aa r X i v : . [ m a t h . AG ] M a y PLANE JACOBIAN CONJECTURE FOR RATIONALPOLYNOMIALS
NGUYEN VAN CHAU
In memory of Professor Carlos Gutierrez
Abstract.
A non-zero constant Jacobian polynomial maps F = ( P, Q ) of C is invertible if P and Q are rational polynomials. Keywords and Phrases:
Jacobian conjecture, Rational polynomial. We shall call a polynomial map F = ( P, Q ) : C −→ C is Keller map if F satisfies the Jacobian condition J ( P, Q ) := P x Q y − P y Q x ≡ c = 0. The Jacobianconjecture, posed first in 1939 by Ott-Heinrich Keller [8] and still opened, assertsthat every Keller map is invertible . We refer the readers to the nice surveys [3] and[4] for the history, the recent developments and the related topics of this mysteriousproblem.One of most simple topological cases of Keller’s question, for which we may hopeto have a complete solution by using the rich knowledge on plane algebraic curves,is the case when one or both of P and Q are rational polynomials , i.e. polynomialswith generic fibre diffeomorphic to the sphere with finite number of punctures.Since 1978 Razar had found the following. Theorem 1 (Razar’s Theorem, [22]) . A Keller map F = ( P, Q ) is invertible if P is a rational polynomial with all irreducible fibres. In attempting to understand the nature of the plane Jacobian conjecture, Razar’s theorem had been reproved by Heitmann [6], Lˆe and Weber [12], Friedland [21],Nemethi and Sigray [14] in several different algebraic and algebro-geometric ap-proaches. In fact, Vistoli [23] and Neumann and Norbury [15] observed that ev-ery rational polynomial with all irreducible fibres is equivalent to the projection( x, y ) x up to algebraic coordinates. Recently, Lˆe ([9], [10]) proved that a Kellermap F = ( P, Q ) is invertible if P is a rational polynomial and, in addition, P issimple, i.e. in regular extensions p : X −→ P of P over a compactification X of C the restriction of p to each irreducible component of the divisor at infinity D := X \ C have degree 0 or 1. As shown in [20], Lˆe’s result is still true withoutthe condition that P is rational.In this paper we would like to note that the plane Jacobian conjecture is truefor the case when both of P and Q are rational. Theorem 2 (Main Theorem) . Suppose F = ( P, Q ) is a Keller map. If P and Q are rational polynomials, then F is invertible. By a polar branch we mean an irreducible branch curve at infinity along which F tends to infinity. An obvious simple fact is that under the Jacobian conditionany irreducible component of any fiber of P must contains some polar branches. Otherwise, the restriction of F to such an exceptional component must be constantmapping that is impossible. In order to prove Theorem 2 we will try to show thateach of fibres of P has only one polar branch. This ensures that all fibres of P areirreducible. Then, invertibility of F follows from Razar’s theorem. Our proof is based on the following facts on Keller maps.Following [7], the non-proper value set A f of a polynomial map f of C is theset of all values a ∈ C such that f ( b i ) → a for a sequence b i → ∞ . This set A f iseither empty or an algebraic curve in C each of whose irreducible components isthe image of a non-constant polynomial map from C into C . If f is a Keller map,the restriction f : C \ f − ( A f ) −→ C \ A f gives a unbranched covering. Theorem 3 ( see Theorem 4.4 in [16]) . Suppose F = ( P, Q ) is a Keller map.The non-proper value set A F , is not empty, is composed of the images of somepolynomial maps t ( α ( t ) , β ( t )) , α, β ∈ C [ t ] , satisfying deg α deg β = deg P deg Q . (1)
In particular, A F can never contains smooth irreducible components to C . This fact was presented in [16] and can be reduced from [2] (see also [18] and [19]for other refine versions). It can be used in assuming that there is a plane algebraiccurve E , composed of some irreducible components parameterized polynomial maps( α ( t ) , β ( t )) satisfying (1), such that the restriction F : C \ F − ( E ) −→ C \ E (2)gives a unbranched covering. Each of irreducible components of such a curve E isa singular curve approaching to ( ∞ , ∞ ) ∈ P × P ⊃ C . In working with genericfibres of P we may use the following convenience, which was presented in [17] in alittle different statement. Proposition 1 (See Theorem 1 in [17]) . Suppose F = ( P, Q ) is a Keller map and E is a plane algebraic curve in (2). If the line L c := { ( c, t ) : t ∈ C } intersectstransversally each irreducible component of the non-proper set E , then P = c is ageneric fiber of P . Below we shall reduce from the results introduced in the previous section someadvance estimations on the polar branches and the genus of generic fibres of P and Q For each c ∈ C let us denote by P ol ( P, c ) the union of polar branches in the fiber P = c . The germ curve P ol ( P, c ) can be realized as the inverse image F − ( V ( c, R ))for enough large number R >
0, where V ( c, R ) := { ( c, t ) : | t | > R } . Theorem 4.
Suppose F = ( P, Q ) is a Keller map. Then, the family of germ curves P ol ( P, c ) , c ∈ C , is equianalytical. Namely, for each c ∈ C there exists a smalldick ∆ ⊂ C centered at c and a number R > such that the restriction P : [ c ∈ ∆ F − ( V ( c, R )) −→ ∆ LANE JACOBIAN CONJECTURE FOR RATIONAL POLYNOMIALS 3 is an analytic fibration. In particular, the number of irreducible components in
P ol ( P, c ) does not depended on c ∈ C . This fact was contained implicitly in Theorem 3.4 in [16] , stated in terms ofNewton-Puiseux expansions. For convenience, we present here a proof by applyingTheorem 3.
Proof of Theorem 4.
In view point of Theorem 3 we can assume that that there is aplane algebraic curve E , composed of some irreducible components parameterizedpolynomial maps ( α ( t ) , β ( t )) withdeg α deg β = deg P deg Q , such that the restriction F : C \ F − ( E ) −→ C \ E gives a unbranched covering.Let c ∈ C be given. Since the components of E approach to the point ( ∞ , ∞ ) ∈ P × P ⊃ C , there exists a small dick ∆ ⊂ C centered at c and a number R > × { t ∈ C : | t | > R } ) ∩ E = ∅ . This ensures that the restriction F : F − (∆ × { t ∈ C : | t | > R } ) −→ ∆ × { t ∈ C : | t | > R } is a unbranched analytic covering. Note that F − (∆ × { t ∈ C : | t | > R } ) = ∪ c ∈ ∆ F − ( V ( c, R )) . This follows that P : [ c ∈ ∆ F − ( V ( c, R )) −→ ∆is an analytic fibration. (cid:3) Let us denote by g h the genus of the generic fiber of a primitive polynomial h ∈ C [ x, y ]. Lemma 1.
Suppose F = ( P, Q ) is a Keller map. If deg P ≤ deg Q , then g P ≤ g Q .Proof. If F is invertible, of course g P = g Q = 0. Consider the situation when F isnot invertible. We need prove only that if deg P ≤ deg Q , then for each t enoughsmall a generic fiber of P can be topologically embedded into a generic fiber of P + tQ . This ensures that g P ≤ g P + tQ for each t enough small that implies thedesired conclusion.Let A F be the non-proper value set of F . Note that A F is a curve in C and therestriction F : C \ F − ( A F ) −→ C \ A F gives a unramified covering. Replacing F by F + p for a generic point p ∈ C if necessary, we can assume that (0 , ∈ C \ A F and for | t | < ǫ the lines L t given by u + tv = 0 intersects transversally A F . ByProposition 1 the late ensures that for | t | < ǫ the curve P + tQ = 0 is a genericfiber of P + tQ .Now, we will construct topological embeddings ( P = 0) ֒ → ( P + tQ = 0) for | t | < ǫ . To do it, we can choose a box B := {| u | < r ; | u | < s } such that ( L ∩ A F ) ⊂ B and A F ∩ B is a smooth manifold. Since the lines L t intersects transversal A F ,by standard arguments we can modify the motion φ t (0 , v ) := ( − tv, v ) such that NGUYEN VAN CHAU φ t ( A F ∩ L ) ⊂ A F ∩ L t and φ t : L ∩ B −→ φ t ( L ∩ B ) ⊂ L t ∩ B are diffeomorphisms.Let Φ t be the lifting map Φ t induced by the covering F : C \ F − ( A F ) −→ C \ A F .Then, Φ t : F − ( L ∩ B ) −→ { P + tQ = 0 } is an embedding of F − ( L ∩ B ) intothe fiber P + tQ = 0. Since ( L ∩ A F ) ⊂ B , it is easy to see that the fiber P = 0can be deformed diffeomorphic to its subset F − ( L ∩ B ). So, we get the desireembeddings. (cid:3) Now, we are ready to prove the main result.
Proof of Theorem 2.
Let F = ( P, Q ) be a given Keller map. Assume that P and Q are rational polynomials. We can assume in addition that the following conditionsholds:a) deg P < deg Q ;b) The curve P = 0 is irreducible;c) For generic λ ∈ C the curve λP + Q is generic fiber of λP + Q .Indeed, if deg P = deg Q , by the Jacobian condition we have P + = cQ + for anumber c = 0, where P + and Q + are leading homogenous components of P and Q ,respectively. Then, deg P − cQ < deg Q and P − cQ is also rational by Lemma 1.So, we can replace P by P − cQ . Further, in view of Theorem 3 and Proposition1, we can choose a generic point ( a, b ) ∈ C such that the curve P = a is a genericfiber of P and the curve λP + Q = λa + b is those of λP + Q for generic λ ∈ C . So,we can replace F by F − ( a, b ).We will show that the fiber P = 0 has only one polar branch. Then, by Theorem5 every fiber of P also has only one polar branch. This ensures that all fibres of P are irreducible. Therefore, by Razar’s theorem F is invertible.To do it, we regard the plane C as a subset of the projective plane P andassociate to F = ( P, Q ) the rational map G F : P −→ P defined by G F ( x, y ) = Q ( x, y ) /P ( x, y ) on the part C . The indeterminacy point set of G F consists ofthe set B := F − (0 ,
0) and a finite number of points in the line at infinity L ∞ ofthe chart C . Note that the restriction of G F to L ∞ is equal to the zero, sincedeg P < deg Q . By blowing-up we can remove indeterminacy points of G F andobtain a blowing-up map π : X −→ P and a regular extension g F : X −→ P overa compactification X of C \ B .Now, suppose g F : X −→ P is such a regular extension of G F . Let D ∞ := π − ( L ∞ ) and D b := π − ( b ), b ∈ B . D ∞ is a connected rational curve with simplenormal crossing and its dual graph is a tree. Each D b is a copy of P . The divisor D := X \ ( C \ B ) then is a distinct union of D ∞ and D b . Observe, for generic λ ∈ P the fiber g F = λ is the closure in X of the portion { ( x, y ) ∈ C \ B : λP ( x, y ) + Q (( x, y ) = 0 } of the fiber λP + Q = 0. Since deg P < deg Q byCondition (a), in view of Lemma 1 the polynomials λP + Q are rational. Therefore,by Condition (c) generic fibres of g F are irreducible rational curves. This meansthat g F : X −→ P is a P -fibration over P .Now, we consider the fiber of g F over ∞ , denoted by C ∞ . Let us denote D ∞ := C ∞ ∩ D ∞ and by Γ the closure in X of the portion { ( x, y ) ∈ C \ B : P ( x, y ) = 0 } .By the conditions (a) and (b) Γ is an irreducible rational curve and D ∞ contains atleast the proper transform of the line at infinity of C . Furthermore, by the Jacobiancondition the multiplicity of g F on Γ is equal to 1. Obviously, C ∞ = D ∞ ∪ Γ andΓ intersects D ∞ at polar branches of the fiber P = 0. By the well-know fact (see LANE JACOBIAN CONJECTURE FOR RATIONAL POLYNOMIALS 5 for example [5] and [1]), that any reducible fiber of a P -fibration over P the canbe contracted by blowing down to any its component of multiplicity 1, the fiber C ∞ can be contracted to Γ. In particular, D ∞ is a blowing-up version of one pointand C ∞ is a blowing-up version of P . Hence, the dual graphs of C ∞ and D ∞ aretree. Therefore, the curve Γ intersects transversally D ∞ at a unique smooth pointof D ∞ . This follows that the fiber P = 0 has only one polar branch. (cid:3) To conclude we would like to note that it remains open the question whether aKeller map F = ( P, Q ) with rational component P is invertible . In view of Theorem2 and its proof, if such a Keller map is not invertible, then deg P < deg Q . In fact,it is possible to show that in such a Keller map deg P does not divide deg Q andthe fibres of P have exactly two polar branches. We will return to discuss on thequestion in a further paper. Acknowledgements.
The first versions of this paper contain some serious mistakesthat was pointed out by Pierete Cassou-Nogues. We would like to express ourthank to her for many valuable discussions. We also thank very much ShreeramAbhyankar, Hyman Bass, Arno Van den Essen, Carlos Gutierrez, Lˆe D˜ung Tr´ang,Walter Neumann, Mutsui Oka, Stepan Orevkov and Nguyen Tien Zung for allvaluable encouragements they spend to us.
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