1 Introduction

1.1 Vector fields and trajectories

A holomorphic vector field X on \({\mathbb {C}}^2\) is a section of the tangent bundle of \({\mathbb {C}}^2\). Set coordinates x, y in \({\mathbb {C}}^{2}\), hence

$$\begin{aligned} X= P_{1}(x,y)\frac{\partial }{\partial x} + P_{2}(x,y)\,\frac{\partial }{\partial y}, \end{aligned}$$

with \(P_{1}\), \(P_{2}\) holomorphic functions. A point which is a common zero of \(P_{1}\) and \(P_{2}\) is called a singular point of X. Take \(z=(x,y)\) in \({\mathbb {C}}^2\) and the differential equation \(\varphi '_{z}(t)=X(\varphi _z(t))\) with \(\varphi _z(0)=z.\) The local solution \(\varphi _z\) can be extended by analytic continuation along paths from \(t=0\) in \({\mathbb {C}}\) to a maximal connected Riemann surface \(\pi _{z}: \Omega _z \rightarrow {\mathbb {C}}\), which is a Riemann domain over \({\mathbb {C}}\). The solution of X through z is \({\varphi }_z:\Omega _z \rightarrow {\mathbb {C}}^{2}\). The (complex) trajectory \(C_{z}\) of X through z is the Riemann surface \(\varphi _{z}(\Omega _z)\) immersed in \({\mathbb {C}}^{2}\).

If \(\Omega _z={\mathbb {C}}\), as domain in \({\mathbb {C}}\) (then, \(\pi _{z}\) is an analytic isomorphism), X is said to be complete on \(C_{z}\). In this case, \(C_{z}\) is uniformized by \({\mathbb {C}}\), and then analytically isomorphic to (= of type) \({\mathbb {C}}\) or \({\mathbb {C}}^{*}\) (maximum principle).

We say that X is complete if it is complete on \(C_{z}\) for any z. In this case, the flow \(\varphi : {\mathbb {C}}\times {\mathbb {C}}^2 \rightarrow {\mathbb {C}}^2\) of X, \((t,z)\mapsto \varphi (t,z)=\varphi _{z}(t)\), defines a holomorphic action of \(({\mathbb {C}},+)\) on \({\mathbb {C}}^2\) by analytic automorphisms and \(X={\frac{\partial }{\partial t} \varphi (t,z)}_{\mid t=0}\).

There are different types of flows:

  • \(\varphi \) is algebraic, if \(\varphi \) is a polynomial map.

  • \(\varphi \) is quasi-algebraic, if \({\varphi }_{t}\), for any \(t\in {\mathbb {C}}\), is a polynomial automorphism.

  • \(\varphi \) is proper, if the topological closure \({\overline{C}}_{z}\) of \(C_{z}\) in \({\mathbb {C}}^2\), for any z, is an analytic curve.

1.2 Vector fields and simply-connected trajectories

Let X be a polynomial vector field (then \(P_{1},\)\(P_{2}\in {\mathbb {C}}[x,y]\)) with simply-connected trajectories. We addressed in [2, 3] the problem of deciding if X is complete or not, under the assumption that there existed a nonconstant primitive polynomial \(P\in {\mathbb {C}}[x,y]\) such that

(i):

\(dP(X)=1\), or

(ii):

\(dP(X)=a\cdot P\) with \(a\ne {0}\).

Concretely, in both cases we proved that \(P=x\) and

$$\begin{aligned} \,\,\,\, X=[ax+d]\frac{\partial }{\partial x} + [b(x)y+c(x)]\frac{\partial }{\partial y} \end{aligned}$$
(1)

after a polynomial automorphism. In fact, it is obtained that \(a=0\) and \(d=1\) in case (i) [2], and \(a\ne 0\), \(d=0\), \(b(0)=0\) and \(c(0)\ne 0\) in case (ii) [3]. In particular, X is complete. There are several motivations to study these problems.

1. If X is a complete polynomial vector field on \({\mathbb {C}}^{2}\) with at most isolated zeros and simply connected trajectories (of type \({\mathbb {C}}\)), by classification of complete polynomial vector fields [1] X is as (1) after a polynomial change of coordinates. Moreover, after performing another polynomial change of coordinates, we can assume that \(d=0\), if \(a\ne {0}\); and \(d=1\), if \(a=0\). Note that these vector fields satisfy one of the following two properties with respect to \(P=x\): either \(dP(X)=ax\), if \(a\ne {0}\); or \(dP(X)=1\), if \(a=0\).

On the other hand, if X has no zeros and flow \(\varphi \), one also knows, according to [13, 16, Théorème 2] and [16, Théorème 4] (see also [2, Introduction]):

  • \(\text {Algebraic}\,\,\varphi \Rightarrow Quasi-algebraic \,\varphi \Rightarrow Proper \,\,\varphi \,\,\,and trajectories of type \,\, {\mathbb {C}}. \)

  • In these three situations for \(\varphi \), after a holomorphic automorphism, there is a polynomial \(P=x\) such that \(dP(X)=1\).

Then, it is natural to study if a reciprocal of Brunella’s result is valid:

If a polynomial vector fieldXon\({\mathbb {C}}^{2}\)with at most isolated zeros and simply-connected trajectories satisfies for a primitive polynomialPthat\(dP(X)=aP\), with\(a\ne {0}\), or\(dP(X)=1\), isXcomplete?

The affirmative answer to this question is given in [2] and [3], and it implies that such an X has no trajectories of type \({\mathbb {D}}\), and they are all of type \({\mathbb {C}}\).

Note that in case (i), the trajectories are always proper in \({\mathbb {C}}^{2}\) and X is the constant horizontal vector field after a holomorphic change of coordinates. However, in case (ii), the trajectories are not necessarily proper.

2. Let X be a polynomial vector field on \({\mathbb {C}}^{2}\), and the \({\mathbb {C}}\,\)-derivation \(D_{X}\) of \({\mathbb {C}}[x,y]\) associated to X:

$$\begin{aligned}&D_{X}:{\mathbb {C}}[x,y]\rightarrow {\mathbb {C}}[x,y]\\&\quad f \quad \mapsto X(f). \end{aligned}$$

A slices of \(D_{X}\) is a polynomial \(s\in {\mathbb {C}}[x,y]\) such that \(D_{X}(s)=1\). Questions about slices and derivations are related to Cancellation Problem in affine spaces [11, Chapter 10]. Moreover, the Jacobian Conjecture can be formulated as a problem in terms of derivations with a slice [11, Chapter 3]. Furthermore, this famous conjecture has been also formulated by Van de Essen, Wright and Zhao [17] in terms of derivations: it holds if the image of every derivation of \({\mathbb {C}}[x,y]\) with zero divergence and having a slice is a Mathieu subspace (we will recall this notion in the last section).

If \(D_{X}\) is surjective, \(1\in \) Im\((D_{X})\) and \(D_{X}\) has a slice. Surjective derivations in \({\mathbb {C}}[x,y]\) are studied in [5], and they are studied too in affine domains B over \({\mathbb {C}}\) with small dimension in [12]. An important property of a surjective derivation \(D_{X}\) of \({\mathbb {C}}[x,y]\) is that X has simply-connected trajectories [5, Proposition 1.6].

Motivated by these facts, we studied in [2] polynomial vector fields X on \({\mathbb {C}}^{2}\) with simply-connected trajectories such that \(D_{X}\) have a slice, and determined X, modulo a polynomial automorphism. Moreover, we applied this result to the study of surjective derivations. In particular we obtained in [2, Theorem 2] an affirmative answer to a conjecture stated by Cerveau in [6]: If \(D_{X}\)is surjective, then, up to a polynomial change of coordinates,

$$\begin{aligned} \,\,\,\, X=\frac{\partial }{\partial x} + by\frac{\partial }{\partial y} \end{aligned}$$
(2)

with\(b\in {\mathbb {C}}.\)

1.3 Main result

In what follows, we will assume that X has at most isolated zeros. We extend the above results to the case of a non-primitive polynomial P with the following theorem:

Theorem 1

LetXbe a polynomial vector field in\({\mathbb {C}}^{2}\). If there is\(P\in {\mathbb {C}}[x,y]\)such that (i) \(dP(X)=1\)or (ii) \(dP(X)= a\cdot P\), with\(a\in {\mathbb {C}}^{*}\); and the trajectories ofXare simply connected, up to a polynomial change of coordinates:

  1. (1.1)

    In case (i), \(P=x\), and

    $$\begin{aligned} X=\frac{\partial }{\partial x} + [b(x)y + c(x)] \frac{\partial }{\partial y}, \end{aligned}$$

    with\(b,\,c\in {\mathbb {C}}[x]\), and

  2. (1.2)

    In case (ii), \(P=x^{n}\), with\(n\in {\mathbb {N}}^{+}\), and

    $$\begin{aligned} X=dx\frac{\partial }{\partial x} + [b(x)y + c(x)] \frac{\partial }{\partial y}, \end{aligned}$$

    with\(d=a/n\), \(b,\,c\in {\mathbb {C}}[x]\), \(b(0)=0\)and\(c(0)\ne {0}\).

In particular,Xis complete and has all its trajectories of type\({\mathbb {C}}\).

2 Proof of Theorem 1

Note that in case (i): \(dP(X)=1\), P is always primitive. Theorem 1, after [2] and [3], follows by this proposition:

Proposition 1

LetXbe a polynomial vector field in\({\mathbb {C}}^{2}\). If there is a non-primitive\(P\in {\mathbb {C}}[x,y]\)such that\(dP(X)= a\cdot P\), with\(a\in {\mathbb {C}}^{*};\)and the trajectories ofXare simply-connected, up to a polynomial change of coordinates,PandXare as in (1.2) with\(n>1\).

Proof

By Stein’s factorization Theorem, we consider a primitive polynomial \(P_{0}\) such that \(P=h(P_{0})\) with h a polynomial in \({\mathbb {C}}[z]\) of degree \(n\ge {2}\).

Lemma 1

The polynomial \(h\in {\mathbb {C}}[z]\) has only one root

Proof

Assume that h(z) has k roots \({\alpha }_{i}\in {\mathbb {C}}\), for i from 1 to k, respectively of multiplicity \(m_{i}\in {\mathbb {N}}^{+}\). Then

$$h(z)=\lambda \prod _{i=1}^{k}{(z-{\alpha }_{i})}^{m_{i}},$$

with \(\lambda \in {\mathbb {C}}^{*}\), and with n equal to \(\sum _{i=1}^{k} m_{i}\). According to \(dP(X)=a P\), it follows that \(h'(P_{0})d{P}_{0}(X)=a h({P}_{0}).\) Then \(h({P}_{0})/h'(P_{0})\in {\mathbb {C}}[x,y]\), and thus \(h(z)/h'(z)\in {\mathbb {C}}[z]\). Because

$$\begin{aligned} \dfrac{h(z)}{h'(z)}=\dfrac{\prod _{i=1}^{k} (z-{\alpha }_{i})}{\sum _{i=1}^{k} m_{i} (\prod _{j\ne i} (z-\alpha _{j}))} \end{aligned}$$

it is clear that if \(k\ge {2}\), the polynomial \(\sum _{i=1}^{k} m_{i} (\prod _{j\ne i} (z-\alpha _{j}))\) has other roots different from \({\alpha }_{i}\) and we obtain a contradiction because \(h(z)/h'(z)\) is not a polynomial. \(\square \)

After Lemma 1, Proposition 1 follows easily from [3] .

Assume that \(h(z)=(z-{\alpha }_{1})^{n}\), with \(n\ge {2}\) (\(\lambda =1\)). Condition \(dP(X)=aP\) can be written as

$$n {(P_{0}-{\alpha }_{1})}^{n-1} dP_{0} (X)= a {(P_{0}-{\alpha }_{1})}^{n}.$$

Hence \(dP_{0} (X)=a/n (P_{0}-{\alpha }_{1})\). As \(dP_{0}=d(P_{0}-{\alpha }_{1})\), if \(Q=P_{0}-{\alpha }_{1}\), one obtains that Q is a primitive polynomial such that \(dQ(X)=a/n Q\). According to [3], we can assume that \(Q=x\) and

$$\begin{aligned} X=(a/n) x\frac{\partial }{\partial x} + [b(x)y + c(x)] \frac{\partial }{\partial y}, \end{aligned}$$

where \(b,\,c\in {\mathbb {C}}[x]\) with \(b(0)=0\) and \(c(0)\ne {0}\) after a polynomial automorphism. Then \(P= x^{n}\), and Proposition 1 is proved.

\(\square \)

3 An application and some questions on the image of derivations

First, we give an application of Theorem 1.

Theorem 2

Let X be a polynomial vector field in \(\mathbb {C}^{2}\) whose trajectories are all simply connected. If there is a nonconstant \(P\in \mathbb {C}[x,y]\) and \(n\in \mathbb {N}^{+}\) satisfying:

  1. a)

    \(X^{n}(P)= 0\) and \(X^{n-1}(P)\ne 0\), or

  2. b)

    \(X^{n+1}(P)= a\cdot X^{n}(P)\) for \(a\in \mathbb {C}^{*}\)

Then, X is complete.

Proof

In case a), suppose first that \(n=1\); then \(X(P)=0\) and as P is not a constant polynomial, according to [15], after a polynomial change of coordinates, \(X=\partial /\partial x\) which is complete. If \(n \ge 2\), take \(\bar{P}:=X^{n-1}(P)\). If \(\bar{P}\) is not constant, then \(X(\bar{P})=0\) and as before [15] implies that after a polynomial change of coordinates \(X=\partial /\partial x\), that is complete. Otherwise, if \(\bar{P}=\lambda \in \mathbb {C}^{*}\), it is enough to apply Theorem 1 to \(\tilde{X}:=(1/\lambda )X\) and P if \(n=2\), and to \(\tilde{X}\) and \(\tilde{P}:=X^{n-2}(P)\) if \(n>2\), because \(\tilde{X}(P)=1\) and \(\tilde{X}(\tilde{P})=1\) respectively, to conclude that \(\tilde{X}\) , and then X, are complete.

In case b), we note that neither \(X^{n}(P)\) nor \(X^{n+1}(P)\) equals a nonzero constant. Denote \(\hat{P}:= X^{n}(P)\). Since \(X(\hat{P})=a\cdot \hat{P}\), Theorem 1 implies that X is complete.\(\square \)

Consider the natural domain \(\Omega \), containing \(\{0\}\times {\mathbb {C}}^{2}\), in \({\mathbb {C}}\times {\mathbb {C}}^2\), where the local flow \(\varphi : \Omega \rightarrow {\mathbb {C}}^2\) of X is defined as \(\varphi (t,x,y)=\varphi _z(t)\) (see Sect. 1).

Take \(P\in {\mathbb {C}}[x,y]\). Then \(P(\varphi (t,x,y))\) can be expressed according to the Lie series as:

$$\begin{aligned} P(\varphi (t,x,y))= P(x,y) + X^2(P)(x,y)\frac{t^2}{2!}+X^3(P)(x,y)\frac{t^3}{3!}+\cdots \end{aligned}$$

Theorem 2 implies the following:

Corollary 1

LetXbe a polynomial vector field in\({\mathbb {C}}^{2}\)whose trajectories are all simply connected. Consider the local flow\(\varphi : \Omega \rightarrow {\mathbb {C}}^2\)ofX. If there is a nonconstant\(P\in {\mathbb {C}}[x,y]\)and\(n\in {\mathbb {N}}^{+}\)satisfying:

  1. (a)
    $$\begin{aligned} P(\varphi (t,x,y))= P+ X(P)t+X^2(P)\frac{t^2}{2!}+\dots +X^{n-1}(P)\frac{t^{n-1}}{(n-1)!},\,\, \end{aligned}$$

    with \(X^{n-1}(P)\ne {0}\), or

  2. (b)
    $$\begin{aligned} \begin{aligned} P(\varphi (t,x,y))&= P+ X(P)t+X^2(P)\frac{t^2}{2!}+\dots +X^{n-1}(P)\frac{t^{n-1}}{(n-1)!}+ \\&\quad +\,a^{-n} X^n(P)\left[ e^{at}-(1+at+\frac{{(at)}^2}{2!}+\dots + \frac{{(at)}^{n-1}}{(n-1)!})\right] , \end{aligned} \end{aligned}$$

    with\(a\in {\mathbb {C}}^{*}\).

Then,\(\varphi \)can be extended to\({\mathbb {C}}\times {\mathbb {C}}^{2}\)andXis complete.

Remark 1

The Lie series is related to r-inflection points of the vector field with respect to curves of degree r, where \(r=\deg (P)\). See [7, 9]; see [10] for an extension of this idea for codimension one foliations.

Finally, we want to point out some ideas related with Mathieu subspaces, recently introduced by Zhao in [18], and the study of derivations of \({\mathbb {C}}[x,y]\). Let us first recall the following

Definition 1

Let R be a commutative k-algebra and M a k-subspace of R. Then M is a Mathieu subspace of R if the following condition holds: if \(a\in R\) is such that \(a^m\in M\) for all \(m\ge 1\), then for any \(b\in R\) there exists and \(N\in {\mathbb {N}}\) such that \(ba^m\in M\) for all \(m\ge N\).

In our situation, \(R={\mathbb {C}}[x,y]\). It is clear that the image of a derivation Im\((D_{X})\) is a \({\mathbb {C}}\)-subspace of \({\mathbb {C}}[x,y]\). However, Im\((D_{X})\) is not necessarily a Mathieu subspace. Indeed, Zhao proved in [18, Lemma 4,5] that if M is a Mathieu subspace of R and \(1\in M\), then \(M=R\). The following example, taken from [17, Example 2.4],

$$\begin{aligned} X=\frac{\partial }{\partial x} - y^2\frac{\partial }{\partial y} \end{aligned}$$

shows that Im\((D_{X})\) is not a Mathieu subspace, as \(1\in \text {Im}\,(D_{X})\) but \(D_{X}\) is not surjective (\(y\notin \text {Im}\,(D_{X})\)).

Recall that \(D_{X}\) is locally finite if for any \(f\in {\mathbb {C}}[x,y]\), the \({\mathbb {C}}\)-vector space spanned by \(\{X^{n}f\,|\,n\ge {0}\}\) has finite dimension. If \(D_{X}\) is locally finite, Im\((D_{X})\) is a Mathieu subspace [17, Theorem 3.1]. In particular, if \(D_{X}\) is locally finite and has a slice, X is surjective, and then of the form (2) after a polynomial automorphism [17, Proposition 3.2].

It would be interesting to determine polynomial vector fieldsXwith all its trajectories simply-connected and such that Im\((D_{X})\)is a Mathieu subspace of\({\mathbb {C}}[x,y]\), up to a polynomial automorphism.

A polynomial vector field X in \({\mathbb {C}}^{2}\) determines a locally finite derivation \(D_{X}\) if and only if its flow \(\varphi : {\mathbb {C}}\times {\mathbb {C}}^2 \rightarrow {\mathbb {C}}^2\) is quasi-algebraic [8, Theorem 3.1]. In particular, X is complete.

As we mentioned before, in [17, Theorem 4.3] it is proved that the Jacobian conjecture in \({\mathbb {C}}^2\) holds if and only if for every derivation D of \({\mathbb {C}}[x,y]\) with zero divergence (where if \(D= p \frac{\partial }{\partial x} + q \frac{\partial }{\partial y}\), \(\text{ Div }\, (D)= \frac{\partial p}{\partial x} + \frac{\partial q}{\partial y}\)) and having a slice, it holds that Im(D) is a Mathieu subspace.

Recall that the jacobian conjecture in \({\mathbb {C}}^2\) affirms that a polynomial map \(F:=(F_1,F_2): {\mathbb {C}}^2 \longrightarrow {\mathbb {C}}^2\) with \(\det J_{F}=1\) is an automorphism. We call a pair of polynomials \(F_1\), \(F_2\in {\mathbb {C}}[x,y]\) with \(\det J_{(F_1,F_2)}=1\) a Jacobian pair. In a joint article with Muciño [4], the authors proved that the invertibility of the map given by the jacobian pair is equivalent to the fact that one of the vector fields

$$\begin{aligned} \frac{\partial }{\partial F_2}:=\frac{\partial F_1}{\partial y}\frac{\partial }{\partial x} - \frac{\partial F_1}{\partial x} \frac{\partial }{\partial y}\,\quad \text {or}\quad \frac{\partial }{\partial F_1}:=\frac{\partial F_2}{\partial y} \frac{\partial }{\partial x} - \frac{\partial F_2}{\partial x}\frac{\partial }{\partial y} \end{aligned}$$

is complete. Hence, the condition that the image of a derivation D with zero divergence and having a slice is a Mathieu subspace is equivalent to the fact that the polynomial vector field inducing D is complete.

Thus, we note that for derivations there is a close relation between having as image a Mathieu subspace and being induced by a complete polynomial vector field. We do not know examples of a derivation\(D_{X}\)determined by a non complete vector fieldXfor which Im\((D_{X})\)is a Mathieu subspace of\({\mathbb {C}}[x,y]\).

Example 1

Let us consider

$$\begin{aligned} X=\frac{\partial }{\partial x} + xy\frac{\partial }{\partial y}. \end{aligned}$$

X is complete with flow \(\varphi (t,x_{0},y_{0})=(t+x_{0},y_{0}e^{\frac{t^2}{2}+x_{0}t})\). Then \(D_{X}\) is not locally finite. Its trajectories are simply-connected. Moreover, \(D_{X}\) has x as slice.

Im\((D_{X})\) is not a Mathieu subspace. Otherwise, as \(1\in \) Im\((D_{X})\), \(D_{X}\) should be surjective as observed above. But this is not possible because \(y\notin \text {Im}\,(D_{X})\), as a simple calculation shows: writing \(Q=a_0(x)+a_1(x)y+\cdots +a_n(n)y^n\), with \(a_i(x)\in {\mathbb {C}}[x]\), \(D_{X}(Q)=y\) implies

$$\begin{aligned} y= a_0^{\prime }(x)+a_1^{\prime }(x)y+\cdots +a_n(x)^{\prime }y^n + xy [a_1(x)+2a_2(x)y+\cdots +na_n(x)y^{n-1}] \end{aligned}$$

hence it should hold that

$$\begin{aligned} a_1^{\prime }(x)+ xa_1(x)=1, \end{aligned}$$

which is impossible.

Example 2

[14, Theorem 2.6] Let us consider

$$\begin{aligned} X=bx^{a}y^{b-1}\frac{\partial }{\partial x} - ax^{a-1}y^{b}\frac{\partial }{\partial y}. \end{aligned}$$

with \(a,b\ge 1\). Then, Im\((D_{X})\) is a Mathieu subspace if and only if \(a=b\).

Note that, when \(a=b\ge 2\), X is a vector field with non isolated singularities, trajectories of type \({\mathbb {C}}^{*}\), and whose image is a Mathieu subspace.