††Keywords: Generalized cancellation problem, Danielewski type varieties,
Additive group action.

# Characterization and Non-cancellation property of Danielewski type varieties

Parnashree Ghosh  and  Dibyendu Mondal  Stat-Math Unit, Indian Statistical
Institute, 203 B.T. Road, Kolkata 700 108, West Bengal, India.
<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

Let $R$ be a factorial affine domain over a field $k$. In this paper we
consider Danielewski and Double Danielewski type rings over $R$. We observe
that these two families produce non-cancellative families of rings over $R$.
When $k$ is a field of characteristic zero, we give characterisation for an
affine variety to be isomorphic to a variety of Double Danielewski type. We
further extend some characterization results of Danielewski type hypersurface
rings containing fields of arbitrary characteristics.

###### 2020 Mathematics Subject Classification:

14R20, 13N15

##  1\. Introduction

All rings considered in this paper are commutative Noetherian rings with
unity. By $k$ we denote a field. For an integral domain $R$, the notation
$R^{[n]}$ denotes a polynomial ring in $n$ variables over $R$ and $Q(R)$
denotes the field of fractions over $R$. The group of units of $R$ is denoted
by $R^{*}$. If $T/L$ is a finite extension of fields then the degree of the
extension is denoted by $[T:L]$. We begin with the following version of
Generalised Cancellation Problem over integral domains (in short, GCP).

######  Question 1.1.

[8] Let $R$ be an integral domain and $A$, $B$ be two $R$-domains such that
$A^{[1]}=_{R}B^{[1]}$. Does this imply that $A\cong_{R}B$ ?

When $R=k$ and $A$, $B$ are one dimensional affine domains, then the answer to
the above question is affirmative ([1]); whereas there are counterexample to
the above question for higher dimensional ($\geq 2$) integral domains (cf.
[5], [3], [7], [13]). Suppose $R$ contains $\mathbb{Q}$, or $R$ is a
seminormal domain, and $B=R^{[1]}$, then Question 1.1 has positive answer
([10]). More generally, in [10], it is shown that if $\mathbb{Q}\subseteq R$
or $R$ is a Noetherian seminormal domain, then $A^{[m]}=R^{[m+1]}$ implies
$A=R^{[1]}$. In this paper we consider Question 1.1 over factorial affine
$k$-domains. In [12], Masuda has extended the family of Danielewski surfaces
to produce non-cancellative families of rings over factorial affine domains
$R$, contating a field of characteristic zero, and thus produced
counterexamples to Question 1.1 over $R$. This family of rings are known as
Danielewski type rings. In this paper we exhibit a larger class of families of
generalized Danielewski type rings over affine factorial domains $R$,
containing fields of arbitrary characteristic, which produce counterexamples
to Question 1.1 over $R$ (cf. Corollary 3.6).

In [4], [11] and [12], several criteria are given for an affine algebra,
containing field of characteristic zero, to be isomorphic to a ring of
Danielewski type $R[X,Y,Z]/(X^{m}Y-F(X,Z))$ where $F(0,Z)\in R[Z]\setminus R$
and $m\geq 1$. In section 4, we show that, under certain hypothesis, those
characterization results can be appropriately extended over arbitrary
characteristics (cf. Theorems 4.4, 4.6, & 4.7). In particular, we obtain the
following extension of [12, Theorem 1.2].

######  Theorem 1.2.

Let $R$ be a factorial affine $k$-domain with $Q(R)$ perfect, and let $B$ be a
factorial affine $R$-domain. Suppose there exists a non-trivial exponential
map $\delta:B\rightarrow B[U]$ such that the following hold.

  1. (1)

$B^{\delta}=R[x]$, where $x\in B$ is trancendental over $R$, and

  2. (2)

there exists an element $z\in B\setminus xB$ such that $\delta(z)=z+x^{m}U$
and $R[\overline{z}]^{[1]}=B/xB$, where $m\in\mathbb{Z}_{\geq 1}$ and
$\overline{z}$ denotes the residue field of $z$ in $B/xB$.

Then $B$ is isomorphic to $R[X,Y,Z]/(X^{m}Y-F(X,Z))$ as $R$-algebra, where
$F(X,Z)\in R[X,Z]$ with $F(0,Z)\in R[Z]\setminus R$.

In [14], Gupta and Sen introduced Double Danielewski surfaces over fields of
arbitrary characteristic. They have shown that these surfaces produce
counterexamples to the GCP over fields. In section 5, extending their examples
we introduce Double Danielewski type rings over affine factorial domains $R$.
We show that these rings provide counterexamples to Question 1.1 over
factorial affine domains (cf. Corollary 5.6).

In section 6, we restrict ourselves to factorial affine domains $R$ containing
a field of characteristics zero. We give the following criteria for an affine
factorial $R$-domain to be isomorphic to a ring of Double Danielewski type
(cf. Theorems 6.5 and 6.6).

######  Theorem 1.3.

Let $R$ be factorial affine domain over a field $k$ of characteristic zero,
and $B$ be a factorial affine $R$-domain. Suppose there exists a non-trivial
locally nilpotent $R$-derivation $\delta$ on $B$ such that the following hold

  1. (1)

$B^{\delta}=R[x]$, where $x\in B$ is trancendental over $R$, and

  2. (2)

there exists an element $z\in B\setminus xB$ such that $\delta(z)=x^{2}$.

If $R[\overline{z}]$ is factorial, then $B$ is isomorphic to
$R[X,Y,Z,T]/(XY-\varphi(Z),XT-\psi(Y,Z))$ as $R$-algebra, where $\varphi(Z)\in
R[Z]\setminus R$, $\psi(Y,0)\in R[Y]\setminus R$ and $\overline{z}$ is the
image of $z$ in $B/xB$.

######  Theorem 1.4.

Let $R$ be factorial affine domain over a field $k$ of characteristic zero,
and $B$ be a factorial affine $R$-domain. Suppose there exists a non-trivial
locally nilpotent $R$-derivation $\delta$ on $B$ such that the following hold

  1. (1)

$B^{\delta}=R[x]$, where $x\in B$ is trancendental over $R$,

  2. (2)

there exists an element $z\in B\setminus xB$ such that $\delta(z)=x^{m}$ for
some $m\geq 2$, and

  3. (3)

there exists $y\in B\setminus xB$ such that $\delta(y)=x^{d}f(x,z)$ and
$R[\overline{y},\overline{z}]^{[1]}=B/xB$, where $0<d<m$, $f(x,z)\in
R[x,z]\setminus xR[x,z]$ is monic in $z$, and $\overline{z},\overline{y}$
denote the residue class of $z,y$ in $B/xB$.

If $\deg_{z}f(0,z)=[Q(R[\bar{z}]):Q(R)]-1$ and $R[\overline{z}]$ is factorial,
then

$B\cong_{R}\frac{R[X,Y,Z,T]}{(X^{m-d}Y-P(X,Z),X^{d}T-Q(X,Y,Z))}$

where $P(0,Z)\in R[Z]\setminus R$ and $Q(0,Y,0)\in R[Y]\setminus R$.

In the next section we recall some concepts on exponential maps on integral
domains and discuss some results which will be used in the subsequent sections
of this paper.

##  2\. Preliminaries

We begin this section with definition and some useful properties of
exponential maps.

######  Definition 2.1.

Let $B$ be an affine $k$-algebra and let $\delta:B\rightarrow B^{[1]}$ be a
$k$-algebra homorphism. For an indeterminant $U$ over $B$, let
$\delta_{U}:B\rightarrow B[U]$ denote the $k$-algebra homomorphism,
$\delta_{U}$ is said to be an _exponential map_ if $\delta_{U}$ satisfies the
following properties:

  1. (i)

For the evaluation map $\epsilon_{0}:B[U]\rightarrow B$ at $U=0$, the
composition $\epsilon_{0}\delta_{U}$ is an identity on $B$.

  2. (ii)

$\delta_{V}\delta_{U}=\delta_{U+V}$, where $\delta_{V}:B\rightarrow B[V]$ is
extended to a homomorphism $\delta_{V}:B[U]\rightarrow B[V,U]$ by setting
$\delta_{V}(U)=U$.

The ring of invariant of $\delta$ is denoted by $B^{\delta}:=\\{b\in
B\,|\,\delta(b)=b\\}$. An exponential map $\delta$ is said to be non-trivial
if $B\neq B^{\delta}$. For each $b\in B$, there exists
$n_{b}\in\mathbb{Z}_{\geq 0}$ such that

$\delta(b)=\sum_{i=0}^{n_{b}}\delta^{(i)}(b)\,U^{i}.$

Note that $\\{\delta^{(i)}\\}_{i\geq 0}$ is a sequence of linear maps on $B$.
This sequence is a locally finite iterative higher derivation, which is
abbreviated as lfihd. By the _Leibniz rule_ $\delta^{(1)}$ is a
$k$-derivation. Let $R$ be a $k$-algebra subring of $B$. By EXP${}_{R}(B)$ we
denote the set of all exponential maps $\delta$ on $B$ with $R\subseteq
B^{\delta}$. Let us denote

$\text{ML}_{R}(B):=\bigcap_{\delta\in\text{EXP}_{R}(B)}B^{\delta}$

Then ML${}_{R}(B)$ is a $k$-subalgebra of $B$. Let $\delta:B\rightarrow B[U]$
be an exponential map, then $\delta$ defines a degree function on $B$ denoted
by $\deg_{\delta}$. Let $b\in B\setminus\\{0\\}$ then
$\deg_{\delta}(b):=\deg_{U}(\delta(b))$, and $\deg_{\delta}(0)=-\infty$.
Suppose $\delta$ is non-trivial then $a\in B$ is called a _preslice_ of
$\delta$ if $\deg_{\delta}(a)=$ min$\\{\deg_{\delta}(b)\,|\,b\in B\setminus
B^{\delta}\\}$.

######  Lemma 2.2.

Let $B$ be an affine domain over a field $k$ with characteristic $p\geq 0$.
Let $\delta$ be a non-trivial exponential map. Let $x\in A$ be a preslice of
$\delta$ and $n=\deg_{\delta}(x)\geq 1$. Then there exists an exponential map
$\widetilde{\delta}$ on $B$ with $B^{\widetilde{\delta}}=B^{\delta}$ and
$\deg_{\widetilde{\delta}}(x)=1$.

The following properties of exponential maps will be used throughout this
paper (cf. [2, p. 1291 & 1292]).

######  Lemma 2.3.

Let $B$ be an affine $k$-domain. Suppose there exists a non-trivial
exponential map $\delta_{U}$ on $B$. Then the following holds.

  1. (1)

$B^{\delta_{U}}$ is a factorially closed subring of $B$.

  2. (2)

$B^{\delta_{U}}$ is algebraically closed in $B$.

  3. (3)

If $x\in B$ is such that $\deg_{U}(\delta(x))$ is of minimal positive degree,
and $b$ is the leading coefficient of $U$ in $\delta_{U}(x)$, then $b\in
B^{\delta_{U}}$ and $B[b^{-1}]=B^{\delta_{U}}[b^{-1}][x]$.

  4. (4)

tr.deg${}_{k}(B^{\delta_{U}})=$ tr.deg${}_{k}(B)-1$.

  5. (5)

If tr.deg${}_{k}(B)=1$ and $\overline{k}$ is the algebraic closure of $k$ in
$B$, then $B=\overline{k}^{[1]}$ and $B^{\delta_{U}}=\overline{k}$.

  6. (6)

For any multiplicatively closed subset $S$ of
$B^{\delta_{U}}\setminus\\{0\\}$, $\delta_{U}$ extends to a non-trivial
exponential map $S^{-1}\delta_{U}$ on $S^{-1}B$ by setting
$(S^{-1}\delta_{U})(b/s):=\delta_{U}(b)/s$ for $b\in B,\,s\in S$; and the ring
of invariant of $S^{-1}\delta_{U}$ is $S^{-1}(B^{\delta_{U}})$.

Next, we recall the definiton of proper and admissible $\mathbb{Z}$-filtration
on an affine domain.

######  Definition 2.4.

Let $k$ be a field and $B$ an affine $k$-domain. A collection
$\\{B_{n}\,\,|\,n\in\mathbb{Z}\\}$ of $k$-linear subspaces of $B$ is said to
be a proper $\mathbb{Z}$-filtration if it satisfies the following properties

  * (i)

$B_{n}\subseteq B_{n+1}$, for every $n\in\mathbb{Z}$.

  * (ii)

$B=\bigcup_{n\in\mathbb{Z}}B_{n}$.

  * (iii)

$\bigcap_{n\in\mathbb{Z}}B_{n}=\\{0\\}$.

  * (iv)

$(B_{n}\setminus B_{n-1}).(B_{m}\setminus B_{m-1})\subseteq B_{m+n}\setminus
B_{m+n-1}$ for all $m,n\in\mathbb{Z}$.

A proper $\mathbb{Z}$-filtration $\\{B_{n}\\}_{n\in\mathbb{Z}}$ of $B$ is said
to be admissible if there is a finite generating set $\Gamma$ of $B$ such that
for each $n\in\mathbb{Z}$, every element in $B_{n}$ can be written as a finite
sum of monomials from $B_{n}\cap k[\Gamma]$.

A proper $\mathbb{Z}$-filtration $\\{B_{n}\\}_{n\in\mathbb{Z}}$ of $B$ defines
an associated graded domain defined by

$gr(B):=\bigoplus_{n\in\mathbb{Z}}\frac{B_{n}}{B_{n-1}}.$

It also defines the natural map $\rho:B\rightarrow gr(B)$ such that
$\rho(b)=b+B_{n-1}$, if $b\in B_{n}\setminus B_{n-1}$.

Suppose the ring $B$ is graded. An exponential map $\delta:B\rightarrow B[U]$
is said to be homogeneous if $\delta$ is homogeneous, where the grading on
$B[U]$ is induced from that of $B$ with $U$ being homogeneous with a certain
weight.

We now record a result on homogenization of exponential maps due to H.
Derksen, O. Hadas and L. Makar-Limanov [6]. The following version can be found
in [2, Theorem 2.6].

######  Theorem 2.5.

Let $B$ be an affine domain over a field $k$ with an admissible proper
$\mathbb{Z}$-filtration and $gr(B)$ be the induced $\mathbb{Z}$-graded domain.
Let $\delta$ be a non-trivial exponential map on $B$. Then $\delta$ induces a
non-trivial homogeneous exponential map $\overline{\delta}$ on $gr(B)$ such
that $\rho(B^{\delta})\subseteq gr(B)^{\overline{\delta}}$.

##  3\. Non cancellative family of Danielewski type

Throughout this section $R$ is a factorial affine domain over a field $k$ and
$K=Q(R)$. The following notation are fixed throughout the rest of this
section. Let $r_{1},\ldots,r_{m}$ be positive integers and
$F:=F(X_{1},\ldots,X_{m},Z)\in R^{[m+1]}$ be such that for
$P(Z):=F(0,\ldots,0,Z)$, $\deg_{Z}P>1$. By $B(\underline{r},F)$ we denote the
ring

$B(\underline{r},F):=B(r_{1},\ldots,r_{m},F)=\frac{R[X_{1},\ldots,X_{m},Y,Z]}{(X_{1}^{r_{1}}\cdots
X_{m}^{r_{m}}Y-F(X_{1},\ldots,X_{m},Z))}.$ (3.1)

where $\underline{r}:=(r_{1},\ldots,r_{m})\in\mathbb{Z}^{m}_{\geqslant 1}$.
Let $x_{1},\ldots,x_{m},y,z$ denote the images of $X_{1},\ldots,X_{m},Y,Z$ in
$B({\underline{r}},F)$. First, we will compute ML${}_{R}(B(\underline{r},F))$
(see, Section 2) and then describe the $R$-isomorphism classes of
$B(\underline{r},F)$.

######  Lemma 3.1.

Let $A$ be an affine $k$-domain and $B$ be a finitely generated $A$-domain.
For any multiplicatively closed subset $S$ of $A$, we have
$\operatorname{ML}_{S^{-1}A}(S^{-1}B)=S^{-1}\operatorname{ML}_{A}(B)$.

###### Proof.

Let $\frac{x}{s}\in\operatorname{ML}_{S^{-1}A}(S^{-1}B)$ where $x\in B$ and
$s\in S$. Let $\phi\in{\rm EXP}_{A}(B)$. Since $A\subseteq B^{\phi}$, $\phi$
induces an exponential map $S^{-1}\phi$ on $S^{-1}B$ such that
$S^{-1}\phi|_{S^{-1}A}=id_{S^{-1}A}$. Then
$(S^{-1}\phi)(\frac{x}{s})=\frac{\phi(x)}{s}=\frac{x}{s}$. Since $B$ is an
integral domain, we have $\phi(x)=x$. The choice of the exponential map
$\phi\in{\rm EXP}_{A}(B)$ is arbitrary. Hence it follows that
$x\in\operatorname{ML}_{A}(B)$. Thus
$\operatorname{ML}_{S^{-1}A}(S^{-1}B)\subseteq
S^{-1}\operatorname{ML}_{A}(B)$.

Conversely, let $\frac{y}{t}\in S^{-1}\operatorname{ML}_{A}(B)$ where
$y\in\operatorname{ML}_{A}(B)$ and $t\in S$. Let $\psi\in{\rm
EXP}_{S^{-1}A}S^{-1}B$ be a non-trivial exponential map of $B$. Since $B$ is a
finitely generated $A$-algebra, $\psi$ induces a non-trivial exponential map
$\psi_{1}$ on $B$ such that $\psi_{1}|_{A}=id_{A}$ and
$S^{-1}(B^{\psi_{1}})=(S^{-1}B)^{\psi}$. As $y\in\operatorname{ML}_{A}(B)$, we
have $\frac{y}{t}\in(S^{-1}B)^{\psi}$. The choice of the non-trivial
exponential map $\psi\in{\rm EXP}_{S^{-1}A}S^{-1}B$ is arbitrary. Hence it
follows that $\frac{y}{t}\in\operatorname{ML}_{S^{-1}A}S^{-1}B$. Thus
$S^{-1}\operatorname{ML}_{A}(B)\subseteq\operatorname{ML}_{S^{-1}A}(S^{-1}B)$.
∎

######  Theorem 3.2.

Let $B({\underline{r},F})$ be the ring as in (3.1). Then the following hold.

  1. (a)

If $m=1$ and $r_{1}>1$ then $B(r_{1},F)=R[x_{1}]$.

  2. (b)

If $m>1$, $\underline{r}\in\mathbb{Z}^{m}_{>1}$ and $F$ is monic in $Z$ as a
polynomial in $K[X_{1},\ldots,X_{m},Z]$,
then $\operatorname{ML}_{R}(B({\underline{r}},F))=R[x_{1},\ldots,x_{m}]$.

  3. (c)

If $\underline{r}\in\mathbb{Z}^{m}_{\geqslant 1}\setminus\mathbb{Z}^{m}_{>1}$,
then $\operatorname{ML}_{R}(B({\underline{r}},F))\subseteq R[x_{i}\mid
r_{i}>1]$, and for $\underline{1}=(1,\ldots,1)$,
$\operatorname{ML}_{R}(B_{\underline{1}})=R$.

###### Proof.

Note that

$R[x_{1},\ldots,x_{m}]\hookrightarrow B({\underline{r}},F)\hookrightarrow
R[x_{1},\ldots,x_{m},x_{1}^{-1},\ldots,x_{m}^{-1},z].$ (3.2)

(a) Let

$B_{1}:=B(r_{1},F)\otimes_{R}K=\dfrac{K[X_{1},Y,Z]}{(X_{1}^{r_{1}}Y-F(X_{1},Z))},$

where $r_{1}>1$. We first show that $\operatorname{ML}(B_{1})=K[x_{1}]$. As
$B_{1}\hookrightarrow C:=K[x_{1},x_{1}^{-1},z]$ and $\deg_{z}F(x_{1},z)>1$,
the subring $K[x_{1}]$ is algebraically closed in $B_{1}$. It is enough to
show that $K[x_{1}]\subseteq B_{1}^{\phi}$ for every non-trivial exponential
map $\phi$ on $B_{1}$. If possible, suppose there exists a non-trivial
exponential map $\phi_{1}$ on $B_{1}$ such that $x_{1}\notin
B_{1}^{\phi_{1}}$.

Firstly, we show that the ring of invariant $B_{1}^{\phi_{1}}\subseteq
K[x_{1},z]$. Suppose, if possible, $b\in B_{1}^{\phi_{1}}\setminus
K[x_{1},z]$. Consider the graded structure of
$C=\bigoplus_{n\in\mathbb{Z}}C_{n}$, where $C_{-n}:=K[z]x_{1}^{n}$. The graded
structure induces a filtration $\\{B_{1n}\\}_{n\in\mathbb{Z}}$ on $B_{1}$
where $B_{1n}:=(\bigoplus_{i\leq n}C_{n})\cap B_{1}$. Note that any element
$a\in B_{1}$ can be uniquely written as

$a=p(x_{1},z)+\sum_{i>0,\,j<r_{1}}\alpha_{ijl}x_{1}^{j}y^{i}z^{l},$ (3.3)

where $p\in K[x_{1},z]$ and $\alpha_{ijl}\in K^{*}$. Thus
$\\{B_{1n}\\}_{n\in\mathbb{Z}}$ is an admissible $\mathbb{Z}$-filtration on
$B_{1}$ with respect to the generating set $\\{x_{1},y,z\\}$. Since
$F(0,Z)\neq 0$, then arguing as in [13, Lemma 3.1] we get that the associated
graded ring $gr(B_{1})$, with respect to $\\{B_{1n}\\}_{n\in\mathbb{Z}}$, is

$\widetilde{B_{1}}:=\dfrac{K[X_{1},Y,Z]}{(X_{1}^{r_{1}}Y-F(0,Z))}.$

By Theorem 2.5, $\phi_{1}$ induces a non-trivial exponential map
$\widetilde{\phi_{1}}$ on $\widetilde{B_{1}}$. For any $a\in B_{1}$, let
$\widetilde{a}$ denote the image of $a$ in $\widetilde{B_{1}}$. Since $b\in
B_{1}^{\phi_{1}}\setminus K[x_{1},z]$, it follows that
$\widetilde{y}\mid\widetilde{b}$ in $\widetilde{B_{1}}$. Using Theorem 2.5, we
have $\widetilde{b}\in\widetilde{B_{1}}^{\widetilde{\phi_{1}}}$. From Lemma
2.3(1), it follows that
$\widetilde{y}\in{\widetilde{B_{1}}}^{\widetilde{\phi_{1}}}$. But this
contradicts the non-triviality of $\widetilde{\phi_{1}}$ (cf. [13, Lemma
3.2]). Thus $B_{1}^{\phi_{1}}\subseteq K[x_{1},z]$.

Now, since $x_{1}\notin{B_{1}}^{\phi_{1}}$, there exists $b(x_{1},z)\in
B_{1}^{\phi_{1}}$ such that $\deg_{z}b(0,z)>1$. Let
$s_{1}=\max\\{\deg_{z}b(x_{1},z),\deg_{z}F(x_{1},z)\\}$. Consider another
graded structure on $C$ with respect to the following weights

$wt(z)=1,~{}~{}wt(x_{1})=-s_{1}.$

Let $\\{B^{\prime}_{1n}\\}_{n\in\mathbb{Z}}$ denotes the filtration on $B_{1}$
induced by the above graded structure on $C$. From the expression (3.3), it
follows that the filtration is admissible with respect to the generating set
$\\{x_{1},y,z\\}$. The associated graded ring of $B_{1}$ with respect to the
filtration $\\{B^{\prime}_{1n}\\}_{n\in\mathbb{Z}}$ is

$\overline{B_{1}}=\dfrac{K[X_{1},Y,Z]}{(X_{1}^{r_{1}}Y-Z^{d})},$ (3.4)

where $d:=\deg_{z}F(0,z)>1$. For every $a\in B_{1}$, let $\overline{a}$ denote
its image in $\overline{B_{1}}$. Now by Theorem 2.5, $\phi_{1}$ induces a non-
trivial homogeneous exponential map $\overline{\phi_{1}}$ on
$\overline{B_{1}}$ such that
$\overline{b}\in\overline{B_{1}}^{\overline{\phi_{1}}}$. Now from the grading
of $\overline{B_{1}}$ we get that $\overline{z}\mid\overline{b}$, and by Lemma
2.3(1) $\overline{z}\in\overline{B_{1}}^{\overline{\phi_{1}}}$. But this
contradicts the non-triviality of $\overline{\phi_{1}}$. Thus we conclude that
for every exponential map $\phi$ on $B_{1}$, $x_{1}\in B_{1}^{\phi}$. Hence,
as desired, $\operatorname{ML}(B_{1})=K[x_{1}]$. Using Lemma 3.1 and the fact
that $\operatorname{ML}_{R}(B(r_{1},F))$ is factorially closed in
$B(r_{1},F)$, it follows that
$\operatorname{ML}_{R}(B(r_{1},F))=\operatorname{ML}(B_{1})\cap
B(r_{1},F)=R[x_{1}]$ (cf. (3.2)).

(b) Consider the ring

$B_{2}:=B({\underline{r}},F)\otimes_{R}K=\frac{K[X_{1},\ldots,X_{m},Y,Z]}{(X_{1}^{r_{1}}\cdots
X_{m}^{r_{m}}Y-F(X_{1},\ldots,X_{m},Z))},$

where $m>1$, $\underline{r}\in\mathbb{Z}^{m}_{>1}$ and $F$ is monic in $Z$ as
a polynomial in $K[X_{1},\ldots,X_{m},Z]$. From [13, Theorem 3.4 (a)] it
follows that $\operatorname{ML}(B_{2})=K[x_{1},\ldots,x_{m}]$. As in (a),
using Lemma 3.1 and the fact that $\operatorname{ML}_{R}(B(\underline{r},F))$
is factorially closed in $B(\underline{r},F)$, we get that
$\operatorname{ML}_{R}(B(\underline{r},F))=\operatorname{ML}(B_{2})\cap
B(\underline{r},F)=R[x_{1},\ldots,x_{m}]$.

(c) Consider the ring

$B_{3}:=B({\underline{r}},F)\otimes_{R}K=\frac{K[X_{1},\ldots,X_{m},Y,Z]}{(X_{1}^{r_{1}}\cdots
X_{m}^{r_{m}}Y-F(X_{1},\ldots,X_{m},Z))},$

where $\underline{r}\in\mathbb{Z}^{m}_{\geqslant
1}\setminus\mathbb{Z}^{m}_{>1}$. For every $i$, where $1\leq i\leq m$, let

$F(x_{1},\ldots,x_{m},z)=F(x_{1},\ldots,x_{i-1},0,x_{i+1},\ldots,x_{m},z)+x_{i}F_{i}(x_{1},\ldots,x_{m},z).$

Arguing as in the proof of [13, Theorem 3.4(b)] we get that for every $i$,
such that $r_{i}=1$, there exists an exponential map $\phi_{i}$ on $B_{3}$
such that
$B_{3}^{\phi_{i}}=K[x_{1},\ldots,x_{i-1},\tilde{y_{i}},x_{i+1},\ldots,x_{m}]$
where $\tilde{y_{i}}:=x_{1}^{r_{1}}\ldots
x_{i-1}^{r_{i-1}}x_{i+1}^{r_{i+1}}\ldots
x_{m}^{r_{m}}y-F_{i}(x_{1},\ldots,x_{m},z)$. Since
$x_{i}\tilde{y_{i}}=F(x_{1},\ldots,x_{i-1},0,x_{i+1},\ldots,x_{m},z)$ in
$B_{3}$ and

$\deg_{z}F(x_{1},\ldots,x_{i-1},0,x_{i+1},\ldots,x_{m},z)>1,$

we have

$\operatorname{ML}(B_{3})\subseteq\bigcap_{\\{i\,|\,r_{i}=1\\}}K\left[x_{1},\ldots,x_{i-1},\dfrac{F(x_{1},\ldots,x_{i-1},0,x_{i},\ldots,x_{m},z)}{x_{i}},x_{i+1}\ldots,x_{m}\right]\subseteq
K[x_{i}\mid r_{i}>1].$

Using Lemma 3.1 and (3.2), it follows that
$\operatorname{ML}_{R}(B(\underline{r},F))\subset R[x_{i}\mid
r_{i}>1]\subsetneq R[x_{1},\ldots,x_{m}]$. Furthermore, if
$\underline{r}=(1,\ldots,1)$, then
$\operatorname{ML}_{R}(B(\underline{r},F))=R$. ∎

######  Theorem 3.3.

Let $m\geq 1$, $(r_{1},\ldots,r_{m})$ and
$(s_{1},\ldots,s_{m})\in\mathbb{Z}^{m}_{>1}$, and $F,G\in
R[X_{1},\ldots,X_{m},Z]$ be polynomials such that $P(Z)=F(0,\ldots,0,Z)$ and
$Q(Z)=G(0,\ldots,0,Z)$ are polynomials with $\deg_{Z}P>1$ and $\deg_{Z}Q>1$.
Suppose

$B:=B(r_{1},\ldots,r_{m},F)=\frac{R[X_{1},\ldots,X_{m},Y,Z]}{(X_{1}^{r_{1}}\cdots
X_{m}^{r_{m}}Y-F(X_{1},\ldots,X_{m},Z))},$

and

$B^{\prime}:=B(s_{1},\ldots,s_{m},G)=\frac{R[X_{1},\ldots,X_{m},Y,Z]}{(X_{1}^{s_{1}}\cdots
X_{m}^{s_{m}}Y-G(X_{1},\ldots,X_{m},Z))}$

. Let $x_{1},\ldots,x_{m},y,z$ and
$x_{1}^{\prime},\ldots,x_{m}^{\prime},y^{\prime},z^{\prime}$ denote the images
of $X_{1},\ldots,X_{m},Y,Z$ in $B$ and $B^{\prime}$ respectively. Further
suppose that $\operatorname{ML}_{R}(B)=R[x_{1},\ldots,x_{m}]$ and
$\operatorname{ML}_{R}(B^{\prime})=R[x_{1}^{\prime},\ldots,x_{m}^{\prime}]$.
If $B(r_{1},\ldots,r_{m},F)\cong_{R}B(s_{1},\ldots,s_{m},G)$ then

  1. (i)

$(r_{1},\ldots,r_{m})=(s_{1},\ldots,s_{m})$ upto a permutation of
$\\{1,\ldots,m\\}$.

  2. (ii)

There exists $\sigma\in Aut_{R}(R[Z])$ such that $\sigma(Q)=\lambda_{1}P$, for
some $\lambda_{1}\in R^{*}$. Thus $\deg_{Z}P=\deg_{Z}Q$.

  3. (iii)

Furthmore, if both $F$ and $G$ are monic in $Z$ when considered as a
polynomials in $K[X_{1},\ldots,X_{m},Z]$, then there exists a permutation
$\sigma^{\prime}$ of $\\{1,\ldots,m\\}$,
$\lambda_{1},\ldots,\lambda_{m},\beta,\gamma\in R^{*}$, $f\in
R[X_{1},\ldots,X_{m}]$ and $\alpha\in R[X_{1},\ldots,X_{m},Z]$ such that

$\displaystyle
G(\lambda_{\sigma^{\prime}(1)}X_{\sigma^{\prime}(1)},\ldots,\lambda_{\sigma^{\prime}(m)}X_{\sigma^{\prime}(m)},\gamma
Z+f(X_{1},\ldots,X_{m}))=$ $\displaystyle X_{1}^{r_{1}}\cdots
X_{m}^{r_{m}}\alpha(X_{1},\ldots,X_{m},Z)+\beta F(X_{1},\ldots,X_{m},Z).$

Furthermore, the conditions (i) and (iii) are sufficient for
$B(r_{1},\ldots,r_{m},F)$ to be isomorphic to $B(s_{1},\ldots,s_{m},G)$ as
$R$-algebras.

###### Proof.

(i) Let $\rho:B\rightarrow B^{\prime}$ be an $R$-isomorphism. Identifying
$\rho(B)$ by $B$, we assume that $B^{\prime}=B$. Then

$\operatorname{ML}_{R}(B)=R[x_{1},\ldots,x_{m}]=R[x_{1}^{\prime},\ldots,x_{m}^{\prime}].$
(3.5)

Furthermore, for $B_{1}^{\prime}:=B^{\prime}\otimes_{R}K$ and
$B_{1}:=B\otimes_{R}K$ we have $B_{1}^{\prime}=B_{1}$, and by Lemma 3.1 and
(3.5) it follows that

$\operatorname{ML}(B_{1})=K[x_{1},\ldots,x_{m}]=K[x_{1}^{\prime},\ldots,x_{m}^{\prime}].$
(3.6)

Now by using similar arguments as in the proof of [13, Theorem 3.6 (i)], it
follows that for each $i\in\\{1,\ldots,m\\}$ there exists $\lambda_{j}\in
K^{*}$ such that

$x_{i}^{\prime}=\lambda_{j}x_{j}$ (3.7)

for some $j\in\\{1,\ldots,m\\}$ with $r_{j}=s_{i}$. Thus assertion (i) holds.
From (3.5), $\lambda_{j}\in R^{*}$ for all $1\leq j\leq m$. Let
$\sigma^{\prime}$ be the premutation of $\\{1,\ldots,m\\}$ such that

$x_{i}^{\prime}=\lambda_{\sigma^{\prime}(i)}x_{\sigma^{\prime}(i)}$ (3.8)

and $s_{i}=r_{\sigma^{\prime}(i)}$ for every $1\leqslant i\leqslant m$.

(ii) From (3.7), we have

$x_{1}^{\prime}\cdots x_{m}^{\prime}=\lambda x_{1}\cdots x_{m}.$ (3.9)

for some $\lambda\in R^{*}$. Now

$B\otimes_{R[x_{1},\ldots,x_{m}]}K(x_{1},\ldots,x_{m})=B\otimes_{R[x_{1}^{\prime},\ldots,x_{m}^{\prime}]}K(x_{1}^{\prime},\ldots,x_{m}^{\prime}),$

and hence

$K(x_{1},\ldots,x_{m})[z]=K(x_{1}^{\prime},\ldots,x_{m}^{\prime})[z^{\prime}]=K(x_{1},\ldots,x_{m})[z^{\prime}].$
(3.10)

Note that $E:=R[x_{1},\ldots,x_{m},z]\hookrightarrow B\hookrightarrow
R[x_{1}^{\pm 1},\ldots,x_{m}^{\pm 1},z]$. Next, we show that $z^{\prime}\in
E$. Suppose $z^{\prime}\in B\setminus E$. Then

$z^{\prime}=g(x_{1},\ldots,x_{m},z)+\sum_{(\underline{i},j)\in\Lambda_{1}}b_{\underline{i}\,j}(z)\,x_{1}^{i_{1}}\cdots
x_{m}^{i_{m}}y^{j},$ (3.11)

where

$\Lambda_{1}:=\biggl{\\{}(\underline{i},j):=(i_{1},\ldots,i_{m},j)\in\mathbb{Z}^{m}_{\geqslant
0}\times\mathbb{Z}_{>0}\mid i_{s}<r_{s}\text{~{}for
some~{}}s\in\\{1,\ldots,m\\}\biggr{\\}},$ (3.12)

$g\in E$ and $b_{\underline{i}\,j}(z)\in R[z]\setminus\\{0\\}$. Since
$y=\frac{F(x_{1},\ldots,x_{m},z)}{x_{1}^{r_{1}}\cdots x_{m}^{r_{m}}}$ and
$\deg_{z}P(z)>1$, from (3.11), it is clear that $\deg_{z}z^{\prime}>1$ when
considered as an element of $K(x_{1},\ldots,x_{m})[z]$. But this contradicts
(3.10). Hence $z^{\prime}\in E$. Similarly, we have $z\in
E^{\prime}:=R[x_{1}^{\prime},\ldots,x_{m}^{\prime},z^{\prime}]$. By (3.5),

$R[x_{1},\ldots,x_{m},z]=R[x_{1}^{\prime},\ldots,x_{m}^{\prime},z^{\prime}]=R[x_{1},\ldots,x_{m},z^{\prime}].$
(3.13)

From (3.10) and (3.13), we get that

$z^{\prime}=\gamma z+f(x_{1},\ldots,x_{m}),$ (3.14)

and

$z=\gamma_{1}z^{\prime}+f_{1}(x_{1},\ldots,x_{m}),$ (3.15)

where $\gamma,\gamma_{1},f,f_{1}\in R[x_{1},\ldots,x_{m}]$. Now (3.14) and
(3.15) implies that $\gamma,\gamma_{1}\in R^{*}$. Also, from (3.9),
$\left(x_{1}\cdots x_{m}\right)B\,\cap\,E=\left(x_{1}^{\prime}\cdots
x_{m}^{\prime}\right)B\,\cap\,E$. Thus

$\left(x_{1}\cdots
x_{m},F(x_{1},\ldots,x_{m},z)\right)E=\left(x_{1}^{\prime}\cdots
x_{m}^{\prime},G(x_{1}^{\prime},\ldots,x_{m}^{\prime},z^{\prime})\right)E.$

Hence we can write

$G(x_{1}^{\prime},\ldots,x_{m}^{\prime},z^{\prime})=(x_{1}\cdots
x_{m})\alpha+F(x_{1},\ldots,x_{m},z)\beta$

for some $\alpha,\beta\in E$. Now using (3.7) and (3.14),

$G(0,\ldots,0,\gamma z+f(0,\ldots,0))=F(0,\ldots,0,z)\beta(0,\ldots,0,z).$
(3.16)

Similarly,

$F(0,\ldots,0,z)=G(0,\ldots,0,\gamma
z+f(0,\ldots,0))\beta^{\prime}(0,\ldots,0,z).$ (3.17)

Thus $\lambda_{1}:=\beta(0,\ldots,0,z)\in R^{*}$. The $R$-homomorphism
$\sigma:R[Z]\rightarrow R[Z]$ defined by $\sigma(Z)=\gamma Z+f(0,\ldots,0)$ is
an automorphism of $R[Z]$. Then, from (3.16), it follows that
$\sigma(Q)=\lambda_{1}P$.

(iii) From (3.8), we have

$(x_{1}^{\prime})^{s_{1}}\cdots(x_{m}^{\prime})^{s_{m}}=\mu
x_{1}^{r_{1}}\cdots x_{m}^{r_{m}}$ (3.18)

where $\mu\in R^{*}$. Thus

$\displaystyle~{}~{}~{}~{}~{}~{}~{}~{}x_{1}^{r_{1}}\cdots x_{m}^{r_{m}}B\cap
E=(x_{1}^{\prime})^{s_{1}}\cdots(x_{m}^{\prime})^{s_{m}}B\cap E,$
$\displaystyle\text{~{}i.e.,~{}}\left(x_{1}^{r_{1}}\cdots
x_{m}^{r_{m}},F(x_{1},\ldots,x_{m},z)\right)E=\left((x_{1}^{\prime})^{s_{1}}\cdots(x_{m}^{\prime})^{s_{m}},G(x_{1}^{\prime},\ldots,x_{m}^{\prime},z^{\prime})\right)E.$
(3.19)

Hence we can write

$G(x_{1}^{\prime},\ldots,x_{m}^{\prime},z^{\prime})=x_{1}^{r_{1}}\cdots
x_{m}^{r_{m}}\alpha(x_{1},\ldots,x_{m},z)+F(x_{1},\ldots,x_{m},z)\beta(x_{1},\ldots,x_{m},z)$

for some $\alpha,\beta\in R^{[m+1]}$ such that no monomial summand of $\beta$
is divisible by $x_{1}^{r_{1}}\cdots x_{m}^{r_{m}}$. Note that both $F$ and
$G$ are monic in $z$ when considered as a polynomials of
$K[x_{1},\ldots,x_{m},z]$, and by (ii) we have $\deg_{z}F=\deg_{z}G$. Hence it
follows that $\beta\in R\setminus\\{0\\}$. Now using (3.18) and the symmetry
in (3) we conclude that $\beta\in R^{*}$. Then the desired assertion follows
from (3.14) and (3.8).

We now prove the converse. Let us define an $R$-algebra homomorphism
$\rho:R[X_{1},\ldots,X_{m},Y,Z]\rightarrow B$ as follows:

$\displaystyle\rho(X_{i})=\lambda_{\sigma^{\prime}(i)}x_{\sigma^{\prime}(i)},\,\,\,\text{for
all}\,\,i\in\\{1,\ldots,m\\}$ $\displaystyle\rho(Z)=\gamma
z+f(x_{1},\ldots,x_{m})$ $\displaystyle\rho(Y)=\mu^{-1}\beta
y+\mu^{-1}\alpha(x_{1},\ldots,x_{m},z),$

where $\mu=\prod_{i=1}^{m}\lambda_{\sigma^{\prime}(i)}^{s_{i}}$. Note that
$\rho$ is surjective, and

$\displaystyle\rho(X_{1}^{s_{1}}\cdots
X_{m}^{s_{m}}Y-G(X_{1},\ldots,X_{m},Z))=\mu\left(\prod_{i=1}^{m}x_{\sigma^{\prime}(i)}^{s_{i}}\right)\left(\mu^{-1}\beta
y+\mu^{-1}\alpha(x_{1},\ldots,x_{m},z)\right)$
$\displaystyle~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}-G(\lambda_{\sigma^{\prime}(1)}x_{\sigma^{\prime}(1)}\ldots\lambda_{\sigma^{\prime}(m)}x_{\sigma^{\prime}(m)},\gamma
z+f(x_{1},\ldots,x_{m})).$

Now, using (3.18) and (iii), it follows that

$\displaystyle\rho(X_{1}^{s_{1}}\cdots
X_{m}^{s_{m}}Y-G(X_{1},\ldots,X_{m},Z))=x_{1}^{r_{1}}\cdots
x_{m}^{r_{m}}\left(\beta y+\alpha(x_{1},\ldots,x_{m},z)\right)$
$\displaystyle~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}-\left(x_{1}^{r_{1}}\cdots
x_{m}^{r_{m}}\alpha(x_{1},\ldots,x_{m},z)+\beta
F(x_{1},\ldots,x_{m},z)\right)$
$\displaystyle~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}=0.$

Thus $\rho$ induces a surjective $R$-algebra homomorphism
$\overline{\rho}:B^{\prime}\rightarrow B$, where $B$ and $B^{\prime}$ both are
affine domains whose dimensions are equal. Hence $\overline{\rho}$ is an
$R$-algebra isomorphism. ∎

######  Remark 3.4.

Let $F\in R[X_{1},\ldots,X_{m},Z]$ be a polynomial with
$\deg_{Z}F(0,\ldots,0,Z)>1$. Then the following hold

  1. (a)

For $m=1$, by Theorem 3.2(a) and Theorem 3.3(i), $B(r,F)\ncong_{R}B(s,F)$ when
$r\neq s$.

  2. (b)

For $m>1$, if $F$ is monic in $Z$ as a polynomial in
$K[X_{1},\ldots,X_{m},Z]$, then by Theorem 3.2(b),(c) and Theorem 3.3(i) we
have $B(\underline{r},F)\ncong_{R}B(\underline{s},F)$ when

     * (i)

$\underline{r}\in\mathbb{Z}^{m}_{>1}$ and
$\underline{s}\in\mathbb{Z}^{m}_{\geq 1}\setminus\mathbb{Z}^{m}_{>1}$, or

     * (ii)

$\underline{r},\underline{s}\in\mathbb{Z}^{m}_{>1}$ but
$\underline{r}\neq\underline{s}$ upto any permutation of $\\{1,\ldots,m\\}$.

The arguments in the proof of the following result is similar to that of [13,
Theorem 3.8]. We have to consider the maps $\phi$ and $\psi$ defined in the
proof of [13, Theorem 3.8] to be an $R$-algebra homorphism. Thus we only state
the theorem.

######  Theorem 3.5.

Let $(r_{1},\ldots,r_{m}),\,(s_{1},\ldots,s_{m})\in\mathbb{Z}^{m}_{\geqslant
1}$. If $\left(F,F_{Z}\right)=R[X_{1},\ldots,X_{m},Z]$, then

$B(r_{1},\ldots,r_{m},F)^{[1]}\cong_{R}B(s_{1},\ldots,s_{m},F)^{[1]}.$

Now, consider the following families of rings

$\displaystyle\Omega_{1}:=\biggl{\\{}$ $\displaystyle B(r_{1},F)\mid
r_{1}\in\mathbb{Z}_{\geqslant 1},F\in R[X_{1},Z]\text{~{}be such
that~{}}\deg_{Z}F(0,Z)>1$
$\displaystyle\text{~{}and~{}}(F,F_{Z})=R[X_{1},Z]\biggr{\\}},$

and

$\displaystyle\Omega_{2}:=\biggl{\\{}$ $\displaystyle
B(\underline{r},F)\mid\underline{r}:=(r_{1},\ldots,r_{m})\in\mathbb{Z}^{m}_{\geqslant
1},F\in R[X_{1},\ldots,X_{m},Z]\text{~{}is monic in~{}}Z\text{~{}as a
polynomial in ~{}}$ $\displaystyle
K[X_{1},\ldots,X_{m},Z]\text{~{}and~{}}(F,F_{Z})=R[X_{1},\ldots,X_{m},Z]\biggr{\\}}.$

######  Corollary 3.6.

The families $\Omega_{1}$ and $\Omega_{2}$ consist of infinitely many rings
which are pairwise non-isomorphic as $R$ algebras and counterexamples to the
Generalized Cancellation Problem over $R$.

###### Proof.

Follows from Remark 3.4 and Theorem 3.5. ∎

##  4\. Characterization of Danielewski type Hypersurfaces

Throughout this section $R$ is an affine factorial domain over a field $k$. In
this section, we consider varieties of type $B(m,F)=R[X,Y,Z]/(X^{m}Y-F(X,Z))$,
where $F(0,Z)\in R[Z]\setminus R$ and $m\in\mathbb{Z}_{\geq 1}$. Let $x,y$ and
$z$ denote the residue classes of $X,Y$ and $Z$, respectively, modulo
$(X^{m}Y-F(X,Z))$. Without loss of generality we can assume that
deg${}_{X}F(X,Z)<m$ in $B(m,F)$. The main aim of this section is to prove
Theorem 4.4 and Corollary 4.8 which characterize such rings and extend the
characterizations of Daigle and Masuda over fields of characteristic zero (cf.
[4], [11], [12]). Let us recall, without proof, some basic facts about these
varieties.

######  Lemma 4.1.

Let $B_{m}:=B(m,F)$ be as above. Then the following hold.

  1. (1)

$B_{m}[x^{-1}]\cong R[X,X^{-1},Z]$, and $B_{m}/xB_{m}\cong R[Y,Z]/(F(0,Z))$.

  2. (2)

Let $\delta:B_{m}\rightarrow B_{m}[T]$ be an $R$-algebra defined by
$\delta(x)=x$, $\delta(z)=z+x^{m}T$ and
$\delta(y)=F(x,z+x^{m}T)/x^{m}=y+f_{1}(x,z,T)$ for some $f_{1}\in R^{[3]}$.
Then $\delta$ is an exponetial map on $B_{m}$ with the ring of invariant
$B_{m}^{\delta}=R[x]$.

  3. (3)

Let $m=1$. Then there exists another exponential map $\gamma:B_{1}\rightarrow
B_{1}[T]$ defined by $\gamma(b)=b$ for all $b\in R[y]$, $\gamma(z)=z+yT$ and
$\gamma(x)=F(z+yT)/y=x+f_{2}(y,z,T)$ for some $f_{2}\in R^{[3]}$.

In view of the above properties and Lemma 2.2 we give some characterization of
affine $R$-algebra $B$ to be $R$-isomorphic to varieties of type $B(m,F)$. The
following lemma’s from [11] and [12] are actually characteristic independent.

######  Lemma 4.2.

[12, Lemma 3.9] Let $R$ be a factorial affine $k$-domain. Let $B$ be an affine
$R$-domain generated over $R$ by the elements $x,y,z\in B$. Suppose that $x$
and $z$ are algebraically independent over $R$ and the generators satisfy
$x^{m}y-F(x,z)$, where $m\geq 1$ and $F(x,z)\in R[x,z]$ such that $F(0,z)\in
R[z]\setminus R$. Then $B$ is isomorphic to $R[X,Y,Z]/(X^{m}Y-F(X,Z))$ as
$R$-algebra.

######  Lemma 4.3.

[11, Lemma 4.1] Let $R$ be a factorial affine $k$-domain. Let $x_{1}$, $x_{2}$
and $z$ be elements of an affine $R$-domain $B$ such that $x_{i}$ and $z$ are
algebraically independent over $R$ for $i=1,2$ and that $R[x_{1}]\neq
R[x_{2}]$. Suppose that

$B\subseteq R[x_{1},f(x_{1})^{-1},z]\cap R[x_{2},g(x_{2})^{-1},z]\subseteq
Q(B)$

where $f(x_{1})$ $($resp. $g(x_{2}))$ is an irreducible element of
$R[x_{1}]\setminus R$ $($resp. $R[x_{2}]\setminus R)$ and $Q(B)$ is the field
of fractions $B$. Then

  1. (1)

One of the following holds:

    1. (i)

$x_{1}=b_{1}x_{2}+b_{0}(z)$ for $b_{1}\in R^{*}$ and $b_{0}(z)\in
R[z]\setminus R$.

    2. (ii)

$f(x_{1})=cx_{1}+c_{2}$, $g(x_{2})=c^{\prime}(cx_{2}+c_{1})$ and
$cx_{1}x_{2}+c_{1}x_{1}+c_{2}x_{2}=h(z)$, where $c^{\prime}\in R^{*}$, $c\in
R\setminus\\{0\\}$, $c_{1},c_{2}\in R$ and $h(z)\in R[z]$.

  2. (2)

In the case $(\rm i)$ of $(1)$, $B=R[x_{1},z]$. In the case $(\rm ii)$ of
$(1)$, $B[c^{-1}]=R[c^{-1}][x_{1},x_{2},z]$.

Firstly, we give a characterization of $B_{1}$. Note that ML${}_{R}(B_{1})=R$
when $\deg_{Z}F(0,Z)\geq 1$ (cf., Theorem 3.2 and Lemma 4.1(2)). Recall that
the locally finite iterative heigher derivations (LFIHD) of an expoential map
$\delta$ is denoted by $\\{\delta^{(i)}\\}_{i\geq 0}$.

######  Theorem 4.4.

Let $R$ be a factorial affine $k$-domain and $B$ be an affine $R$-domain.
Suppose there exist exponential maps $\delta_{1}$ and $\delta_{2}$ on $B$ such
that the following hold.

  1. (i)

$B^{\delta_{i}}=R[x_{i}]$, where $x_{i}\in B$ is trancendental over $R$ for
$i=1,2$,

  2. (ii)

$B^{\delta_{1}}\neq B^{\delta_{2}}$,

  3. (iii)

There exists $z\in B$ such that $\delta_{i}^{(1)}(z)$ is an irreducible
element of $R[x_{i}]\setminus R$ for $i=1,2$.

Then there exists $r\in R\setminus\\{0\\}$ and $\varphi(Z)\in
R[r^{-1}][Z]\setminus R[r^{-1}]$ such that $B[r^{-1}]$ is isomorphic to
$R[r^{-1}][X_{1},X_{2},Z]/(X_{1}X_{2}-\varphi(Z))$ as an $R[r^{-1}]$-algebra.
Furthermore, if $B^{\delta_{i}}=R[\delta_{i}^{(1)}(z)]$ for $i=1,2$, then $B$
is $R$-algebra isomorphic to $R[X_{1},X_{2},Z]/(X_{1}X_{2}-\varphi(Z))$ for
some $\varphi(Z)\in R[Z]\setminus R$.

###### Proof.

Let $\delta_{1}^{(1)}(z)=f(x_{1})$ and $\delta_{2}^{(1)}(z)=g(x_{2})$. By the
assumption, $f(x_{1})$ and $g(x_{2})$ are irreducible elements of
$R[x_{1}]\setminus R$ and $R[x_{2}]\setminus R$, respectively. Note that $z$
is algebraically independent over $R[x_{i}]$ for $i=1,2$ (Lemma 2.3 (2)). From
Lemma 2.3(3), it follows that

$B\subseteq R[x_{1},f(x_{1})^{-1},z]\cap R[x_{2},g(x_{2})^{-1},z]$

By Lemma 4.3(2), we have $B=R[x_{1},z]=R[x_{2},z]$, or
$B[c^{-1}]=R[c^{-1}][x_{1},x_{2},z]$ with
$cx_{1}x_{2}+c_{1}x_{1}+c_{2}x_{2}=h(z)$, where $c\in R\setminus\\{0\\}$,
$c_{1},c_{2}\in R$ and $h(z)\in R[z]$.

If $B=R[x_{1},z]$ then it is isomorphic to a $B(1,\varphi)$ with
$\varphi(X,Z)=Z$. Thus we consider the case when
$B[c^{-1}]=R[c^{-1}][x_{1},x_{2},z]$ with
$cx_{1}x_{2}+c_{1}x_{1}+c_{2}x_{2}=h(z)$. We claim that $h(z)\in R[z]\setminus
R$. On the contrary, suppose $h(z)\in R$. Now applying $\delta_{1}$ to the
equation $cx_{1}x_{2}+c_{1}x_{1}+c_{2}x_{2}=h(z)$, we have
$cx_{1}\delta_{1}(x_{2})+c_{1}x_{1}+c_{2}\delta_{1}(x_{2})=h(z)$. Let
$\delta_{1}:B\rightarrow B[T]$. Equating the coefficients of $T^{i}$, for
$i\geq 1$, we have $(\delta_{1}^{(i)}(x_{2}))(cx_{1}+c_{2})=0$. Since
$cx_{1}+c_{2}\neq 0$, $\delta_{1}^{(i)}(x_{2})=0$ for all $i\geq 1$, that is,
$x_{2}\in B^{\delta_{1}}=R[x_{1}]$. Similarly, $x_{1}\in
B^{\delta_{2}}=R[x_{2}]$. Thus $B^{\delta_{1}}=B^{\delta_{2}}$, which is a
contradiction. Hence, $h(z)\in R[z]\setminus R$. By replacing $x_{1}$ and
$x_{2}$ by $x_{1}-c^{-1}c_{2}$ and $x_{2}-c^{-1}c_{1}$, respectively, we have
$x_{1}x_{2}=\varphi(z)$ in $B[c^{-1}]=R[c^{-1}][x_{1},x_{2},z]$ where
$\varphi(z)=h(z)+c^{-1}c_{1}c_{2}\in R[c^{-1}][z]\setminus R[c^{-1}]$. Then
the assertion follows from Lemma 4.2.

Furthermore, suppose $B^{\delta_{i}}=R[\delta_{i}^{(1)}(z)]$ for $i=1,2$.
Recall that $\delta_{1}^{(1)}(z)=f(x_{1})$ and $\delta_{2}^{(1)}(z)=g(x_{2})$.
As before, it sufficies to consider the case that $f(x_{1})=cx_{1}+c_{2}$,
$g(x_{2})=c^{\prime}(cx_{2}+c_{1})$ and $B[c^{-1}]=R[c^{-1}][x_{1},x_{2},z]$
with $cx_{1}x_{2}+c_{1}x_{1}+c_{2}x_{2}=h(z)$, where $c^{\prime}\in R^{*}$,
$c\in R\setminus\\{0\\}$, $c_{1},c_{2}\in R$ and $h(z)\in R[z]$. As observed
in (1), $h(z)\notin R$. Since $R[x_{1}]=R[f(x_{1})]$, it follows that $c\in
R^{*}$. Again, by replacing $x_{1}$ and $x_{2}$ by $x_{1}-c^{-1}c_{2}$ and
$x_{2}-c^{-1}c_{1}$, respectively, we have $B=R[x_{1},x_{2},z]$ with
$x_{1}x_{2}=\varphi(z)$, where $\varphi(z)\in R[z]\setminus R$. The assertion
follows from Lemma 4.2. ∎

Next, we give another characterization of rings $B(1,F)$.

######  Lemma 4.5.

Let $R$ be a factorial affine $k$-domain and $B$ be a factorial affine
$R$-domain. Suppose there exists a non-trivial exponential map
$\delta:B\rightarrow B[T]$ such that the following holds.

  1. (1)

$B^{\delta}=R[x]$, where $x\in B$ is trancendental over $R$, and

  2. (2)

there esists $z\in B\setminus xB$ such that $\delta(z)=z+x^{m}T$ for some
$m\in\mathbb{Z}_{\geq 1}$.

If $B\neq R[x,z]$, then $R[z]\cap xB=\varphi(z)R[z]$ where $\varphi(z)$ is an
irreducible element of $R[z]$ with $\deg_{z}\varphi(z)\geq 1$.

###### Proof.

From Lemma 2.3(1), it follows that $x$ is an irreducible element of $B$, and
hence a prime element of $B$. Also, by Lemma 2.3(3), $B\subseteq
B[x^{-1}]=R[x,x^{-1},z]$. Thus $x$ and $z$ are algebraically independent over
$R$, and $R[z]\cap xB$ is a prime ideal of $R[z]$. Since $R[x,z]\subsetneq B$,
there exists an element $b\in B$ such that $x^{n}b=F(x,z)$ where $F(x,z)\in
R[x,z]\setminus xR[x,z]$ and $n\geq 1$. Write $F(x,z)=xG(x,z)+H(z)$, where
$G(x,z)\in R[x,z]$ and $H(z)\in R[z]$. Then $H(z)$ is non-zero and belongs to
$R[z]\cap xB$, and hence $R[z]\cap xB\neq(0)$. Let us consider the ideal
$K[z]\cap x(B\otimes_{R}K)$ of $K[z]$. Since $K[z]$ is a principal ideal
domain and $K[z]\cap x(B\otimes_{R}K)$ is a zon-zero prime ideal of $K[z]$, we
have $K[z]\cap x(B\otimes_{R}K)=\varphi(z)K[z]$ for some non-constant
irrducible element $\varphi(z)$ of $K[z]$ with $\deg_{z}\varphi(z)\geq 1$. As
$R$ is factorial, we can assume that $\varphi(z)$ is an irreducible element of
$R[z]$. Also, $\varphi(z)K[z]\cap R[z]=\varphi(z)R[z]$. Hence, $R[z]\cap
xB=\varphi(z)R[z]$ where $\varphi(z)$ is an irreducible element of $R[z]$ with
$\deg_{z}\varphi(z)\geq 1$. ∎

######  Theorem 4.6.

Let $R$, $B$ and $\delta$ be as in Lemma 4.5. Further suppose that $m=1$. If
$\partial_{z}\varphi(z)$ is non-zero, then $B$ is isomorphic to
$R[X,Y,Z]/(XY-F(Z))$ as $R$-algebra, where $F(Z)\in R[Z]\setminus R$.

###### Proof.

Suppose $B=R[x,z]$, then $B$ is isomorphic to $R[X,Y,Z]/(XY-F(Z))$, where
$F(Z)=Z\in R[Z]\setminus R$, as an $R$-algebra. Henceforth, we only consider
the case when $R[x,z]\subsetneq B$. Note that

$R[z]\cap xB=\varphi(z)R[z]$ (4.1)

where $\varphi(z)$ is an irreducible element of $R[z]$ with
$\deg_{z}\varphi(z)\geq 1$ (cf., Lemma 4.5). From (4.1), we can write
$\varphi(z)=xy$ for some $y\in B$. Suppose that $y\in xB$, say, $y=xb$ for
some $b\in B$. Applying $\delta^{(1)}$ to $\varphi(z)=x^{2}b$, we get that
$(\partial_{z}\varphi(z))x=x^{2}(\delta^{(1)}(b))$. Then it follows that
$\partial_{z}\varphi(z)\in xB\cap R[z]=\varphi(z)R[z]$, which is a
contradiction. Hence

$\varphi(z)=xy$ (4.2)

where $y\in B\setminus xB$. Applying $\delta$ on the equation (4.2), we have
$\delta(y)=y+(\partial_{z}\varphi(z))T+g_{1}(x,z,T)T^{2}$ where
$g_{1}(x,z,T)\in R[x,z][T]$.

Let $\overline{B}:=B/xB$, and let $\overline{\delta}$ denote the induced
exponential map on $\overline{B}$. By $\overline{y}$ and $\overline{z}$ we
denote the residue class of $y$ and $z$, respectively, in $\overline{B}$.
Since $\partial_{z}\varphi(z)\notin xB$, it follows that
$\overline{y}\notin{\overline{B}}^{\overline{\delta}}$. Hence
$\overline{\delta}$ is a nontrivial exponential map on $\overline{B}$. The
subring $R[\overline{z}]$ of $\overline{B}$ is contained in
${\overline{B}}^{\overline{\delta}}$. By Lemma 2.3(2), $\overline{y}$ is
algebraically independent over ${\overline{B}}^{\overline{\delta}}$.

Let $b\in B$. Then there esists some $n\in\mathbb{Z}_{\geq 0}$ such that
$x^{n}b\in R[x,z]\subset R[x,y,z]$. By induction on $n$ we show that $b\in
R[x,y,z]$. For $n=0$ the assertion holds trivially. Suppose $n\geq 1$ and
$x^{n}b=f(x,y,z)$ for some $f(x,y,z)\in R[x,y,z]$. Write

$x^{n}b=f(x,y,z)=xg(x,y,z)+\sum_{i=0}^{s}h_{i}(z)y^{i}$ (4.3)

where $s\in\mathbb{Z}_{\geq 0}$, $g(x,y,z)\in R[x,y,z]$, and $h_{i}(z)\in
R[z]$ for all $0\leq i\leq s$. Considering equation (4.3) in $\overline{B}$,
we get that

$h_{s}(\overline{z}){\overline{y}}^{s}+h_{s-1}(\overline{z}){\overline{y}}^{s-1}+\ldots+h_{1}(\overline{z})\overline{y}+h_{0}({\overline{z}})=0.$
(4.4)

where $h_{i}(\overline{z})\in R[\overline{z}]$ for all $0\leq i\leq s$. But
$\overline{y}$ is algebraically independent over $R[\overline{z}]$. Hence
$h_{i}(z)\in xB$ for all $0\leq i\leq s$. Using (4.1) and (4.2), it follows
that there exists $\alpha_{i}(z)\in R[z]$ such that $h_{i}(z)=xy\alpha_{i}(z)$
for all $0\leq i\leq s$. Substituting $h_{i}(z)^{\prime}s$ in the equation
(4.3), we get that

$x^{n-1}b=g(x,y,z)+\sum_{i=0}^{s}\alpha_{i}(z)y^{i+1}\in R[x,y,z].$ (4.5)

By induction hypothesis, $b\in R[x,y,z]$. Hence $B=R[x,y,z]$ with the
generators satisfying the relation $xy=\varphi(z)$, where $\varphi(z)\in
R[z]\setminus R$. The assertion follows from Lemma 4.2. ∎

We give a similar characterization to that of [12, Theorem 1.2] for the family
of Danielewski type varieties $B(m,F)$.

######  Theorem 4.7.

Let $R$, $B$ and $\delta$ be as in Lemma 4.5. If $\partial_{z}\varphi(z)$ is
non-zero and $R[\overline{z}]^{[1]}=B/xB$, then $B$ is isomorphic to
$R[X,Y,Z]/(X^{m}Y-F(X,Z))$ as $R$-algebra, where $F(X,Z)\in R[X,Z]$ such that
$F(0,Z)\in R[Z]\setminus R$ and $\overline{z}$ denotes the residue class of
$z$ in $B/xB$.

###### Proof.

We assume that $m$ is the smallest positive integer such that there exists a
non-trivial exponential map on $B$ with preslice satisying conditions of Lemma
4.5 and the theorem. As before, we only consider the case when
$R[x,z]\subsetneq B$. Let $\overline{B}:=B/xB$, the induced exponential map
from $\delta$ on $\overline{B}$ is denoted by $\overline{\delta}$, and for any
$b\in B$ its residue class in $\overline{B}$ is denoted by $\overline{b}$.
Note that

$R[z]\cap xB=\varphi(z)R[z]$ (4.6)

where $\varphi(z)$ is an irreducible element of $R[z]$ with
$\deg_{z}\varphi(z)\geq 1$ (cf., Lemma 4.5). Since $\varphi(z)\in xB$, we can
write

$\varphi(z)=x^{l_{1}}u_{1}$ (4.7)

where $u_{1}\in B\setminus xB$ and $l_{1}\geq 1$. Applying $\delta^{(1)}$, it
follows that $\partial_{z}\varphi(z)x^{m}=x^{l_{1}}\delta^{(1)}(u_{1})$.
Suppose $l_{1}>m$, then
$\partial_{z}\varphi(z)=x^{l_{1}-m}\delta^{(1)}(u_{1})\in xB$. Hence
$\partial_{z}\varphi(z)\in R[z]\cap xB=\varphi(z)R[z]$, which is a
contradiction by $\deg_{z}(\partial_{z}\varphi(z))<\deg_{z}\varphi(z)$. Thus
$l_{1}\leq m$, and

$\delta(u_{1})=u_{1}+x^{m-l_{1}}(\partial_{z}\varphi(z))T+g_{1}(x,z,T)T^{2}$
(4.8)

where $g_{1}(x,z,T)\in R[x,z][T]$.

Suppose $m=l_{1}$. Then
$\delta(u_{1})=u_{1}+\partial_{z}\varphi(z)T+g(x,z,T)T^{2}$. Since
$\overline{\partial_{z}\varphi(z)}$ is non-zero in $\overline{B}$, the element
$\overline{u_{1}}\notin\overline{B}^{\overline{\delta}}$. Now arguing as in
the proof of Theorem 4.6 we can show that $B$ is isomorphic to
$R[X,Y,Z]/(X^{m}Y-\varphi(Z))$ as an $R$-algebra, where $\varphi(Z)\in
R[Z]\setminus R$. Henceforth, we assume that $m>l_{1}$.

First we show that $\overline{u_{1}}\in R[\overline{z}]$. On the contrary,
suppose $\overline{u_{1}}\notin R[\overline{z}]$. Since
$\overline{B}=R[\overline{z}]^{[1]}$, $\overline{u_{1}}$ is algebraically
independent over $R[\overline{z}]$. For any $b\in B$, there exists
$n\in\mathbb{Z}_{\geq 0}$ such that $x^{n}b\in R[x,z]\subseteq R[x,u_{1},z]$.
By induction on $n$ we show that $B=R[x,u_{1},z]$. For $n=0$ the assertion
holds. Suppose $n\geq 1$, then $x^{n}b=f(x,u_{1},z)$ for some $f(x,u_{1},z)\in
R[x,u_{1},z]$. Write

$x^{n}b=f(x,u_{1},z)=xg(x,u_{1},z)+\sum_{i=0}^{s}h_{i}(z)\,u_{1}^{i}$ (4.9)

where $s\in\mathbb{Z}_{\geq 0}$, $g(x,u_{1},z)\in R[x,u_{1},z]$, and
$h_{i}(z)\in R[z]$ for all $0\leq i\leq s$. Considering the equation (4.9) in
$\overline{B}$, we get that

$h_{s}(\overline{z}){\overline{u_{1}}}^{s}+h_{s-1}(\overline{z}){\overline{u_{1}}}^{s-1}+\ldots+h_{1}(\overline{z})\overline{u_{1}}+h_{0}({\overline{z}})=0.$
(4.10)

Since $\overline{u_{1}}$ is algebraically independent over $R[\overline{z}]$,
it follows that $h_{i}(z)\in xB$ for all $0\leq i\leq s$. Using (4.6) and
(4.7), there exists $\alpha_{i}(z)\in R[z]$ such that
$h_{i}(z)=x^{l_{1}}u_{1}\alpha_{i}(z)$ for all $0\leq i\leq s$. Substituting
$h_{i}(z)$’s in (4.9) we have

$x^{n-1}b=g(x,u_{1},z)+\sum_{i=0}^{s}x^{l_{1}-1}\alpha_{i}(z){u_{1}}^{i+1}\in
R[x,u_{1},z].$ (4.11)

By induction hypothesis $b\in R[x,u_{1},z]$. Hence $B=R[x,u_{1},z]$ with
$\varphi(z)=x^{l_{1}}u_{1}$ where $\varphi(z)\in R[z]\setminus R$. Therefore
$B\cong R[X,U_{1},Z]/(X^{l_{1}}U_{1}-\phi(Z))$, but this contradicts the
minimality of $m$ (c.f. Lemma 4.1(2) and Lemma 4.2). Thus $\overline{u_{1}}\in
R[\overline{z}]$.

We can write $\overline{u_{1}}=\beta_{1}(\overline{z})$ for some
$\beta_{1}(z)\in R[z]$ with $\beta_{1}(z)\notin xB$. Then
$u_{1}=\beta_{1}(z)+x^{l_{2}}u_{2}$ where $u_{2}\in B\setminus xB$ and
$l_{2}\geq 1$. From (4.7), we get that

$\varphi(z)=x^{l_{1}}\beta_{1}(z)+x^{l_{1}+l_{2}}u_{2}.$ (4.12)

Applying $\delta^{(1)}$ it follows that

$(\partial_{z}\varphi(z))x^{m}=x^{m+l_{1}}(\partial_{z}\beta_{1}(z))+x^{l_{1}+l_{2}}\delta^{(1)}(u_{2}).$
(4.13)

Suppose $l_{1}+l_{2}>m$, then $\partial_{z}\varphi(z)\in R[z]\cap
xB=\varphi(z)R[z]$. Since $(\partial_{z}\varphi(z))\neq 0$ and
$\deg_{z}(\partial_{z}\varphi(z))<\deg_{z}\varphi(z)$, we get a contradiction.
Hence $l_{1}+l_{2}\leq m$.

If $l_{1}+l_{2}=m$, then
$\delta^{(1)}(u_{2})=\partial_{z}\varphi(z)+x^{l_{1}}(\partial_{z}\beta_{1}(z))$.
Hence $\overline{u_{2}}\notin\overline{B}^{\overline{\delta}}$. Thus the
induced exponential map $\overline{\delta}$ on $\overline{B}$ is non-trivial.
By $R[\overline{z}]\subseteq\overline{B}^{\overline{\delta}}$ and Lemma
2.3(2), it follows that $\overline{u_{2}}$ is algebraically independent over
$R[\overline{z}]$. Using the induction process as above we can show that
$B=R[x,u_{2},z]$ where the generators satisfy the relations
$\varphi(z)-x^{l_{1}}\beta_{1}(z)=x^{m}u_{2}$. Then the assertion follows from
Lemma 4.2 with $F(x,z)=\varphi(z)-x^{l_{1}}\beta_{1}(z)$.

Suppose $l_{1}+l_{2}<m$. Repeating the above process, we can find $y\in
B\setminus xB$ satisfying the following equation

$\varphi(z)=x^{m_{1}}\beta_{1}(z)+x^{m_{2}}\beta_{2}(z)+\ldots+x^{m_{s}}\beta_{s}(z)+x^{m}y$
(4.14)

where $m>m_{r}>\ldots>m_{2}>m_{1}\geq 1$ and $\beta_{i}(z)\in R[z]$ such that
$\beta_{i}(z)\notin xB$ for all $1\leq i\leq s$. Applying $\delta^{(1)}$ to
(4.14), we have

$\delta^{(1)}(y)=\partial_{z}\varphi(z)-x^{m_{1}}(\partial_{z}\beta_{1}(z))-x^{m_{2}}(\partial_{z}\beta_{2}(z))-\ldots-x^{m_{s}}(\partial_{z}\beta_{s}(z))$
(4.15)

Since $\partial_{z}\varphi(z)\notin xB$, it follows that
$\overline{y}\notin\overline{B}^{\overline{\delta}}$. Hence
$\overline{\delta}$ is a non-trivial exponential map on $\overline{B}$. Also,
$R[\overline{z}]\subseteq\overline{B}^{\overline{\delta}}$. From Lemma 2.3(2),
it follows that $\overline{y}$ is algebraically independent over
$R[\overline{z}]$. Let
$F(x,z):=\varphi(z)-x^{m_{1}}\beta_{1}(z)-x^{m_{2}}\beta_{2}-\ldots-x^{m_{s}}\beta_{s}(z)$.
Then arguing inductively as before we can show that $B=R[x,y,z]$ where the
generators satisfy the relation $x^{m}y=F(x,z)$. By Lemma 4.2 the assertion
follows. ∎

######  Corollary 4.8.

Let $R$ be a factorial affine $k$-domain with $Q(R)$ perfect, and let $B$ be a
factorial affine $R$-domain. Suppose there exists a non-trivial exponential
map $\delta:B\rightarrow B[T]$ such that the following hold.

  1. (1)

$B^{\delta}=R[x]$, where $x\in B$ is trancendental over $R$, and

  2. (2)

there exists an element $z\in B\setminus xB$ such that $\delta(z)=z+x^{m}T$
and $R[\overline{z}]^{[1]}=B/xB$, where $m\in\mathbb{Z}_{\geq 1}$ and
$\overline{z}$ denotes the residue field of $z$ in $B/xB$.

Then $B$ is isomorphic to $R[X,Y,Z]/(X^{m}Y-F(X,Z))$ as $R$-algebra, where
$F(X,Z)\in R[X,Z]$ with $F(0,Z)\in R[Z]\setminus R$.

###### Proof.

From Lemma 4.5, $R[z]\cap xB=\varphi(z)R[z]$ where $\varphi(z)$ is
irreducible. Since $Q(R)$ is perfect $\partial_{z}\varphi(z)$ is non-zero. The
assertion follows from Theorem 4.6 and 4.7. ∎

##  5\. Non-cancellation of Double Danielewski type varieties

Throughout this section the following notations are fixed. Recall that $R$ is
a factorial affine $k$-domain with field of fraction $K$. Let $X,Y,Z$ and $T$
denote indeterminates over $R$. Let

$B_{d,e}:=\dfrac{R[X,Y,Z,T]}{(X^{d}Y-P(X,Z),X^{e}T-Q(X,Y,Z))},$ (5.1)

where $d,e\geq 1$, $r:=\deg_{Z}P(0,Z)\geq 1$, $s:=\deg_{Y}Q(X,Y,Z)\geq 1$, $Q$
is monic in $Y$ as a polynomial in $K[X,Y,Z]$ and $d,e,r,s$ satisfy the
following conditions

$\displaystyle\text{~{}either~{}}r\geq 2\text{~{}and~{}}s\geq 2,$ (5.2)
$\displaystyle\text{~{}or~{}}r\geq 2\text{~{}and~{}}s=1,$
$\displaystyle\text{~{}or~{}}r=1,s\geq 2\text{~{}and~{}}e\geq 2.$

Let $x,y,z,t$ be the images of $X,Y,Z,T$, respectively, in $B_{d,e}$, and
$B_{d,e}^{K}:=B_{d,e}\otimes_{R}K$. Note that $B_{d,e}$ is an integral domain
by (cf., [14, Lemma 3.2]).

######  Remark 5.1.

In [14, Theorem 3.9], Gupta and Sen have shown that if $P\in K[X,Z]$ is monic
in $Z$, then

$\operatorname{ML}(B_{d,e}^{K})=K[x].$

However, from their proof it can be observed that the result holds without the
assumption on monicness of $P$.

######  Lemma 5.2.

$\operatorname{ML}_{R}(B_{d,e})=R[x]$.

###### Proof.

Note that $\operatorname{ML}(B_{d,e}^{K})=K[x]$ (cf. Remark 5.1). Now since
$\operatorname{ML}_{R}(B_{d,e})$ is factorially closed in $B_{d,e}$, by Lemma
3.1, we have
$\operatorname{ML}_{R}(B_{d,e})=\operatorname{ML}(B_{d,e}^{K})\cap
B_{d,e}=K[x]\cap B_{d,e}$. As $B_{d,e}\hookrightarrow R[x,x^{-1},z]$, it
follows that $\operatorname{ML}_{R}(B_{d,e})=R[x]$. ∎

The following result is a classification of varieties $B_{d,e}$ upto
$R$-isomorphism. It is motivated from a result of Gupta and Sen which gives a
similar classification where $R=k$ (c.f. [14, Theorem 3.11]). Throughout their
classification it was assumed that $P\in k[X,Z]$ is monic in $Z$ with
$r=\deg_{Z}P$ and $r$ satisfies the conditions as in (5.2). For $r>1$ the
following theorem relaxes the condition on monicness of $P$ in $Z$ at some
places.

######  Theorem 5.3.

Let $R$ be an affine $k$-domain,

$B_{1}:=\dfrac{R[X,Y,Z,T]}{(X^{d_{1}}Y-P_{1}(X,Z),X^{e_{1}}T-Q_{1}(X,Y,Z))},$

and

$B_{2}:=\dfrac{R[X,Y,Z,T]}{(X^{d_{2}}Y-P_{2}(X,Z),X^{e_{2}}T-Q_{2}(X,Y,Z))}$

be such that $r_{i}=\deg_{Z}P_{i}(0,Z)>1$, $s_{i}=\deg_{Y}Q_{i}(X,Y,Z)$ where
$Q_{i}$ is monic in $Y$ as polynomials in $K[X,Y,Z]$ for $i=1,2$. Let
$x,y,z,t$ and $x^{\prime},y^{\prime},z^{\prime},t^{\prime}$ be the images of
$X,Y,Z,T$ in $B_{1}$ and $B_{2}$ respectively. If $\rho:B_{1}\rightarrow
B_{2}$ be an $R$-isomorphism then the following hold.

  * (i)

$(d_{1},e_{1},r_{1},s_{1})=(d_{2},e_{2},r_{2},s_{2})=(d,e,r,s)$.

  * (ii)

If $P_{1}$ and $P_{2}$ are both monic in $Z$ as polynomials in $K[X,Z]$, then

$\displaystyle
P_{2}(\lambda_{1}X,\mu_{1}Z+\delta_{1}(X))=\lambda_{1}^{d}\widetilde{\beta_{1}}P_{1}(X,Z)+X^{d}\lambda_{1}^{d}\widetilde{\alpha_{1}}(X,Z),$

and,

$\displaystyle
Q_{2}(\lambda_{1}X,\widetilde{\beta_{1}}Y+\widetilde{\alpha_{1}}(X,Z),\mu_{1}Z+\delta_{1}(X))$
$\displaystyle=X^{e}g_{1}^{\prime}(X,Y,Z)+g_{2}^{\prime}Q_{1}(X,Y,Z)+(X^{d}Y-P_{1}(X,Z))g_{3}^{\prime}(X,Y,Z)$

for some $\lambda_{1},\mu_{1},\widetilde{\beta_{1}},g_{2}^{\prime}\in R^{*}$,
$\delta_{1}(X)\in R[X]$, $\widetilde{\alpha_{1}}(X,Z)\in R[X,Z]$,
$g_{1}^{\prime}(X,Y,Z),g_{3}^{\prime}(X,Y,Z)\in R[X,Y,Z]$.

  * (iii)

The isomorphism $\rho$ is given as follows

$\displaystyle\rho(x)=\lambda_{1}^{-1}x^{\prime}$
$\displaystyle\rho(z)=\mu_{1}^{-1}z^{\prime}+\delta(x^{\prime})$
$\displaystyle\rho(y)=\widetilde{\beta_{1}}^{-1}y^{\prime}+\alpha(x^{\prime},z^{\prime})$
$\displaystyle\rho(t)=\lambda_{1}^{e}(g_{2}^{\prime})^{-1}t^{\prime}+g(x^{\prime},y^{\prime},z^{\prime}).$

for some $\delta\in R[x^{\prime}]$, $\alpha\in R[x^{\prime},z^{\prime}]$ and
$g\in R[x^{\prime},y^{\prime},z^{\prime}]$.

Conversely, if $(i)$, $(ii)$ hold, then $B_{1}\cong_{R}B_{2}$.

###### Proof.

(i) Identifying $B_{1}$ with $\rho(B_{1})$ we assume that $B_{1}=B_{2}=B$. By
Lemma 5.2, $\operatorname{ML}_{R}(B)=R[x]=R[x^{\prime}]$. Hence

$x^{\prime}=\lambda_{1}x+\lambda_{2}$ (5.3)

for some $\lambda_{1}\in R^{*}$ and $\lambda_{2}\in R$. Also,
$B\otimes_{R[x]}K(x)=B\otimes_{R[x^{\prime}]}K(x^{\prime})$. Thus

$K(x)[z]=K(x^{\prime})[z^{\prime}].$ (5.4)

Therefore, we have $z^{\prime}=\dfrac{\alpha_{1}z+\beta_{1}}{\gamma_{1}}$ for
$\alpha_{1},\beta_{1}\in R[x]$, $\gamma_{1}\in R[x]\setminus\\{0\\}$. Suppose
either $\gamma_{1}\nmid\alpha_{1}$, or $\gamma_{1}\nmid\beta_{1}$. From
$B\hookrightarrow R[x,x^{-1},z]$ it follows that $\gamma_{1}=\gamma_{2}x^{i}$
for some $\gamma_{2}\in R^{*}$ and $i\geq 0$. If $i>0$, then
$\alpha_{1}(0)z+\beta_{1}(0)\in xB\cap R[z]=(P_{1}(0,z))R[z]$. This is a
contradiction as $\deg_{z}P_{1}(0,z)>1$. Thus $\gamma_{1}\in R^{*}$, and hence
$z^{\prime}\in R[x,z]$. Write

$z^{\prime}=\mu_{1}z+\delta_{1}(x)$ (5.5)

for some $\mu_{1},\delta_{1}\in R[x]$.

Similarly, using the facts that $B\hookrightarrow
R[x^{\prime},(x^{\prime})^{-1},z^{\prime}]$ and
$\deg_{z^{\prime}}P_{2}(0,z^{\prime})>1$ we have

$z=\mu_{2}z^{\prime}+\delta_{2}(x)$ (5.6)

for some $\mu_{2},\delta_{2}\in R[x]$. From (5.5) and (5.6) we get that
$\mu_{1},\mu_{2}\in R^{*}$ and

$E:=R[x,z]=R[x^{\prime},z^{\prime}].$ (5.7)

Note that $y=\dfrac{P_{1}(x,z)}{x^{d_{1}}}\in B\setminus E$ and
$B\hookrightarrow E[(x^{\prime})^{-1}]$. Then there exists $n>0$ such that
$(x^{\prime})^{n}y=\dfrac{(x^{\prime})^{n}P_{1}(x,z)}{x^{d_{1}}}\in E$. Since
$x\nmid P_{1}(x,z)$ in $E$, we have $x\mid x^{\prime}$, and hence

$x^{\prime}=\lambda_{1}x.$ (5.8)

Suppose that $d_{1}>d_{2}$. Then

$\displaystyle~{}~{}~{}~{}~{}~{}x^{d_{1}}B\cap E=(x^{\prime})^{d_{1}}B\cap E$
$\displaystyle
i.e.,(x^{d_{1}},P_{1}(x,z))E=((x^{\prime})^{d_{1}},(x^{\prime})^{d_{1}-d_{2}}P_{2}(x^{\prime},z^{\prime}))E\subseteq
x^{\prime}E=xE$

This is a contradiction as $P_{1}(0,z)\neq 0$. Hence $d_{1}\leq d_{2}$. By
symmetry, we have $d_{2}\leq d_{1}$. Thus

$d_{1}=d_{2}=d.$ (5.9)

Next, we consider the ideal

$x^{d}B\cap E=(x^{d},P_{1}(x,z))E=(x^{d},P_{2}(x^{\prime},z^{\prime}))E$
(5.10)

Using (5.5) and (5.8),

$P_{2}(x^{\prime},z^{\prime})=P_{2}(\lambda_{1}x,\mu_{1}z+\delta_{1}(x))=x^{d}\alpha_{1}(x,z)+P_{1}(x,z)\beta_{1}(x,z),$
(5.11)

where the monomials of $\beta_{1}(x,z)$ are not divisible by $x^{d}$. Putting
$x=0$ in (5.11) we have

$P_{2}(0,\mu_{1}z+\delta_{1}(0))=P_{1}(0,z)\beta_{1}(0,z).$ (5.12)

Similarly using the symmetry in (5.10),

$P_{1}(x,z)=x^{d}\alpha_{2}(x,z)+P_{2}(x^{\prime},z^{\prime})\beta_{2}(x,z)=x^{d}\alpha_{2}(x,z)+P_{2}(\lambda_{1}x,\mu_{1}z+\delta_{1}(x))\beta_{2}(x,z),$
(5.13)

where monomials of $\beta_{2}(x,z)\in E$ are not divisible by $x^{d}$, and

$P_{1}(0,z)=P_{2}(0,\mu_{1}z+\delta_{1}(0))\beta_{2}(0,z).$ (5.14)

Hence from (5.12) and (5.14) it follows that

$r_{1}=r_{2}=r,$ (5.15)

and $\beta_{1}(0,z),\beta_{2}(0,z)\in R^{*}$. Now, using (5.8), from (5.11)
and (5.13) we have

$\displaystyle
y^{\prime}=\widetilde{\alpha_{1}}(x,z)+y\widetilde{\beta_{1}}(x,z)$ (5.16)
$\displaystyle
y=\widetilde{\alpha_{2}}(x,z)+y^{\prime}\widetilde{\beta_{2}}(x,z),$

where $\widetilde{\alpha_{1}}=\lambda_{1}^{-d}\alpha_{1}$,
$\,\widetilde{\alpha_{2}}=\lambda_{1}^{d}\alpha_{2}$,
$\,\widetilde{\beta_{1}}=\lambda_{1}^{-d}\beta_{1}$,
$\,\widetilde{\beta_{2}}=\lambda_{1}^{d}\beta_{2}$. From (5.7),

$S:=R[x,y,z]=R[x^{\prime},y^{\prime},z^{\prime}].$

Now suppose $e_{1}>e_{2}$. Note that

$x^{e_{1}}B\cap
S=(x^{e_{1}},Q_{1}(x,y,z))S=(x^{e_{1}},x^{e_{1}-e_{2}}Q_{2}(x^{\prime},y^{\prime},z^{\prime}))S.$

Thus $Q_{1}(X,Y,Z)\in(X,X^{d}Y-P_{1}(X,Z))R[X,Y,Z]$ which contradicts the
monicness of $Q_{1}$ in $Y$ as a polynomial in $K[X,Y,Z]$. Hence $e_{1}\leq
e_{2}$. Similarly we have $e_{2}\leq e_{1}$. Thus

$e_{1}=e_{2}=e.$ (5.17)

Now consider the ideal

$x^{e}\cap
S=(x^{e},Q_{1}(x,y,z))S=(x^{e},Q_{2}(x^{\prime},y^{\prime},z^{\prime}))S.$
(5.18)

Now using (5.6), (5.8) and (5.16),

$\displaystyle(X^{e},Q_{1}(X,Y,Z),X^{d}Y-P_{1}(X,Z))R[X,Y,Z]$
$\displaystyle=(X^{e},Q_{2}(\lambda_{1}X,\widetilde{\alpha_{1}}(X,Z)+\widetilde{\beta_{1}}(X,Z)Y,\mu_{1}Z+\delta_{1}(X)),X^{d}Y-P_{1}(X,Z))R[X,Y,Z]$
(5.19)

In the above equation putting $X=0$, we have

$(Q_{1}(0,Y,Z),P_{1}(0,Z))R[Y,Z]\\\
=(Q_{2}(0,\widetilde{\alpha_{1}}(0,Z)+\widetilde{\beta_{1}}(0,Z)Y,\mu_{1}Z+\delta_{1}(0)),P_{1}(0,Z))R[Y,Z].$
(5.20)

Recall that $\widetilde{\beta_{1}}(0,Z),\widetilde{\beta_{2}}(0,Z)\in R^{*}$.
Since $\widetilde{\beta_{1}}(0,Z)\in R^{*}$ and $Q_{2}$ is monic in $Y$ as a
polynomial in $K[X,Y,Z]$, it follows that
$Q_{3}(Y,Z):=Q_{2}(0,\widetilde{\alpha_{1}}(0,Z)+\widetilde{\beta_{1}}(0,Z)Y,\mu_{1}Z+\delta_{1}(0))$
is monic in $Y$ as a polynomial in $K[X,Y,Z]$ and
$\deg_{Y}Q_{3}=\deg_{Y}Q_{2}=s_{2}$. Now, let $r(Z)$ be a prime divisor of
$P_{1}(0,Z)$ in $R[Z]$, and let $D:=\dfrac{R[Z]}{(r(Z))}$. From (5.20), we get
that

$(Q_{1}(0,Y,\overline{Z}))D[Y]=(Q_{3}(Y,\overline{Z}))D[Y],$ (5.21)

where $\overline{Z}$ is the image of $Z$ in $D$. Then it follows that

$s_{1}=s_{2}=s.$ (5.22)

(ii) We now assume that $P_{1},P_{2}\in R[X,Z]$ are monic in $Z$ as
polynomials in $K[X,Z]$. Since $r_{1}=r_{2}$, then from (5.11) and (5.13) we
get that $\beta_{1}(x,z),\beta_{2}(x,z)\in R\setminus\\{0\\}$. Furthermore,

$(1-\beta_{1}\beta_{2})P_{1}(x,z)\in xR[x,z].$ (5.23)

Using the fact that $P_{1}(0,Z)\neq 0$, we have $\beta_{1},\beta_{2}\in
R^{*}$. Thus $\widetilde{\beta_{1}}=\lambda_{1}^{-d}\beta_{1}$ and
$\widetilde{\beta_{2}}=\lambda_{1}^{d}\beta_{2}\in R^{*}$. Then from (5.11) it
follows that

$P_{2}(\lambda_{1}X,\mu_{1}Z+\delta_{1}(X))=X^{d}\alpha_{1}(X,Z)+P_{1}(X,Z)\beta_{1}.$
(5.24)

Now (5.16) implies

$y^{\prime}=\widetilde{\alpha_{1}}(x,z)+y\widetilde{\beta_{1}}.$ (5.25)

From (5.19) we can write

$Q_{1}(X,Y,Z)=X^{e}g_{1}+Q_{2}(\lambda_{1}X,\widetilde{\alpha_{1}}(X,Z)+\widetilde{\beta_{1}}Y,\mu_{1}Z+\delta_{1}(X))g_{2}+(X^{d}Y-P_{1}(X,Z))g_{3}$

for some $g_{1},g_{2},g_{3}\in R[X,Y,Z]$ such that $g_{2}$ has no monomial
divisible by $X^{e}$ and the coefficients of $Y$ in $g_{2}$ are not divisible
by $X^{d}$. Now using the fact that $Q_{1},Q_{2}$ are monic in $Y$ as
polynomials in $K[X,Y,Z]$ and $s_{1}=s_{2}$, we have $g_{2}\in R$. Similarly,

$Q_{2}(\lambda_{1}X,\widetilde{\alpha_{1}}(X,Z)+\widetilde{\beta_{1}}Y,\mu_{1}Z+\delta_{1}(X))=X^{e}g_{1}^{\prime}+Q_{1}(X,Y,Z)g_{2}^{\prime}+(X^{d}Y-P_{1}(X,Z))g_{3}^{\prime}$
(5.26)

for some $g_{1}^{\prime},g_{3}^{\prime}\in R[X,Y,Z]$ and $g_{2}^{\prime}\in
R$. Therefore, we have

$(1-g_{2}g_{2}^{\prime})Q_{1}(X,Y,Z)\in(X^{e},X^{d}Y-P_{1}(X,Z))R[X,Y,Z].$

Since $Q_{1}(0,Y,Z)$ is monic in $Y$ as a polynomial in $K[X,Y,Z](=K^{[3]})$
it is not divisible by $P(0,Z)$, we have $g_{2}g_{2}^{\prime}=1$, that is,
$g_{2},g_{2}^{\prime}\in R^{*}$. Thus the assertion follows from (5.24) and
(5.26).

Now from (5.26) and (5.8),

$t^{\prime}=\lambda_{1}^{-e}g_{2}^{\prime}t+\lambda_{1}^{-e}g_{1}^{\prime}(x,y,z).$
(5.27)

(iii) The result follows from (5.5), (5.8), (5.25) and (5.27).

We now show the converse. Consider the following surjective map
$\phi:R[X,Y,Z,T]\rightarrow B_{1}$ defined as follows.

$\displaystyle\phi\mid_{R}=id_{R}$ $\displaystyle\phi(X)=\lambda_{1}x$
$\displaystyle\phi(Y)=\tilde{\beta_{1}}y+\tilde{\alpha_{1}}(x,z)$
$\displaystyle\phi(Z)=\mu_{1}z+\delta_{1}(x)$
$\displaystyle\phi(T)=\lambda_{1}^{e}(g_{2}^{\prime}t-g_{1}^{\prime}(x,y,z)).$

Note that
$(X^{d}Y-P_{2}(X,Z),X^{e}T-Q_{2}(X,Y,Z))R[X,Y,Z]\subseteq\ker(\phi)$, and
hence $\phi$ induces a surjective map

$\overline{\phi}:B_{2}=\dfrac{R[X,Y,Z,T]}{(X^{d}Y-P_{2}(X,Z),X^{e}T-Q_{2}(X,Y,Z))}\rightarrow
B_{1}.$

The ring $B_{2}$ is an integral domain. Note that $R$ is an affine domain, and
$B,B_{1}$ are also affine. Since
$\operatorname{tr.deg}_{k}B_{2}=\operatorname{tr.deg}_{k}B_{1}$ and $\dim
B=\dim B_{1}$, it follows that $\overline{\phi}$ is an $R$-isomorphism. ∎

######  Remark 5.4.

(i) Note that for the classification over fields in [14, Theorem 3.11(i)] the
fact that “$P$ is monic in $Z$” is heavily used, but our result shows that the
equality of the exponents i.e., Theorem 5.3(i) holds without the assumption of
monicness of $P$ in $Z$ as a polynomial in $K[X,Z]$.
(ii) If $B_{d,e}=\dfrac{R[X,Y,Z,T]}{(X^{d}Y-P(X,Z),X^{e}T-Q(X,Y,Z))}$ be such
that $\deg_{Y}Q=1$, then it is isomorphic to the Danielewski type varieties
$B(d+e,F)$. Also, by Theorem 5.3(i), the family $\\{B_{d,e}\\}$ consists of
infinitely many varieties which are not isomorphic to Danielewski type
varieties over $R$.

######  Theorem 5.5.

Let $R$ be a factorial affine $k$-domain and $B_{d,e}$ be the integral domain
as in (5.1). Let

$P(X,Z)=a_{0}(X)+a_{1}(X)Z+\cdots+a_{m}(X)Z^{m}$

and

$Q(X,Y,Z)=b_{0}(X,Z)+b_{1}(X,Z)Y+\cdots+b_{s}Y^{s}$

where $a_{i}\in R^{[1]}$ for $1\leq i\leq m$, $b_{i}\in R^{[2]}$ for $1\leq
i\leq s-1$ and $b_{s}\in R\setminus\\{0\\}$. We define the following
polynomials

$P^{\prime}(X,Z)=a_{1}(X)+2a_{2}(X)Z+\cdots+ma_{m}(X)Z^{m-1}$

and

$Q^{\prime}(X,Y,Z)=b_{1}(X,Z)+2b_{2}(X,Z)Y+\cdots+sb_{s}Y^{s-1}.$

Suppose that $e>1$, $\deg_{Z}P(X,Z)>1$, $(P(0,Z),P^{\prime}(0,Z))=R[Z]$ and
$(P(0,Z),Q(0,Y,Z),Q^{\prime}(0,Y,Z))$ $=R[Y,Z]$. Then

$B_{d,e}^{[1]}\cong_{R}B_{d,e-1}^{[1]}.$

###### Proof.

Let the exponential map $\phi$ considered in the proof of [14, Theorem 3.15]
be an $R$-algebra homorphism. Then the desired isomorphism is obtained by
arguing as in the proof of [14, Theorem 3.15]. ∎

Let us consider the following family of affine domains from (5.1). The family
consist of varieties which are different from the Danielewski type varieties
considered in the family $\Omega_{1}$ (cf., Remark 5.4(ii)).

$\displaystyle\Omega_{3}:=\biggl{\\{}B_{d,e}\mid d,e\geq
1,\deg_{Z}P(0,Z)>1,Q\text{~{}is monic in~{}}Y\text{~{}as a polynomial
in~{}}K[X,Y,Z]$
$\displaystyle\text{~{}with~{}}\deg_{Y}Q>1,(P(0,Z),P^{\prime}(0,Z))=R[Z],\text{~{}and~{}}$
$\displaystyle(P(0,Z),Q(0,Y,Z),Q^{\prime}(0,Y,Z))=R[Y,Z]\biggr{\\}}.$

######  Corollary 5.6.

Let $R$ be an affine factorial $k$-domain. Then the family $\Omega_{3}$
consists of infinitely many rings which are pairwise non-isomorphic as
$R$-algebras and counterexamples to the Generalized Cancellation Problem over
$R$.

###### Proof.

Follows from the Theorems 5.3(i) and 5.5. ∎

##  6\. Characterization of Double Danielewski type varieties

Throughout this section $R$ is a factorial affine domain over a field $k$ of
characteristics zero. Let $B$ be an affine $R$-domain. Recall that $K:=Q(R)$.
In terms of locally nilpotent derivations, we give a characterization of
factorial $B$ to be $R$-isomorphic to a variety of type

$B_{d,e}:=R[X,Y,Z,T]/(X^{d}Y-P(X,Z),X^{e}T-Q(X,Y,Z))$ (6.1)

where $d,e\in\mathbb{N}$, $P(X,Z)\in R[X,Z]$ such that $P(0,Z)$ is irreducible
in $R[Z]$ with $\deg_{Z}P(0,Z)\geq 1$, and $Q(X,Y,Z)\in R[X,Y,Z]$ such that
$Q(0,Y,0)\in R[Y]\setminus R$. We denote by $x,y,z$ and $t$ the residue class
of $X,Y,Z$ and $T$ in $B$. Hence $B=R[x,y,z,t]$ with generators satisfying the
relations $x^{d}y=P(x,z)$ and $x^{e}t=Q(x,y,z)$.

We first state two lemmas.

######  Lemma 6.1.

Let $A$ be an integral domain and $a,b\in A\setminus\\{0\\}$, if $a$ is not a
zero divisor in $A/(b)$ then $b^{n}$ is not a zero divisor in $A/(a)$ for
every $n\geq 1$.

######  Lemma 6.2.

Let $A$ be an integral domain and $a,b\in A\setminus\\{0\\}$. If $b$ is not a
zero divisor in $A/(a)$ then the ring $A[T]/(bT-a)$ is an integral domain.

Using the above two lemmas we now prove the following.

######  Lemma 6.3.

The ring $B_{d,e}$, as in (6.1), is an integral domain.

###### Proof.

The ring $R[X,Y,Z,T]$ is a factorial. Consider $S:=R[X,Y,Z]/(X^{d}Y-P(X,Z))$.
Since $P(0,Z)\in R[Z]\setminus R$, $X^{d}Y-P(X,Z)$ is an irreducible, and
hence prime, element of $R[X,Y,Z]$. Thus $S$ is an integral domain. With the
notations defined above $S=R[x,y,z]$, a subring of $B_{e,d}$. Then
$B_{d,e}=S[T]/(x^{e}T-Q(x,y,z))$.

Now $\frac{S}{(x)}\cong\left(\frac{R[Z]}{(P(0,Z))}\right)[Y]$. Since $P(0,Z)$
is irreducible, it follows that $S/(x)$ is an integral domain. Then $Q(x,y,z)$
is a non-zero divisor of $S/(x)$. By Lemma 6.1, $x^{e}$ is a non-zero divisor
of $S/(Q(x,y,z))$. The assertion follows from Lemma 6.2. ∎

Let $D$ be a locally nilpotent derivation on $R[X,Y,Z,T]$ defined by

$D=X^{d+e}\frac{\partial}{\partial
Z}+\,X^{e}\partial_{Z}P(X,Z)\frac{\partial}{\partial
Y}+\,(X^{d}\partial_{Z}Q(X,Y,Z)+\partial_{Y}Q(X,Y,Z)\,\partial_{Z}P(X,Z))\frac{\partial}{\partial
T}.$

Then $D$ induces a locally nilpotent $R$-derivation $\delta$ on $B_{d,e}$ with
kernel $R[x]$, $\delta(z)=x^{d+e}$ and $\delta(y)=x^{e}\partial_{z}P(x,z)$. If
$R=k$, $P$ and $Q$ are monic in $Z$ and $Y$, respectively, satisfying (5.2),
then Gupta and Sen determined all locally nilpotent derivations of $B_{d,e}$
[15]. They proved that all locally nilpotent derivation of $B_{d,e}$ is
generated by the derivation $\delta$. In view of this we give some
characterization of factorial affine domains to be $R$-isomorphic to varieties
of type (6.1). First we state the following preparatory lemma.

######  Lemma 6.4.

Let $R$ be a factorial affine $k$-domain. Let $B$ be an affine $R$-domain
generated over $R$ by the elements $x,y,z,t\in B$ with tr.deg${}_{k}B=$
tr.deg${}_{k}R+2$. Suppose that $x$ and $z$ are algebraically independent over
$R$ and the generators satisfy relations $x^{d}y-F(x,z)$ and
$x^{e}t=G(x,y,z)$, where $e,d\geq 1$, $F(x,z)\in R[x,z]$ such that $F(0,z)\in
R[z]\setminus R$ is irreducible, and $G(x,y,z)\in R[x,y,z]$ such that
$G(0,y,0)\in R[y]\setminus R$. Then $B$ is isomorphic to
$R[X,Y,Z,T]/(X^{d}Y-F(X,Z),X^{e}T-G(X,Y,Z))$ as $R$-algebra.

###### Proof.

Let $\Phi:R[X,Y,Z,T]\rightarrow B$ be an $R$-algebra homomorphism defined by
$\Phi(X)=x$, $\Phi(Y)=y$, $\Phi(Z)=z$ and $\Phi(T)=t$. Then
$(X^{d}Y-F(X,Z),X^{e}T-G(X,Y,Z))\subseteq$ Ker$\Phi$. Let
$S:=R[X,Y,Z,T]/(X^{d}Y-F(X,Z),X^{e}T-G(X,Y,Z))$. Note that $S$ is an integral
domain (cf., Lemma 6.3). Then the induced homomorphism
$\overline{\Phi}:S\rightarrow B$ is a surjective $R$-algebra homomorphism
between integral domains with same trancendence degree over $k$. Hence
$\overline{\Phi}$ is an $R$-algebra isomorphism. ∎

######  Theorem 6.5.

Let $R$ be factorial affine domain over a field $k$ of characteristic zero,
and $B$ be a factorial affine $R$-domain. Suppose there exists a non-trivial
locally nilpotent $R$-derivation $\delta$ on $B$ such that the following hold

  1. (1)

$B^{\delta}=R[x]$, where $x\in B$ is trancendental over $R$, and

  2. (2)

there exists an element $z\in B\setminus xB$ such that $\delta(z)=x^{2}$.

If $R[\overline{z}]$ is factorial, then $B$ is isomorphic to
$R[X,Y,Z,T]/(XY-\varphi(Z),XT-\psi(Y,Z))$ as $R$-algebra, where $\varphi(Z)\in
R[Z]\setminus R$, $\psi(Y,0)\in R[Y]\setminus R$ and $\overline{z}$ is the
image of $z$ in $B/xB$.

###### Proof.

By Lemma 2.3(1), $x$ is irreducible, and hence prime in $B$. From Lemma
2.3(3), we get that $B\subseteq R[x,x^{-1},z]$. Thus $x$ and $z$ are
algebraically independent over $R$. Suppose $B=R[x,z]$, then $B$ is isomorphic
to $R[X,Y,Z,T]/(XY-\varphi(Z),XT-\psi(Y,Z))$ as an $R$-algebra, where
$\varphi(Z)=Z\in R[Z]\setminus R$ and $\psi(Y,Z)=Y\in R[Y]\setminus R$.
Henceforth, we only consider the case when $R[x,z]\subsetneq B$. We adhere to
these notations for rest of the proof, $\overline{B}:=B/xB$,
$\overline{\delta}$ denotes the induced locally nilpotent derivation on
$\overline{B}$, and $\overline{b}$ denotes the residue class of $b\in B$ in
$\overline{B}$. Note that

$R[z]\cap xB=\varphi(z)R[z]$ (6.2)

where $\varphi(z)$ is an irreducible polynomial of $R[z]$ with
$\deg_{z}\varphi(z)\geq 1$ (c.f. Lemma 4.5). Since $\varphi(z)\in
xB\setminus\\{0\\}$, we can write

$\varphi(z)=x^{l_{1}}y$ (6.3)

where $y\in B\setminus xB$ and $l_{1}\geq 1$. Applying $\delta$, it follows
that $\partial_{z}\varphi(z)x^{2}=x^{l_{1}}\delta(y)$. Suppose $l_{1}>2$, then
$\partial_{z}\varphi(z)=x^{l_{1}-2}\delta(y)\in xB$. Hence
$\partial_{z}\varphi(z)\in R[z]\cap xB=\varphi(z)R[z]$, which is a
contradiction as $\deg_{z}(\partial_{z}\varphi(z))<\deg_{z}\varphi(z)$. Thus
$l_{1}\leq 2$, and

$\delta(y)=x^{2-l_{1}}(\partial_{z}\varphi(z)).$ (6.4)

Suppose $l_{1}=2$. Since $\partial_{z}\varphi(z)\neq 0$ in $\overline{B}$, the
element $\overline{y}\notin\bar{B}^{\overline{\delta}}$ and is algebraically
independent over $R[\overline{z}]$. Hence $\overline{\delta}$ is a non-trivial
locally nilpotent derivation on $\overline{B}$. The subring $R[\overline{z}]$
of $\overline{B}$ is contained in ${\overline{B}}^{\overline{\delta}}$. Let
$b\in B$. Then there exists some $n\in\mathbb{Z}_{\geq 0}$ such that
$x^{n}b\in R[x,z]\subset R[x,y,z]$. By induction on $n$ we show that $b\in
R[x,y,z]$. For $n=0$ the assertion holds trivially. Suppose $n\geq 1$ and
$x^{n}b=f(x,y,z)$ for some $f(x,y,z)\in R[x,y,z]$. Write

$x^{n}b=f(x,y,z)=xg(x,y,z)+\sum_{i=0}^{s}h_{i}(z)y^{i}$ (6.5)

where $s\in\mathbb{Z}_{\geq 0}$, $g(x,y,z)\in R[x,y,z]$, and $h_{i}(z)\in
R[z]$ for all $0\leq i\leq s$. Considering equation (6.5) in $\overline{B}$,
we get that

$h_{s}(\overline{z}){\overline{y}}^{s}+h_{s-1}(\overline{z}){\overline{y}}^{s-1}+\ldots+h_{1}(\overline{z})\overline{y}+h_{0}({\overline{z}})=0.$
(6.6)

where $h_{i}(\overline{z})\in R[\overline{z}]$ for all $0\leq i\leq s$. But
$\overline{y}$ is algebraically independent over $R[\overline{z}]$. Hence
$h_{i}(z)\in xB$ for all $0\leq i\leq s$. Using (6.2) and (6.3), it follows
that there exists $\alpha_{i}(z)\in R[z]$ such that
$h_{i}(z)=x^{2}y\alpha_{i}(z)$ for all $0\leq i\leq s$. Substituting
$h_{i}(z)$ in the equation (6.5), we get that

$x^{n-1}b=g(x,y,z)+\sum_{i=0}^{s}\alpha_{i}(z)xy^{i+1}\in R[x,y,z].$ (6.7)

By induction hypothesis, $b\in R[x,y,z]$. Hence $B=R[x,y,z]$ with the
generators satisfying the relation $x^{2}y=\varphi(z)$, where $\varphi(z)\in
R[z]\setminus R$. From Lemma 4.2, $B$ is $R$-isomorphic to
$R[X,Y,Z]/(X^{2}Y-\varphi(Z))$, where $\varphi(Z)\in R[z]\setminus R$.

Now, we assume that $l_{1}=1$ i.e.,

$\phi(z)=xy.$ (6.8)

Then

$\delta(y)=x(\partial_{z}\varphi(z)),$ (6.9)

and hence $\overline{y}\in\overline{B}^{\overline{\delta}}$. The elements $y$
and $z$ are algebraically independent over $R$. Next, we consider the prime
ideal $\mathfrak{p}:=R[y,z]\cap xB$ of $R[y,z]$. Note that
$\varphi(z)R[z]=R[z]\cap xB\subseteq R[y,z]\cap xB$, and hence $\mathfrak{p}$
is non-zero prime ideal of $R[y,z]$. Let $\overline{\mathfrak{p}}$ denote the
prime ideal $\mathfrak{p}/\varphi(z)R[y,z]$ of $R[\overline{z}][y]$.

Let $\delta_{K}$ be the extension of $\delta$ to $B\otimes_{R}K$. By Lemma
2.3, $\delta_{K}$ is a locally nilpotent derivation on $B\otimes_{R}K$. Let
$\overline{\delta_{K}}$ be the inducted locally nilpotent derivation on
$\overline{B}\otimes_{R}K$. Note that
$K[\overline{y},\overline{z}]\subseteq{(\overline{B}\otimes_{R}K)}^{\overline{\delta_{K}}}$.
Consider the prime ideal $\overline{\mathfrak{p}}_{K}=K[y,z]\cap
xB/\varphi(z)K[y,z]$ of $K[\overline{z}][y]$. Suppose, if
$\mathfrak{p}=\varphi(z)R[y,z]$, that is, $\overline{\mathfrak{p}}_{K}=0$ then
$\overline{y}$ is algebraically independent over $R[\overline{z}]$, which is a
contradiction. Hence it follows that $\overline{\mathfrak{p}}\neq 0$ in
$R[\overline{z}][y]$.

Since $K[\overline{z}][y]$ is a principal ideal domain, it follows that
$\overline{\mathfrak{p}}_{K}=\psi(y,\overline{z})K[\overline{z}][y]$ where
$\psi(y,\overline{z})$ is an irreducible polynomial of $K[\overline{z}][y]$
with $\deg_{y}\psi(y,\overline{z})\geq 1$. We can assume that
$\psi(y,\overline{z})$ is an element of $R[\overline{z}][y]$ and the highest
power of $y$ has coefficient in $R$. Since $R[\overline{z}]$ is factorial then
$\psi(y,\overline{z})K[\overline{z}][y]\cap
R[\overline{z}][y]=\psi(y,\overline{z})R[\overline{z}][y]$, and hence we
deduce that $\overline{\mathfrak{p}}=\psi(y,\overline{z})R[\overline{z}][y]$.
Thus $\mathfrak{p}=(\psi(y,z),\varphi(z))R[y,z]$, where $\psi(y,z)\in R[y,z]$
is monic in $y$ as a polynomial in $K[y,z]$.

Write $\psi(y,z)=x^{n}t$ where $t\in B\setminus xB$ and $n\geq 1$. Applying
$\delta$ we get that

$x^{n}\delta(t)=x^{2}(\partial_{z}\psi(y,z))+x(\partial_{y}\psi(y,z))(\partial_{z}\varphi(z)).$

Suppose $n>1$. Then $(\partial_{y}\psi(y,z))(\partial_{z}\varphi(z))\in
R[y,z]\cap xB=(\varphi(z),\psi(y,z))R[y,z]$. Now,
$\partial_{y}\psi(y,\bar{z})$
$\partial_{z}\varphi(\bar{z})\in\psi(y,\bar{z})K[\bar{z}][y]$. Since
$\deg_{y}\psi(y,\bar{z})\geq 1$, $\partial_{y}\psi(y,\bar{z})$ is non-zero in
$R[\bar{z}][y]$. Hence we obtain a contradiction as
$\deg_{y}(\partial_{y}\psi(y,\bar{z}))<\deg_{y}\psi(y,\bar{z})$. Thus $n=1$
and

$\psi(y,z)=xt$ (6.10)

where $t\notin B$. Now, by (6.9),
$\delta(t)=x(\partial_{z}\psi(y,z))+(\partial_{y}\psi(y,z))(\partial_{z}\varphi(z))$,
which implies that $\bar{t}\notin\bar{B}^{\bar{\delta}}$. Since
$R[\bar{z},\bar{y}]\subseteq\bar{B}^{\bar{\delta}}$, it follows from Lemma 2.3
that $\bar{t}$ is algebraically independent over $R[\bar{y},\bar{z}]$.

Let $b\in B$, then there exists $m\in\mathbb{Z}_{\geq 0}$ such that $x^{m}b\in
R[x,z]\subset R[x,y,z,t]$. Using induction on $m$ we show that $b\in
k[x,y,z,t]$. For $m=0$ the assertion follows trivially. Suppose $m\geq 1$. Let
$x^{m}b=f(x,y,z,t)$ for some $f(x,y,z,t)\in R[x,y,z,t]$. Write

$x^{m}b=f(x,y,z,t)=xg(x,y,z,t)+\sum_{i=0}^{s}\alpha_{i}(y,z)t^{i}$ (6.11)

where $\alpha_{i}(y,z)\in R[y,z]$. Considering equation (6.11) in
$\overline{B}$ we get that

$\alpha_{s}(\overline{y},\overline{z})\overline{t}^{s}+\alpha_{s-1}(\overline{y},\overline{z})\overline{t}^{s-1}+\ldots+\alpha_{1}(\overline{y},\overline{z})\overline{t}+\alpha_{k}(\overline{y},\overline{z})=0$
(6.12)

where $\alpha_{i}(\overline{y},\overline{z})\in R[\overline{y},\overline{z}]$.
But $\overline{t}$ is algebraically independent over
$R[\overline{y},\overline{z}]$. Hence $\alpha_{i}(y,z)\in R[y,z]\cap
xB=(\varphi(z),\psi(y,z))R[y,z]$ for all $0\leq i\leq s$. Using above
equations (6.8) and (6.10), we can write
$\alpha_{i}(y,z)=\beta_{i}(y,z)xt+\gamma_{i}(y,z)xy$ where
$\beta_{i}(y,z),\gamma_{i}(y,z)\in R[y,z]$ for all $0\leq i\leq s$.
Substituting in equation (6.11), we get that

$x^{m-1}b=g(x,y,z,t)+\sum_{i=0}^{s}\left(\beta_{i}(y,z)t^{i+1}+\gamma_{i}(y,z)yt^{i}\right)\in
R[x,y,z,t].$ (6.13)

By induction hypothesis $b\in R[x,y,z,t]$. Hence $B=R[x,y,z,t]$ where the
generators satisfy the equations (6.8) and (6.10). Then the assertion follows
from Lemma 6.4. ∎

######  Theorem 6.6.

Let $R$ be factorial affine domain over a field $k$ of characteristic zero,
and $B$ be a factorial affine $R$-domain. Suppose there exists a non-trivial
locally nilpotent $R$-derivation $\delta$ on $B$ such that the following hold

  1. (1)

$B^{\delta}=R[x]$, where $x\in B$ is trancendental over $R$,

  2. (2)

there exists an element $z\in B\setminus xB$ such that $\delta(z)=x^{m}$ for
some $m\geq 2$, and

  3. (3)

there exists $y\in B\setminus xB$ such that $\delta(y)=x^{d}f(x,z)$ and
$R[\overline{y},\overline{z}]^{[1]}=B/xB$, where $0<d<m$, $f(x,z)\in
R[x,z]\setminus xR[x,z]$ is monic in $z$, and $\overline{z},\overline{y}$
denote the residue class of $z,y$ in $B/xB$.

If $\deg_{z}f(0,z)=[Q(R[\bar{z}]):K]-1$ and $R[\overline{z}]$ is factorial,
then

$B\cong_{R}R[X,Y,Z,T]/(X^{m-d}Y-F(X,Z),X^{d}T-G(X,Y,Z))$

where $F(0,Z)\in R[Z]\setminus R$ and $G(0,Y,0)\in R[Y]\setminus R$.

###### Proof.

Note that $x$ is prime in $B$, $B\subseteq R[x,x^{-1},z]$, and $x$ and $z$ are
algebraically independent over $R$. As befor, we only consider the case
$B\subsetneq R[x,z]$. As before, $\overline{B}:=B/xB$, $\overline{\delta}$
denotes the induced locally nilpotent derivation on $\overline{B}$, and
$\overline{b}$ denotes the residue class of $b\in B$ in $\overline{B}$.
Furthermore, we can assume that $\delta$ is an irreducible locally nilpotent
derivation (cf., [9, Proposition 2.2(c)]).

There exists $n\geq 0$ such that $x^{n}y=F(x,z)$ for some $F(x,z)\in
R[x,z]\setminus xR[x,z]$. Applying $\delta$, we have
$x^{n}\delta(y)=x^{n+d}f(x,z)=x^{m}(\partial_{z}F(x,z))$. If $m>n+d$, then
$f(x,z)\in xR[x,z]$, which is a contradiction. Now, suppose $m<n+d$, then
$\partial_{z}F(x,z)\in xR[x,z]$. Let $F(x,z)=\sum_{i=0}^{s}H_{i}(x)z^{i}$ for
some $s\in\mathbb{Z}_{\geq 0}$ and $H_{i}(x)\in R[x]$ for all $0\leq i\leq s$.
Since $R[x,z]\cong R^{[2]}$, $H_{i}(x)\in xR[x]$ for all $1\leq i\leq s$.
Since $F(x,z)=x^{n}y$, by Lemma 2.3(2) it follows that $H_{0}(x)\in xR[x]$.
Hence, $F(x,z)\in xR[x,z]$, which is a contradiction. Thus $n=m-d$, and
therefore

$x^{m-d}y=F(x,z)$ (6.14)

with $\partial_{z}F(x,z)=f(x,z)$. Note that the monicness of $f$ in $z$
implies that $F$ is monic in $z$. Also,

$R[z]\cap xB=\varphi(z)R[z]$ (6.15)

where $\varphi(z)$ is an irreducible polynomial of $R[z]$ with
$\deg_{z}\varphi(z)\geq 1$ (c.f. Lemma 4.5). Next, we show that
$F(0,z)=\alpha\varphi(z)$ for some $\alpha\in R^{\star}$. Since
$R[\overline{z}]=R[z]/\varphi(z)R[z]$, we have
$\deg_{z}f(0,z)=\deg_{z}\varphi(z)-1$. Therefore,
$\deg_{z}F(0,z)=\deg_{z}\varphi(z)$. Also, $F(0,z)\in R[z]\cap
xB=\varphi(z)R[z]$. Hence

$F(0,z)=\beta\varphi(z)$ (6.16)

for some $\beta\in R\setminus\\{0\\}$. By monicness of $F(0,z)$, we get that
$\beta\in R^{*}$. Thus

$\varphi(z)=x^{m-d}\beta^{-1}y+xH_{1}(x,z)$ (6.17)

for some $H_{1}(x,z)\in R[x,z]$. Since
$R[\overline{y},\overline{z}]^{[1]}=\overline{B}$ and
$R[\overline{y},\overline{z}]\subseteq\overline{B}^{\overline{\delta}}$, it
follows from Lemma 2.3(2) that
$R[\overline{y},\overline{z}]=\overline{B}^{\overline{\delta}}$. Also, $y$ and
$z$ are algebraically independent over $R$. Next, we consider the prime ideal
$\mathfrak{p}:=R[y,z]\cap xB$ of $R[y,z]$. Arguing as in the proof of Theorem
6.5, we have $\mathfrak{p}=(\varphi(z),\psi(y,z))R[y,z]$ where $\psi(y,z)\in
R[y,z]$ is monic in $y$ as a polymomial in $K[y,z]$. Furthermore,
$\psi(y,\overline{z})$ is an irreducible element of $R[\overline{z}][y]$ with
$\deg_{y}\psi(y,\overline{z})\geq 1$. Since $\psi(y,z)\in xB$, we can write

$\psi(y,z)=x^{t_{1}}v_{1}$ (6.18)

where $v_{1}\in B\setminus xB$ and $t_{1}\geq 1$. Applying $\delta$ we obtain
the following

$x^{t_{1}}\delta(v_{1})=x^{m}(\partial_{z}\psi(y,z))+x^{d}(\partial_{y}\psi(y,z))(\partial_{z}\varphi(z)).$
(6.19)

Similarly as in Theorem 6.5, if $t_{1}>d$ then we get a contradiction by
$\deg_{y}\psi(y,\overline{z})<\deg_{y}\psi(y,\overline{z})$. Hence $t_{1}\leq
d$. Suppose $t_{1}<d$, then
$\overline{v_{1}}\in\overline{B}^{\overline{\delta}}$. We can write
$\overline{v_{1}}=\beta_{1}(\overline{y},\overline{z})$ for some
$\beta_{1}(y,z)\in R[y,z]$ with $\beta_{1}(y,z)\notin xB$. Then
$v_{1}=\beta_{1}(y,z)+x^{t_{2}}v_{2}$ for some $v_{2}\notin xB$ and $t_{2}\geq
1$. From (6.18),

$\psi(y,z)=x^{t_{1}}\beta_{1}(y,z)+x^{t_{1}+t_{2}}v_{2}.$ (6.20)

Applying $\delta$ it follows that

$x^{m}(\partial_{z}\psi(y,z))+x^{d}f(x,z)(\partial_{y}\psi(y,z))\
=x^{m+t_{1}}(\partial_{z}\beta_{1}(y,z))+x^{d+t_{1}}f(x,z)(\partial_{y}\beta_{1}(y,z))+x^{t_{1}+t_{2}}\delta(v_{2}).$
(6.21)

Suppose $t_{1}+t_{2}>d$, then
$\partial_{y}\psi(y,\overline{z})\in\overline{\mathfrak{p}}=\psi(y,\overline{z})R[y,\overline{z}]$,
which is a contradiction as
$\deg_{y}(\partial_{y}\psi(y,\overline{z}))<\deg_{y}\psi(y,\overline{z})$.
Hence $t_{1}+t_{2}\leq d$. Suppose $t_{1}+t_{2}=d$, then

$\delta(v_{2})=x^{m-d}(\partial_{z}\psi(y,z))+f(x,z)(\partial_{y}\psi(y,z))\
-x^{m-t_{2}}(\partial_{z}\beta_{1}(y,z))-x^{t_{1}}f(x,z)(\partial_{y}\beta_{1}(y,z)).$
(6.22)

Since $f(x,z)(\partial_{y}\psi(y,z))\notin xB$, we get that
$\overline{v_{2}}\notin\overline{B}^{\overline{\delta}}$. Hence
$\overline{v_{2}}$ is algebraically independent over
$R[\overline{y},\overline{z}]$. For any $b\in B$ there exists
$n\in\mathbb{Z}_{\geq 0}$ such that $x^{n}b\in R[x,z]\subset R[x,y,z,v_{2}]$.
Arguing inductively we show that $b\in R[x,y,z,v_{2}]$. If $n=0$ then the
assertion holds trivially. Suppose $n\geq 1$. Let $x^{n}b=f(x,y,z,v_{2})$ for
some $f(x,y,z,v_{2})\in R[x,y,z,v_{2}]$. Write

$x^{n}b=f(x,y,z,v_{2})=xg(x,y,z,v_{2})+\sum_{i=0}^{s}\alpha_{i}(y,z)\,v_{2}^{i}$
(6.23)

for some $g(x,y,z,v_{2})\in R[x,y,z,v_{2}]$ and $\alpha_{i}(y,z)\in R[y,z]$
for all $0\leq i\leq s$. Considering (6.23) in $\overline{B}$, we have

$\alpha_{s}(\overline{y},\overline{z})\overline{v_{2}}^{s}+\alpha_{s-1}(\overline{y},\overline{z})\overline{v_{2}}^{s-1}+\ldots+\alpha_{1}(\overline{y},\overline{z})\overline{v_{2}}+\alpha_{0}(\overline{y},\overline{z})=0.$
(6.24)

Since $\overline{v_{2}}$ is algebraically independent over
$R[\overline{y},\overline{z}]$, it follows that
$\alpha_{i}(y,z)\in(\varphi(z),\psi(y,z))R[y,z]$. Let
$\alpha_{i}(y,z)=\varphi(z)\gamma_{i}(y,z)+\psi(y,z)\zeta_{i}(y,z)$ where
$\gamma_{i}(y,z),\,\zeta_{i}(y,z)\in R[y,z]$ for all $0\leq i\leq s$.
Substituting $\alpha_{i}(y,z)$ in (6.23) and then using (6.17) and (6.20), we
get that

$x^{n-1}b=g(x,y,z,v_{2})+\sum_{i=0}^{s}\left((x^{m-d-1}\beta^{-1}y+H_{1}(x,z))\gamma_{i}(y,z)\,v_{2}^{i}+x^{t_{1}-1}\beta_{1}(y,z)\zeta_{i}(y,z)v_{2}^{i}+x^{d-1}v_{2}^{i+1}\right)$
(6.25)

where $\beta\in R^{*}$. Hence $x^{n-1}b\in R[x,y,z,v_{2}]$. Thus by induction
hypothesis $b\in R[x,y,z,v_{2}]$. Then the assertion follows from Lemma 6.4.

Suppose $t_{1}+t_{2}<d$. Repeating the above process, we can find $v\in
B\setminus xB$ satisfying the following equation

$\psi(y,z)=x^{d_{1}}\beta_{1}(y,z)+x^{d_{2}}\beta_{2}(y,z)+\ldots+x^{d_{s}}\beta_{s}(y,z)+x^{d}v$
(6.26)

where $d>d_{s}>\ldots>d_{2}>d_{1}\geq 1$ and $\beta_{i}(y,z)\in R[y,z]$ such
that $\beta_{i}(y,z)\notin xB$ for all $1\leq i\leq s$. Applying $\delta$ to
(6.26), we have

$\delta(v)=x^{m-d}(\partial_{z}\psi(y,z))+f(x,z)(\partial_{y}\psi(y,z))-x^{d_{1}}\delta(\beta_{1}(y,z))-\ldots-x^{d_{s}}\delta(\beta_{s}(y,z))$
(6.27)

Since $f(x,z)(\partial_{y}\psi(y,z))\notin xB$, it follows that
$\overline{v}\notin\overline{B}^{\overline{\delta}}$. Hence
$\overline{\delta}$ is a non-trivial locally nilpotent derivation on
$\overline{B}$. From Lemma 2.3(2), it follows that $\overline{v}$ is
algebraically independent over $R[\overline{y},\overline{z}]$. Let
$G(x,y,z):=\psi(y,z)-x^{d_{1}}\beta_{1}(y,z)-x^{d_{2}}\beta_{2}(y,z)-\ldots-x^{d_{s}}\beta_{s}(y,z)$.
Then arguing inductively as before we can show that $B=R[x,y,z,v]$. The
$R$-algebra generators of $B$ satisfy the relations (6.14) and (6.26). Note
that $F(0,z)=\beta\varphi(z)$ and $G(0,y,z)=\psi(y,z)$. By Lemma 6.4 the
assertion follows. ∎

######  Remark 6.7.

(i) Note that, in Theorem 6.5 and 6.6, if $R=k$ then the condition that
$k[\overline{z}]$ is factorial is redundant.
(ii) The polynomial $F(X,Z)$ of Theorem 6.6 is monic in $Z$, and $G(X,Y,Z)\in
R[X,Y,Z]$ is such that $G(0,Y,Z)$ is monic in $Y$ as a polynomial in $K[Y,Z]$.

## Acknowledgment

The second author gratefully acknowledges the support provided by the NBHM
(National Board of Higher Mathematics) postdoctoral fellowship (Number:
0204/16(9)/2020/R&D-II/10922) from the Department of Atomic Energy (DAE),
Government of India.

## References

  * [1] S. S. Abhyankar, P. Eakin and W. Heinzer, _On the uniqueness of the coefficient ring in a polynomial ring_ , J. Algebra 23 (1972), 310–342.
  * [2] A.J. Crachiola, _T_ he hypersurface $x+x^{2}y+z^{2}+t^{3}=0$ over a field of arbitrary characteristic, Proc. Amer. Math. Soc., 134 (2005), 1289–1298.
  * [3] A. J. Crachiola, On automorphisms of Danielewski surfaces J. Algebraic Geom. 15(1) (2006), no. 1, 111–132.
  * [4] D. Daigle, _Locally nilpotent derivations and Danielewski surfaces_ , Osaka J. Math, 41 (2004), 37–80.
  * [5] W. Danielewski, _On a cancellation problem and automorphism group of affine algebraic varieties_ , preprint, 1989.
  * [6] H. Derksen, O. Hadas and L. Makar-Limanov, _N_ ewton polytopes of invariants of additive group actions, J. Pure Appl. Algebra 156 (2001), 187–197.
  * [7] A. Dubouloz, _Additive group actions on Danielewski varieties and the cancellation problem_ , Math Z., 255 (1) (2007), 77–93.
  * [8] P. Eakin and W. Heinzer, _A cancellation problem for rings_ , In: Conference on Commutative Algebra (Brewer, J.W.; Rutter, E.A., eds.). Lecture Notes in Mathematics vol. 311, Springer Verlag, berlin-Heidelberg-New York, 1973, pp. 61–77.
  * [9] G. Freudenburg, _Algebraic theory of locally nilpotent derivations,in: Encyclopaedia of Mathematical Sciences_ , vol. 136, Springer-Verlag, Berlin Heidelberg, 2006.
  * [10] E. Hamann, _On the $R$-invariance of $R[x]$_, J. Algebra, 35 (1975), 1–16.
  * [11] K. Masuda, _Characterizations of hypersurfaces of a Danielewski type_ , J. Pure Appl. Algebra, 218 (2014), 624–633.
  * [12] K. Masuda, _Families of hypersurfaces with noncancellation property_ , Proc. the American Math. Soc., 145, 4 (2017), 1439–1452.
  * [13] P. Ghosh, N. Gupta, _On generalised Danielewski and Asanuma varieties_ , Journal of Algebra, 632 (2023), 226–250.
  * [14] N. Gupta and S. Sen, _On double Danielewski surfaces and the Cancellation Problem_ , Journal of Algebra, 533 (2019), 25–43.
  * [15] N. Gupta and S. Sen, _Locally nilpotent derivations of double Danielewski surfaces_ , J. Pure Appl. Algebra, 224 (2020), 106208.