On simple Shamsuddin derivations in two variables

Abstract

We study the subgroup of $k$-automorphisms of $k[x,y]$ which commute with a simple derivation $d$ of $k[x,y]$. We prove, for instance, that this subgroup is trivial when $d$ is a shamsuddin simple derivation. in the general case of simple derivations, we obtain properties for the elements of this subgroup.

dynamical degree; isotropy group; Shamsuddin derivations; simple derivations

Introduction

Let $k$ be an algebraically closed field of zero characteristic and $k[x,y]$ be the ring of polynomials over $k$ in two variables.

A $k$-derivation$d:k[x,y]\to k[x,y]$ of $k[x,y]$ is a $k$-linear map such that

$d(ab)=d(a)b+ad(b),$

for any $a,b\in k[x,y]$. We denote by ${\mathrm{Der}}_{\mathrm{k}}(k[x,y])$ the set of all $k$-derivations of $k[x,y]$. Let $d\in {\mathrm{Der}}_{\mathrm{k}}(k[x,y])$. An ideal $I$ of $k[x,y]$ is called $d$-stable if $d(I)\subset I$. For example, the ideals $0$ and $k[x,y]$ are always $d$-stable. If these are the only $d$-stable ideals, we say $k[x,y]$ is $d$-simple. Even in the case of two variable polynomials, only a few examples of simple derivations are known (see, for instance, Brumatti et al. ( ²⁰⁰³, Saraiva ²⁰¹², Nowicki ²⁰⁰⁸, Baltazar and Pan ²⁰¹⁵, Kour and Maloo ²⁰¹³, Lequain ²⁰¹¹)).

We denote by $\mathrm{Aut}(k[x,y])$ the group of $k$-automorphisms of $k[x,y]$. Let $\mathrm{Aut}(k[x,y])$ act on ${\mathrm{Der}}_{\mathrm{k}}(k[x,y])$ by:

$(\rho ,D)\mapsto {\rho }^{-1}\circ D\circ \rho ={\rho }^{-1}D\rho .$

Fix a derivation $d\in {\mathrm{Der}}_{\mathrm{k}}(k[x,y])$. The isotropy subgroup,with respect to this group action, is defined as

$\mathrm{Aut}{(k[x,y])}_d:=\{\rho \in \mathrm{Aut}(k[x,y])/\rho d=d\rho \}.$

We are interested in the following question proposed by I.Pan (see Baltazar ( ²⁰¹⁴)):

Conjecture 1 If $d$ is a simple derivation of $k\mathit{}\mathrm{[}x\mathrm{,}y\mathrm{]}$, then $\mathrm{Aut}\mathit{}{\mathrm{(}k\mathit{}\mathrm{[}x\mathrm{,}y\mathrm{]}\mathrm{)}}_d$ is finite.

Initially, in Section 2, we prove Theorem 6, which shows that the conjecture is true for a family of derivations, namely Shamsuddin derivations. For this purpose, we use a theorem due to Shamsuddin ( ¹⁹⁷⁷) (see also ^{Nowicki 1994}, Theorem 13.2.1.) that gives a necessary and sufficient condition for a derivation to be extended to $R[t]$, with $t$ an indeterminate, and preserving simplicity. We observe Shamsuddin derivations is a reasonable class of objects. For instance, they have been previously used by Lequain ( ²⁰¹¹) in order to establish a conjecture about the Weyl algebra $\mathrm{\backslash mathbb}{A}_n$ over $k$.

In Section 3, to understand the isotropy of a simple derivation of $k[x,y]$, we give necessary conditions for an automorphism to belong to the isotropy of a simple derivation. We prove in Proposition 7 that if such an automorphism has a fixed point, then it is the identity. Next, we present the definition of dynamical degree of a polynomial map and prove in Corollary 9 that for $k=\mathrm{\backslash mathbb}C$, the elements of $\mathrm{Aut}{(\mathrm{\backslash mathbb}C[x,y])}_d$, with $d$ a simple derivation, have dynamical degree $1$. More precisely, the condition that the dynamical degree is greater than 1corresponds to exponential growth of the degree under iteration, and this may be viewed as a complexity of the automorphism in the isotropy (see Friedland and Milnor ( ¹⁹⁸⁹)).

SHAMSUDDIN DERIVATIONS

The aim of this section is study the isotropy group of a Shamsuddin derivation in $k[x,y]$. In Nowicki ( ¹⁹⁹⁴), there are numerous examples of these derivations and a criterion for determining the simplicity. Furthermore, Lequain ( ²⁰⁰⁸) introduced an algorithm for determining whether a Shamsuddin derivation is simple. We begin with an example that shows that the isotropy of an arbitrary derivation can be quite complicated.

Example 1 Let $d\mathrm{=}{\mathrm{\partial }}_x\mathrm{\in }{\mathrm{Der}}_{\mathrm{k}}\mathit{}\mathrm{(}k\mathit{}\mathrm{[}x\mathrm{,}y\mathrm{]}\mathrm{)}$ and $\rho \mathrm{\in }\mathrm{Aut}\mathit{}{\mathrm{(}k\mathit{}\mathrm{[}x\mathrm{,}y\mathrm{]}\mathrm{)}}_d$. Note that $d$ is not a simple derivation. Indeed, for any $u\mathit{}\mathrm{(}y\mathrm{)}\mathrm{\in }k\mathit{}\mathrm{[}y\mathrm{]}$, the ideal generated by $u\mathit{}\mathrm{(}x\mathrm{)}$ is always invariant. Consider

$\rho (x)=f(x,y)=a_0(x)+a_1(x)y+\mathrm{\dots }+a_t(x)y^t$

$\rho (y)=g(x,y)=b_0(x)+b_1(x)y+\mathrm{\dots }+b_s(x)y^s.$

Since $\rho \mathrm{\in }\mathrm{Aut}\mathit{}{\mathrm{(}k\mathit{}\mathrm{[}x\mathrm{,}y\mathrm{]}\mathrm{)}}_d$, we obtain two conditions:

1) $\mathrm{\rho }\mathrm{}\mathrm{(}\mathrm{d}\mathrm{}\mathrm{(}\mathrm{x}\mathrm{)}\mathrm{)}\mathrm{=}\mathrm{d}\mathrm{}\mathrm{(}\mathrm{\rho }\mathrm{}\mathrm{(}\mathrm{x}\mathrm{)}\mathrm{)}\mathrm{.}$

Thus,

$1=d(a_0(x)+a_1(x)y+\mathrm{\dots }+a_t(x)y^t)=d(a_0(x))+d(a_1(x))y+\mathrm{\dots }+d(a_t(x))y^t.$

Then, $d\mathit{}\mathrm{(}{a}_{\mathrm{0}}\mathit{}\mathrm{(}x\mathrm{)}\mathrm{)}\mathrm{=}\mathrm{1}$ and $d\mathit{}\mathrm{(}{a}_j\mathit{}\mathrm{(}x\mathrm{)}\mathrm{)}\mathrm{=}\mathrm{0}$, $j\mathrm{=}\mathrm{1}\mathrm{,}\mathrm{\dots }\mathrm{,}t$. We conclude that $\rho \mathit{}\mathrm{(}x\mathrm{)}$ is of the type

$\rho (x)=x+c_0+c_{1}y+\mathrm{\dots }+c_t{y}^t,c_i\in k.$

2) $\mathrm{\rho }\mathrm{}\mathrm{(}\mathrm{d}\mathrm{}\mathrm{(}\mathrm{y}\mathrm{)}\mathrm{)}\mathrm{=}\mathrm{d}\mathrm{}\mathrm{(}\mathrm{\rho }\mathrm{}\mathrm{(}\mathrm{y}\mathrm{)}\mathrm{)}\mathrm{.}$

Analogously,

$0=d(b_0(x)+b_1(x)y+\mathrm{\dots }+b_s(x)y^s)=d(b_0(x))+d(b_1(x))y+\mathrm{\dots }+d(b_s(x))y^s,$

that is, $b_i\mathit{}\mathrm{(}x\mathrm{)}\mathrm{=}{d}_i$, with $d_i\mathrm{\in }k$. We also infer that $\rho \mathit{}\mathrm{(}y\mathrm{)}$ is of the type

$\rho (y)=d_0+d_{1}y+\mathrm{\dots }+d_s{y}^s,d_i\in k.$

Thus, $\mathrm{Aut}\mathit{}{\mathrm{(}k\mathit{}\mathrm{[}x\mathrm{,}y\mathrm{]}\mathrm{)}}_d$ contains the affine automorphisms

$(x+uy+r,uy+s),$

with $u\mathrm{,}r\mathrm{,}s\mathrm{\in }k$. In particular, the isotropy group $\mathrm{Aut}\mathit{}{\mathrm{(}k\mathit{}\mathrm{[}x\mathrm{,}y\mathrm{]}\mathrm{)}}_d$ of a derivation which is not simple can be infinite. Indeed, $\mathrm{Aut}\mathit{}{\mathrm{(}k\mathit{}\mathrm{[}x\mathrm{,}y\mathrm{]}\mathrm{)}}_d$ contains all automorphisms of the type $\mathrm{(}x\mathrm{+}p\mathit{}\mathrm{(}y\mathrm{)}\mathrm{,}y\mathrm{)}$, with $p\mathit{}\mathrm{(}y\mathrm{)}\mathrm{\in }k\mathit{}\mathrm{[}y\mathrm{]}$. Actually, these are all the elements of $\mathrm{Aut}\mathit{}{\mathrm{(}k\mathit{}\mathrm{[}x\mathrm{,}y\mathrm{]}\mathrm{)}}_d$. By conditions $\mathrm{1}$ and $\mathrm{2}$,

$\rho =(x+p(y),q(y)),$

with $p\mathit{}\mathrm{(}y\mathrm{)}\mathrm{,}q\mathit{}\mathrm{(}y\mathrm{)}\mathrm{\in }k\mathit{}\mathrm{[}y\mathrm{]}$. Since $\rho$ is an automorphism, the determinant of the Jacobian matrix must be nonzero. Thus, $\mathrm{|}{J}_{\rho }\mathrm{|}\mathrm{=}{q}^{\mathrm{\prime }}\mathit{}\mathrm{(}y\mathrm{)}\mathrm{=}c$, $c\mathrm{\in }{k}^{\mathrm{*}}$. Therefore, $\rho \mathrm{=}\mathrm{(}x\mathrm{+}p\mathit{}\mathrm{(}y\mathrm{)}\mathrm{,}a\mathit{}y\mathrm{+}c\mathrm{)}$, with $p\mathit{}\mathrm{(}y\mathrm{)}\mathrm{\in }k\mathit{}\mathrm{[}y\mathrm{]}$ and $a\mathrm{,}b\mathrm{\in }k$. Consequently, $\mathrm{Aut}\mathit{}{\mathrm{(}k\mathit{}\mathrm{[}x\mathrm{,}y\mathrm{]}\mathrm{)}}_d$ is not finite and its first component has elements with any degree.

The following is a well known lemma.

Lemma 2 Let $R$ be a commutative ring, $d$ a derivation of $R$, and $h\mathit{}\mathrm{(}t\mathrm{)}\mathrm{\in }R\mathit{}\mathrm{[}t\mathrm{]}$, with $t$ an indeterminate. Then, we can also extend $d$ to a unique derivation $\stackrel{\mathrm{~}}{d}$ of $R\mathit{}\mathrm{[}t\mathrm{]}$ such that $\stackrel{\mathrm{~}}{d}\mathit{}\mathrm{(}t\mathrm{)}\mathrm{=}h\mathit{}\mathrm{(}t\mathrm{)}$.

We also use the following result of Shamsuddin ( ¹⁹⁷⁷).

Theorem 3 Let $R$ be a ring containing $\mathrm{\backslash mathbb}\mathit{}Q$ and let $d$ be a simple derivation of $R$. Extend the derivation $d$ to a derivation $\stackrel{\mathrm{~}}{d}$ of the polynomial ring $R\mathit{}\mathrm{[}t\mathrm{]}$ by setting $\stackrel{\mathrm{~}}{d}\mathit{}\mathrm{(}t\mathrm{)}\mathrm{=}a\mathit{}t\mathrm{+}b$ where $a\mathrm{,}b\mathrm{\in }R$. Then the following two conditions are equivalent:

$(1)$ $\stackrel{\mathrm{~}}{d}$ is a simple derivation.

$(2)$ There exist no elements $r\mathrm{\in }R$ such that $d\mathit{}\mathrm{(}r\mathrm{)}\mathrm{=}a\mathit{}r\mathrm{+}b$.

Proof See (^{Nowicki 1994}, Theorem 13.2.1.) for a detailed proof.

A derivation $d$ of $k[x,y]$ is said to be a Shamsuddin derivation if $d$ is of the form

$d={\partial }_x+(a(x)y+b(x)){\partial }_y,$

where $a(x),b(x)\in k[x]$.

Example 4 Let $d$ be a derivation of $k\mathit{}\mathrm{[}x\mathrm{,}y\mathrm{]}$ as follows

$d={\partial }_x+(xy+1){\partial }_y.$

Writing $R\mathrm{=}k\mathit{}\mathrm{[}x\mathrm{]}$, we know that $R$ is ${\mathrm{\partial }}_x$-simple and, taking $a\mathrm{=}x$ and $b\mathrm{=}\mathrm{1}$, we are exactly in the conditions of Theorem 3. Thus, we know that $d$ is simple if, and only if, there exist no elements $r\mathrm{\in }R$ such that ${\mathrm{\partial }}_x\mathit{}\mathrm{(}r\mathrm{)}\mathrm{=}x\mathit{}r\mathrm{+}\mathrm{1}$; but the right hand side of the equivalence is satisfied by the degree of $r$. Therefore, by Theorem 3, $d$ is a simple derivation of $k\mathit{}\mathrm{[}x\mathrm{,}y\mathrm{]}$.

Lemma 5. (^{Nowicki 1994}, Proposition. 13.3.2) Let $d\mathrm{=}{\mathrm{\partial }}_x\mathrm{+}\mathrm{(}a\mathit{}\mathrm{(}x\mathrm{)}\mathit{}y\mathrm{+}b\mathit{}\mathrm{(}x\mathrm{)}\mathrm{)}\mathit{}{\mathrm{\partial }}_y$ be a Shamsuddin derivation, where $a\mathit{}\mathrm{(}x\mathrm{)}\mathrm{,}b\mathit{}\mathrm{(}x\mathrm{)}\mathrm{\in }k\mathit{}\mathrm{[}x\mathrm{]}$. Thus, if $d$ is a simple derivation, then $a\mathit{}\mathrm{(}x\mathrm{)}\mathrm{\ne }\mathrm{0}$ and $b\mathit{}\mathrm{(}x\mathrm{)}\mathrm{\ne }\mathrm{0}$.

Proof If $b(x)=0$, then the ideal $(y)$ is $d$-invariante. If $a(x)=0$, let $h(x)\in k[x]$ such that $h^{\prime }=b(x)$, then the ideal $(y-h)$ is $d$-invariante.

One can determine the simplicity of the a Shamsuddin derivation according the polynomials $a(x)$ and $b(x)$ (see ^{Nowicki 1994}, §13.3).

Theorem 6 Let $D\mathrm{\in }{\mathrm{Der}}_{\mathrm{k}}\mathit{}\mathrm{(}k\mathit{}\mathrm{[}x\mathrm{,}y\mathrm{]}\mathrm{)}$ be a Shamsuddin derivation. If $D$ is a simple derivation, then $\mathrm{Aut}\mathit{}{\mathrm{(}k\mathit{}\mathrm{[}x\mathrm{,}y\mathrm{]}\mathrm{)}}_D\mathrm{=}\mathrm{\{}i\mathit{}d\mathrm{\}}$.

Proof Let us denote $\rho (x)=f(x,y)$ and $\rho (y)=g(x,y)$. Let $D$ be a Shamsuddin derivation and

$D(x)=1,$

$D(y)=a(x)y+b(x),$

where $a(x),b(x)\in k[x]$. Since $\rho \in \mathrm{Aut}{(k[x,y])}_D$, we obtain two conditions:

$(1)$ $\rho (D(x))=D(\rho (x)),$

$(2)$ $\rho (D(y))=D(\rho (y)).$

Then, by condition $(1)$, $D(f(x,y))=1$ and since $f(x,y)$ can be written in the form

$f(x,y)=a_0(x)+a_1(x)y+\mathrm{\dots }+a_s(x)y^s,$

with $s\ge 0$, we obtain

$D(a_0(x))+D(a_1(x))y+a_1(x)(a(x)y+b(x))+\mathrm{\dots }$

$+D(a_s(x))y^s+s{a}_s(x)y^{s-1}(a(x)y+b(x))=1.$

By comparing the coefficients of $y^s$,

$D(a_s(x))=-s{a}_s(x)a(x),$

which can not occur by simplicity. More explicitly, Lemma 5 implies $a(x)=0$. Thus, $s=0$, that is, $f(x,y)=a_0(x)$. Therefore, $D(a_0(x))=1$ and $f=x+c$, with $c$ constant.

By using condition $(2)$,

By the previous part, we can suppose that $t>0$, because $\rho$ is a automorphism. Now, write $g(x,y)=b_0(x)+b_1(x)y+\mathrm{\dots }+b_t(x)y^t$. Thus,

By comparing the coefficients of $y^t$, we obtain

$D(b_t(x))+t{b}_t(x)a(x)=a(x+c)b_t(x).$

Then $D(b_t(x))=b_t(x)(-ta(x)+a(x+c))$. In this way, $b_t(x)$ is a constant and, consequently, $a(x+c)=ta(x)$. Comparing the coefficients in the last equality, we obtain $t=1$ and then $b_1(x)=b_1$ is constant. Moreover, if $a(x)$ is not a constant, since $a(x+c)=a(x)$, it is easy to see that $c=0$. Indeed, if $c\ne 0$, we obtain that the polynomial $a(x)$ has infinite distinct roots. If $a(x)$ is constant, then $a(x)\cdot D$ is not a simple derivation (this is a consequence of ^{Lequain 2008}, Lemma.2.6 and Theorem.3.2); thus, we obtain $c=0$.

Note that $g(x,y)=b_0(x)+b_{1}y$ and, using the condition $(2)$,

Considering the independent term of $y$, we have

$D(b_0(x))=b_0(x)a(x)+b(x)(1-b_1).$ (1)

By (^{Nowicki 1994}, Proposition. 13.3.3), if $b_1\ne 1$, we have that $D$ is a simple derivation if and only if $D^{\prime }$, defined by

$D^{\prime }(x)=1,D^{\prime }(y)=a(x)y+b(x)(1-b_1),$

is a simple derivation. Furthermore, by Theorem 3, there exist no elements $h(x)$ in $K[x]$ such that

$D(h(x))=h(x)a(x)+b(x)(1-b_1).$

This contradicts equation “eqrefeq1.1. Then, $b_1=1$ and $D(b_0(x))=b_0(x)a(x)$. Since $D$ is a simple derivation, we know that $a(x)\ne 0$ and consequently $b_0(x)=0$. This shows that $\rho =id$.

ON THE ISOTROPY OF THE SIMPLE DERIVATIONS

The purpose of this section is to study the isotropy in the general case of a simple derivation. More precisely, we obtain results that reveal nice features of the elements of $\mathrm{Aut}{(k[x,y])}_D$. For this, we use some concepts presented in the previous sections and the concept of dynamical degree of a polynomial map.

In Baltazar and Pan ( ²⁰¹⁵), which was inspired by Brumatti et al. ( ²⁰⁰³), the authors introduce and study a general notion of solution associated to a Noetherian differential $k$-algebra and its relationship with simplicity.

The following proposition has a geometrical flavour: it says that if an element in the isotropy of a simple derivation has fixed point, then it is the identity automorphism.

Proposition 7 Let $D\mathrm{\in }{\mathrm{Der}}_{\mathrm{k}}\mathit{}\mathrm{(}k\mathit{}\mathrm{[}{x}_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{x}_n\mathrm{]}\mathrm{)}$ be a simple derivation and $\rho \mathrm{\in }\mathrm{Aut}\mathit{}{\mathrm{(}k\mathit{}\mathrm{[}{x}_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{x}_n\mathrm{]}\mathrm{)}}_D$ be an automorphism in the isotropy. Suppose that there exists a maximal ideal $\mathrm{\backslash mathfrak}\mathit{}m\mathrm{\subset }k\mathit{}\mathrm{[}{x}_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{x}_n\mathrm{]}$ such that $\rho \mathit{}\mathrm{(}\mathrm{\backslash mathfrak}\mathit{}m\mathrm{)}\mathrm{=}\mathrm{\backslash mathfrak}\mathit{}m$, then $\rho \mathrm{=}i\mathit{}d$.

Proof Let $\phi$ be a solution of $D$ passing through $\mathrm{\backslash mathfrak}m$ (see ^{Baltazar and Pan 2015, Definition 1}). We know that $\mathrm{\backslash dfrac}\partial \partial t\phi =\phi D$ and ${\phi }^{-1}((t))=\mathrm{\backslash mathfrak}m$. If $\rho \in \mathrm{Aut}{(k[x_1,\mathrm{\dots },x_n])}_D$, then

$\mathrm{\backslash dfrac}\partial \partial t\phi \rho =\phi D\rho =\phi \rho D.$

In other words, $\phi \rho$ is a solution of $D$ passing through ${\rho }^{-1}(\mathrm{\backslash mathfrak}m)=\mathrm{\backslash mathfrak}m$. Then, by the uniqueness of the solution (^{Baltazar and Pan 2015}, Theorem.7.(c)), $\phi \rho =\phi$. Because $k[x_1,\mathrm{\dots },x_n]$ is $D$-simple and $\phi$ is a nontrivial solution, we have that $\phi$ is one-to-one. Therefore, $\rho =id$.

Lane ( ¹⁹⁷⁵) proved that every $k$-automorphism $\rho$ of $k[x,y]$ leaves a nontrivial proper ideal $I$ invariant over an algebraically closed field, that is, $\rho (I)\subseteq I$. In Shamsuddin ( ¹⁹⁸²), Shamsuddin proved that this result does not extend to $k[x,y,z]$, proving that the $k$-automorphism given by $\chi (x)=x+1$, $\chi (y)=y+xz+1$, and $\chi (z)=y+(x+1)z$ has no nontrivial invariant ideal.

In addition, since $k[x,y]$ is Noetherian, $\rho$ leaves a nontrivial proper ideal $I$ invariant if, and only if, $\rho (I)=I$. In fact, the ascending chain

$I\subset {\rho }^{-1}(I)\subset {\rho }^{-2}(I)\subset \mathrm{\dots }\subset {\rho }^{-l}(I)\subset \mathrm{\dots }$

must stabilize; thus, there exists a positive integer $n$ such that ${\rho }^{-n}(I)={\rho }^{-n-1}(I)$. Hence, $\rho (I)=I$.

Suppose that $\rho \in \mathrm{Aut}{(k[x,y])}_D$ and that $D$ is a simple derivation of $k[x,y]$. By Proposition 7, if this invariant ideal $I$ is maximal, we have $\rho =id$. Suppose that $I$ is radical and let $I=(\mathrm{\backslash mathfrak}{m}_1\cap \mathrm{\dots }\cap \mathrm{\backslash mathfrak}{m}_s)\cap (\mathrm{\backslash mathfrak}{p}_1\cap \mathrm{\dots }\cap \mathrm{\backslash mathfrak}{p}_t)$ be a primary decomposition, where the ideals $\mathrm{\backslash mathfrak}{m}_i$ are maximal and $\mathrm{\backslash mathfrak}{p}_j$ are prime ideals with height $1$ such that $\mathrm{\backslash mathfrak}{p}_j=(f_j)$, with $f_j$ irreducible (see ^{Kaplansky 1974}, Theorem 5). If

$\rho (\mathrm{\backslash mathfrak}{m}_1\cap \mathrm{\dots }\cap \mathrm{\backslash mathfrak}{m}_s)=\mathrm{\backslash mathfrak}{m}_1\cap \mathrm{\dots }\cap \mathrm{\backslash mathfrak}{m}_s,$

we claim that ${\rho }^N$ leaves invariant one maximal ideal for some $N\in \mathrm{\backslash mathbb}N$. Indeed, we know that $\rho (\mathrm{\backslash mathfrak}{m}_1)\supset \mathrm{\backslash mathfrak}{m}_1\cap \mathrm{\dots }\cap \mathrm{\backslash mathfrak}{m}_s$ and since $\rho (\mathrm{\backslash mathfrak}{m}_1)$ is a prime ideal, we deduce that $\rho (\mathrm{\backslash mathfrak}{m}_1)\supseteq \mathrm{\backslash mathfrak}{m}_i$, for some $i=1,\mathrm{\dots },s$ (^{Atiyah and Macdonald 1969}, Prop.11.1.(ii)). Then, $\rho (\mathrm{\backslash mathfrak}{m}_1)=\mathrm{\backslash mathfrak}{m}_i$, that is, ${\rho }^N$ leaves invariant the maximal ideal $\mathrm{\backslash mathfrak}{m}_1$, for some $N\in \mathrm{\backslash mathbb}N$. Thus, it follows from Proposition 7 that ${\rho }^N=id$.

Note that $\rho (\mathrm{\backslash mathfrak}{p}_1\cap \mathrm{\dots }\cap \mathrm{\backslash mathfrak}{p}_t)=\mathrm{\backslash mathfrak}{p}_1\cap \mathrm{\dots }\cap \mathrm{\backslash mathfrak}{p}_t$. In fact, writing $\mathrm{\backslash mathfrak}{p}_1\cap \mathrm{\dots }\cap \mathrm{\backslash mathfrak}{p}_t=(f_1\mathrm{\dots }{f}_t)$, with $f_i$ irreducible, we would like to choose $h\in \mathrm{\backslash mathfrak}{m}_1\cap \mathrm{\dots }\cap \mathrm{\backslash mathfrak}{m}_s$ such that $\rho (h)\notin \mathrm{\backslash mathfrak}{p}_1$. If such $h$ does not exist, we would obtain $\mathrm{\backslash mathfrak}{m}_1\cap \mathrm{\dots }\cap \mathrm{\backslash mathfrak}{m}_s\subset \mathrm{\backslash mathfrak}{p}_1$, then $\mathrm{\backslash mathfrak}{p}_1\supseteq \mathrm{\backslash mathfrak}{m}_i$, for some $i=1,\mathrm{\dots },s$ (^{Atiyah and Macdonald 1969}, Prop.11.1.(ii)): a contradiction. Thus, since $h{f}_1\mathrm{\dots }{f}_t\in I$, we obtain $\rho (h)\rho (f_1)\mathrm{\dots }\rho (f_t)\in I\subset \mathrm{\backslash mathfrak}{p}_1$. Therefore, $\rho (f_1\mathrm{\dots }{f}_t)\in \mathrm{\backslash mathfrak}{p}_1$. Likewise, the same conclusion holds for the other prime ideals $\mathrm{\backslash mathfrak}{p}_i$, $i=1,\mathrm{\dots },t$. Finally, $\rho (\mathrm{\backslash mathfrak}{p}_1\cap \mathrm{\dots }\cap \mathrm{\backslash mathfrak}{p}_t)=\mathrm{\backslash mathfrak}{p}_1\cap \mathrm{\dots }\cap \mathrm{\backslash mathfrak}{p}_t$.

In the next corollary, we obtain consequences on the case of radical ideals.

Corollary 8 Let $\rho \mathrm{\in }\mathrm{Aut}\mathit{}{\mathrm{(}k\mathit{}\mathrm{[}x\mathrm{,}y\mathrm{]}\mathrm{)}}_D$, $D$ a simple derivation of $k\mathit{}\mathrm{[}x\mathrm{,}y\mathrm{]}$, and $I\mathrm{=}\mathrm{(}f\mathrm{)}$ an ideal with height $\mathrm{1}$ such that $\rho \mathit{}\mathrm{(}I\mathrm{)}\mathrm{=}I$, with $f$ reduced. If $V\mathit{}\mathrm{(}f\mathrm{)}$ is singular or some irreducible component $C_i$ of $V\mathit{}\mathrm{(}f\mathrm{)}$ has genus greater than two, then $\rho$ is an automorphism of finite order.

Proof Suppose that $V(f)$ is not a smooth variety and let $q$ be a singularity of $V(f)$. Since the set of singular points is invariant by $\rho$, there exists $N\in \mathrm{\backslash mathbb}N$ such that ${\rho }^N(q)=q$. Using that $\rho \in \mathrm{Aut}{(k[x,y])}_D$, we obtain, by Proposition 7, ${\rho }^N=id$.

Let $C_i$ be a component irreducible of $V(f)$ that has genus greater than two. Note that there exists $M\in \mathrm{\backslash mathbb}N$ such that ${\rho }^M(C_i)=C_i$. By (^{Farkas and Kra 1992}, Theorem Hunvitz, p.241), the number of elements in $\mathrm{Aut}(C_i)$ is finite; in fact, , where $g_i$ is the genus of $C_i$. Then, we deduce that $\rho$ is an automorphism of finite order.

In the rest of this section, we let $k=\mathrm{\backslash mathbb}C.$

Consider a polynomial map $f(x,y)=(f_1(x,y),f_2(x,y)):\mathrm{\backslash mathbb}{C}^2\to \mathrm{\backslash mathbb}{C}^2$ and define the degree of $f$ by $\mathrm{deg}(f):=\max (\mathrm{deg}(f_1),\mathrm{deg}(f_2))$. Thus, we may define the dynamical degree (see Blanc and J. (), Friedland and Milnor ( ¹⁹⁸⁹), Silverman ( ²⁰¹²)) of $f$ as

$\delta (f):=\underset{n\to \mathrm{\infty }}{lim}{(\mathrm{deg}(f^n))}^{\frac{1}{n}}.$

Corollary 9 If $\rho \mathrm{\in }\mathrm{Aut}\mathit{}{\mathrm{(}\mathrm{\backslash mathbb}\mathit{}C\mathit{}\mathrm{[}x\mathrm{,}y\mathrm{]}\mathrm{)}}_D$ and $D$ is a simple derivation of $\mathrm{\backslash mathbb}\mathit{}C\mathit{}\mathrm{[}x\mathrm{,}y\mathrm{]}$, then $\delta \mathit{}\mathrm{(}\rho \mathrm{)}\mathrm{=}\mathrm{1}$.

Proof Suppose $\delta (\rho )>1$. By (^{Friedland and Milnor 1989}, Theorem 3.1.), ${\rho }^n$ has exactly $\delta {(\rho )}^n$ fixed points counted with multiplicities. Then, by Proposition 7, $\rho =id$, which shows that the dynamical degree of $\rho$ is 1.

ACKNOWLEDGMENTS

I would like to thank Ivan Pan for his comments and suggestions. Research of R. Baltazar was partially supported by Coordenação de Aperfeiçoamento de Pessoal de Nível Superior (CAPES).

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  » http://www-users.mat.umk.pl/anow/psdvi/pol-der.pdf
• ^r13
  NOWICKI A. 2008. An Example of a Simple Derivation in Two Variables. Colloq Math 113(1): 25-31.
• ^r15
  SARAIVA C. 2012. Sobre Derivações Simples e Folheações holomorfas sem Solução Algébrica, Tese de Doutorado.
• ^r17
  SHAMSUDDIN A. 1977. Automorphisms and Skew Polynomial Rings. Ph.D. thesis, Univesity of Leeds.
• ^r18
  SHAMSUDDIN A. 1982. Rings with automorphisms leaving no nontrivial proper ideals invariant. Canadian Math Bull 25: 478-486.
• ^r19
  SILVERMAN JH. 2012. Dynamical Degrees, Arithmetic Degrees, and Canonical Heights for Dominant Rational Self-Maps of Projective Space. Ergodic Theory and Dynamical Systems 34(2): 647-678.

Publication Dates

• Publication in this collection
  Dec 2016

History

• Received
  26 Jan 2015
• Accepted
  29 Sept 2015