(j_{1},\cdots,j_{k})}\binom{n}{k}\delta_{j_{1}}\cdots\delta_{j_{k}}(a)T^{j_{1}+\cdots+j_{k}-k+n},$
$aT^{m}\cdot
bT^{n}=abT^{m+n}+\sum_{k>0}\sum_{(j_{1},\cdots,j_{k})}\binom{m}{k}a\delta_{j_{1}}\cdots\delta_{j_{k}}(b)T^{j_{1}+\cdots+j_{k}-k+m+n}.$
(3.5)

###### Proof.

First we prove (3.4) in the case $n\geq 1$, by induction on $n$.

The case $n=1$ is clear. Assume $n>1$ and (3.4) holds for all positive
integers $\leq n$. Then

$\displaystyle T^{n+1}\cdot a$ $\displaystyle=T(T^{n}\cdot
a)=T\cdot\left(\sum_{k\geq
0}\sum_{(j_{1},\cdots,j_{k})}\binom{n}{k}\delta_{j_{1}}\cdots\delta_{j_{k}}(a)T^{j_{1}+\dots+j_{k}-k+n}\right)$
$\displaystyle=\sum_{k\geq
0}\sum_{(j_{1},\dots,j_{k})}\binom{n}{k}(T\cdot\delta_{j_{1}}\dots\delta_{j_{k}}(a))T^{j_{1}+\cdots+j_{k}-k+n}$
$\displaystyle=\sum_{k\geq
0}\sum_{(j_{1},\dots,j_{k})}\binom{n}{k}\left(\delta_{j_{1}}\cdots\delta_{j_{k}}(a)T+\sum_{l\leq
0}\delta_{l}(\delta_{j_{1}}\cdots\delta_{j_{k}}(a))T^{l}\right)\cdot
T^{j_{1}+\cdots+j_{k}-k+n}$ $\displaystyle=\sum_{k\geq
0}\sum_{(j_{1},\dots,j_{k})}\binom{n}{k}\delta_{j_{1}}\cdots\delta_{j_{k}}(a)T^{j_{1}+\cdots+j_{k}-k+n+1}$
$\displaystyle+\sum_{k\geq
0}\binom{n}{k}\sum_{(l,j_{1},\cdots,j_{k})}\delta_{l}\delta_{j_{1}}\cdots\delta_{j_{k}}(a)T^{j_{1}+\cdots+j_{k}+l-(k+1)+(n+1)}$
$\displaystyle=aT^{n+1}+\sum_{k>0}\sum_{(j_{1},\cdots,j_{k})}[\binom{n}{k}+\binom{n}{k-1}]\delta_{j_{1}}\cdots\delta_{j_{k}}(a)T^{j_{1}+\cdots+j_{k}-k+(n+1)}$
$\displaystyle=aT^{n+1}+\sum_{k>0}\sum_{(j_{1},\cdots,j_{k})}\binom{n+1}{k}\delta_{j_{1}}\cdots\delta_{j_{k}}(a)T^{j_{1}+\cdots+j_{k}-k+(n+1)}.$

Thus (3.4) holds for all $n\geq 1$.

Next we consider the case $n=-1$. One has

$\displaystyle T\left(aT^{-1}+\sum_{k\geq
1}\sum_{(j_{1},\dots,j_{k})}\binom{-1}{k}\delta_{j_{1}}\cdots\delta_{j_{k}}(a)T^{j_{1}+\cdots+j_{k}-k-1}\right)$
$\displaystyle=(aT+\sum_{i\leq 0}\delta_{i}(a)T^{i})T^{-1}+\sum_{k\geq
1}\sum_{(j_{1},\dots,j_{k})}\binom{-1}{k}T\cdot\delta_{j_{1}}\cdots\delta_{j_{k}}(a)T^{j_{1}+\cdots+j_{k}-k-1}$
$\displaystyle=a+\sum_{i\leq 0}\delta_{i}(a)T^{i-1}+\sum_{k\geq
1}\sum_{(j_{1},\dots,j_{k})}\binom{-1}{k}\left[\delta_{j_{1}}\cdots\delta_{j_{k}}(a)\cdot
T+\sum_{l\leq
0}\delta_{l}(\delta_{j_{1}}\cdots\delta_{j_{k}}(a))T^{l}\right]T^{j_{1}+\cdots+j_{k}-k-1}$
$\displaystyle=a+\sum_{i\leq 0}\delta_{i}(a)T^{i-1}+\sum_{k\geq
1}\sum_{(j_{1},\dots,j_{k})}\binom{-1}{k}\delta_{j_{1}}\cdots\delta_{j_{k}}(a)T^{j_{1}+\cdots+j_{k}-k}$
$\displaystyle+\sum_{k\geq
1}\sum_{(l,j_{1},\dots,j_{k})}\binom{-1}{k}\delta_{l}\delta_{j_{1}}\cdots\delta_{j_{k}}(a)T^{l+j_{1}+\cdots+j_{k}-k-1}$
$\displaystyle=a+\sum_{i\leq
0}[1+\binom{-1}{1}]\delta_{i}(a)T^{i-1}+\sum_{k\geq
2}\sum_{(j_{1},\dots,j_{k})}[\binom{-1}{k}+\binom{-1}{k-1}]\delta_{j_{1}}\cdots\delta_{j_{k}}(a)T^{j_{1}+\cdots+j_{k}-k}$
$\displaystyle=a$

Multiply $T^{-1}$ on both sides from the left, one has

$\displaystyle T^{-1}a=aT^{-1}+\sum_{k\geq
1}\sum_{(j_{1},\dots,j_{k})}\binom{-1}{k}\delta_{j_{1}}\cdots\delta_{j_{k}}(a)T^{j_{1}+\cdots+j_{k}-k-1}.$

So (3.4) holds for $n=-1$. Assume (3.4) holds for $n=-m$ with $m>0$, i.e.,

$\displaystyle T^{-m}a=\sum_{k\geq
0}\sum_{(l_{1},\dots,l_{k})}\binom{-m}{k}\delta_{l_{1}}\cdots\delta_{l_{k}}(a)T^{l_{1}+\cdots+l_{k}-k-m}.$
$\displaystyle T^{-1-m}a$ $\displaystyle=T^{-1}(T^{-m}a)=T^{-1}\sum_{k\geq 0}\
\sum_{(l_{1},\dots,l_{k})}\binom{-m}{k}\delta_{l_{1}}\cdots\delta_{l_{k}}(a)T^{l_{1}+\cdots+l_{k}-k-m}$
$\displaystyle=\sum_{k\geq 0}\
\sum_{(l_{1},\dots,l_{k})}\binom{-m}{k}\left[T^{-1}\delta_{l_{1}}\cdots\delta_{l_{k}}(a)\right]T^{l_{1}+\cdots+l_{k}-k-m}$
$\displaystyle=\sum_{k\geq 0}\
\sum_{(l_{1},\dots,l_{k})}\binom{-m}{k}\left(\sum_{u\geq 0}\
\sum_{(j_{1},\dots,j_{u})}\binom{-1}{u}\delta_{j_{1}}\cdots\delta_{j_{u}}(\delta_{l_{1}}\cdots\delta_{l_{k}}(a))T^{j_{1}+\cdots+j_{u}-u-1}\right)T^{l_{1}+\cdots+l_{k}-k-m}$
$\displaystyle=\sum_{k\geq 0,u\geq 0}\
\sum_{(j_{1},\cdots,j_{u};l_{1},\dots,l_{k})}\binom{-m}{k}\binom{-1}{u}\delta_{j_{1}}\cdots\delta_{j_{u}}\delta_{l_{1}}\cdots\delta_{l_{k}}(a)T^{j_{1}+\cdots+j_{u}+l_{1}+\cdots+l_{k}-(u+1)-(m+k)}$
$\displaystyle=\sum_{p\geq 0}\ \sum_{k,u\geq 0}^{k+u=p}\
\sum_{(s_{1},\dots,s_{p})}\binom{-m}{k}\binom{-1}{u}\delta_{s_{1}}\cdots\delta_{s_{p}}(a)T^{s_{1}+\cdots+s_{p}-p-(m+1)}$
$\displaystyle=\sum_{p\geq 0}\
\sum_{(s_{1},\dots,s_{p})}\binom{-(m+1)}{p}\delta_{s_{1}}\cdots\delta_{s_{p}}(a)T^{s_{1}+\cdots+s_{p}-p-(m+1)}.$

So (3.4) holds for $-(m+1)$, thus it holds for all $n\in\mathbb{Z}_{<0}$. So
(3.4) holds for all $n\in\mathbb{Z}$.

It is clear that (3.5) follows from (3.4). ∎

The lemma assumes $\delta_{1}=1$, if $1\neq\delta_{1}\in\text{Aut}(L)$, the
corresponding results can be obtained similarly.

If $\delta_{i}=0$ for all $i\leq 0$, then $L((T^{-1};\tilde{\delta}))$ is the
usual field of Laurent series. Henceforth we will always assume that
$\delta_{1}=1$ and $\delta_{i}\neq 0$ for some $i\leq 0$, unless otherwise
stated.

Henceforth we will always assume that the characteristic of the field $L$ is
0.

For a ring $A$ and $\emptyset\neq S,D\subseteq A$, let

$C(S,D)=\\{b\in D|ab=ba,\forall a\in S\\}$

be the centralizer of $S$ in $D$, and $C(A):=C(A,A)$ the center of $A$.

######  Lemma 3.5.

Let $L((T^{-1};\tilde{\delta}))$ be a deformed Laurent series ring with the
standard degree map satisfying (3.3), and $\delta_{i}\neq 0$ for some $i\leq
0$. Let $A=L((T^{-1};\tilde{\delta}))$. Then $C(L,A)=L$; the center $C(A)$ is
$\cap_{i\leq 0}\text{Ker}(\delta_{i})$, which is a field.

###### Proof.

Let $i_{0}=max\\{i|i\leq 0,\delta_{i}\neq 0\\}.$ Let $a\in
L\setminus\text{Ker}(\delta_{i_{0}})$. Then $C(a,A)\supseteq L$. If
$C(a,A)\neq L$, one can assume that there exists $\sum_{i\leq n}b_{i}T^{i}\in
C(a,A)\setminus L$ with $n\neq 0$ and $b_{n}\neq 0$.

$\displaystyle\sum_{i\leq n}b_{i}T^{i}\cdot a$ $\displaystyle=\sum_{i\leq
n}b_{i}(T^{i}\cdot a)$ $\displaystyle=\sum_{i\leq n}b_{i}\left(a\cdot
T^{i}+\sum_{k>0}\sum_{(j_{1},\cdots
j_{k})}\binom{i}{k}\delta_{j_{1}}\cdots\delta_{j_{k}}(a)T^{j_{1}+\cdots+j_{k}-k+i}\right)$
$\displaystyle=\sum_{i\leq
n}(b_{i}a)T^{i}+\binom{n}{1}b_{n}\delta_{i_{0}}(a)T^{i_{0}-1+n}+\quad terms\
of\ degree<i_{0}-1+n$ $\displaystyle a\sum_{i\leq n}b_{i}T^{i}$
$\displaystyle=\sum_{i\leq n}(ab_{i})T^{i}.$

As $\sum_{i\leq n}b_{i}T^{i}\cdot a=a\cdot\sum_{i\leq n}b_{i}T^{i}$ and
$b_{n}\neq 0$, $\binom{n}{1}\delta_{i_{0}}(a)=0$. As the characteristic of $L$
is 0, $\delta_{i_{0}}(a)=0$, which is a contradiction. Thus $C(a,A)\subseteq
L$, one has $C(a,A)=L$ as $L\subseteq C(a,A)$.

By $T\cdot a=aT+\sum_{i\leq 0}\delta_{i}(a)T^{i}$,

$C(T,L)=\bigcap_{i\leq 0}\text{Ker}(\delta_{i}).$

So $C(A)=\bigcap_{i\leq 0}\text{Ker}(\delta_{i})$.

As $L((T^{-1};\tilde{\delta}))$ is a division ring, its center $C(A)$ must be
a field.

∎

Let $K=\cap_{i\leq 0}\text{Ker}(\delta_{i})$. Then $A$ is a $K$-algebra.

######  Lemma 3.6.

Let $L((T^{-1};\tilde{\delta}))$ be a deformed Laurent series ring with the
standard degree map satisfying (3.3). Then for $i\leq 0$,

$\delta_{i}(ab)=a\delta_{i}(b)+\delta_{i}(a)b+\sum_{k>0}\sum_{(l,j_{1},\cdots,j_{k})}^{j_{1}+\cdots+j_{k}-k+l=i}\binom{l}{k}\delta_{l}(a)\delta_{j_{1}}\cdots\delta_{j_{k}}(b),a,b\in
L.$ (3.6)

(See (1.3) for the meaning of
$\sum_{(l,j_{1},\cdots,j_{k})}^{j_{1}+\cdots+j_{k}-k+l=i}$.)

Let $i_{0}=max\\{i\in\mathbb{Z}_{\leq 0}|\delta_{i}\neq 0\\}.$ Then
$\delta_{i_{0}}$ is a $K$-linear derivation. Each $\delta_{i},i\leq 0,$ is a
$K$-linear differential operator of $L$.

###### Proof.

One has

$T\cdot(ab)=abT+\sum_{i\leq 0}\delta_{i}(ab)T^{i},a,b\in L,$

and

$\displaystyle(Ta)b=$ $\displaystyle(aT+\sum_{i\leq 0}\delta_{i}(a)T^{i})b$
$\displaystyle=$ $\displaystyle a(Tb)+\sum_{i\leq 0}\delta_{i}(a)(T^{i}b)$
$\displaystyle=$ $\displaystyle a(bT+\sum_{i\leq
0}\delta_{i}(b)T^{i})+\sum_{i\leq 0}\delta_{i}(a)\left[bT^{i}+\sum_{k\geq
1}\sum_{(j_{1},\cdots,j_{k})}\binom{i}{k}\delta_{j_{1}}\cdots\delta_{j_{k}}(b)T^{j_{1}+\cdots+j_{k}-k+i}\right]$
$\displaystyle=$ $\displaystyle abT+\sum_{i\leq
0}\left(a\delta_{i}(b)T^{i}+\delta_{i}(a)bT^{i}\right)+\sum_{l\leq
0}\sum_{k\geq
1}\sum_{(j_{1},\cdots,j_{k})}\binom{l}{k}\delta_{l}(a)\delta_{j_{1}}\cdots\delta_{j_{k}}(b)T^{j_{1}+\cdots+j_{k}-k+l}$

by $T(ab)=(Ta)b$, compare the coefficient of each $T^{i}$, one has

$\displaystyle\delta_{i}(ab)=a\delta_{i}(b)+\delta_{i}(a)b+\sum_{k\geq
1}\sum_{(l,j_{1},\cdots,j_{k})}^{l+j_{1}+\cdots+j_{k}-k=i}\binom{l}{k}\delta_{l}(a)\delta_{j_{1}}\cdots\delta_{j_{k}}(b),a,b\in
L,i\leq 0.$

It follows that for $a\in K,b\in L$, $\delta_{i}(ab)=a\delta_{i}(b)$. Thus
each $\delta_{i}$ is $K$-linear.

It is clear that $\delta_{i_{0}}$ is a $K$-linear derivation. By induction one
can easily show that for those $i$ with $\delta_{i}\neq 0$, $\delta_{i}$ is a
(high order) $K$-linear differential operator on $L$.

∎

As the characteristic of $L$ is 0, the characteristic of its center $K$ is
also 0.

All the tensor products below will be over $K$.

Let $B=End_{K-lin}(L)$ be the $K$-algebra of $K$-linear operators on $L$. Then
the tensor product $B\otimes B$ is also a $K$-algebra, which can be
canonically identified with a subalgebra of $End_{K-lin}(L\otimes L)$.

Let

$\mathbb{m}:L\otimes L\rightarrow L,a\otimes b\mapsto ab$

be the multiplication map. Then $\mathbb{m}$ is a homomorphism of rings, whose
kernel is the ideal $I$ of $L\otimes L$ generated by all such elements
$a\otimes 1-1\otimes a,a\in L$.

Let

$\Gamma=\\{(\phi,\tau)\in(B\otimes B)\times
B|\mathbb{m}\cdot\phi=\tau\cdot\mathbb{m}\\},$

i.e. $(\phi,\tau)\in\Gamma$ iff $(\phi,\tau)$ make the following diagram
commutative.

${L\otimes L}$${L}$${L\otimes
L}$${L}$$\scriptstyle{\mathbb{m}}$$\scriptstyle{\phi}$$\scriptstyle{\tau}$$\scriptstyle{\mathbb{m}}$

Let

$C=\\{\psi\in B\otimes B|\psi(I)\subseteq I\\}.$

Then $C$ is a subalgebra of $B\otimes B$. It is clear that
$(\phi,\tau)\in\Gamma$ iff $\phi\in C$. For $\phi\in C$, it is clear that
there exists a unique $\tau$ such that $(\phi,\tau)\in\Gamma$, and denote the
$\tau$ by $\overline{\phi}$.

The map

$\Psi:C\rightarrow B,\phi\mapsto\overline{\phi},$

is a ring homomorphism. It is clear that

$\text{Ker}(\Psi)=\\{\phi\in B\otimes B|\phi(L\otimes L)\subseteq I\\}.$

Let $B_{0}=\Psi(C)$, which is a subalgebra of $B$. The homomorphism $\Psi$
induces a ring isomorphism $C/\text{Ker}(\Psi)\rightarrow B_{0}.$

Let

$\Delta:B_{0}\rightarrow C/\text{Ker}(\Psi)$

be the its inverse. Then for $\tau\in B_{0}$,
$\Delta(\tau)=\phi+\text{Ker}(\Psi)$ iff $(\phi,\tau)\in\Gamma$.

By (3.6), one has for $i\leq 0$,

$\displaystyle\Delta\delta_{i}$
$\displaystyle=1\otimes\delta_{i}+\sum_{k>0}[\sum_{(l,j_{1},\cdots,j_{k})}^{j_{1}+\cdots+j_{k}+l=i+k}\binom{l}{k}\delta_{l}\otimes\delta_{j_{1}}\cdots\delta_{j_{k}}]+\delta_{i}\otimes
1+\text{Ker}\ \Psi,$ (3.7)

or,

$\displaystyle\Delta\delta_{i}=1\otimes\delta_{i}+\sum_{k\geq
0}[\sum_{(l,j_{1},\cdots,j_{k})}^{j_{1}+\cdots+j_{k}+l=i+k}\binom{l}{k}\delta_{l}\otimes\delta_{j_{1}}\cdots\delta_{j_{k}}]+\text{Ker}\
\Psi,$ (3.8)

where $\delta_{i}\otimes 1$ in (3.7) is combined into the summation (regarded
as $k=0$). Another equivalent formulation of (3.7) is

$\Delta\delta_{i}=1\otimes\delta_{i}+\sum_{i<l\leq
0}\delta_{l}\otimes\sum_{k=1}^{l-i}\binom{l}{k}[\sum_{(j_{1},\cdots,j_{k})}^{j_{1}+\cdots+j_{k}=i+k-l}\delta_{j_{1}}\cdots\delta_{j_{k}}]+\delta_{i}\otimes
1+\text{Ker}\ \Psi.$ (3.9)

Note that $1\otimes\delta_{i}+\sum_{i<l\leq
0}\delta_{l}\otimes\sum_{k=1}^{l-i}\binom{l}{k}[\sum_{(j_{1},\cdots,j_{k})}^{j_{1}+\cdots+j_{k}=i+k-l}\delta_{j_{1}}\cdots\delta_{j_{k}}]+\delta_{i}\otimes
1$ is always a finite sum.

In particular $\delta_{i}\in B_{0}$ for all $i\leq 1$.

######  Remark 3.7.

For $i=0,1,2,3$, one has

$\displaystyle\Delta\delta_{0}=1\otimes\delta_{0}+\delta_{0}\otimes
1+\text{Ker}\ \Psi,$
$\displaystyle\Delta\delta_{-1}=1\otimes\delta_{-1}+\delta_{-1}\otimes
1+\binom{0}{1}\delta_{0}\otimes\delta_{0}=1\otimes\delta_{-1}+\delta_{-1}\otimes
1+\text{Ker}\ \Psi,$
$\displaystyle\Delta\delta_{-2}=1\otimes\delta_{-2}+\delta_{-2}\otimes
1+\binom{-1}{1}\delta_{-1}\otimes\delta_{0}+\text{Ker}\ \Psi,$
$\displaystyle\Delta\delta_{-3}=1\otimes\delta_{-3}+\delta_{-3}\otimes
1+\binom{-1}{1}\delta_{-1}\otimes\delta_{-1}+\binom{-2}{1}\delta_{-2}\otimes\delta_{0}+\binom{-1}{2}\delta_{-1}\otimes\delta_{0}^{2}+\text{Ker}\
\Psi.$

We will use the following notation in later proof. According to the expression
(3.7), we write

$\delta_{l}\otimes\delta_{j_{1}}\cdots\delta_{j_{k}}\in\Delta\delta_{i},if\
j_{1}+\cdots+j_{k}+l=i+k.$

Next we will show that if (3.7) holds then (3.5) defines an associative
product on $L((T^{-1}))$.

By (3.5) and (3.7), for $a,b,c\in L\setminus\\{0\\},m,n,l\in\mathbb{Z}$, one
has

$(aT^{m}\cdot bT^{n})\cdot cT^{l}=\sum_{u\geq 0,k\geq
0}\sum_{(j_{1},\cdots,j_{k};s_{1},\cdots,s_{u})}\binom{m}{k}\binom{\sum
j_{i}-k+m+n}{u}a\delta_{j_{1}}\cdots\delta_{j_{k}}(b)\delta_{s_{1}}\cdots\delta_{s_{u}}(c)T^{m+n+l+\sum
j_{i}+\sum s_{v}-(k+u)},$

where $\sum j_{i}$ means $\sum_{i=1}^{k}j_{i}$, which equals 0 if $k=0$.
Similarly $\sum s_{v}=\sum_{v=1}^{u}s_{v}$, which equals 0 if $u=0$. Such
conventions will be used henceforth.

Similarly one has

$aT^{m}(bT^{n}\cdot cT^{l})=\sum_{u\geq 0,k\geq
0}\sum_{(j_{1},\cdots,j_{k};s_{1},\cdots,s_{u})}\binom{m}{k}\binom{n}{u}a\delta_{j_{1}}\cdots\delta_{j_{k}}(b\delta_{s_{1}}\cdots\delta_{s_{u}}(c))T^{m+n+l+\sum
j_{i}+\sum s_{v}-(k+u)},$

So, $(aT^{m}\cdot bT^{n})\cdot cT^{l}=aT^{m}(bT^{n}\cdot cT^{l})$ if for any
fixed $p\in\mathbb{Z}_{\leq 0}$,

$\sum\binom{m}{k}\binom{\sum
j_{i}-k+m+n}{t}\delta_{j_{1}}\cdots\delta_{j_{k}}(b)\delta_{s_{1}}\cdots\delta_{s_{t}}(c)=\sum\binom{m}{u}\binom{n}{r}\delta_{l_{1}}\cdots\delta_{l_{u}}\left(b\delta_{q_{1}}\cdots\delta_{q_{r}}(c)\right),$
(3.10)

where the summation on the left (resp. on the right) is over $\Omega_{1}(p)$
(resp.$\Omega_{2}(p)$), where

$\Omega_{1}(p)=\\{(j_{1},\cdots,j_{k};s_{1},\cdots,s_{t})|k,t\geq
0;j_{i},s_{v}\leq 0;\sum j_{i}+\sum s_{v}-(k+t)=p\\},$
$\Omega_{2}(p)=\\{(l_{1},\cdots,l_{u};q_{1},\cdots,q_{r})|u,r\geq
0;l_{i},q_{v}\leq 0;\sum l_{i}+\sum q_{v}-(u+r)=p\\}.$

The left hand side of (3.10) is

$\sum\binom{m}{k}\binom{\sum
j_{i}-k+m+n}{t}\mathbb{m}\cdot(\delta_{j_{1}}\cdots\delta_{j_{k}}\otimes\delta_{s_{1}}\cdots\delta_{s_{t}})(b\otimes
c),$

The right hand side of (3.10) is

$\sum\binom{m}{u}\binom{n}{r}\mathbb{m}\cdot\Delta(\delta_{l_{1}}\cdots\delta_{l_{u}})\cdot(1\otimes\delta_{q_{1}}\cdots\delta_{q_{r}})(b\otimes
c).$

Recall that $B=End_{K-lin}(L)$. Let

$L(p)=\sum\binom{m}{k}\binom{\sum
j_{i}-k+m+n}{t}\delta_{j_{1}}\cdots\delta_{j_{k}}\otimes\delta_{s_{1}}\cdots\delta_{s_{t}}\in
B\otimes_{K}B,$ (3.11)

where the summation is over $\Omega_{1}(p)$, and

$R(p)=\sum\binom{m}{u}\binom{n}{r}\Delta(\delta_{l_{1}}\cdots\delta_{l_{u}})\cdot(1\otimes\delta_{q_{1}}\cdots\delta_{q_{r}})\in
B\otimes_{K}B,$ (3.12)

where the summation is over $\Omega_{2}(p)$. Then $(aT^{m}\cdot bT^{n})\cdot
cT^{l}=aT^{m}(bT^{n}\cdot cT^{l})$ if $L(p)=R(p)$ for all fixed
$p\in\mathbb{Z}_{\leq 0}$ and (3.5) holds.

######  Lemma 3.8.

Assume there are additive maps $\delta_{i}:L\rightarrow L,i\in\mathbb{Z}_{\leq
0},$ which satisfy (3.7). Let $m,n,l\in\mathbb{Z}$. Then $L(p)=R(p)$ for all
$p\in\mathbb{Z}_{\leq 0}$. Thus if (3.5) holds, then $aT^{m}(bT^{n}\cdot
cT^{l})=(aT^{m}\cdot bT^{n})\cdot cT^{l}$ for $a,b,c\in L$.

###### Proof.

As the 2nd statement follows form the 1st one, one only needs to prove the 1st
statement.

Fix $p\in\mathbb{Z}_{\leq 0}$.

Note that if (3.7) holds, then
$\Delta(\delta_{l_{1}}\cdots\delta_{l_{u}})\cdot(1\otimes\delta_{q_{1}}\cdots\delta_{q_{r}})=\Delta(\delta_{l_{1}})\cdots\Delta(\delta_{l_{u}})\cdot(1\otimes\delta_{q_{1}}\cdots\delta_{q_{r}})$
in $R(p)$ can further be expanded as the sum of terms of the form
$\lambda\delta_{j_{1}}\cdots\delta_{j_{k}}\otimes\delta_{s_{1}}\cdots\delta_{s_{t}}$
with $(j_{1},\cdots,j_{k};s_{1},\cdots,s_{t})\in\Omega_{1}$ (and
$\lambda\in\mathbb{Z}$), which is in $L(p)$. So we only need to check that the
coefficient of any fixed term in $L(p)$ equals the coefficient of the same
term in the expansion of $R(p)$. There are 3 steps.

1\. Let $k,t\geq 0$ and
$j_{1},\cdots,j_{k},s_{1},\cdots,s_{t}\in\mathbb{Z}_{\leq 0}$. Let

$\phi_{0}({j_{1}},\cdots,{j_{k}};{s_{1}},\cdots,{s_{t}})$

be the sum of coefficients of
$\delta_{j_{1}}\cdots\delta_{j_{k}}\otimes\delta_{s_{1}}\cdots\delta_{s_{t}}$
in the expansion of all the terms in $R(p)$ of the form
$\binom{m}{u}\binom{n}{0}\Delta(\delta_{l_{1}}\cdots\delta_{l_{u}})\cdot(1\otimes
1)$ (the terms with $r=0$ in (3.12)).

For example, if $t=0$ then
$\phi_{0}({j_{1}},\cdots,{j_{k}};\emptyset)=\binom{m}{k}$, if $k=0$ then
$\phi_{0}(\emptyset;{s_{1}},\cdots,{s_{t}})=\binom{m}{t}$.

First we will show

$\displaystyle\phi_{0}(j_{1},\cdots,j_{k};s_{1},\cdots,s_{t})=\sum_{i=0}^{t}\binom{m}{k+i}\binom{k+i}{k}\binom{j_{1}+\cdots+j_{k}}{t-i}.$
(3.13)

To avoid complicated subscripts, we illustrate it in the case $k=2,t=3$ below.
The general case can be treated similarly.

We will compute the coefficient of
$\delta_{j_{1}}\delta_{j_{2}}\otimes\delta_{s_{1}}\delta_{s_{2}}\delta_{s_{3}}$
in the expansion of terms of the type
$\binom{m}{u}\binom{n}{0}\Delta(\delta_{l_{1}}\cdots\delta_{l_{u}})\cdot(1\otimes
1)$, which equals
$\binom{m}{u}\Delta(\delta_{l_{1}})\cdots\Delta(\delta_{l_{u}})$. One must
have $2\leq u\leq 2+3$. We will compute the coefficient of
$\delta_{j_{1}}\delta_{j_{2}}\otimes\delta_{s_{1}}\delta_{s_{2}}\delta_{s_{3}}$
in the expansion of those terms of the type
$\Delta(\delta_{l_{1}})\cdots\Delta(\delta_{l_{2+i}})$ with $i=0,1,2,3$.

Assume $i=0$. By (3.8) one has

$\displaystyle\Delta\delta_{l_{1}}=\sum_{k\geq
0}\sum_{(u,p_{1},\cdots,p_{k})}^{p_{1}+\cdots+p_{k}+u=l_{1}+k}\binom{u}{k}\delta_{u}\otimes\delta_{p_{1}}\cdots\delta_{p_{k}}+1\otimes\delta_{l_{1}}+\text{Ker}\
\Psi,$

$\displaystyle\Delta\delta_{l_{2}}=\sum_{r\geq
0}\sum_{(v,q_{1},\cdots,q_{r})}^{q_{1}+\cdots+q_{r}+v=l_{2}+r}\binom{v}{r}\delta_{v}\otimes\delta_{q_{1}}\cdots\delta_{q_{r}}+1\otimes\delta_{l_{2}}+\text{Ker}\
\Psi.$

If
$\delta_{j_{1}}\delta_{j_{2}}\otimes\delta_{s_{1}}\delta_{s_{2}}\delta_{s_{3}}$
appears in the expansion of $\Delta\delta_{l_{1}}\cdot\Delta\delta_{l_{2}}$,
then for some $u=0,1,2,3$,
$\delta_{j_{1}}\otimes\delta_{s_{1}}\cdots\delta_{s_{u}}$ is in the expansion
of $\Delta\delta_{l_{1}}$, and
$\delta_{j_{2}}\otimes\delta_{s_{u+1}}\cdots\delta_{s_{3}}$ is in the
expansion of $\Delta\delta_{l_{2}}$. Thus the coefficient of
$\delta_{j_{1}}\delta_{j_{2}}\otimes\delta_{s_{1}}\delta_{s_{2}}\delta_{s_{3}}$
in those $\binom{m}{2}\Delta\delta_{l_{1}}\cdot\Delta\delta_{l_{2}}$ is

$\binom{m}{2}\sum_{u=0}^{3}\binom{j_{1}}{u}\binom{j_{2}}{3-u}=\binom{m}{2}\binom{j_{1}+j_{2}}{3}.$
(3.14)

Note that (3.14) is just the summand of the right side of (3.13) with $i=0$ in
the case $k=2,t=3$.

Assume $i=1$. Let
$\alpha=\delta_{j_{1}}\delta_{j_{2}}\otimes\delta_{s_{1}}\delta_{s_{2}}\delta_{s_{3}}$.
If $\alpha$ appears in the expansion of
$\Delta\delta_{l_{1}}\cdot\Delta\delta_{l_{2}}\cdot\Delta\delta_{l_{3}}$, then
there are 3 cases:

(a)
$\alpha=(\delta_{j_{1}}\otimes\delta_{s_{1}}\cdots\delta_{s_{u}})\cdot(\delta_{j_{2}}\otimes\delta_{s_{u+1}}\cdots\delta_{s_{2}})\cdot(1\otimes\delta_{s_{3}}),u=0,1,2$,
with

$\delta_{j_{1}}\otimes\delta_{s_{1}}\cdots\delta_{s_{u}}\in\Delta\delta_{l_{1}},\delta_{j_{2}}\otimes\delta_{s_{u+1}}\cdots\delta_{s_{2}}\in\Delta\delta_{l_{2}},1\otimes\delta_{s_{3}}\in\Delta\delta_{l_{3}};$

(b)
$\alpha=(\delta_{j_{1}}\otimes\delta_{s_{1}}\cdots\delta_{s_{u}})\cdot(1\otimes\delta_{s_{u+1}})\cdot(\delta_{j_{2}}\otimes\delta_{s_{u+2}}\cdots\delta_{s_{3}}),\
u=0,1,2$, with

$\delta_{j_{1}}\otimes\delta_{s_{1}}\cdots\delta_{s_{u}}\in\Delta\delta_{l_{1}},1\otimes\delta_{s_{u+1}}\in\Delta\delta_{l_{2}},\delta_{j_{2}}\otimes\delta_{s_{u+2}}\cdots\delta_{s_{3}}\in\Delta\delta_{l_{3}};$

(c)
$\alpha=(1\otimes\delta_{s_{1}})\cdot(\delta_{j_{1}}\otimes\delta_{s_{2}}\cdots\delta_{s_{u}})\cdot(\delta_{j_{2}}\otimes\delta_{s_{u+1}}\cdots\delta_{s_{3}}),u=1,2,3$,
with

$1\otimes\delta_{s_{1}}\in\Delta\delta_{l_{1}},\delta_{j_{1}}\otimes\delta_{s_{2}}\cdots\delta_{s_{u}}\in\Delta\delta_{l_{2}},\delta_{j_{2}}\otimes\delta_{s_{u+1}}\cdots\delta_{s_{3}}\in\Delta\delta_{l_{3}}.$

Thus the sum of coefficient of
$\delta_{j_{1}}\delta_{j_{2}}\otimes\delta_{s_{1}}\delta_{s_{2}}\delta_{s_{3}}$
in $\binom{m}{3}\Delta\delta_{l_{1}}\Delta\delta_{l_{2}}\Delta\delta_{l_{3}}$
is

$\binom{m}{3}\binom{3}{2}\sum_{u=0}^{2}\binom{j_{1}}{u}\binom{j_{2}}{3-1-u}=\binom{m}{2+1}\binom{2+1}{2}\binom{j_{1}+j_{2}}{3-1},$

which is just the summand of the right side of (3.13) with $i=1$ in the case
$k=2,t=3$.

If $i=2,3$, by similar computations, one finds that the sum of coefficient of
$\delta_{j_{1}}\delta_{j_{2}}\otimes\delta_{s_{1}}\delta_{s_{2}}\delta_{s_{3}}$
in $\binom{m}{2+i}\Delta\delta_{l_{1}}\cdots\Delta\delta_{l_{2+i}}$ is
$\binom{m}{2+i}\binom{2+i}{2}\binom{j_{1}+j_{2}}{3-i}.$ So (3.13) holds in the
case $k=2,t=3$.

In the general case,
$\delta_{j_{1}}\cdots\delta_{j_{k}}\otimes\delta_{s_{1}}\cdots\delta_{s_{t}}$
occurs in the expansion of
$\binom{m}{k+i}\Delta(\delta_{l_{1}})\cdots\Delta(\delta_{l_{k+i}})$ for
$i=0,\cdots,t$. For some fixed $i=0,\cdots,t$, there are $\binom{k+i}{k}$
cases to write
$\delta_{j_{1}}\cdots\delta_{j_{k}}\otimes\delta_{s_{1}}\cdots\delta_{s_{t}}$
as a product of items of the form
$\delta_{j_{u}}\otimes\delta_{s_{w}}\cdots\delta_{s_{w+p}}$ ($u=1,\cdots,k$)
and $1\otimes\delta_{s_{q}}$ (there are $i$ such $1\otimes\delta_{s_{q}}$ ).
In each case, there are also sub-cases as in above example, and the sum of
coefficient of
$\delta_{j_{1}}\cdots\delta_{j_{k}}\otimes\delta_{s_{1}}\cdots\delta_{s_{t}}$
in all the sub-cases (subordinate to a given case) equals

$\sum_{w_{1},\cdots,w_{k}\geq 0}^{w_{1}+\cdots
w_{k}=t-i}\binom{j_{1}}{w_{1}}\cdots\binom{j_{k}}{w_{k}}=\binom{j_{1}+\cdots+j_{k}}{t-i}.$

So for a fixed $i$, the sum of coefficient of
$\delta_{j_{1}}\cdots\delta_{j_{k}}\otimes\delta_{s_{1}}\cdots\delta_{s_{t}}$
in the expansion of
$\binom{m}{k+i}\Delta(\delta_{l_{1}})\cdots\Delta(\delta_{l_{k+i}})$ is
$\binom{m}{k+i}\binom{k+i}{k}\binom{j_{1}+\cdots+j_{k}}{t-i}$, thus (3.13)
holds in the general case.

2\. Let

$\phi({j_{1}},\cdots,{j_{k}};{s_{1}},\cdots,{s_{t}}),\ \ (k,t\geq 0)$

be the sum of coefficient of
$\delta_{j_{1}}\cdots\delta_{j_{k}}\otimes\delta_{s_{1}}\cdots\delta_{s_{t}}$
in the expansion of all the terms in $R(p)$ of the form
$\binom{m}{u}\binom{n}{r}\Delta(\delta_{l_{1}}\cdots\delta_{l_{u}})\cdot(1\otimes\delta_{q_{1}}\cdots\delta_{q_{r}})$
where $u\geq 0,r\geq 0,{(l_{1},\cdots,l_{u};q_{1},\cdots,q_{r})}$ satisfying
$\sum l_{i}+\sum q_{i}-(u+r)=p$. see (3.12).

For example, if $k=0$ then
$\phi(\emptyset;{s_{1}},\cdots,{s_{t}})=\binom{m+n}{t}$.

Now we illustrate the equality

$\phi({j_{1}},\cdots,{j_{k}};{s_{1}},\cdots,{s_{t}})=\sum_{i=0}^{t}\binom{n}{i}\phi_{0}({j_{1}},\cdots,{j_{k}};{s_{1}},\cdots,{s_{t-i}})$
(3.15)

in the case $t=2$:

$\delta_{j_{1}}\cdots\delta_{j_{k}}\otimes\delta_{s_{1}}\delta_{s_{2}}$ occurs
in
$\binom{m}{u}\binom{n}{0}\Delta(\delta_{l_{1}}\cdots\delta_{l_{u}})\cdot(1\otimes
1)$ with coefficient $\binom{n}{0}\phi_{0}(j_{1},\cdots,j_{k};s_{1},s_{2})$;

$\delta_{j_{1}}\cdots\delta_{j_{k}}\otimes\delta_{s_{1}}\delta_{s_{2}}=\delta_{j_{1}}\cdots\delta_{j_{k}}\otimes\delta_{s_{1}}\cdot
1\otimes\delta_{s_{2}}$ occurs in
$\binom{m}{u}\binom{n}{1}\Delta(\delta_{l_{1}}\cdots\delta_{l_{u}})\cdot(1\otimes\delta_{s_{2}})$
with coefficient $\binom{n}{1}\phi_{0}(j_{1},\cdots,j_{k};s_{1})$;

$\delta_{j_{1}}\cdots\delta_{j_{k}}\otimes\delta_{s_{1}}\delta_{s_{2}}=\delta_{j_{1}}\cdots\delta_{j_{k}}\otimes
1\cdot 1\otimes\delta_{s_{1}}\delta_{s_{2}}$ occurs in
$\binom{m}{u}\binom{n}{2}\Delta(\delta_{l_{1}}\cdots\delta_{l_{u}})\cdot(1\otimes\delta_{s_{1}}\delta_{s_{2}})$
with coefficient $\binom{n}{2}\phi_{0}(j_{1},\cdots,j_{k};\emptyset)$.

These exhaust all the possibilities for
$\delta_{j_{1}}\cdots\delta_{j_{k}}\otimes\delta_{s_{1}}\delta_{s_{2}}$ to
occur in the expansion of the terms in $R(p)$ of the form
$\binom{m}{u}\binom{n}{r}\Delta(\delta_{l_{1}}\cdots\delta_{l_{u}})\cdot(1\otimes\delta_{q_{1}}\cdots\delta_{q_{r}})$,
thus (3.15) holds in the case $t=2$. It is easy to see that (3.15) holds in
the general case .

3\. Now we compute $\phi({j_{1}},\cdots,{j_{k}};{s_{1}},\cdots,{s_{t}})$.

Recall that

$\binom{u}{v+w}\binom{v+w}{w}=\binom{u}{v}\binom{u-v}{w},v,w\in\mathbb{Z}_{\geq
0},u\in\mathbb{Z}.$ (3.16)

Let $J=j_{1}+\cdots+j_{k}$. Then

$\displaystyle\phi({j_{1}},\cdots,{j_{k}};{s_{1}},\cdots,{s_{t}})$
$\displaystyle=\sum_{q=0}^{t}\phi_{0}({j_{1}},\cdots,{j_{k}};{s_{1}},\cdots,{s_{t-q}})\cdot\binom{n}{q}$
(3.17)
$\displaystyle=\sum_{q=0}^{t}[\sum_{i=0}^{t-q}\binom{m}{k+i}\binom{k+i}{k}\binom{J}{t-q-i}]\binom{n}{q}$
$\displaystyle=\sum_{q=0}^{t}[\sum_{i=0}^{t-q}\binom{m}{k}\binom{m-k}{i}\binom{J}{t-q-i}]\binom{n}{q}\
\ \ (by\ (\ref{e27}))$
$\displaystyle=\binom{m}{k}\sum_{q=0}^{t}[\sum_{i=0}^{t-q}\binom{m-k}{i}\binom{J}{t-q-i}]\binom{n}{q}$
$\displaystyle=\binom{m}{k}\sum_{q=0}^{t}\binom{m-k+J}{t-q}\binom{n}{q}$
$\displaystyle=\binom{m}{k}\binom{m+n-k+J}{t}.$

Thus one has

$\phi({j_{1}},\cdots,{j_{k}};{s_{1}},\cdots,{s_{t}})=\binom{m}{k}\binom{m+n-k+(j_{1}+\cdots
j_{k})}{t}$

and $L(p)=R(p)$ (see (3.11) and (3.12)). ∎

######  Corollary 3.9.

Assume there are additive maps $\delta_{i}:L\rightarrow L,i\in\mathbb{Z}_{\leq
0},$ which satisfy (3.7). Then there exists some unique product
$\tilde{\delta}$ on $L((T^{-1}))$ satisfying (3.5) such that
$L((T^{-1};\tilde{\delta}))$ is a deformed Laurent series ring with the
standard degree map.

###### Proof.

It has been show in last lemma that if (3.7) and (3.5) holds then for any
$a,b,c\in L,m,n,l\in\mathbb{Z}$, one has $(aT^{m}\cdot bT^{n})\cdot
cT^{l}=aT^{m}\cdot(bT^{n}\cdot cT^{l})$.

Extending the product (3.5) to $L((T^{-1}))$ by distributive law and
continuity. Then for $z_{1},z_{2},z_{3}\in L((T^{-1}))$ with
$z_{1}=\sum_{i\leq m}a_{i}T^{i},z_{2}=\sum_{j\leq
n}b_{j}T^{j},z_{3}=\sum_{k\leq l}c_{k}T^{k}$, one has

$(z_{1}z_{2})z_{3}=\sum_{i\leq m,j\leq n,k\leq l}(a_{i}T^{i}\cdot
b_{j}T^{j})c_{k}T^{k}=\sum_{i\leq m,j\leq n,k\leq l}a_{i}T^{i}(b_{j}T^{j}\cdot
c_{k}T^{k})=z_{1}(z_{2}z_{3}).$

(One has $(z_{1}z_{2})z_{3}=z_{1}(z_{2}z_{3})=0$ if one of $z_{1},z_{2},z_{3}$
is 0.)

Then it is easy to see that $L((T^{-1}))$ with this product satisfies all the
conditions in Definition 3.1, thus is a deformed Laurent series ring with the
standard degree map. The uniqueness of the product is clear.

∎

To sum up, here is the main result of this section.

######  Theorem 3.10.

Let $L$ be a field of characteristic 0. Let $L((T^{-1}))$ have the standard
degree map and the usual addition. Assume there are additive maps
$\delta_{i}:L\rightarrow L,i\in\mathbb{Z}_{\leq 0}.$ Then the following are
equivalent.

(1) There exists some unique multiplication $\tilde{\delta}$ on $L((T^{-1}))$
such that $L((T^{-1};\tilde{\delta}))$ is a deformed Laurent series ring with
the standard degree map, satisfying

$T\cdot a=aT+\sum_{i=0}^{-\infty}\delta_{i}(a)T^{i},a\in L,$ (3.18)

(2) For all $i\in\mathbb{Z}_{\leq 0}$,

$\delta_{i}(ab)=a\delta_{i}(b)+\delta_{i}(a)b+\sum_{k>0}\sum_{(l,j_{1},\cdots,j_{k})}^{j_{1}+\cdots+j_{k}-k+l=i}\binom{l}{k}\delta_{l}(a)\delta_{j_{1}}\cdots\delta_{j_{k}}(b),a,b\in
L.$ (3.19)

Assume (1) or (2) holds, then the formula for the multiplication
$\tilde{\delta}$ is

$(\sum_{i\leq m}a_{i}T^{i})(\sum_{j\leq n}b_{j}T^{j})=\sum_{i\leq m,j\leq
n}a_{i}T^{i}\cdot b_{j}T^{j},where$ $a_{i}T^{i}\cdot
b_{j}T^{j}=a_{i}b_{j}T^{i+j}+\sum_{k\geq
1}\sum_{(j_{1},\cdots,j_{k})}\binom{i}{k}a_{i}\cdot\delta_{j_{1}}\cdots\delta_{j_{k}}(b_{j})T^{j_{1}+\cdots+j_{k}-k+i+j}.$
(3.20)

Moreover, $L((T^{-1};\tilde{\delta}))$ is a topological division ring which is
complete with respect to the topology induced by the standard degree map, and
the its center is the field $K=\cap_{i\leq 0}\text{Ker}(\delta_{i})$.

###### Proof.

(1) implies (2) by Lemma 3.6. (2) implies (1) by Lemma 3.8 and Corollary 3.9.
The formula (3.20) is proven in Lemma 3.4. ∎

It is clear by (3.20) that the multiplication $\tilde{\delta}$ in
$L((T^{-1};\tilde{\delta}))$ is determined by the set of $\delta_{i},i\leq 0$
in (3.18). So one can write $\tilde{\delta}=(\delta_{i})_{i\leq 0}.$

By (3.20), the associated graded algebra of $L((T^{-1};\tilde{\delta}))$ with
respect to the standard degree map $deg$ is just the usual field $L((T^{-1}))$
of Laurent series.

The proof of the following result is direct and is omitted.

######  Lemma 3.11.

Let $L((T^{-1};\widetilde{\delta}))$ be a deformed Laurent series ring and
$\widetilde{\delta}=(\delta_{i})_{i\leq 0}$. Assume that
$w:L\rightarrow\mathbb{R}_{-\infty}$ is a degree map such that
$w(\delta_{i}(a))\leq w(a)$ for all $a\in L$ and all $i\leq 0$. Then the set
of elements $\sum_{i\leq n}a_{i}T^{i}$ in $L((T^{-1};\widetilde{\delta}))$,
such that $w(a_{i})$ is a bounded function of $i$, forms a subring of
$L((T^{-1};\widetilde{\delta}))$.

We will denote this ring by $L((T^{-1};\widetilde{\delta}))^{b}$.

Now we convert Theorem (3.10) to right $L$-vector space of Laurent seires.

Let $L_{\textbf{r}}((T^{-1}))=\\{\sum_{i\leq k}T^{i}a_{i}|a_{i}\in
L,k\in\mathbb{Z}\\}$, the right $L$-vector space of Laurent seires. One has
the map

$\phi:L((T^{-1}))\rightarrow L_{\textbf{r}}((T^{-1})),\ \ \sum_{i\leq
k}a_{i}T^{i}\mapsto\sum_{i\leq k}T^{i}a_{i}.$ (3.21)

The map

$deg_{r}:L_{\textbf{r}}((T^{-1}))\rightarrow\mathbb{Z_{-\infty}},\ \
\sum_{i\leq k}T^{i}a_{i}\mapsto k\ (a_{k}\neq 0),0\mapsto-\infty,$ (3.22)

is called the standard degree map on $L_{\textbf{r}}((T^{-1}))$. Note that
$deg_{r}=deg\cdot\phi$.

Analogous to Def. 3.1, if there exists a multiplication $\widetilde{\eta}$ on
$L_{\textbf{r}}((T^{-1}))$ such that the corresponding axioms 1-4 with the
obvious modification (for example, in Axiom 4, the degree map $deg$ is
replaced by $deg_{r}$) are satisfied, then $L_{\textbf{r}}((T^{-1}))$ is
called a right deformed Laurent series ring (with the standard degree map),
and denoted by $L_{\textbf{r}}((T^{-1};\widetilde{\eta}))$.

Assume that $L_{\textbf{r}}((T^{-1};\tilde{\eta}))$ is a right deformed
Laurent series ring. Then there exist maps $\eta_{i}:L\rightarrow
L,i\in\mathbb{Z}_{\leq 1},$ such that

$a\cdot T=T\eta_{1}(a)+\sum_{i=0}^{-\infty}T^{i}\eta_{i}(a),a\in L.$

Then $\eta_{i}$ preserve addition for all $i$, and $\eta_{1}:L\rightarrow L$
is an injective ring homomorphism.

We will always assume $\eta_{1}=1$. As $L_{\textbf{r}}((T^{-1};\tilde{\eta}))$
is associative, one has $(ab)T=a(bT)$, from which one can get following
equations (as in Lemma 3.6):

$\eta_{i}(ab)=a\eta_{i}(b)+\eta_{i}(a)b+\sum_{k>0}\sum_{(l,j_{1},\cdots,j_{k})}^{j_{1}+\cdots+j_{k}-k+l=i}\binom{l}{k}\eta_{j_{1}}\cdots\eta_{j_{k}}(a)\eta_{l}(b),a,b\in
L,i\leq 0.$ (3.23)

Now we have the following result analogous to Theorem 3.10.

######  Corollary 3.12.

Let $L$ be a field of characteristic 0. Let $L_{\textbf{r}}((T^{-1}))$ have
the standard degree map $deg_{r}$ and the usual addition. Assume there are
additive maps $\eta_{i}:L\rightarrow L,i\in\mathbb{Z}_{\leq 0}.$ Then the
following are equivalent.

(1) There exists some unique multiplication $\tilde{\eta}$ on
$L_{\textbf{r}}((T^{-1}))$ such that

$a\cdot T=Ta+\sum_{i=0}^{-\infty}T^{i}\eta_{i}(a),a\in L.$ (3.24)

(2) (3.23) holds for all $i\leq 0$ and $a,b\in L$.

Assume (3.23) holds, then one has

$(\sum_{i\leq m}T^{i}a_{i})(\sum_{j\leq n}T^{j}b_{j})=\sum_{i\leq m,j\leq
n}T^{i}a_{i}\cdot T^{j}b_{j},where$ $T^{i}a_{i}\cdot
T^{j}b_{j}=T^{i+j}a_{i}b_{j}+\sum_{k>0}\sum_{(j_{1},\cdots,j_{k})}\binom{j}{k}T^{j_{1}+\cdots+j_{k}-k+i+j}\eta_{j_{1}}\cdots\eta_{j_{k}}(a_{i})b_{j}.$

Moreover, $L_{\textbf{r}}((T^{-1};\tilde{\eta}))$ is a topological division
ring which is complete with respect to the topology induced by the degree map
$deg_{r}$, and the center of $L_{\textbf{r}}((T^{-1};\tilde{\eta}))$ is the
field $K=\cap_{i\leq 0}\text{Ker}(\eta_{i})$.

The sketch of the proof is as follows. Given
$L_{\textbf{r}}((T^{-1};\tilde{\eta}))$, there exists some
$L((T^{-1};\tilde{\delta}))$ such that the map
$\phi:L_{\textbf{r}}((T^{-1};\tilde{\eta}))\rightarrow
L((T^{-1};\tilde{\delta})),\sum_{i\leq k}T^{i}a_{i}\mapsto\sum_{i\leq
k}a_{i}T^{i}$ is an anti-isomorphism of rings, i.e. $\phi$ is a bijection
preserving addition and $\phi(f\cdot g)=\phi(g)\phi(f)$.

Applying $\phi$ on both sides of (3.24) one has

$Ta=aT+\sum_{i=0}^{-\infty}\eta_{i}(a)T^{i},a\in L.$

So, $\delta_{i}=\eta_{i}$ for all $i\leq 0.$ The above corollary can be proven
using this anti-isomorphism and Theorem 3.10.

##  4 Construction of $\widetilde{D}_{r,s}$ and $\check{D}_{r,s}$ and
embedding of $\widehat{D}_{r,s}$

Let $K$ be a field of characteristic 0.

Assume that $(r,s)\in\mathbb{Z}^{2}$ with $s\neq 0$, where $r,s$ are not
required to be coprime. Let $\nu=r+s$. Assume $\nu>0$.

Let $\alpha_{r,s},T_{r,s}$ be two indeterminants associated with $(r,s)$, and
denote $\alpha=\alpha_{r,s},T=T_{r,s}$ for simplicity.

For $k\geq 0$, let

$d_{k}=[1,\nu]_{k}/s^{k}=(\frac{\nu}{s})^{k}[\nu^{-1}]_{k}.$

In particular, $d_{0}=1,d_{1}=1/s$.

Let $L=K(\alpha)$, and $\delta_{i}\in End_{K-lin}(L)$ be defined as follows.

$\delta_{i}=\left\\{\begin{array}[]{ll}\frac{d_{k}}{k!}(\frac{d}{d\alpha})^{k},&i=1-k\nu\text{
with }k\geq 0;\\\ 0,&i\leq 1\text{ and }i\not\equiv 1\ (mod\
\nu).\end{array}\right.$ (4.1)

Note that $\delta_{1}=1$ is the identity map, and the set of linear operators
$\delta_{1-k\nu},k=0,1,2,\cdots,$ are linearly independent, and commute with
each other.

######  Proposition 4.1.

The set of linear operators $\delta_{i},i\in\mathbb{Z}_{\leq 0}$, satisfy
(3.19), so there exists a multiplication $\tilde{\delta}$ on $L((T^{-1}))$
such that

$T\cdot f=f\cdot T+\sum_{k=1}^{\infty}\delta_{1-k\nu}(f)\cdot
T^{1-k\nu}=f\cdot
T+s^{-1}f^{\prime}T^{1-\nu}+\frac{1}{2!}\frac{1-\nu}{s^{2}}f^{(2)}T^{1-2\nu}+\cdots,f\in
L,$ (4.2)

and $L((T^{-1};\tilde{\delta}))$ is a deformed Laurent series ring. The
multiplication formula in this ring is as follows:

$\sum_{i\leq m}a_{i}T^{i}\cdot\sum_{j\leq n}b_{j}T^{j}=\sum_{i\leq m,j\leq
n}a_{i}T^{i}\cdot b_{j}T^{j},\ where$ $\displaystyle a_{i}T^{i}\cdot
b_{j}T^{j}=a_{i}(T^{i}\cdot b_{j})T^{j}=a_{i}(\sum_{k\geq
0}\lambda_{i}^{k}b_{j}^{(k)}T^{i-k\nu})T^{j}=\sum_{k\geq
0}\lambda_{i}^{k}a_{i}b_{j}^{(k)}T^{i-k\nu+j},and$
$\lambda_{i}^{k}=(\frac{\nu}{s})^{k}\sum_{l=1}^{k}(-1)^{k-l}\binom{i}{l}\binom{i-l-1}{k-l}\binom{l\nu^{-1}}{k}\
for\ k>0;\lambda_{i}^{0}=1.$ (4.3)

In particular, if $(r,s)=(0,1)$ then $\lambda_{i}^{k}=\binom{i}{k}$.

Denote this ring by $\widetilde{D}_{r,s}$, and also by
$K(\alpha)((T^{-1};\tilde{\delta}_{r,s}))$.

It follows from (4.2) and (3.4) that for $f\in K(\alpha)$ and
$n\in\mathbb{Z}$, one has

$[T^{n},f]=\binom{n}{1}\sum_{k>0}\frac{d_{k}}{k!}(\frac{d}{d\alpha})^{k}f\cdot
T^{n-k\nu}+\binom{n}{2}\sum_{k_{1},k_{2}>0}\frac{d_{k_{1}}d_{k_{2}}}{k_{1}!k_{2}!}(\frac{d}{d\alpha})^{k_{1}+k_{2}}f\cdot
T^{n-(k_{1}+k_{2})\nu}+\cdots.$ (4.4)

In particular,

$[T^{n},\alpha]=ns^{-1}T^{n-\nu}.$ (4.5)

We need some preparation before we prove Proposition 4.1.

######  Lemma 4.2.

One has for $a\in\mathbb{R},m,t\geq 0$,

$\displaystyle\sum_{j=0}^{m}(-1)^{j}\binom{a}{j}\binom{a-j}{t}=(-1)^{m}\binom{a}{t}\binom{a-t-1}{m}.$
(4.6)

###### Proof.

Assume that $a\in\mathbb{Z}_{\geq m+t}$. Then

$\displaystyle\
\sum_{j=0}^{m}(-1)^{j}\binom{a}{j}\binom{a-j}{t}=\sum_{j=0}^{m}(-1)^{j}\binom{a}{a-j}\binom{a-j}{t}$
$\displaystyle=\sum_{j=0}^{m}(-1)^{j}\binom{a}{t}\binom{a-t}{a-j-t}=\binom{a}{t}\sum_{j=0}^{m}(-1)^{j}\binom{a-t}{j}$
$\displaystyle=(-1)^{m}\binom{a}{t}\binom{a-t-1}{m}.$

Thus (4.6), as a polynomial in $a$, holds for any $a\in\mathbb{Z}_{\geq m+t}$.
So it holds for any $a\in\mathbb{R}$. ∎

######  Lemma 4.3.

For any $l\geq 0$ and $p(z)\in K\left[z\right]$ with $deg\ p\leq l$, one has

$\displaystyle\sum_{t=0}^{l}(-1)^{l-t}\binom{z}{t}\binom{z-1-t}{l-t}p(t)=p(z).$
(4.7)

###### Proof.

Fix some $z\in\left\\{0,1,\cdots,l\right\\}$.

As $z-1-t<l-t$, if $\binom{z-1-t}{l-t}\neq 0$ then $z-1-t<0$, i.e., $t>z-1$.
If $\binom{z}{t}\neq 0$ then $t\leq z$.

Then $(-1)^{l-t}\binom{z}{t}\binom{z-1-t}{l-t}p(t)\neq 0$ only when $t=z,$ so
the left hand side of (4.7) equals
$(-1)^{l-z}\binom{z}{z}\binom{z-1-z}{l-z}p(z)=p(z)$.

So (4.7) holds for $z=0,1,\cdots,l$, then the lemma holds as $deg\ p\leq l$. ∎

Recall that

$\binom{a}{n}\binom{n}{k}=\binom{a}{k}\binom{a-k}{n-k},n,k\in\mathbb{Z},n\geq
k\geq 0;a\in\mathbb{R},$ (4.8)

and

$\sum_{j=0}^{n}(-1)^{j}\binom{a}{j}=(-1)^{n}\binom{a-1}{n},n\geq
0,a\in\mathbb{R}.$ (4.9)

For $(i,m)\in\mathbb{Z}^{2}$ with $m\geq i\geq 1$, let

$T(i,m)=\left\\{(j_{1},\cdots,j_{i})\in(\mathbb{Z}_{\geq
1})^{i}|\sum_{t=1}^{i}j_{t}=m\right\\}.$

######  Lemma 4.4.

For $(i,l)\in\mathbb{Z}^{2}$ with $1\leq i\leq l$, let

$\varphi(i,l)=\sum_{\left(j_{1},\cdots,j_{i}\right)\in
T(i,l)}\binom{\nu^{-1}}{j_{1}}\cdots\binom{\nu^{-1}}{j_{i}}.$

Then

$\displaystyle\varphi(i,l)=\sum_{j=0}^{i-1}(-1)^{j}\binom{i}{j}\binom{(i-j)\nu^{-1}}{l}.$
(4.10)

###### Proof.

We prove it by induction on $i$.

For $i=1,\quad\varphi(1,l)=\binom{\nu^{-1}}{l}$ for any $l\geq 1,$ then (4.10)
holds.

For $m,i\geq 1$, one has

$\displaystyle\sum_{t=0}^{m-1}(-1)^{t}\binom{i+1}{m-t}\binom{i+1-(m-t)}{t}$
$\displaystyle=\sum_{t=0}^{m-1}(-1)^{t}\binom{i+1}{m}\binom{m}{m-t}\ \ \ (by\
\ref{e60})$ $\displaystyle=\binom{i+1}{m}\sum_{t=0}^{m-1}(-1)^{t}\binom{m}{t}$
$\displaystyle=(-1)^{m-1}\binom{i+1}{m}.\ \ \ \ (by\ \ref{e61})$ (4.11)

Assume that $l>i\geq 1$,and (4.10) holds for all $\varphi(j,l)$ with $1\leq
j\leq i.$ We will show that $\varphi(i+1,l)$ satisfies (4.10).

Let $S=\\{(j_{1},\cdots,j_{i+1})\in\mathbb{Z}_{\geq
0}|j_{1}+\cdots+j_{i+1}=l\\}.$ For $k=0,1,\cdots,i$, let

$S_{k}=\\{(j_{1},\cdots,j_{i+1})\in S|\text{exactly k elements in}\
j_{1},\cdots,j_{i+1}\ \text{are 0}\\}.$

One has

$\displaystyle\binom{(i+1)\nu^{-1}}{l}=\sum_{(j_{1},\cdots,j_{i+1})\in
S}\binom{\nu^{-1}}{j_{1}}\cdots\binom{\nu^{-1}}{j_{i+1}}$
$\displaystyle=(\sum_{(j_{1},\cdots,j_{i+1})\in
S_{0}}+\sum_{(j_{1},\cdots,j_{i+1})\in
S_{1}}+\cdots+\sum_{(j_{1},\cdots,j_{i+1})\in
S_{i}})\binom{\nu^{-1}}{j_{1}}\cdots\binom{\nu^{-1}}{j_{i+1}}$
$\displaystyle=\binom{i+1}{0}\varphi(i+1,l)+\binom{i+1}{1}\varphi(i,l)+\cdots+\binom{i+1}{i}\varphi(1,l).$

Then

$\displaystyle\
\varphi(i+1,l)=\binom{(i+1)\nu^{-1}}{l}-\sum_{j=1}^{i}\binom{i+1}{j}\varphi(i+1-j,l)$
$\displaystyle=\binom{(i+1)\nu^{-1}}{l}-\sum_{j=1}^{i}\binom{i+1}{j}\cdot\sum_{t=0}^{i-j}(-1)^{t}\binom{i+1-j}{t}\binom{(i+1-j-t)\nu^{-1}}{l}$
$\displaystyle=\binom{(i+1)\nu^{-1}}{l}-\sum_{j=1}^{i}\sum_{t=0}^{i-j}(-1)^{t}\binom{i+1}{j}\binom{i+1-j}{t}\binom{(i+1-j-t)\nu^{-1}}{l}$
( let $m=j+t$)
$\displaystyle=\binom{(i+1)\nu^{-1}}{l}-\sum_{m=1}^{i}\left[\sum_{t=0}^{m-1}(-1)^{t}\binom{i+1}{m-t}\binom{i+1-(m-t)}{t}\right]\binom{(i+1-m)\nu^{-1}}{l}$
$\displaystyle=\binom{(i+1)\nu^{-1}}{l}-\sum_{m=1}^{i}(-1)^{m+1}\binom{i+1}{m}\cdot\binom{(i+1-m)\nu^{-1}}{l}\
\ (by\ \ref{e62})$
$\displaystyle=\sum_{m=0}^{i}(-1)^{m}\binom{i+1}{m}\binom{(i+1-m)\nu^{-1}}{l},$

so $\varphi(i+1,l)$ satisfies (4.10). ∎

Now we prove Proposition 4.1.

###### Proof.

By (3.19) (see also (3.9)) we need to show that for $n\leq 0,a,b\in L$,

$\begin{aligned} \delta_{n}(ab)=a\delta_{n}(b)+\sum_{n<l\leq
0}\delta_{l}(a)\cdot\sum_{k=1}^{l-n}\binom{l}{k}[\sum_{(j_{1},\cdots,j_{k})}^{j_{1}+\cdots+j_{k}=n+k-l}\delta_{j_{1}}\cdots\delta_{j_{k}}(b)]+\delta_{n}(a)b\end{aligned}.$
(4.12)

First assume that $n\not\equiv 1\ (mod\ \nu)$. If some
$\delta_{l}(a)\delta_{j_{1}}\cdots\delta_{j_{k}}(b)$ in right hand side of
(4.12) with $l+j_{1}+\cdots+j_{k}=n+k$ is not identically 0, then
$l+j_{1}+\cdots+j_{k}\equiv k+1\ (mod\ \nu)$, while $n+k\not\equiv k+1\ (mod\
\nu)$, which is impossible. So in the case $n\not\equiv 1\ (mod\ \nu)$,
$\delta_{n}=0$ and both sides of (4.12) are identically 0, thus (4.12) is
fulfilled.

From now on we assume that $n=1-k\nu$ for some $k\geq 1$.

For $k\geq 1,a,b\in L$, one has

$\displaystyle D_{k}(ab)=\sum_{j=0}^{k}D_{j}(a)D_{k-j}(b),where\
D_{j}=\frac{1}{j!}(d/d\alpha)^{j}.$

As $D_{k}={d_{k}}^{-1}\cdot\delta_{1-k\nu}$, one has

$\delta_{1-k\nu}(ab)=\sum_{j=0}^{k}\frac{d_{k}}{d_{j}\cdot
d_{k-j}}\delta_{1-j\nu}(a)\delta_{1-(k-j)\nu}(b).$ (4.13)

Assume $\sum_{i=1}^{k}j_{i}=u$ with each $j_{i}\geq 1$. Then by (4.1),

$\displaystyle\delta_{1-j_{1}\nu}\cdots\delta_{1-j_{k}\nu}=(\frac{\nu}{s})^{u}\binom{\nu^{-1}}{j_{1}}\cdots\binom{\nu^{-1}}{j_{k}}(\frac{d}{d\alpha})^{u}.$
(4.14)

We need to show that (4.12) holds for $n=1-k\nu$ for $k\geq 1$. As the case
$k=1$ is clear, we need to prove the following identity for $k\geq 2,a,b\in
L$:

$\delta_{1-k\nu}(ab)=\left(a\delta_{1-k\nu}(b)+\delta_{1-k\nu}(a)b\right)+\sum_{l=1}^{k-1}\delta_{1-l\nu}(a)\left[\sum_{i=1}^{k-l}\binom{1-l\nu}{i}\sum_{(j_{1},\cdots,j_{i})\in
T(i,k-l)}\delta_{1-j_{1}\nu}\cdots\delta_{1-j_{i}\nu}(b)\right].$

By (4.1) the right hand side equals

$\left(a\delta_{1-k\nu}(b)+\delta_{1-k\nu}(a)b\right)+\sum_{l=1}^{k-1}\delta_{1-l\nu}(a)\sum_{i=1}^{k-l}\binom{1-l\nu}{i}(\frac{\nu}{s})^{k-l}\left(\sum_{(j_{1},\cdots,j_{i})\in
T(i,k-l)}\binom{\nu^{-1}}{j_{1}}\cdots\binom{\nu^{-1}}{j_{i}}\right)\left(\frac{d}{d\alpha}\right)^{k-l}(b).$

Then by (4.13) one only need to show for $k\geq 2$ and $1\leq l\leq k-1$,

$\sum_{i=1}^{k-l}\binom{1-l\nu}{i}(\frac{\nu}{s})^{k-l}\left(\sum_{(j_{1},\cdots,j_{i})\in
T(i,k-l)}\binom{\nu^{-1}}{j_{1}}\cdots\binom{\nu^{-1}}{j_{i}}\right)\left(\frac{d}{d\alpha}\right)^{k-l}(b)=\frac{d_{k}}{d_{l}\cdot
d_{k-l}}\delta_{1-(k-l)\nu}(b),$

or,

$\displaystyle\sum_{i=1}^{k-l}\binom{1-l\nu}{i}(\frac{\nu}{s})^{k-l}\left(\sum_{(j_{1},\cdots,j_{i})\in
T(i,k-l)}\binom{\nu^{-1}}{j_{1}}\cdots\binom{\nu^{-1}}{j_{i}}\right)\left(\frac{d}{d\alpha}\right)^{k-l}=\frac{d_{k}}{d_{l}\cdot
d_{k-l}}\delta_{1-(k-l)\nu}.$ (4.15)

As $\delta_{1-(k-l)\nu}=d_{k-l}\cdot D_{k-l},$ right hand side of (4.15) is

$\displaystyle(\frac{\nu}{s})^{k-l}\frac{\left[\nu^{-1}\right]_{k}}{\left[\nu^{-1}\right]_{l}}\cdot\frac{1}{(k-l)!}\left(\frac{d}{d\alpha}\right)^{k-l}=(\frac{\nu}{s})^{k-l}\binom{\nu^{-1}-l}{k-l}\left(\frac{d}{d\alpha}\right)^{k-l},$

so (4.15) is equivalent to

$\displaystyle\sum_{i=1}^{k-l}\binom{1-l\nu}{i}\cdot\sum_{\left(j_{1},\cdots,j_{i}\right)\in
T(i,k-l)}\binom{\nu^{-1}}{j_{1}}\cdots\binom{\nu^{-1}}{j_{i}}=\binom{\nu^{-1}-l}{k-l}.$
(4.16)

By Lemma 4.4, the left hand side of (4.16) is

$\sum_{i=1}^{k-l}\binom{1-l\nu}{i}\sum_{j=0}^{i-1}(-1)^{j}\binom{i}{j}\binom{(i-j)\nu^{-1}}{k-l}=\sum_{i=1}^{k-l}\sum_{j=0}^{i-1}(-1)^{j}\binom{1-l\nu}{i}\binom{i}{j}\binom{(i-j)\nu^{-1}}{k-l}.$

So (4.16) is equivalent to

$\sum_{i=1}^{k-l}\sum_{j=0}^{i-1}(-1)^{j}\binom{1-l\nu}{i}\binom{i}{j}\binom{(i-j)\nu^{-1}}{k-l}=\binom{\nu^{-1}-l}{k-l}.$

Replacing $k-l$ by $u$, it is

$\displaystyle\sum_{i=1}^{u}\sum_{j=0}^{i-1}(-1)^{j}\binom{1-l\nu}{i}\binom{i}{j}\binom{(i-j)\nu^{-1}}{u}=\binom{\nu^{-1}-l}{u},\
\ u\geq 1.$ (4.17)

The left hand side of (4.17) equals

$\displaystyle\sum_{i=1}^{u}\sum_{j=0}^{i-1}(-1)^{j}\binom{1-l\nu}{j}\binom{1-l\nu-j}{i-j}\binom{(i-j)\nu^{-1}}{u}\
\ \ \ \text{(Let $t=i-j$)}$
$\displaystyle=\sum_{t=1}^{u}\left(\sum_{j=0}^{u-t}(-1)^{j}\binom{1-l\nu}{j}\binom{1-l\nu-j}{t}\right)\binom{t\nu^{-1}}{u}$
$\displaystyle=\sum_{t=1}^{u}(-1)^{u-t}\binom{1-l\nu}{t}\binom{-l\nu-t}{u-t}\binom{t\nu^{-1}}{u}\
\ (\text{by Lemma }\ref{fl3}\text{ with $m=u-t$})$
$\displaystyle=\sum_{t=0}^{u}(-1)^{u-t}\binom{1-l\nu}{t}\binom{-l\nu-t}{u-t}\binom{t\nu^{-1}}{u}.$

So (4.17) is equivalent to

$\displaystyle\sum_{t=0}^{u}(-1)^{u-t}\binom{1-l\nu}{t}\binom{-l\nu-t}{u-t}\binom{t\nu^{-1}}{u}=\binom{\nu^{-1}-l}{u},\
\ u\geq 1.$ (4.18)

Set $z=1-l\nu,\ p(z)=\binom{\nu^{-1}z}{u}$ in (4.7), we get (4.18).

Now we prove the formula (4.3) in Proposition 4.1.

By (3.4), for $b\in K(\alpha),n\in\mathbb{Z}$ one has

$\displaystyle T^{n}b$ $\displaystyle=bT^{n}+\binom{n}{1}\sum_{k\geq
1}\frac{d_{k}}{k!}b^{(k)}T^{n-k\nu}+\binom{n}{2}\sum_{k_{1},k_{2}\geq
1}\frac{d_{k_{1}}}{k_{1}!}\frac{d_{k_{2}}}{k_{2}!}b^{(k_{1}+k_{2})}T^{n-(k_{1}+k_{2})\nu}+\cdots$
$\displaystyle=\sum_{k\geq 0}\lambda_{n}^{k}b^{(k)}T^{n-k\nu},\ \
\lambda_{n}^{0}=1.$

(Note that $\lambda_{n}^{k}$ is not the $k$-th power of $\lambda_{n}$.)
Comparing the coefficients of $T^{n-k\nu}$, one finds that if $k>0$ then

$\displaystyle\lambda_{n}^{k}$
$\displaystyle=\sum_{r=1}^{k}\left[\binom{n}{r}\cdot\sum_{\begin{subarray}{c}s_{1}+s_{2}+...+s_{r}=k\\\
s_{1},s_{2},...,s_{r}\geq
1\end{subarray}}\frac{d_{s_{1}}}{s_{1}!}\frac{d_{s_{2}}}{s_{2}!}\cdots\frac{d_{s_{r}}}{s_{r}!}\right]=\sum_{r=1}^{k}\binom{n}{r}\cdot\left(\dfrac{\nu}{s}\right)^{k}\cdot\sum_{\begin{subarray}{c}s_{1}+...+s_{r}=k\\\
s_{1},...,s_{r}\geq
1\end{subarray}}\binom{\nu^{-1}}{s_{1}}\cdots\binom{\nu^{-1}}{s_{r}}$
$\displaystyle=\left(\dfrac{\nu}{s}\right)^{k}\sum_{r=1}^{k}\sum_{j=0}^{r-1}(-1)^{j}\binom{n}{r}\binom{r}{j}\binom{(r-j)\nu^{-1}}{k}\
\ \ \ (by\ lemma\ 4.4)$
$\displaystyle=\left(\dfrac{\nu}{s}\right)^{k}\sum_{i=1}^{k}\sum_{j=0}^{k-i}(-1)^{j}\binom{n}{i+j}\binom{i+j}{j}\binom{i\nu^{-1}}{k}\
\ \ \ \ \ (i=r-j)$
$\displaystyle=\left(\dfrac{\nu}{s}\right)^{k}\sum_{i=1}^{k}\sum_{j=0}^{k-i}(-1)^{j}\binom{n}{i}\binom{n-i}{j}\binom{i\nu^{-1}}{k}$
$\displaystyle=\left(\dfrac{\nu}{s}\right)^{k}\sum_{i=1}^{k}\left[\binom{n}{i}\binom{i\nu^{-1}}{k}\sum_{j=0}^{k-i}(-1)^{j}\binom{n-i}{j}\right]$
$\displaystyle=\left(\dfrac{\nu}{s}\right)^{k}\sum_{i=1}^{k}(-1)^{k-i}\binom{n}{i}\binom{n-i-1}{k-i}\binom{i\nu^{-1}}{k}.\
\ \ \ (by\ (\ref{e61})).$

∎

Note that this proposition also holds for $L=K((\alpha^{-1}))$.

Assume $\left(r,s\right)\in\mathbb{Z}^{2},s\neq 0,\nu=r+s,\nu>0.$ Let
$\alpha=\alpha_{r,s},T=T_{r,s}$ and
$\widetilde{D}_{r,s}=L((T^{-1};\widetilde{\delta}_{r,s}))$ with
$L=K\left(\alpha\right).$ We also write
$\widetilde{D}_{r,s}=K(\alpha)\left(\left(T^{-1};\widetilde{\delta}_{r,s}\right)\right)$.
Next we extend the maps $\mathsf{v}_{r,s}$ and $\mathsf{f}_{r,s}$ to
$\widetilde{D}_{r,s}$.

Let

$\mathsf{v}_{r,s}:\ \widetilde{D}_{r,s}\to\mathbb{Z}_{-\infty},\
0\mapsto-\infty;T\mapsto 1;z=\sum_{i\leq
n}a_{i}\left(\alpha\right)T^{i}\mapsto n.$

Note that when we write some element in $\widetilde{D}_{r,s}$ as
$z=\sum_{i\leq n}a_{i}(\alpha)T^{i}$, we will always assume that $a_{n}\neq
0$. Similar conventions will be used henceforth. It is clear that
$\mathsf{v}_{r,s}$ is a degree map on $\widetilde{D}_{r,s}$.

One has $\mathsf{v}_{r,s}\left(T\right)=1$ and
$\mathsf{v}_{r,s}\left(a(\alpha)\right)=0$ for $0\neq a(\alpha)\in K(\alpha)$.

Let $\widetilde{D}_{r,s}^{\leq i}=\mathsf{v}_{r,s}^{-1}(\mathbb{Z}_{\leq
i}\cup\\{-\infty\\})$ for $i\in\mathbb{Z}$. There is a natural filtration on
$\widetilde{D}_{r,s}$

$\cdots\subset\widetilde{D}_{r,s}^{\leq i-1}\subset\widetilde{D}_{r,s}^{\leq
i}\subset\widetilde{D}_{r,s}^{\leq i+1}\subset\cdots$

By (4.4) one has for $i,j\in\mathbb{Z}$,

$[f(\alpha)T^{i},g(\alpha)T^{j}]=f(\alpha)[T^{i},g(\alpha)]T^{j}+g(\alpha)[f(\alpha),T^{j}]T^{i}\in
K(\alpha)\cdot T^{i+j-\nu},$

thus

$[\widetilde{D}_{r,s}^{\leq i},\widetilde{D}_{r,s}^{\leq
j}]\subseteq\widetilde{D}_{r,s}^{\leq i+j-\nu}.$

As $\nu\geq 1$, the associated graded algebra
$gr(\widetilde{D}_{r,s})=\oplus_{i\in\mathbb{Z}}\
gr_{i}(\widetilde{D}_{r,s}),gr_{i}(\widetilde{D}_{r,s})=\widetilde{D}_{r,s}^{\leq
i}/\widetilde{D}_{r,s}^{\leq i-1}$, is commutative.

Now we show that (See [A2009] for the definition of non-commutative Henselian
rings.)

######  Proposition 4.5.

The ring $\widetilde{V}_{r,s}$ is a Henselian ring.

This result is a consequence of the following lemma.

Assume $D$ is a division ring, $u:D\to\mathbb{Z}_{-\infty}$ is a degree map.
Let $V=D^{\leq 0}$ be its valuation ring. Then $I=D^{\leq-1}$ is the unique
maximal ideal of $V$, thus $V$ is a local ring. The degree map $u$ induces a
filtration on $D$:

$\cdots\subset D^{\leq i-1}\subset D^{\leq i}\subset D^{\leq
i+1}\subset\cdots,i\in\mathbb{Z},$

with the associated graded algebra
$gr(D)=\bigoplus_{i\in\mathbb{Z}}gr_{i}(D),gr_{i}(D)=D^{\leq i}/D^{\leq i-1}$.

######  Lemma 4.6.

Assume that

(1) $D$ is a division ring complete with respect to $u$;

(2) $[D^{\leq i},D^{\leq j}]\subseteq D^{\leq i+j-1}$, $\forall
i,j\in\mathbb{Z}$.

Then the valuation ring $V$ is a Henselian ring.

###### Proof.

One has that $V$ is a local ring with the unique maximal ideal of $I$,
$\cap_{n\in\mathbb{Z}}I^{n}=\\{0\\}$. The degree map $u$ restricts to the
degree map $u|_{V}$ on $V$, which induces the corresponding filtration on $V$.
By (2), $gr(D)$ is commutative, thus $gr(V)$ is also commutative. It follows
that the residue field $V/I$ is also commutative.

As $D$ is complete with respect to $u$, $V$ is also complete with respect to
$u|_{V}$.

By Theorem 2.1 of [A2009], $V$ is Henselian.∎

Let

$\mathsf{f}_{r,s}:\widetilde{D}_{r,s}\to K(X,Y^{1/s}),z=\sum_{i\leq
n}a_{i}(\alpha)T^{i}\mapsto a_{n}(XY^{-r/s})Y^{n/s},0\mapsto 0.$

Its image is
$\mathsf{f}_{r,s}(\widetilde{D}_{r,s})=\bigcup_{n\in\mathbb{Z}}K(XY^{-r/s})Y^{n/s}$,
and $\mathsf{f}_{r,s}(\widetilde{D}_{r,s})\setminus\\{0\\}$ is a
multiplicative group.

######  Proposition 4.7.

For $z_{1},z_{2}\in\widetilde{D}_{r,s}$, one has

(1)
$\mathsf{f}_{r,s}(z_{1}z_{2})=\mathsf{f}_{r,s}(z_{1})\mathsf{f}_{r,s}(z_{2}).$

(2) For $\lambda\in
K,\mathsf{f}_{r,s}(\lambda)=\lambda;\mathsf{f}_{r,s}(\lambda
z)=\lambda\mathsf{f}_{r,s}(z).$

(3)$\mathsf{f}_{r,s}:\widetilde{D}_{r,s}\to K(X,Y^{1/s})$ is a multiplicative
homomorphism. In particular,

$\mathsf{f}_{r,s}(z^{-1})={\mathsf{f}_{r,s}(z)}^{-1},z\neq 0.$

(4) If
$\mathsf{v}_{r,s}(z_{1})=\mathsf{v}_{r,s}(z_{2})=\mathsf{v}_{r,s}(z_{1}+z_{2})$,
then
$\mathsf{f}_{r,s}(z_{1}+z_{2})=\mathsf{f}_{r,s}(z_{1})+\mathsf{f}_{r,s}(z_{2}).$

###### Proof.

(1) Assume $z_{1}=\sum_{i\leq n}a_{i}(\alpha)T^{i},z_{2}=\sum_{j\leq
m}b_{j}(\alpha)T^{j}$. Then

$z_{1}z_{2}=a_{n}(\alpha)b_{m}(\alpha)T^{n+m}+\sum_{i<n+m}c_{i}(\alpha)T^{i},$

thus
$\mathsf{f}_{r,s}(z_{1}z_{2})=\mathsf{f}_{r,s}(z_{1})\mathsf{f}_{r,s}(z_{2})$.

(2) is clear.

(3) By (2) $\mathsf{f}_{r,s}(1)=1$. Then it follows from (1) that
$\mathsf{f}_{r,s}$ is a multiplicative homomorphism.

(4) is clear.

∎

Now we embed $D_{1}$ into $\widetilde{D}_{r,s}$. By (4.5) one has

$[T,\alpha]=s^{-1}T^{1-\nu},\ so$ $\left[T^{s},\alpha
T^{r}\right]=[T^{s},\alpha]T^{r}=sT^{s-1}[T,\alpha]T^{r}=sT^{s-1}(s^{-1}T^{1-\nu})T^{r}=T^{s-\nu+r}=1.$

So there exists a unique $K$-algebra homomorphism

$\eta:A_{1}\to\widetilde{D}_{r,s},\textbf{q}\mapsto
T^{s},\textbf{p}\mapsto\alpha T^{r},$

which is injective as $A_{1}$ is simple . Since $\widetilde{D}_{r,s}$is a
division ring , $\eta$ can be extented to an injective $K$-homomorphism of
division algebras

$\eta:D_{1}\to\widetilde{D}_{r,s},\ zw^{-1}\mapsto\eta(z)\eta(w)^{-1}.$

Henceforth we will identify $D_{1}$ as a subalgebra of $\widetilde{D}_{r,s}$
by $\eta$, and we denote

$T=\textbf{q}^{1/s},\alpha=\textbf{p}T^{-r}=\textbf{pq}^{-r/s}.$

Let $\widehat{D}_{r,s}$ be the completion of the $K$-division algebra $D_{1}$
in $\widetilde{D}_{r,s}$. As $\widetilde{D}_{r,s}$ is complete,
$\widehat{D}_{r,s}$ is a complete $K$-division subalgebra of
$\widetilde{D}_{r,s}$. Thus one has

$A_{1}\subset D_{1}\subset\widehat{D}_{r,s}\subseteq\widetilde{D}_{r,s}.$

The map $\mathsf{f}_{r,s}$ restricts to a multiplicative homomorphism
$\mathsf{f}_{r,s}:D_{1}\to K(X,Y^{1/s}).$ One has

$\mathsf{f}_{r,s}(q)=\mathsf{f}_{r,s}(T^{s})=Y,\mathsf{f}_{r,s}(p)=\mathsf{f}_{r,s}(\alpha
T^{r})=XY^{-r/s}Y^{r/s}=X.$

If $z\in A_{1},\mathsf{f}_{r,s}(z)\in K[X,Y]$. Thus if $z,w\in A_{1}$ with
$w\neq 0$, then $\mathsf{f}_{r,s}(zw^{-1})\in K(X,Y).$So we have the following
commutative diagram:

${D_{1}}$${K(X,Y)}$${\widetilde{D}_{r,s}}$${K(X,Y^{1/s})}$$\scriptstyle{\mathsf{f}_{r,s}}$$\scriptstyle{\mathsf{f}_{r,s}}$

Note that if $(r,s)=(0,1)$, then
$\widetilde{D}_{0,1}=\widehat{D}_{0,1}=K(\textbf{p})((\textbf{q}^{-1};\delta))$.

It is clear that $\mathsf{v}_{r,s}$ restrict to a degree map on
$\widehat{D}_{r,s}$, the associated graded algebra $gr(\widehat{D}_{r,s})$ of
$\widehat{D}_{r,s}$ with respect to this degree map is commutative, and the
valuation ring $\widehat{V}_{r,s}=\widehat{D}_{r,s}\cap\widetilde{V}_{r,s}.$
By Lemma 4.6, one has

######  Corollary 4.8.

The valuation ring $\widehat{V}_{r,s}$ of $\widehat{D}_{r,s}$ is a Henselian
ring.

Next we give the result parallel to Proposition 4.1.

Assume that $(r,s)\in\mathbb{Z}^{2}$ with $r\neq 0$, where $r,s$ are not
required to be coprime. Let $\nu=r+s$. Assume $\nu>0$.

Let $d^{\prime}_{k}=(\nu/r)^{k}[\nu^{-1}]_{k}$ for $k>0$, and
$d^{\prime}_{0}=1$. Let $L=K(\beta)$ with $\beta$ being an indeterminant,
$\eta_{i}\in End_{K-lin}(L)$ be defined as follows:

$\eta_{i}=\left\\{\begin{array}[]{ll}\frac{d^{\prime}_{k}}{k!}(\frac{d}{d\beta})^{k},&i=1-k\nu\text{
with }k\geq 0;\\\ 0,&i\leq 1\text{ and }i\not\equiv 1\ (mod\
\nu).\end{array}\right.$

######  Corollary 4.9.

The set of linear operators $\eta_{i},i\in\mathbb{Z}_{\leq 0}$, satisfy
(3.19), so there exists a multiplication $\tilde{\eta}$ on
$L_{\textbf{r}}((T^{-1}))$ such that

$f\cdot T=T\cdot f+\sum_{k=1}^{\infty}T^{1-k\nu}\cdot\eta_{1-k\nu}(f)=T\cdot
f+r^{-1}T^{1-\nu}f^{\prime}+\frac{1}{2!}\frac{1-\nu}{r^{2}}T^{1-2\nu}f^{(2)}+\cdots,f\in
L,$

and $L_{\textbf{r}}((T^{-1};\tilde{\eta}))$ is a deformed Laurent series ring,
with the multiplication

$\sum_{i\leq m}T^{i}a_{i}\cdot\sum_{j\leq n}T^{j}b_{j}=\sum_{i\leq m,j\leq
n}T^{i}a_{i}\cdot T^{j}b_{j},\ where$ $\displaystyle T^{i}a_{i}\cdot
T^{j}b_{j}=\sum_{k\geq 0}\lambda_{j}^{k}T^{j-k\nu+i}a_{i}^{(k)}b_{j},$ (4.19)

and $\lambda_{j}^{k}$ is defined as in (4.3).

Denote this ring by $\check{D}_{r,s}$, and also by
$K(\beta)_{\textbf{r}}\left(\left(T^{-1};\tilde{\eta}_{r,s}\right)\right)$.

It satisfies that for $f\in L,n\in\mathbb{Z}$,

$[f,T^{n}]=\binom{n}{1}\sum_{k>0}\frac{d^{\prime}_{k}}{k!}T^{n-k\nu}\cdot(\frac{d}{d\beta})^{k}f+\binom{n}{2}\sum_{k_{1},k_{2}>0}\frac{d^{\prime}_{k_{1}}d^{\prime}_{k_{2}}}{k_{1}!k_{2}!}T^{n-(k_{1}+k_{2})\nu}\cdot(\frac{d}{d\beta})^{k_{1}+k_{2}}f+\cdots$

In particular,

$[\beta,T^{n}]=nr^{-1}T^{n-\nu}.$

Then

$[T^{s}\beta,T^{r}]=T^{s}[\beta,T^{r}]=T^{s}rr^{-1}T^{r-\nu}=1.$

So there exists a unique $K-$algebra homomorphism

$\xi:A_{1}\rightarrow K(\beta)_{\textbf{r}}((T^{-1};\tilde{\eta}_{r,s})),\ \
\textbf{p}\mapsto T^{r},\textbf{q}\mapsto T^{s}\beta,$

which can be extended to an injective $K$-homomorphism from $D_{1}$ to
$\check{D}_{r,s}$. We identify $D_{1}$ as a subalgebra of
$K(\beta)_{\textbf{r}}((T^{-1};\tilde{\eta}_{r,s}))$ by $\xi$, and we denote

$T=\textbf{p}^{1/r},\beta=\textbf{p}^{-s/r}\textbf{q}.$

Analogously, one defines

$u_{r,s}:\ \check{D}_{r,s}\to\mathbb{Z}_{-\infty},\ 0\mapsto-\infty;T\mapsto
1;z=\sum_{i\leq n}T^{i}b_{i}\left(\beta\right)\mapsto n,$

which is a degree map on $\check{D}_{r,s}$. The associated graded algebra
$gr(\check{D}_{r,s})$ of $\check{D}_{r,s}$ with respect to $u_{r,s}$ is
commutative, and the valuation ring
$\check{V}_{r,s}:={u_{r,s}}^{-1}(\mathbb{Z}_{\leq 0}\cup\\{-\infty\\})$ of
$\check{D}_{r,s}$ is a Henselian ring.

It is clear that the completion of $D_{1}$ in $\check{D}_{r,s}$ with respect
to $u_{r,s}$ is isomorphic to $\widehat{D}_{r,s}$.

##  5 Topological generators of $\widehat{D}_{r,s}$

Let $K$ be a field of characteristic 0. Let $A_{1}=A_{1}(K),D_{1}=D_{1}(K)$.

Assume that $(r,s)\in\mathbb{Z}^{2}$, $s\neq 0$, and $r$ , $s$ are coprime.
Let $\nu=r+s>0$. Choose $i,j\in\mathbb{Z}$ such that $ri+sj=1$. Let

$T_{0}=\textbf{p}^{i}\textbf{q}^{j},\ \
\alpha_{0}=\textbf{p}^{s}\textbf{q}^{-r}.$

One has

$\binom{s\ \ \ i}{-r\ \ j}\binom{j\ \ -i}{r\ \ s}=\binom{1\ \ 0}{0\ \ 1},$
(5.1)

which can be applied to simplify some of the calculations later.

One has $\mathsf{v}_{r,s}(T_{0})=1,\mathsf{v}_{r,s}(\alpha_{0})=0.$ Let

$R=\left\\{\ \sum_{i\leq n}a_{i}(\alpha_{0})T_{0}^{i}\ |\ a_{i}\in
K(X),n\in\mathbb{Z}.\right\\}\cup\\{0\\}.$

One has $R\subseteq\widetilde{D}_{r,s}$, and it is clear that $R$ is complete.
We will show that $R=\widehat{D}_{r,s}$, thus $\alpha_{0},T_{0}$ can be viewed
as a pair of topological generators of $\widehat{D}_{r,s}$.

######  Lemma 5.1.

Assume $z_{i}=\textbf{p}^{l_{i}}\textbf{q}^{k_{i}}\in A_{1}$, $i=1,\cdots,t$
and $t\geq 2$. Let $l=\sum_{i=1}^{t}l_{i},k=\sum_{i=1}^{t}k_{i}$. Then

$\Pi_{i=1}^{t}z_{i}=\textbf{p}^{l}\textbf{q}^{k}+\sum_{m\geq
1}\mu_{m}\textbf{p}^{l-m}\textbf{q}^{k-m},\mu_{m}\in\mathbb{Q}.$

This follows from (2.2).

######  Lemma 5.2.

$R$ is a complete sub-division-ring of $\widetilde{D}_{r,s}$.

###### Proof.

We only need to prove that $R$ is closed under multiplication and inversion.
First we show that $R$ is closed under multiplication.

One has

$\displaystyle\textbf{p}^{i}\textbf{q}^{j}\cdot\textbf{p}^{u}\textbf{q}^{t}=\textbf{p}^{i+u}\textbf{q}^{j+t}+\sum_{k\geq
1}\frac{[j]_{k}[u]_{k}}{k!}\textbf{p}^{i+u-k}\textbf{q}^{j+t-k},i,j,u,t\in\mathbb{Z}.$
(5.2)

Then for $k,l\in\mathbb{Z}$,

$\alpha_{0}^{k}T_{0}^{l}=\textbf{p}^{ks+li}\textbf{q}^{-kr+lj}+\sum_{m\geq
1}\lambda_{m}\textbf{p}^{ks+li-m}\textbf{q}^{-kr+lj-m},$
$T_{0}^{l}\alpha_{0}^{k}=\textbf{p}^{ks+li}\textbf{q}^{-kr+lj}+\sum_{m\geq
1}\mu_{m}\textbf{p}^{ks+li-m}\textbf{q}^{-kr+lj-m},\lambda_{m},\mu_{m}\in\mathbb{Q}.$

Then

$T_{0}^{l}\alpha_{0}^{k}=\alpha_{0}^{k}T_{0}^{l}+\sum_{m\geq
1}\xi_{m}\textbf{p}^{ks+li-m}\textbf{q}^{-kr+lj-m},\xi_{m}=\mu_{m}-\lambda_{m}.$
(5.3)

As

$\alpha_{0}^{-(j-i)}T_{0}^{-\nu}=\textbf{p}^{-1}\textbf{q}^{-1}+\sum_{m\geq
1}\mu_{m}\textbf{p}^{-1-m}\textbf{q}^{-1-m},$

one has

$\alpha_{0}^{k-m(j-i)}T_{0}^{l-m\nu}=\textbf{p}^{ks+li-m}\textbf{q}^{-kr+lj-m}+\sum_{s\geq
1}\nu_{s}\textbf{p}^{ks+li-m-s}\textbf{q}^{-kr+lj-m-s},$

and

$\textbf{p}^{ks+li-m}\textbf{q}^{-kr+lj-m}=\alpha_{0}^{k-m(j-i)}T_{0}^{l-m\nu}-\sum_{s\geq
1}\nu_{s}\textbf{p}^{ks+li-m-s}\textbf{q}^{-kr+lj-m-s},\nu_{s}\in\mathbb{Q}.$
(5.4)

Substitute (5.4) into (5.3) for $m=1,2,3,\cdots,$ successively, one has

$T_{0}^{l}\alpha_{0}^{k}=\alpha_{0}^{k}T_{0}^{l}+\sum_{m\geq
1}\epsilon_{m}\alpha_{0}^{k-m(j-i)}T_{0}^{l-m\nu},\epsilon_{m}\in\mathbb{Q}.$

Thus for any $f\in K[X],n\in\mathbb{Z}$, there exist $f_{n,m}\in K(X)$ such
that

$T_{0}^{n}f(\alpha_{0})=f(\alpha_{0})T_{0}^{n}+\sum_{m\geq
1}f_{n,m}(\alpha_{0})T_{0}^{n-m\nu}.$ (5.5)

Now assume $f\neq 0$. Then multiply $f(\alpha_{0})^{-1}$ on both sides of
(5.5) and reorder the terms, one has

$T_{0}^{n}f(\alpha_{0})^{-1}=f(\alpha_{0})^{-1}T_{0}^{n}+\sum_{m\geq
1}g_{n,m}(\alpha_{0})T_{0}^{n-m\nu}f(\alpha_{0})^{-1},$ (5.6)

where $g_{n,m}\in
K(X),g_{n,m}(\alpha_{0})=-f(\alpha_{0})^{-1}f_{n,m}(\alpha_{0})$. Then one has

$T_{0}^{n-m\nu}f(\alpha_{0})^{-1}=f(\alpha_{0})^{-1}T_{0}^{n-m\nu}+\sum_{k\geq
1}g_{n-m\nu,k}(\alpha_{0})T_{0}^{n-(m+k)\nu}f(\alpha_{0})^{-1},$ (5.7)

substitute $T_{0}^{n-m\nu}f(\alpha_{0})^{-1}$ in (5.6) by the right hand side
of (5.7), one has

$\displaystyle T_{0}^{n}f(\alpha_{0})^{-1}$
$\displaystyle=f(\alpha_{0})^{-1}T_{0}^{n}+\sum_{m\geq
1}g_{n,m}(\alpha_{0})\left(f(\alpha_{0})^{-1}T_{0}^{n-m\nu}+\sum_{k\geq
1}g_{n-m\nu,k}(\alpha_{0})T_{0}^{n-(m+k)\nu}f(\alpha_{0})^{-1}\right)$
$\displaystyle=f(\alpha_{0})^{-1}T_{0}^{n}+\sum_{m\geq
1}g_{n,m}(\alpha_{0})f(\alpha_{0})^{-1}T_{0}^{n-m\nu}+\sum_{m,k\geq
1}g_{n,m}(\alpha_{0})g_{n-m\nu,k}(\alpha_{0})T_{0}^{n-(m+k)\nu}f(\alpha_{0})^{-1}.$

Then substitute $T_{0}^{n-(m+k)\nu}f(\alpha_{0})^{-1}$ in the right hand side
of above equation by the right hand side of (5.7) (with $m$ replaced by
$m+k$), and continue this process. As $\mathsf{v}_{r,s}(T_{0})=1$ and $\nu>0$,
eventually one gets

$T_{0}^{n}f(\alpha_{0})^{-1}=f(\alpha_{0})^{-1}T_{0}^{n}+\sum_{m\geq
1}a_{n,m}(\alpha_{0})T_{0}^{n-m\nu},a_{n,m}\in K(X).$ (5.8)

By (5.5) and (5.8) ,we know that for any $h\in K(X)$ and $n\in\mathbb{Z}$,
there exist $h_{n,m}\in K(X)$ such that

$T_{0}^{n}h(\alpha_{0})=h(\alpha_{0})T_{0}^{n}+\sum_{m\geq
1}h_{n,m}(\alpha_{0})T_{0}^{n-m\nu}\in R.$

Then for any $f,g\in K(X),i,j\in\mathbb{Z},$

$f(\alpha_{0})T_{0}^{i}\cdot
g(\alpha_{0})T_{0}^{j}=f(\alpha_{0})g(\alpha_{0})T_{0}^{i+j}+\sum_{m\geq
1}f(\alpha_{0})g_{i,m}(\alpha_{0})T_{0}^{i+j-m\nu}\in R.$

For any $z,w\in R^{\times}$, $z=\sum_{i\leq
n}a_{i}(\alpha_{0})T_{0}^{i},w=\sum_{j\leq m}b_{j}(\alpha_{0})T_{0}^{j},$ one
has

$zw=\sum_{k\leq m+n}\sum_{i+j=k}a_{i}(\alpha_{0})T_{0}^{i}\cdot
b_{j}(\alpha_{0})T_{0}^{j}=\sum_{k\leq
m+n}\sum_{i+j=k}\left(a_{i}(\alpha_{0})b_{j}(\alpha_{0})T_{0}^{k}+\sum_{t\geq
1}a_{i}(\alpha_{0})b_{j,i,t}(\alpha_{0})T_{0}^{k-t\nu}\right)\in R.$

So $R$ is closed under multiplication.

Assume $z\in R^{\times}$. Then

$z=\sum_{i\leq
n}a_{i}(\alpha_{0})T_{0}^{i}=a_{n}(\alpha_{0})T_{0}^{n}(1+w),w=\sum_{i\leq
n-1}T_{0}^{-n}a_{n}(\alpha_{0})^{-1}a_{i}(\alpha_{0})T_{0}^{i}\in R.$

Then $\mathsf{v}_{r,s}(w)\leq-1$. Let

$u=(1+\sum_{i\geq 1}(-1)^{i}w^{i})T_{0}^{-n}a_{n}(\alpha_{0})^{-1}\in R.$

Then $u=z^{-1}$. Thus $R$ is closed under inversion. ∎

######  Proposition 5.3.

$R=\widehat{D}_{r,s}$.

###### Proof.

As $\alpha_{0},T_{0}\in D_{1},a_{i}(\alpha_{0})T_{0}^{i}\in D_{1}$. As
$\widehat{D}_{r,s}$ is a ring complete with respect to $\mathsf{v}_{r,s}$, for
any $z=\sum_{i\leq n}a_{i}(\alpha_{0})T_{0}^{i}\in R,$ one has
$z\in\widehat{D}_{r,s}.$ Thus $R\subseteq\widehat{D}_{r,s}.$

On the other hand,

$\alpha_{0}^{j}T_{0}^{r}=\textbf{p}+\sum_{m\geq
1}\lambda_{m}\textbf{p}^{1-m}\textbf{q}^{-m},\alpha_{0}^{-i}T_{0}^{s}=\textbf{q}+\sum_{m\geq
1}\mu_{m}\textbf{p}^{1-m}\textbf{q}^{-m},\lambda_{m},\mu_{m}\in\mathbb{Q}.$

As in the proof of Lemma 5.2,one has

$\textbf{p}=\alpha_{0}^{j}T_{0}^{r}+\sum_{m\geq
1}\omega_{m}\alpha_{0}^{j-m(j-i)}T_{0}^{r-m\nu},\textbf{q}=\alpha_{0}^{-i}T_{0}^{s}+\sum_{m\geq
1}\xi_{m}\alpha_{0}^{-i-m(j-i)}T_{0}^{s-m\nu},\omega_{m},\xi_{m}\in\mathbb{Q}.$

So $\textbf{p},\textbf{q}\in R$. As $R$ is a skew field complete with respect
to $\mathsf{v}_{r,s}$, one has $\widehat{D}_{r,s}\subseteq R$. ∎

##  6 $\widehat{D}_{r,s}$ satisfies the commutative centralizer condition

Recall that $D=L((T^{-1};\widetilde{\delta}))$ is a deformed Laurent series
ring with the standard degree map $deg$ as defined in (3.1). Assume $char\
L=0$. Let $D=L((T^{-1};\widetilde{\delta}))$. Then $K=\cap_{i\leq
0}\text{Ker}(\delta_{i})$ is a field of characteristic 0, and is the center of
$D$. Assume that there exists $i\in\mathbb{Z}_{\leq 0}$ with $\delta_{i}\neq
0$. Let $i_{0}=\max\\{i\in\mathbb{Z}_{\leq 0}|\delta_{i}\neq 0\\}$. Then
$\delta_{i_{0}}\in\text{Der}_{K}(L)$. For $z\in D$, let $C(z)$ be the
centralizer of $z$ in $D$.

Recall that a noncommutative algebra $A$ is said to satisfy the commutative
centralizer condition (the ccc, for short) if the centralizer of each element
not in the center of $A$ is a commutative algebra.

######  Proposition 6.1.

Assume $K=\overline{K}$ (i.e. $K$ is algebraically closed) and
$\text{Ker}(\delta_{i_{0}})=K$. Assume $D=L((T^{-1};\widetilde{\delta}))$ and
$z\in D\backslash K$. Then

1\. If $z\in L\backslash K$, then $C(z)=L$.

2\. If $z\in D$ and $deg(z)\neq 0$, then $C(z)=K((v^{-1}))$, where $v$ is an
element in $C(z)$ with the least positive degree. One has
$[C(z):K((z^{-1}))]=|\frac{deg(z)}{deg(v)}|$ if $deg(z)>0$, and
$[C(z):K((z))]=|\frac{deg(z)}{deg(v)}|$ if $deg(z)<0$.

3\. If $z\in D$ and $deg(z)=0$, then write $z=a_{0}+a_{-1}T^{-1}+\cdots$.

3.1. If $a_{0}\in K$, then $C(z)=C(z-a_{0})$, $z-a_{0}$ is in Case 2.

3.2. If $a_{0}\in L\backslash K$, then $C(z)\rightarrow
L,w=b_{0}+b_{-1}T^{-1}+\cdots\mapsto b_{0}$, is a ring isomorphism.

So $D$ satisfies the ccc. For all $z\in D\backslash K$, $C(z)$ is a complete
subfield of D.

###### Proof.

1\. This follows from Lemma 3.5.

2\. Assume $z=\sum_{i\leq n}a_{i}T^{i}$, and $w=\sum_{j\leq m}b_{j}T^{j}\in
C(z)$. One has

$\displaystyle a_{i}T^{i}\cdot
b_{j}T^{j}=a_{i}b_{j}T^{i+j}+a_{i}\binom{i}{1}\delta_{i_{0}}(b_{j})T^{i+j-1+i_{0}}+lower\
degree\ terms,\ \ and$ (6.1) $\displaystyle b_{j}T^{j}\cdot
a_{i}T^{i}=a_{i}b_{j}T^{i+j}+b_{j}\binom{j}{1}\delta_{i_{0}}(a_{i})T^{i+j-1+i_{0}}+lower\
degree\ terms.$ (6.2)

Then

$[a_{i}T^{i},b_{j}T^{j}]=\left(ia_{i}\delta_{i_{0}}(b_{j})-jb_{j}\delta_{i_{0}}(a_{i})\right)T^{i+j-1+i_{0}}+lower\
degree\ terms.$ (6.3)

For $a\in L$, write $\delta_{i_{0}}(a)=a^{\prime}$. Then

$[z,w]=\sum_{i\leq n,j\leq
m}[a_{i}T^{i},b_{j}T^{j}]=(na_{n}b_{m}^{\prime}-mb_{m}a_{n}^{\prime})T^{n+m-1+i_{0}}+lower\
degree\ terms.$

As $[z,w]=0$,
$na_{n}b_{m}^{\prime}-mb_{m}a_{n}^{\prime}=a_{n}^{m+1}b_{m}^{1-n}\delta_{i_{0}}(a_{n}^{-m}b_{m}^{n})=0$,
i.e. $a_{n}^{-m}b_{m}^{n}\in\text{Ker}\delta_{i_{0}}=K$, then
$b_{m}^{n}=\lambda a_{n}^{m}$ for some $\lambda\in K^{\times}$. For
$\widetilde{w}\in C(z)$ with $deg(\widetilde{w})=m$ and leading term
$\widetilde{b_{m}}T^{m}$, one also has
$\widetilde{b_{m}}^{n}=\lambda^{\prime}a_{n}^{m}$, where $\lambda^{\prime}\in
K^{\times}$. Then $\widetilde{b_{m}}^{n}=\mu b_{m}^{n},\mu\in K^{\times}$. As
$K=\overline{K}$, $\widetilde{b_{m}}/b_{m}\in K^{\times}$.

Let $S=\\{i\in\mathbb{Z}|\exists w\in C(z),deg(w)=i\\}$. Then $S$ is a
subgroup of $\mathbb{Z}$. As $K[z,z^{-1}]\subseteq C(z)$, one has
$n\mathbb{Z}\subset S$, one has $S=l\mathbb{Z}$ for some $l|n$ and $l>0$. As
it is shown that for a given degree, the leading term of $x\in C(z)$ is unique
up to a factor of $K^{\times}$. Then, by the choice of $v$, we can choose
$\lambda_{1}\in K^{\times}$ such that $w$ and $\lambda_{1}v^{k_{1}}$ have the
same leading term, where $k_{1}=m/l$. Let $w_{1}=w-\lambda_{1}v^{k_{1}}$, then
$w_{1}\in C(z)$ and $deg(w_{1})<deg(w)$. If $w_{1}\neq 0$, repeat the process
for $w_{1}$. Continuing in this way, we can find infinitely many elements of
$C(z)$,

$w_{s}:=w_{s-1}-\lambda_{s}v^{k_{s}},\lambda_{s}\in K^{\times},s\geq
1,k_{s}\in\mathbb{Z},where\quad w_{0}:=w,$

such that $deg(w_{1})>deg(w_{2})>\cdots$ and $k_{1}>k_{2}>\cdots$. Hence
$w=\sum_{s\geq 1}\lambda_{s}v^{k_{s}}$. Thus $C(z)\subseteq K((v^{-1}))$. The
reverse inclusion is clear. Hence one has $C(z)=K((v^{{-1}}))$.

The last statement of Case 2 follows from a simple degree argument.

3\. Recall that $z=a_{0}+a_{-1}T^{-1}+\cdots\in D$ and $deg(z)=0$. Thus
$a_{0}\neq 0$.

3.1. This is clear.

3.2. Assume $w=\sum_{j\leq m}b_{j}T^{j}\in C(z)$. By (6.3) one has
$m\delta_{i_{0}}(a_{0})b_{m}=0$, as $K=\text{Ker}(\delta_{i_{0}})$ and
$a_{0}\in L\setminus K$, one has $\delta_{i_{0}}(a_{0})b_{m}\neq 0$, thus $m$
must be $0$. Then we define

$\phi:C(z)\rightarrow L,w=b_{0}+b_{-1}T^{-1}+\cdots\mapsto b_{0},$

which is a ring homomorphism. Next we will show that $\phi$ is bijective,
i.e., for any $b_{0}\in L$, there exists some unique
$w=b_{0}+b_{-1}T^{-1}+\cdots\in C(z)$ such that $[z,w]=0$.

By (3.20) one has

$[z,w]=\sum_{i,j\leq 0}[a_{i}T^{i},b_{j}T^{j}]=\sum_{i,j\leq 0}\sum_{k\geq
1}\sum_{(j_{1},\cdots,j_{k})}[\binom{i}{k}a_{i}\cdot\delta_{j_{1}}\cdots\delta_{j_{k}}(b_{j})-\binom{j}{k}b_{j}\cdot\delta_{j_{1}}\cdots\delta_{j_{k}}(a_{i})]T^{j_{1}+\cdots+j_{k}-k+i+j},$
(6.4)

which can be written as $\sum_{l}f_{l}T^{l}$ by collecting the terms of the
same power, $f_{l}$ is an algebraic expression of those $a_{i},b_{j}$ and
their images under (composition of) certain $\delta_{k}$. As $[z,w]=0$, one
has a series of equations by equating the coefficients $f_{l}$ of $T^{l}$ to
be 0, from which one can solve $b_{i}$ for $i=-1,-2,\cdots,$ successively, as
follows. By (6.4), one has

$[z,w]=\left(b_{-1}\delta_{i_{0}}(a_{0})-a_{-1}\delta_{i_{0}}(b_{0})\right)T^{-2+i_{0}}+lower\
degree\ terms.$

As $\delta_{i_{0}}(a_{0})\neq 0$, there is a unique solution for $b_{-1}$
satisfying $b_{-1}\delta_{i_{0}}(a_{0})-a_{-1}\delta_{i_{0}}(b_{0})=0$. Assume
$s<-1$ and the solutions for $b_{k}$, $s<k\leq-1$, have been found and are all
unique. Now we consider $b_{s}$, who firstly occurs in the term
$-sb_{s}\delta_{i_{0}}(a_{0})T^{s-1+i_{0}}$ which is contained in the
expansion of $[a_{0},b_{s}T^{s}]$. Collecting all the terms of $T^{s-1+i_{0}}$
and equate the coefficient to be 0, one has

$-s\delta_{i_{0}}(a_{0})b_{s}+d_{s}=0,$

where $d_{s}$ is an algebraic expression involves only
$b_{i},\delta_{j}(b_{i}),a_{k},\delta_{j}(a_{k}),s+1\leq i\leq 0,s\leq k\leq
0$. As $char\ L=0$, one has $char\ K=0$. So $s\delta_{i_{0}}(a_{0})\neq 0$,
thus there is a unique solution for $b_{s}$ in $L$. It follows by induction
that for any $b_{0}\in L$, there exists some unique $w=\sum_{s\leq
0}b_{s}T^{s}$ such that $[z,w]=0$. Thus $\phi$ is an isomorphism. ∎

It is clear that
$\widetilde{D}_{r,s}=K(\alpha)\left(\left(T^{-1};\widetilde{\delta}_{r,s}\right)\right)$,
where $K$ is an algebraically closed field of characteristic 0, satisfies the
condition of above proposition, thus satisfies the ccc. As
$\widehat{D}_{r,s}\subset\widetilde{D}_{r,s}$, $\widehat{D}_{r,s}$ also
satisfies the ccc.

The result for $L_{\textbf{r}}((T^{-1};\widetilde{\delta}))$ analogous to
Proposition 6.1 can be obtained similarly. Thus
$\check{D}_{r,s}=K(\beta)_{\textbf{r}}\left(\left(T^{-1};\widetilde{\eta}_{r,s}\right)\right)$,
where $K$ is an algebraically closed field of characteristic 0, also satisfies
the ccc. So one has

######  Corollary 6.2.

The division rings $\widetilde{D}_{r,s},\widehat{D}_{r,s}$ and
$\check{D}_{r,s}$ all satisfy the ccc.

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