Jacobian conjecture via differential Galois theory

We prove that a polynomial map is invertible if and only if some associated differential ring homomorphism is bijective. To this end, we use a theorem of Crespo and Hajto linking the invertibility of polynomial maps with Picard-Vessiot extensions of partial differential fields, the theory of strongly normal extensions as presented by Kovacic and the characterization of Picard-Vessiot extensions in terms of tensor products given by Levelt.


Introduction
The Jacobian Conjecture originates from the problem posed by Keller in [7]. It is the 16th problem in Stephen Smale's list of mathematical problems for the twenty-first century (cf. [11]). Let us recall the precise statement of the Jacobian Conjecture.
Let C denote an algebraically closed field of characteristic zero. Let n > 0 be a fixed integer and let F = (F 1 , . . . , F n ) : C n → C n be a polynomial map, i.e. F i ∈ C[X], for i = 1, . . . , n, where X = (X 1 , . . . , X n ). We consider the Jacobian matrix J F = [ ∂F i ∂X j ] 1≤i,j≤n . The Jacobian Conjecture states that if det(J F ) is a non-zero constant, then F has an inverse map, which is also polynomial.
In spite of many different approaches involving various mathematical tools the question is still open. In 1980 S.S.-S. Wang proved the Jacobian Conjecture for quadratic maps. There are a lot of partial results concerning the general case (see [6]). Recently, in [2] and [3] we have consider the class of Pascal finite automorphisms. On the other hand the conjecture holds under strong additional assumptions. As an example let us recall the result of L. A. Campbell (see [4]), which states that the thesis holds if C(X)/C(F ) is a Galois extension.
In a previous paper using Picard-Vessiot theory of partial differential fields Crespo and Hajto obtained a differential version of the classical theorem of Campbell. Let us consider the field C(X) with the differential structure given by the Nambu derivations (see Section 3). Then Crespo and Hajto proved that if the differential extension C(X)/C is Picard-Vessiot, then F is invertible (cf. [5], Theorem 2). A computational approach to this results has been given in [1].
In [10] Theorem 1, Levelt proved a necessary condition for a differential field extension K/k to be Picard-Vessiot in terms of tensor products.
In this paper, we prove a partial converse of Levelt's theorem. To this end, we use the theory of strongly normal extensions as presented by Kovacic in [8] and [9]. By using the converse of Levelt's theorem together with the above mentioned result by Crespo and Hajto we obtain that a polynomial map is invertible if and only if some associated differential ring homomorphism is bijective. This provides a criterion to check the invertibility of polynomial maps.

Inverting a theorem of A. H. M. Levelt
In this section we prove that Levelt's necessary condition for a differential field extension K/k to be Picard-Vessiot is sufficient for K/k to be strongly normal in the case in which the base field is a field of constants. In the next section we shall apply this result to the extension C(X)/C endowed with the Nambu derivations associated to a polynomial map F in order to obtain an equivalent condition to the invertibility of F .
Let C be an algebraically closed field of characteristic zero. Let R be an integral domain containing C with the differential ring structure given by a finite set ∆ of commuting derivations. Let K be the field of fractions of R with the differential structure given by extending the derivations in ∆ in the standard way. Let us assume that C is the field of constants of K and that K is differentially finitely generated over C. Any derivation δ ∈ ∆ extends to K ⊗ C R by the formula δ(x ⊗ y) = δ(x) ⊗ y + x ⊗ δ(y) on elementary tensors and by linearity to the whole tensor product. Let E denote the differential subring of constants in K ⊗ C R. We consider the differential ring homomorphism Our aim is to prove that if φ is an isomorphism, then the extension K/C is strongly normal. We shall prove first that under the above hypothesis, φ is injective. To this end, we need the following lemma.
induces a bijection between the set I(E) of ideals of E and the set I(K ⊗ C E) of differential ideals of K ⊗ C E.
Proof. To an ideal a of E we associate its extension a e = K ⊗ C a and to an The inclusion a ec ⊃ a is well known. Let us prove a ec ⊂ a. We take a basis Λ of the C-vector space a and extend it to a basis M of the C-vector space E. Then 1 ⊗ M is a basis of the K-vector space K ⊗ C E. Let d be any element in a ec . Then 1 ⊗ d ∈ a ece = a e , so On the other hand d = µ∈M c µ µ, c µ ∈ C, so Comparing the coefficients in both expressions, we get that c µ = 0 for µ ∈ M \ Λ and r λ = c λ , for λ ∈ Λ. Hence d ∈ a. The We take a C-vector space basis Λ of b c and extend it to a basis M of the C-vector space E. Let us choose an element a ∈ b \ b ce such that its representation in the form a = µ∈M r µ ⊗ µ, r µ ∈ K has the smallest number of nonzero terms. First let us consider the case when a is an elementary tensor, i.e. a = r⊗µ, for a ∈ K, µ ∈ M. If µ ∈ Λ, then a ∈ b ce and we have a contradiction. So let us assume that µ ∈ M \ Λ. Then we multiply a by r −1 ∈ K and obtain that 1 ⊗ µ ∈ b and consequently µ ∈ b c , hence again a ∈ b ce and we have a contradiction.
Let us assume now that the representation a = µ∈M r µ ⊗ µ, r µ ∈ K, has at least two nonzero terms. Since b is a differential ideal, then for any differential operator δ ∈ ∆ we have Since a = 0, we can choose µ 0 such that r µ 0 = 0. Then The last equality follows from δr µ 0 r µ − r µ 0 δr µ = 0 for µ ∈ M \ Λ, which holds by the choice of a. Hence δ( rµ rµ 0 ) = 0 for µ ∈ M \ Λ. This means rµ rµ 0 is a constant in K. Hence there exists c µ ∈ C such that r µ = c µ r µ 0 for µ ∈ M \ Λ. We obtain a = µ∈M \Λ Since K is a field, we get that 1 And we have a contradiction with the assumption a ∈ b \ b ce .
Proof. Denote b = ker φ. Using lemma 1 we can assume that b = K ⊗ c, where c ∈ I(E). Take c ∈ c. Then So c = 0 and b = (0).
We recall the notion of almost constant differential ring which will be used in the sequel.
Definition 3 ([8] 5.1). Let A be a differential ring and C A its ring of constants. We say that A is almost constant if the inclusion C A ⊂ A induces a bijection between the set of radical ideals of C A and the set of radical differential ideals of A.
Proposition 4. Let C be an algebraically closed field of characteristic zero. Let R be an integral differential ring containing C and let K be the field of fractions of R. We assume that C is the field of constants of K and that K is differentially finitely generated over C. If the differential morphism is an isomorphism, then the differential ring K ⊗ C R is almost constant.
Proof. If φ is a differential isomorphism, there is a bijection between the set of radical differential ideals of K ⊗ C R and the set of radical differential ideals of K ⊗ C E. By Lemma 1 and [8] Proposition 3.4, this last set is in bijection with the set of radical ideals of the ring of constants E of K ⊗ C R.
Theorem 5. Let C be an algebraically closed field of characteristic zero. Let R be an integral differential ring containing C and let K be the field of fractions of R. We assume that C is the field of constants of K and that K is differentially finitely generated over C. If the differential morphism is an isomorphism, then K/C is a strongly normal extension.
Proof. To prove that K/C is strongly normal, we shall apply [8], Proposition 12.5. Let σ be an arbitrary ∆-isomorphism of K over C. We put σ : K → M, where M is any differential field extension of K and denote by D σ the field of constants of M. Define σ : Because E consists of constants, then ψ(1⊗E) ⊂ D σ (regardless of the choice of the differential isomorphism σ). So We have then the commutative diagram We can then use [8], Proposition 12.5 and conclude that K/C is a strongly normal extension.
Remark 6. Let us observe that in order to prove that K/C is a Picard-Vessiot extension it would be sufficient to know that R is a differentially simple ring. In this case, the fact that K/C is strongly normal and K ⊗ C R is almost constant imply that K/C is Picard-Vessiot.
Theorem 7. Let C be an algebraically closed field of characteristic zero and let F = (F 1 , . . . , F n ) : C n → C n be a polynomial map such that det(J F ) ∈ C \{0}. Let R (respectively K) denote the polynomial ring C[X] (respectively the rational function field C(X)) with the partial differential structure given by the Nambu derivations. We extend these derivations to the tensor product K ⊗ C R and denote by E the field of constants of K ⊗ C R. If the differential ring homomorphism φ : K ⊗ C E → K ⊗ C R, φ(a ⊗ d) = (a ⊗ 1)d is an isomorphism, then F is invertible and its inverse is a polynomial map.
Proof. By Theorem 5, the differential field extension K/C is a strongly normal extension. If we consider the intermediate differential field k = C(F 1 , . . . , F n ), then K/k is again strongly normal. But obviously it is an algebraic extension, so it is a Galois extension. Then by Campbell's theorem [4], F is invertible and its inverse is a polynomial map.
Remark 8. By Proposition 2, the map φ is injective. In order to prove that it is also surjective, it is enough to prove that the elements 1 ⊗ X i , 1 ≤ i ≤ n lie in the image of φ. Hence Theorem 7 provides an effective criterion to check the invertibility of polynomial maps. Finally when we know that F has a polynomial inverse, the ring C[X] is the same as C[F ] and therefore it is the Picard-Vessiot ring over C.