ON THE LND CONJECTURE

Abstract

Let k be a field of characteristic zero and $k^{[n]}$ the polynomial algebra in n variables over k. The LND conjecture concerning the images of locally nilpotent derivations arose from the Jacobian conjecture. We give a positive answer to the LND conjecture in several cases. More precisely, we prove that the images of rank-one locally nilpotent derivations of $k^{[n]}$ acting on principal ideals are MZ-subspaces for any $n\geq 2$, and that the images of a large class of locally nilpotent derivations of $k^{[3]}$ (including all rank-two and homogeneous rank-three locally nilpotent derivations) acting on principal ideals are MZ-subspaces.

1 Introduction

Throughout the paper, k is a field of characteristic zero and $k^{[n]}:= k[x_{1},x_{2},\ldots ,x_{n}]$ the polynomial algebra in n variables over k. The well-known Jacobian conjecture asserts that any polynomial map of $k^{[n]}$ with nonzero constant determinant is invertible (see [1, 11]).

In the last decade, it was found that the Jacobian conjecture is closely related to the study of the images of derivations and differential operators of polynomial algebras. More precisely, the Jacobian conjecture is associated with the problem of whether the images of some derivations or differential operators are MZ-subspaces (see [14–17]). Now we recall the notion of MZ-subspaces, which is a natural generalisation of ideals.

Definition 1.1 [18, 19].

Let A be a commutative k-algebra. A k-subspace M of A is called a Mathieu–Zhao subspace (MZ-subspace for short) if for each pair $f,g \in A$ with $f^{m} \in M$ , for all $m \geq 1$ , we have $gf^m \in M$ , for all $m \gg 0$ , that is, for all sufficiently large m.

The notion of MZ-subspace was first introduced by Zhao in [18] (named after Mathieu [8]) when he investigated the Jacobian conjecture, and originally named Mathieu subspace. Later, van den Essen proposed the change of name to Mathieu–Zhao subspace (see [12]).

In 2011, van den Essen et al. [15] asked the following question (called the EWZ problem): for a k-derivation D of $k^{[2]}$ with divergence zero, is the image Im $D:=D(k^{[2]})$ an MZ-subspace of $k^{[2]}$ ? They proved that the two-dimensional Jacobian conjecture is equivalent to the assertion that the EWZ problem has an affirmative answer when $1 \in $ Im D.

In 2017, the first author [9] proved that the EWZ problem has a negative answer in general by investigating monomial preserving derivations, and subsequently van den Essen and the first author [13] completely solved the problem of whether the images of monomial preserving derivations of $k^{[n]}$ are MZ-subspaces.

Since locally nilpotent derivations of $k^{[n]}$ are of divergence zero, it is natural to ask if the images of locally nilpotent derivations of $k^{[n]}$ are MZ-subspaces. In fact, Zhao [20] formulated the following LND conjecture.

Conjecture 1.2 (LND conjecture).

Let D be a locally nilpotent derivation of $k^{[n]}$ . Then for any ideal I of $k^{[n]}$ , the image $D(I)$ is an MZ-subspace of $k^{[n]}$ .

A more general conjecture called the LNED conjecture was proposed in [20]. It asserts that the same conclusion holds even if we replace $k^{[n]}$ by any commutative k-algebra, and includes a variant of the conjecture where derivations are replaced by $\epsilon $ -derivations. However, even the LND conjecture is still open for any $n\geq 2$ .

We focus on the LND conjecture. In 2017, Zhao [20] proved that the LND conjecture holds for $k^{[1]}$ . Van den Essen et al. [15] proved that the image Im $D:=D(k^{[2]})$ of any locally finite derivation D of $k^{[2]}$ is an MZ-subspace using the technique of Newton polytopes. Liu, Zeng and the first author [7] proved that the LND conjecture holds for principal ideals and some other ideals of $k^{[2]}$ . In 2020, Liu and the first author [6] proved that the images of linear locally nilpotent derivations of $k^{[3]}$ are MZ-subspaces using the technique of integrals. In 2021, Liu and the first author [10] proved that Im D is an MZ-subspace of $k^{[3]}$ for any rank-two or homogeneous rank-three locally nilpotent derivation D of $k^{[3]}$ , by improving some results on local slice constructions. In conclusion, the LND conjecture was only verified for some special cases and remains unsolved in general for any dimension $n\geq 2$ .

In this paper, we prove in Section 2 that the LND conjecture holds for rank-one locally nilpotent derivations acting on principal ideals of $k^{[n]}$ for any $n\geq 2$ using the technique of integrals. For higher-rank locally nilpotent derivations, the problem becomes more complicated. We prove by the technique of local slices in Section 3 that the images of a large class of locally nilpotent derivations of $k^{[3]}$ (including all rank-two and homogeneous rank-three locally nilpotent derivations) acting on principal ideals are MZ-subspaces; the crucial point is that for these derivations, we reduce the discussion successfully to a local slice. Whether the LND conjecture can be treated in this way for general locally nilpotent derivations deserves study, but this question seems hard.

2 Images of rank-one locally nilpotent derivations acting on principal ideals

We begin by recalling some basic concepts and properties of locally nilpotent derivations.

Let A be a commutative k-algebra over a field k of characteristic zero. A k-derivation D of A is a k-linear map $D : A\rightarrow A$ that satisfies $D(ab) = D(a)b+aD(b)$ for all $a,b \in A$ . A derivation D is locally nilpotent if for each $a \in A$ , there exists some $m_a \ge 1$ such that $D^{m_a}(a) = 0$ . For a derivation D, the kernel of D is $\ker D:=\{a\in A\mid D(a)=0\}$ . For brevity, we use the following notation:

1. (1) Der $_k(A$ ) = $\{D \mid D$ is a k-derivation of $A\}$ ;

2. (2) LND(A) = $\{D \in $ Der $_k(A$ ) $\mid \ D$ is locally nilpotent $\}$ .

When A is a k-domain and $0\neq a\in A$ , a derivation $aD_1$ is locally nilpotent if and only if $a\in \ker D_1$ and $D_1$ is locally nilpotent (see [11, Corollary 1.3.34]).

Now we focus on the LND conjecture (Conjecture 1.2). It is a standard technique (using Lefschetz’s principle) to reduce the base field k to the complex field $\mathbb {C}$ . For completeness, we give a proof here.

Lemma 2.1.

1. (1) If the LND conjecture holds over $\mathbb {C}$ , then it holds over any field k of characteristic zero.

2. (2) If the LND conjecture holds for principal ideals over $\mathbb {C},$ then it holds for principal ideals over k.

Proof. (1) Suppose the LND conjecture holds over $\mathbb {C}$ . Let $k^{[n]}=k[x_1,\ldots ,x_n]$ , $D\in \mathrm{LND}(k^{[n]})$ and $I=(u_1,\ldots ,u_s)$ be any ideal of $k^{[n]}$ . Given $f, g\in k^{[n]}$ with $f^m\in D(I)$ , for all $m\geq 1$ , it suffices to show that $gf^m\in D(I)$ , for all $m\gg 0$ .

Let L be the extension field over $\mathbb {Q}$ generated by the coefficients of f, g, $D(x_i)$ for $1\leq i\leq n$ , and $u_j$ for $1\leq j\leq s$ . By Lefschetz’s principle (see [11, Lemma 1.1.13]), L can be seen as a subfield of $\mathbb {C}$ . Then D induces a locally nilpotent derivation $\bar {D}$ on $\mathbb {C}^{[n]}:=\mathbb {C}[x_1,\ldots ,x_n]$ by $\bar {D}(x_i)=D(x_i)$ for each i. Denote by $\bar {I}$ the ideal of $\mathbb {C}^{[n]}$ generated by $u_1, u_2,\ldots ,u_s$ . Both D and $\bar {D}$ restrict to $L^{[n]}:=L[x_1,\ldots ,x_n]$ .

One may verify that $D(I)\cap L^{[n]}=\bar {D}(\bar {I}) \cap L^{[n]}$ . In fact, for any $u\in D(I)\cap L^{[n]}$ , there exist $w_i=\sum _{\alpha _i}c_{\alpha _{i}} x^{\alpha _i}\in k^{[n]}, i=1,2,\ldots ,s$ , such that

$$ \begin{align*}u=D\bigg(\sum_{i=1}^su_iw_i\bigg)=\sum_{i=1}^sD(u_i)w_i+\sum_{i=1}^su_iD(w_i)\in L^{[n]},\end{align*} $$

and this equality is equivalent to a system of linear equations over k with solution $c_{\alpha _i}$ . Noticing that all coefficients of the system are in L, we may take the solution $c_{\alpha _i}$ in L, which implies that $u\in \bar {D}(\bar {I})\cap L^{[n]}$ . Hence, $D(I)\cap L^{[n]}\subseteq \bar {D}(\bar {I}) \cap L^{[n]}$ . Similarly, $\bar {D}(\bar {I}) \cap L^{[n]}\subseteq D(I)\cap L^{[n]}$ .

Since $f^m\in D(I)\cap L^{[n]}\subseteq \bar {D}(\bar {I})$ , for all $m\geq 1$ , and $\bar {D}(\bar {I})$ is an MZ-subspace, we have $gf^m\in \bar {D}(\bar {I})\cap L^{[n]}\subseteq D(I)$ , for all $m\gg 0,$ as desired.

(2) The conclusion follows from the fact that when I is a principal ideal of $k^{[n]}$ , $\bar {I}$ is a principal ideal of $\mathbb {C}^{[n]}$ .

Lemma 2.2. Let B be a commutative k-domain, $D = aD_{1} \in \mathrm{LND}(B)$ and I an ideal of B. If $D_{1}(I)$ is an MZ-subspace of B, then so is $D(I)$ .

Proof. Note that $D(I)=aD_1(I)=D_1(aI)\subseteq D_1(I)$ . Let $f\in B$ be such that $f^m\in D(I)~(\subseteq D_1(I))$ , for all $m\geq 1$ . Certainly, $f\in D(I)=aD_1(I)$ , say $f=af_1$ where $f_1\in D_1(I)$ . For any $g\in B$ , since $D_1(I)$ is an MZ-subspace, $(gf_1)f^m\in D_1(I)$ , for all $m\gg 0$ , and thus $gf^{m+1}=agf_1f^m\in aD_1(I)= D(I)$ , for all $m\gg 0.$ Hence, $D(I)$ is an MZ-subspace.

Lemma 2.3. Let $A[x]$ be a polynomial algebra in one variable over a k-domain A, I an ideal of $A[x]$ and $D=\partial _{x}$ . If $f(x) \in D(I)$ , then

$$ \begin{align*} x\int_{0}^{1}f(xt) \,{d}t \in A + I. \end{align*} $$

Proof. Write $f(x)=\sum _{i}a_ix^i$ , where $a_i\in A$ . Since $\int x^i\,{d}x={x^{i+1}}/({i+1})=x\int _0^1(xt)^i\,{d}t$ for any $i\in \mathbb {N}$ , we have

$$ \begin{align*}w:=\int f(x)\,{d}x=\int \sum_ia_ix^i\,{d}x=x\int_0^1\sum_ia_i(xt)^i\,{d}t=x\int_0^1 f(xt)\,{d}t. \end{align*} $$

Since $D(w)=f(x)\in D(I)$ , it follows that $w\in \ker D+I$ , that is, $x\int _0^1 f(xt)\,{d}t\in A+I.$

We also need the following property of moments of polynomial functions.

Lemma 2.4 [3].

Let $a \ne b \in k$ . If $f \in k^{[1]} = k[t]$ is such that $\int _{a}^{b}f(t)^m \,{d}t=0$ , for all $m \ge 1$ , then $f = 0$ .

An n-tuple $f_1, f_2, \ldots , f_n$ in $k^{[n]}= k[x_{1}, \ldots , x_{n}]$ is called a system of coordinates if $k^{[n]}= k[f_{1}, f_{2}, \ldots , f_{n}]$ . For a derivation D of $k^{[n]}$ , the rank of D, denoted by rank(D), is the least integer $r \geq 0$ for which there exists a system of coordinates $f_1, f_2, \ldots , f_n$ of B such that $k[f_{r+1},f_{r+2},\ldots ,f_{n}] \subseteq \ker D$ . When $k = \overline {k}$ , for a polynomial $f \in k^{[n]}$ , we denote by $V(f)$ the set of zero points of f in the affine space $\mathbb {A}_k^n$ .

Theorem 2.5. Let I be a principal ideal of $k^{[n]} = k[x_{1},x_{2},\ldots ,x_{n}]$ and $D \in $ LND( $k^{[n]}$ ) with rank $\,D = 1$ . Then $DI$ is an MZ-subspace of $k^{[n]}$ .

Proof. It is well known that D is conjugate by an automorphism of $k^{[n]}$ to $a\partial _{x_{1}}$ , where $a \in k[x_{2},\ldots ,x_{n}]$ . So we may assume that $D=a\partial _{x_{1}}$ . From Lemma 2.1, we may assume that $k=\mathbb {C}$ , and from Lemma 2.2 that $D=\partial _{x_{1}}$ .

Let $I = (u(x_{1},\ldots ,x_{n}))$ be a nonzero principal ideal of $k^{[n]}$ . We may assume that $u(x_{1},\ldots ,x_{n})\notin k[x_2,\ldots ,x_n]$ for otherwise, $\partial _{x_1}(I)=I$ is an MZ-subspace.

If $u(x_{1},\ldots ,x_{n})$ has a divisor in $k[x_{2},\ldots ,x_{n}] \backslash k$ , say $u(x_{1},\ldots ,x_{n}) = b\tilde {u}(x_{1},\ldots ,x_{n})$ , $b\in k[x_{2},\ldots ,x_{n}] \backslash k$ , then

$$ \begin{align*}\begin{aligned} \partial _{x_{1}}I = \partial _{x_{1}}(u(x_{1},\ldots,x_{n})k^{[n]}) = b \partial _{x_{1}}(\tilde{u}(x_{1},\ldots,x_{n})k^{[n]}). \end{aligned}\end{align*} $$

So we may assume $u(x_{1},\ldots ,x_{n})$ has no divisor in $k[x_{2},\ldots ,x_{n}] \backslash k$ due to Lemma 2.2.

1. (i) Suppose that $u(x_{1},\ldots ,x_{n})$ has at least two distinct roots over $\overline {k(x_{2},\ldots ,x_{n})}$ , say a and b. For any $h \in \partial _{x_{1}}I$ , there exists $v(x_{1},\ldots ,x_{n}) \in k^{[n]}$ such that $h = \partial _{x_{1}}(u(x_{1},\ldots ,x_{n})v(x_{1},\ldots ,x_{n}))$ , and then

   $$ \begin{align*} \int_{a}^{b}\!\! h \,{d} x_{1} \!\! = \int_{a}^{b}\!\! \partial_{x_{1}}(u(x_{1},\ldots,x_{n})v(x_{1},\ldots,x_{n})) \,{d}x_{1} = u(x_{1},\ldots,x_{n})v(x_{1},\ldots,x_{n})\bigg|_{a}^{b}\!\! = 0. \end{align*} $$

   Let $f \in k^{[n]}$ be such that $f^m \in \partial _{x_{1}}I$ , for all $m \ge 1$ . Then, $\int _{a}^{b} f^m \,{d}x_{1} = 0$ , for all  $m \ge 1$ , and by Lemma 2.4, we have $f = 0$ . Then for any $g \in k^{[n]}$ , we have ${0 = gf^m \in \partial _{x_{1}}I}$ , for all $m \ge 1$ . Thus, $\partial _{x_{1}}I$ is an MZ-subspace of $k^{[n]}$ .

2. (ii) Suppose that $u(x_{1},\ldots ,x_{n})$ does not have a nonzero root over $\overline {k(x_{2},\ldots ,x_{n})}$ . Then, $u(x_{1},\ldots ,x_{n}) = cx_{1}^n$ , where $c\in k\backslash \{0\}$ . In this case, $\partial _{x_{1}}I = x_{1}^{n-1}k^{[n]}$ is an ideal and thus an MZ-subspace.

Following items (i) and (ii), we now assume that $u(x_{1},\ldots ,x_{n})$ has a unique root over $\overline {k(x_{2},\ldots ,x_{n})}$ and the root is nonzero. Then one may verify that

$$ \begin{align*}u(x_{1},\ldots,x_{n}) = (c_{1}x_{1} - c_{2})^p, \end{align*} $$

where $c_{1}, c_{2}\in k[x_{2},\ldots ,x_{n}] \backslash 0$ and $c_1, c_2$ are coprime. When $c_1\in k$ , we have $\partial _{x_1}I=(x_1-c_1^{-1}c_2)^{p-1}k^{[n]}$ is an MZ-subspace, so we assume that $c_1\notin k$ .

Let $c_3$ be the square-free part of $c_1$ . Then for $f\in k^{[n]}\backslash \{0\}$ ,

$$ \begin{align*}\tilde{s}(f):=\min\{i\in \mathbb{Z}\mid c_3^if\in k[c_1x_1,x_2,\ldots,x_n]\}\end{align*} $$

exists.

Claim 1. If $f \in k^{[n]}$ such that $f^m \in \partial _{x_{1}}I$ , for all $m \ge 1$ , then $\tilde {s}(f)<0$ .

Suppose that $f \in k^{[n]}$ such that $f^m \in \partial _{x_{1}}I$ , for all $m \ge 1$ , and that $t:=\tilde {s}(f)\geq 0$ . Let $\tilde {f}=c_3^tf.$ Then, $\tilde {f}^m\in \partial _{x_1}I,$ for all $m\geq 1$ , and $\tilde {s}(\tilde {f})=0.$ As $\tilde {s}(\tilde {f})\leq 0$ , $\tilde {f}$ has the form $\sum _ia_i(c_1x_1)^i$ , where $a_i\in k[x_2,\ldots ,x_n].$

By Lemma 2.3, since $\tilde {f}^m \in \partial _{x_{1}}I$ , for all $m \ge 1$ ,

$$ \begin{align*}x_1\int_{0}^{1}\tilde{f}(x_{1}t)^m \,{d}t \in k[x_{2},\ldots,x_{n}] + I= k[x_{2},\ldots,x_{n}] +(c_{1} x_{1} - c_{2} )^pk^{[n]}.\end{align*} $$

Taking $x_1 = {c_{2} }/{c_{1} }$ , we obtain $c_{2} \int _{0}^{1}\tilde {f}({c_{2} }/{c_{1} }t)^m \,{d}t \in c_{1} k[x_{2},\ldots ,x_{n}],$ that is,

$$ \begin{align*}c_{2} \int_{0}^{1}\bigg(\sum_ia_i (c_2 t)^i\bigg)^m \,{d}t \in c_{1} k[x_{2},\ldots,x_{n}].\end{align*} $$

For any $(\xi _2,\ldots ,\xi _n) \in V(c_{1} ) \backslash V(c_{2} )$ ,

$$ \begin{align*}\int_{0}^{1}\bigg(\sum_ia_i(\xi_{2},\ldots,\xi_{n})(c_2(\xi_{2},\ldots,\xi_{n})t)^i\bigg)^m \,{d}t = 0 \quad\mbox{for all } m\geq 1.\end{align*} $$

By Lemma 2.4, $\sum _ia_i(\xi _{2},\ldots ,\xi _{n})(c_2(\xi _{2},\ldots ,\xi _{n})t)^i = 0$ , and thus $a_i(\xi _{2}, \ldots ,\xi _{n}) = 0$ for all $i.$

Hence, $V(c_{1} ) \backslash V(c_{2} ) \subseteq V(a_{i} )$ . From $\gcd (c_1,c_2)=1$ , it follows that $V(c_{1} ) \backslash V(c_{2} )$ is dense in $V(c_{1} )$ for the Zariski topology. Therefore,

$$ \begin{align*}V(c_{1} ) = \overline{V(c_{1} ) \backslash V(c_{2} )} \subseteq V(a_{i} ).\end{align*} $$

Consequently, $c_3\mid a_i$ for all i. This contradicts $\tilde {s}(\tilde {f})=0,$ so Claim 1 has been proved.

Claim 2. If $h \in k^{[n]}$ such that $x_1h\in k[c_1x_1,x_2,\ldots ,x_n]$ and $(c_{1}x_1 - c_{2} )^{p-1}\mid h$ , then ${h \in \partial _{x_{1}}I}$ .

Suppose that $h \in k^{[n]}$ such that $x_1h\in k[c_1x_1,x_2,\ldots ,x_n]$ . Then,

$$ \begin{align*} h\in c_1k[c_1x_1,x_2,\ldots,x_n]=c_1k[c_1x_1-c_2,x_2,\ldots,x_n]. \end{align*} $$

Suppose in addition that $(c_{1}x_1 - c_{2} )^{p-1}\mid h$ . Then,

$$ \begin{align*} h\in c_1(c_{1}x_1 - c_{2} )^{p-1}k[c_1x_1-c_2,x_2,\ldots,x_n]. \end{align*} $$

Note that when $j\geq p-1,$

$$ \begin{align*} c_1(c_1x_1 - c_2 )^j = \partial_{x_1}\bigg(\frac{1}{j+1}(c_1x_1 - c_2 )^{j+1}\bigg) \in \partial_{x_1}I. \end{align*} $$

It follows that $h \in \partial _{x_{1}}I$ . Thus, Claim 2 has been proved.

Finally, given $f,g \in k^{[n]}$ with $f^m \in \partial _{x_{1}}I$ , for all $m \ge 1$ , we obtain by Claim 1 that $x_1gf^m\in k[c_1x_1,x_2,\ldots ,x_n]$ , for all $ m\geq \max \{\tilde {s}(x_1g),1\}$ . Combining this with

$$ \begin{align*} f \in \partial_{x_{1}}I \subseteq (c_{1}x_1 - c_{2} )^{p-1}k^{[n]}, \end{align*} $$

we obtain by Claim 2 that $gf^m \in \partial _{x_{1}}I$ , for all $m \gg 0$ . So $\partial _{x_{1}}I$ is an MZ-subspace of $k^{[n]}$ .

3 Images of locally nilpotent derivations acting on ideals of $k^{[3]}$

To study the LND conjecture for higher-rank locally nilpotent derivations, we recall the notion and basic properties of local slices for locally nilpotent derivations (see [4, Sections 1.1, 2.2] for details).

Let B be a commutative k-domain over a field k of characteristic zero, $D \in $ Der $_k(B)$ and $A= \ker D$ . An element $r \in B$ with $Dr \neq 0$ and $D^2r = 0$ is called a local slice of D. Any nonzero $D \in $ LND(B) has a local slice. A local slice r of D is called minimal if $A[r]$ is a maximal element in $\{A[r']\mid r'$ is a local slice of $D\}$ with respect to inclusion. Denote by $\min D$ the set of all minimal local slices of D.

If B satisfies the ascending chain condition (ACC) on principal ideals, then any nonzero $D \in $ LND(B) has a minimal local slice. Take any $r \in \min (D)$ . Let $B_0 = A[r]$ and $s = {r}/{Dr}$ . Then we have $B_{Dr} = A_{Dr}[s] = (B_0)_{Dr}$ . Thus, for any $f \in B$ , there exists some $i \in \mathbb {N}$ such that $(Dr)^if \in B_0$ . For $f \in B \backslash \{0\}$ , define

$$ \begin{align*} s(f) := \min \{i \in \mathbb{Z} \mid (Dr)^if \in B_0\}. \end{align*} $$

One may verify easily the following lemma.

Lemma 3.1. Set $D_0 := D|_{B_0}$ . Then,

$$ \begin{align*} s(f) \leq 0 &\Longleftrightarrow f \in B_0, \\ s(f) < 0 &\Longleftrightarrow f \in (Dr)B_0 = \mathrm{Im}D_0, \end{align*} $$

and

$$ \begin{align*} s(fg) &\leq s(f) + s(g), \\ s(f + g) &\leq \max \{s(f),s(g)\}, \\ s((Dr)^jf) &= s(f) - j,\quad j \in \mathbb{N}. \end{align*} $$

Remark 3.2. Let B be a commutative k-domain. Given $D \in $ LND(B), we consider the condition:

1. (*) If $f \in B$ is such that $f^m \in \ D(B)$ , for all $m \geq 1$ , then $s(f^l) < 0$ for some $l \geq 1$ .

It was proved in [10, Theorems 3.9, 5.3] that most locally nilpotent derivations of $k^{[3]}$ satisfy the condition ( $\ast $ ); more precisely, all rank-one, rank-two and homogeneous rank-three derivations in LND $(k^{[3]})$ satisfy condition $(*)$ .

We will prove that the LND conjecture holds for all locally nilpotent derivations of $k^{[3]}$ with the condition $(\ast )$ . The first step is to show that, for these derivations, we may reduce the LND conjecture to a local slice.

Theorem 3.3. Suppose that B is a commutative k-domain and I is an ideal of B. Let $0\neq D \in $ LND(B) which satisfies the condition $(\ast )$ and let $A = \ker D$ . Take a minimal local slice r of D and let $B_0 = A[r]$ , $D_0:= D|_{B_0}$ and $I_0:= I \cap B_0$ . If $D_0I_0$ is an MZ-subspace of $B_0$ , then $DI$ is an MZ-subspace of B.

Proof. Note that $D_0$ acts as $(Dr)\partial _r$ on $B_0=A[r]$ and $\text {Im}\, D_0=(Dr)B_0$ . First, we show that if $h \in \text {Im}\, D_0$ , then $h \in DI$ if and only if $h \in D_0I_0$ . The if part follows from the fact that $D_0I_0 \subseteq DI$ . Write $h = D_0(w), ~w \in B_0$ . Since $h \in DI$ , we have $h = D_0(w) = D(v)$ for some $v\in I$ . It follows that

$$ \begin{align*} v - w \in \ker D = A \subseteq B_0 \end{align*} $$

and thus $v \in B_0 \cap I = I_0$ . Hence, $h \in D_0I_0$ . The only if part is proved.

Given $f, g \in B$ with $f^m \in DI$ , for all $m \geq 1$ , it suffices to show that $gf^m \in DI$ , for all $m \gg 0$ .

Since $f^m \in DI\subseteq \text {Im}\,D$ , for all $m \geq 1$ and $D \in $ LND(B) satisfies condition ( $\ast $ ), $s(f^{n_0}) < 0$ for some $n_0 \geq 1$ and thus $s((f^{n_0})^m) < 0$ , for all $m\geq 1$ . It follows that $(f^{n_0})^m \in (Dr)B_0$ = Im $D_0$ . Then by the discussion in the first paragraph, $(f^{n_0})^m \in D_0I_0$ , for all $m\geq 1$ .

For $g \in B$ , take an $n_1 \geq 1$ such that $s(gf^{n_0n_1}) < 0$ , and so $gf^{n_0n_1} \in (Dr)B_0 \subseteq B_0$ . Then, because $D_0I_0$ is an MZ-subspace of $B_0$ , we have

$$ \begin{align*} g(f^{n_0})^{m + n_1} = (gf^{n_0n_1})(f^{n_0})^m \in D_0I_0 \subseteq DI, \quad\mbox{for all } m \gg 0. \end{align*} $$

Therefore, there exists $m_0 \geq 1$ such that $g(f^{n_0})^m \in DI$ , for all $m \geq m_0$ .

For each $j=1,2,\ldots ,n_0-1$ , replacing g by $gf^j$ , we see that there exists $m_j \geq 1$ such that $gf^j(f^{n_0})^m \in DI$ , for all $m \geq m_j$ . Let $\overline {m}:= \max \{m_j\mid 0\leq j\leq n_0-1\}$ . Then,

$$ \begin{align*} gf^{n_0m + j} = gf^j(f^{n_0})^m \in DI, \quad\mbox{for all } m \geq \overline{m}\mbox{ and } 0\leq j\leq n_0 - 1. \end{align*} $$

Then $gf^m \in DI$ , for all $m \geq n_0\overline {m}.$ Therefore, Im D is an MZ-subspace of B.

A gcd-domain is a commutative domain for which any two elements have a greatest common divisor (gcd), that is, there is a unique minimal principal ideal containing the ideal generated by the two elements. In a gcd-domain, the gcd of finitely many elements exists and is unique up to a unit factor. A polynomial ring $A[x]$ over a gcd-domain A is still a gcd-domain (see [5, page 172]). A gcd-domain is a Schreier domain (see [2, Theorem 2.4]). A Schreier domain is an integrally closed commutative domain of which every element f is primal: whenever $f \mid gh$ , f can be written as $f = f_1 f_2$ such that $f_1 \mid g$ and $f_2 \mid h$ .

Theorem 3.4. Suppose that B is a k-gcd-domain and $I=Bu$ is a principal ideal of B. Let $0\neq D \in $ LND(B) and $A=\ker D$ . Suppose that r is a minimal local slice of D, and let $B_0 = A[r]$ . If $\gcd (u, Dr)=1$ , then $I_0: = I \cap B_0$ is also a principal ideal of $B_0$ .

Proof. Suppose that $\gcd (u, Dr) = 1$ . The case $u = 0$ is trivial, so assume that $u\neq 0$ . Since $A = \ker D$ is factorially closed, it does not matter if we take the gcd of elements of A over A or over B: the result will be the same. Furthermore, A is a gcd-domain as well. Consequently, $B_0 = A[r]$ is also a gcd-domain. From $\gcd (u, Dr) = 1$ , it follows that $t:= s(u)\geq 0$ . As $(Dr)^tu \in B_0$ , we can write $(Dr)^tu =\sum _i\tilde {a}_ir^i$ with $\tilde {a}_i\in A$ . Since $(Dr) ^t \in A$ as well, we infer that $f := \gcd ((Dr)^t, \tilde {a}_0, \tilde {a}_1, \tilde {a}_2,\ldots )\in A.$ We can write $\hat {u} := f^{-1}(Dr)^tu =\sum _ia_ir^i$ with $a_i=f^{-1}\tilde {a}_i\in A.$

We show that $\gcd (\hat {u}, Dr) = 1$ over $B_0 = A[r]$ . Suppose that $d \mid \gcd (\hat {u}, Dr)$ over $A[r]$ . As $Dr \in A$ and A is factorially closed, it follows that $d\in A$ . So $d\mid \hat {u}$ over $A[r]$ gives $d \mid a_i$ over A for all i, and therefore $df \mid \tilde {a}_i$ over A for all i. Since $d \mid Dr$ and $\gcd (u, Dr) = 1$ over B, we infer that $\gcd (u, d) = 1$ over B. So by way of primality of d, we infer from $d \mid \hat {u} = (f^{-1}(Dr)^t)u$ over B that $d\mid f^{-1}(Dr)^t$ and $df \mid (Dr)^t$ over B. As $(Dr)^t \in A$ and A is factorially closed, $df \mid (Dr)^t$ over A. So $df \mid \gcd ((Dr)^t, \tilde {a}_0, \tilde {a}_1, \tilde {a}_2,\ldots ) = f$ and $d \mid 1$ over A. Hence, $\gcd (\hat {u}, Dr) = 1$ over $B_0$ .

Let $b\in I_0 = I \cap B_0$ , and write $b = uv$ with $v \in B$ . Take $t'\in \mathbb {N}$ such that $t'\geq s(v).$ Then, $(Dr)^tu =\hat {u}f$ and $(Dr)^{t'}v\in B_0$ , so $(Dr)^{t+t'}b\in \hat {u}fB_0$ . Hence, $\hat {u}\mid (Dr)^{t+t'}b$ over $B_0$ . Since $\gcd (\hat {u}, Dr) = 1$ over $B_0$ , we infer by way of primality of $\hat {u}$ and induction that $\hat {u} \mid b$ over $B_0$ . So $I_0\subseteq B_0\hat {u}$ . However, $B_0\hat {u}\subseteq B_0$ and $B_0\hat {u} = B_0(f^{-1}(Dr)^t)u \subseteq I.$ So $I_0 = B_0\hat {u}$ is a principal ideal of $B_0$ .

Theorem 3.5. Suppose that $B = k^{[3]} = k[x,y,z]$ , I is a principal ideal of B, and $D \in \mathrm{LND}(B)$ satisfies the condition $(\ast )$ . Then $DI$ is an MZ-subspace of B.

Proof. By Miyanishi’s theorem [4, Theorem 5.1], there exist $F, G \in B$ such that $A: = \ker D = k[F,G]\cong k^{[2]}$ . Take a minimal local slice r of D and let $B_0 = A[r]$ and $D_0:= D|_{B_0}$ . Then, $B_0= k[F,G,r]\cong k^{[3]}$ and $D_0$ acts as $(Dr)\partial _{r}$ on $B_0=A[r]$ , where $Dr\in A$ .

Let $I =Bu$ and $I_0= I \cap B_0$ . If u has a divisor in $A\backslash k$ , say $u = au_1$ for some $a\in k[F,G] \backslash k$ and $u_1\in B$ . Then,

$$ \begin{align*} DI = D(uB) = D(au_1B) =aD(u_1B). \end{align*} $$

By Lemma 2.2, we may assume that u has no divisor in $A\backslash k$ . Since A is factorially closed and $Dr\in A$ , we have $\gcd (u, Dr)\in A$ . Hence, $\gcd (u, Dr)=1$ . Then by Theorem 3.4, $I_0$ is a principal ideal of $B_0$ . It follows by Theorem 2.5 that $D_0I_0$ is an MZ-subspace of $B_0$ . Therefore, from Theorem 3.3, $DI$ is an MZ-subspace of B.