On image ideals of nice and quasi-nice derivations over a UFD

TL;DR

This paper characterizes the image ideals of certain derivations on polynomial rings over a UFD, providing explicit generators for these ideals in the context of nice and quasi-nice derivations.

Contribution

It introduces a set of generators for the image ideals of irreducible nice and quasi-nice derivations over polynomial rings with a UFD base.

Findings

Explicit generators for image ideals of nice derivations

Extension of results to quasi-nice derivations

Applicable to polynomial rings over UFDs

Abstract

In this paper, for a field $k$ of characteristic zero and a finitely generated $k$-algebra $R$, we give a set of generators for the image ideals of irreducible nice and quasi-nice $R$-derivations on the polynomial ring $R[X,Y]$, where $R$ is a UFD.

Equations

$$I_n=A\cap D^{n}B\text{ for }n⩾1.$$

$$\theta (p)=\mathrm{Max}\{\theta (m):m\in M(p)\},$$

$$1=D^n(\frac{s^n}{n!})\in A\cap D^{n}B=I_n.$$

$$\eta (\alpha )=\{\begin{array}{ll}\inf \{\lambda (h):h\in {\pi }^{-1}(\alpha )\},& \text{if }\alpha \ne 0,\\ -\infty ,& \text{if }\alpha =0,\end{array}$$

$$\mathcal{G}_{j}=\underset{\underset{j_{1},\dots,j_{m+n}\geqslant 0}{j_{1}+\dots+j_{m}+u_{1}j_{m+1}+\dots+u_{n}j_{m+n}=j}}{\text{\Large$\sum$}}\big{(}s_{1}^{j_{1}}\dots s_{m}^{j_{m}}X_{1}^{j_{m+1}}\dots X_{n}^{j_{m+n}}\big{)}\mathcal{F}_{0}.$$

$$\eta (\alpha )=\inf \{\lambda (h):h\in {\pi }^{-1}(\alpha )\},\text{where }\pi :C\to B\text{ is the canonical map}.$$

$${\mathbb{bG}}_{j}=\underset{\underset{j_{1},\dots,j_{m+n}\geqslant 0}{j_{1}+\dots+j_{m}+u_{1}j_{m+1}+\dots+u_{n}j_{m+n}=j}}{\text{\Large$\bigoplus$}}\big{(}S_{1}^{j_{1}}\dots S_{m}^{j_{m}}X_{1}^{j_{m+1}}\dots X_{n}^{j_{m+n}}\big{)}R[Z_{1},\dots,Z_{r}].$$

$${\mu }_1:=\mathrm{Max}\{\lambda (x):x\in M\setminus M_0\}\text{ and }{M}_1:=\{x\in M\setminus M_0:\lambda (x)={\mu }_1\}.$$

$${\mu }_2:=\mathrm{Max}\{\lambda (x):x\in (M\setminus M_0)\setminus M_1\}$$

$$\text{and }{M}_2:=\{x\in (M\setminus M_0)\setminus M_1:\lambda (x)={\mu }_2\}.$$

$$P(n):\text{ If }\sum _{x\in M_j}x=0\text{ for all }j<n,\text{ then, for }{t}_m>0,x(m;s_m;t_m)\in M_n\text{ only if }x(m;s_m;t_m)\in J_m.$$

$$f_{1}h=X_{1}P(u)+Q_1(u)\text{ and }{f}_{2}h=X_{2}P(u)+Q_2(u)\text{ for some }{Q}_1(u),Q_2(u)\in R[u].$$

$$J=\{i:1⩽i⩽n\text{ and the induced }{R}_{(p_i)}\text{-lnd }{D}_{p_i}\text{ is not fixed point free}\}.$$

$$I_j=\{\begin{array}{ll}(\prod _{i\in J}{p_i}^{r_i}{)}^{m}A,& \text{if }J\ne \varnothing \\ A,& \text{if }J=\varnothing ,\end{array}$$

Full text

However, there has not been much study about how a set of generators for each of the image ideals would look like. Therefore, it becomes interesting to find a explicit set of generators for each of the image ideals when $B$ is a polynomial algebra over a ring (atleast for the case when $D$ is of some special type and $R$ is a UFD or PID). In this paper, for a field $k$ of characteristic zero and a finitely generated $k$-algebra $R$, we give a set of generators for the image ideals of irreducible nice and quasi-nice $R$-derivations on the polynomial ring $R[X,Y]$, when $R$ is a UFD.

Keywords. Locally nilpotent derivation, polynomial ring, filtration.

2022 MSC. Primary: 14R20; Secondary: 13A50.

1 Introduction

We will assume all rings to be commutative containing unity. By $k$, we will always denote a field of characteristic zero. The set of all non-negative integers will be denoted by $\mathbb{N}$. The notation $R^{[n]}$ will be used to denote an $R$-algebra isomorphic to a polynomial algebra in $n$ variables over $R$. Unless otherwise stated, capital letters like $X_{1},\dots,X_{n},Z_{1},\dots,Z_{m},X,Y,Z,U,V,W$ will be used as variables in the polynomial ring.

Let $R$ be a $k$-algebra and $B$ an $R$-algebra. An $R$-linear map $D:B\longrightarrow B$ is said to be an $R$-derivation, if it satisfies the Leibnitz rule: $D(ab)=aDb+bDa,~{}\forall a,b\in B$. An $R$-derivation $D$ is said to be a locally nilpotent derivation (abbrev. $R$-lnd) if, for each $b\in{B}$, there exists $n(b)\in{\mathbb{N}}$ such that $D^{n(b)}b=0$. The set of all locally nilpotent $R$-derivations on $B$ will be denoted by ${\rm LND}_{R}(B)$. The kernel of $D$, denoted by ${\rm Ker}(D)$, is defined to be the set $\{b\in B~{}|~{}Db=0\}$. The image ideals of $D$ are the ideals of ${\rm Ker}(D):=A$, defined by

[TABLE]

$I_{1}$ is called the plinth ideal of $D$.

An important problem in the area of locally nilpotent derivations is to study the minimal number of generators of the image ideals $I_{n}$. When $B=k^{[3]}$, it is well-known that the plinth ideal $I_{1}$ is principal ([9, Theorem 5.12], [6, Theorem 1]). Freudenburg conjectured that all the image ideals are also principal for $D\in{\rm LND}_{k}(k^{[3]})$ ([9, 11.2]). This conjecture is called the Freeness Conjecture.

In this paper, we consider affine $k$-algebras $R$ which are UFDs and investigate the image ideals of irreducible nice and quasi-nice $R$-derivations (see Definition 2.2) on $R^{[2]}$. For any $k$-domain $B$ and an lnd $D$ having a slice (i.e., $\exists~{}s\in B$ such that $Ds=1$), it is easy to observe (see Lemma 2.8) that all the image ideals are equal to the whole ring and hence principal. In fact, if $R$ is an affine $k$-domain, then any fixed point free $R$-lnd $D$ (i.e., $(DB)B=B$) on $B=R^{[2]}$ has a slice in $B$ and hence all the image ideals are principal (Corollary 2.10). So, it is interesting to investigate the situation when $D$ is not fixed point free. Here after in this section we assume $D$ is not fixed point free.

In Theorem 3.5 of this paper, for a UFD $R$, we study the structure of the image ideals of an irreducible $R$-lnd $D$ on $R[X_{1},X_{2}]$ satisfying $D^{2}X_{1}=D^{2}X_{2}=0$ and gave an explicit set of $n+1$ generators of $I_{n}$. As a corollary, when $R$ is a PID, we also obtain similar kind of results for irreducible $D\in{\rm LND}_{R}(R[X,Y,Z])$ satisfying $D^{2}X=D^{2}Y=D^{2}Z=0$ (see Theorem 3.7). It follows from our results that over a UFD $R$, the image ideals of irreducible nice $R$-derivations on $R^{[2]}$ are principal if and only if the derivation is fixed point free.

In Theorem 3.12, we show that over a UFD $R$, an irreducible $R$-lnd $D$ on $R[X_{1},X_{2}]$ satisfying $D^{2}X_{1}=0$ and $DX_{1}$ irreducible, has all the image ideals principal if and only if $D$ is not a nice derivation. In fact, each image ideal is generated by a power of $DX_{1}$. We have showed that when $R$ is a PID, the condition “$DX_{1}$ is irreducible” can be removed. In fact, we have proved that if $DX_{1}={\prod_{i=1}^{n}p_{i}^{r_{i}}}~{}(\in R)$, then there exists $J\subseteq\{1,2,\dots,n\}$ such that each image ideal is a power of the ideal $({\underset{i\in J}{\prod}}p_{i}^{r_{i}})A$ (see Theorem 3.17).

Corollary 3.3 and Proposition 3.4 are the main tools used to calculate the higher image ideals in case of both nice and quasi-nice derivations. Corollary 3.3 gives an explicit set of generators for each $I_{n}$ under the primality of an ideal $\widetilde{J}$, and Proposition 3.4 is used to find the generators of $\widetilde{J}$. In fact, given an ideal $I$ in a polynomial ring over an affine $k$-domain, equipped with a weighted degree map, using Proposition 3.4 one can compute the ideal generated by the top degree terms of elements of $I$, provided at most one of the generators of $I$ is non-homogeneous. Proof of Corollary 3.3 uses some techniques of LND-filtration introduced by B. Alhajjar in [1].

2 Preliminaries

First, we will recall some useful definitions.

Definition 2.1**.**

A derivation $D$ on $B$ is said to be irreducible if there does not exist any non-unit $b$ in $B$ such that $DB\subseteq bB$.**

Definition 2.2**.**

Let $R$ be a $k$-domain, $B=R^{[n]}$ and $m~{}(\leqslant n)$ be a positive integer. An $R$-lnd $D$ on $B$ is said to be m-quasi-nice if there exists a coordinate system $(X_{1},X_{2},\dots,X_{n})$ in $B$ such that $D^{2}(X_{i})=0$ for all $i\in\{1,\dots,m\}$. If $m=n$, then $D$ is called nice.**

Definition 2.3**.**

An $m$-quasi-nice derivation is said to be strictly $m$-quasi-nice if it is not $r$-quasi-nice for any positive integer $r>m$. **

Definition 2.4**.**

Let $B$ be a ring. A family of additive subgroups $\{\mathcal{B}_{i}\}_{i\in\mathbb{N}}$ of $B$ is said to be an $\mathbb{N}$-filtration* of $B$ if the following conditions hold:*

1. (i)

$\mathcal{B}_{i}\subseteq\mathcal{B}_{i+1}$ for each $i\in\mathbb{N}$ 2. (ii)

$\underset{{i\in\mathbb{N}}}{\bigcup}\mathcal{B}_{i}=B$, 3. (iii)

$\mathcal{B}_{i}\mathcal{B}_{j}\subseteq{\mathcal{B}_{i+j}}$ for all $i,j\in\mathbb{N}$.

An $\mathbb{N}$-filtration $\{\mathcal{B}_{i}\}_{i\in\mathbb{N}}$ of $B$ is said to be proper if the following additional conditions hold:

1. (iv)

$\bigcap_{i\in\mathbb{N}}\mathcal{B}_{i}=\{0\}$, 2. (v)

$a\in\mathcal{B}_{i}\setminus\mathcal{B}_{i-1}$, $b\in\mathcal{B}_{j}\setminus\mathcal{B}_{j-1}$ will imply $ab\in\mathcal{B}_{i+j}\setminus\mathcal{B}_{i+j-1}$.

Definition 2.5**.**

Let $B$ be a ring. A map $\theta:B\longrightarrow\mathbb{N}\cup\{-\infty\}$ is said to be an $\mathbb{N}$-semi-degree map* on $B$ if the following conditions hold:*

1. (i)

$\theta(a)=-\infty$ if and only if $a=0$, 2. (ii)

$\theta(ab)\leqslant\theta(a)+\theta(b)$ for all $a,b\in B,$ 3. (iii)

$\theta(a+b)\leqslant{\rm Max}~{}\{\theta(a),\theta(b)\}$ for all $a,b\in B.$

An $\mathbb{N}$-semi-degree map $\theta$ is called a degree map if in condition (iii) equality occurs for all $a,b\in B$.**

There is a one-one correspondence between proper $\mathbb{N}$-filtrations of $B$ and $\mathbb{N}$-degree maps on $B$. In fact, if $\mathcal{B}:=\{\mathcal{B}_{i}\}_{i\in\mathbb{N}}$ is a proper $\mathbb{N}$-filtration of $B$, then it induces a degree map $\theta_{\mathcal{B}}$ on $B$ defined by $\theta_{\mathcal{B}}(a)=\begin{cases}-\infty,&\text{if }a=0,\\ n,&\text{if }a\in{\mathcal{B}_{n}\setminus\mathcal{B}_{n-1}}.\end{cases}$

Conversely, if $\theta$ is a degree map on $B$, then $B=\underset{i\in\mathbb{N}}{\bigcup}\mathcal{B}_{i}$, where $\mathcal{B}_{i}=\{a\in B:\theta(a)\leqslant i\}$. It is easy to check that $\{\mathcal{B}_{i}\}_{i\in\mathbb{N}}$ is a proper $\mathbb{N}$-filtration of $B$.

Definition 2.6**.**

Let $R$ be a ring and $B=R[X_{1},\dots,X_{n}]$. A real valued degree map $\theta$ on $B$ is said to be a weighted degree map on $B$, if, for any $p\in B$,

[TABLE]

*where $M(p)$ is the set of all monomials occurring in the expression of $p$.

---

*If $M^{\prime}(p):=\{m\in M(p):\theta(m)=\theta(p)\}$, then we will denote $\sum\limits_{m\in{M^{\prime}(P)}}m$ by by $\widetilde{p}$. For an ideal $I$ of $B$, $\widetilde{I}$ will denote the ideal of $B$ generated by $\{\widetilde{p}:p\in I\}$.

---

Now, we will record some elementary facts and some important results which will be used throughout the rest of the paper. First, we state and prove an elementary result which follows from the higher product rule for a derivation ([9, Proposition 1.6]).

Lemma 2.7**.**

Let $R$ be a $k$-algebra, $B$ an $R$-algebra, $D\in{\rm LND}_{R}(B)$ and $f_{1},\dots,f_{n}\in B$ with ${\rm deg}_{D}(f_{i})=m_{i}$. If $m=\sum_{i=1}^{n}m_{i}$, then we have the following:

1. (i)

$D^{m}(f_{1}\dots f_{n})=\dfrac{m!}{\prod_{i=1}^{n}m_{i}!}\prod_{i=1}^{n}D^{m_{i}}f_{i}$, 2. (ii)

$D^{m+t}(f_{1}\dots f_{n})=0$, where $t\geqslant 1$. In particular, ${\rm deg}_{D}(f_{1}\dots f_{n})=m$.

Proof.

Proof of (i) follows from the higher product rule, $D^{m}(fg)=\sum\limits_{i+j=m}{m\choose i}D^{i}fD^{j}g$. The proof of (ii) follows from the fact that for each $i\in\{1,\dots,n\}$, $D^{l}f_{i}=0$ if and only if $l>m_{i}$. ∎

Next, we state an observation describing the image ideals of a locally nilpotent derivation with a slice.

Lemma 2.8**.**

Let $R$ be a $k$-algebra, $B$ an $R$-algebra, $D\in{\rm LND}_{R}(B)$ and $A={\rm Ker}(D)$. Then the following are equivalent.

1. (i)

$D$* has a slice.* 2. (ii)

$I_{1}=A$. 3. (iii)

$I_{n}=A$* for all $n\in\mathbb{N}$.* 4. (iv)

$I_{n}=A$* for some $n\in\mathbb{N}$.*

Proof.

(i)$\implies$(iii): Let $s\in B$ be such that $Ds=1$. Then, for each $n\in\mathbb{N}$, we have

[TABLE]

(ii)$\implies$(i): If $I_{1}=A$, then $1\in I_{1}\subseteq DB$.

The implications (iii)$\implies$(ii), (iii)$\implies$(iv) and (iv)$\implies$(ii) are obvious. ∎

For a Noetherian $k$-domain $R$ and an irreducible $D\in{\rm LND}_{R}(R[X,Y])$, the structure of ${\rm Ker}(D)$ is given in [4, Theorem 4.7].

Theorem 2.9**.**

Let $R$ be a Noetherian domain containing $k$ and $B=R[X,Y]$. Let $D$ be an irreducible $R$-lnd on $B$ and $A={\rm Ker}(D)$. Then $A=R^{[1]}$ if and only if one of the following conditions hold:

1. (i)

$DX$* and $DY$ form a regular sequence in $B$.* 2. (ii)

$(DX,DY)B=B$.

Moreover, if (ii) holds, then $B=A^{[1]}$.

As an immediate corollary to Theorem 2.9 we have the following result.

Corollary 2.10**.**

Let $R$ be a Noetherian domain containing $k$, $B=R^{[2]}$ and $D\in{\rm LND}_{R}(B)$ be fixed point free. Then $I_{n}=A$ for all $n\in\mathbb{N}$.

Proof.

The proof follows from Lemma 2.8 and Theorem 2.9. ∎

The following result of Z. Wang ([11, Lemma 4.2]) describes ${\rm Ker}(D)$ for an irreducible nice (or quasi-nice) $R$-lnd $D$ on $R[X,Y]$, when $R$ is a UFD containing $k$.

Lemma 2.11**.**

Let $R$ be a UFD containing $k$, $B=R[X,Y]$ and $D$ an irreducible $R$-lnd. Then, the following hold:

1. (i)

If $D^{2}X=0$, then ${\rm Ker}(D)=R[bY+f(X)]$, 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 $D=b\dfrac{\partial}{\partial X}-a\dfrac{\partial}{\partial Y}$ for some $a,b\in R$. Moreover, ${\rm Ker}(D)=R[aX+bY]$. 3. (iii)

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

The following result ([7, Theorem 3.6]) describes the structure of the kernels of nice derivations on $R^{[3]}$, where $R$ is a PID containing $k$ .

Theorem 2.12**.**

Let $R$ be a PID containing $k$ and $B=R[X,Y,Z]$. Let $D$ be an irreducible $R$-lnd on $B$ with $D^{2}X=D^{2}Y=D^{2}Z=0$. Let $A={\rm Ker}(D)$. Then there exists a coordinate system $(U,V,W)$ in $B$ related to $(X,Y,Z)$ by a linear change such that the following hold:

1. (i)

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

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

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

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

Next, we state a generalization of a result proved by S. Kaliman and L. Makar-Limanov ([10]). The proof given by the authors in [10] for $R=\mathbb{C}$ works for any affine $k$-domain $R$ with a slight modification of the hypotheses.

Lemma 2.13**.**

*Let $R$ be an affine $k$-domain, $C=R[X_{1},\dots,X_{n}]$ and $\lambda$ a weighted degree map on $C$. Suppose $J$ is an ideal of $C$ contained in the ideal $(X_{1},\dots,X_{n})C$ and

$\pi:C\rightarrow C/J$ is the canonical map. Define $\eta:C/J\longrightarrow\mathbb{R}\cup\{-\infty\}$ by*

[TABLE]

Then, for $\alpha\neq 0$, the following hold:

1. (i)

there exists $h\in\pi^{-1}(\alpha)$ such that $\widetilde{h}\notin{\widetilde{J}}$, 2. (ii)

$\eta(\alpha)=\lambda(h)$* for some $h\in\pi^{-1}(\alpha)$ if and only if $\widetilde{h}\notin\widetilde{J}$,* 3. (iii)

$\eta$* is a semi-degree map on $C/J$. If $\widetilde{J}$ is a prime ideal of $C$, then $\eta$ is a degree map.*

3 Main Results

In this section we will prove our main results. Throughout rest of the section $R$ will denote a finitely generated $k$-domain. Let $B=R[X_{1},\dots,X_{n}]$. Let $D\in{\rm LND}_{R}(B)\setminus\{0\}$, $A={\rm Ker}(D)$ and $I_{1}=(Ds_{1},\dots,Ds_{m})A$, for some $s_{1},\dots,s_{m}\in B$. Without loss of generality, we can assume that $s_{i}(0,\dots,0)=0$, for each $i$.

Consider the filtration $\{\mathcal{F}_{j}\}_{j\in{\mathbb{N}}}$ defined by $\mathcal{F}_{j}={\rm Ker}(D^{j+1})$. Then it is easy to see the following:

1. (i)

$\mathcal{F}_{0}=A$, 2. (ii)

$\mathcal{F}_{1}=\sum_{l=1}^{m}s_{l}\mathcal{F}_{0}+\mathcal{F}_{0}$, 3. (iii)

I_{j}=\big{(}D^{j}(\mathcal{F}_{j})\big{)}A for each $j\in\mathbb{N}$.

Let $u_{l}:={\rm deg}_{D}(X_{l})$ for $l=1,\dots,n.$ For each $j\in\mathbb{N}$, define $\mathcal{G}_{j}$ in the following way:

[TABLE]

We note the following easy observations:

1. (i)

$\mathcal{G}_{0}=\mathcal{F}_{0}$, 2. (ii)

$\mathcal{G}_{i}\mathcal{G}_{j}\subseteq\mathcal{G}_{i+j}$, for all $i,j\in\mathbb{N}$.

For each $j\in\mathbb{N}$, let $\mathcal{H}_{j}=\sum_{i=0}^{j}\mathcal{G}_{i}$. Clearly, for each $j\in\mathbb{N}$, $\mathcal{H}_{j}\subseteq\mathcal{H}_{j+1}$. For $i,j,q,r\in\mathbb{N}$ with $0\leqslant q\leqslant i$ and $0\leqslant r\leqslant j$, since $\mathcal{G}_{q}\mathcal{G}_{r}\subseteq\mathcal{G}_{q+r}\subseteq{\mathcal{H}_{i+j}}$, we have $\mathcal{H}_{i}\mathcal{H}_{j}\subseteq\mathcal{H}_{i+j}$. Let $\alpha=\sum a_{i_{1},\dots,i_{n}}{X_{1}}^{i_{1}}\dots X_{n}^{i_{n}}\in{B}$ and $S=\{(i_{1},\dots,i_{n})\in{\mathbb{N}}^{n}:a_{i_{1},\dots,i_{n}}\neq 0\}$.

If $n_{0}={\rm Max}~{}\{u_{1}i_{1}+\dots+u_{n}i_{n}:(i_{1},\dots,i_{n})\in S\}$, then $\alpha\in\mathcal{H}_{n_{0}}$. So, $B=\bigcup_{j\in\mathbb{N}}\mathcal{H}_{j}$. Hence $\{\mathcal{H}_{j}\}_{j\in\mathbb{N}}$ is an $\mathbb{N}$-filtration of $B$. When $\{\mathcal{H}_{j}\}_{j\in\mathbb{N}}$ is a proper $\mathbb{N}$-filtration, the following lemma gives a generating set for each $I_{j}$.

Lemma 3.1**.**

If $\{\mathcal{H}_{j}\}_{j\in\mathbb{N}}$ is a proper $\mathbb{N}$-filtration, then I_{j}=\big{(}D^{j}(\mathcal{G}_{j})\big{)}A for all $j\in{\mathbb{N}}$.

Proof.

By Lemma 2.7, $\mathcal{H}_{j}\subseteq\mathcal{F}_{j}$ for each $j\in\mathbb{N}$. It is clear that $\mathcal{H}_{0}=\mathcal{G}_{0}=\mathcal{F}_{0}$ and $\mathcal{F}_{1}~{}(=\sum_{l=1}^{m}s_{l}\mathcal{F}_{0}+\mathcal{F}_{0})\subseteq{\mathcal{G}_{1}}+\mathcal{G}_{0}=\mathcal{H}_{1}$. Hence $\mathcal{F}_{j}=\mathcal{H}_{j}$ for $j=0,1$. Let $\alpha\in\mathcal{F}_{j}$ for some $j\in\mathbb{N}~{}(j\geqslant 2)$ and $\theta$ be the degree map on $B$ corresponding to the filtration $\{\mathcal{H}_{j}\}_{j\in\mathbb{N}}$ of $B$. Let $s\in{\mathcal{F}_{1}\setminus\mathcal{F}_{0}}$ and $f=Ds$. Since $B[1/f]=A[1/f][s]$, $\exists~{}\epsilon\in\mathbb{N}$ and $a_{0},\dots a_{j}\in A$ such that $f^{\epsilon}\alpha=a_{0}+a_{1}s+\dots+a_{j}s^{j}$. Since $\mathcal{F}_{j}=\mathcal{H}_{j}$ for $j=0,1$, we have $\theta(s)=1$ and $\theta(f)=0=\theta(a_{i})$ for all $i=0,\dots,j$. Hence $\theta(\alpha)\leqslant j$. So, $\alpha\in\mathcal{H}_{j}$. Thus $\mathcal{H}_{j}=\mathcal{F}_{j}$ for each $j\in\mathbb{N}$. Since $\mathcal{H}_{j}=\mathcal{G}_{j}+\mathcal{H}_{j-1}$, we have $\mathcal{F}_{j}=\mathcal{G}_{j}+\mathcal{F}_{j-1}$ for all $j\in\mathbb{N}$. Hence I_{j}=\big{(}D^{j}(\mathcal{G}_{j})\big{)}A for all $j\in\mathbb{N}$. ∎

Now, we will consider the situation when ${\rm Ker}(D)=R[z_{1},\dots,z_{r}]$ for some $z_{1},\dots,z_{r}$ in $R[X_{1},\dots,X_{n}]$ with $z_{i}(0,\dots,0)=0$ for all $i$, $1\leqslant i\leqslant r$. We define a weighted $\mathbb{N}$-degree map $\lambda$ on $C:=R[X_{1},\dots,X_{n},Z_{1},\dots,Z_{r},S_{1},\dots,S_{m}]~{}(=R^{[n+r+m]})$, by setting $\lambda(r)=0$ for all $r\in R$, $\lambda(X_{l})=u_{l}$ for $1\leqslant l\leqslant n$, $\lambda(Z_{l})=0$ for $1\leqslant l\leqslant r$ and $\lambda(S_{l})=1$ for $1\leqslant l\leqslant m$. Let $J$ be the ideal of $C$ generated by the set $\{Z_{1}-z_{1},\dots,Z_{r}-z_{r},S_{1}-s_{1},\dots,S_{m}-s_{m}\}$. Then $C/J\cong B$. By Lemma 2.13, $\lambda$ induces a semi-degree map $\eta$ on $B$ defined for each $\alpha\in B\setminus\{0\}$,

[TABLE]

The next result shows that the filtration induced by $\eta$ is $\{\mathcal{H}_{j}\}_{j\in\mathbb{N}}$.

Lemma 3.2**.**

$\{\mathcal{H}_{i}\}_{i\in\mathbb{N}}$* is the filtration induced by the semi-degree map $\eta$ on $B$.*

Proof.

Let $\{\mathbb{H}_{i}\}_{i\in\mathbb{N}}$ is the filtration of $C$ induced by the weighted degree map $\lambda$ on $C$. Then, \mathbb{H}_{i}=\text{\Large\bigoplus}_{j=0}^{i}{\mathbb{bG}}_{j}, where, for each $j$ with $0\leqslant j\leqslant i$,

[TABLE]

Clearly, for each $j$ with $0\leqslant j\leqslant i$, we have $\pi(\mathbb{bG}_{j})=\mathcal{G}_{j}$ and hence for each $i\in\mathbb{N}$, $\pi(\mathbb{H}_{i})=\mathcal{H}_{i}$.

Let $\alpha\in B\setminus\{0\}$ be such that $\eta(\alpha)\leqslant i$. By part (i) and (ii) of Lemma 2.13, there exists $h\in\pi^{-1}(\alpha)$ such that $\lambda(h)=\eta(\alpha)$ and hence $h\in\mathbb{H}_{i}$. So, \alpha~{}\big{(}=\pi(h)\big{)}\in\pi(\mathbb{H}_{i})=\mathcal{H}_{i}.

Since $\eta(X_{l})\leqslant\lambda(X_{l})=u_{l}$ for all $l\in\{1,\dots,n\}$, $\eta(s_{l})\leqslant\lambda(S_{l})=1$ for all $l\in\{1,\dots,m\}$, $\eta(z_{l})\leqslant\lambda(Z_{l})=0$ for all $l\in\{1,\dots,r\}$ and $\eta$ is a semi-degree map on $C$, we have $\eta(\alpha)\leqslant i$ for all $\alpha\in\mathcal{H}_{i}$. ∎

From Lemma 2.13(iii), Lemma 3.2 and Lemma 3.1, we immediately deduce the following corollary.

Corollary 3.3**.**

If $\widetilde{J}$ is a prime ideal of $C$, then for each $j\geqslant 0$ we have I_{j}=\big{(}D^{j}(\mathcal{G}_{j})\big{)}A.

Suppose $\lambda$ is a weighted degree map on $B~{}(=R^{[n]})$ and $J=(f_{1},\dots,f_{m})$ is an ideal of $B$. It is interesting to investigate conditions under which $\widetilde{J}=(\widetilde{f_{1}},\dots,\widetilde{f_{m}})$. If $m=1$, then it is easy to see that $\widetilde{J}=(\widetilde{f_{1}})$. But for $m\geqslant 2$ this is not the case. The following proposition gives sufficient conditions under which $\widetilde{J}=(\widetilde{f_{1}},\dots,\widetilde{f_{m}})$.

Proposition 3.4**.**

Let $\lambda$ be a weighted degree map on $B~{}(=R^{[n]})$ and $J=(f_{1},\dots,f_{m})$ an ideal of $B$. Suppose the following conditions hold:

1. (i)

$\widetilde{f_{i}}=f_{i}$* for $1\leqslant i\leqslant m-1$,* 2. (ii)

$\widetilde{f_{m}}$* is a non-zerodivisor in $B/(f_{1},\dots,f_{m-1})$.*

Then $\widetilde{J}=(f_{1},\dots,f_{m-1},\widetilde{f_{m}})$.

Proof.

Since for each $i$, $1\leqslant i\leqslant m$, $\widetilde{f_{i}}\in\widetilde{J}$, we clearly have $(f_{1},\dots,f_{m-1},\widetilde{f_{m}})\subseteq\widetilde{J}$.

Let $I=\{1,\dots,m\}$ and $0\neq g={\underset{i\in I}{\sum}}r_{i}f_{i}$, where $r_{i}\in B$ for all $i\in I$. Let $J_{m}:=(f_{1},\dots,f_{m-1})$ and for each $i\in I$, $\lambda(r_{i})=u_{i}$ and $\lambda(f_{i})=v_{i}$. For $f\in B$, by $f^{(l)}$ we will denote the $l$-degree homogeneous component of $f$ and for $0\leqslant s_{i}\leqslant u_{i}$, $0\leqslant t_{i}\leqslant v_{i}$, we define $x(i;s_{i};t_{i}):=r_{i}^{(u_{i}-s_{i})}f_{i}^{(v_{i}-t_{i})}$. Then $g={\underset{0\leqslant t_{i}\leqslant v_{i}}{\underset{0\leqslant s_{i}\leqslant u_{i}}{\underset{i\in I}{\sum}}}}x(i;s_{i};t_{i}).$

Let $M=\{x(i;s_{i};t_{i}):i\in I,0\leqslant s_{i}\leqslant u_{i},0\leqslant t_{i}\leqslant v_{i}\}$, $\mu_{0}:={\rm Max}\{\lambda(x):x\in M\}$ and $M_{0}:=\{x\in M:\lambda(x)=\mu_{0}\}$. Clearly, if $x(i;s_{i};t_{i})\in M_{0}$, then $s_{i}=0=t_{i}$.

If ${\underset{x\in{M_{0}}}{\sum}}x\neq 0$, then $\widetilde{g}={\underset{x\in{M_{0}}}{\sum}}x={\underset{M_{0}}{\sum}}\widetilde{r_{i}}\widetilde{f_{i}}\in(J_{m},\widetilde{f_{m}}).$ If ${\underset{x\in{M_{0}}}{\sum}}x=0$, then we will define $\mu_{1}$ and $M_{1}$ in the following way:

[TABLE]

If ${\underset{x\in{M_{1}}}{\sum}}x\neq 0$, then $\widetilde{g}={\underset{x\in{M_{1}}}{\sum}}x$. In this situation we note that for some $t_{i}>0$, $x(i;s_{i};t_{i})\in M_{1}$ will imply $i=m$. Clearly, $x(m;s_{m};t_{m})\in M_{1}$ for some $t_{m}>0$ only if $x(m;s_{m},0)\in M_{0}$ and hence $s_{m}=0$. Since ${\underset{x\in{M_{0}}}{\sum}}x=0$, by hypothesis (ii) of the proposition, $x(m;0;0)\in M_{0}$ only if $r_{m}^{(u_{m})}\in J_{m}$. So, for some $t_{m}>0$, $x(m;s_{m};t_{m})\in M_{1}$ only if x(m;s_{m};t_{m})~{}\big{(}=x(m;0;t_{m})=r_{m}^{(u_{m})}f_{m}^{(v_{m}-t_{m})}\big{)}\in J_{m}. Hence, ${\underset{x\in{M_{1}}}{\sum}}x\neq 0$ will imply $\widetilde{g}\in(J_{m},\widetilde{f_{m}})$.

If ${\underset{M_{1}}{\sum}}x=0$, then we will define $\mu_{2}$ and $M_{2}$ in the following way and proceed similarly.

[TABLE]

[TABLE]

Now, we consider the following statement:

[TABLE]

We have already observe that $P(1)$ is true. Assume that $P(1),P(2),\dots,P(n)$ is true for some $n\geqslant 2$. We will show that $P(n+1)$ is also true. Let ${\underset{x\in M_{j}}{\sum}}x=0$ for all $j\leqslant n$ and $x(m;s_{m}^{{}^{\prime}};t_{m}^{{}^{\prime}})\in M_{n+1}$ for some fixed $s_{m}^{{}^{\prime}},t_{m}^{{}^{\prime}}$ with $0\leqslant s_{m}^{{}^{\prime}}\leqslant u_{m}$ and $1\leqslant t_{m}^{{}^{\prime}}\leqslant v_{m}$. Then $x(m;s_{m}^{{}^{\prime}};0)\in M_{j_{0}}$ for some $0\leqslant j_{0}\leqslant n$. We observe the following:

$0={\underset{M_{j_{0}}}{\sum}}x(i;s_{i};t_{i})={\underset{i\neq m}{\underset{M_{j_{0}}}{\sum}}}x(i;s_{i};0)+{\underset{M_{j_{0}}}{\sum}}x(m;s_{m};0)+{\underset{t_{m}>0}{\underset{M_{j_{0}}}{\sum}}}x(m;s_{m};t_{m})$, 2. 2.

${\underset{i\neq m}{\underset{M_{j_{0}}}{\sum}}}x(i;s_{i};0)\in J_{m}$, 3. 3.

Since $P(j_{0})$ is true, ${\underset{t_{m}>0}{\underset{M_{j_{0}}}{\sum}}}x(m;s_{m};t_{m})\in J_{m}$, 4. 4.

Since $x(m;s_{m}^{{}^{\prime}};0)\in M_{j_{0}}$, $x(m;s_{m};0)\in M_{j_{0}}$ if and only if $s_{m}=s_{m}^{{}^{\prime}}$.

Hence $x(m,s_{m}^{{}^{\prime}},0)\in J_{m}$ and by hypothesis (ii), $r_{m}^{(u_{m}-s_{m}^{{}^{\prime}})}\in J_{m}$. So, $x(m;s_{m}^{{}^{\prime}};t_{m}^{{}^{\prime}})\in J_{m}$. Thus, $P(n+1)$ is true and hence $P(n)$ is true for all $n\in\mathbb{N}$.

Now, for $n\in\mathbb{N}$, if ${\underset{x\in M_{j}}{\sum}}x=0$ for all $j<n$ and ${\underset{x\in M_{n}}{\sum}}x\neq 0$, then $\widetilde{g}={\underset{x\in M_{n}}{\sum}}x\in(J_{m},\widetilde{f_{m}})$. If ${\underset{x\in M_{j}}{\sum}}x=0$ for all $j\leqslant n$, then we will consider $\mu_{n+1},M_{n+1}$ and repeat the arguments for $M_{n+1}$. Since $M$ is a finite set this process will stop after a finite number of steps. ∎

We now study the structure of the image ideals of an irreducible nice derivation on $R^{[2]}$, where $R$ is a UFD.

Theorem 3.5**.**

Let $R$ be a UFD and $B=R[X_{1},X_{2}]$. Let $D$ be an irreducible $R$-lnd on $B$ such that $D^{2}X_{1}=D^{2}X_{2}=0$. Let $A={\rm Ker}(D)$. Then, for each $j\in\mathbb{N},I_{j}={(DX_{1},DX_{2})}^{j}A$.

Proof.

By (ii) of Lemma 2.11, there exist $f_{1},f_{2}$ in $R$ such that the following hold:

• •

$DX_{1}=f_{1}$ and $DX_{2}=f_{2}$,

• •

$A=R[f_{2}X_{1}-f_{1}X_{2}]~{}(=R^{[1]})$.

First, we will show that $I_{1}=(f_{1},f_{2})A$. It is clear that $(f_{1},f_{2})A\subseteq I_{1}.$ For the converse, let $h\in B$ be such that $Dh\in A$. Let $u:=f_{2}X_{1}-f_{1}X_{2}$. Since $D$ is irreducible, $f_{1},f_{2}$ are mutually coprime and hence $u$ is irreducible in $B$. Let $Dh=P(u)\in R[u]$. Since $D(f_{1}h-X_{1}P(u))=0=D(f_{2}h-X_{2}P(u))$, we have

[TABLE]

Therefore, $uh=(f_{2}h)X_{1}-(f_{1}h)X_{2}=X_{1}Q_{2}(u)-X_{2}Q_{1}(u)$.

Let $c_{1},c_{2}\in R$ and $P_{1}(u),P_{2}(u)\in R[u]$ be such that $Q_{i}(u)=c_{i}+uP_{i}(u)$ for $i=1,2.$ Then $uh=u(X_{1}P_{2}(u)-X_{2}P_{1}(u))+(c_{2}X_{1}-c_{1}X_{2})$ and hence $u|(c_{2}X_{1}-c_{1}X_{2})$. Let $c_{2}X_{1}-c_{1}X_{2}=cu$ for some $c\in B$. Then $h=X_{1}P_{2}(u)-X_{2}P_{1}(u)+c$. It is clear that $c\in R$ and hence $Dh=P_{2}(u)f_{1}-P_{1}(u)f_{2}\in(f_{1},f_{2})A$.

Let $C=R[X_{1},X_{2},Z,S_{1},S_{2}]~{}(=R^{[5]})$ and $\lambda$ a weighted degree map on $C$ defined by $\lambda(X_{1})=\lambda(X_{2})=\lambda(S_{1})=\lambda(S_{2})=1$, $\lambda(Z)=0$. Suppose $J:=(Z-u,S_{1}-X_{1},S_{2}-X_{2})C$. By Proposition 3.4, $\widetilde{J}=(u,S_{1}-X_{1},S_{2}-X_{2})$. Since $\widetilde{J}$ is a prime ideal of $C$, by Corollary 3.3 $I_{j}=\underset{i_{1}+i_{2}=j}{\sum}D^{j}(X_{1}^{i_{1}}X_{2}^{i_{2}})A$. Hence, by Lemma 2.7, $I_{j}=\underset{i_{1}+i_{2}=j}{\sum}{f_{1}}^{i_{1}}{f_{2}}^{i_{2}}A$. Thus, $I_{j}={(f_{1},f_{2})}^{j}A$. ∎

The following results are immediate corollaries of Theorem 2.9 and Theorem 3.5.

Corollary 3.6**.**

Let $R$ be a UFD and $B=R[X,Y]$. Let $D$ be an irreducible $R$-lnd on $B$ such that $D^{2}X=D^{2}Y=0$. Let $A={\rm Ker}(D)$. Then, for each $n\geq 1$ the following hold.

1. (i)

If $D$ is fixed point free, then $I_{n}=A$; 2. (ii)

If $D$ is not fixed point free, then $I_{n}$ is generated by $n+1$ elements of $R$ with ${\rm grade}_{A}(I_{n})=2$ 111for a Noetherian ring $R$ and an ideal $I$ of $R$, ${\rm grade}_{A}(I)$ denotes the length of the maximal $R$-regular sequence contained in $I$., i.e., $I_{n}$ is not principal.

Theorem 3.7**.**

Let $R$ be a PID and $B=R[X,Y,Z]$. Let $D$ be an irreducible $R$-lnd on $B$ such that $D^{2}X=D^{2}Y=D^{2}Z=0$. Let $A={\rm Ker}(D)$. Then, for each $n\geqslant 1$ the following hold:

1. (i)

If $D$ is fixed point free, then $D$ has a slice and hence $I_{n}=A$; 2. (ii)

If $D$ is not fixed point free, then $I_{n}$ is generated by $n+1$ elements with ${\rm grade}_{A}(I_{n})=2$, i.e., $I_{n}$ is not principal.

Proof.

The proof follows from Theorem 2.12 and Theorem 3.5. ∎

Remark 3.8**.**

Part (i) of Theorem 3.7 holds even if $R$ is a Dedekind domain. In fact, in this case too $D$ has a slice (see [7, Proposition 3.8]). 2. 2.

Part (ii) of Theorem 3.7 need not be true when dimension of $R$ is more than one, even when $R$ is a UFD. Example 3.9, gives a nice $R$-lnd on $R^{[3]}$, for $R=k^{[2]}$, which is not fixed point free and hence has no slice in $B$. It follows from Lemma 2.8 that $I_{1}\neq A$. In fact, we will show that $I_{1}$ is generated by three elements and has grade $2$. 3. 3.

Part (ii) of Theorem 3.7 need not be true when $R$ is a Dedekind domain. In Example 3.10, we have given an example of a nice $R$-lnd on $R^{[3]}$, for $R=\mathbb{R}{[U,V]/(U^{2}+V^{2}-1)}$, which is not fixed point free, and $I_{1}$ is generated by three elements of $A$ with ${\rm grade}_{A}(I_{1})=1$.

Example 3.9**.**

Let $R=k[a,b]~{}(=k^{[2]})$ and define an $R$-lnd $D$ on $B=R[X,Y,Z]$ by setting $DX=a$, $DY=b$ and $DZ=bX-aY$. Clearly, $D$ is a nice $R$-lnd, which is not fixed point free. Let $A={\rm Ker}(D)$, $u=bX-aY$, $v=bZ-uY$ and $w=aZ-uX$. From [7, Example 3.10 ]), we have $A=R[u,v,w]$.

It is easy to see that there exists an $R$-algebra isomorphism from $R[u,v,w]$ onto $R[U,V,W]/(aV-bW-U^{2})$ sending $u,v,w$ to the respective images of $U,V,W$. Let $J=(a,b,u)A$. Then $A/J\cong{k[V,W]}~{}(=k^{[2]})$. The canonical projection map $\eta:B\longrightarrow B/(a,b)B$ induces a map $\sigma:A/J\longrightarrow B/(a,b)B$ with $\sigma(A/J)=\eta(A)\cong k$.

We will show $I_{1}=J$. Clearly, $J\subseteq I_{1}$. Let $f\in I_{1}\setminus\{0\}$. For each $g\in A$, let $\bar{g}$ denote the image of $g$ in $A/J$. Enough to show that $\bar{f}=0$ in $A/J$. Otherwise, there exists $P(V,W)~{}(\neq 0)$ such that $\bar{f}=P(V,W)$. Let $h\in B$ be such that $Dh=f$. Let $G:=ZDh-hDZ$. Then $G\in A$ and hence $\eta(G)~{}(=\eta(Zf))\in k$. So, $\eta(Z)\sigma(\bar{f})\in k$, i.e., $\eta(Z)\sigma(P(V,W))\in k$. We can choose $\lambda,\mu\in k$ be such that $P(\lambda,\mu)\in k\setminus\{0\}$. Then $\eta(Z)\in k$, which is a contradiction. Therefore, $I_{1}=(a,b,u)A$. **

Example 3.10**.**

Let $R=\dfrac{\mathbb{R}[U,V]}{(U^{2}+V^{2}-1)}$, $B=R[X,Y,Z]$ and $u,v$ denote the images of $U,V$ respectively in $R$. Define an $R$-lnd $D$ on $B$ by $DX=u$, $DY=v-1$ and $DZ=(1-v)X+uY$. Clearly, $D$ is not fixed point free and nice. Let $A:={\rm Ker}(D)$ and $f=(1-v)X+uY$, $g=(1+v)Y+uX$ and $h=2Z+fY-gX$. In [7, Example 3.3 ]), it has been proved that $A=R[f,g,h]$.

It is easy to see that there exists an $R$-algebra isomorphism from $R[f,g,h]$ onto $\dfrac{R[F,G]}{((1+v)F-uG)}[H]$ sending $f,g,h$ to the respective images of $F,G,H$. Let $J=(u,v-1,f)A$. Then $A/J\cong{\mathbb{R}[G,H]}~{}(=\mathbb{R}^{[2]})$. The canonical projection map $\eta:B\longrightarrow B/(u,v-1)B$ induces a map $\sigma:A/J\longrightarrow B/(u,v-1)B$ with $\sigma(A/J)=\eta(A)\cong\mathbb{R}[Y,Z-XY]$. Since $\eta(Z)\notin\mathbb{R}[Y,Z-XY]$, we can conclude as in Example 3.9 that $I_{1}=J=(u,v-1,f)$.

We now turn to the case of $1$-quasi-nice derivations on $R^{[2]}$, where $R$ is a UFD. If the derivation is fixed point free, it follows from Corollary 2.10 that each image ideal is the whole ring. The following result describes the plinth ideal of a strictly $1$-quasi-nice derivation $D$ on $R[X_{1},X_{2}]$, which is not fixed point free and $D^{2}X_{1}=0$ with $DX_{1}$ irreducible.

Proposition 3.11**.**

Let $R$ be a UFD and $B=R[X_{1},X_{2}]$. Let $D$ be an irreducible $R$-lnd on $B$ which is not fixed point free. If $D^{2}X_{1}=0$ and $DX_{1}$ is irreducible then the following are equivalent.

1. (I)

$I_{1}$* is principal.* 2. (II)

$I_{1}=(DX_{1})A$. 3. (III)

$D$* is strictly $1$-quasi-nice.*

Proof.

Since (II)$\implies$(I) is obvious, it is enough to show that (I)$\implies$(III) and (III)$\implies$(II).

(I) $\Rightarrow$ (III): Suppose that $D$ is a nice derivation. Then, there exists a coordinate system $\{U,V\}$ in $B$ such that $D^{2}U=D^{2}V=0$. By Lemma 2.11(ii), there exist $p,q\in R$, such that $DU=p$ and $DV=q$. Since $D$ is not fixed point free, by Theorem 2.9 $p,q$ form a regular sequence in $B$ and hence in $A$. So $I_{1}$ is not principal.

(III) $\Rightarrow$ (II): Assume (III) holds. By Lemma 2.11(i), there exist $b\in R$ and $f(X_{1})\in{R[X_{1}]}$ such that the following hold:

• •

$DX_{1}=b$ and $DX_{2}=-f^{\prime}(X_{1})~{}(:=\frac{df}{dX_{1}})$.

• •

$A=R[bX_{2}+f(X_{1})]~{}(=R^{[1]})$.

Without loss of generality, we can assume that $f$ has no constant term. Since $D$ is not fixed point free, it follows from Theorem 2.9 that $b,f^{\prime}$ form a regular sequence in $B$ and hence $b,f$ is a regular sequence in $B$. So, there exists $~{}u(X_{1}),~{}v(X_{1})$ in $R[X_{1}]$ such that $f(X_{1})=u(X_{1})+bv(X_{1})$ and none of the coefficients of $u(X_{1})$ is divisible by $b$. Since $f(0)=0$, ${\rm deg}_{X_{1}}u(X_{1})\geqslant 1$. Let $\sigma:B\longrightarrow{B/bB}$ be the natural projection map and set $\bar{A}:=\sigma(A),\bar{R}:=\sigma(R)$. Then $\bar{A}=\bar{R}[\sigma(u)]$.

We will show that $I_{1}\subseteq bA$. Otherwise, there exists $h\in{B}$ be such that $Dh\in{A\setminus bA}$. Since $bB\cap A=bA$, $\sigma(Dh)\neq 0$. Let $\sigma(Dh)=P(\sigma(u))\in{\bar{R}[\sigma(u)]}$. Since $X_{1}Dh-hDX_{1}\in{A}$, there exists $0\neq Q(\sigma(u))$ such that $\sigma(X_{1})P(\sigma(u))=Q(\sigma(u))$. Let $K$ be the field of fractions of the integral domain $R/bR$. Then, $K(\sigma(X_{1}))=K(\sigma(u))$. Then ${\rm deg}_{\sigma(X_{1})}(\sigma(u))=1$ and hence ${\rm deg}_{X_{1}}(u)=1$. Let $u(X_{1})=ax_{1}+c$ for some $a,c\in R$. Setting $X_{1}^{\prime}=X_{1}$ and $X_{2}^{\prime}=X_{2}+v(X_{1})$ we get $D^{2}X_{1}^{\prime}=D^{2}X_{2}^{\prime}=0$, contradicting the fact that $D$ is strictly $1$-quasi-nice. ∎

Now, we describe the higher image ideals of $D$ under the hypotheses of Proposition 3.11.

Theorem 3.12**.**

Let $R$ be a UFD and $B=R[X_{1},X_{2}]$. Let $D$ be a not fixed point free irreducible $R$-lnd on $B$ such that the following conditions hold:

1. (i)

$DX_{1}~{}(=b)\in R$* is irreducible,* 2. (ii)

$DX_{2}=-f^{\prime}(X_{1})$* and {\rm deg}_{X_{1}}\big{(}f(X_{1})\big{)}=d,* 3. (iii)

$D$* is strictly $1$-quasi-nice.*

Then, for each $j\geqslant 1$, $I_{j}=b^{m}A$, where $m={\rm Min}\{i_{1}+(d-1)i_{2}:i_{1},i_{2}\in\mathbb{N},i_{1}+di_{2}=j\}$.

Proof.

The case $j=1$ follows from Proposition 3.11. Let $C=R[X_{1},X_{2},Z,S]$ and $\lambda$ a weighted degree map on $C$ defined by $\lambda(X_{1})=1,\lambda(X_{2})=d,\lambda(S)=1$ and $\lambda(Z)=0$. Suppose $J:=(Z-bX_{2}-f(X_{1}),S-X_{1})C$. By Proposition 3.4, $\widetilde{J}=(bX_{2}+f(X_{1}),S-X_{1})$. Since $\widetilde{J}$ is a prime ideal of $C$, by Corollary 3.3, we have $I_{j}=\underset{i_{1}+di_{2}=j}{\sum}D^{j}(X_{1}^{i_{1}}X_{2}^{i_{2}})A$. Hence $I_{j}=\underset{i_{1}+di_{2}=j}{\sum}{(DX_{1})}^{i_{1}}{(D^{d}X_{2})}^{i_{2}}A$ by Lemma 2.7. Since $D^{d}X_{2}=D^{d-1}(f^{\prime}(X_{1}))$ and ${\rm deg}_{X_{1}}(f^{\prime}(X_{1}))=d-1$, $D^{d}X_{2}\in b^{d-1}R$. Thus, $I_{j}=\underset{i_{1}+di_{2}=j}{\sum}b^{i_{1}+(d-1)i_{2}}A$. ∎

The next proposition shows that we can remove the condition “$DX_{1}$ is irreducible” from Proposition 3.11 when $R$ is a DVR.

Proposition 3.13**.**

Let $(R,p)$ be a DVR with parameter $p$ and $B=R[X_{1},X_{2}]$. Let $D$ be an irreducible $R$-lnd on $B$, which is not fixed point free. If $D^{2}X_{1}=0$, then $I_{1}=(DX_{1})A$.

Proof.

By Lemma 2.11(i), there exist $b\in R$ and $f(X_{1})\in{R[X_{1}]}$ such that the following hold:

• •

$DX_{1}=b$ and $DX_{2}=-f^{\prime}(X_{1})~{}(:=\frac{df}{dX_{1}})$.

• •

$A=R[bX_{2}+f(X_{1})]~{}(=R^{[1]})$.

Without loss of generality, we can assume that $f$ has no constant term. Since $D$ is not fixed point free, we can assume that $b=p^{n}~{}(n\geqslant 1)$; and it follows from Theorem 2.9 that $b,f^{\prime}$ form a regular sequence in $B$ and hence $b,f$ is a regular sequence in $B$. Then $p,f$ form a regular sequence in $B$. So, there exists $u(X_{1}),~{}v(X_{1})$ in $R[X_{1}]$ such that $f(X_{1})=u(X_{1})+pv(X_{1})$ and none of the coefficients of $u(X_{1})$ is divisible by $p$. Since $f(0)=0$, ${\rm deg}_{X_{1}}u(X_{1})\geqslant 1$. Let $\sigma:B\longrightarrow{B/pB}$ be the natural projection map and set $\bar{A}:=\sigma(A),\bar{R}:=\sigma(R)$. Then $\bar{A}=\bar{R}[\sigma(u)]$.

Let $h\in B$ be such that $Dh\in I_{1}$. Then $Dh={p}^{m}g$ where $g\in A\setminus pA$ and $m\geqslant 0$. If $m\geqslant n$, then $Dh\in bA$. If possible, suppose that $m<n$. Since $XDh-bh\in A$, we have $X_{1}g-p^{n-m}h\in A$. Let $\sigma(g)=P(\sigma(u))\in{\bar{R}[\sigma(u)]}$. Since $\sigma(g)\neq 0$, there exists $Q(\sigma(u))~{}(\neq 0)$ such that $\sigma(X_{1})P(\sigma(u))=Q(\sigma(u))$. Then, $\frac{R}{pR}(\sigma(X_{1}))=\frac{R}{pR}(\sigma(u))$ and hence $\frac{R}{pR}[\sigma(X_{1})]=\frac{R}{pR}[\sigma(u)]$. So, $B\otimes_{R}\frac{R}{pR}=\frac{R}{pR}[\sigma(F)]^{[1]}$, where $F=bX_{2}+f(X_{1})$. Again, $B\otimes_{R}R[\frac{1}{p}]=R[\frac{1}{p}][F,X_{1}]$. Hence, $F$ is a residual coordinate in $R[X_{1},X_{2}]$ and hence a coordinate in $R[X_{1},X_{2}]$ (see [3, Theorem 3.2]). So, $D$ has a slice in $B$, contradicting the fact that $D$ is not fixed point free. ∎

Remark 3.14**.**

Let $R$ be a PID and $D$ an irreducible $R$-lnd on $B=R[X_{1},X_{2}]$, which is not fixed point free. Then $D$ is quasi-nice if and only if it is strictly $1$-quasi-nice.

The next proposition shows that Proposition 3.13 can be extended to a PID $R$ under some additional hypotheses.

Proposition 3.15**.**

Let $R$ be a PID, $B=R[X_{1},X_{2}]$ and $D$ an irreducible $R$-lnd on $B$ such that $DX_{1}=\prod_{i=1}^{n}{p_{i}}^{r_{i}}~{}(\in R)$, where each $p_{i}$ is a prime element in $R$. Suppose $J=\{i:1\leqslant i\leqslant n\text{ and the induced }R_{(p_{i})}\text{-lnd }D_{p_{i}}\text{ is not fixed point free}\}$. Then $I_{1}=\begin{cases}({\underset{i\in J}{\prod}}{p_{i}}^{r_{i}})A,&\text{if }J\neq\emptyset;\\ A,&\text{if }J=\emptyset.\end{cases}$

Proof.

Let ${I_{1}}({i})$ denote the plinth ideal of $D_{p_{i}}$ for $1\leqslant i\leqslant n$. Then $I_{1}(i)=I_{1}A_{(p_{i})}$, where $A_{(p_{i})}$ is the localisation of $A$ under the multiplicatively closed set $R\setminus p_{i}R$. By Proposition 3.13, $I_{1}A_{(p_{i})}={p_{i}}^{r_{i}}A_{(p_{i})}$ for each $i\in J$; and $I_{1}A_{(p_{i})}=A_{(p_{i})}$ if $i\notin J$. Let $\mathpzc{m}$ be a maximal ideal of $A$ and $\mathpzc{p}=\mathpzc{m}\cap R$. If $\mathpzc{p}=(0)$, then $I_{1}A{m}=A{m}$. If $p_{i}\notin\mathpzc{p}$ for any $i$, then also $I_{1}A{m}=A{m}$. If there exists $i$ such that $p_{i}\in\mathpzc{p}$, then $\mathpzc{p}=p_{i}R$ and hence $I_{1}A{m}~{}(=(I_{1}A{p})A{m})=\begin{cases}{p_{i}}^{r_{i}}A{m},&\text{if }i\in J;\\ A{m},&\text{if }i\notin J.\end{cases}$

Since $I_{1}={\underset{\mathpzc{m}\in{\rm maxSpec}(A)}{\bigcap}}I_{1}A{m}$, we have $I_{1}={\underset{i\in J}{\bigcap}}{p_{i}}^{r_{i}}A{m}$. Since $I_{1}\subseteq A$ and ${p_{i}}^{r_{i}}A{m}\cap A={p_{i}}^{r_{i}}A$ for each $i$, we have $I_{1}={\underset{i\in J}{\bigcap}}{p_{i}}^{r_{i}}A$. Since $A$ is a UFD, $I_{1}=({\underset{i\in J}{\prod}}{p_{i}}^{r_{i}})A$. ∎

Corollary 3.16**.**

Let $R$ be a PID, $B=R[X_{1},X_{2}]$ and $D$ an irreducible $R$-lnd on $B$. Suppose that $DX_{1}=\prod_{i=1}^{n}{p_{i}}^{r_{i}}~{}(\in R)$, where each $p_{i}$ is a prime element in $R$. If, for each $i$, $1\leqslant i\leqslant n$, the induced $R_{(p_{i})}$-lnd $D_{p_{i}}$ on $B\otimes_{R}R_{(p_{i})}$ is not fixed point free, then $I_{1}=(DX_{1})A$.

Now, we describe the higher image ideals of $D$ under the hypotheses of Proposition 3.15.

Theorem 3.17**.**

Let $R$ be a PID, $B=R[X_{1},X_{2}]$ and $D$ an irreducible $R$-lnd on $B$ such that $DX_{1}=\prod_{i=1}^{n}{p_{i}}^{r_{i}}~{}(\in R)$, where $p_{i}$ be a prime element of $R$ for each $i$. Suppose $DX_{2}=-f^{\prime}(X_{1})$, {\rm deg}_{X_{1}}\big{(}f(X_{1})\big{)}=d and

[TABLE]

Then, for each $j\geqslant 1$

[TABLE]

where $m={\rm Min}\{i_{1}+(d-1)i_{2}:i_{1},i_{2}\in\mathbb{N},i_{1}+di_{2}=j\}$.

Proof.

The case $j=1$ follows from Proposition 3.15 and rest of the proof is similar to the proof of theorem 3.12. ∎

The following example shows that in Corollary 3.16, the condition “for each $i$, $1\leqslant i\leqslant n$, the induced $R_{(p_{i})}$-lnd $D_{p_{i}}$ is not fixed point free” is necessary.

Example 3.18**.**

Let $R=k[t]~{}(=k^{[1]})$ and $B=R[X_{1},X_{2}]$. $D\in{\rm LND}_{R}(B)$ is defined by $DX_{1}=t(1-t)$ and $DX_{2}=-tX_{1}+1-t$. Clearly, $D$ is an irreducible quasi-nice $R$-derivation with ${\rm Ker}(D)=R[F]$, where $F=t(1-t)X_{2}+(t/2)X_{1}^{2}-X_{1}+tX_{1}$. Let $h=(1/2)X_{1}^{2}+(1-t)X_{2}$. Then $Dh=(1-t)^{2}$. So, $I_{1}\neq t(1-t)A$. By Proposition 3.17, $I_{1}=(1-t)A$. Here, we note that the induced $R_{(t)}$-lnd $D_{t}$ is fixed point free, whereas the $R_{(1-t)}$-lnd $D_{1-t}$ is not fixed point free. In fact, if we suppose that $D_{1-t}$ is fixed point free, then, by Theorem 2.9, $D_{1-t}$ would have a slice in $\widetilde{B}:=B\otimes_{R}R_{(1-t)}$ and hence the image of $F$ would be a coordinate in $\widetilde{B}$, which is not true as we can see that the image of $F$ in $\widetilde{B}/(1-t)\widetilde{B}$ is $X_{1}^{2}$, which is not a coordinate in $\widetilde{B}/(1-t)\widetilde{B}$.

Acknowledgement

The authors thank Dr. Prosenjit Das for his valuable comments while going through the earlier drafts and suggesting improvements. This research is supported by the Indo-Russia Project DST/INT/RUS/RSF/P-48/2021 with TPN 64842.