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Anais da Academia Brasileira de Ciências

versão impressa ISSN 0001-3765versão On-line ISSN 1678-2690

An. Acad. Bras. Ciênc. vol.88 no.4 Rio de Janeiro out./dez. 2016 

Mathematical Sciences

On simple Shamsuddin derivations in two variables


1Instituto de Matematica, Estatística e Física, Universidade Federal do Rio Grande/FURG, Santo Antônio da Patrulha, Rua Barão do Caí, 125, 95500-000 Rio Grande, RS, Brazil


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.

Key words: dynamical degree; isotropy group; Shamsuddin derivations; simple derivations


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]k[x,y] of k[x,y] is a k -linear map such that


for any a,bk[x,y] . We denote by Derk(k[x,y]) the set of all k -derivations of k[x,y] . Let dDerk(k[x,y]) . An ideal I of k[x,y] is called d -stable if d(I)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 Aut(k[x,y]) the group of k -automorphisms of k[x,y] . Let Aut(k[x,y]) act on Derk(k[x,y]) by:


Fix a derivation dDerk(k[x,y]) . The isotropy subgroup,with respect to this group action, is defined as


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

Conjecture 1 If d is a simple derivation of k[x,y] , then Aut(k[x,y])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 \mathbbAn 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=\mathbbC , the elements of Aut(\mathbbC[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)).


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=xDerk(k[x,y]) and ρAut(k[x,y])d . Note that d is not a simple derivation. Indeed, for any u(y)k[y] , the ideal generated by u(x) is always invariant. Consider


Since ρAut(k[x,y])d , we obtain two conditions:

1) ρ(d(x))=d(ρ(x)).



Then, d(a0(x))=1 and d(aj(x))=0 , j=1,,t . We conclude that ρ(x) is of the type


2) ρ(d(y))=d(ρ(y)).



that is, bi(x)=di , with dik . We also infer that ρ(y) is of the type


Thus, Aut(k[x,y])d contains the affine automorphisms


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


with p(y),q(y)k[y] . Since ρ is an automorphism, the determinant of the Jacobian matrix must be nonzero. Thus, |Jρ|=q(y)=c , ck* . Therefore, ρ=(x+p(y),ay+c) , with p(y)k[y] and a,bk . Consequently, Aut(k[x,y])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(t)R[t] , with t an indeterminate. Then, we can also extend d to a unique derivation d~ of R[t] such that d~(t)=h(t) .

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

Theorem 3 Let R be a ring containing \mathbbQ and let d be a simple derivation of R . Extend the derivation d to a derivation d~ of the polynomial ring R[t] by setting d~(t)=at+b where a,bR . Then the following two conditions are equivalent:

(1) d~ is a simple derivation.

(2) There exist no elements rR such that d(r)=ar+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


where a(x),b(x)k[x] .

Example 4 Let d be a derivation of k[x,y] as follows


Writing R=k[x] , we know that R is x -simple and, taking a=x and b=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 rR such that x(r)=xr+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[x,y] .

Lemma 5. (Nowicki 1994, Proposition. 13.3.2) Let d=x+(a(x)y+b(x))y be a Shamsuddin derivation, where a(x),b(x)k[x] . Thus, if d is a simple derivation, then a(x)0 and b(x)0 .

Proof If b(x)=0 , then the ideal (y) is d -invariante. If a(x)=0 , let h(x)k[x] such that h=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 DDerk(k[x,y]) be a Shamsuddin derivation. If D is a simple derivation, then Aut(k[x,y])D={id} .

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


where a(x),b(x)k[x] . Since ρAut(k[x,y])D , we obtain two conditions:

(1) ρ(D(x))=D(ρ(x)),

(2) ρ(D(y))=D(ρ(y)).

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


with s0 , we obtain


By comparing the coefficients of ys ,


which can not occur by simplicity. More explicitly, Lemma 5 implies a(x)=0 . Thus, s=0 , that is, f(x,y)=a0(x) . Therefore, D(a0(x))=1 and f=x+c , with c constant.

By using condition (2) ,


By the previous part, we can suppose that t>0 , because ρ is a automorphism. Now, write g(x,y)=b0(x)+b1(x)y++bt(x)yt . Thus,


By comparing the coefficients of yt , we obtain


Then D(bt(x))=bt(x)(-ta(x)+a(x+c)) . In this way, bt(x) is a constant and, consequently, a(x+c)=ta(x) . Comparing the coefficients in the last equality, we obtain t=1 and then b1(x)=b1 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 c0 , we obtain that the polynomial a(x) has infinite distinct roots. If a(x) is constant, then a(x)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)=b0(x)+b1y and, using the condition (2) ,


Considering the independent term of y , we have

D(b0(x))=b0(x)a(x)+b(x)(1-b1). (1)

By (Nowicki 1994, Proposition. 13.3.3), if b11 , we have that D is a simple derivation if and only if D , defined by


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


This contradicts equation “eqrefeq1.1. Then, b1=1 and D(b0(x))=b0(x)a(x) . Since D is a simple derivation, we know that a(x)0 and consequently b0(x)=0 . This shows that ρ=id .


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 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 DDerk(k[x1,,xn]) be a simple derivation and ρAut(k[x1,,xn])D be an automorphism in the isotropy. Suppose that there exists a maximal ideal \mathfrakmk[x1,,xn] such that ρ(\mathfrakm)=\mathfrakm , then ρ=id .

Proof Let φ be a solution of D passing through \mathfrakm (see Baltazar and Pan 2015, Definition 1). We know that \dfractφ=φD and φ-1((t))=\mathfrakm . If ρAut(k[x1,,xn])D , then


In other words, φρ is a solution of D passing through ρ-1(\mathfrakm)=\mathfrakm . Then, by the uniqueness of the solution (Baltazar and Pan 2015, Theorem.7.(c)), φρ=φ . Because k[x1,,xn] is D -simple and φ is a nontrivial solution, we have that φ is one-to-one. Therefore, ρ=id .

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

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


must stabilize; thus, there exists a positive integer n such that ρ-n(I)=ρ-n-1(I) . Hence, ρ(I)=I .

Suppose that ρAut(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 ρ=id . Suppose that I is radical and let I=(\mathfrakm1\mathfrakms)(\mathfrakp1\mathfrakpt) be a primary decomposition, where the ideals \mathfrakmi are maximal and \mathfrakpj are prime ideals with height 1 such that \mathfrakpj=(fj) , with fj irreducible (see Kaplansky 1974, Theorem 5). If


we claim that ρN leaves invariant one maximal ideal for some N\mathbbN . Indeed, we know that ρ(\mathfrakm1)\mathfrakm1\mathfrakms and since ρ(\mathfrakm1) is a prime ideal, we deduce that ρ(\mathfrakm1)\mathfrakmi , for some i=1,,s (Atiyah and Macdonald 1969, Prop.11.1.(ii)). Then, ρ(\mathfrakm1)=\mathfrakmi , that is, ρN leaves invariant the maximal ideal \mathfrakm1 , for some N\mathbbN . Thus, it follows from Proposition 7 that ρN=id .

Note that ρ(\mathfrakp1\mathfrakpt)=\mathfrakp1\mathfrakpt . In fact, writing \mathfrakp1\mathfrakpt=(f1ft) , with fi irreducible, we would like to choose h\mathfrakm1\mathfrakms such that ρ(h)\mathfrakp1 . If such h does not exist, we would obtain \mathfrakm1\mathfrakms\mathfrakp1 , then \mathfrakp1\mathfrakmi , for some i=1,,s (Atiyah and Macdonald 1969, Prop.11.1.(ii)): a contradiction. Thus, since hf1ftI , we obtain ρ(h)ρ(f1)ρ(ft)I\mathfrakp1 . Therefore, ρ(f1ft)\mathfrakp1 . Likewise, the same conclusion holds for the other prime ideals \mathfrakpi , i=1,,t . Finally, ρ(\mathfrakp1\mathfrakpt)=\mathfrakp1\mathfrakpt .

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

Corollary 8 Let ρAut(k[x,y])D , D a simple derivation of k[x,y] , and I=(f) an ideal with height 1 such that ρ(I)=I , with f reduced. If V(f) is singular or some irreducible component Ci of V(f) has genus greater than two, then ρ 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 ρ , there exists N\mathbbN such that ρN(q)=q . Using that ρAut(k[x,y])D , we obtain, by Proposition 7, ρN=id .

Let Ci be a component irreducible of V(f) that has genus greater than two. Note that there exists M\mathbbN such that ρM(Ci)=Ci . By (Farkas and Kra 1992, Theorem Hunvitz, p.241), the number of elements in Aut(Ci) is finite; in fact, #(Aut(Ci))<84(gi-1) , where gi is the genus of Ci . Then, we deduce that ρ is an automorphism of finite order.

In the rest of this section, we let k=\mathbbC.

Consider a polynomial map f(x,y)=(f1(x,y),f2(x,y)):\mathbbC2\mathbbC2 and define the degree of f by deg(f):=max(deg(f1),deg(f2)) . Thus, we may define the dynamical degree (see Blanc and J. (4), Friedland and Milnor ( 1989), Silverman ( 2012)) of f as


Corollary 9 If ρAut(\mathbbC[x,y])D and D is a simple derivation of \mathbbC[x,y] , then δ(ρ)=1 .

Proof Suppose δ(ρ)>1 . By (Friedland and Milnor 1989, Theorem 3.1.), ρn has exactly δ(ρ)n fixed points counted with multiplicities. Then, by Proposition 7, ρ=id , which shows that the dynamical degree of ρ is 1.


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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Received: January 26, 2015; Accepted: September 29, 2015

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