Nice derivations over principal ideal domains

TL;DR

This paper explores the extension of results on nice derivations from polynomial rings over fields to those over principal ideal domains, revealing that certain kernels are polynomial rings over the base field.

Contribution

It extends the theory of nice derivations to polynomial rings over PIDs, showing kernels are polynomial rings under specific conditions.

Findings

Kernel of a nice derivation in four variables over a field is a polynomial ring.

Results generalize previous work from fields to principal ideal domains.

Provides conditions under which derivation kernels are polynomial rings.

Abstract

In this paper we investigate to what extent the results of Z. Wang and D. Daigle on nice derivations of the polynomial ring in three variables over a field k of characteristic zero extend to the polynomial ring over a PID R, containing the field of rational numbers. One of our results shows that the kernel of a nice derivation on the polynomial ring in four variables over k of rank at most three is a polynomial ring over k.

Equations

$${\pi }_r(f)=\sum _{i⩾0}\frac{(-1{)}^i}{i!}{D}^{i}f\frac{r^i}{(Dr{)}^i}.$$

$$D=f(X^{{}^{\prime }})\frac{\partial }{\partial Y^{{}^{\prime }}}+g(X^{{}^{\prime }})\frac{\partial }{\partial Z^{{}^{\prime }}}$$

$$DX=a,DY=b-1\text{and}DZ=aY+(1-b)X.$$

$$DX=0,DY=X-t\text{and}DZ=X+t.$$

$$\overline{D}=p(U)\frac{\partial }{\partial V^{{}^{\prime }}}+q(U)\frac{\partial }{\partial W^{{}^{\prime }}}.$$

$$DX=b,DY=-a\text{and}DZ=aX+bY.$$

$$DX=0,DY=X,DZ=Y.$$

$$DX=t,DY=tZ+X^2\text{and}DZ=-2X.$$

$$B=k[X,Y,Z],DX=0,DY\in k[X]\text{and}DZ\in k[X,Y].$$

$$DX=t,DY=X\text{and}DZ=Y.$$

$$\frac{\partial F}{\partial X}=-2X,\frac{\partial G}{\partial X}=-3tY+3X^2\text{and}\frac{\partial H}{\partial X}=-18tYZ-6X{Y}^2+18X^{2}Z.$$

$$\frac{\partial F}{\partial Y}=-2t,\frac{\partial G}{\partial Y}=-3tX,\text{and}\frac{\partial H}{\partial Y}=16tY-18tXZ-6X^{2}Y,$$

Full text

Keywords. Polynomial Rings, Locally Nilpotent Derivation, Nice Derivation.

**2010 MSC. Primary: 13N15; Secondary: 14R20, 13A50. **

1 Introduction

By a ring, we will mean a commutative ring with unity. Let $R$ be a ring and $n(\geqslant 1)$ be an integer. For an $R$-algebra $A$, we use the notation $A=R^{[n]}$ to denote that $A$ is isomorphic to a polynomial ring in $n$ variables over $R$. We denote the group of units of $R$ by $R^{*}$.

Let $k$ be a field of characteristic zero, $R$ a $k$-domain, $B:=R^{[n]}$ and $m$ is a positive integer $\leq n$. In this paper, we consider locally nilpotent derivations $D$ on $B$, which satisfy $D^{2}(T_{i})=0$ for all $i\in\{1,\dots,m\}\subseteq\{1,\dots,n\}$ for some coordinate system $(T_{1},T_{2},\dots,T_{n})$ of $B$. For convenience, we shall call such a derivation $D$ as a quasi-nice derivation. In the case $m=n$, such a $D$ is called a nice derivation (Thus a nice derivation is also a quasi-nice derivation). We investigate the rank of $D$ when $n=3$ and $R$ is a PID (see Section $2$ for the definition of rank of $D$).

The case when $B=k^{[3]}$ was investigated by Z. Wang in [14]. He showed that $rank$ $D$ is less than 3 for the cases $(m,n)=(2,3),(3,3)$ and that $rank$ $D$ $=1$ when $D$ is a nice derivation (i.e., for $(m,n)=(3,3)$). In [6], Daigle proved that the rank of $D$ is less than $3$ even in the case $(m,n)=(1,3)$ ([6, Theorems 5.1 and 5.2]).

Now let $R$ be a Noetherian domain containing $\mathbb{Q}$, say $R$ is regular. It is natural to ask how far we can extend the above results to $R[X,Y,Z](=R^{[3]})$. In particular, we consider the following question for nice derivations.

Question 1. If $D$ is a nice derivation of $R[X,Y,Z]$, then is rank $D$ $=1$, or, at least, is rank $D$ $<3$?

In Section $3$, we show that when $R$ is a PID, the rank of $D$ is indeed less than 3 (Theorem 3.6) and construct a nice derivation $D$ over $k^{[1]}$ with $rank$ $D=2$ (Example 3.5). Moreover, we construct a nice derivation $D$ over $k^{[2]}$ with $rank$ $D=3$ (Example 3.10) showing that Theorem 3.6 does not extend to two-dimensional regular or factorial domains.

An important open problem in Affine Algebraic Geometry asks whether the kernel of any $D\in LND_{k}(k^{[4]})$ is necessarily finitely generated. In the case when $rank~{}D\leq 3$, Bhatwadekar and Daigle had shown that the kernel is indeed finitely generated [3, Theorem 1]. However Daigle and Freudenburg had constructed an example to show that the kernel need not be $k^{[3]}$ [8]. Under the additional hypothesis that the kernel is regular, Bhatwadekar, Gupta and Lokhande showed that the kernel is indeed $k^{[3]}$ [5, Theorem 3.5]. A consequence of Theorem 3.6 of our paper is that in the case $rank~{}D\leq 3$, the kernel of any nice derivation $D$ is necessarily $k^{[3]}$ (Corollary 3.8).

The following question on quasi-nice derivations arises in view of Wang’s result that $rank$ $D$ is less than $3$ for $(m,n)=(2,3)$.

Question 2. *If $D$ is a locally nilpotent derivation of $R[X,Y,Z]$, such that $D$ is irreducible and $D^{2}X=D^{2}Y=0$, is then $rank$ $D$ $<3?$ *

In Section $4$, we investigate this question and obtain some partial results when $R$ is a PID (Proposition 4.4) and a Dedekind domain (Proposition 4.6). Example 4.5 shows that Question 2 has a negative answer in general, even when $R$ is a PID. We shall also construct a strictly $1$-quasi derivation (defined in Section $4$) on $R^{[3]}$ over a PID $R$ (Example 4.9). By a result of Daigle (quoted in Section $4$ as Theorem 4.8), there does not exist such a derivation on $k^{[3]}$, where $k$ is a field of characteristic zero.

2 Preliminaries

For a ring $A$ and a nonzerodivisor $f\in A$, we use the notation $A_{f}$ to denote the localisation of $A$ with respect to the multiplicatively closed set $\{1,f,f^{2},\dots\}$.

Let $A\subseteq B$ be integral domains. Then the transcendence degree of the field of fractions of $B$ over that of $A$ is denoted by $\operatorname{tr.deg}_{A}{B}$.

A subring $A\subseteq B$ is defined to be factorially closed in $B$ if, given nonzero $f,g\in B$, the condition $fg\in A$ implies $f\in A$ and $g\in A$. When the ambient ring $B$ is understood, we will simply say that $A$ is factorially closed. A routine verification shows that a factorially closed subring of a UFD is a UFD. If $A$ is a factorially closed subring of $B$, then $A$ is algebraically closed in $B$; further if $S$ is a multiplicatively closed set in $A$ then $S^{-1}A$ is a factorially closed subring of $S^{-1}B$.

Let $k$ be a field of characteristic zero, $R$ a $k$-domain, and $B$ an $R$-domain. The set of locally nilpotent $R$-derivations of $B$ is denoted by $LND_{R}(B)$. When $R$ is understood from the context (e.g. when $R=k$), we simply denote it by $LND(B)$. We denote the kernel of a locally nilpotent derivation $D$ by $Ker$ $D$.

Let $D\in LND_{R}(B)$ and $A:=Ker$ $D$. It is well-known that $A$ is a factorially closed subring of $B$ [7, 1.1(1)]. For any multiplicatively closed subset $S$ of $A\setminus\{0\}$, $D$ extends to a locally nilpotent derivation of $S^{-1}B$ with kernel $S^{-1}A$ and $B\cap S^{-1}A=A$ [7, 1.1(2)]. Moreover if $D$ is non-zero, then $\operatorname{tr.deg}_{A}B=1$ [7, 1.1(4)]. A locally nilpotent derivation $D$ is said to be reducible if there exists a non-unit $b\in B$ such that $DB\subseteq(b)B$; otherwise $D$ is said to be irreducible. An element $s\in B$ is called a slice if $Ds=1$, and a local slice if $Ds\in A$ and $Ds\neq 0$. Moreover $D$ is said to be fixed-point free if the $B$ ideal $(DB)=B$.

When $B:=R^{[n]}$ and $D\in LND_{R}(B)$, the rank of $D$, denoted by $rank$ $D$, is defined to be the least integer $i$ for which there exists a coordinate system $(X_{1},X_{2},\dots,X_{n})$ of $B$ satisfying $R[X_{i+1},\dots,X_{n}]\subseteq A$.

Now let $B$ be a $k$-domain and $D$ an element of $LND(B)$ with a local slice $r\in B$. The Dixmier map induced by $r$ is defined to be the $k$-algebra homomorphism $\pi_{r}:B\rightarrow B_{Dr}$, given by

[TABLE]

The following important result is known as the Slice Theorem [10, Corollary 1.22].

Theorem 2.1**.**

Let $k$ be a field of characteristic zero and $B$ a $k$-domain. Suppose $D\in LND(B)$ admits a slice $s\in B$, and let $A=Ker$ $D$. Then

1. (a)

$B=A[s]$* and $D=\frac{\partial}{\partial s}$.* 2. (b)

$A=\pi_{s}(B)$* and $Ker$ $\pi_{s}=sB$.* 3. (c)

If $B$ is affine, then $A$ is affine.

The following theorem of Daigle and Freudenburg characterizes locally nilpotent derivations of $R^{[2]}$, where $R$ is a UFD containing $\mathbb{Q}$ [7, Theorem 2.4].

Theorem 2.2**.**

Let $R$ be a UFD containing $\mathbb{Q}$ with field of fractions $K$ and let $B=R[X,Y]=R^{[2]}$. For an $R$-derivation $D\neq 0$ of $B$, the following are equivalent:

1. (i)

$D$* is locally nilpotent.* 2. (ii)

$D=\alpha(\frac{\partial F}{\partial Y}\frac{\partial}{\partial X}-\frac{\partial F}{\partial X}\frac{\partial}{\partial Y})$, for some $F\in B$ which is a variable of $K[X,Y]$ satisfying

$gcd_{B}(\frac{\partial F}{\partial X},\frac{\partial F}{\partial Y})=1$, and for some $\alpha\in R[F]\setminus\{0\}$.

Moreover, if the above conditions are satisfied, then $Ker$ $D=R[F]=R^{[1]}$.

With the same notation as above, the following lemma gives interesting results when $D$ satisfies some additional hypothesis [14, Lemma 4.2].

Lemma 2.3**.**

Let $R$ be a UFD containing $\mathbb{Q}$, $B=R[X,Y](=R^{[2]})$ and $D\in LND_{R}(B)$ such that $D$ is irreducible. Then the following hold:

1. (i)

If $D^{2}X=0$, then $Ker$ $D=R[bY+f(X)]$, where $b\in R$ and $f(X)\in R[X]$. Moreover, $DX\in R$ and $DY\in R[X]$. 2. (ii)

*If $D^{2}X=D^{2}Y=0$, then $D=b\frac{\partial}{\partial X}-a\frac{\partial}{\partial Y}$ for some $a,b\in R$. Moreover, *

$Ker$* $D=R[aX+bY]$.* 3. (iii)

If $R$ is a PID and $D^{2}X=D^{2}Y=0$, then $D$ has a slice.

Over a Noetherian domain containing $\mathbb{Q}$, a necessary and sufficient condition for the kernel of a nonzero irreducible $D\in LND_{R}(R[X,Y])$ to be a polynomial ring is given by [4, Theorem 4.7].

Theorem 2.4**.**

Let $R$ be a Noetherian domain containing $\mathbb{Q}$ and let $D$ be a non-zero irreducible locally nilpotent derivation of the polynomial ring $R[X,Y]$. Then the kernel $A$ of $D$ is a polynomial ring in one variable over $R$ if and only if $DX$ and $DY$ either form a regular $R[X,Y]$-sequence or are comaximal in $R[X,Y]$. Moreover if $DX$ and $DY$ are comaximal in $R[X,Y]$, then $R[X,Y]$ is a polynomial ring in one variable over $A$.

An important result on fixed-point free locally nilpotent derivations is the following [10, Theorem 4.16].

Theorem 2.5**.**

Let $R$ be any $\mathbb{Q}$-algebra, and let $B=R[X,Y]=R^{[2]}$. Given $D\in LND_{R}(R[X,Y])$, the following conditions are equivalent:

1. (1)

$D$* is fixed-point free, i.e., $(DB)=B$, where $(DB)$ is the $B$-ideal generated by $DB$.* 2. (2)

There exists $s\in B$ with $Ds=1$.

In addition, when these conditions hold, $Ker$ $D=R^{[1]}$.

For a ring containing $\mathbb{Q}$, the following cancellation theorem was proved by Hamann [11, Theorem 2.8].

Theorem 2.6**.**

Let $R$ be a ring containing $\mathbb{Q}$ and $A$ be an $R$-algebra such that $A^{[1]}=R^{[2]}$. Then $A=R^{[1]}$.

The following is a well-known result of Abhyankar, Eakin and Heinzer [1, Proposition 4.8].

Theorem 2.7**.**

Let $C$ be a UFD and let $X_{1},\dots,X_{n}$ be indeterminates over $C$. Suppose that $A$ is an integral domain of transcendence degree one over $C$ and that $C\subseteq A\subseteq C[X_{1},\dots,X_{n}]$. If $A$ is a factorially closed subring of $C[X_{1},\dots,X_{n}]$, then $A=C^{[1]}$.

The following local-global theorem was proved by Bass, Connell and Wright [2] and independently by Suslin [13].

Theorem 2.8**.**

Let $R$ be a ring and $A$ a finitely presented $R$-algebra. Suppose that for all maximal ideals $\mathpzc{m}$ of $R$, the $R_{\mathpzc{m}}$-algebra $A_{\mathpzc{m}}$ is isomorphic to the symmetric algebra of some $R_{\mathpzc{m}}$-module. Then $A\cong Sym_{R}{(L)}$ for some finitely presented $R$-module $L$.

The following result is known as Serre’s Splitting Theorem [12, Theorem 7.1.8].

Theorem 2.9**.**

*Let $A$ be a Noetherian ring of finite Krull dimension. Let $P$ be a finitely generated projective $A$-module of rank greater than dimension of $A$. Then $P$ has a unimodular element. *

Following is the famous Cancellation Theorem of Hyman Bass [12, Theorem 7.1.11].

Theorem 2.10**.**

Let $R$ be a Noetherian ring of dimension $d$ and $P$ a finitely generated projective $R$-module of rank $>d$. Then $P$ is “cancellative”, i.e., $P\oplus Q\cong P^{{}^{\prime}}\oplus Q$ for some finitely generated projective $R$-module $Q$ implies that $P\cong P^{{}^{\prime}}$.

We now state a local-global result for a graded ring [12, Theorem 4.3.11].

Theorem 2.11**.**

Let $S=S_{0}\oplus S_{1}\oplus S_{2}\dots$ be a graded ring and let $M$ be a finitely presented $S$-module. Assume that for every maximal ideal $\mathpzc{m}$ of $S_{0}$, $M_{\mathpzc{m}}$ is extended from $(S_{0})_{\mathpzc{m}}$. Then $M$ is extended from $S_{0}$.

For convenience, we state below an elementary result.

Lemma 2.12**.**

Let $A$ and $B$ be integral domains with $A\subseteq B$. If there exists $f$ in $A$, such that $A_{f}=B_{f}$ and $fB\cap A=fA$, then $A=B$.

Proof.

Let $b\in B$. Suppose, if possible $b\notin A$. Now since $B_{f}=A_{f}$, we have $b\in A_{f}$. Hence there exist $a\in A$ and an integer $n>0$ such that $b=a/f^{n}$. We may assume that $n$ is the least possible. But then $a\in fB\cap A=fA$. Let $a=fa_{1}$ for some $a_{1}\in A$. Then $b=a_{1}/f^{n-1}$, contradicting the minimality of $n$. ∎

3 Nice Derivations

In this section, we shall explore generalisations of the following theorem of Z. Wang [14, Proposition 4.6].

Theorem 3.1**.**

Let $K$ be a field of characteristic zero and $K[X,Y,Z]=K^{[3]}$. Suppose that $D(\neq 0)\in LND_{K}(K[X,Y,Z])$ satisfies $D^{2}X=D^{2}Y=D^{2}Z=0$. Then the following hold:

1. (i)

$Ker$* $D$ contains a nonzero linear form of {$X,Y,Z$}.* 2. (ii)

$rank$* $D=1$.* 3. (iii)

If $D$ is irreducible, then for some coordinate system ($X^{{}^{\prime}},Y^{{}^{\prime}},Z^{{}^{\prime}}$) of $K[X,Y,Z]$ related to ($X,Y,Z$) by a linear change,

[TABLE]

where $f$, $g\in K[X^{{}^{\prime}}]$ and ${\rm gcd}_{K[X^{{}^{\prime}}]}(f,g)=1$.

We first observe the following result.

Lemma 3.2**.**

Let $R$ be a UFD containing $\mathbb{Q}$ and $D$($\neq 0$) $\in LND_{R}(R[X,Y,Z])$, where $R[X,Y,Z]=R^{[3]}$ and $rank$ $D$ $<3$. Then $Ker$ $D=R^{[2]}$.

Proof.

Let $A:=Ker$ $D$. Since $rank$ $D$ $<3$, there exists $X^{{}^{\prime}}\in R[X,Y,Z]$ such that $R[X,Y,Z]=R[X^{{}^{\prime}}]^{[2]}$ and $X^{{}^{\prime}}\in A$. Then taking $C=R[X^{{}^{\prime}}]$, it follows from Theorem 2.7 that $A(=Ker$ $D)=R[X^{{}^{\prime}}]^{[1]}=R^{[2]}$. ∎

The following example shows that Lemma 3.2 does not extend to a Noetherian normal domain $R$ which is not a UFD.

Example 3.3**.**

Let $\mathbb{R}[a,b]=\mathbb{R}^{[2]}$ and $R:=\frac{\mathbb{R}[a,b]}{(a^{2}+b^{2}-1)}$. Let $B:=R[X,Y,Z]=R^{[3]}$ and $D$ be an $R$-linear $LND$ of $B$, such that

[TABLE]

Setting $u=aY+(1-b)X$, $v=(1+b)Y+aX$ and $w=2Z+uY-vX$, we see that $Du=Dv=Dw=0$ and $D^{2}X=D^{2}Y=D^{2}Z=0$.

Let $A:=Ker$ $D$. Now $B_{(1+b)}=R_{(1+b)}[v,w,X]$ and $B_{(1-b)}=R_{(1-b)}[u,w,Y]$. Thus it follows that $A_{(1+b)}=R_{(1+b)}[v,w]={R_{(1+b)}}^{[2]}$ and $A_{(1-b)}=R_{(1-b)}[u,w]={R_{(1-b)}}^{[2]}$. Since $(1+b)$ and $(1-b)$ are comaximal elements of $R$, $A=R[u,v,w]$ and $A_{\mathpzc{m}}={R_{\mathpzc{m}}}^{[2]}$ for every maximal ideal $\mathpzc{m}$ of $R$.

Now $B=R[X,Y,Z]=R[X,Y,w]$ and $w\in A$; so $rank$ $D$ $<3$. Setting $T=\frac{u}{a}$, we see that $A=R[aT,(1+b)T,w]$. By Theorems 2.8 and 2.9, $A=Sym_{R}(F\oplus P)$, where $F$ is a free $R$-module of rank $1$ and $P$ is a rank $1$ projective $R$-module given by the ideal $(a,1+b)R$, which is not principal. Hence $P$ is not stably free and so $A\neq R^{[2]}$

[9, Lemma 1.3]. **

Remark 3.4**.**

In Proposition 3.9, we will see that over any Dedekind domain $R$, the kernel of a nice derivation of $R^{[3]}$ is generated by (at most) three elements. **

The following example shows that Part (ii) of Theorem 3.1 does not hold when $K$ is replaced by a PID $R$.

Example 3.5**.**

Let $k$ be a field of characteristic zero, $R=k{[t]}=k^{[1]}$ and $B:=R[X,Y,Z](=R^{[3]})$. Let $D\in LND_{R}(B)$ be such that

[TABLE]

Let $A=Ker$ $D$ and $G:=(X-t)Z-(X+t)Y$. We will show that

1. (i)

$A=R[X,G]$. 2. (ii)

$B\neq A^{[1]}$; in fact, $B$ is not $A$-flat. 3. (iii)

$rank$ $D=2$.

Proof.

(i) Let $C:=R[X,G]$. We show that $C=A$. Clearly $C\subseteq A$. Set $f:=X-t$. Then $B_{f}={R[X,G,Y]}_{f}={C_{f}}^{[1]}$. Hence, as both $C_{f}(\subseteq A_{f})$ and $A_{f}$ are factorially closed subrings of $B_{f}$ and as $\operatorname{tr.deg}_{C_{f}}{B_{f}}=1=\operatorname{tr.deg}_{A}{B}$, we have $C_{f}=A_{f}$.

Now $B/fB$ may be identified with $R[Y,Z](=R^{[2]})$. Clearly $C/fC=R^{[1]}$ and the image of $C/fC$ in $B/fB$ is $R[tY](=R^{[1]})$. Thus the natural map $C/fC\rightarrow B/fB$ is injective, i.e, $fB\cap C=fC$. Since $A$ is factorially closed in $B$, we also have $fB\cap A=fA$ and hence $fA\cap C=fB\cap A\cap C=fB\cap C=fC$. Therefore as $C_{f}=A_{f}$, we have $C=A$ by Lemma 2.12.

(ii) $(X-t,X+t)B$ is a prime ideal of height $2$ in $B$ and $(X-t,X+t)B\cap A=(X,t,G)A$ is a prime ideal of height $3$ in $A$, violating the going-down principle. Hence $B$ is not $A$-flat and therefore $B\neq A^{[1]}$.

(iii) Since $DX=0$, $rank$ $D<3$. If $rank$ $D=1$, then clearly $B=A^{[1]}$ contradicting (ii). Hence $rank$ $D=2$. ∎

We now prove an extension of Theorem 3.1 over a PID.

Theorem 3.6**.**

*Let $R$ be a PID containing $\mathbb{Q}$ with field of fractions $L$ and

$B:=R[X,Y,Z]=R^{[3]}$. Let $D(\neq 0)\in LND_{R}(B)$, and $A:=Ker$ $D$. Suppose that $D$ is irreducible and $D^{2}X=D^{2}Y=D^{2}Z=0$. Then there exists a coordinate system $(U,V,W)$ of $B$ related to $(X,Y,Z)$ by a linear change such that the following hold:*

1. (i)

$A$* contains a nonzero linear form of $\{X,Y,Z\}$.* 2. (ii)

$rank$* $D\leq 2$. In particular, $A=R^{[2]}$.* 3. (iii)

$A=R[U,gV-fW]$, where $DV=f$, $DW=g$, and $f,g\in R[U]$ such that $gcd_{R[U]}(f,g)=1$. 4. (iv)

Either $f$ and $g$ are comaximal in $B$ or they form a regular sequence in $B$. Moreover if they are comaximal, (i.e., $D$ is fixed-point free) then $B=A^{[1]}$ and $rank$ $D=1$; and if they form a regular sequence, then $B$ is not $A$-flat and $rank$ $D=2$.

Proof.

(i) $D$ extends to an $LND$ of $L[X,Y,Z]$ which we denote by $\overline{D}$. By Theorem 3.1 there exists a coordinate system $(U,V^{{}^{\prime}},W^{{}^{\prime}})$ of $L[X,Y,Z]$ related to $(X,Y,Z)$ by a linear change and mutually coprime polynomials $p(U)$, $q(U)$ in $L[U]$ for which

[TABLE]

Multiplying by a suitable nonzero element of $R$, we can assume $U\in R[X,Y,Z]$. Clearly $A=Ker$ $\overline{D}\cap R[X,Y,Z]$ and $U\in A$. Moreover without loss of generality we can assume that there exist $l,m,n\in R$ with $gcd_{R}(l,m,n)=1$ such that $U=lX+mY+nZ$. As $R$ is a PID, $(l,m,n)$ is a unimodular row of $R^{3}$ and hence can be completed to an invertible matrix $M\in GL_{3}(R)$. Let $\begin{pmatrix}U\\ V\\ W\end{pmatrix}=M\begin{pmatrix}X\\ Y\\ Z\end{pmatrix}$.

Then $R[U,V,W]=R[X,Y,Z]$ and as $U\in A$, $A$ contains a nonzero linear form in ${X,Y,Z}$.

(ii) Follows from (i) and Lemma 3.2.

(iii) $R[U]$ is a UFD and $B=R[U,V,W]={R[U]}^{[2]}$. So $D$ is a locally nilpotent $R[U]$-derivation of $B$. Now the proof follows from Part (ii) of Lemma 2.3.

(iv) Since $B=R[U,V,W]={R[U]}^{[2]}$, the first part follows from Theorem 2.4. Moreover when $f$ and $g$ are comaximal in $B$, it also follows from Theorem 2.4 that $B=A^{[1]}$. Hence in this case $rank$ $D=1$.

If $f$ and $g$ form a regular sequence in $B$ (and hence in $A$ since $A$ is factorially closed in $B$), $(f,g)B\cap A=(f,g,gV-fW)A$. But $(f,g,gV-fW)A$ is an ideal of height $3$, while $(f,g)B$ is an ideal of height $2$, violating the going-down principle. It follows that in this case $B$ is not $A$-flat. In this case indeed $rank$ $D=2$, or else if $rank$ $D=1$, we would have $B=A^{[1]}$. ∎

The proof of Theorem 3.6 shows the following:

Corollary 3.7**.**

With the notation as above, the following are equivalent:

1. (i)

$B=A^{[1]}$. 2. (ii)

$rank$* $D=1$.* 3. (iii)

$B$* is $A$-flat.*

Proof.

(i)$\Leftrightarrow$(ii) and (ii)$\Rightarrow$(iii) are trivial. (iii)$\Rightarrow$(i) follows from Theorem 3.6(iv). ∎

As mentioned in the Introduction, Theorem 3.6 shows that the kernel of an irreducible nice derivation of $k^{[4]}$ of $rank\leq 3$ is $k^{[3]}$. More precisely, we have:

Corollary 3.8**.**

Let $K$ be a field of characteristic zero and let $K[X_{1},X_{2},X_{3},X_{4}]=K^{[4]}$. Let $D\in LND_{K}(K[X_{1},X_{2},X_{3},X_{4}])$, be such that $D$ is irreducible and $DX_{1}=0$ and $D^{2}X_{i}=0$ for $i=2,3,4$. Then $Ker$ $D=K^{[3]}$.

By a result of Bhatwadekar and Daigle [3, Proposition 4.13], we know that over a Dedekind domain $R$ containing $\mathbb{Q}$, the kernel of any locally nilpotent $R$-derivation of $R^{[3]}$ is necessarily finitely generated. We now show that if $D$ is a nice derivation, then the kernel is generated by at most three elements.

Proposition 3.9**.**

Let $R$ be a Dedekind domain containing $\mathbb{Q}$ and $B:=R[X,Y,Z]=R^{[3]}$. Let $D\in LND_{R}(B)$ such that $D$ is irreducible and $D^{2}X=D^{2}Y=D^{2}Z=0$. Let $A:=Ker$ $D$. Then the following hold:

1. (i)

$A$* is generated by at most 3 elements.* 2. (ii)

Moreover, if $D$ is fixed-point free, then $rank$ $D$ $<3$ and $D$ has a slice. In particular, $rank$ $D=1$.

Proof.

(i) By Theorem 3.6, $A_{\mathpzc{p}}={R_{\mathpzc{p}}}^{[2]}$ for all $\mathpzc{p}\in Spec(R)$. Hence by Theorem 2.8, $A\cong Sym_{R}(Q)$ for some rank $2$ projective $R$-module $Q$. Since $R$ is a Dedekind domain by Theorem 2.9, $Q\cong Q_{1}\oplus M$ where $Q_{1}$ is a rank $1$ projective $R$-module and $M$ is a free $R$-module of rank $1$. Again since $R$ is a Dedekind domain $Q_{1}$ is generated by at most 2 elements. Hence $A$ is generated by at most 3 elements.

(ii) Now assume $D$ is fixed-point free. Let $DX=f_{1}$, $DY=f_{2}$ and $DZ=f_{3}$. Then, by Theorem 2.1, $B_{f_{i}}={A_{f_{i}}}^{[1]}$ for each $i\in\{1,2,3\}$. Since $(f_{1},f_{2},f_{3})B=B$ we have $B_{\tilde{\mathpzc{p}}}={A_{\tilde{\mathpzc{p}}}}^{[1]}$, for each $\tilde{\mathpzc{p}}\in Spec(A)$. Hence, by Theorem 2.8, $B=Sym_{A}(P)$, where $P$ is a projective $A$-module of rank $1$. Now for each $\mathpzc{p}\in Spec(R)$, $P_{\mathpzc{p}}$ is an $A_{\mathpzc{p}}$-module and as $A_{\mathpzc{p}}={R_{\mathpzc{p}}}^{[2]}$, we have $P_{\mathpzc{p}}$ is a free $A_{\mathpzc{p}}$-module since $R_{\mathpzc{p}}$ is a discrete valuation ring and hence extended from $R_{\mathpzc{p}}$. Therefore, by Theorem 2.11, $P$ is extended from $R$. Let $P=P_{1}\otimes_{R}A$, where $P_{1}$ is a projective $R$-module of rank $1$. Hence $B=Sym_{A}(P)=Sym_{R}(M\oplus Q_{1}\oplus P_{1})$, where $M$ is a free $R$-module of rank $1$. Since $B=R^{[3]}$, $M\oplus Q_{1}\oplus P_{1}$ is a free $R$-module of rank $3$ [9, Lemma 1.3]. By Theorem 2.10, $Q_{1}\oplus P_{1}$ is free of rank $2$. Let $M=Rf$ and set $S:=R[f]$. Then $B={R[f]}^{[2]}$ and as $f\in A$, we have $rank$ $D$ $<3$. Now $B=S^{[2]}$ and $D\in LND_{S}(B)$ such that $D$ is fixed-point free. Hence, by Theorem 2.5, $D$ has a slice.

Let $B=R[f,g,h](=R^{[3]})$ and $s\in B$ be such that $Ds=1$. Then by Theorem 2.1, $B(=S^{[2]})=A[s](=A^{[1]})$. Hence by Theorem 2.6, $A=S^{[1]}$. Let $A=R[f,t]$. Then $B=R[f,g,h]=R[f,t,s]$ and $f,t\in A$. So $rank$ $D=1$. ∎

The following example shows that Theorem 3.6 does not extend to a higher-dimensional regular UFD, not even to $k^{[2]}$.

Example 3.10**.**

Let $k$ be a field of characteristic zero and $R=k[a,b]=k^{[2]}$. Let $B=R[X,Y,Z](=R^{[3]})$ and $D$($\neq 0$)$\in LND_{R}(B)$ be such that

[TABLE]

Let $u=aX+bY$, $v=bZ-uX$, and $w=aZ+uY$. Then $Du=Dv=Dw=0$, $D$ is irreducible and $D^{2}X=D^{2}Y=D^{2}Z=0$. Let $A=Ker$ $D$. We show that

1. (i)

$A=R[u,v,w]$. 2. (ii)

$A=R[U,V,W]/(bW-aV-U^{2})$, where $R[U,V,W]=R^{[3]}$ and hence $A\neq R^{[2]}$. 3. (iii)

$rank$ $D=3$.

Proof.

(i) Let $C:=R[u,v,w]$. We show that $C=A$. Clearly $C\subseteq A$. Note that, $B_{a}={C_{a}}^{[1]}$, so $C_{a}$ is algebraically closed in $B_{a}$. But $A$ is algebraically closed in $B$. So $A_{a}=C_{a}$. Similarly $A_{b}=C_{b}$. Since $a,b$ is a regular sequence in $C$, $C_{a}\cap C_{b}=C$. Therefore $A\subseteq A_{a}\cap A_{b}=C_{a}\cap C_{b}=C$.

(ii) Let $\phi:R[U,V,W](=R^{[3]})\twoheadrightarrow A$ be the $R$-algebra epimorphism such that $\phi(U)=u$, $\phi(V)=v$ and $\phi(W)=w$. Then $(bW-aV-U^{2})\subseteq Ker$ $\phi$ and $bW-aV-U^{2}$ is an irreducible polynomial of the UFD $R[U,V,W]$. Now $\operatorname{tr.deg}_{R}{(R[U,V,W]/(bW-aV-U^{2}))}=2=\operatorname{tr.deg}_{R}{A}$. Hence $A\cong R[U,V,W]/(bW-aV-U^{2})$. Let $F=bW-aV-U^{2}$. Now $(\frac{\partial F}{\partial U},\frac{\partial F}{\partial V},\frac{\partial F}{\partial W},F)R[U,V,W]\neq R[U,V,W]$. So $A$ is not a regular ring, in particular, $A\neq R^{[2]}$.

(iii) $rank$ $D$ $=3$ by Lemma 3.2. ∎

4 Quasi-nice Derivations

In this section we discuss quasi-nice derivations. Let $k$ be a field of characteristic zero, $R$ a $k$-domain, $B:=R^{[n]}$ and $m$ be a positive integer $\leq n$. We shall call a quasi-nice $R$-derivation of $B$ to be $m$-quasi if, for some coordinate system $(T_{1},T_{2},\dots,T_{n})$ of $B$, $D^{2}(T_{i})=0$ for all $i\in\{1,\dots,m\}$. Thus for any two positive integers $r$ and $m$ such that $1\leq m<r\leq n$, it is easy to see that an $r$-quasi derivation is also an $m$-quasi derivation. We shall call an $m$-quasi derivation to be strictly $m$-quasi if it is not $r$-quasi for any positive intger $r>m$.

Over a field $K$, Z. Wang [14, Theorem 4.7 and Remark 5] has proved the following result for $2$-quasi derivations.

Theorem 4.1**.**

Let $K$ be a field of characteristic zero and $K[X,Y,Z]=K^{[3]}$. Let $D(\neq 0)\in LND_{K}(K[X,Y,Z])$ be such that $D$ is irreducible and $D^{2}X=D^{2}Y=0$. Then one of the following holds:

1. (I)

There exists a coordinate system ($L_{1},L_{2},Z$) of $K[X,Y,Z]$, where $L_{1}$ and $L_{2}$ are linear forms in $X$ and $Y$ such that

1. (i)

$DL_{1}=0$. 2. (ii)

$DL_{2}\in K[L_{1}]$. 3. (iii)

$DZ\in K[L_{1},L_{2}]=K[X,Y]$.

In this case, $rank$ $D$ can be either $1$ or $2$. 2. (II)

There exists a coordinate system ($V,X,Y$) of $K[X,Y,Z]$, such that $DV=0$ and $DX,DY\in K[V]$. In particular, $rank$ $D=1$.

Conversely if $D\in Der_{K}(K[X,Y,Z])$ satisfies (I) or (II), then $D\in LND_{K}(K[X,Y,Z])$ and $D^{2}X=D^{2}Y=0$.

The following two examples illustrate the cases $rank$ $D=1$ and $rank$ $D=2$ for Part (I) of Theorem 4.1.

Example 4.2**.**

Let $D\in LND_{K}(K[X,Y,Z])$ be such that $DX=DY=0$ and $DZ=1$. Then $rank$ $D=1$. **

Example 4.3**.**

Let $D\in LND_{K}(K[X,Y,Z])$ such that

[TABLE]

Setting $R=K[X]$, we see that $D\in LND_{R}(R[Y,Z])$ and $D$ is irreducible. By Theorem 2.2, $D=\frac{\partial F}{\partial Z}\frac{\partial}{\partial Y}-\frac{\partial F}{\partial Y}\frac{\partial}{\partial Z}$ for some $F\in R[Y,Z]=K[X,Y,Z]$ such that $K(X)[Y,Z]=K(X)[F]^{[1]}$, $gcd_{R[Y,Z]}(\frac{\partial F}{\partial Y},\frac{\partial F}{\partial Z})=1$. Moreover $Ker$ $D=R[F]=R^{[1]}$. Setting $F=XZ-\frac{Y^{2}}{2}$ we see $\frac{\partial F}{\partial Y}=-Y=-DZ$ and $\frac{\partial F}{\partial Z}=X=DY$.

Therefore $Ker$ $D=K[X,F]$. But $F$ is not a coordinate in $K[X,Y,Z]$ since $(\frac{\partial F}{\partial X},\frac{\partial F}{\partial Y},\frac{\partial F}{\partial Z})K[X,Y,Z]\neq K[X,Y,Z]$. So $rank$ $D=2$. **

We now address Question $2$ of the Introduction, which gives a partial generalisation of Theorem 4.1.

Proposition 4.4**.**

Let $R$ be a PID containing $\mathbb{Q}$ with field of fractions $K$. Let $D\in LND_{R}(R[X,Y,Z])$, where $R[X,Y,Z]=R^{[3]}$ such that $D$ is irreducible and $D^{2}X=D^{2}Y=0$. Let $\overline{D}\in LND(K[X,Y,Z])$ denote the extension of $D$ to $K[X,Y,Z]$. Let $A:=Ker$ $D$. Suppose $\overline{D}$ satisfies condition (I) of Theorem 4.1. Then the following hold:

1. (i)

$rank$* $D<3$.* 2. (ii)

There exists a coordinate system $(L_{1},L_{2},Z)$ of $B$, such that $L_{1},L_{2}$ are linear forms in $X$ and $Y$, $DL_{1}=0$, $DL_{2}\in R[L_{1}]$ and $DZ\in R[L_{1},L_{2}]=R[X,Y]$. Moreover, $A=R[L_{1},bZ+f(L_{2})]$, where $b\in R[L_{1}]$ and $f(L_{2})\in R[L_{1},L_{2}]$.

Proof.

(i) Let $(\overline{L_{1}},\overline{L_{2}},Z)$ be the coordinate system of $K[X,Y,Z]$ such that $\overline{D}$ satisfies condition (I) of Theorem 4.1. Multiplying by a suitable nonzero constant from $R$, we can assume $\overline{L_{1}}\in R[X,Y]$. Let $\overline{L_{1}}=aX+bY$ where $a,b\in R$. Without loss of generality we can assume $gcd_{R}(a,b)=1$. Since $R$ is a PID, $(a,b,0)$ is a unimodular row in $R^{3}$ and hence can be completed to an invertible matrix (say $N$) in $GL_{3}(R)$. Thus $\overline{L_{1}}$ is a coordinate in $R[X,Y,Z]$. As $\overline{L_{1}}\in KerD=Ker\overline{D}\cap R[X,Y,Z]$, $rank$ $D$ is at most $2$ and hence $rank$ $D\leq 2<3$.

(ii) Now set $L_{1}=\overline{L_{1}}$. Since $gcd_{R}(a,b)=1$, there exist $c,d\in R$ such that $ad-bc=1$. Hence we can choose $N$ as $\begin{pmatrix}a&b&0\\ c&d&0\\ 0&0&1\end{pmatrix}$. Then $N\begin{pmatrix}X\\ Y\\ Z\end{pmatrix}=\begin{pmatrix}L_{1}\\ L_{2}\\ Z\end{pmatrix}$.

Now the proof follows from Part (i) of Lemma 2.3. ∎

With the notation as above, if $\overline{D}$ satisfies condition (II) of Theorem 4.1, $rank$ $D$ need not be $1$. The following example shows that $rank$ $D$ can even be $3$.

Example 4.5**.**

Let $k$ be a field of characteristic zero, $R=k[t](=k^{[1]})$ with field of fractions $L$ and $B:=R[X,Y,Z](=R^{[3]})$. Let $D\in LND_{R}(B)$ be defined by

[TABLE]

Then $D$ is irreducible and $D^{2}X=D^{2}Y=0$. Let $\overline{D}$ denote the extension of $D$ to $L[X,Y,Z]$. Let $F=-GX+tY$ where $G=tZ+X^{2}$. Then $F^{2}-G^{3}=tH$, where $H=tY^{2}-2tX^{2}Z^{2}-2tXYZ-2X^{3}Y-X^{4}Z-t^{2}Z^{3}\in R[X,Y,Z]$. Set $C:=R[F,G,H]$. We show that

1. (i)

$\overline{D}$ satisfies condition (II) of Theorem 4.1. 2. (ii)

Then $C\cong R[U,V,W]/(U^{2}-V^{3}-tW)$, where $R[U,V,W]=R^{[3]}$ and hence $C\neq R^{[2]}$. 3. (iii)

$Ker$ $D=C$. 4. (iv)

$rank$ $D=3$.

Proof.

(i) $L[X,Y,Z]=L[X,Y,G]$, $DG=\overline{D}G=0$ and $\overline{D}X,\overline{D}Y\in L[G]$. By Theorem 2.2, $Ker$ $\overline{D}=L[F,G](=L^{[2]})$.

(ii) Consider the $R$-algebra epimorphism $\phi:R[U,V,W]\twoheadrightarrow R[F,G,H]$($=C$), given by $\phi(U)=F$, $\phi(V)=G$ and $\phi(W)=H$. Clearly ($U^{2}-V^{3}-tW$) $\subseteq Ker$ $\phi$. Since $U^{2}-V^{3}-tW$ is an irreducible polynomial in $R[U,V,W]=k[t,U,V,W]=k^{[4]}$, and $\operatorname{tr.deg}_{k}R[U,V,W]/(U^{2}-V^{3}-tW)=3$ we have $Ker$ $\phi=(U^{2}-V^{3}-tW)$ and hence $C\cong R[U,V,W]/(U^{2}-V^{3}-tW)$.

Set $f:=U^{2}-V^{3}-tW$. Then $C$ is not regular, since $(\frac{\partial f}{\partial t},\frac{\partial f}{\partial U},\frac{\partial f}{\partial V},\frac{\partial f}{\partial W},f)\subseteq(t,U,V,W)k[t,U,V,W]$. Thus $C$ is not regular; in particular, $C\neq R^{[2]}$.

(iii) Let $A:=Ker$ $D$. Then $A=L[F,G]\cap R[X,Y,Z]$. We note that since $Ker~{}\overline{D}$ is factorially closed, $H\in Ker$ $\overline{D}$ and hence $H\in A$.

$C_{t}=R_{t}[F,G,H]=R_{t}[F,G]$. Also $R_{t}[X,Y,Z]=R_{t}[X,Y,G]=R_{t}[X,F,G]$. $D$ extends to a locally nilpotent $R_{t}$-derivation (say $\tilde{D}$) of $R_{t}[X,Y,Z]$ and $\tilde{D}X\in{R_{t}}^{*}$. So by Theorem 2.1, $C_{t}=A_{t}$.

Clearly $C\subseteq A$. By Lemma 2.12, it is enough to show that the map $C/tC\rightarrow A/tA$ is injective. Since $A$ is factorially closed in $B$, there exists an inclusion $A/tA\hookrightarrow B/tB$. So we will be done if we show the composite map $\psi:C/tC\rightarrow B/tB$ is injective. For $g\in B$, let $\overline{g}$ denote the image of $g$ in $B/tB$. In $\psi(C/tC)$, $\overline{G}=\overline{X}^{2}$, $\overline{F}=-\overline{X}^{3}$ and $\overline{H}=-2\overline{X}^{3}\overline{Y}-\overline{X}^{4}\overline{Z}$. Since $\overline{X}$ and $\overline{Z}$ are algebraically independent over $k$, $\operatorname{tr.deg}_{k}\psi(C/tC)=2$. From (ii) it follows that $C/tC$ is an integral domain and $\operatorname{tr.deg}_{k}C/tC=\operatorname{tr.deg}_{k}\psi(C/tC)=2$. Hence $\psi$ is injective. So $C/tC\hookrightarrow A/tA$ and hence $C=A$ as desired.

(iv) By Lemma 3.2, $rank$ $D=3$. ∎

Over a Dedekind domain $R$, we have the following generalisation of Proposition 3.9 and Proposition 4.4.

Proposition 4.6**.**

Let $R$ be a Dedekind domain containing $\mathbb{Q}$ with field of fractions $K$, and $B:=R[X,Y,Z](=R^{[3]})$. Let $D\in LND_{R}(B)$ be irreducible and $D^{2}X=D^{2}Y=0$ and $\overline{D}$ denote the extension of $D$ to $K[X,Y,Z]$. Let $A:=Ker$ $D$. If $\overline{D}$ satisfies condition (I) of Theorem 4.1, then the following hold:

1. (i)

$A$* is generated by at most $3$ elements.* 2. (ii)

Moreover, if $D$ is fixed-point free, then $rank$ $D$ $<3$ and $D$ has a slice. In particular, $rank$ $D=1$.

Proof.

(i) By Proposition 4.4 and Lemma 3.2, $A_{\mathpzc{p}}={R_{\mathpzc{p}}}^{[2]}$ for all $\mathpzc{p}\in Spec(R)$. Now the proof follows from the proof of Part (i) of Proposition 3.9.

(ii) For each $\mathpzc{p}\in Spec(R)$, let $D_{\mathpzc{p}}$ denote the extension of $D$ to $B_{\mathpzc{p}}$. Then by Proposition 4.4 and Theorem 2.5, $D_{\mathpzc{p}}$ has a slice. Thus $B_{\mathpzc{p}}={A_{\mathpzc{p}}}^{[1]}$ for all $\mathpzc{p}\in Spec(R)$. Now the proof follows from the proof of Part (ii) of Proposition 3.9. ∎

Remark 4.7**.**

Example 3.10 shows that Proposition 4.4 does not extend to a higher-dimensional UFD, not even to $k^{[2]}$, where $k$ is a field of characteristic zero. In fact, in that example, taking $L_{1}=u$ and $L_{2}=cX+dY$ for some $c,d\in k[a,b]$ such that $ad-bc\neq 0$, we find that $\overline{D}$ satisfies condition (I) of Theorem 4.1 and considering the coordinate system $(u,X,Z)$, of $K[X,Y,Z]$ where $K$ is the field of fractions of $k[a,b]$, we also see that $\overline{D}$ satisfies condition (II). **

The following theorem of Daigle shows that over a field $k$ of characteristic zero, there does not exist any strictly $1$-quasi derivation of $k^{[3]}$ [6, Theorem 5.1].

Theorem 4.8**.**

Let $k$ be a field of characteristic zero , $B=k^{[3]}$ and $D:B\rightarrow B$ be an irreducible locally nilpotent derivation. We assume that some variable $Y$ of $B$ satisfies $DY\neq 0$ and $D^{2}Y=0$. Then there exist $X,Z$ such that

[TABLE]

We now present an example of a strictly $1$-quasi derivation of $R^{[3]}$ over a PID $R$ containing $\mathbb{Q}$. Thus Theorem 4.8 does not extend to a PID.

Example 4.9**.**

Let $k$ be a field of characteristic zero, $R=k[t](=k^{[1]})$ and $B=R[X,Y,Z](=R^{[3]})$. Let $D\in LND_{R}(B)$ be such that

[TABLE]

Then $D$ is irreducible and $D^{2}X=0$. Let $F:=2tY-X^{2}$, $G=3t^{2}Z-3tXY+X^{3}$ and $H=8tY^{3}+9t^{2}Z^{2}-18tXYZ-3X^{2}Y^{2}+6X^{3}Z$. Then $F^{3}+G^{2}=t^{2}H$. Set $C:=R[F,G,H]$. We now show the following:

1. (i)

Then $C\cong R[U,V,W]/(U^{3}+V^{2}-t^{2}W)$,

where $R[U,V,W]=R^{[3]}$. 2. (ii)

$Ker$ $D=C$. 3. (iii)

There does not exist any coordinate system $(U_{1},U_{2},U_{3})$ of $B$, such that $D^{2}(U_{1})=D^{2}(U_{2})=0$.

Proof.

(i) Consider the $R$-algebra epimorphism $\phi:R[U,V,W]\twoheadrightarrow R[F,G,H](=C)$, given by $\phi(U)=F$, $\phi(V)=G$ and $\phi(W)=H$. Clearly $(U^{3}+V^{2}-t^{2}W)\subseteq Ker$ $\phi$. Since $U^{3}+V^{2}-t^{2}W$ is an irreducible polynomial in $R[U,V,W]=k[t,U,V,W]=k^{[4]}$, and $\operatorname{tr.deg}_{k}R[U,V,W]/(U^{3}+V^{2}-t^{2}W)=3$ we have $Ker$ $\phi=(U^{3}+V^{2}-t^{2}W)$ and hence $C\cong R[U,V,W]/(U^{3}+V^{2}-t^{2}W)$.

(ii) Let $A:=Ker$ $D$. Since $A$ is factorially closed in $B$, $H\in A$. $C_{t}=R_{t}[F,G,H]=R_{t}[F,G]$. Also $R_{t}[X,Y,Z]=R_{t}[X,F,G]$. $D$ extends to a locally nilpotent $R_{t}$-derivation (say $\tilde{D}$) of $R_{t}[X,Y,Z]$ and $\tilde{D}X\in{R_{t}}^{*}$. So by Theorem 2.1, $C_{t}=A_{t}$.

Clearly $C\subseteq A$. By Lemma 2.12, it is enough to show that the map $C/tC\rightarrow A/tA$ is injective. Since $A$ is factorially closed in $B$, there exists an inclusion $A/tA\hookrightarrow B/tB$. So we will be done if we show the composite map $\psi:C/tC\rightarrow B/tB$ is injective. For $g\in B$, let $\overline{g}$ denote the image of $g$ in $B/tB$. In $\psi(C/tC)$, $\overline{F}=-\overline{X}^{2}$, $\overline{G}=\overline{X}^{3}$ and $\overline{H}=-3\overline{X^{2}}\overline{Y^{2}}+6\overline{X^{3}}\overline{Z}$. Since $\overline{X}$, $\overline{Y}$ and $\overline{Z}$ are algebraically independent over $k$, $\operatorname{tr.deg}_{k}\psi(C/tC)=2$. From (ii) it follows that $C/tC$ is an integral domain and $\operatorname{tr.deg}_{k}C/tC=\operatorname{tr.deg}_{k}\psi(C/tC)=2$. Hence $\psi$ is injective. So $C/tC\hookrightarrow A/tA$ and hence $C=A$ as desired.

(iii) For $f\in B$, let $\alpha:=$ coefficient of $X$ in $f$, $\beta:=$ coefficient of $Y$ in $f$ and $deg$ $f:=$ total degree of $f$. Then $\frac{\partial f}{\partial X}(0,0,0)=\alpha$ and $\frac{\partial f}{\partial Y}(0,0,0)=\beta$. If $f\in A$, let $f=p(F,G,H)$ for some $p\in R^{[3]}$. Then $\frac{\partial f}{\partial X}=\frac{\partial p}{\partial F}\frac{\partial F}{\partial X}+\frac{\partial p}{\partial G}\frac{\partial G}{\partial X}+\frac{\partial p}{\partial H}\frac{\partial H}{\partial X}$. We also have

[TABLE]

Thus $\alpha=0$. Again, since

[TABLE]

similarly we have $\beta=\lambda t$, for some $\lambda\in R$. Let $U$ be a coordinate in $B$ such that $D^{2}U=0$. Since $U$ is a coordinate, there exist $a,b,c\in R$, not all [math] and $V\in B$ with $deg$ $V\geqslant 2$ and no linear term, such that $U=aX+bY+cZ+V$. Then $DU=at+bX+cY+(\frac{\partial V}{\partial X})t+(\frac{\partial V}{\partial Y})X+(\frac{\partial V}{\partial Z})Y$. Let $\gamma:=$ coefficient of $X$ in $\frac{\partial V}{\partial X}$ and $\delta:=$ coefficient of $Y$ in $\frac{\partial V}{\partial X}$. Then coefficient of $X$ in $DU=b+\gamma t$ and coefficient of $Y$ in $DU=c+\delta t$ (We can ignore terms from $(\frac{\partial V}{\partial Y})X$ and $(\frac{\partial V}{\partial Z})Y$ since neither of them has any linear term). Thus $b+\gamma t=0$ and $c+\beta t=\lambda t$ for some $\lambda\in R$. Therefore, $b\in(t)R$ and $c\in(t)R$.

Let $(U_{1},U_{2},U_{3})$ be a coordinate system of $B$ such that $D^{2}(U_{1})=D^{2}(U_{2})=0$ and let $U_{i}=a_{i}X+b_{i}Y+c_{i}Z+V_{i}$ where $a_{i},b_{i},c_{i}\in R$, not all [math] and $V_{i}\in B$ with $deg$ $V_{i}\geqslant 2$ and no linear term, for $i=1,2$. Thus for each $i$, we have $b_{i},c_{i}\in(t)R$. For $f\in B$, let $\overline{f}$ denote its image in $B/tB(=k^{[3]})$. Since $B/tB=k[\overline{U_{1}},\overline{U_{2}},\overline{W}]$, $\overline{U_{1}},\overline{U_{2}}$ form a partial coordinate system in $B/tB$. But $\overline{U_{i}}=\overline{a_{i}}\overline{X}+V_{i}$ and since $\overline{V_{i}}$ has no linear term, $\overline{a_{i}}\neq 0$, for each $i$. Then $\overline{a_{2}}\overline{U_{1}}-\overline{a_{1}}\overline{U_{2}}$ has no linear term, but it is a coordinate in $B/tB$. Hence we have a contradiction. ∎

Acknowledgement: The authors thank Professor Amartya K. Dutta for asking Questions 1 and 2, going through the earlier drafts and suggesting improvements.The second author also acknowledges Department of Science and Technology for their Swarnajayanti award.