aa r X i v : . [ m a t h . A C ] D ec ON SIMPLE SHAMSUDDIN DERIVATIONS IN TWO VARIABLES
RENE BALTAZAR
Abstract.
We study the subgroup of k -automorphisms of k [ x, y ] which commute witha simple derivation D of k [ x, y ] . We prove, for example, that this subgroup is trivialwhen D is a Shamsuddin simple derivation. In the general case of simple derivations, weobtain properties for the elements of this subgroup. introduction Let k be an algebraically closed field of characteristic zero and the ring k [ x, y ] of poly-nomials 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 d ( ab ) = d ( a ) b + ad ( b )for any a, b ∈ k [ x, y ]. Denoting by Der k ( k [ x, y ]) the set of all k -derivations of k [ x, y ].Let d ∈ Der k ( 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 k [ x, y ] has no other d -stable ideal it iscalled d - simple . Even in the case of two variables, a few examples of simple derivationsare known (see for explanation [BLL2003], [Cec2012], [No2008], [BP2015], [KM2013] and[Leq2011]).We denote by Aut( k [ x, y ]) the group of k -automorphisms of k [ x, y ]. Let Aut( k [ x, y ])act on Der k ( k [ x, y ]) by: ( ρ, D ) ρ − ◦ D ◦ ρ = ρ − Dρ.
Fixed a derivation d ∈ Der k ( k [ x, y ]). The isotropy subgroup, with respect to this groupaction, is Aut( k [ x, y ]) D := { ρ ∈ Aut( k [ x, y ]) /ρD = Dρ } . We are interested in the following question proposed by I.Pan (see [B2014]):
Conjecture 1. If d is a simple derivation of k [ x, y ] , then Aut( k [ x, y ]) d is finite. Research of R. Baltazar was partially supported by CAPES.
At a first moment, in the §
2, we show that the conjecture is true for a family ofderivations, named Shamsuddin derivations (Theorem 6). For this, we use a theorem ofthe Shamsuddin [Sh1977], mentioned in [No1994, Theorem 13.2.1.], that determines acondition that would preserve the simplicity by extending, in some way, the derivation to R [ t ], with t an indeterminate. The reader may also remember that Y.Lequain [Leq2011]showed that these derivations check a conjecture about the A n , the Weyl algebra over k .In order to understand the isotropy of a simple derivation of the k [ x, y ], in §
3, weanalysed necessary conditions for an automorphism to belong to the isotropy of a simplederivation. For example, we prove that if such an automorphism has a fixed point, thenit is the identity (Proposition 7). Following, we present the definition of dynamical degree of a polynomial application and thus proved that in the case k = C , the elements inAut( C [ x, y ]) d , with d a simple derivation, has dynamical degree 1 (Corollary 9). Moreprecisely, the condition dynamical degree > Shamsuddin derivation
The main aim of this section is study the isotropy group of the a Shamsuddin derivationin k [ x, y ]. In [No1994, § Example . Let be d = ∂ x ∈ Der k ( k [ x, y ]) and ρ ∈ Aut( k [ x, y ]) d . Note that d is nota simple derivation; indeed, for any u ( y ) ∈ k [ y ], the ideal generated by u ( x ) is alwaysinvariant. Consider ρ ( x ) = f ( x, y ) = a ( x ) + a ( x ) y + . . . + a t ( x ) y t ρ ( y ) = g ( x, y ) = b ( x ) + b ( x ) y + . . . + b s ( x ) y s . Since ρ ∈ Aut( k [ x, y ]) d , we obtain two conditions: ρ ( d ( x )) = d ( ρ ( x )) . Thus,1 = d ( a ( x ) + a ( x ) y + . . . + a t ( x ) y t ) = d ( a ( x )) + d ( a ( x )) y + . . . + d ( a t ( x )) y t . Then, d ( a ( x )) = 1 and d ( a j ( x )) = 0, j = 1 , . . . , t . We conclude that ρ ( x ) is of the type ρ ( x ) = x + c + c y + . . . + c t y t , c i ∈ k. ρ ( d ( y )) = d ( ρ ( y )) . N SOLUTIONS FOR DERIVATIONS 3
Analogously,0 = d ( b ( x ) + b ( x ) y + . . . + b s ( x ) y s ) = d ( b ( x )) + d ( b ( x )) y + . . . + d ( b s ( x )) y s . That is, b i ( x ) = d i with d i ∈ k . We conclude also that ρ ( y ) is of the type ρ ( y ) = d + d y + . . . + d s y s , d i ∈ k. Thus, Aut( k [ x, y ]) d contains the affine automorphisms( x + uy + r, uy + s ) , with u, r, s ∈ k . In particular, Aut( k [ x, y ]) d is not finite.Notice that Aut( k [ x, y ]) d contains also the automorphisms of the type ( x + p ( y ) , y ),with p ( y ) ∈ k [ y ].Now, we determine indeed the isotropy. Using only the conditions 1 and 2, ρ = ( x + p ( y ) , q ( y ))with p ( y ) , q ( y ) ∈ k [ y ]. However, ρ is an automorphism, in other words, the determinantof the Jacobian matrix must be a nonzero constant. Thus, | J ρ | = q ′ ( y ) = c , c ∈ k ∗ .Therefore, ρ = ( x + p ( y ) , ay + c ), with p ( y ) ∈ k [ y ] and a, b ∈ k . Consequently, Aut( k [ x, y ]) d is not finite and, more than that, the first component has elements with any degree.The following lemma is a well known result. Lemma 2.
Let R be a commutative ring, d a derivation of R and h ( t ) ∈ R [ t ] , with t anindeterminate. Then, we can also extend d to a unique derivation ˜ d of the R [ t ] such that ˜ d ( t ) = h ( t ) . We will use the following result of Shamsuddin [Sh1977].
Theorem 3.
Let R be a ring containing Q and let d be a simple derivation of R . Extendthe derivation d to a derivation ˜ d of the polynomial ring R [ t ] by setting ˜ d ( t ) = at + b where a, b ∈ R . Then the following two conditions are equivalent: (1) ˜ d is a simple derivation. (2) There exist no elements r ∈ R such that d ( r ) = ar + b .Proof. See [No1994, Theorem 13.2.1.] for a demonstration in details. (cid:3)
A derivation d of k [ x, y ] is said to be a Shamsuddin derivation if d is of the form d = ∂ x + ( a ( x ) y + b ( x )) ∂ y , where a ( x ) , b ( x ) ∈ k [ x ]. RENE BALTAZAR
Example . Let d be a derivation of k [ x, y ] as follows d = ∂ x + ( xy + 1) ∂ y . Writing R = k [ x ], we know that R is ∂ x -simple and, taking a = x and b = 1, we areexactly the conditions of Theorem 3. Thus, we know that d is simple if, and only if, thereexist no elements r ∈ R such that ∂ x ( r ) = xr + 1; but the right side of the equivalence issatisfied by the degree of r . Therefore, by Theorem 3, d is a simple derivation of k [ x, y ]. Lemma 5. ( [No1994, Proposition. 13.3.2] ) Let d = ∂ x +( a ( x ) y + b ( x )) ∂ y be a Shamsuddinderivation, 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. (cid:3) One can determine the simplicity of the a Shamsuddin derivation according the poly-nomials a ( x ) and b ( x ) (see ([No1994, § Theorem 6.
Let D ∈ Der k ( k [ x, y ]) be a Shamsuddin derivation. If D is a simple deriva-tion, 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 derivationand D ( x ) = 1 ,D ( y ) = a ( x ) y + b ( x ) , where a ( x ) , b ( x ) ∈ k [ x ]Since ρ ∈ Aut( 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 f ( x, y ) = a ( x ) + a ( x ) y + ... + a s ( x ) y s , with s ≥
0, we obtain D ( a ( x )) + D ( a ( x )) y + a ( x )( a ( x ) y + b ( x )) + . . . + D ( a s ( x )) y s + sa s ( x ) y s − ( a ( x ) y + b ( x )) = 1Comparing the coefficients in y s , D ( a s ( x )) = − sa s ( x ) a ( x ) , N SOLUTIONS FOR DERIVATIONS 5 which can not occur by the simplicity. More explicitly, the Lemma 5 implies that a ( x ) = 0.Thus s = 0, this is f ( x, y ) = a ( x ). Therefore D ( a ( x )) = 1 and f = x + c , with c constant.Using the condition (2), D ( g ( x, y )) = ρ ( a ( x ) y + b ( x ))= ρ ( a ( x )) ρ ( y ) + ρ ( b ( x ))= a ( x + c ) g ( x, y ) + b ( x + c )Writing g ( x, y ) = b ( x ) + b ( x ) y + . . . + b t ( x ) y t ; wherein, by the previous part, we cansuppose that t >
0, because ρ is a automorphism. Thus a ( x + c ) g ( x, y ) + b ( x + c ) = D ( b ( x )) + D ( b ( x )) y + b ( x )( a ( x ) y + b ( x ))++ . . . + D ( b t ( x )) y t + tb t ( x ) y t − ( a ( x ) y + b ( x )) . Comparing the coefficients in y t , we obtain D ( b t ( x )) + tb 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, conse-quently, a ( x + c ) = ta ( x ). Comparing the coefficients in the last equality, we obtain t = 1and then b ( x ) = b a constant. Moreover, if a ( x ) is not a constant, since a ( x + c ) = a ( x ),is easy to see that c = 0. Indeed, if c = 0 we obtain that the polynomial a ( x ) has infinitedistinct roots. If a ( x ) is a constant, then a ( x ) D is not a simple derivation (a consequenceof [Leq2008, Lemma.2.6 and Theorem.3.2]; thus, we obtain c = 0.Note that g ( x, y ) = b ( x ) + b y and, using the condition (2) again, D ( g ( x, y )) = D ( b ( x )) + b ( a ( x ) y + b ( x ))= a ( x )( b ( x ) + b y ) + b ( x ) . Considering the independent term of y , D ( b ( x )) = b ( x ) a ( x ) + b ( x )(1 − b ) (1)If b = 1, we consider the derivation D ′ such that D ′ ( x ) = 1 , D ′ ( y ) = a ( x ) y + b ( x )(1 − b ) . In [No1994, Proposition. 13.3.3], it is noted that D is a simple derivation if and only if D ′ is a simple derivation. Furthermore, by the Theorem 3, there exist no elements h ( x )in K [ x ] such that D ( h ( x )) = h ( x ) a ( x ) + b ( x )(1 − b ) :what contradicts the equation (1). Then, b = 1 and D ( b ( x )) = b ( x ) a ( x ), since D is a simple derivation we know that a ( x ) = 0, consequently b ( x ) = 0. This shows that ρ = id . (cid:3) RENE BALTAZAR On the isotropy of the simple derivations
The purpose of this section is to study the isotropy in the general case of a simplederivation. More precisely, we obtain results that reveal some characteristics of the ele-ments in Aut( k [ x, y ]) D . For this, we use some concepts presented in the previous sectionsand also the concept of dynamical degree of a polynomial application.In [BP2015], which was inspired by [BLL2003], we introduce and study a general notionof solution associated to a Noetherian differential k -algebra and its relationship withsimplicity.The following proposition geometrically says that if an element in the isotropy of asimple derivation has fixed point then it is the identity automorphism. Proposition 7.
Let D ∈ Der k ( k [ x , ..., x n ]) be a simple derivation and ρ ∈ Aut( k [ x , ..., x n ]) D an automorphism in the isotropy. Suppose that there exist a maximal ideal m ⊂ k [ x , ..., x n ] such that ρ ( m ) = m , then ρ = id .Proof. Let ϕ be a solution of D passing through m (see [BP2015, Definition.1.]). We knowthat ∂∂t ϕ = ϕD and ϕ − (( t )) = m . If ρ ∈ Aut( k [ x , ..., x n ]) D , then ∂∂t ϕρ = ϕDρ = ϕρD. In other words, ϕρ is a solution of D passing through ρ − ( m ) = m . Therefore, by theuniqueness of the solution ([BP2015, Theorem.7.(c)]), ϕρ = ϕ . Note that ϕ is one toone, because k [ x , ..., x n ] is D -simple and ϕ is a nontrivial solution. Then, we obtain that ρ = id . (cid:3) F. Lane, in [Lane75], proved that every k -automorphism ρ of k [ x, y ] leaves a nontrivialproper ideal I invariant, over an algebraically closed field; this is, ρ ( I ) ⊆ I . Em [Sh1982],A. 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 e χ ( z ) = y + ( x + 1) z has nonontrivial invariant ideal.Note that, in addition, ρ leaves a nontrivial proper ideal I invariant if and only if ρ ( I ) = I , because k [ x, y ] is Noetherian. In fact, the ascending chain I ⊂ ρ − ( I ) ⊂ ρ − ( I ) ⊂ . . . ⊂ ρ − l ( I ) ⊂ . . . must stabilize; thus, there exists a positive integer n such that ρ − n ( I ) = ρ − n − ( I ), then ρ ( I ) = I . N SOLUTIONS FOR DERIVATIONS 7
Suppose that ρ ∈ Aut( k [ x, y ]) D and D is a simple derivation of k [ x, y ]. If this invariantideal I is maximal, by the Proposition 7, we have ρ = id .Suppose that I , this invariant ideal, is radical. Let I = ( m ∩ . . . ∩ m s ) ∩ ( p ∩ . . . ∩ p t ) bea primary decomposition where each ideal m i is a maximal ideal and p j are prime idealswith height 1 such that p j = ( f j ), with f j irreducible (by [Kaplan74, Theorem 5.]). If ρ ( m ∩ . . . ∩ m s ) = m ∩ . . . ∩ m s , we claim that ρ N leaves invariant one maximal ideal for some N ∈ N : suppose m thisideal. Indeed, we know that ρ ( m ) ⊃ m ∩ . . . ∩ m s , since ρ ( m ) is a prime ideal, we deducethat ρ ( m ) ⊇ m i , for some i = 1 , . . . , s ([AM1969, Prop.11.1.(ii)]). Then, ρ ( m ) = m i ;that is, ρ N leaves invariant the maximal ideal m for some N ∈ N . Thus follows fromProposition 7 that ρ N = id .Note that ρ ( p ∩ . . . ∩ p t ) = p ∩ . . . ∩ p t . In fact, writing p ∩ . . . ∩ p t = ( f . . . f t ),with f i irreducible, we can choose h ∈ m ∩ . . . ∩ m s such that ρ ( h ) p . We observethat there exists h . Otherwise, we obtain m ∩ . . . ∩ m s ⊂ p , then p ⊇ m i , for some i = 1 , . . . , s ([AM1969, Prop.11.1.(ii)]): a contradiction. Thus, since hf . . . f t ∈ I , weobtain ρ ( h ) ρ ( f ) . . . ρ ( f t ) ∈ I ⊂ p . Therefore, ρ ( f . . . f t ) ∈ p . Likewise, we concludethe same for the other primes p i , i = 1 , . . . , t . Finally, ρ ( p ∩ . . . ∩ p t ) = p ∩ . . . ∩ p t .With the next corollary, we obtain some consequences on the last case. Corollary 8.
Let ρ ∈ Aut( k [ x, y ]) D , D a simple derivation of k [ x, y ] and I = ( f ) , with f reduced, a ideal with height such that ρ ( I ) = I . If V ( f ) is singular or some irreduciblecomponent C i of V ( f ) has genus greater than two, then ρ is a automorphism of finiteorder.Proof. Suppose that V ( f ) is not a smooth variety and let q be a singularity of V ( f ).Since the set of the singular points is invariant by ρ , then there exist N ∈ N such that ρ N ( q ) = q . Using that ρ ∈ Aut( k [ x, y ]) D , we obtain, by Proposition 7, ρ N = id .Let C i be a component irreducible of V ( f ) that has genus greater than two. Note thatthere exist M ∈ N such that ρ M ( C i ) = C i . By [FK1992, Thm. Hunvitz, p.241], thenumber of elements in Aut( C i ) is finite; in fact, C i )) < g i − g i is thegenus of C i . Then, we deduce that ρ is a automorphism of finite order. (cid:3) We take for the rest of this section k = C . Consider a polynomial application f ( x, y ) = ( f ( x, y ) , f ( x, y )) : C → C and definethe degree of f by deg( f ) := max(deg( f ) , deg( f )). Thus we may define the dynamicaldegree (see [BD2012], [FM1989], [Silv12]) of f as δ ( f ) := lim n →∞ (deg( f n )) n . RENE BALTAZAR
Corollary 9. If ρ ∈ Aut( C [ x, y ]) D and D is a simple derivation of C [ x, y ] , then δ ( ρ ) = 1 .Proof. Suppose δ ( ρ ) >
1. By [FM1989, Theorem 3.1.], ρ n has exactly δ ( ρ ) n fix pointscounted with multiplicities. Then, by Proposition 7, ρ = id , which shows that dynamicaldegree of ρ is 1. (cid:3) Acknowledgements.
I would like to thank Ivan Pan for his comments and suggestions.
References [AM1969] M. F. Atiyah, I. G. Macdonald,
Introduction to Commutative Algebra , Addison-Wesley Pub-lishing Company, Massachusetts, 1969.[BP2015] R. Baltazar, I. Pan,
On solutions for derivations of a Noetherian k -algebra and local simplicity ,To appear in Communications in Algebra (2015).[B2014] R. Baltazar, Sobre solues de derivaes em k-lgebras Noetherianas e simplici-dade , Tese de Doutorado, Universidade Federal do Rio Grande do Sul, 2014. athttp://hdl.handle.net/10183/97877[BD2012] J. Blanc and J. Dserti,
Degree Growth of Birational Maps of the Plane ∼ deserti/articles/degreegrowth.pdf, 2012.[BLL2003] P. Brumatti, Y. Lequain and D. Levcovitz, Differential simplicity in Polynomial Rings andAlgebraic Independence of Power Series , J. London Math. Soc. (2), 68, 615-630, 2003.[FK1992] H.M. Farkas, I. Kra,
Riemann Surfaces , 2nd ed., Graduate Texts in Mathematics, Springer,1992.[FM1989] S. Friedland, J. Milnor,
Dynamical properties of plane polynomial automorphisms , ErgodicTheory Dyn. Syst. 9., 67-99, 1989.[Kaplan74] I. Kaplansky,
Commutative Rings , Chicago, 1974, (2nd edition).[KM2013] S. Kour, A.K. Maloo,
Simplicity of Some Derivations of k[x, y] , Communications in Algebra,41, 4, 2013.[Lane75] D.R. Lane,
Fixed points of affine Cremona transformations of the plane over an algebraicallyclosed field , Amer. J. Math., 97, 3, 707-732, 1975.[Leq2008] Y. Lequain,
Simple Shamsuddin derivations of K [ X, Y , ..., Y n ] : An algorithmic characterizar-ion , J. Pure Appl. Algebra, 212, 2008.[Leq2011] Y. Lequain, Cyclic irreducible non-holonomic modules over the Weyl algebra: An algorithmiccharacterization , J. Pure Appl. Algebra, 215, 2011.[No1994] A. Nowicki,
Polynomial derivations and their rings of constants ∼ anow/ps-dvi/pol-der.pdf[No2008] A. Nowicki, An Example of a Simple Derivation in Two Variables , Colloq. Math., 113, No.1,25-31, 2008.[Cec2012] C. Saraiva,
Sobre Derivaes Simples e Folheaes holomorfas sem Soluo Algbrica , Tese deDoutorado, 2012.[Sei1967] A. Seidenberg,
Differential ideals in rings of finitely generated type , American Journal of Math-ematics, 89, No. 1, 22-42, 1967.[Sh1977] A. Shamsuddin,
Ph.D. thesis , Univesity of Leeds, 1977.[Sh1982] A. Shamsuddin,
On Automorphisms of Polynomial Rings , Bull. London Math. Soc. 14, 1982.[Sh1982’] A. Shamsuddin,
Rings with automorphisms leaving no nontrivial proper ideals invariant , Cana-dian Math. Bull. 25, 1982.
N SOLUTIONS FOR DERIVATIONS 9 [Silv12] J.H. Silverman,
Dynamical Degrees, Arithmetic Degrees, and Canonical Heights for DominantRational Self-Maps of Projective Space , Ergodic Theory and Dynamical Systems, 2012. (Onlinepublication)
Instituto de Matematica, Estatistica e Fisica, Universidade Federal do Rio Grande -FURG, Santo Antonio da Patrulha, 95500-000
E-mail address , R. Baltazar:, R. Baltazar: