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bstract. It is show that a polynomial map F = (P, Q) of C2 is a polynomial automorphism of C2 if J(P, Q) := PxQy − PyQx ≡ c 6= 0 and, in addition, both of polynomials P and Q are rational, i.e. the generic fibres of P and of Q are irreducible rational curves.
JACOBIAN PAIRS OF TWO RATIONAL POLYNOMIALS ARE AUTOMORPHISMS NGUYEN VAN CHAU In memory of Professor Carlos Guti´errez Abstract. It is show that a polynomial map F = (P, Q) of C2 is a polynomial automorphism of C2 if J(P, Q) := Px Qy − Py Qx ≡ c = 0 and, in addition, both of polynomials P and Q are rational, i.e. the generic fibres of P and of Q are irreducible rational curves. Keywords and Phrases: Jacobian conjecture, Rational polynomial. Mathematics Subject Classification 2000: 14R15, 14H20. 1. We will call a polynomial function h : C2 −→ C to be rational if the generic fibres of h are irreducible rational curves ( and hence, all fiberes of h are rational curves). The plane Jacobian conjecture (JC2)([3]), was posed firstly in 1939 by Ott-Heinrich Keller [8] and still opened, claims that a polynomial map F = (P, Q) : C2 −→ C2 is a polynomial automorphism of C2 if it satisfies the Jacobian condition J(P, Q) := Px Qy − Py Qx ≡ c = 0. In usual, such a pair F = (P, Q) satisfying the Jacobian condition is called a Jacobian pair. One of simple topological cases of (JC2) is when one or both of components P and Q are rational. Since 1978, in an attempt to understand the nature of (JC2) Razar [16] had found the following fact: A Jacobian pair F = (P, Q) is a polynomial automorphism of C2 if P is rational and all of its fibres are irreducible. Later, this fact had been re-proved by some authors in several different algebraic and algebro-geometric approaches ([6],[9],[4],[10]). In fact, it is shown in [17], [11] and [15], that if h : C2 −→ C is a polynomial function whose fibres are irreducible curves of same a genus, then h must be equivalent to the projection (x, y) → x up to algebraic coordinates. Our result here is the following. Theorem 1. A Jacobian pair F = (P, Q) is a polynomial automorphism of C2 if P and Q are rational On the other words, (JC2) would be proved if one could show that the Jacobian condition implies that both of P and Q are rational. In the proof presented in the next sections, Theorem 1 will be reduced to the result mentioned in above by showing that among polynomials aP + bQ, (0, 0) = (a, b) ∈ C2 , the member of minimal degree is rational and all of it’s fibres are irreducible. Our argument is based on the basic properties of the rational fibrations and some early well-known observations on (JC2). This work is supposed in partial by NAFOSTED and VIASM, Vietnam. 1 2 NGUYEN VAN CHAU 2. Recall from [7], the non-proper value set AF of a polynomial map F : C2 −→ C2 is the set of all values a ∈ C2 such that F (bi ) → a for a sequence bi → ∞. This set AF is either empty or an algebraic curve in C2 composed of the images of some non-constant polynomial maps from C into C2 . When F does not have singular points, the restriction F : C2 \ F −1 (AF ) −→ C2 \ AF gives a unbranched covering and F has a polynomial inverse if and only if AF = ∅. Recall, a number c ∈ C is called a regular value of a polynomial function h : C2 −→ C if there is an open neighborhood U of c such that the restriction h : h−1 (U ) −→ U determines a locally trivial fibration. A fiber h−1 (c) at a regular value c of h is called a regular fiber of h. Proposition 1. Suppose F = (P, Q) is a Jacobian pair. Then, a) (Theorem 4.4 in [12]) The non-proper value set AF , if not empty, is composed of the images of some polynomial maps t → (α(t), β(t)), α, β ∈ C[t], satisfying deg P deg α = . deg β deg Q In particular, AF can never contains irreducible components isomorphic to C. b) (see [13]) A number c ∈ C is a regular value of P if and only if the vertical line Lc := {(c, t) : t ∈ C} intersects transversally every irreducible local branh of AF located at Lc ∩ AF . Conclution (a) was presented in [12] and can be deduced from [2] (see also [14] for other refined versions). Conclution (b) was presented in [13] in a little different statement. In view of Proposition 1, in working with a Jacobian pair F = (P, Q), we can assume that there is a curve E composed of the images of some polynomial maps t → (α(t), β(t)), α, β ∈ C[t], such that deg α deg P = , deg β deg Q (1) F : C2 \ F −1 (E) −→ C2 \ E (2) and gives a unbranched covering. Then, AF ⊂ E and c ∈ C is a regular value of P if the line Lc intersects transversally every irreducible local branh of E located at Lc ∩ E. Observe, when deg P > 0 and deg Q > 0, by (1) all the branches at infinity of E approach to (∞, ∞). So, one can find an open set U ⊂ C2 such that i) U ∩ E = ∅ ii) The projection π : U −→ C, (u, v) → u, gives a trivial fibration and for each c ∈ C the fiber π −1 (c) is of the form {(c, v) ∈ C2 : |v| > R} for a number R > 0. Then, by (2) one obtains a trivial fibration ψ : F −1 (U ) −→ C, ψ := π ◦ F . For each c ∈ C the fiber ψ −1 (c) is the distinct union of all irreducible branches at infinity γ of the curve P = c such that along γ the function Q tends to infinity. For convenience, we will call such an irreducible branch γ polar branh of F on the curve P = c. These observations alllows us to obtain JACOBIAN PAIRS OF TWO RATIONAL POLYNOMIALS ARE AUTOMORPHISMS 3 Corollary 1. Suppose F = (P, Q) is a Jacobian pair. Then, the number of the polar branches of F on a fiber P = c, c ∈ C, does not depended on c. Corollary 2. Suppose F = (P, Q) is a Jacobian pair, deg P > 1 and deg Q > 1. ¯ := (P − cQ − a, Q − b) we Then, there are numbers a, b, c ∈ C such that for (P¯ , Q) have ¯ a) deg P¯ < deg Q, ¯ for all but a finite number b) the zero is a regular value of P¯ and of αP¯ + β Q 2 of (α, β) ∈ C . Proof. If deg P < deg Q, we let c = 0. If deg P = deg Q, by the Jacobian condition, we have P+ = cQ+ for a number c = 0, where P+ and Q+ are leading homogeneous components of P and Q respectively. Then, deg P − cQ < deg Q. Next, considering a unbranched covering (2), we can always choose a point (a, b) ∈ C2 \ E such that the line La intersects transversally the curve E. Obviously, the curve E intersects transversally all but a finite number of the lines passing through (a, b). Let P¯ := ¯ := Q−b. Then, by Proposition 1 we get all desired conclutions. P −cQ−a and Q 3. Let us denote by gh the genus of the generic fiber of a primitive polynomial h ∈ C[x, y]. Lemma 1. Suppose F = (P, Q) is a Jacobian pair. If deg P ≤ deg Q, then gP ≤ gQ . Proof. We only need to show that if deg P ≤ deg Q, then a regular fiber of P can be topologically embedded into a regular fiber of P + tQ for small enough t = 0. This ensures that gP ≤ gP +tQ that implies the desired conclusion. Again, we consider a unbranched covering (2). Replacing F by F −p for a generic point p ∈ C2 \ E if necessary, we can assume that (0, 0) ∈ C2 \ E and that for small enough ε > 0 the lines Lt , given by u + tv = 0, |t| < ε, intersects transversally E. By Proposition 1 the late ensures that for |t| < ε the curve P + tQ = 0 is a regular fiber of P + tQ. Now, we will construct topological embeddings (P = 0) → (P + tQ = 0) for |t| < ε. We can choose a box B := {(u, v) ∈ C2 : |u| < r; |v| < s} such that (L0 ∩ E) ⊂ B and E ∩ B is a smooth manifold. Since the lines Lt , |t| < , intersects transversally E, by standard arguments, we can modify the motion φt (0, v) := (−tv, v) such that φt (E ∩ L0 ) ⊂ E ∩ Lt and φt : L0 ∩ B −→ φt (L0 ∩ B) ⊂ Lt ∩ B are diffeomorphisms. Let Φt be the lifting map induced by the covering F : C2 \ F −1 (E) −→ C2 \ E. Then, each restriction Φt : F −1 (L0 ∩ B) −→ {P + tQ = 0} is an embedding of F −1 (L0 ∩ B) into the fiber P + tQ = 0. Since (L0 ∩ E) ⊂ B, it is easy to see that the fiber P = 0 can be deformed diffeomorphically to its subset F −1 (L0 ∩ B). So, we get the desired embedding. 4. Proof of Theorem 1. Assume that F = (P, Q) is a Jacobian pair and P and Q are rational. In view of Corollary 2, without a loss of generality, we can assume (a) deg P < deg Q, (b) the curve P = 0 is a regular fiber of P and (c) the curve αP + βQ = 0 is a regular fiber of aP + bQ for all but a finite number of (α, β) ∈ C2 . From Lemma 1 we have (d) αP + βQ is rational for all (0, 0) = (α, β) ∈ C2 . 4 NGUYEN VAN CHAU Claim 1. If F has only one polar branch on the fiber P = 0, then F is an automorphism of C2 . Proof. Corollary 1 says that the number of polar branch of F on the fiber P = c does not depended on c ∈ C. One can easy verify that under the Jacobian condition a fiber P = c is irreducible if F has only one polar branch on P = c. Thus, if F has only one polar branch on the curve P = 0, all of the fibres of P are irreducible rational curves. Then, by the results mentioned in the first section, P is equivalent to the projection (x, y) → x up to polynomial automorphisms of C2 , and hence, F = (P, Q) is a polynomial automorphism of C2 . Claim 2. F has only one polar branch on the fiber P = 0. Proof. We will regard C2 as the set of points (x : y : 1) of the projective plane P2 and consider a rational map Φ : P2 −→ P1 , given by Φ(x : y : 1) := (P (x, y) : Q(x, y)) for (x, y) ∈ C2 \ B, where B := F −1 (0, 0) ⊂ C2 . The indeterminacy point set of Φ consists of the finite set B and a finite number of points in the line at infinity L∞ of C2 . Let π : X −→ P2 be a blow-up of P2 , which is obtained by a minimal number of blow-ups removing all indeterminacy points of Φ and of the resulted blow-up versions of Φ. Consider the corresponding regular extension ϕ : X −→ P1 of Φ. It follows from (c-d) that the extension ϕ : X −→ P1 gives a rational fibration, i.e. a fibration with generic fiberes isomorphic to P1 . Let us denote D := X \ (C2 \ B), D∞ := π −1 (L∞ ), Db := π −1 (b), b ∈ B, and by ∞ the proper transform of L∞ . The minimality of π guarantees that among irreducible components of D the only component ∞ and the horizontal components of ϕ may have self-intersection −1. Then, by the Jacobian condition every divisor Db is an irreducible rational curve and is a horizontal component of ϕ. Let Γ be the proper transform of the closure in P2 of the fiber P = 0, C be the fiber of ϕ at (1 : 0) ∈ P1 and Γ∞ := C ∩ D∞ . It follows from (a-b) that C = Γ ∪ Γ∞ , Γ is an irreducible rational curve and ∞ is a component of Γ∞ . Further, an irreducible branch at infinity γ ⊂ C2 is a polar branch of F = (P, Q) on the affine curve P = 0 if and only if the proper transform of γ is an irreducible branch of Γ located at Γ ∩ Γ∞ . Moreover, by the Jacobian condition the divisor Γ occurs in the fiber C with multiply 1. It is a well-known fact that every reducible fiber of a rational fibration is a tree curve composed of irreducible rational curves with simple normal crossing and can be blown-down to each of its irreducible components of multiply 1 (see, for example, [1] and [5]). In our situation, the curve Γ∞ intersects transversally with Γ and each connected component V of Γ∞ can be blown-down to the unique intersection point of V and Γ. As observed in above, by the minimality of π : X −→ P2 among the irreducible components of Γ∞ ⊂ D∞ the only component ∞ may have self-intersection −1 (and hence, 2 = −1 in this case). Furthermore, any process of blowing-down each componet V must begin by blowing-down an irreducible component of V having self-intersection −1. It follows that Γ∞ is connected and intersects Γ at one point. The late implies that F = (P, Q has only one polar branch on the affine curve P = 0. 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Institute of Mathematics, 18 Hoang Quoc Viet, 10307 Hanoi, Vietnam E-mail address: nvchau@math.ac.vn ... there exists a Jacobian pair F = (P, Q) with non-trivial rational polynomial JACOBIAN PAIRS OF TWO RATIONAL POLYNOMIALS ARE AUTOMORPHISMS P In view of Theorem and its proof, for such a Jacobian pair,... alllows us to obtain JACOBIAN PAIRS OF TWO RATIONAL POLYNOMIALS ARE AUTOMORPHISMS Corollary Suppose F = (P, Q) is a Jacobian pair Then, the number of the polar branches of F on a fiber P = c,... So, we get the desired embedding Proof of Theorem Assume that F = (P, Q) is a Jacobian pair and P and Q are rational In view of Corollary 2, without a loss of generality, we can assume (a) deg