On simple Shamsuddin derivations in two variables

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 example, 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.

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.
At a first moment, in the §2, we show that the conjecture is true for a family of derivations, named Shamsuddin derivations (Theorem 6). For this, we use a theorem of the Shamsuddin [Sh1977], mentioned in [No1994, Theorem 13.2.1.], that determines a condition 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, we analysed necessary conditions for an automorphism to belong to the isotropy of a simple derivation. For example, we prove that if such an automorphism has a fixed point, then it 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 in Aut(C[x, y]) d , with d a simple derivation, has dynamical degree 1 (Corollary 9). More precisely, the condition dynamical degree > 1 corresponds to exponential growth of degree under iteration, and this may be viewed as a complexity of the automorphism in the isotropy (see [FM1989]).

Shamsuddin derivation
The main aim of this section is study the isotropy group of the a Shamsuddin derivation in k[x, y]. In [No1994, §13.3], there are numerous examples of these derivations and also shown a criterion for determining the simplicity; furthermore, Y.Lequain [Leq2008] introduced an algorithm for determining when an Shamsuddin derivation is simple. However, before this, the following example shows the isotropy of an arbitrary derivation can be complicated.
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, with p(y), q(y) ∈ k[y]. However, ρ is an automorphism, in other words, the determinant of the Jacobian matrix must be a nonzero constant. Thus, 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 an indeterminate. Then, we can also extend d to a unique derivationd 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. Extend the derivation d to a derivationd of the polynomial ring R[t] by settingd(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.
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 the conditions of Theorem 3. Thus, we know that d is simple if, and only if, there exist no elements r ∈ R such that ∂ x (r) = xr + 1; but the right side of the equivalence is satisfied by the degree of r. Therefore, by Theorem 3, d is a simple derivation of k[x, y].
One can determine the simplicity of the a Shamsuddin derivation according the polynomials a(x) and b(x) (see ([No1994, §13.3])).
+D(a s (x))y s + sa s (x)y s−1 (a(x)y + b(x)) = 1 Comparing the coefficients in y s , D(a s (x)) = −sa s (x)a(x), 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 0 (x). Therefore D(a 0 (x)) = 1 and f = x+c, with c constant.
Using the condition (2), Writing g(x, y) = b 0 (x) + b 1 (x)y + . . . + b t (x)y t ; wherein, by the previous part, we can suppose that t > 0, because ρ is a automorphism. Thus Comparing the coefficients in y t , we obtain + 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 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 infinite distinct roots. If a(x) is a constant, then a(x) D is not a simple derivation (a consequence of [Leq2008, 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) again, Considering the independent term of y, If b 1 = 1, we consider the derivation D ′ such that 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( what contradicts the equation (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) = 0, consequently b 0 (x) = 0. This shows that ρ = 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 some characteristics of the elements in Aut(k[x, y]) D . For this, we use some concepts presented in the previous sections and also the concept of dynamical degree of a polynomial application.
In [BP2015], which was inspired by [BLL2003], we introduce and study a general notion of solution associated to a Noetherian differential k-algebra and its relationship with simplicity.
The following proposition geometrically says that if an element in the isotropy of a simple derivation has fixed point then it is the identity automorphism. In other words, ϕρ is a solution of D passing through ρ −1 (m) = m. Therefore, by the uniqueness of the solution ([BP2015, Theorem.7.(c)]), ϕρ = ϕ. Note that ϕ is one to one, because k[x 1 , ..., x n ] is D-simple and ϕ is a nontrivial solution. Then, we obtain that ρ = id.
F. Lane, in [Lane75], proved that every k-automorphism ρ of k[x, y] leaves a nontrivial proper 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 kautomorphism given by χ(x) = x + 1, χ(y) = y + xz + 1 e χ(z) = y + (x + 1)z has no nontrivial invariant ideal. Suppose that I, this invariant ideal, is radical. Let I = (m 1 ∩ . . . ∩ m s ) ∩ (p 1 ∩ . . . ∩ p t ) be a primary decomposition where each ideal m i is a maximal ideal and p j are prime ideals with height 1 such that p j = (f j ), with f j irreducible (by [Kaplan74,Theorem 5.]). If we claim that ρ N leaves invariant one maximal ideal for some N ∈ N: suppose m 1 this ideal. Indeed, we know that ρ(m 1 ) ⊃ m 1 ∩ . . . ∩ m s , since ρ(m 1 ) is a prime ideal, we deduce that ρ(m 1 ) ⊇ m i , for some i = 1, . . . , s ([AM1969, Prop.11.1.(ii)]). Then, ρ(m 1 ) = m i ; that is, ρ N leaves invariant the maximal ideal m 1 for some N ∈ N. Thus follows from Proposition 7 that ρ N = id.
Let C i be a component irreducible of V (f ) that has genus greater than two. Note that there exist M ∈ N such that ρ M (C i ) = C i . By [FK1992, Thm. Hunvitz, p.241], the number of elements in Aut(C i ) is finite; in fact, #(Aut(C i )) < 84(g i − 1), where g i is the genus of C i . Then, we deduce that ρ is a automorphism of finite order.
We take for the rest of this section k = C.