Injectivity on One Line

Let k be an algebraically closed field of characteristic zero. Let H : k → k 2 be a polynomial mapping such that the Jacobian JacH is a non-zero constant. In this note we prove, that if there is a line l ⊂ k such that H|l : l → k 2 is an injection, then H is a polynomial automorphism. 1. Main result Let k be an algebraically closed field of characteristic zero. Put k = k \ {0}. By Aut k we denote the group of polynomial automorphisms i.e. all mappings H = (f, g) : k → k, f , g ∈ k[x, y] for which there exists an inverse polynomial mapping. Remind that for H ∈ Aut k the Jacobian JacH is a non-zero constant. The famous Keller conjecture states that any polynomial mapping which has the non-zero constant Jacobian is a polynomial automorphism. Theorem 1.1. Let H : k → k be a polynomial mapping such that JacH ∈ k. If there exists a line l ⊂ k such that H |l : l → k 2 is injective then H is a polynomial automorphism. The proof of the above theorem is given in Section 2 of this note. Let us note here that a weaker result (injectivity on three lines) has been proved recently in [4]. 2. Proof of the theorem The proof of our result is based on the Abhyankar–Moh theorem and on some properties of Newton’s polygon which we quote below. If f is a polynomial in k[x, y] then Sf denotes the support of f , that is, Sf is the set of integer points (i, j) such that the monomial xy appears in f with a non-zero coefficient. We denote by Nf the convex hull (in the real space R ) of Sf ∪ {(0, 0)}. The set Nf is called (see [2]) Newton’s polygon of f . Theorem 2.1. ( [2], [3, theorem 3.4.]) Let f , g be polynomials in k[x, y]. Assume that Jac(f, g) is a non-zero constant and deg f > 1, deg g > 1. Then the polygons Nf and Ng are similar, that is Ng = deg g deg f Nf . The following lemma is easy to check, so we omit the proof. Lemma 2.2. Let f , g be polynomials in k[x, y] with Jac(f, g) ∈ k. If deg f ≤ 1 or deg g ≤ 1 then the mapping (f, g) is a polynomial automorphism. Typeset by AMS-TEX 1 2 BY JANUSZ GWOŹDZIEWICZ Lemma 2.3. Let H = (f, g) be a polynomial mapping such that JacH is a non-zero constant. If H(x, 0) = (x, 0) then H ∈ Aut k. Proof. First suppose that deg f > 1, deg g > 1. We have f(x, 0) = x (1) g(x, 0) = 0. (2) From (1) the point (1, 0) belongs to the polygon Nf , so by theorem 2.1 ( deg g deg f , 0) ∈ Ng. This means that the polynomial g contains some monomials of the form x , i > 0 with non-zero coefficients but this is a cotradiction with (2). We have deg f ≤ 1 or deg g ≤ 1 and by lemma 2.2 H is a polynomial automorphism. Proof of theorem 1.1. Without loss of generality we may assume that the line l has an equation y = 0. Otherwiese we replace H by H ◦ L where L is an affine automorphism such that L({y = 0}) = l. Put γ(x) = H(x, 0). By assumptions of the theorem the mapping γ : k → k is an injection and γ(x) 6= 0 for x ∈ k, hence γ is an embedding of the line in the plane. By Abhyankar-Moh theorem [1] there exist an automorphism H1 ∈ Aut k 2 such that γ(x) = H1(x, 0). Let G = H1 ◦H . We get JacG = JacH −1 1 JacH is a non-zero constant and G(x, 0) = (x, 0), so by lemma 2.3 G ∈ Aut k. Therefore H = H1 ◦G is a polynomial automorphism.