The Cremona problem in dimension 2

The Cremona conjecture, also called Jacobi problem, claims that a polynomial morphism Cn⟶Cn\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${{\mathbb C}^n \longrightarrow {\mathbb C}^n}$$\end{document} is invertible as a polynomial morphism if its Jacobian is constant and not zero. In this paper, we show that the conjecture is true for n=2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$n = 2$$\end{document}. The starting point of our proof is an important result of Shreeram Abhyankar. Then we use a computation in rigid geometry to achieve the result.


Introduction.
A polynomial map (f, g) : C 2 −→ C 2 is given by two polynomials f and g in two variables X and Y with complex scalars. We write f and g as sums of their homogenous components f = f m + · · · + f m and g = g n + · · · + g n , where f μ respectively g ν are linear combinations of the terms of total degree μ respectively ν. The forms f m respectively g n of highest degree are called the leading forms.
It was shown by Abhyankar that, for a given counterexample (f, g) to the Jacobian conjecture in dimension 2, one can assume that, after a suitable transformation of variables, the leading forms of f and g have the following shape f m = X m1 Y m2 and g n = X n1 Y n2 ; cf. [1,Theorem 8.7] or [5,Corollary 10.2.22]. In this paper, we will show that this assumption leads to a contradiction. Thus the Jacobian conjecture is true in dimension 2.

Division algorithm.
In this section, let K be an algebraically closed field of characteristic 0. The K-algebra L := K[X, Y ] XY consists of all Laurent polynomials in two variables. It carries a canonical graduation of type Z given by the total degree function. Let H n be the subspace of all homogenous Laurent polynomials of degree n including the zero polynomial. For f = ν∈Z f ν ∈ L and f = 0, we set On L, we have a filtration (L n ; n ∈ N) where The completion with respect to this filtration is denoted by cf. [5,Prop. 10.2.8]. The degree, the multiplication, and the filtration on A are declared as on L. The K-algebra A represents the formal functions on a neighborhood of the twice punctured projective line at infinity which behave like meromorphic functions there. The algebra A has similar properties as the algebra R ∼ defined in [5,Prop. 10.2.8]. In this section, we consider these functions without conditions of convergence; in Section 2, we will focus on that by means of rigid geometry.

Lemma 1.1. An element g ∈ A is a unit in A if and only if g is of the form
Such a representation is unique. Such a unit g admits a k-th root for 0 = k ∈ Z if and only if k divides both numbers n 1 and n 2 .
Proof. The proof can be left to the reader. For example, we have for the inverse For c = 1 and k ∈ N, the k-th root is given by if k divides n 1 and n 2 . We consider this as the canonical k-th root of g.
Vol. 119 (2022) The Cremona problem in dimension 2 55 and r ∈ Q be such that r · n 1 and r · n 2 belong to Z, then In the following, we denote by ∂/∂X respectively ∂/∂Y the partial derivatives of Laurent series. Obviously they give rise to K-derivations on the Kalgebra A. They satisfy the usual rules for K-derivations. Since the field K has characteristic 0, we have ker(∂/∂X , ∂/∂Y ) = K.
is constant and not 0.
As for polynomials in two variables, we also have the notion of a leading form for a Laurent series in A. Proposition 1.4. Consider a Jacobian couple (f, g) as introduced above, where m := deg f and n := deg g. Assume that the leading form of g has the shape g n = X n1 Y n2 .
(a) Then we always have m + n ≥ 2.
Proof. (a) The homogenous components of the Jacobian of degree m + n > 2 vanish and for m + n = 2 it is given by Since that the Jacobian is constant and the degree of a constant is 0, we see that m + n ≥ 2.
For homogenous polynomials, we have Euler's differential equation Then the term in parentheses of equation (2) is equal to This vanishes due to (1) since m + n > 2. Thus we see that the left hand term of equation (2) is equal to 0. Analogously, one shows So the total differential of f n m · g −m n vanishes. Thus, the function f n m · g −m n is constant.
Therefore the following expression is well defined Proof. The first assertion follows from 1.4(b). For the second assertion, we use So we see |m 1 − m 2 | = |n 1 − n 2 | if |m| = |n|. The formula for g m/n follows from 1.2.
Now we turn to the division algorithm. Proposition 1.6. Let (f, g) be as in 1.5. Then there exists a rational number r ∈ Q such that r · n 1 ∈ Z and r · n 2 ∈ Z are integers, and a constant c ∈ K × with deg(f − c · g r ) < m. The couple (c, r) is uniquely determined; actually we have r = m/n and f m = c · X m1 Y m2 . If, in addition, deg(f − c · g r ) = 2 − n, then n 1 = n 2 and the leading form of h := f − c · g r is given by Proof. The first assertion follows from 1.5.
For the supplement, set m := deg(f − c · g r ) = 2 − n. The Jacobian d of the couple (h, g) is equal to the Jacobian of (f, g). Then we have For (i, j) = (1 − n 1 , 1 − n 2 ), it follows that Thus, we see n 1 = n 2 and c 1−n1,1−n2 = 0. For all the other indices, we have If c i,j = 0, then Moreover we know i + j = 2 − n and n = n 1 + n 2 . This yields and i · n 2 = (2 − n − i) · n 1 and hence i · n = i · (n 2 + n 1 ) = (2 − n) · n 1 Corollary 1.7. Keep the assumptions of 1.6. Then we have n 1 = n 2 and there exist a natural number s ∈ N, constants c σ ∈ K, and rational numbers r σ ∈ Q satisfying belongs to A and the leading term of (f − G) fulfills Proof. Apply 1.6 inductively. Note that deg(f − G) is always an integer and that (f − G , g) is a Jacobian couple. Therefore the procedure stops after finitely many steps until we arrive at the situation deg(f − G) = 2 − n since there is at each step at most one term which has to be cancelled. In the case deg(f − G) = 2 − n, we apply the additional claim of 1.6. Then we obtain for the leading form where i/n 1 = j/n 2 = (2 − n)/n as follows from 1.6. Then we subtract (2−n)/n which cancels the term c i,j X i Y j . Thus the assertion is proved.

Convergence of the division algorithm.
In the following, we make use of some elementary results in rigid geometry; for a general reference, we cite [3] or [4]. We consider an algebraically closed field K which is complete with respect to a non-Archimedean valuation and which has residue characteristic 0. We assume that K contains the field K of characteristic 0 as a subfield, where K is the algebraically closed field over which the Jacobian problem is posed. Such a field can be constructed in the following way: Consider the field of fractions K of K[[T ]] and define K as the topological algebraic closure of K . The canonical valuation on K[[T ]] extends to a valuation of K. Note that we write valuations in the multiplicative way. So we obtain on K 2 a canonical structure of rigid space in the sense of Tate. On each subset V ⊂ K 2 , we have the spectral norm of functions f In particular, we have the notion of an affinoid domain V ⊂ K 2 ; for example, bounded domains described by finitely many inequalities Affinoid functions on such a domain are functions which can be uniformly approximated by rational functions without poles in V . Such functions are bounded and take their maximal absolute value in V . Thus the spectral norm |f | V is always a non-negative real number which actually lies in the value group of K. We are mainly interested in domains of the following shape for values ε ≤ ρ belonging to the value group of K. The affinoid functions on W ε,ρ are exactly the Laurent series which converge on W ε,ρ . Of particular interest will be the following domains These subsets are also affinoid and they are open subsets in the rigid analytic sense.

Lemma 2.1. Keep the above notations. Let ε , ρ be elements of the value group
(a) If v is an affinoid function on U := U ε,ρ with |v| U < 1, then the series for any r ∈ Q, converges uniformly on U ε,ρ and gives rise to an affinoid function there. In particular, (1 + v) r is well-defined and affinoid on U ε,ρ . (b) Let g = g n + · · · + g 0 ∈ K[X, Y ] be a polynomial with homogenous components g ν of degree ν. Assume g n = X n1 Y n2 . Then there exists an ε in |K × | such that |g ν (x, y)| < |g n (x, y)| for all (x, y) ∈ U ε,ρ and all ν = 0, . . . , n − 1 and ρ ≥ ε. Especially, for any r ∈ Q with n 1 · r ∈ Z and Vol. 119 (2022) The Cremona problem in dimension 2 59 n 2 · r ∈ Z, the function g r is well-defined and affinoid on U ε,ρ for all ρ with ρ ≥ ε.
For the last assertion, note that (X n1 Y n2 ) r = X m1 Y m2 , where n 1 · r = m 1 and n 2 · r = m 2 with m 1 , m 2 ∈ Z . Then it follows from (a). (f, g) be a Jacobian couple of polynomials with homogenous decompositions
If we apply the division algorithm of 1.6 and 1.7 to f and set v := n−1 ν=0 g ν g −1 n , then there exists an ε ∈ |K × | such that the formal series G defined in 1.7 converges on every affinoid domain U ε,ρ for all ρ ≥ ε and gives rise to an affinoid function there.
Proof. The claim follows from Lemma 2.1.
In the following, we will compute the cardinality of the fibers of (f − G , g) on U ε,ρ . Set k := gcd(n 1 , n 2 ). Then, for any domain V := U ε ,ρ ⊂ U ε,ρ , the fibers of the morphism f − G , g 1/k | V consist of exactly |n 1 − n 2 |/k points. The fibers of (f , g) | V consist of exactly |n 1 − n 2 | points.
Due to the construction, all numbers k · r σ are integers by 1.6 since r σ · n 1 ∈ Z and r σ · n 2 ∈ Z. Obviously, this map is injective. So, for every (x 0 , y 0 ) ∈ V , the map Ψ induces a mapping