# Recursion operators in the cotangent covering of the rdDym equation

I.S. Krasil′shchik  Trapeznikov Institute of Control Sciences, 65
Profsoyuznaya street, Moscow 117997, Russia<EMAIL_ADDRESS>and  A.M.
Verbovetsky  Independent University of Moscow, Bolshoy Vlasyevskiy Pereulok
11, Moscow, 119002, Russia<EMAIL_ADDRESS>
###### Abstract.

We describe a general method of constructing nonlocal recursion operators for
symmetries of PDEs. As an example, the cotangent equation to the 3D rdDym
equation $u_{yt}=u_{x}u_{xy}-u_{y}u_{xx}$ for which two mutually inverse
operators are found. The exposition includes a rigorous criterion to check the
hereditary property.

###### Key words and phrases:

Partial differential equations, integrable linearly degenerate equations,
nonlocal symmetries, recursion operators

###### 2010 Mathematics Subject Classification:

35B06

The work ISK was partially supported by the RFBR Grant 18-29-10013 and RSF
Grant 21-71-20034.

###### Contents

  1. 1 Preliminaries and notation
  2. 2 The main result
  3. 3 Conclusions

## Introduction

The rdDym equation ($r$th dispersionless Dym equation)

$u_{yt}=u_{x}u_{xy}-u_{y}u_{xx}$ (1)

belongs to the class of linearly degenerate equations [8] and arose in the
Martínez Alonso-Shabat hierarchy [16]. It is Lax integrable and its symmetry
properties were studied in [1, 3]. Recursion operators for symmetries of
Equation (1) was found by O. Morozov in [19].

In [22], V. Ovsienko considered (in slightly different notation) the system

$\begin{array}[]{rcl}v_{yt}&=&2(v_{y}u_{xx}-v_{x}u_{xy})+u_{x}v_{xy}-u_{y}v_{xx}-2(u_{y}u_{xx}+2u_{x}u_{xy}),\\\\[2.0pt]
u_{yt}&=&u_{x}u_{xy}-u_{y}u_{xx},\end{array}$ (2)

which is a two-component extension of the rdDym equation (another two-
component generalization of (1) was studied in [20]). Treating (2) as the
Euler equation on the Virasoro algebra, its bi-Hamiltonian structure was
established.

System (2) may be also considered in the following way. Recall (see [12, 13])
that to any equation $\mathscr{E}=\\{F[u]=0\\}$ in unknowns $u$ its cotangent
equation $\mathscr{T}^{*}\mathscr{E}$

$F[u]=0,\qquad\ell_{F}^{*}[u,p]=0$

is naturally associated, $\ell_{F}^{*}$ is the operator adjoint to the
linearisation of $F$ (here and below $[u]$, $[u,p]$, etc. denotes the
variables together with their derivatives up to certain order). Then (2) may
be understood as the cotangent equation of (1), with $v$ instead of $p$,
‘deformed’ by the Lagrangian term

$\frac{\delta L}{\delta u}\ \text{ with }\ L=u_{y}u_{x}^{2}.$

System (2) is an Euler-Lagrange equations.

Lagrangian deformations of some linearly degenerate equations were studied in
[2]. In particular, the system

$\begin{array}[]{rcl}v_{yt}&=&2(v_{y}u_{xx}-v_{x}u_{xy})+u_{x}v_{xy}-u_{y}v_{xx}-2(2ku_{xy}u_{x}+(ku_{y}+l)u_{xx}),\\\\[2.0pt]
u_{yt}&=&u_{x}u_{xy}-u_{y}u_{xx},\end{array}$ (3)

which is considered below, is the deformation of the cotangent to rdDym
equation with the deforming Lagrangian $L=(ku_{y}+l)u_{x}^{2}$, $k$,
$l\in\mathbb{R}$. System (2) is obtained from (3) when $k=1$, $l=0$. We find
below recursion operators for symmetries of (3).

######  Remark 1 (courtesy O.I. Morozov).

Actually, both parameters $k$ and $l$ are fake in a sense: the change of
variables $v\mapsto v-2kxu_{x}-ku+ly$ kills them and transforms System (3) to
the pure cotangent equation of (1). In particular, the Ovsienko system (2)
also reduces to the cotangent equation. Though the computations in the general
case are not much harder that in a particular one, the result look more
complicated. For that reason we set $k=l=0$ everywhere below, i.e., the
equation we study below will be of the form

$\begin{array}[]{rcl}v_{yt}&=&2(v_{y}u_{xx}-v_{x}u_{xy})+u_{x}v_{xy}-u_{y}v_{xx},\\\\[2.0pt]
u_{yt}&=&u_{x}u_{xy}-u_{y}u_{xx}.\end{array}$ (4)

There exist several approaches to construct recursion operators (see, for
example, [21, 25]). Our method is based on the technology described in [13]
and originating from [11]. It is based on geometrical interpretation of
recursion operators for symmetries as Bäcklund auto-transformations of the
equation tangent to the given one (cf. [17]) and essentially uses the theory
of differential coverings [14].

######  Remark 2.

Recursion operators as Bäcklund auto-transformations of the tangent equations
initially aroused in M. Marvan’s paper [17], though implicitly the tangent
equation as an instrument for computation of recursion operators appears in
[9]. Informally, tangent and cotangent bundles to PDEs are discussed in the
paper [15] by B. Kupershmidt. A detailed discussion of the tangent covering
and its role in the theory of Hamiltonian (Poisson) structures can be found in
[12] and especially in [10].

The necessary theoretical matters needed for computations are shortly
described in Section 1, while Section 2 contains the main results.

##  1\. Preliminaries and notation

The exposition here is based on [5, 13, 14].

Let $\pi\colon E\to M$ be a vector bundle over a smooth manifold $M$, $\dim
M=n$, $\operatorname{rank}\pi=m$, and $\pi_{\infty}\colon J^{\infty}(\pi)\to
M$ be the bundle of its infinite jets. Consider an infinitely prolonged
differential equation $\mathscr{E}\in J^{\infty}(\pi)$ and assume that
$\mathscr{E}=\\{F=0\\}$, where $F\in\Gamma(\pi_{\infty}^{*}(\xi))$, where
$\xi\colon N\to M$ is another vector bundle. Any $\mathscr{E}$ is naturally
endowed with an $n$-dimensional Frobenius integrable distribution
$\mathscr{C}$ (the Cartan distribution) that defines its geometry. The Cartan
distribution is $\pi_{\infty}$-horizontal and, consequently, defines a flat
connection in $\pi_{\infty}\colon\mathscr{E}\to M$.

A morphism of equations $f\colon\tilde{\mathscr{E}}\to\mathscr{E}$ is a smooth
map, such that
$f_{*}(\tilde{\mathscr{C}}_{\theta})\subset\mathscr{C}_{f(\theta)}$ for any
$\theta\in\tilde{\mathscr{E}}$. A morphism is called a (differential) covering
if it is a submersion and the restriction
$\left.f_{*}\right|_{\mathscr{C}_{\theta}}$ of its differential to any Cartan
plane is an isomorphism. Two coverings
$f_{i}\colon\mathscr{E}_{i}\to\mathscr{E}$, $i=1$, $2$, are called equivalent
if there exists a diffeomorphism $g\colon\mathscr{E}_{1}\to\mathscr{E}_{2}$
that preserves the Cartan distributions and such that $f_{1}=f_{2}\circ g$.

Denote by $\mathscr{F}=\mathscr{F}(\mathscr{E})$ the algebra of smooth
functions on $\mathscr{E}$. A (higher infinitesimal) symmetry of $\mathscr{E}$
is a $\pi_{\infty}$-vertical derivation $S\colon\mathscr{F}\to\mathscr{F}$ (a
vector field on $\mathscr{E}$) that preserves the Cartan distribution. There
exists a one-to-one correspondence between symmetries and solutions of the
equation

$\ell_{\mathscr{E}}(\varphi)=0,$

where $\ell_{\mathscr{E}}$ is the restriction of the linearisation
$\ell_{F}\colon\varkappa\to P$ to $\mathscr{E}$ and $\varkappa$ denotes the
module of sections $\Gamma(\pi_{\infty}^{*}(\pi))$. A conservation law of
$\mathscr{E}$ is a $\pi_{\infty}$-horizontal form
$\omega\in\Lambda_{h}^{n-1}(\mathscr{E})$ closed with respect to the
horizontal de Rham differential $d_{h}$. Trivial conservation laws are those
of the form $\omega=d_{h}\rho$, $\rho\in\Lambda^{n-2}(\mathscr{E})$.

Given a covering $\tau\colon\tilde{\mathscr{E}}\to\mathscr{E}$, symmetries and
conservation laws of $\tilde{\mathscr{E}}$ are called nonlocal for
$\mathscr{E}$. Let
$S\colon\mathscr{F}(\mathscr{E})\to\mathscr{F}(\tilde{\mathscr{E}})$ be a
$\pi_{\infty}$-vertical derivation. It is called a $\tau$-shadow if

$\tilde{\mathscr{C}}_{X}\circ S=S\circ\mathscr{C}_{X}$

for any vector field $X$ on $M$. We say that a symmetry $\tilde{S}$ is a lift
of the shadow $S$ if $\left.\tilde{S}\right|_{\mathscr{E}}=S$.

Let an equation $\mathscr{E}$ be given by $\\{F=0\\}$. Then the system
$\mathscr{T}\mathscr{E}$

$F[u]=0,\qquad\ell_{F}[u,q]=0$

is called the tangent equation of $\mathscr{E}$, while the projection
$\mathrm{t}\colon[u,q]\mapsto[u]$ is the tangent covering over $\mathscr{E}$.
It must be stressed that $q$ is considered as an odd variable of degree $1$.
Sections of $\mathrm{t}$ that preserve the Cartan distributions are identified
with symmetries of $\mathscr{E}$ and consequently Bäcklund auto-
transformations $\mathscr{R}_{\tau,\tau^{\prime}}$ of the form

$\textstyle{W\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\tau}$$\scriptstyle{\tau^{\prime}}$$\textstyle{\mathscr{T}\mathscr{E}}$$\textstyle{\mathscr{T}\mathscr{E}\hbox
to0.0pt{,\hss}}$ (5)

where $\tau$, $\tau^{\prime}$ are coverings, are naturally interpreted as
recursion operators for symmetries of $\mathscr{E}$.

To construct these operators, we use a scheme consisting of two steps. In the
first step, we are looking for two-component conservation laws $\omega$ of
$\mathscr{T}\mathscr{E}$ linear on the fibers of the projection $\mathrm{t}$.
Provided such a law was found, we consider the covering $\tau=\tau_{\omega}$
associated with it and, as the second step, try to find
$\tau_{\omega}$-shadows (also linear on fibers both of $\mathrm{t}$ and
$\tau_{\omega}$). If such a shadow $S$ exists, it delivers the covering
$\tau^{\prime}=\tau_{S}$ and the desired recursion operator
$\mathscr{R}_{\tau_{\omega},\tau_{S}}$.

Finally, few words about the type of equations that are the subject of our
study. Assume $\mathscr{E}=\\{F=0\\}$, $F\in P$, and consider the operator
$\ell_{F}^{*}\colon\hat{P}\to\hat{\varkappa}$ formally adjoint to $\ell_{F}$,
where $\hat{\bullet}=\hom(\bullet,\Lambda_{h}^{n})$. Then the system
$\mathscr{T}^{*}\mathscr{E}$

$\ell_{F}^{*}[u,p]=0,\qquad F[u]=0$ (6)

is said to be the cotangent equation of $\mathscr{E}$, while the projection
$\mathrm{t}^{*}\colon\mathscr{T}^{*}\mathscr{E}\to\mathscr{E}$,
$(u,p)\mapsto(u)$, is called the cotangent covering. Similar to the case of
the tangent equation, the variable $p$ is odd. System (6) is always an Euler-
Lagrange equation with the Lagrangian density $L=\sum p^{j}F^{j}$ and we are
interested in equations

$\ell_{F}^{*}[u,p]+G[u,p]=0,\qquad F[u]=0,$

(‘Lagrangian deformations’ of (6)), such that (a) they are Lagrangian as well
and (b) they possess a nontrivial recursion operator. Equation (2) is exactly
of this type. Note that for Euler-Lagrange equations $\mathscr{T}\mathscr{E}$
and $\mathscr{T}^{*}\mathscr{E}$ coincide, since $\ell_{F}=\ell_{F}^{*}$ in
this case.

### Coordinates

Let $\mathscr{U}\subset M$ be a coordinate neighborhood with local coordinates
$x=(x^{1},\dots,x^{n})$ and
$E\supset\pi^{-1}(\mathscr{U})\simeq\mathscr{U}\times\mathbb{R}^{m}$ be a
trivialization of $\pi$ with fiber-wise coordinates $u=(u^{1},\dots,u^{m})$.
Then adapted coordinates $u_{\sigma}^{j}$ arise on $J^{\infty}(\pi)$, where
$u_{\sigma}^{j}$ correspond to the partial derivative
$\partial^{\left|\sigma\right|}u^{j}/\partial x^{\sigma}$. The Cartan
distribution is spanned by the total derivatives

$D_{i}=\frac{\partial}{\partial x^{i}}+\sum u_{\sigma
i}^{j}\frac{\partial}{\partial u_{\sigma}^{j}},$

while the Cartan connection takes $\partial/\partial x^{i}$ to $D_{i}$.
Equation $\mathscr{E}$ is given by the system of relations
$F^{l}(x,\dots,u_{\sigma}^{j},\dots)=0$, $l=1,\dots,r$. To restrict necessary
objects to $\mathscr{E}$, one needs to choose internal coordinates on
$\mathscr{E}$ (cf. [18]) and express these objects in terms of internal
coordinates. In particular, the Cartan distribution on $\mathscr{E}$ is got by
rewriting the operators $D_{i}$ in internal coordinates.

The coordinate presentation of the linearisation operator acts on
$\varphi=(\varphi^{1},\dots,\varphi^{m})$ by

$\ell_{F}(\varphi)=\sum_{\sigma,j}\frac{\partial F^{l}}{\partial
u_{\sigma}^{j}}D_{\sigma}(\varphi^{j}),$

where $D_{\sigma}$ is the composition of the total derivatives corresponding
to the multi-index $\sigma$, while for the adjoint one we have

$\ell_{F}^{*}(\psi)=\sum_{\sigma,l}(-1)^{\left|\sigma\right|}D_{\sigma}\left(\frac{\partial
F^{l}}{\partial u_{\sigma}^{j}}\psi^{l}\right),$

where $\psi=(\psi^{1},\dots,\psi^{r})$. Thus, the tangent equation of
$\mathscr{E}$ is

$\sum_{\sigma,j}\frac{\partial F^{l}}{\partial
u_{\sigma}^{j}}q_{\sigma}^{j}=0,\quad F(x,\dots,u_{\sigma}^{j},\dots)=0,\qquad
q=(q^{1},\dots,q^{m}),$

while while the cotangent one is given by

$\sum_{\sigma,l}(-1)^{\left|\sigma\right|}D_{\sigma}\left(\frac{\partial
F^{l}}{\partial u_{\sigma}^{j}}p^{l}\right)=0,\quad
F(x,\dots,u_{\sigma}^{j},\dots)=0,\qquad p=(p^{1},\dots,p^{r}).$

Without loss of generality, we can define two-component conservation laws as
the forms

$\omega=(X_{1}\,dx^{1}+X_{2}\,dx^{2})\,dx^{3}\wedge\dots\wedge\,dx^{n},\qquad
D_{1}(X_{2})=D_{2}(X_{1}),$

where $X_{1}$ and $X_{2}$ are smooth functions on $\mathscr{E}$. Triviality of
$\omega$ means existence of a potential $Y$, such that $D_{i}(Y)=X_{i}$,
$i=1$, $2$.

A covering structure in a locally trivial vector bundle $\tau\colon
W\to\mathscr{E}$ is determined by the system of pair-wise commuting vector
fields

$\tilde{D}_{i}=D_{i}+X_{i},\qquad i=1,\dots,n,$ (7)

where $X_{i}=\sum X_{i}^{\alpha}\partial/\partial w^{\alpha}$ are
$\tau$-vertical fields and $w^{\alpha}$ are coordinates in the fiber of $\tau$
(nonlocal variables). The equalities $[\tilde{D}_{i},\tilde{D}_{j}]=0$ amount
to the fact that the system

$w_{x^{i}}^{\alpha}=X_{i}^{\alpha},\qquad
i=1,\dots,n,\quad\alpha=1,\dots,\operatorname{rank}\tau,$

is compatible modulo $\mathscr{E}$. Note that (7) allows to lift any
differential operator $\Delta$ in total derivatives from $\mathscr{E}$ to an
operator $\tilde{\Delta}$ on the covering equation.

Let $\omega$ be a two-component conservation law like above. Then one can
construct the covering $\tau_{\omega}$ as follows. In the case $\dim M=2$ we
set

$w_{x^{1}}=X_{1},\qquad w_{x^{2}}=X_{2},$

and $\operatorname{rank}\tau_{\omega}=1$. When $\dim M>2$, we introduce
infinite number of nonlocal variables $w^{\sigma}$, where $\sigma$ is a multi-
index consisting of integers $2,\dots,n$, $\left|\sigma\right|\geq 0$, and set

$w_{x^{i}}^{\sigma}=\begin{cases}D_{\sigma}(X_{i}),&i=1,2,\\\ w^{\sigma
i},&i>2.\end{cases}$

Obviously, we obtain covering structures in both cases.

Consider an $m$-component function $S=(S^{1},\dots,S^{m})$. It defines a
$\tau$-shadow if and only if

$\tilde{\ell}_{\mathscr{E}}(S)=0,$

where $\tilde{\ell}_{\mathscr{E}}$ is the natural lift of the operator
$\ell_{\mathscr{E}}$ to the covering equation (see above). Let now $S$ be a
$\tau$-shadow, where $\tau=\tau_{\omega}$ in Diagram (5). Denote by
$p_{\sigma}$ the fiber coordinates in the left copy of
$\mathscr{T}\mathscr{E}$ and by $\bar{p}_{\sigma}$ the same coordinates in the
right one and set $\bar{p}_{\sigma}^{j}=\tilde{D}_{\sigma}(S^{j})$, where
$\tilde{D}_{\sigma}$ is the composition of total derivatives in $W$. This
gives the covering $\tau^{\prime}=\tau_{S}$. The desired recursion operator is
obtained, when we set

$S^{j}=\sum_{l,\sigma}S_{l}^{j,\sigma}p_{\sigma}^{l}+\sum_{\sigma}S_{\sigma}^{j}w^{\sigma},$

where $S_{l}^{j,\sigma}$, $S_{\sigma}^{j}$ are smooth functions on
$\mathscr{E}$.

##  2\. The main result

We now pass to the equation $\mathscr{E}$ given by (4) and choose the
functions $x$, $y$, $t$ and

$u_{x_{i}}=u_{\underbrace{x\dots x}_{i\text{ times}}},\
u_{x_{i},y_{j}}=u_{\underbrace{x\dots x}_{i\text{ times}}\underbrace{y\dots
y}_{j\text{ times}}},\ u_{x_{i},y_{j}}=u_{\underbrace{x\dots x}_{i\text{
times}}\underbrace{t\dots t}_{j\text{ times}}},\quad i\geq 0,\ j>0,$

and similar for $v$. Then the total derivative on $\mathscr{E}$ acquire the
form

$\displaystyle D_{x}$ $\displaystyle=\frac{\partial}{\partial
x}+\sum_{i,j}\left(u_{x_{i+1}}\frac{\partial}{\partial
u_{x_{i}}}+u_{x_{i+1},y_{j}}\frac{\partial}{\partial
u_{x_{i},y_{j}}}+u_{x_{i+1},t_{j}}\frac{\partial}{\partial
u_{x_{i},t_{j}}}\right)$
$\displaystyle+\sum_{i,j}\left(v_{x_{i+1}}\frac{\partial}{\partial
v_{x_{i}}}+v_{x_{i+1},y_{j}}\frac{\partial}{\partial
v_{x_{i},y_{j}}}+v_{x_{i+1},t_{j}}\frac{\partial}{\partial
v_{x_{i},t_{j}}}\right),$ $\displaystyle D_{y}$
$\displaystyle=\frac{\partial}{\partial
y}+\sum_{i,j}\left(u_{x_{i},y}\frac{\partial}{\partial
u_{x_{i}}}+u_{x_{i},y_{j+1}}\frac{\partial}{\partial
u_{x_{i},y_{j}}}+D_{x}^{i}D_{t}^{j-1}(U)\frac{\partial}{\partial
u_{x_{i},t_{j}}}\right)$
$\displaystyle+\sum_{i,j}\left(v_{x_{i},y}\frac{\partial}{\partial
v_{x_{i}}}+v_{x_{i},y_{j+1}}\frac{\partial}{\partial
v_{x_{i},y_{j}}}+D_{x}^{i}D_{t}^{j-1}(V)\frac{\partial}{\partial
v_{x_{i},t_{j}}}\right),$ $\displaystyle D_{t}$
$\displaystyle=\frac{\partial}{\partial
t}+\sum_{i,j}\left(u_{x_{i},t}\frac{\partial}{\partial
u_{x_{i}}}+D_{x}^{i}D_{y}^{j-1}(U)\frac{\partial}{\partial
u_{x_{i},y_{j}}}+u_{x_{i},t_{j+1}}\frac{\partial}{\partial
u_{x_{i},t_{j}}}\right)$
$\displaystyle+\sum_{i,j}\left(v_{x_{i},t}\frac{\partial}{\partial
v_{x_{i}}}+D_{x}^{i}D_{y}^{j-1}(V)\frac{\partial}{\partial
v_{x_{i},y_{j}}}+v_{x_{i},t_{j+1}}\frac{\partial}{\partial
v_{x_{i},t_{j}}}\right),$

where $V=2(v_{y}u_{xx}-v_{x}u_{xy})+u_{x}v_{xy}-u_{y}v_{xx}$ and
$U=u_{x}u_{xy}-u_{y}u_{xx}$ are right-hand sides of the first and second
equations in (4), respectively.

### Symmetries

The defining equation $\ell_{\mathscr{E}}(S)=0$ for symmetries of (4) is

$\begin{array}[]{rcl}D_{y}D_{t}(\psi)&=&2v_{y}D_{x}^{2}(\varphi)-2v_{x}D_{x}D_{y}(\varphi)+v_{xy}D_{x}(\varphi)-v_{xx}D_{y}(\varphi)\\\\[2.0pt]
&&\qquad-2u_{xy}D_{x}(\psi)-u_{y}D_{x}^{2}(\psi)+u_{x}D_{x}D_{y}(\psi)+2u_{x}D_{y}(\psi),\\\\[3.0pt]
D_{y}D_{t}(\varphi)&=&u_{xy}D_{x}(\varphi)-u_{y}D_{x}^{2}(\varphi)+u_{x}D_{x}D_{y}(\varphi)-u_{xx}D_{y}(\varphi).\end{array}$
(8)

Solving System (8) for $\varphi$ and $\psi$ of order $\leq 3$, we get the
symmetries

$\displaystyle S_{1}$ $\displaystyle=\big{(}xu_{x}-2u,xv_{x}+2v\big{)},$
$\displaystyle S_{2}(T)$ $\displaystyle=\big{(}T,0\big{)},$ $\displaystyle
S_{3}(T)$ $\displaystyle=\big{(}Tu_{x}+\dot{T}x,Tv_{x}\big{)},$ $\displaystyle
S_{4}(T)$
$\displaystyle=\big{(}\frac{1}{2}\ddot{T}x^{2}+\dot{T}(xu_{x}-u)+Tu_{t},(xv_{x}+2v)\dot{T}+Tv_{t}\big{)},$
$\displaystyle S_{5}(Y)$ $\displaystyle=\big{(}Yu_{y},Yv_{y}\big{)},$
$\displaystyle S_{6}$ $\displaystyle=\big{(}0,u_{xxx}\big{)},$ $\displaystyle
S_{7}$
$\displaystyle=\big{(}0,\frac{1}{2}u_{xx}^{2}+u_{x}u_{xxx}-u_{xxt}\big{)},$
$\displaystyle S_{8}$
$\displaystyle=\big{(}0,u_{x}u_{xx}^{2}-u_{xt}u_{xx}+u_{x}^{2}u_{xxx}-2u_{x}u_{xxt}+u_{xtt}\big{)},$
$\displaystyle S_{9}$
$\displaystyle=\big{(}0,\frac{3}{2}u_{x}^{2}u_{xx}^{2}-3u_{x}u_{xt}u_{xx}+\frac{3}{2}u_{xt}^{2}-3u_{x}^{2}u_{xxt}+3u_{x}u_{xtt}+u_{x}^{3}u_{xxx}-u_{ttt}\big{)},$
$\displaystyle S_{10}$
$\displaystyle=\big{(}0,\frac{u_{xy}^{2}-2u_{y}u_{xxy}}{u_{y}^{2}}\big{)},$
$\displaystyle S_{11}$
$\displaystyle=\big{(}0,\frac{u_{xy}u_{yy}-u_{y}u_{xyy}}{u_{y}^{3}}\big{)},$
$\displaystyle S_{12}$ $\displaystyle=\big{(}0,v\big{)},$ $\displaystyle
S_{13}(T)$ $\displaystyle=\big{(}0,Tu_{x}-\frac{1}{2}\dot{T}x\big{)},$
$\displaystyle S_{14}(T)$
$\displaystyle=\big{(}0,\frac{1}{6}\ddot{T}x^{2}+Tu_{x}^{2}+(T-\frac{2}{3}\dot{T}x)u_{x}-\frac{1}{6}(2u+3x)\dot{T}-\frac{2}{3}Tu_{t}\big{)},$
$\displaystyle S_{15}(T)$ $\displaystyle=\big{(}0,T\big{)},$ $\displaystyle
S_{16}(Y)$ $\displaystyle=\big{(}0,\frac{Y}{u_{y}^{2}}\big{)}.$

where $Y=Y(y)$ and $T=T(t)$ are arbitrary smooth functions and ‘dot’ denotes
the $t$-derivative. These symmetries commute as follows111We omit zero
commutators. ‘Prime’ denotes the $y$-derivative.

$\displaystyle[S_{1},S_{2}(T)]$
$\displaystyle=2S_{2}(T),\quad[S_{1},S_{3}(T)]=S_{3}(T),$
$\displaystyle[S_{1},S_{6}]$
$\displaystyle=-S_{6},\quad[S_{1},S_{7}]=-2S_{7},\quad[S_{1},S_{8}]=-3S_{8},\quad[S_{1},S_{9}]=-4S_{9},$
$\displaystyle[S_{1},S_{11}]$
$\displaystyle=S_{11},\quad[S_{1},S_{13}(T)]=-3S_{13}(T),\quad[S_{1},S_{14}(T)]=4S_{14}(T)-S_{13}(T),$
$\displaystyle[S_{1},S_{15}(T)]$
$\displaystyle=-2S_{15}(T),\quad[S_{1},S_{16}(Y)]=2S_{16}(Y);$
$\displaystyle[S_{2}(T_{1}),S_{4}(T_{2})$
$\displaystyle=-S_{2}(\dot{T}_{1}T_{2}-T_{1}\dot{T}_{2}),\quad[S_{2}(T),S_{9}]=-S_{15}(\dddot{T}),$
$\displaystyle[S_{2}(T_{1}),S_{14}(T_{2})]$
$\displaystyle=-\frac{1}{3}S_{15}(T_{1}\dot{T}_{2}+2\dot{T}_{1}T_{2});$
$\displaystyle[S_{3}(T_{1}),S_{3}(T_{2})]$
$\displaystyle=S_{2}(\dot{T}_{1}-T_{1}\dot{T}_{2})-T_{1}\dot{T}_{2}),$
$\displaystyle[S_{3}(T_{1}),S_{4}(T_{2})]$
$\displaystyle=S_{3}(\dot{T}_{1}T_{2}-\dot{T}_{2}T_{1}),$
$\displaystyle[S_{3}(T),S_{8}]$
$\displaystyle=S_{15}(\dddot{T}),\quad[S_{3}(T),S_{9}]=2S_{13}(\dddot{T}),$
$\displaystyle[S_{3}(T_{1}),S_{13}(T_{2})]$
$\displaystyle=S_{15}(\dot{T}_{1}T_{2}+\frac{1}{2}T_{1}\dot{T}_{2}),$
$\displaystyle[S_{3}(T_{1}),S_{14}(T_{2})]$
$\displaystyle=\frac{2}{3}S_{13}(2\dot{T}_{1}T_{2}+T_{1}\dot{T_{2}})-S_{15}(\dot{T}_{1}T_{2}+\frac{1}{2}T_{1}\dot{T}_{2});$
$\displaystyle[S_{4}(T_{1}),S_{4}(T_{2})]$
$\displaystyle=S_{4}(\dot{T}_{1}T_{2}-T_{1}\dot{T}_{2}),\quad[S_{4}(T),S_{7}]=-S_{15}(\dddot{T}),$
$\displaystyle[S_{4}(T),S_{8}]$
$\displaystyle=-2S_{13}(\dddot{T}),\quad[S_{4}(T),S_{9}]=3S_{13}(\dddot{T})+3S_{14}(\dddot{T}),$
$\displaystyle[S_{4}(T_{1}),S_{13}(T_{2})]$
$\displaystyle=-S_{13}(2\dot{T_{1}}T_{2}+T_{1}\dot{T}_{2}),\quad[S_{4}(T_{1}),S_{14}(T_{2})]=-S_{14}(2\dot{T}_{1}T_{2}+T1\dot{T}_{2}),$
$\displaystyle[S_{4}(T_{1}),S_{15}(T_{2})]$
$\displaystyle=S_{15}(2\dot{T}_{1}T_{2}+T_{1}\dot{T}_{2});$
$\displaystyle[S_{5}(Y_{1}),S_{5}(Y_{2})]$
$\displaystyle=S_{5}(Y_{1}^{\prime}Y_{2}-Y_{2}^{\prime}Y_{1}),\quad[S_{5}(Y_{1}),S_{16}(Y_{2})]=-S_{16}(Y_{2}^{\prime}Y_{1}+2Y_{1}^{\prime}Y_{2});$
$\displaystyle[S_{6},S_{12}]$
$\displaystyle=S_{6};\quad[S_{7},S_{12}]=S_{7};\quad[S_{8},S_{12}]=S_{8};\quad[S_{9},S_{12}]=S_{9};$
$\displaystyle[S_{10},S_{12}]$
$\displaystyle=S_{10};\quad[S_{11},S_{12}]=S_{11};\quad[S_{12},S_{13}(T)]=-S_{13}(T),$
$\displaystyle[S_{12},S_{15}(T)]$
$\displaystyle=-S_{15}(T),\quad[S_{12}S_{16}(Y)]=-S_{16}(Y).$

### Tangent equation

Due to (8), the tangent equation is obtained by adding to System (4) the
equations

$\begin{array}[]{rcl}q_{yt}&=&2v_{y}p_{xx}-2v_{x}p_{xy}+v_{xy}p_{x}\\\\[2.0pt]
&&\qquad-
v_{xx}p_{y}\par-2u_{xy}q_{x}-u_{y}q_{xx}+u_{x}q_{xy}+2u_{x}q_{y},\\\\[3.0pt]
p_{yt}&=&u_{xy}p_{x}-u_{y}p_{xx}+u_{x}p_{xy}-u_{xx}p_{y}.\end{array}$ (9)

### Conservation laws

There exist four conservation laws on $\mathscr{T}\mathscr{E}$ of order $\leq
2$ and linear with respect to the variables $p_{\sigma}$, $q_{\sigma}$:

$\displaystyle\omega_{1}$
$\displaystyle=(X_{1}\,dx+T_{1}\,dt)\wedge\,dy,\quad\omega_{2}=(X_{2}\,dx+T_{2}\,dt)\wedge\,dy,$
$\displaystyle\omega_{3}$
$\displaystyle=(X_{3}\,dx+Y_{3}\,dy)\wedge\,dt,\quad\omega_{4}=(X_{4}\,dx+Y_{4}\,dy)\wedge\,dt,$

where

$\displaystyle X_{1}=-\frac{Yp_{y}}{u_{y}^{2}},\quad
T_{1}=\frac{Y(u_{y}p_{x}-u_{x}p_{y})}{u_{y}^{2}};$ $\displaystyle
X_{2}=Y(v_{y}p_{y}+u_{y}q_{y}),\quad
T_{2}=Y(v_{y}u_{y}p_{x}-(2v_{x}u_{y}-u_{x}v_{y})p_{y}-u_{y}^{2}q_{x}+u_{y}u_{x}q_{y});$
$\displaystyle X_{3}=T(2u_{x}p_{x}-p_{t}),\quad
Y_{3}=T(u_{y}p_{x}+u_{x}p_{y});$ $\displaystyle
X_{4}=T(v_{x}p_{x}+u_{x}q_{x}+q_{t}),\quad
Y_{4}=-T(v_{x}p_{y}+2v_{y}p_{x}+u_{y}q_{x}-2u_{x}q_{y}).$

Setting $Y=T=1$ and after slight relabeling, we obtain the following nonlocal
variables associated with the above conservation laws:

$\displaystyle\begin{array}[]{rcl}w_{1,x}&=&\dfrac{p_{y}}{u_{y}^{2}},\\\
w_{1,t}&=&\dfrac{u_{x}p_{y}-u_{y}p_{x}}{u_{y}^{2}};\\\\[8.0pt]
w_{2,x}&=&v_{y}p_{y}+u_{y}q_{y},\\\
w_{2,t}&=&v_{y}u_{y}p_{x}+(u_{x}v_{y}-2v_{x}u_{y})p_{y}u_{y}^{2}q_{x}+u_{y}u_{x}q_{y};\end{array}$
(14) $\displaystyle\begin{array}[]{rcl}w_{3,x}&=&2u_{x}p_{x}-p_{t},\\\
w_{3,y}&=&u_{y}p_{x}+u_{x}p_{y};\\\\[8.0pt]
w_{4,x}&=&v_{x}p_{x}+u_{x}q_{x}+q_{t},\\\
w_{4,y}&=&2v_{y}p_{x}-v_{x}p_{y}-u_{y}q_{x}+2u_{x}q_{y}.\end{array}$ (19)

All the nonlocal variables $w_{i}$ are odd of degree $1$.

### Shadows

Direct computations reveal two shadows of symmetries that linearly depend on
the variables $p_{\sigma}$, $q_{\sigma}$, and $w_{i,\sigma}$ up to second
order:

$\displaystyle s_{0}:\qquad$ $\displaystyle\bar{p}_{0}=p,\quad\bar{q}_{0}=q;$
$\displaystyle s_{1}:\qquad$
$\displaystyle\bar{p}_{1}=u_{x}p-w_{3},\quad\bar{q}_{1}=v_{x}p-2u_{x}q+w_{4};$
$\displaystyle s_{2}:\qquad$
$\displaystyle\bar{p}_{2}=u_{y}w_{1},\quad\bar{q}_{2}=v_{y}w_{1}+\frac{w_{2}}{u_{y}^{2}}.$

The shadow $s_{0}$ is responsible for the identical operator and thus is of no
interest, while the other two ones provide nontrivial results.

### Recursion operators

Using the last formulas, we obtain the following expressions for the nonlocal
variables $w_{i}$:

$\displaystyle w_{1}$ $\displaystyle=\frac{\bar{p}_{2}}{u_{y}},\quad
w_{2}=u_{y}(u_{y}\bar{q}_{2}-v_{y}\bar{p}_{2});$ (20) $\displaystyle w_{3}$
$\displaystyle=u_{x}p-\bar{p}_{1},\quad w_{4}=-v_{x}p+2u_{x}q+\bar{q}_{1}.$
(21)

Let us now substitute these expressions to the defining equations of the
coverings (14), (19). As the result, we obtain two Bäcklund auto-
transformations of $\mathscr{T}\mathscr{E}$:

$\begin{array}[]{rcl}\bar{p}_{1,x}&=&u_{xx}p-u_{x}p_{x}+p_{t},\\\\[3.0pt]
\bar{p}_{1,y}&=&u_{xy}p-u_{y}p_{x},\\\\[3.0pt]
\bar{q}_{1,x}&=&2v_{x}p_{x}-u_{x}q_{x}-2u_{xx}q+v_{xx}p+q_{t},\\\\[3.0pt]
\bar{q}_{1,y}&=&2v_{y}p_{x}-u_{y}q_{x}-2u_{xy}q+v_{xy}p;\end{array}$ (22)

and

$\begin{array}[]{rcl}\bar{p}_{2,x}&=&\dfrac{1}{u_{y}}(u_{xy}\bar{p}_{2}+p_{y}),\\\\[8.0pt]
\bar{p}_{2,t}&=&\dfrac{1}{u_{y}}((u_{x}u_{xy}-u_{y}u_{xx})\bar{p}_{2}-u_{y}p_{x}+u_{x}p_{y}),\\\
\bar{q}_{2,x}&=&\dfrac{1}{u_{y}^{2}}((2v_{y}u_{xy}+u_{y}v_{xy})\bar{p}_{2}+2v_{y}p_{y})-\dfrac{1}{u_{y}}(2u_{xy}\bar{q}_{2}-q_{y}),\\\
\bar{q}_{2,t}&=&\Big{(}\dfrac{u_{x}v_{xy}-2v_{x}u_{xy}}{u_{y}}+\dfrac{2u_{x}v_{y}u_{xy}}{u_{y}^{2}}-v_{xx}\Big{)}\bar{p}_{2}\\\
&+&\Big{(}\dfrac{2v_{y}u_{x}}{u_{y}^{2}}-\dfrac{2v_{x}}{u_{y}}\Big{)}p_{y}+\dfrac{2(u_{y}u_{xx}-u_{x}u_{xy})}{u_{y}}\bar{q}_{2}+\dfrac{u_{x}}{u_{y}}q_{y}-q_{x}.\end{array}$
(23)

Thus, relations (22) and (23) define recursion operators for symmetries of
Equation (4). The $p$-components of these operators give recursion operators
for the rdDym equation (cf. [3]).

######  Remark 3.

Though formulas (22) and (23) look different, essentially they present the
same object. Namely, if we resolve (23) with respect to the variables $p_{x}$,
$p_{y}$, $q_{x}$, and $q_{y}$, we shall obtain relations (22) up to some
relabeling. Consequently, the above presented Bäcklund transformations are
mutually inverse.

######  Remark 4.

Note also that (22), understood as a covering over $\mathscr{T}\mathscr{E}$
with the nonlocal variables $\bar{p}$ and $\bar{q}$ by the very construction
is equivalent to the one with the nonlocal variables $w_{3}$ and $w_{4}$.
Explicitly, the corresponding gauge transformation is given by formulas (21).
In a similar way, (20) delivers equivalence between (23) and the covering with
the variables $w_{1}$ and $w_{2}$.

### Actions

Let us indicate how the operator (22) acts on symmetries of the equation
$\mathscr{E}$; due to Remark 3, the action of (23) is the opposite. Note also
that of (22) is defined up to the image of $0$, which is $S_{2}(T)$. Keeping
in mind these remarks, we have:

$\displaystyle 0\mapsto S_{2}(T)\mapsto S_{3}(T)\mapsto S_{4}(T)\mapsto\star$
$\displaystyle S_{1}\mapsto\star$ $\displaystyle S_{5}(Y)\mapsto 0$
$\displaystyle S_{11}\mapsto-\frac{1}{2}S_{10}\mapsto 2S_{6}\mapsto\star$
$\displaystyle S_{7}\mapsto-S_{8}\mapsto\star$ $\displaystyle
S_{9}\mapsto\star$ $\displaystyle S_{12}\mapsto\star$ $\displaystyle
S_{15}(T)\mapsto-2S_{13}(T)\mapsto\frac{3}{2}\big{(}S_{13}(T)-S_{14}(T)\big{)}$
$\displaystyle S_{14}(T)\mapsto\star$ $\displaystyle S_{16}(Y)\mapsto 0.$

Here $\star$ denotes a nonlocal result.

### Lifts of shadows

Note finally that any nonlocal symmetry in the coverings (14) and (19) is
determined by the coefficients at $\partial/\partial u$, $\partial/\partial
v$, $\partial/\partial p$, $\partial/\partial q$, $\partial/\partial w_{1}$,
$\partial/\partial w_{2}$ (in the case of (14)), and$~{}\partial/\partial
w_{3}$, $\partial/\partial w_{4}$ (for (19)). Denote these coefficients by

$U,\quad V,\quad P,\quad Q,\quad W_{1},\quad W_{2},\quad W_{3},\quad W_{4},$

respectively. We state that there exist nonlocal symmetries that are the lifts
of the shadows $s_{1}$ and $s_{2}$ to the corresponding coverings. Namely,
these symmetries are

$\displaystyle\sigma_{1}\colon\quad$ $\displaystyle U=u_{y}w_{1},$
$\displaystyle V=\frac{w_{2}}{u_{y}^{2}}+v_{y}w_{1},$ $\displaystyle
P=w_{1}p_{y},$ $\displaystyle Q=w_{1}q_{y}+2\frac{p_{y}w_{2}}{u_{y}^{3}},$
$\displaystyle W_{1}=w_{1}w_{1,y},$ $\displaystyle
W_{2}=2w_{1,y}w_{2}-w_{2,y}w_{1}$ and $\displaystyle\sigma_{2}\colon\quad$
$\displaystyle U=u_{x}p-w_{3},$ $\displaystyle V=-2u_{x}q+v_{x}p-w_{4},$
$\displaystyle P=p_{x}p,$ $\displaystyle Q=2qp_{x}+q_{x}p,$ $\displaystyle
W_{3}=pq_{t}+2p_{t}q-v_{x}pp_{x}+u_{x}pq_{x}+2u_{x}p_{x}q,$ $\displaystyle
W_{4}=pp_{t}-2u_{x}pp_{x}.$

Moreover, $\sigma_{i}$ are odd vector fields of degree $1$. A direct
computation shows that the super-commutators $[\sigma_{1},\sigma_{1}]$ and
$[\sigma_{2},\sigma_{2}]$ of these fields vanish, which means that the
constructed recursion operators are hereditary, [13].

##  3\. Conclusions

The method used here to construct recursion operators for symmetries of
Equation (4) seems to be of a universal nature. In particular, it allows to
test rigorously the hereditary property of nonlocal recursion operators. It
would be interesting to apply it to other two-component extensions of linearly
degenerate equations constructed in [2], as well as to the Dunajski equation
[7], etc.

It would be also interesting to describe (similar to how it was done in [3])
the algebra of nonlocal symmetries (and the corresponding coverings) under the
action of recursion operators described above.

## Acknowledgments

Computations were done using the Jets [4] and Cadabra [6, 23, 24] software.
The authors are grateful to O.I. Morozov for discussion.

## References

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# Valuations and nonzero torsion in module of differentials

Sarasij Maitra<EMAIL_ADDRESS>Department of Mathematics, Indian
Institute of Technology Delhi, Hauz Khas, India.  and  Vivek Mukundan
<EMAIL_ADDRESS>
###### Abstract.

Let $(R,\mathfrak{m}_{R},\mathbb{k})$ be a one-dimensional complete local
reduced $\mathbb{k}$-algebra over a field of characteristic zero. Berger
conjectured that $R$ is regular if and only if the universally finite module
of differentials $\Omega_{R}$ is torsion free. When $R$ is a domain, we prove
the conjecture in several cases. Our techniques are primarily reliant on
making use of the valuation semi-group of $R$. First, we establish a method of
verifying the conjecture by analyzing the valuation semi-group of $R$ and
orders of units of the integral closure of $R$. We also prove the conjecture
in the case when certain monomials are missing from the monomial support of
the defining ideal of $R$. These monomials are based on the smallest power of
$\mathfrak{m}_{R}$ that is contained within the conductor ideal. This also
generalizes a previous result of [CnGW98].

###### Key words and phrases:

module of differentials, Berger Conjecture, reduced curves

###### 2010 Mathematics Subject Classification:

Primary: 13A15. Secondary: 13H05

##  1\. Introduction

Let $X$ be a (affine or projective) curve over a field $\mathbb{k}$ of
characteristic zero. The classification of circumstances when $X$ is smooth is
of importance to both algebraists and geometers. Though there are other
smoothness characterizations, the one involving the differential sheaf
$\Omega_{X,x}$, where $x$ is a closed point on $X$, has eluded researchers.
Consider the completion of the stalk of the structure sheaf $\mathcal{O}_{X}$
at $x$ and denote this by $\widehat{\mathcal{O}_{X,x}}$. R.W. Berger
conjectured that $X$ is smooth at $x$ if and only if the sheaf of
differentials $\widehat{\mathcal{O}_{X,x}}\otimes\Omega_{X,x}$ is torsion
free.

Let
$R=\widehat{\mathcal{O}_{X,x}}\cong\mathbb{k}\llbracket{X_{1},\dots,X_{n}}\rrbracket/I$
with $I\subseteq(X_{1},\dots,X_{n})^{2}$. For a curve, the differential sheaf
mentioned above is $R\otimes\Omega_{X,x}$ where $\Omega_{X,x}$ is the module
of Kähler differentials and $R\otimes\Omega_{X,x}$ is the universally finite
module of differentials, henceforth denoted by $\Omega_{R}$. For the purposes
of this article, we can define the universally finite module of differentials
$\Omega_{R}$ to be the module whose presentation is as follows:

$R^{\mu(I)}\stackrel{{\scriptstyle\mathcal{J}}}{{\to}}R^{n}\to\Omega_{R}\to 0$

where $\mu(I)$ is the minimal set of generators of $I$ and $\mathcal{J}$ is
the Jacobian matrix on a generating set for $I$, considered as a matrix over
$R$. For further details, we refer the reader to the excellent resource
[Kun86]. One of the first pieces of evidence on the smoothness of the curve
$X$ via the torsion freeness of its differential sheaf was studied by E. Kunz
([Kun61]) when the ring is of characteristic $p$ and the base field is
imperfect. Berger ([Ber63]) in 1963 conjectured the following for perfect
fields:

######  Conjecture A (R. W. Berger [Ber63]).

Let $\mathbb{k}$ be a perfect field and let $R$ be a reduced one-dimensional
analytic $\mathbb{k}$-algebra. Then $R$ is regular if and only if the torsion
submodule $\displaystyle\tau(\Omega_{R})$ of $\Omega_{R}$ vanishes.

It follows from the above presentation that if $R$ is regular (i.e., $n=1$),
then $\Omega_{R}$ is free. Hence it is the converse direction that is of
interest. Thus we assume $n\geqslant 2$ throughout this article.

A direct generalization of this conjecture for higher dimensional surfaces
does not hold (see for example [Mat67, Example Pg 160], [Sch70, Satz 9.4]).
But with added constraints on $R$ (such as Cohen-Macaulayness of $R$), a
generalized version does exist due to R. Hübl and D. Arapura independently
([H9̈0, Ara89]). There have been numerous instances in which A has been
established by a variety of different approaches (see [Sch70], [HK71], [Her78]
[BG80], [Ulr81] [Koc83], [HW84], [HW86], [Ber88], [Poh89], [Güt90], [Iso91],
[Poh91], [CnGW98], [CK99], [HMM21], [Mai22a]). An excellent exposition on the
earlier ones among the above listed results can be found in [Ber94]. An
Artinian variant of the conjecture, named the Artin-Berger conjecture, was
explored in [CnGW98] recovering some of the earlier cases.

Recently, the authors of this article along with C. Huneke [HMM21]
demonstrated a new approach for explicitly constructing torsion in
$\Omega_{R}$, where $R$ is assumed to be a domain with residue field
$\mathbb{k}$ of characteristic $0$ and embedding dimension $n$. In that
article, we provided a strategy for first constructing an over-ring $S$ such
that $R\subseteq S\subseteq\overline{R}=\mathbb{k}\llbracket{t}\rrbracket$,
and embedding dimension of $S$ is $n+s$. Here $\overline{R}$ denotes the
integral closure of $R$ in its field of fractions $\operatorname{Quot(R)}$. We
then demonstrated that constructing “sufficiently many” torsion elements in
$\Omega_{S}$ enables us to pull back a torsion element to $\Omega_{R}$. In
Proposition 3.3 of this article, we show that

$\lambda(\Omega_{S})\geqslant ns+{s\choose 2}$

(where $\lambda(M)$ denotes the length of a module $M$), which is one less
than the required number of torsion elements meeting the above sufficiency
criteria. This settles [HMM21, Remark 4.10].

Identifying $R$ within $\overline{R}=\mathbb{k}\llbracket{t}\rrbracket$ allows
us to write
$R=\mathbb{k}\llbracket{\alpha_{1}t^{a_{1}},\dots,\alpha_{n}t^{a_{n}}}\rrbracket$
where $\alpha_{i}$’s are units in $\overline{R}$. The valuation semi-group of
$R$, namely $v(R)$, is critical for our methods throughout this article. In
fact, the precise valuations $a_{i}$ and the valuation of the conductor ideal,
$\mathfrak{C}_{R}=R:_{\text{Quot}(R)}\overline{R}$, are critical in our work
of constructing torsion elements. Moreover, the “order” of the unit
$\alpha_{d}$ in the description of $x_{d}=\alpha_{d}t^{a_{d}}$ also plays a
role. Given $\alpha_{d}=\sum_{i=0}^{\infty}u_{di}t^{i}$, a unit in
$\mathbb{k}\llbracket{t}\rrbracket$, its order is defined to be

$o(\alpha_{d})=v(\alpha_{d}-u_{d0})=\min_{i\geqslant 1}\\{i~{}|~{}u_{ji}\neq
0\\}.$

We state our first result here which is quite technical, but we illustrate
with examples that the conditions are often satisfied (see Theorem 4.4 and
Theorem 4.6).

######  Theorem B.

Let
$R=\mathbb{k}\llbracket{\alpha_{1}t^{a_{1}},\dots,\alpha_{n}t^{a_{n}}}\rrbracket\subseteq\overline{R}=k[[t]]$
be as above with $\mathfrak{C}_{R}=(t^{c_{R}})\overline{R}$ and $\mathbb{k}$,
an algebraically closed field of characteristic zero. Then $\tau(\Omega_{R})$
is nonzero in the following cases:

  1. (1)

the valuations $a_{n},a_{n-1}$ satisfy $a_{n}+a_{n-1}\geqslant c_{R}$.

  2. (2)

there exist $i,d$ such that $\displaystyle o(\alpha_{d})+a_{i}\geqslant
c_{R}$, $\displaystyle o(\alpha_{d})<\infty$ and any one of the following
statements holds.

    1. (a)

$o(\alpha_{d})\leqslant o(\alpha_{i})<\infty$,

    2. (b)

$d\geqslant i$.

It is possible to get hold of a torsion element by mapping $\Omega_{R}$ to a
rank one torsion free module (see Section 2.2). However, proving that it is
non-zero is the most challenging part. Our discussion in Proposition 4.7
provides an approach towards achieving this goal, by going modulo a monomial
ideal. We implement this method and obtain our second main result. This is the
broadest class where we establish the truth of Berger’s Conjecture (see
Theorem 4.9).

######  Theorem C.

Let $R=\mathbb{k}\llbracket{X_{1},\dots,X_{n}}\rrbracket/I$ be a one-
dimensional local domain where $\mathbb{k}$ is algebraically closed of
characteristic zero and $I\subseteq(X_{1},\dots,X_{n})^{2}$. Let $x_{i}$
denote the image of $X_{i}$ in $R$ and let the conductor be
$\mathfrak{C}_{R}$. Suppose $\mathfrak{m}^{N}\subseteq\mathfrak{C}_{R}$ and
$x_{1}^{N-1},x_{1}^{N-2}x_{2}$, $N\geqslant 2$, are not in the monomial
support ideal of $I$. Then the torsion submodule $\tau(\Omega_{R})$ is
nonzero.

The monomial support ideal of $I$ is the same as $\operatorname{Mono}(I)$
appearing in [PUV07]. The theorem recovers the main theorem of [CnGW98]. We
also present other interesting examples that satisfy the hypothesis of the
above theorem.

This raises the question if condition (2) in C can be relaxed. So far our
methods have been unable to resolve this. Getting a complete answer even in
the case when $\mathfrak{C}_{R}=\mathfrak{m}^{N}$ for some large enough $N$
would be interesting. The methods used in this article are innovative, in
that, they explicitly compute torsion elements in $\Omega_{R}$, whereas most
of the previous approaches to the A have been unable to do so.

### Acknowledgements

Both authors are extremely grateful to C. Huneke for insightful talks about
the article’s content. Numerous results in this article are ascribed to the
concepts he imparted to us. Additionally, we are obliged to D. Arapura for
providing us with a geometric viewpoint on Berger’s Conjecture.

##  2\. Setting and Preliminaries

Throughout this article we will assume that $\mathbb{k}$ is an algebraically
closed field of characteristic $0$, and $(R,\mathfrak{m}_{R},\mathbb{k})$ is a
one-dimensional complete equicharacteristic local $\mathbb{k}$-algebra which
is a domain with embedding dimension $n$ ($\mu_{R}(\mathfrak{m}_{R})=n$).
Choosing $t$ to be a uniformizing parameter of the integral closure
$\overline{R}$ of $R$, we may assume that
$\overline{R}=\mathbb{k}\llbracket{t}\rrbracket$. As a result, we may write
$R=\mathbb{k}\llbracket{\alpha_{1}t^{a_{1}},\dots,\alpha_{n}t^{a_{n}}}\rrbracket$
where $\alpha_{i}$’s are units in $\overline{R}$ and
$a_{1}<a_{2}<\cdots<a_{n}$. Using the technique as in the first paragraph of
the proof of [HMM21, Theorem 3.1], we can alternatively write
$R=\mathbb{k}\llbracket{t^{a_{1}},\alpha_{2}t^{a_{2}},\dots,\alpha_{n}t^{a_{n}}}\rrbracket$.
We briefly explain the process. First we multiply by a nonzero element of
$\mathbb{k}$ to assume that the constant term of $\alpha_{1}$ is $1$. Let
$s=\beta t$ where $\beta^{a_{1}}=\alpha_{1}$. Such a $\beta$ exists due to
Hensel’s Lemma [Eis95, THeorem 7.3]. The characterisitic of $\mathbb{k}$ being
zero is important here. Notice that under this change of variables,
$\mathbb{k}\llbracket{t}\rrbracket=\mathbb{k}\llbracket{s}\rrbracket$ and
$s=\beta t=t+\beta_{1}t^{2}+\beta_{2}t^{3}+\cdots$ where
$\beta=1+\sum_{i=1}^{\infty}\beta_{i}t^{i}$. Notice that $s^{a_{1}}=(\beta
t)^{a_{1}}=\alpha_{1}t^{a_{1}}$. Also
$\alpha_{i}t^{a_{i}}=\alpha_{i}^{\prime}s^{a_{i}}$. Then
$R=\mathbb{k}\llbracket{\alpha_{1}t^{a_{1}},\dots,\alpha_{n}t^{a_{n}}}\rrbracket\cong\mathbb{k}\llbracket{s^{a_{1}},\alpha_{2}^{\prime}s^{a_{2}},\dots,\alpha_{n}^{\prime}s^{a_{n}}}\rrbracket$.

We also define the valuation $v$ on $\overline{R}$ given by $v(p(t))=a$ if
$p(t)=t^{a}\alpha$ where $\alpha$ is a unit in
$\mathbb{k}\llbracket{t}\rrbracket$ (also called $\operatorname{ord}(-)$, see
for example [SH06, Example 6.7.5]). Let $v(R)$ denote the valuation semi-group
of $R$.

The conductor $\mathfrak{C}_{R}$ is the largest common ideal of $R$ and its
integral closure, $\overline{R}$. Since
$\displaystyle\overline{R}=k\llbracket{t}\rrbracket$, we have that
$\mathfrak{C}_{R}=(t^{i})_{i\geqslant c_{R}}\overline{R}$ where $c_{R}$ is the
least integer such that $t^{c_{R}-1}\not\in R$, and $t^{c_{R}+i}\in R$ for all
$i\geqslant 0$. The number $c_{R}$ is characterized as the least valuation in
$\mathfrak{C}_{R}$. It is clear from this discussion that there cannot be any
element $r\in R$, such that $v(r)=c_{R}-1$. Since $\displaystyle\overline{R}$
is finitely generated over $R$ ([SH06, Theorem 4.3.4]), $\mathfrak{C}_{R}\neq
0$, and it is never all of $R$ unless $R$ is regular.

We define an epimorphism

$\displaystyle\Phi:\mathbb{k}\llbracket{X_{1},\dots,X_{n}}\rrbracket$
$\displaystyle\twoheadrightarrow R$ $\displaystyle\Phi(X_{i})$
$\displaystyle=\alpha_{i}t^{a_{i}}\text{ for }1\leqslant i\leqslant n.$

We denote the kernel of $\Phi$ by $I=(f_{1},\dots,f_{m})$ and hence have the
natural isomorphism
$R\cong\mathbb{k}\llbracket{X_{1},\dots,X_{n}}\rrbracket/I$. Since
$\operatorname{edim}R=n$, $I$ is contained in $\mathfrak{m}_{R}^{2}$ where
$\mathfrak{m}_{R}=(X_{1},\dots,X_{n})$, the maximal ideal of
$\mathbb{k}\llbracket{X_{1},\dots,X_{n}}\rrbracket$. Such rings are called
analytic $\mathbb{k}$-algebras. We will interchangeably use
$\alpha_{i}t^{a_{i}}$ for $x_{i}$, the images of $X_{i}$ in the quotient
$\mathbb{k}\llbracket{X_{1},\dots,X_{n}}\rrbracket/I$.

###  2.1. Universally Finite Module of Differentials

######  Definition 2.1.

Let $R$ be an analytic one-dimensional $\mathbb{k}$-algebra as above, which is
a domain. Let $I=(f_{1},\dots,f_{m})$ where $f_{j}\in
P=\mathbb{k}\llbracket{X_{1},\dots,X_{n}}\rrbracket,n\geqslant 2$. We assume
that $I\subseteq\mathfrak{m}_{P}^{2}$ where
$\mathfrak{m}_{P}=(X_{1},\dots,X_{n})$. Then the universally finite module of
differentials over $\mathbb{k}$, denoted by $\Omega_{R}$, has a (minimal)
presentation given as follows:

$R^{m}\xrightarrow{\left[\frac{\partial f_{j}}{\partial
x_{i}}\right]}R^{n}\to\Omega_{R}\to 0$

where $\left[\frac{\partial f_{j}}{\partial x_{i}}\right]$ is the Jacobian
matrix of $I$, with entries in $R$.

We refer the reader to the excellent resource [Kun86] for more information.
Let $\displaystyle\tau(\Omega_{R})$ denote the torsion submodule of
$\Omega_{R}$.

###  2.2. Computing Torsion

It is well-known that
$\operatorname{rank}_{R}(\Omega_{R})=\operatorname{rank}_{R}(\Omega_{\overline{R}})=1$
($R$ and $\bar{R}$ have the same fraction field) and $\Omega_{\overline{R}}$
is torsion-free. Let $f:\Omega_{R}\to\Omega_{\overline{R}}$ be the natural map
induced by the inclusion $R\hookrightarrow\overline{R}$. Recall that the
torsion submodule of $\Omega_{R}$ is the kernel of the homomorphism
$\Omega_{R}\rightarrow\Omega_{R}\otimes Q$ where $Q=\text{Quot}(R)$. Since
$\Omega_{\overline{R}}$ is torsion-free, we have
$\Omega_{\overline{R}}\hookrightarrow\Omega_{R}\otimes Q$. Thus $\ker f$ is
the torsion submodule of $\Omega_{R}$ and hence $\tau(\Omega_{R})=\ker f$. The
map $f$ acts as follows:

$f(dx_{i})=\frac{dx_{i}}{dt}dt.$

Since $\Omega_{\overline{R}}$ is isomorphic to $\overline{R}$, $\Omega_{R}$
surjects to an $R$-submodule $\sum_{i=1}^{n}R\frac{dx_{i}}{dt}$ of
$\overline{R}=\mathbb{k}\llbracket{t}\rrbracket$. Thus the torsion submodule,
$\tau(\Omega_{R})$, consists of the column vectors $\begin{bmatrix}r_{1}\ldots
r_{n}\end{bmatrix}^{t}$ such that $\sum_{i=1}^{n}r_{i}\frac{dx_{i}}{dt}=0$.
Evidently, $\tau(\Omega_{R})$ is nonzero precisely when the tuples
$\begin{bmatrix}r_{1}\ldots r_{n}\end{bmatrix}^{t}$ are not in the image of
the presentation matrix (Jacobian matrix of $I$) of $\Omega_{R}$, all entries
written in terms of the uniformizing parameter $t$.

One of the main issues while constructing torsion is to check whether the
column vectors $[r_{1},\dots,r_{n}]^{t}$ is nonzero in $\Omega_{R}$. One
technique is to show that this column vector is nonzero in $\Omega_{R/J}$ for
some suitable ideal $J$ in $R$ which will then imply that it is nonzero in
$\Omega_{R}$. Though we have a more general theorem for this purpose in
Proposition 4.7, a version which we use in some of the results in the results
in the sequel is the following theorem.

######  Theorem 2.2.

[CnGW98, Proposition 2.6] Suppose
$R=\mathbb{k}\llbracket{\alpha_{1}t^{a_{1}},\dots,\alpha_{n}t^{a_{n}}}\rrbracket$
with maximal ideal $\mathfrak{m}_{R}=(x_{1},\dots,x_{n})$. Then the torsion
$\tau(\Omega_{R/\mathfrak{m}_{R}^{2}})$ has ${n\choose 2}$
$\mathbb{k}$-linearly independent torsion elements
$\overline{x_{i}}d\overline{x_{j}}$ for $i<j$. Here $\overline{x_{i}}$ denotes
the image of $x_{i}$ in $R/\mathfrak{m}_{R}^{2}$.

We often avoid using the $\overline{\cdot}$ for notational convenience.

###  2.3. Quasi-homogeneous rings

These rings form an important class which satisfies the Berger conjecture. A
ring $R$ is said to be quasi-homogeneous, if there is a surjection
$\Omega_{R}\twoheadrightarrow\mathfrak{m}_{R}$. Kunz and Ruppert in [KR77]
showed that such rings are completions of graded rings. Scheja ([Sch70])
showed that the torsion submodule of the module of differentials of such rings
are always nonzero. In fact, he showed that
$\mu(\tau(\Omega_{R}))\geqslant{\operatorname{edim}R\choose 2}$ when $R$ is
quasi-homogeneous.

###  2.4. The extension $R[\frac{\mathfrak{C}_{R}}{x_{1}}]$

The basic facts about the finite birational extension
$S=R[\frac{\mathfrak{C}_{R}}{x_{1}}]$ of $R$ are discussed in detail in
[HMM21, Section 4].

We write $\mathfrak{C}_{R}/{x_{1}}$ to denote the set of elements of the form
${c}/{x_{1}}$ where $c\in\mathfrak{C}_{R}$. Using [HK71, Proposition 2.9] and
[HMM21, Lemma 4.1], we get that $\displaystyle\mathfrak{C}_{R}/x_{1}$ is
generated as an $R$-submodule of $\overline{R}$ (in fact, it is the ideal
$(t^{c-a_{1}})\overline{R}$) by $t^{b_{i}}$ with $b_{1}<\cdots<b_{s}$ where
$b_{i}\in[c-a_{1},c-1]\setminus v(R)$. For further details, see [HMM21,
Notation 4.2]. It turns out that $\mathfrak{C}_{R}/x_{1}$ is the conductor of
the ring $S$. In other words the conductor $\mathfrak{C}_{S}$ equals the ideal
$(t^{c-a_{1}})\overline{R}$. A complete description of the ring $S$ is as
follows:

######  Theorem 2.3.

[HMM21, Thoerem 4.6] Let
$\displaystyle\mathfrak{C}_{R}\subseteq\mathfrak{m}_{R}^{2}$. Set
$P=\mathbb{k}\llbracket{X_{1},...,X_{n}}\rrbracket$. Choose
$S,b_{1},\dots,b_{s}$ as described above. Then there exists a presentation of
$S$ as follows:

$S=R\Big{[}\frac{\mathfrak{C}_{R}}{x_{1}}\Big{]}=\frac{\mathbb{k}\llbracket{X_{1},\dots,X_{n},T_{1},\dots,T_{s}}\rrbracket}{\ker\Phi+\left(X_{i}T_{j}-g_{ij}(X_{1},\dots,X_{n}),T_{k}T_{l}-h_{kl}(X_{1},\dots,X_{n})\right)_{{1\leqslant
i\leqslant n,1\leqslant j\leqslant s\atop 1\leqslant k\leqslant l\leqslant
s}}}$

where
$g_{ij}(X_{1},\dots,X_{n}),h_{kl}(X_{1},\dots,X_{n})\in(X_{1},\dots,X_{n})^{2}P$
for all $i,j,k,l$.
Moreover,
$g_{ij}(x_{1},\dots,x_{n}),h_{kl}(x_{1},\dots,x_{n})\in\mathfrak{C}_{R}$. In
particular, when $\mathfrak{C}_{R}\subseteq\mathfrak{m}_{R}^{2}$, it follows
that $\operatorname{edim}S=n+s$.

##  3\. Finer computations of Torsion elements in $\Omega_{S}$

One of the main results in this section is the existence of torsion elements
in $\Omega_{S}$. We show that we can in fact construct $ns+{s\choose 2}$
$\mathbb{k}$-linearly independent torsion elements in $\Omega_{S}$ which
completes [HMM21, Remark 4.10].

The following theorem establishes that once $ns+{s\choose 2}+1$ torsion
elements are constructed with entries in $\mathfrak{m}_{S}$, then we can pull
back a nonzero torsion element from $\Omega_{S}$ to $\Omega_{R}$.

######  Theorem 3.1.

[HMM21, Theorem 4.9] Let
$R=\mathbb{k}\llbracket{\alpha_{1}t^{a_{1}},\dots,\alpha_{n}t^{a_{n}}}\rrbracket$
with conductor $\mathfrak{C}_{R}\subseteq\mathfrak{m}_{R}^{2}$. Construct
$S=R\left[\frac{\mathfrak{C}_{R}}{x_{1}}\right]=R[T_{1},\dots,T_{s}]$ as in
Theorem 2.3. Let $\tau(\Omega_{R}),\tau(\Omega_{S})$ represent the torsion
submodules of $\Omega_{R},\Omega_{S}$ respectively. If
$\lambda(\tau(\Omega_{S}))\geqslant ns+{s\choose 2}+1$ and all these torsion
elements have non-units in the last $s$ rows (corresponding to
$dT_{1},\dots,dT_{s}$), then a $\mathbb{k}$-linear combination of these
torsion elements can be pulled back to a nonzero torsion element in
$\tau(\Omega_{R})$. In particular, $\tau(\Omega_{R})\neq 0$.

######  Remark 3.2.

An obvious consequence of this theorem is that if
$\lambda(\tau(\Omega_{S}))\geqslant ns+{s\choose 2}+k,k\geqslant 1$ and all
these torsions have non-units in the last $s$ rows, then we can pull back $k$
torsion elements to $\Omega_{R}$, each of which are $\mathbb{k}$-linear
combinations of the $ns+{s\choose 2}+k$ torsion elements. Notice that these
$k$ torsion elements are $\mathbb{k}$-linearly independent in $\Omega_{R}$ as
they are $\mathbb{k}$-linearly independent in $\Omega_{S}$. Thus
$\lambda(\tau(\Omega_{R}))\geqslant k$.

It is easy to see that we can construct ${s\choose 2}$ torsion elements,
namely $\Gamma_{ij}:=b_{j}T_{j}dT_{i}-b_{i}T_{i}dT_{j}$. This can be clearly
seen as $T_{i}$ are monomials $t^{b_{i}}$ in $S$. Now Proposition 3.3 shows
that we can indeed construct $ns$ torsion elements in $\Omega_{S}$ too.

If $x_{i}$ is a monomial $t^{a_{i}}$ to begin with, then we can construct the
torsion element $a_{i}x_{i}dT_{j}-b_{j}T_{j}dx_{i}$. This is clearly nonzero
in $\Omega_{S}$. But often $x_{i}$ are not of this form and hence the need for
the following theorem.

######  Proposition 3.3.

Let
$R=\mathbb{k}\llbracket{\alpha_{1}t^{a_{1}},\dots,\alpha_{n}t^{a_{n}}}\rrbracket$
with conductor $\mathfrak{C}_{R}$. Construct
$S=R\left[\frac{\mathfrak{C}_{R}}{x_{1}}\right]=R[T_{1},\dots,T_{s}]$. Then

(3.16) $\displaystyle[x_{i},T_{j}]=\begin{bmatrix}-a_{i}x_{i}\partial
f_{ij}/\partial x_{1}\\\ \vdots\\\
b_{j}(T_{j}+f_{ij}(\underline{x})+\sum\limits_{k=1}^{s}\delta_{k}T_{k})-a_{i}x_{i}\partial
f_{ij}/\partial x_{i}\\\ \vdots\\\ -a_{i}x_{i}(1+\delta_{j})\\\ \vdots\\\
-a_{i}x_{i}\delta_{s}\end{bmatrix}\begin{array}[]{c;{2pt/2pt}c}{}\hfil&{}\\\
{}\hfil&{}\\\ {}\hfil&{}\\\ {}\hfil&{}\\\ {}\hfil&{}\\\ {}\hfil&{}\\\
{}\hfil&{}\\\ {}\hfil&{}\\\ \end{array}\overbrace{\begin{array}[]{c}dx_{1}\\\
\vdots\\\ dx_{i}\\\ \vdots\\\ dT_{j}\\\ \vdots\\\
dT_{s}\end{array}}^{\text{basis}}$

where $f_{ij}\in(x_{1},\dots,x_{n})$ and $\delta_{r}\in\mathbb{k}$, are
$\mathbb{k}$-linearly independent nonzero torsion elements in
$\tau(\Omega_{S})$.

###### Proof.

We follow the process of “monomialization” as in [HMM21, Theorem 3.1] (see
Section 2). Let $\beta=\alpha_{i}^{1/a_{i}}$ and $s=\beta t$ (since the
characteristic of $\mathbb{k}$ is $0$, by Hensel’s lemma [Eis95, Theorem 7.3],
there exists such a $\beta\in\overline{R}$). Thus
$R\cong\mathbb{k}\llbracket{\alpha_{1}^{\prime}s^{a_{1}},\alpha_{2}^{\prime}s^{a_{2}},\dots,s^{a_{i}},\dots,\alpha_{n}^{\prime}s^{a_{n}}}\rrbracket$.
Under this isomorphism (change of variables), let the new variables be
$x_{1}^{\prime},\dots,x_{n}^{\prime},T_{1}^{\prime},\dots,T_{s}^{\prime}$. The
valuations do not change under this isomorphism. Since
$T_{1}^{\prime}\dots,T_{s}^{\prime}\in\mathfrak{C}_{S}$, we assume that,
without loss of generality, under another change of variables,
$T_{1}^{\prime},\dots,T_{s}^{\prime}$ are monomials
$s^{b_{1}},\dots,s^{b_{s}}$. Notice that under this change of variables
$x_{i}^{\prime}=s^{a_{i}}$. Clearly, for $j=1,\dots,s$ we have

(3.17) $\displaystyle
b_{j}T_{j}^{\prime}d(x_{i}^{\prime})-a_{i}x_{i}^{\prime}d(T_{j}^{\prime})$

is a nonzero torsion element of $\Omega_{S}$ (the nonzero part can be obtained
by showing that
$\overline{b_{j}T_{j}^{\prime}d(x_{i}^{\prime})-a_{i}x_{i}^{\prime}d(T_{j}^{\prime})}\equiv(b_{j}-a_{i})\overline{T_{j}^{\prime}}d\overline{x_{i}^{\prime}}$
in $\Omega_{S/\mathfrak{m}_{S}^{2}}$, which is nonzero using [CnGW98,
Proposition 2.6]).

Now we convert it back to the original variables
$x_{1},\dots,x_{n},T_{1},\dots,T_{s}$. Under the change of variables, $x_{i}$
is mapped to $x_{i}^{\prime}$. So we just need to convert $T_{j}^{\prime}$
back to $T_{j}$. We have
$T_{j}^{\prime}=s^{b_{j}}=\beta^{b_{j}}t^{b_{j}}=t^{b_{j}}+\sum_{k}\gamma_{k}t^{b_{j}+k}$.
Since $\mathfrak{C}_{S}=(t^{c_{R}-a_{1}})\overline{R}$ and $c_{R}-a_{1}$ is
smaller than $b_{j}$, we can replace $t^{b_{j}+k}$ as a function of
$x_{1},\dots,x_{n},T_{1},\dots,T_{s}$. Thus we have

$\displaystyle
T_{j}^{\prime}=t^{b_{j}}+\sum_{k}\gamma_{k}t^{b_{j}+k}=T_{j}+g_{ij}(x_{1},\dots,x_{n},T_{1},\dots,T_{s}).$

Due to the nature of the defining ideal of $S$,

(3.18) $\displaystyle
g_{ij}(x_{1},\dots,x_{n},T_{1},\dots,T_{s})=f_{ij}(x_{1},\dots,x_{n})+\sum\limits_{k>j}^{s}\delta_{ik}T_{k}$

where $\delta_{ik}\in\mathbb{k}$. In the latter equality, only $T_{k}$ with
$k>j$ appear because the only nonzero terms which can appear in
$g_{ij}(\underline{x},\underline{T})$ are those which have valuations more
than that of $T_{j}$.

Therefore

$\displaystyle dT_{j}^{\prime}=dT_{j}+\sum_{u=1}^{n}\frac{\partial
f_{ij}}{\partial x_{u}}dx_{u}+\sum_{k>j}^{s}\delta_{ik}dT_{k}$

Putting the above equation in (3.17) we have

(3.37) $\displaystyle[x_{i},T_{j}]=\begin{bmatrix}-a_{i}x_{i}\partial
f_{ij}/\partial x_{1}\\\ \vdots\\\
b_{j}(T_{j}+f_{ij}(\underline{x})+\sum\limits_{k>j}^{s}\delta_{ik}T_{k})-a_{i}x_{i}\partial
f_{ij}/\partial x_{i}\\\ \vdots\\\ -a_{i}x_{i}\\\ -a_{i}x_{i}\delta_{ij+1}\\\
\vdots\\\
-a_{i}x_{i}\delta_{is}\end{bmatrix}\begin{array}[]{c;{2pt/2pt}c}{}\hfil&{}\\\
{}\hfil&{}\\\ {}\hfil&{}\\\ {}\hfil&{}\\\ {}\hfil&{}\\\ {}\hfil&{}\\\
{}\hfil&{}\\\ {}\hfil&{}\\\ {}\hfil&{}\\\ {}\hfil&{}\\\
\end{array}\overbrace{\begin{array}[]{c}dx_{1}\\\ \vdots\\\ dx_{i}\\\
\vdots\\\ dT_{j}\\\ dT_{j+1}\\\ \vdots\\\ dT_{s}\end{array}}^{\text{basis}}.$

Next we show that $[x_{i},T_{j}],1\leqslant i\leqslant n,1\leqslant j\leqslant
s$ are $\mathbb{k}$-linearly independent. First let
$f_{ij}(x_{1},\dots,x_{n})=\sum_{k}\beta_{ijk}x_{k}+f_{ij}^{\prime}(x_{1},\dots,x_{n})$
where
$f_{ij}^{\prime}(x_{1},\dots,x_{n})\in\mathfrak{m}_{R}^{2},\beta_{ijk}\in\mathbb{k}$.
Now consider the torsion elements $[x_{i},T_{j}]$ in
$\Omega_{S/\mathfrak{m}_{S}^{2}}$. These would take the form

(3.38) $\displaystyle\begin{bmatrix}-a_{i}x_{i}\beta_{ij1}\\\ \vdots\\\
-a_{i}x_{i}\beta_{iji-1}\\\ b_{j}(T_{j}+\sum_{k\neq
i}\beta_{ijk}x_{k}+\sum\limits_{k>j}^{s}\delta_{ijk}T_{k})\\\
-a_{i}x_{i}\beta_{iji+1}\\\ \vdots\\\ -a_{i}x_{i}\beta_{ijn}\\\ 0\\\ \vdots\\\
0\\\ -a_{i}x_{i}\\\ \vdots\\\ -a_{i}x_{i}\delta_{ijs}\end{bmatrix}$
$\displaystyle=\begin{bmatrix}0\\\ \vdots\\\ 0\\\
(b_{j}-a_{i})T_{j}+\sum\limits_{k>j}^{s}\delta_{ijk}(b_{j}-a_{i})T_{k}-(b_{j}+a_{i})\sum_{k\leqslant
i}\beta_{ijk}x_{k}\\\ -(b_{j}+a_{i})x_{i}\beta_{iji+1}\\\ \vdots\\\
-(b_{j}+a_{i})x_{i}\beta_{ijn}\\\ 0\\\ \vdots\\\ 0\end{bmatrix},$

where the equality is due to the fact that
$x_{1}x_{i},\dots,x_{i-1}x_{i},x_{i}T_{1},\dots,x_{i}T_{s}\in\mathfrak{m}_{S}^{2}$
(the latter containment implies that $x_{j}dx_{i}=-x_{i}dx_{j},j\leqslant i$
and $x_{i}dT_{j}=-T_{j}dx_{i}$ in $\Omega_{S/\mathfrak{m}_{S}^{2}}$). So to
show that $[x_{i},T_{j}],1\leqslant i\leqslant n,1\leqslant j\leqslant s$ are
$\mathbb{k}$-linearly independent, it is enough to show that these torsions
are $\mathbb{k}$-linearly independent in $\Omega_{S/\mathfrak{m}_{S}^{2}}$.

We first show that $[x_{i},T_{j}],1\leqslant j\leqslant s$, for a fixed $i$,
are $\mathbb{k}$-linearly independent. It is known that
$x_{k}dx_{l},T_{j}dx_{i},1\leqslant k<l\leqslant n,1\leqslant j\leqslant s$
are $\mathbb{k}$-linearly independent in $\Omega_{S/\mathfrak{m}_{S}^{2}}$
[CnGW98, Proposition 2.6, Corollary 2.7]. Now writing the coefficients of part
of the ordered basis $x_{k}dx_{l},T_{j}dx_{i},1\leqslant k<l\leqslant
n,1\leqslant j\leqslant s$ in (3.38), we get a $s\times({n\choose 2}+ns)$
matrix

Assume the first block is $A_{i}$ which is a $s\times{n\choose 2}$ matrix, and
the second block is $[0\cdots 0~{}B_{i}~{}0\cdots 0]$ which is a $s\times{ns}$
matrix, where $B_{i}$ is an upper triangular $s\times s$ matrix. The columns
of $A_{i}$ correspond to the ordered basis $x_{k}dx_{l},1\leqslant
k<l\leqslant n$ and that of $[0\cdots 0~{}B_{i}~{}0\cdots 0]$ correspond to
the ordered basis $T_{j}dx_{i},1\leqslant j\leqslant s$. Clearly the matrix
$B_{i}$ has rank $s$ as $a_{i},b_{1},\dots,b_{s}$ are distinct and nonzero.
Since the rows are $\mathbb{k}$-linearly independent,
$[x_{i},T_{j}]=\left((b_{j}-a_{i})T_{j}+\sum\limits_{k>j}^{s}\delta_{ijk}(b_{j}-a_{i})T_{k}-(b_{j}+a_{i})\sum_{k\leqslant
i}\beta_{ijk}x_{k}\right)dx_{i}-\sum_{k=i+1}^{n}a_{i}x_{i}\beta_{ijk}dx_{k}$
are $\mathbb{k}$-linearly independent for a fixed $i$ and $1\leqslant
j\leqslant s$.

If we consider all the torsion elements $[x_{i},T_{j}]$ for arbitrary $i,j$,
then we get the following matrix

$\displaystyle\begin{bmatrix}[x_{1},T_{1}]\\\ \vdots\\\
[x_{n},T_{s}]\end{bmatrix}=\begin{bmatrix}A_{1}&B_{1}&0&0&\cdots&0\\\
A_{2}&0&B_{2}&0&\cdots&0\\\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\\
A_{n}&0&0&0&\cdots&B_{n}\end{bmatrix}\left[\begin{array}[]{c}x_{1}dx_{2}\\\
\vdots\\\ x_{n-1}dx_{n}\\\ \hline\cr T_{1}dx_{1}\\\ \vdots\\\
T_{s}dx_{n}\end{array}\right]$

where each of the $B_{i}$ are upper triangular invertible $s\times s$ matrix.
Thus the rows are $\mathbb{k}$-linearly independent and hence
$[x_{i},T_{j}]$’s are $\mathbb{k}$-linearly independent. ∎

######  Theorem 3.4.

With the hypothesis as in Proposition 3.3, we have

$\lambda(\tau(\Omega_{S}))\geqslant ns+{s\choose 2}.$

###### Proof.

We observe that $\Gamma_{ij}=b_{j}T_{j}dT_{i}-b_{i}T_{i}dT_{j}$ and
$[x_{i},T_{j}]$ are $\mathbb{k}$-linearly independent: suppose

$0=\sum_{i,j}k_{ij}[x_{i},T_{j}]+\sum_{r,s}k^{\prime}_{rs}\Gamma_{rs},k_{ij},k^{\prime}_{rs}\in\mathbb{k}.$

Now let $J=\langle x_{1},\dots,x_{n}\rangle+\langle
T_{1},\dots,T_{s}\rangle^{2}$. Then in $\Omega_{S/J}$, $[x_{i},T_{j}]=0$
(using (3.38)). Thus in $\Omega_{S/J}$, we have
$0=\sum_{r,s}k^{\prime}_{rs}\Gamma_{rs}$. Since $\Gamma_{rs}$ are
$\mathbb{k}$-linearly independent ([CnGW98, Proposition 2.6]), we have
$k^{\prime}_{rs}=0$. Thus we have $0=\sum_{i,j}k_{ij}[x_{i},T_{j}]$ in
$\Omega_{S}.$ Hence $k_{ij}=0$ by Proposition 3.3. Thus
$\lambda(\tau(\Omega_{S}))\geqslant ns+{s\choose 2}$.∎

######  Remark 3.5.

In [HMM21, Theorem 4.11], we established that $\tau(\Omega_{R})$ is nonzero if
$S=R\left[\frac{\mathfrak{C}_{R}}{x_{1}}\right]$ is quasi-homogeneous. Recall
that quasi-homogeneous means that there is a surjection
$\Omega_{R}\twoheadrightarrow\mathfrak{m}_{R}$. In fact, the proof mentioned
above proceeded by showing that there are ${n+s\choose 2}$
$\mathbb{k}$-linearly independent torsion elements in $\Omega_{S}$ and none of
these have units in the last $s$ rows. However, notice that ${n+s\choose
2}=ns+{s\choose 2}+\binom{n}{2}$. Thus we can pull back ${n\choose 2}$
torsions to $\Omega_{R}$ using Theorem 3.1 and Remark 3.2, i.e.,
$\lambda(\tau(\Omega_{R}))\geqslant\binom{n}{2}$.

##  4\. Value semi-group of $R$ and nonzero torsion

In this section, we provide partial answers to Berger’s conjecture by
carefully observing the value semi-group of $R$ as well as order of the units
$\alpha_{j}$ in $\mathbb{k}\llbracket{t}\rrbracket$.

We begin by stating a useful test of checking the nonvanishing of torsion in
$\Omega_{R}$. First notice that units in $\mathbb{k}\llbracket{t}\rrbracket$
are power series of the form $\sum_{i=0}^{\infty}u_{i}t^{i}$ where $u_{0}\neq
0$.

######  Notation 4.1.

Let
$\alpha_{j}=\sum_{i=0}^{\infty}u_{ji}t^{i}\in\mathbb{k}\llbracket{t}\rrbracket$
be a unit. We denote

$\displaystyle o(\alpha_{j})=v(\alpha_{j}-u_{j0})=\min_{i\geqslant
1}\\{i~{}|~{}u_{ji}\neq 0\\},$

where $v(\cdot)$ denotes the order valuation (also called
$\operatorname{ord}(-)$, see for example [SH06, Example 6.7.5]). If
$\alpha_{j}=u_{j0}\in\mathbb{k}$, then notice that $o(\alpha_{j})=\infty$.

######  Theorem 4.2.

Let the conductor of $R$ be $\mathfrak{C}_{R}=(t^{c_{R}})\overline{R}$.
Suppose there exists $i,d$ such that $\displaystyle
o(\alpha_{d})+a_{i}\geqslant c_{R}$ and $\displaystyle o(\alpha_{d})<\infty$.
Then the torsion $\tau(\Omega_{R})$ is nonzero if any one of the following
statements holds.

  1. (1)

$o(\alpha_{d})\leqslant o(\alpha_{i})<\infty$,

  2. (2)

$d\geqslant i$.

###### Proof.

As in [HMM21, Theorem 3.1], we write
$R=\mathbb{k}\llbracket{t^{a_{1}},\alpha_{2}t^{a_{2}},\dots,\alpha_{n}t^{a_{n}}}\rrbracket$
with conductor $\mathfrak{C}_{R}=(t^{c_{R}})\overline{R}$. Notice that the
$o(\alpha_{i})$ remain the same by applying the above change of variables. For
$(1)$, observe that $o(\alpha_{i})+a_{i}\geqslant o(\alpha_{d})+a_{i}\geqslant
c_{R}$. Now

$\displaystyle
x_{i}=\alpha_{i}t^{a_{i}}=\alpha_{i0}t^{a_{i}}+t^{o(\alpha_{i})+a_{i}}b$

where $b\in\overline{R}$. Since $o(\alpha_{i})+a_{i}\geqslant c_{R}$,
$t^{o(\alpha_{i})+a_{i}}b\in\mathfrak{C}_{R}\subseteq R$. Thus,
$t^{a_{i}}=\frac{1}{\alpha_{i0}}(x_{i}-t^{o(\alpha_{i})+a_{i}}b)\in R$. Thus,
we get that
$R=k\llbracket{t^{a_{1}},\alpha_{2}t^{a_{2}},\dots,t^{a_{i}},\dots,\alpha_{n}t^{a_{n}}}\rrbracket$.
The proof is now complete using [HMM21, Remark 3.3].

For $(2)$, first observe that by our setup, $o(\alpha_{d})+a_{d}\geqslant
o(\alpha_{d})+a_{i}\geqslant c_{R}$. Apply the same argument as above to see
that
$R=k\llbracket{t^{a_{1}},\alpha_{2}t^{a_{2}},\dots,t^{a_{d}},\dots,\alpha_{n}t^{a_{n}}}\rrbracket$.
This finishes the proof.

∎

The following is an example of a non-quasi-homogeneous ring $R$ where the
above result can be quickly applied.

######  Example 4.3.

Let
$R=\mathbb{k}\llbracket{t^{8}+t^{9},t^{9}+t^{15},t^{12}+t^{20},t^{14}}\rrbracket=\mathbb{k}\llbracket{t^{8}(1+t),t^{9}(1+t^{6}),t^{12}(1+t^{8}),t^{14}}\rrbracket$.
Macaulay2 computations show that the conductor is
$(t^{c_{R}})\overline{R}=(t^{20})\overline{R}$. With $i=3,d=3$, we see that
$a_{i}+o(\alpha_{d})=12+8=20=c_{R}$. Thus by the above theorem, the torsion
$\tau(\Omega_{R})$ is nonzero. Moreover, Macaulay 2 computations show that
$v(\text{trace}(\Omega_{R}))=16$ where $\text{trace}(\cdot)$ refers to the
trace ideal (see [Lin17] for details). Hence, $R$ is not quasi-homogeneous
[Mai22a, Conclusion 2].

The second part of the next result appears as [Mai22b, Theorem 4.9].

######  Theorem 4.4.

Suppose the valuations $a_{1},\dots,a_{n}$ of $x_{1},\dots,x_{n}$ of $R$
satisfies one of the following:

  1. (1)

$a_{1}+a_{n}\geqslant c_{R}$,

  2. (2)

$a_{n-1}+a_{n}\geqslant c_{R}+a_{1}$

Then the torsion $\tau(\Omega_{R})$ is nonzero.

###### Proof.

Throughout the proof, we can assume that
$\mathfrak{C}_{R}\subseteq\mathfrak{m}_{R}^{2}$ as otherwise the conclusion
follows from [HMM21, Theorem 3.1]. Further applying the same “monomialization”
as in the proof of Theorem 4.2, we can assume that
$R=k\llbracket{t^{a_{1}},\alpha_{2}t^{a_{2}},\dots,\alpha_{n}t^{a_{n}}}\rrbracket$.

(1): Consider $S=R\left[\frac{\mathfrak{C}_{R}}{x_{1}}\right]$ with conductor
$\mathfrak{C}_{S}=(t^{c-a_{1}})\overline{R}$. Since $a_{n}\geqslant c-a_{1}$,
$x_{n}$, by change of variables, can be made into monomial $t^{a_{n}}$. Thus
$t^{a_{n}}=x_{n}^{\prime}=x_{n}-\sum\delta_{i}T_{i}-g(\underline{x}),\delta_{i}\in\mathbb{k}$.
Then
$S=\mathbb{k}\llbracket{x_{1},\dots,x_{n-1},x_{n}^{\prime},T_{1},\dots,T_{s}}\rrbracket=\mathbb{k}\llbracket{t^{a_{1}},\alpha_{2}t^{a_{2}},\dots,\alpha_{n-1}t^{a_{n-1}},t^{a_{n}},t^{b_{1}},\dots,t^{b_{s}}}\rrbracket$.
Clearly $\tau=a_{n}x_{n}^{\prime}dx_{1}-a_{1}x_{1}dx_{n}^{\prime}$ is a
nonzero torsion in $\Omega_{S}$. Hence (using the notations as in Section 3)
$\tau,[x_{i},T_{j}],\Gamma_{ij}$ are $ns+{s\choose 2}+1$ torsion elements in
$\Omega_{S}$. If we show that these torsion elements are $\mathbb{k}$-linearly
independent, then the result follows from Theorem 3.1.

Consider

(4.1) $\displaystyle{\gamma\tau+\sum k_{ij}[x_{i},T_{j}]+\sum
k_{ij}^{\prime}\Gamma_{ij}=0.}$

First let $J=\langle x_{1},\dots,x_{n}^{\prime}\rangle+\mathfrak{m}^{2}_{S}$;
then in $\Omega_{S/J}$, (4.1) takes the form $\sum
k^{\prime}_{ij}\Gamma_{ij}=0$. But these are $\mathbb{k}$-linearly independent
and hence $k_{ij}^{\prime}=0$. Now let $J_{1}=\langle
x_{n}^{\prime}\rangle+\mathfrak{m}_{S}^{2}$. Then in $\Omega_{S/J_{1}}$,
$\tau=0$ as $x_{n}^{\prime}\in J_{1}$ and $dx_{n}^{\prime}=0$ in
$\Omega_{S/J_{1}}$. Thus (4.1) now becomes $\sum_{i\neq
n}k_{ij}[x_{i},T_{j}]=0$. Since we already proved $[x_{i},T_{j}]$’s are
$\mathbb{k}$-linearly independent, we have $k_{ij}=0$ for $i\neq n,1\leqslant
j\leqslant s$. Now (4.1) becomes
$\gamma\tau+\sum_{j}k_{nj}[x_{n}^{\prime},T_{j}]=0$. Now let $J_{2}=\langle
x_{1}\rangle+\mathfrak{m}^{2}_{S}$. Then in $\Omega_{S/J_{2}}$, we have again
$\tau=0$, but
$0\neq[x_{n}^{\prime},T_{j}]=a_{n}x_{n}^{\prime}dT_{j}-b_{j}T_{j}dx_{n}^{\prime}=(a_{n}+b_{j})x_{n}^{\prime}dT_{j}$.
Thus we have $k_{nj}=0$ as well. We now have $\gamma\tau=0$ which implies
$\gamma=0$. Thus $\tau,[x_{i},T_{j}],\Gamma_{ij}$ are $\mathbb{k}$-linearly
independent.

(2): In $\Omega_{\overline{R}}\cong\mathbb{k}\llbracket{t}\rrbracket$ where
$\overline{R}=\mathbb{k}\llbracket{t}\rrbracket$, we have

$\displaystyle\frac{dx_{n}}{dt}\left(\frac{dx_{1}}{dt}\right)$
$\displaystyle-\frac{dx_{1}}{dt}\left(\frac{dx_{n}}{dt}\right)=0$ (4.2)
$\displaystyle\left(a_{n}\alpha_{n}t^{a_{n}-1}+t^{a_{n}}\frac{d(\alpha_{n})}{dt}\right)\left(\frac{dx_{1}}{dt}\right)$
$\displaystyle-a_{1}t^{a_{1}-1}\left(\frac{dx_{n}}{dt}\right)=0$

Multiplying (4.2) by $\alpha_{n-1}t^{a_{n-1}-a_{1}+1}$, we get

$\displaystyle\left(a_{n}\alpha_{n}\alpha_{n-1}t^{a_{n}-1+a_{n-1}-a_{1}+1}+\alpha_{n-1}t^{a_{n}+a_{n-1}-a_{1}+1}\frac{d\alpha_{n}}{dt}\right)\frac{dx_{1}}{dt}$
$\displaystyle-
a_{1}\alpha_{n-1}t^{a_{1}-1+a_{n-1}-a_{1}+1}\frac{dx_{n}}{dt}=0$
$\displaystyle\left(a_{n}\alpha_{n}\alpha_{n-1}t^{a_{n}+a_{n-1}-a_{1}}+\alpha_{n-1}t^{a_{n}+a_{n-1}-a_{1}+1}\frac{d\alpha_{n}}{dt}\right)\frac{dx_{1}}{dt}$
$\displaystyle-a_{1}\alpha_{n-1}t^{a_{n-1}}\frac{dx_{n}}{dt}=0$ (4.3)
$\displaystyle\left(a_{n}\alpha_{n}\alpha_{n-1}t^{a_{n}+a_{n-1}-a_{1}}+\alpha_{n-1}t^{a_{n}+a_{n-1}-a_{1}+1}\frac{d\alpha_{n}}{dt}\right)\frac{dx_{1}}{dt}$
$\displaystyle-a_{1}x_{n-1}\frac{dx_{n}}{dt}=0$

The order valuation of
$u=a_{n}\alpha_{n}\alpha_{n-1}t^{a_{n}+a_{n-1}-a_{1}}+\alpha_{n-1}t^{a_{n}+a_{n-1}-a_{1}+1}\frac{d\alpha_{n}}{dt}$
is at least $a_{n}+a_{n-1}-a_{1}$. Since $a_{n}+a_{n-1}-a_{1}\geqslant c_{R}$,
we get that $u\in\mathfrak{C}_{R}\subseteq R$. Thus (4.3), as a member of
$\Omega_{R}$, takes the form

(4.4) $\displaystyle\begin{bmatrix}u\\\ 0\\\ \vdots\\\ 0\\\
a_{1}x_{n-1}\end{bmatrix}$

Since $\mathfrak{C}_{R}\subseteq\mathfrak{m}_{R}^{2}$, consider the torsion
element (4.4) in $\Omega_{R/\mathfrak{m}_{R}^{2}}$. It takes the from
$a_{1}x_{n-1}d(x_{n})$ which is nonzero due to [CnGW98, Proposition 2.6,
Corollary 2.7]. ∎

######  Remark 4.5.

An immediate consequence of the above theorem is that, henceforth, we can
always assume $a_{n}<c_{R}-a_{1}$. This in turn results in the simplification
of the description of $[x_{i},T_{j}]$ (3.38) in
$\Omega_{S/\mathfrak{m}_{S}^{2}}$. First, consider the description of
$f_{ij}(x_{1},\dots,x_{n})$ in (3.18). The valuations of the terms in
$f_{ij}(x_{1},\dots,x_{n})$ are more than that of $T_{j}$, or more than
$c_{R}-a_{1}$. Since $a_{n}<c_{R}-a_{1}$, no linear terms can appear in
$f_{ij}(x_{1},\dots,x_{n})$. Thus
$f_{ij}(x_{1},\dots,x_{n})\in\mathfrak{m}_{R}^{2}$ and hence $\beta_{ijk}=0$.
Thus (3.38) takes the form

(4.22) $\displaystyle\begin{bmatrix}0\\\ \vdots\\\ 0\\\
(b_{j}-a_{i})T_{j}+\sum\limits_{k>j}^{s}\delta_{ik}(b_{j}-a_{i})T_{k}\\\ 0\\\
\vdots\\\ 0\end{bmatrix}\begin{array}[]{c;{2pt/2pt}c}{}\hfil&{}\\\
{}\hfil&{}\\\ {}\hfil&{}\\\ {}\hfil&{}\\\ {}\hfil&{}\\\ {}\hfil&{}\\\
{}\hfil&{}\\\ {}\hfil&{}\\\ {}\hfil&{}\\\
{}\hfil&{}\end{array}\overbrace{\begin{array}[]{c}dx_{1}\\\ \vdots\\\
dx_{i-1}\\\ dx_{i}\\\ dx_{i+1}\\\ \vdots\\\ dT_{s}\end{array}}^{\text{basis}}$

in $\Omega_{S/\mathfrak{m}_{S}^{2}}$. This simpler form will help us to prove
$\mathbb{k}$-linear independence of the torsion elements.

######  Theorem 4.6.

Suppose the valuations $a_{1},\dots,a_{n}$ of $x_{1},\dots,x_{n}$ of $R$
satisfy $a_{n}+a_{n-1}\geqslant c_{R}$, then the torsion $\tau(\Omega_{R})$ is
nonzero.

###### Proof.

Throughout the proof, we can assume that
$\mathfrak{C}_{R}\subseteq\mathfrak{m}_{R}^{2}$ as otherwise the conclusion
follows from [HMM21, Theorem 3.1].

Consider $S=R\left[\frac{\mathfrak{C}_{R}}{x_{1}}\right]$ with conductor
$\mathfrak{C}_{S}=(t^{c_{R}-a_{1}})\overline{R}$. If
$x_{n}\in\mathfrak{C}_{s}$, then $a_{n}\geqslant c_{S}=c_{R}-a_{1}$ and hence
we are done by Theorem 4.4(1).

So it is enough to assume $x_{n}\not\in\mathfrak{C}_{S}$. Since
$a_{n-1}+a_{n}\geqslant c_{S}+a_{1}=(c_{R}-a_{1})+a_{1}$, we use the same
steps as in Theorem 4.4(2) and end up at the torsion element (4.4), now in
$\Omega_{S}$ (we put zeros in the last $s$ rows corresponding to
$dT_{1},\dots,dT_{s}$). This torsion element is nonzero as it is nonzero in
$\Omega_{S/(\mathfrak{C}_{S}+\mathfrak{m}_{S}^{2})}$.

Let $\tau$ denote this torsion element. Next we show that
$\tau,\mathfrak{r}_{ij}=[x_{i},T_{j}],{\Gamma_{ij}=[T_{i},T_{j}]}$ (as in
Section 3) are $\mathbb{k}$-linearly independent. In the description of
$\tau$, let
$u=f_{u}(x_{1},\dots,x_{n})+\sum\limits_{k=1}^{s}\gamma_{k}T_{k},\gamma_{k}\in\mathbb{k}$.
Since $x_{n}\not\in\mathfrak{C}_{S}$ and $u\in\mathfrak{C}_{S}$, we have
$f_{u}(x_{1},\dots,x_{n})\in\mathfrak{m}_{R}^{2}$. Thus $\tau$ in
$\Omega_{S/\mathfrak{m}_{S}^{2}}$, will take the form

$\displaystyle\tau=\begin{bmatrix}\sum\limits_{k=1}^{s}\gamma_{k}T_{k}\\\ 0\\\
\vdots\\\ 0\\\ a_{1}x_{n-1}\\\ 0\\\ \vdots\\\
0\end{bmatrix}\begin{array}[]{c;{2pt/2pt}c}{}\hfil&{}\\\ {}\hfil&{}\\\
{}\hfil&{}\\\ {}\hfil&{}\\\ {}\hfil&{}\\\ {}\hfil&{}\\\ {}\hfil&{}\\\
{}\hfil&{}\\\ {}\hfil&{}\\\ {}\hfil&{}\\\
\end{array}\overbrace{\begin{array}[]{c}{}\hfil\\\ dx_{1}\\\ dx_{2}\\\
\vdots\\\ dx_{n-1}\\\ dx_{n}\\\ dT_{1}\\\ \vdots\\\
dT_{s}\end{array}}^{\text{basis}}.$

Let $J=\mathfrak{m}_{S}^{2}+\langle T_{1},\dots,T_{s}\rangle$. In
$\Omega_{S/J}$, the above torsion takes the from $a_{1}x_{n-1}dx_{n}$ which is
nonzero by [CnGW98, Proposition 2.6]. Now suppose
$\tau=\sum_{i,j}k_{ij}[x_{i},T_{j}]{+\sum_{i,j}k^{\prime}_{ij}\Gamma_{ij}}$,
$k_{ij},k^{\prime}_{ij}\in\mathbb{k}$. In $\Omega_{S/J}$,
$[x_{i},T_{j}]={\Gamma_{ij}=0}$ (see (4.22)) and this contradicts the fact
that $\tau$ is nonzero in $\Omega_{S/J}$. Thus $\tau$ is $\mathbb{k}$-linearly
independent with $[x_{i},T_{j}],\Gamma_{ij}$. Now the result follows from
[HMM21, Theorem 4.9]. ∎

######  Proposition 4.7.

(Monomial Case) Let $K=(m_{1},\dots,m_{s})$ be a monomial ideal minimally
generated by monomials $m_{i}=X_{1}^{b_{i1}}\cdots X_{n}^{b_{in}}$ in
$\mathbb{k}\llbracket{X_{1},\dots,X_{n}}\rrbracket$ and
$R^{\prime}=\mathbb{k}\llbracket{X_{1},\dots,X_{n}}\rrbracket/K$. Suppose
$m=X_{1}^{b_{1}}\cdots X_{n}^{b_{n}}\not\in K$ where $b_{u}\geqslant 1$ for
some $1\leqslant u\leqslant n$. Then $mdx_{u}=0$ in $\Omega_{R^{\prime}}$ if
and only if $X_{u}^{b_{u}+1}\in K$ or $\frac{\partial(mX_{u})}{\partial
X_{i}}\in K$ for all $1\leqslant i\neq u\leqslant n$. Here $x_{i}$ denotes the
image of $X_{i}$ in $R^{\prime}$.

###### Proof.

First, let $mdx_{u}=0$. Without loss of generality we assume that $u=1,i=2$.
Suppose by contradiction, $X_{1}^{b_{1}+1}\not\in K$ and
$\frac{mX_{1}}{X_{2}}\not\in K$. Here we replaced
$\frac{\partial(mX_{u})}{\partial X_{i}}=\frac{\partial(mX_{1})}{\partial
X_{2}}$ by $\frac{mX_{1}}{X_{2}}$.

Consider the monomial ideal
$L=(X_{1}^{b_{1}+1}X_{2}^{b_{2}})+K=(X_{1}^{b_{1}+1}X_{2}^{b_{2}},m_{1},\dots
m_{s})$. First suppose that $m_{i}$, for some $1\leqslant i\leqslant s$,
divides $X_{1}^{b_{1}+1}X_{2}^{b_{2}}$. Then $m_{i}=X_{1}^{u}X_{2}^{v}$ where
$u\leqslant b_{1}+1,v\leqslant b_{2}$. If $u<b_{1}+1,v\leqslant b_{2}$, then
$m\in(m_{i})\subseteq K$, a contradiction. Now suppose $u=b_{1}+1,v<b_{2}$,
then $\frac{mX_{1}}{X_{2}}\in(m_{i})\subseteq K$, a contradiction. Therefore,
in these cases, none of the $m_{i}$’s divide $X_{1}^{b_{1}+1}X_{2}^{b_{2}}$.
Finally if $u=b_{1}+1,v=b_{2}$, then $m_{i}=X_{1}^{b_{1}+1}X_{2}^{b_{2}}$ and
hence $K=L$. Let $(X_{1}^{b_{1}+1}X_{2}^{b_{2}},m_{1},\dots m_{l}),l\leqslant
s$ be a minimal generating set for $L$. Also, clearly
$\frac{mX_{1}}{X_{2}}=X_{1}^{b_{1}+1}X_{2}^{b_{2}-1}X_{2}^{b_{3}}\cdots
X_{n}^{b_{n}}\not\in L$.

Now since $mdx_{1}=0$ in $\Omega_{R^{\prime}}$, we also have $mdx_{1}=0$ in
$\Omega_{\mathbb{k}[[X_{1},\dots,X_{n}]]/L}$. Thus

$\displaystyle
mdx_{1}=rd(x_{1}^{b_{1}+1}x_{2}^{b_{2}})+\sum_{i=1}^{l}r_{i}dm_{i},r,r_{i}\in{\mathbb{k}[[X_{1},\dots,X_{n}]]/L}$

(here we have again used $x_{i}$ to denote the images in the suitable
quotient). Therefore,

(4.23) $\displaystyle m$
$\displaystyle=r\frac{\partial(x_{1}^{b_{1}+1}x_{2}^{b_{2}})}{\partial
x_{1}}+\sum_{i=1}^{l}r_{i}\frac{\partial m_{i}}{\partial
x_{1}}=r(b_{1}+1)x_{1}^{b_{1}}x_{2}^{b_{2}}+\sum_{i=1}^{l}r_{i}b_{i1}x_{1}^{b_{i1}-1}x_{2}^{b_{i2}}\cdots
x_{n}^{b_{in}}.$ (4.24) $\displaystyle 0$
$\displaystyle=r\frac{\partial(x_{1}^{b_{1}+1}x_{2}^{b_{2}})}{\partial
x_{k}}+\sum_{i=1}^{l}r_{i}\frac{\partial m_{i}}{\partial x_{k}}\text{ for
}2\leqslant k\leqslant n.$

First we analyze (4.23). Suppose $\frac{\partial m_{i}}{\partial x_{1}}$
divides $m$, then we have $b_{i1}-1\leqslant b_{1}$ and $b_{ik}\leqslant
b_{k}$ for $2\leqslant k\leqslant n$. If $b_{i1}-1<b_{1}$, then $m\in K$ which
leads to a contradiction. Therefore, $b_{i1}=b_{1}+1$. Now if $b_{i2}<b_{2}$,
then $\frac{mX_{1}}{X_{2}}\in K$ which is again a contradiction. Thus
$b_{i2}=b_{2}$. But now this forces $m_{i}\in(X_{1}^{b_{1}+1}X_{2}^{b_{2}})$
which is also contradiction as
$(X_{1}^{b_{1}+1}X_{2}^{b_{2}},m_{1},\dots,m_{l})$ is a minimal generating set
for $L$. Thus $\frac{\partial m_{i}}{\partial x_{1}}$ does not divide $m$.
This shows that $m_{i}$’s have the property that either $b_{i1}>b_{1}+1$ or
$b_{ik}>b_{k}$ for some $2\leqslant k\leqslant n$.

From the above discussion and (4.23), we have
$r=\frac{1}{b_{1}+1}x_{3}^{b_{3}}\cdots x_{n}^{b_{n}}$. Using $k=2$ in (4.24),
we have

$\displaystyle 0$
$\displaystyle=r\frac{\partial(x_{1}^{b_{1}+1}x_{2}^{b_{2}})}{\partial
x_{2}}+\sum_{i=1}^{s}r_{i}\frac{\partial m_{i}}{\partial x_{2}}$
$\displaystyle=rb_{2}x_{1}^{b_{1}+1}x_{2}^{b_{2}-1}+\sum_{i=1}^{s}r_{i}b_{i2}x_{1}^{b_{i1}}x_{2}^{b_{i2}-1}\cdots
x_{n}^{b_{in}}$
$\displaystyle=\frac{b_{2}}{b_{1}+1}\frac{mx_{1}}{x_{2}}+\sum_{i=1}^{s}r_{i}b_{i2}x_{1}^{b_{i1}}x_{2}^{b_{i2}-1}\cdots
x_{n}^{b_{in}}$

By our choice of $m_{i}$ (preceeding discussion on choice of $b_{ik}$), we
have that the term $\frac{b_{2}}{b_{1}+1}\frac{mx_{1}}{x_{2}}$ cannot cancel
with any term in the sum
$\sum_{i=1}^{s}r_{i}b_{i2}x_{1}^{b_{i1}}x_{2}^{b_{i2}-1}\cdots
x_{n}^{b_{in}}$. Thus $\frac{b_{2}}{b_{1}+1}\frac{mx_{1}}{x_{2}}=0$ or in
other words, $\frac{b_{2}}{b_{1}+1}\frac{mX_{1}}{X_{2}}\in L$. Since
$\frac{b_{2}}{b_{1}+1}\in\mathbb{k}$, this forces $\frac{mX_{1}}{X_{2}}\in L$,
which leads to a contradiction. Thus we have either $X_{1}^{b_{u}+1}\in K$ or
$\frac{\partial(mX_{u})}{\partial X_{i}}\in K$.

Conversely, suppose $\frac{\partial(mX_{u})}{\partial X_{i}}\in K$ for all
$1\leqslant i\neq u\leqslant n$. This implies $mX_{u}\in K$ and hence
$d(mX_{u})=0$ in $\Omega_{R^{\prime}}$. Hence
$\displaystyle(b_{u}+1)mdx_{u}+\sum_{i=1,i\neq
u}^{n}b_{i}\frac{mx_{u}}{x_{i}}dx_{i}\in KdR^{\prime}+dK$. Using the
assumption, we have $(b_{u}+1)mdx_{u}\in KdR^{\prime}+dK$. Also, clearly, if
$X_{u}^{b_{u}+1}\in K$, then $mdx_{u}=0$, completing the proof. ∎

In the following, we use the notion of monomial support of an ideal which is
the same as $\operatorname{Mono}(I)$ in [PUV07]. The ideal
$\operatorname{Mono}(I)$ is defined to be the smallest monomial ideal
containing $I$ and is generated by the monomial support of a generating set
for $I$.

######  Example 4.8.

If $I=(X^{3}-YZ,X^{2}Y-Z^{2},XZ-Y^{2})\subseteq
S=\mathbb{k}\llbracket{X,Y,Z}\rrbracket$, then
$\operatorname{Mono}(I)=(X^{3},YZ,X^{2}Y,Z^{2},XZ,Y^{2})$. Notice that $I$ is
the defining ideal of $\mathbb{k}\llbracket{t^{3},t^{4},t^{5}}\rrbracket$.
Clearly, we have $\tau=3xdz-5zdx\in\tau(\Omega_{S/I})$. We can now use
Proposition 4.7 with $K=\operatorname{Mono}(I)$ and $mdx_{u}=xdz$ to see that
$\tau$ is nonzero in $\Omega_{S/\operatorname{Mono}(I)}$. Hence, it is nonzero
in $\Omega_{S/I}$.

######  Theorem 4.9.

Let
$R=\mathbb{k}\llbracket{\alpha_{1}t^{a_{1}},\dots,\alpha_{n}t^{a_{n}}}\rrbracket\cong\mathbb{k}\llbracket{X_{1},\dots,X_{n}}\rrbracket/I$
with conductor $\mathfrak{C}_{R}=(t^{c_{R}})\overline{R}$. Let
$\tau(\Omega_{R})$ denote the torsion submodule of $\Omega_{R}$ and $x_{i}$
denote the images of $X_{i}$ in $R$. Let $N\geqslant 2$ be the integer such
that $\mathfrak{m}^{N}\subseteq\mathfrak{C}_{R}$ and suppose that
$x_{1}^{N-1},x_{1}^{N-2}x_{2}\not\in\operatorname{Mono}(I)$ where $I$ is the
defining ideal of $R$. Then the torsion submodule $\tau(\Omega_{R})$ is
nonzero.

We can always choose $N$ to be the least integer satisfying all the conditions
in the hypothesis. For example, if $k$ is the least integer such that
$\mathfrak{m}^{k}\subseteq\mathfrak{C}_{R}$ and $k\leqslant N$, then
$x_{1}^{k-1},x_{1}^{k-2}x_{2}\not\in\operatorname{Mono}(I)$ (as
$x_{1}^{n-1},x_{1}^{N-2}x_{2}\not\in\operatorname{Mono}(I)$) as well. Thus, we
will assume $N$ is least henceforth. The hypothesis now guarantees that
$x_{1}^{N-1}\not\in\mathfrak{C}_{R}$. Before we go to the proof of the
theorem, we present a small lemma which helps us reduce to the case that
$x_{1}^{N-2}x_{2}\not\in\mathfrak{C}_{R}$.

######  Lemma 4.10.

Under the conditions of the above theorem, if
$x_{1}^{N-2}x_{2}\in\mathfrak{C}_{R}$, then the torsion $\tau(\Omega_{R})$ is
nonzero.

###### Proof.

Since $x_{1}^{N-2}x_{2}\in\mathfrak{C}_{R}$, we have
$t^{(N-2)a_{1}+a_{2}}\in\mathfrak{C}_{R}$. Thus there exists
$f(\underline{x})$ such that
$t^{(N-2)a_{1}+a_{2}}=x_{1}^{N-2}x_{2}-f(\underline{x})$. Since the valuation
$v(x_{1}^{N-2}x_{2})=a_{1}(N-2)+a_{2}$, the valuation of $f(\underline{x})$ is
more than $(N-2)a_{1}+a_{2}$. Now we know that
$a_{1}(t^{a_{1}})dt^{(N-2)a_{1}+a_{2}}-((N-2)a_{1}+a_{2})t^{(N-2)a_{1}+a_{2}}dt^{a_{1}}=0$.
Thus we have the torsion element

$a_{1}x_{1}d(x_{1}^{N-2}x_{2}-f(\underline{x}))-\left((N-2)a_{1}+a_{2})x_{1}^{N-2}x_{2}-f(\underline{x})\right)dx_{1}\\\
=a_{1}x_{1}\left((N-2)x_{1}^{N-3}x_{2}dx_{1}+x_{1}^{N-2}dx_{2}-\sum_{i}\frac{\partial
f}{\partial
x_{i}}dx_{i}\right)-\left((N-2)a_{1}+a_{2})x_{1}^{N-2}x_{2}-f(\underline{x})\right)dx_{1}\\\
=(N-2)a_{1}x_{1}^{N-2}x_{2}dx_{1}+a_{1}x_{1}^{N-1}dx_{2}-\sum_{i}a_{1}x_{1}\frac{\partial
f}{\partial
x_{i}}dx_{i}-\left((N-2)a_{1}+a_{2})x_{1}^{N-2}x_{2}-f(\underline{x})\right)dx_{1}\\\
\hfilneg=\left(-a_{2}x_{1}^{N-2}x_{2}+f(\underline{x})\right)dx_{1}+a_{1}x_{1}^{N-1}dx_{2}-\sum_{i}a_{1}x_{1}\frac{\partial
f}{\partial x_{i}}dx_{i}.\hskip 10000.0pt$

Let $J=\langle x_{3},\dots,x_{n}\rangle+\operatorname{Mono}\langle p\in
R~{}|~{}v(p)>(N-2)a_{1}+a_{2}\rangle+\operatorname{Mono}(I)$ (we treat the
last part as an ideal in $R$; we keep writing $\operatorname{Mono}(I)$ for
notational convenience) where $I$ is the defining ideal of $R$. Note that $J$
is a monomial ideal not containing $x_{1}^{N-1},x_{1}^{N-2}x_{2}$ as these are
not in $\operatorname{Mono}(I)$ and its valuations are less than or equal to
$(N-2)a_{1}+a_{2}$. Also notice that $f(\underline{x})\in J$ as the valuation
of $f(\underline{x})$ is more than $(N-2)a_{1}+a_{2}$. Now in $\Omega_{R/J}$,
we show that the above torsion element takes the form
$\left(-a_{2}x_{1}^{N-2}x_{2}\right)dx_{1}+a_{1}x_{1}^{N-1}dx_{2}$. First,
$dx_{i}=0$ for $i\geqslant 3$ and hence it is enough to show that
$\sum_{i=1}^{2}a_{1}x_{1}\frac{\partial f}{\partial x_{i}}dx_{i}=0$ in
$\Omega_{R/J}$.

Since $f(\underline{x})$ has valuation more than $(N-2)a_{1}+a_{2}$,
$a_{1}x_{1}\frac{\partial f}{\partial x_{1}}$ is either $0$ or has valuation
more than $a_{1}+(N-3)a_{1}+a_{2}=(N-2)a_{1}+a_{2}$. Thus it vanishes in
$R/J$. So all that remains is $a_{1}x_{1}\frac{\partial f}{\partial x_{2}}$
which is either $0$ or has valuation more than
$a_{1}+(N-2)a_{1}=(N-1)a_{1}=v(x_{1}^{N-1})$. Since
$x_{1}^{N-1}\not\in\operatorname{Mono}(I)$ and $x_{i}\in J$ for $i\geqslant
3$, we need to only focus on the case when one of the terms in the product
$a_{1}x_{1}\frac{\partial f}{\partial x_{2}}$ is possibly $x_{1}^{N-2}x_{2}$.
This in turn shows that $f(\underline{x})$ has a term $\beta
x_{1}^{N-3}x_{2}^{2},\beta\in\mathbb{k}$. But the valuation shows that
$x_{1}^{N-3}x_{2}^{2}\in J$. Hence $a_{1}x_{1}\frac{\partial f}{\partial
x_{1}}dx_{1}+a_{1}x_{1}\frac{\partial f}{\partial x_{2}}dx_{2}=0+\beta
a_{1}x_{1}d(x_{1}^{N-3}x_{2}^{2})=0$ in $\Omega_{R/J}$, i.e.,
$\sum_{i}a_{1}x_{1}\frac{\partial f}{\partial x_{i}}dx_{i}=0$ in
$\Omega_{R/J}$ as well.

Now in $\Omega_{R/J}$, the torsion element
$\left(-a_{2}x_{1}^{N-2}x_{2}\right)dx_{1}+a_{1}x_{1}^{N-1}dx_{2}$ simplifies
to $\omega=-(a_{2}+a_{1}(N-1))x_{1}^{N-2}x_{2}dx_{1}$. This is because
$x_{1}^{N-1}x_{2}\in J$ and hence
$x_{1}^{N-1}dx_{2}+(N-1)x_{1}^{N-2}x_{2}dx_{1}=0$ in $\Omega_{R/J}$. Now
Proposition 4.7 can be applied with $m=x_{1}^{N-2}x_{2}$ to see that $\omega$
is nonzero. ∎

###### Proof of Theorem 4.9.

Due to the previous lemma, we assume
$x_{1}^{N-1},x_{1}^{N-2}x_{2}\not\in\mathfrak{C}_{R}$. Also, we can safely
assume that we have “monomialized” $x_{1}$ so that
$R=k\llbracket{t^{a_{1}},\alpha_{2}t^{a_{2}},\dots,\alpha_{n}t^{a_{n}}}\rrbracket$.

Consider

$\displaystyle\frac{dx_{2}}{dt}\left(\frac{dx_{1}}{dt}\right)-\frac{dx_{1}}{dt}\left(\frac{dx_{2}}{dt}\right)=0$
$\displaystyle(\alpha_{2}a_{2}t^{a_{2}-1}+t^{a_{2}}\frac{\partial\alpha_{2}}{\partial
t})\left(\frac{dx_{1}}{dt}\right)-a_{1}t^{a_{1}-1}\left(\frac{dx_{2}}{dt}\right)=0$

Multiply both sides by $t^{(N-2)a_{1}+1}$ to get

$\displaystyle(\alpha_{2}a_{2}t^{(N-2)a_{1}+a_{2}}+t^{(N-2)a_{1}+a_{2}+1}\frac{\partial\alpha_{2}}{\partial
t})\left(\frac{dx_{1}}{dt}\right)-a_{1}t^{(N-1)a_{1}}\left(\frac{dx_{2}}{dt}\right)=0$
(4.25)
$\displaystyle(a_{2}x_{2}x_{1}^{N-2}+t^{(N-2)a_{1}+a_{2}+1}\frac{\partial\alpha_{2}}{\partial
t})\left(\frac{dx_{1}}{dt}\right)-a_{1}x_{1}^{N-1}\left(\frac{dx_{2}}{dt}\right)=0$

We have $S=R[T_{1},\dots,T_{s}]$ and
$\mathfrak{C}_{S}=(t^{c-a_{1}})\overline{R}$. Since $(N-2)a_{1}+a_{2}\geqslant
c-a_{1}$ ($\mathfrak{m}^{N}\subseteq\mathfrak{C}_{R}$), we have that the above
torsion element is a member of $\tau(\Omega_{S})$.

The above torsion element takes the form:

$\displaystyle\tau=\begin{bmatrix}a_{2}x_{2}x_{1}^{N-2}+t^{(N-2)a_{1}+a_{2}+1}\frac{\partial\alpha_{2}}{\partial
t}\\\ -a_{1}x_{1}^{N-1}\\\ 0\\\ \vdots\\\ 0\end{bmatrix}.$

For the defining ideal $L$ of $S$, let the monomial support be denoted by
$\operatorname{Mono}(L)$.

We first show that $\tau$ is nonzero in $\Omega_{S}$. Let $J^{\prime}=\langle
x_{3},\dots,x_{n},T_{1},\dots,T_{s}\rangle+\operatorname{Mono}(L)+\operatorname{Mono}(p\in
S~{}|~{}v(p)>(N-2)a_{1}+a_{2})$. It is a monomial ideal. Now consider $\tau$
in $\Omega_{S/J^{\prime}}$. Since the valuation of
$t^{(N-2)a_{1}+a_{2}+1}\frac{\partial\alpha_{2}}{\partial t}$ is more than
$(N-2)a_{1}+a_{2}$, it will be of the form

$\displaystyle\overline{\tau}=\begin{bmatrix}a_{2}x_{1}^{N-2}x_{2}\\\
-a_{1}x_{1}^{N-1}\\\ 0\\\ \vdots\\\ 0\end{bmatrix}\in\Omega_{S/J^{\prime}}.$

By hypothesis, $x_{1}^{N-2}x_{2},x_{1}^{N-1}$ are not in the monomial support
of the defining ideal of $R$. Now the rest of the defining ideal of $S$ is of
the form $x_{i}T_{j}-g_{ij}(\underline{x}),T_{i}T_{j}-h_{ij}(\underline{x})$.
The valuation of $x_{i}T_{j}$ is $a_{i}+b$ where $b\geqslant c_{R}-a_{1}$.
Since $x_{1}^{N-1}\not\in\mathfrak{C}_{R}$, the valuation $c_{R}>(N-1)a_{1}$.
Thus for $i\geqslant 2$, we have $a_{i}+b\geqslant
a_{i}+c_{R}-a_{1}>a_{i}+(N-2)a_{1}\geqslant
a_{2}+(N-2)a_{1}=v(x_{1}^{N-2}x_{2})>v(x_{1}^{N-1})$. Thus the monomial
support of $g_{ij}(\underline{x}),i\geqslant 2$ can never have
$x_{1}^{N-2}x_{2},x_{1}^{N-1}$ as terms. Also, the valuation of $x_{1}T_{j}$
is at least $c_{R}>(N-2)a_{1}+a_{2}$ (as
$x_{1}^{N-2}x_{2}\not\in\mathfrak{C}_{R}$). Thus $g_{1j}(\underline{x})$
cannot have $x_{1}^{N-2}x_{2},x_{1}^{N-1}$ as terms as well. Applying similar
arguments to analyze the elements $T_{i}T_{j}-h_{ij}(\underline{x})$, we
conclude that $x_{1}^{N-2}x_{2},x_{1}^{N-1}$ cannot appear in the monomial
support of $L$. Thus these terms do not lie in $J^{\prime}$ too.

Since $x_{1}^{N-1}x_{2}\in J^{\prime}$, we have $d(x_{1}^{N-1}x_{2})=0$ in
$\Omega_{S/J^{\prime}}$. Thus we have
$(N-1)x_{1}^{N-2}x_{2}dx_{1}=-x_{1}^{N-1}dx_{2}$. Using this equality in the
above torsion element we have

$\displaystyle\overline{\tau}$
$\displaystyle=(a_{2}x_{1}^{N-2}x_{2})dx_{1}-a_{1}x_{1}^{N-1}dx_{2}$
$\displaystyle=(a_{2}x_{1}^{N-2}x_{2})dx_{1}+a_{1}(N-1)x_{1}^{N-2}x_{2}dx_{1}$
(4.26) $\displaystyle=(a_{2}+a_{1}(N-1))x_{1}^{N-2}x_{2}dx_{1}.$

Clearly $(a_{2}+a_{1}(N-1))\neq 0$, and $x_{1}^{N-2}x_{2}\not\in J^{\prime}$.
Using Proposition 4.7, we get that the above torsion element in nonzero in
$\Omega_{S/J^{\prime}}$ and hence $\tau$ is nonzero in $\Omega_{S}$.

Next we show that $\tau$ is $\mathbb{k}$-linearly independent with the torsion
elements $[x_{i},T_{j}],{\Gamma_{ij}}$ (using the notations as in Section 3).
Suppose

(4.27)
$\displaystyle\tau=\sum_{i,j}k_{ij}[x_{i},T_{j}]+\sum_{i,j}k^{\prime}_{ij}\Gamma_{ij}$

where $k_{ij},k^{\prime}_{ij}\in\mathbb{k}$.

Let $J^{\prime\prime}=\langle x_{1}\rangle+\mathfrak{m}_{S}^{2}$. Then in
$\Omega_{S/J^{\prime\prime}}$, the torsion element $\tau$ (4.25) is zero.
Since $[x_{1},T_{j}]=a_{1}x_{1}dT_{j}-b_{j}T_{j}dx_{1}$, this element is also
zero in $\Omega_{S/J^{\prime\prime}}$. Thus in $\Omega_{S/J^{\prime\prime}}$,
Equation 4.27 is of the form $0=\sum_{i\geqslant 2,j\geqslant
1}k_{ij}\overline{[x_{i},T_{j}]}+\sum_{i,j}k^{\prime}_{ij}\overline{\Gamma_{ij}}$.
Now consider the description of $\overline{[x_{i},T_{j}]},i\geqslant
2,j\geqslant 1$ in (4.5). Following the same proof as in Theorem 3.4, it is
easy to see that $\overline{[x_{i},T_{j}]},i\geqslant 2,j\geqslant 1$ and the
$\overline{\Gamma_{ij}}$ are still $\mathbb{k}$-linearly independent in
$\Omega_{S/J^{\prime\prime}}$. It follows that $k_{ij}=0$ for $i\geqslant
2,j\geqslant 1$ and also $k^{\prime}_{ij}=0$ for $1\leqslant i,j\leqslant s$.
Thus we have $\tau=\sum_{j\geqslant 1}k_{ij}[x_{1},T_{j}]$ in $\Omega_{S}$.
Now consider the ideal $J^{\prime}=\langle
x_{3},\dots,x_{n},T_{1},\dots,T_{s}\rangle+\operatorname{Mono}(L)+\operatorname{Mono}(p\in
S~{}|~{}v(p)>(N-2)a_{1}+a_{2})$ from the previous part of the proof. In
$\Omega_{S/J^{\prime}}$, $\tau$ is nonzero which we already proved above. But
$\overline{[x_{1},T_{j}]}=0$ in $\Omega_{S/J^{\prime}}$ as $T_{j}\in
J^{\prime}$. Thus we arrive at $\tau=0$ in $\Omega_{S/J^{\prime}}$, a
contradiction. This proves the $\mathbb{k}$-linear independence of
$\tau,[x_{i},T_{j}],\Gamma_{ij}$. Thus, $\lambda(\tau(\Omega_{S}))\geqslant
ns+{s\choose 2}+1$.

Hence, a $\mathbb{k}$-linear combination of $\tau,[x_{i},T_{j}],\Gamma_{ij}$
can be pulled back to a nonzero torsion element in $\Omega_{R}$ by [HMM21,
Theorem 4.9]. ∎

The above theorem recovers [CnGW98, Theorem 2.13].

######  Corollary 4.11.

[CnGW98, Theorem 2.13] Under the conditions of Theorem 4.9, if
$\mathfrak{m}^{3}\subseteq\mathfrak{C}_{R}$, then the torsion
$\tau(\Omega_{R})$ is nonzero.

###### Proof.

Since $\mathfrak{m}^{3}\subseteq\mathfrak{C}_{R}$, we only need to check if
$x_{1}^{2},x_{1}x_{2}\not\in\operatorname{Mono}(I)$ to use the above theorem.
This is equivalent to checking if
$X_{1}^{2}-f(\underline{X}),X_{1}X_{2}-g(\underline{X})\in I$ or
$x_{1}^{2}=f(\underline{x}),x_{1}x_{2}=g(\underline{x})$ in $R$.

We have $v(x_{1}^{2})=2a_{1},v(x_{1}x_{2})=a_{1}+a_{2}$. Also, note that
$v(f)$ or $v(g)$ is of the form $\sum_{i=1}^{n}j_{i}a_{i},j_{i}\in\mathbb{N}$.
Since $I\subseteq\mathfrak{m}^{2}$, $f(\underline{X})$ and $g(\underline{X})$
cannot have any linear terms in $X_{i}$. Thus, either $j_{i}\geqslant 2$ for
some $i\geqslant 2$ or there exist at least two values of $i$, say
$\\{i_{1},i_{2}\\}\neq\\{1,2\\}$, such that $j_{i}\geqslant 1$. Since we have
$a_{1}<a_{2}<\cdots<a_{n}$, it is clear that the valuations
$2a_{1},a_{1}+a_{2}$ cannot be attained in either of these situations. Hence,
$x_{1}^{2},x_{1}x_{2}\not\in\operatorname{Mono}(I)$. ∎

######  Example 4.12.

Let
$R=\mathbb{k}\llbracket{t^{22},t^{23}+t^{27},t^{24}+t^{27},t^{25}+t^{27},t^{26}+t^{27}}\rrbracket\cong\mathbb{k}\llbracket{x,y,z,w,u}\rrbracket$.
Here $\mathfrak{C}_{R}=(t^{110})\overline{R}$ and hence
$\mathfrak{m}^{5}\subseteq\mathfrak{C}_{R}$. M2 computations show that
$x^{4},x^{3}y\not\in\operatorname{Mono}(I)$. Thus the torsion submodule
$\tau(\Omega_{R})\neq 0$ by the above theorem.

######  Example 4.13.

Let
$R=\mathbb{k}\llbracket{t^{30},t^{31}+t^{36},t^{32}+t^{36},t^{33}+t^{36},t^{34}+t^{36}}\rrbracket\cong\mathbb{k}\llbracket{x,y,z,w,u}\rrbracket$.
Here $\mathfrak{C}_{R}=(t^{180})\overline{R}$ and hence
$\mathfrak{m}^{6}\subseteq\mathfrak{C}_{R}$. M2 computations show that
$x^{5},x^{4}y\not\in\operatorname{Mono}(I)$. Thus the torsion submodule
$\tau(\Omega_{R})\neq 0$ by the above theorem.

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