WEAKLY NILPOTENT AND WEAKLY SEMISIMPLE POLYNOMIALS ON THE PLANE

YOSEF STEIN

Abstract. In this paper we introduce weakly nilpotent and weakly semisimple polynomials in K[x, y]. We study the connection between these polynomials and the Jacobian Conjecture as well as the canonical form of the Newton polygon for these polynomials.

1. Introduction

Let K be an algebraically closed field of characteristic zero. For f, g ∈ K[x, y] let:

[f, g] = \frac{∂f}{∂x} \frac{∂g}{∂y} − \frac{∂f}{∂y} \frac{∂g}{∂x}

For a fixed A ∈ K[x, y] and f ∈ K(x, y) let D_{A}(f) = [A, f]; D_{A} is a derivation on both K[x, y] and K(x, y). The operation [ , ] imposes a Lie-algebra structure on both K[x, y] and K(x, y).

The main goal of this paper is to classify polynomials according to their Lie-algebra properties. The derivations D_{A} play a prominent role in studying seemingly purely algebraic properties of K[x, y] (see [7]). They also play an important role in the analytic case K = ℂ, especially in studying problems related to the Jacobian Conjecture (see [1], [3], [5], [6], [8]). Some of the questions studied here were raised and some of the results obtained in the context of Weyl algebras (see [2]). There are also facts and results which are part of ”folklore” or are scattered throughout various papers. This paper will, hopefully, serve two purposes: proving new results and arranging the old ones in some kind of order, making them accessible to mathematicians working on the Jacobian Conjecture and related topics.

We will start with some definitions. Let S, v ∈ K[x, y]; S ≠ 0, v ≠ 0 and let λ ∈ K be such that D_{S}(v) = λv. Such a number λ will be called an eigenvalue of S and v - a λ-eigenfunction of S.

Definition 1.1. S ∈ K[x, y] is called weakly semisimple (w.s.) if it has a non-zero eigenvalue.

Date: December 20, 1999.

This work was supported in part by the EC TMR network ”Algebraic Lie Representations”, grant number ERB FMRX-CT97-0100.

I would also like to thank the Max-Planck-Institut für Mathematik for inviting me and for providing excellent working conditions.

Definition 1.2. A ∈ K[x, y] is called weakly nilpotent (w.n.) if there exists B ∈ K[x, y] such that D_{A}(B) ≠ 0 and D_{A}^{2}(B) = 0.

For a w.s. polynomial S let E(S) denote the set of all eigenvalues of S. For λ ∈ E(S) we can consider the set of all λ-eigenfunctions of S. If we add 0 to this set, we obtain a vector space, which will be denoted by V_{λ}(S).

Now we are able to state some of the more important results and to formulate several conjectures.

(1) If a w.s. polynomial S has a non-trivial rational λ-eigenfunction, then it has a non-trivial polynomial λ-eigenfunction (Corollary to Lemma 2.1).

(2) For a w.s. S, E(S) is a group isomorphic to ℤ (Theorem 2.1).

(3) If a polynomial A is weakly nilpotent on the field K(x, y), then it is w.n. on K[x, y] (Proposition 2.4).

(4) A polynomial is w.n. iff it has a rational Jacobian mate (Proposition 3.1). If, in addition to this, all its fibers are irreducible, then it has a polynomial Jacobian mate (Corollary to Theorem 3.1).

So, if the Jacobian Conjecture is true, then every totally irreducible w.n. polynomial must be a coordinate. In other words, it would be possible to disprove the Jacobian Conjecture, if we were able to construct a totally irreducible w.n. polynomial which is not a coordinate. This provides, in my opinion, a good way to try to construct a counter-example, if such a thing exists, of course. Using the techniques developed in the end of Section 2, it is possible to construct a variety of more and more complicated w.n. polynomials as eigenfunctions of w.s. polynomials, and we can try to do it in such a way, that the end result is not a polynomial of a coordinate (which is easily done by applying the algorithm for reduction to the quasimonic form, developed in Section 4), and is totally irreducible (which is much more difficult). The main advantage of such an approach is that it is necessary to construct only one polynomial in order to disprove the Jacobian Conjecture and not a pair of Jacobian mates. The degree of this polynomial does not have to be especially large, but the polynomial is guaranteed to have a Jacobian mate, which may have a very large degree. This situation is not covered by the famous result of Moh ( [4]), which proves the Jacobian Conjecture in the case when the degrees of both mates are ≤ 100.

At this point I would like to formulate three conjectures:

Conjecture 1.

All irreducible components of a fiber of a w.n. or a w.s. polynomial are rational curves.

Proving this conjecture leads, in a standard way, to proving the Jacobian Conjecture.

Let us call both w.n. and w.s. polynomials special.

Conjecture 2.

All irreducible factors of a special polynomial are special.

Conjecture 3.

At least one of the irreducible factors of a w.n. polynomial is w.n.

Proving these two conjectures will, in my opinion, be a significant progress towards proving the Jacobian Conjecture.

Section 4 deals with a ”canonical form” of the Newton polygon for w.s. and w.n. polynomials.

The motivation for the results comes from J. Dixmier [2], who obtained several similar results for the Weyl algebras. Our methods, however, are quite different. The main result of Section 4 is that the Newton polygon of a w.n. polynomial can be reduced by an automorphism of K[x, y] to a special form (quasi-monic), unless the polynomial in question is a polynomial of a coordinate.

Remark. If a polynomial P has a Jacobian mate, then P is w.n. It is possible, therefore, to reduce P to the quasi-monic form, unless P is a coordinate. This fact was known for a long time and proofs were published in various papers. Unfortunately, many unsuccessful attempts to prove the Jacobian Conjecture were based on nothing more than this single fact, without realizing that this property is shared by all w.n. polynomials, e.g. those which have rational and not necessarily polynomial Jacobian mates.

2. Basic properties

We will, first of all, prove the following simple result:

Proposition 2.1. Let S ∈ K[x, y] be w.s. Then S is primitive.

Proof. Assume, on the contrary, that S = R(A); where A ∈ K[x, y], R(A) ∈ K[A] and deg R ≥ 2. Since S is w.s., there exist v ∈ K[x, y]; v ≠ 0 and λ ∈ K; λ ≠ 0 such that D_{S}(v) = λv. Then: λv = D_{S}(v) = R′(A)D_{A}(v). Applying D_{S} to v n times and using the fact that D_{S} and D_{A} commute we obtain by induction on n: λ^{n}v = D_{S}^{n}v = (R′(A))^{n}D_{A}^{n}(v). Thus D_{A}^{n}(v) ≠ 0 and deg v ≥ n deg R′(A) for all n. This is impossible since deg R ≥ 2.

Here is the place to recall some of the results obtained in [7]. The main object of study in the second section of [7] was the equation D_{A}(f) = T f with A, T ∈ K[x, y]; f ∈ K(x, y), f ≠ 0. So, our present notion of eigenvalue is a particular case of a more general concept. The set of all polynomials T, satisfying the equation D_{A}(f) = T f for a given polynomial A and some none-0 rational function f, was denoted by Γ̃(A) and the main result was that Γ̃(A) is a free abelian group of finite rank ρ(A) ( [7], Theorem 2.4), where ρ(A) is the total reducibility order of A ([S4], p.113). There is one additional result which can be easily deduced from the proofs of various results in [7], but we will prove it here to make it easier for the reader:

Lemma 2.1. Let A, T, B ∈ K[x, y]; B ≠ 0 satisfy the equation D_{A}(B) = T B. Then there exists G ∈ K[x, y]; G ≠ 0 such that GB = R(A) ∈ K[A].

Proof. Let B be an irreducible polynomial satisfying the equation D_{A}(B) = T B for a polynomial T. Then, by Lemma 2.1. of [7], B is a divisor of A − α for some α ∈ K. Let now B = B_{0}^{n}, where n is a positive integer and B_{0} is an irreducible polynomial and assume that B satisfies the equation D_{A}(B) = T B. Then nB_{0}^{n−1}D_{A}(B_{0}) = T B_{0}^{n} or nD_{A}(B_{0}) = T B_{0}. It follows from the previous result that B is a divisor of (A − α)^{n}. Let now B = B_{1}B_{2} … B_{m}, where the polynomials B_{1}, …, B_{m} do not have common factors, and assume that D_{A}(B) = T B. Then

∑_{i=1}^{m} B_{1} … B_{i−1}D_{A}(B_{i})B_{i+1} … B_{m} = T B_{1} … B_{m}

This means that B_{i} divides D_{A}(B_{i}). So, if we consider now the prime decomposition:

B = ∏_{i=1}^{m}(B_{i} − α_{i})^{k_{i}}

and assume that D_{A}(B) = T B, we can conclude that B divides some polynomial R(A) ∈ K[A].

Corollary. Let A, T ∈ K[x, y], f ∈ K(x, y); f ≠ 0 satisfy the equation D_{A}(f) = T f. Then there exists F ∈ K[x, y] satisfying the same equation.

Proof. Let f = \frac{U}{V}, where U, V ∈ K[x, y] and do not have common factors. Then:

D_{A}(U)V − D_{A}(V)U = T UV

This means that V divides D_{A}(V). There exists,therefore, G ∈ K[x, y] such that GV = R(A) ∈ K[A]. Thus f = \frac{GU}{R(A)} and D_{A}(f) = \frac{D_{A}(GU)}{R(A)} = \frac{T GU}{R(A)} or D_{A}(GU) = T GU. So the polynomial F = GU satisfies the same equation as does the rational function f.

Proposition 2.2. Let S ∈ K[x, y] be w.s. Then

(1) E(S) is a free abelian group of finite rank.

(2) For λ, μ ∈ E(S) it follows that V_{λ}V_{μ} ⊂ V_{λ+μ} and [V_{λ}, V_{μ}] ⊂ V_{λ+μ}.

(3) Let {λ_{1}, …, λ_{n}} ⊂ E(S); λ_{i} ≠ λ_{j} for i ≠ j. Let {v_{1}, …, v_{n}} be a collection of nontrivial eigenfunctions of S such that v_{i} ∈ V_{λ_{i}}. Then the collection {v_{1}, …, v_{n}} is linearly independent.

Proof. (1) Let λ, μ ∈ E(S); F_{1} ∈ V_{λ}, F_{1} ≠ 0; F_{2} ∈ V_{μ}, F_{2} ≠ 0. D_{S}(F_{1}F_{2}) = (λ + μ)F_{1}F_{2} and D_{S}(\frac{F_{1}}{F_{2}} ) = (λ − μ)\frac{F_{1}}{F_{2}} since D_{S} is a derivation of K(x, y). So, the sum of eigenvalues is an eigenvalue, and it follows from the Corollary to Lemma 2.1 that the difference of eigenvalues is an eigenvalue too. The result now follows from Theorem 2.4 of [7].

(2) Follows from the fact that D_{S} is a derivation of K[x, y] as an algebra and as a Lie-algebra.

(3) Straightforward.

Corollary. Set V (S) = ∑_{λ∈E(S)} V_{λ}(S). Then V (S) = ⨁_{λ∈E(S)} V_{λ}(S) and is a subalgebra of K[x, y] as an algebra and as a Lie-algebra.

Proof. Follows from (3) and (2).

Lemma 2.2. Let S ∈ K[x, y] be w.s. and let λ, μ ∈ E(S); λ ≠ 0, μ ≠ 0. Then \frac{λ}{μ} ∈ ℚ.

Proof. Let v ∈ V_{λ}(S), u ∈ V_{μ}(S); v ≠ 0, u ≠ 0. Assume that \frac{λ}{μ} ∉ ℚ. The polynomials S, v, u belong to K[x, y] and are, therefore, algebraically dependent. This fact can be expressed as ∑_{i,j} R_{ij}(S)v^{i}u^{j} = 0, where R_{ij}(S) ∈ K[S] and are not all trivial. Note that R_{ij}(S)v^{i}u^{j} ∈ V_{iλ+jμ} by Proposition 2.2.

If the pairs (i, j) and (i_{1}, j_{1}) are distinct, then the spaces V_{iλ+jμ}(S) and V_{i_{1}λ+j_{1}μ}(S) cannot coincide since, by our assumption, \frac{λ}{μ} ∉ ℚ. Therefore those ”monomials” R_{ij}(S)v^{i}u^{j} which are nontrivial, are linearly independent - a contradiction.

Theorem 2.1. Let S ∈ K[x, y] be w.s. Then the group E(S) is isomorphic to ℤ.

Proof. E(S) is a free abelian group of finite rank (by Proposition 2.2). Let {λ_{1}, …, λ_{n}} be a basis of E(S) over ℤ. Choose a non-zero λ_{0} ∈ E(S). Then, by Lemma 2.2, every λ_{k} = \frac{p_{k}}{q_{k}} λ_{0} where p_{k} ∈ ℤ; q_{k} ∈ ℕ. Let q be the least common denominator of \{\frac{p_{1}}{q_{1}} , …, \frac{p_{n}}{q_{n}} \}. Then every λ ∈ E(S) can be written as λ = \frac{p_{λ}}{q} λ_{0}; p_{λ} ∈ ℤ. The map λ → p_{λ} is, obviously, an isomorphism between E(S) and a subgroup of ℤ, but all subgroups of ℤ are isomorphic to ℤ.

Since E(S) is isomorphic to ℤ, it is always possible to multiply S by a constant c so that E(cS) = ℤ. A w.s. polynomial S such that E(S) = ℤ will be called a normalized w.s. polynomial. Now we will establish some additional properties of w.s. polynomials.

Proposition 2.3. Let S be w.s. Then S − α has no multiple factors for all α ∈ K.

Proof. Fix α ∈ K and assume S − α = A^{k}B; k ≥ 2; A, B ∈ K[x, y]. Let v ∈ K[x, y], v ≠ 0; λ ∈ K, λ ≠ 0 be such that D_{S}(v) = λv. Let v = A^{n}u; n ≥ 0; u ∈ K[x, y] and A does not divide u. Then:

λA^{n}u = [A^{k}B, A^{n}u] = A^{n+k−1}L

for some L ∈ K[x, y]. Therefore λu = A^{k−1}L, which is impossible since k ≥ 2 and A does not divide u.

Theorem 2.2. Let S be normalized w.s. Then there exist v_{1} ∈ V_{1}(S) and v_{−1} ∈ V_{−1}(S) such that V (S) = K[S][v_{1}] ⊕ K[S] ⊕ K[S][v_{−1}].

Proof. First recall that S is primitive by Proposition 2.1. Let Ṽ_{0}(S) = {f ∈ K(x, y) : D_{S}(f) = 0}. It is well known (see, for example, Theorem 1.7. of [7]) that V_{0}(S) = K[S] and Ṽ_{0}(S) = K(S) if S is primitive. Let v_{n} be a polynomial of minimal degree in V_{n}(S) and let v ∈ V_{n}(S). Then \frac{v}{v_{n}} ∈ Ṽ_{0}. So there exist F (S), G(S) ∈ K(S) such that F (S) and G(S) do not have common factors and v = \frac{F (S)}{G(S)}v_{n}. Or

G(S)v = F (S)v_{n}

Since F (S) and G(S) are coprime, G(S) must divide v_{n}. But then the polynomial \frac{v_{n}}{G(S)} ∈ V_{n}(S) and deg \frac{v_{n}}{G(S)} < deg v_{n} unless G(S) = const. So V_{n}(S) = K[S]v_{n}. This also proves that v_{n} is unique up to a constant multiplier. Assume now n > 0. Then v_{1}^{n} ∈ V_{n}(S) and, therefore, v_{1}^{n} = F (S)v_{n} for some F (S) ∈ K[S]. It follows now from Proposition 2.3 that if F (S) divides v_{1}^{n}, then it must divide v_{1}. This would contradict our choice of v_{1}, unless F (S) = const. Thus V_{n}(S) = K[S]v_{1}^{n}. Similarly for the negative values of n: V_{n}(S) = K[S]v_{−1}^{−n} and the result follows from Proposition 2.2.

Corollary. D_{v_{1}} and D_{v_{−1}} are locally nilpotent on V (S).

Proof. Straightforward.

These results hold (with obvious modifications) for all w.s. polynomials. For a w.n. polynomial A let N_{0}(A) = ker D_{A} and let N(A) = {B ∈

K[x, y] : ∃n ∈ ℕ / D_{A}^{n}(B) = 0}. Both N_{0}(A) and N(A) are, obviously, subalgebras of K[x, y] and D_{A} is locally nilpotent on N(A).

Proposition 2.4. Let A ∈ K[x, y], f ∈ K(x, y) be such that D_{A}(f) ≠ 0; D_{A}^{2}(f) = 0. Then A is w.n.

Proof. It is well known that for any A, ker D_{A} = K[B] for some primitive B (see, for example, Corollary to Theorem 1.7 of [7]). So N_{0}(A) = K[B] where B is primitive. A = R(B) ∈ K[B] and, therefore, D_{A}(f) = R′(B)D_{B}(f) and D_{A}^{2}(f) = (R′(B))^{2}D_{B}^{2}(f). Hence D_{B}(f) ≠ 0 and D_{B}^{2}(f) = 0. Since B is primitive and D_{B}(D_{B}(f)) = 0, it follows that D_{B}(f) = \frac{F (B)}{G(B)} where F (B), G(B) ∈ K[B] and are coprime. Let f = \frac{U}{V} where U, V ∈ K[x, y] and have no common factors. Then D_{B}(\frac{U}{V}) = \frac{F (B)}{G(B)} or D_{B}(\frac{UG(B)}{V}) = F (B). Let

\frac{UG(B)}{V} = \frac{U_{1}}{V_{1}} , where U_{1}, V_{1} have no common factors. Then

D_{B}(U_{1})V_{1} − D_{B}(V_{1})U_{1} = F (B)V_{1}^{2}

This means that V_{1} divides D_{B}(V_{1}) and it follows from Lemma 2.1 that there exists L ∈ K[x, y], L ≠ 0 such that LV_{1} = H(B) ∈ K[B]. Then D_{B}(\frac{LU_{1}}{H(B)}) = F (B) or D_{B}(LU_{1}) = H(B)F (B). This proves that B is w.n. Then A = R(B) is, obviously, also w.n.

Lemma 2.3. Let S be w.s. and let A be an eigenfunction of S. Then:

(1) A is w.n.

(2) If B ∈ N_{0}(A), then S + B is w.s.

Proof. (1) D_{A}(S) = −D_{S}(A) = −λA and D_{A}^{2}(S) = 0.

(2) Straightforward.

So, if A is an eigenfunction of a w.s. S, then A is w.n. Then, of course, every non-constant polynomial in N_{0}(A) is also w.n. Can all w.n. polynomials be obtained in this way? The answer is no.

Example 2.1. Let A = x^{3}y + x^{2} + x. Then A is w.n. and primitive and no element of K[A] is an eigenfunction of a w.s. polynomial.

Proof. Let P = x(xy + 1). Then A = (P + 1)x and it is easy to check that D_{A}((P + 1)^{2}) = 2A^{2}. A is, obviously, primitive. We will prove now that A + c cannot be an eigenfunction of a w.s. polynomial for any c ∈ K. Assume, on the contrary, that D_{S}(A) = A + c for some S ∈ K[x, y]; c ∈ K. Now: [A, \frac{(P +1)^{2}}{2A}] = A and [A, \frac{c(P +1)^{2}}{2A^{2}} ] = c. Therefore

[A, S + \frac{(P + 1)^{2}}{2A} + \frac{c(P + 1)^{2}}{2A^{2}} ] = 0

Since A = (P + 1)x, we obtain that [A, S + \frac{P +1}{2x} + \frac{c}{2x^{2}} ] = 0 Since A is primitive, it follows that

S + \frac{P + 1}{2x} + \frac{c}{2x^{2}} = \frac{F (A)}{G(A)}

where F (A), G(A) ∈ K[A] and are coprime. This implies that G(A) divides x^{2}. Hence G(A) = const. So:

2x^{2}S + x(P + 1) + c = 2x^{2}F (A)

Then c = 0 and 2xS + P + 1 = 2xF (A). Thus x divides P + 1, which is impossible. If we assume now that a polynomial of A is an eigenfuction of a w.s. polynomial, then it is very easy prove that some A + c must also be an eigenfunction. Thus no element of N_{0}(A) is an eigenfunction of a w.s. polynomial.

Remark. This example was suggested to me by Pierrette Cassou-Nogues.

Definition 2.1. A w.n. polynomial A is called natural if an element of N_{0}(A) is an eigenfunction of a w.s. polynomial.

The previous example shows that not all w.n. polynomials are natural. So, is it possible to find a good description of the set of all natural w.n. polynomials? I don’t know how to do it.

Definition 2.2. A w.s. polynomial S is called semisimple if V (S) = K[x, y].

Definition 2.3. A w.n. polynomial A is called locally nilpotent if N(A) = K[x, y].

Now, using Lemma 2.3, we can construct increasingly more complicated w.s. and w.n. polynomials. We start with a very basic semisimple polynomial like S = xy and one of its eigenfunctions and apply the two steps, described in Lemma 2.3, in succession.

Let S_{1} = xy. Then A_{1} = xy^{3} ∈ V_{2}(S_{1}). Now let S_{2} = \frac{xy+xy^{3}}{2} . Then, by Lemma 2.3, S_{2} is w.s. (it is not semisimple anymore!) and A_{1} ∈ V_{1}(S_{2}). Let A_{2} = S_{2}^{2}A_{1}. Then A_{2} ∈ V_{1}(S_{2}). Let S_{3} = S_{2} + A_{2} for instance. Then S_{3} is w.s. and so on. This machinery enables us to produce more and more complicated examples of w.s. and w.n. polynomials and this is important in the context of the Jacobian Conjecture.

3. w.s. and w.n. polynomials in the context of the Jacobian Conjecture

Let us recall first what the Jacobian Conjecture is.

Definition 3.1. Two rational functions f, g ∈ K(x, y) are called Jacobian mates if [f, g] = const ≠ 0.

Definition 3.2. A ∈ K[x, y] is called a coordinate if there exists B ∈ K[x, y] such that K[A, B] = K[x, y].

The Jacobian Conjecture can be formulated as follows:

A polynomial has a polynomial Jacobian mate iff it is a coordinate.

Proposition 3.1. A ∈ K[x, y] is w.n. iff it has a rational Jacobian mate.

Proof. Assume first that A is w.n. Then there exists B ∈ K[x, y] such that D_{A}(B) ≠ 0 and D_{A}(B) ∈ N_{0}(A). But then D_{A}(\frac{B}{D_{A}(B)}) = 1.

Assume now that D_{A}(f) = 1 for some f ∈ K(x, y). Let f = \frac{F}{G}; F, G ∈ K[x, y] and do not have common factors. Then

D_{A}(F)G − D_{A}(G)F = G^{2}

This implies that G divides D_{A}(G) and by Lemma 2.1, there exists B ∈ K[x, y]; B ≠ 0 such that BG = R(A) ∈ K[A]. Then D_{A}(F B) = R(A) and A is w.n.

Definition 3.3. A ∈ K[x, y] is called totally irreducible if A − λ is irreducible for each λ ∈ K.

Theorem 3.1. Let P be totally irreducible and let f ∈ K(x, y) be such that D_{P} (f) = U ∈ K[x, y]; U ≠ 0. Then there exists a polynomial F such that D_{P} (F) = U.

Proof. Let f = \frac{A}{B}; A, B ∈ K[x, y] and do not have common factors. Then

D_{P} (A)B − D_{P} (B)A = UB^{2}

This implies that B divides D_{P} (B) and, by Lemma 2.1, there exists a polynomial H ≠ 0 such that HB = R(P) ∈ K[P]. Then D_{P} (HA) = UR(P). Let W (P, U) denote the set of all G ∈ K[x, y] such that D_{P} (G) ≠ 0 and \frac{D_{P} (G)}{U} ∈ K[P]. We have just proved that W (P, U) ≠ ∅. Let G_{0} ∈ W (P, U) be such that

deg G_{0} = min_{G∈W (P,U)} deg G

So, D_{P} (G_{0}) = UR_{0}(P) for some R_{0}(P) ∈ K[P]. If R_{0}(P) ∈ K, then we are done. Assume that deg R_{0} > 0 and let α ∈ K be a root of the polynomial R_{0}, R_{0}(P) = R_{1}(P)(P −α). Then P −α divides D_{G_{0}}(P) and, by Lemma 2.1, P −α must divide a polynomial of G_{0}. P −α is irreducible, since P is totally irreducible, and this implies that P −α divides G_{0} −c for some c ∈ K. Thus \frac{G_{0}−c}{P −α} = G_{1} ∈ K[x, y] and

D_{P} (G_{0}) = D_{P} (G_{1})(P − α) = UR_{1}(P)(P − α)

Therefore D_{P} (G_{1}) = UR_{1}(P), but deg G_{1} < deg G_{0}, contradicting our choice of G_{0}. Thus R_{0}(P) = const ≠ 0 and this concludes the proof.

Corollary. Let P be w.n. and totally irreducible. Then P has a polynomial Jacobian mate.

Proof. Obvious.

4. The Newton polygon

In this section we will construct the ”canonical” form of the Newton polygons of w.s. and w.n. polynomials. Similar results can be found in [2] in the context of Weyl algebras. Our methods are, however, somewhat different and based on the results obtained in the previous sections.

Definition 4.1. Let r, s ∈ ℕ. Let

A = ∑_{i,j} α_{ij}x^{i}y^{j} ∈ K[x, y]

The (r, s)-degree is defined as follows:

deg_{r,s} A = max_{α_{ij}≠0}(ri + sj)

Remark. The same definition was used in [2] with r, s ∈ ℝ. In this paper we will need only r, s ∈ ℕ. Polynomials homogeneous with respect to this degree are sometimes called quasi-homogeneous. We will call them (r, s)- homogeneous.

We will now list some well known properties of the (r, s)-degree without proof (which is a very easy one) in:

Proposition 4.1. (1) deg_{r,s}(AB) = deg_{r,s} A + deg_{r,s} B (2) deg_{r,s}(A + B) ≤ max(deg_{r,s} A, deg_{r,s} B)

(3) Every A ∈ K[x, y] can be decomposed into a sum of (r, s)-homogeneous polynomials:

A = ∑_{i=1}^{k} A_{i}

where each A_{i} is (r, s)-homogeneous, deg_{r,s} A_{i} = n_{i} and n_{1} < ⋯ < n_{k}. A_{k} is called the (r, s)-leading term of A and is denoted by Ā_{r,s}.

(4) deg_{r,s}[A, B] ≤ deg_{r,s} A + deg_{r,s} B − (r + s)

and deg_{r,s}[A, B] < deg_{r,s} A + deg_{r,s} B − (r + s) iff [Ā_{r,s}, B̄_{r,s}] = 0.

(5) If A is an (r, s)-homogeneous polynomial with deg_{r,s} A = n, then A satisfies the Euler’s equation

rx\frac{∂A}{∂x} + sy \frac{∂A}{∂y} = nA

(6) Let A be a (r, s)-homogeneous polynomial with deg_{r,s} A = n and let m = min_{B∈N_{0}(A),B∉K} deg_{r,s}(B). Then \frac{n}{m} = k ∈ ℕ and N_{0}(A) = K[A_{0}], where A_{0} is primitive (r, s)-homogeneous, deg_{r,s} A_{0} = m and A = const · A_{0}^{k}. (This follows easily from (5)).

Lemma 4.1. Let r, s ∈ ℕ and let W ⊂ K[x, y] be a subalgebra satisfying the following condition: for each pair of polynomials A, B ∈ W [Ā_{r,s}, B̄_{r,s}] = 0. Then W ⊂ ker D_{A} for all A ∈ W.

Proof. We can assume, of course, that W ≠ K. Let W_{A} = {B ∈ W : D_{A}(B) ≠ 0}. Suppose that W_{A} ≠ ∅ for some A ∈ W. We will define now a function d(B) on the set W_{A} as follows:

d(B) = deg_{r,s} D_{A}(B) − deg_{r,s} B

d(B) ≠ −∞, is integer-valued, and it follows from (4) of Proposition 4.1 that d(B) ≤ deg_{r,s} A − (r + s), so that d(B) is bounded from above. There exists, therefore, M ∈ W_{A} such that

d(M) = max_{B∈W_{A}} d(B)

It is easy to see that d(B^{n}) = d(B) for n ∈ ℕ. Indeed:

d(B^{n}) = (n − 1) deg_{r,s} B + deg_{r,s} D_{A}(B) − n deg_{r,s} B = d(B)

Since [Ā_{r,s}, M̄_{r,s}] = 0 by our assumption, it follows from (6) of Proposition 4.1 that there exist n, m ∈ ℕ and c ∈ K such that M̄_{r,s}^{n} − cĀ_{r,s}^{m} = 0. Let H = M^{n} − cA^{m}. Then, obviously, H ∈ W_{A}, deg_{r,s} H < deg_{r,s} M^{n} and

d(H) = deg_{r,s} D_{A}(M^{n}) − deg_{r,s} H > d(M^{n}) = d(M)

This contradicts our choice of M and, therefore, W_{A} = ∅.

Remark. This kind of proof, using the defect function, was, as far as I know, first suggested be Leonid Makar-Limanov about 30 years ago.

Lemma 4.2. Let F ∈ K[x, y] be irreducible and let a non-constant rational function h = \frac{A}{B} ∈ K(x, y) be such that A and B do not have common factors and D_{F} (h) = 0 on the curve {F = 0}. Then there exist a, b ∈ K such that at least one of them ≠ 0 and F divides aA − bB.

Proof. If D_{F} (h) = D_{F} (\frac{A}{B}) = 0 on the curve F = 0, then F divides BD_{F} (A)− AD_{F} (B).

Case 1. A and B are algebraically dependent.

In this case there exists a primitive P ∈ K[x, y] such that A = R(P) ∈ K[x, y]; B = G(P) ∈ K[x, y]. Then F divides (G(P)R′(P) − R(P)G′(P))D_{F} (P). Since F is irreducible, it must divide either D_{F} (P) or G(P)R′(P)−R(P)G′(P) or both. If F divides D_{F} (P), then, by Lemma 2.1, it must divide a polynomial of P. Since F is irreducible, this implies that F divides P − c for some constant c. Then F divides G(c)R(P) − R(c)G(P). Note that G(c) and R(c) cannot both be 0, since R(P) and G(P) do not have common factors. Assume now that F divides G(P)R′(P)−R(P)G′(P). This polynomial cannot be 0, since in this case h = \frac{R(P)}{G(P)} would be a constant. Neither can it be a non-0 constant, since F cannot divide a non-0 constant. Therefore again F must divide P − c for some constant c and this concludes the proof for this case.

Case 2. A and B are algebraically independent.

Since D_{F} (h) = 0 on the curve F = 0, we obtain that F divides BD_{F} (A)− AD_{F} (B). Since A, B, F ∈ K[x, y] and A and B are algebraically independent, there must exist a nontrivial polynomial R(A, B) ∈ K[A, B] such that F divides R(A, B). We can assume that this polynomial is of minimal degree in (A, B). Let’s define a derivation δ on K(x, y) in the following way:

δ(g) = B[g, A] − A[g, B]

Then F divides δ(F), δ(A) = −A[A, B] and δ(B) = B[B, A]. Applying δ to R(A, B), we obtain:

(A\frac{∂R}{∂A} + B \frac{∂R}{∂B})[A, B] = UF^{n}

where n > 0 and F does not divide U.

Let L(A, B) = A\frac{∂R}{∂A} + B \frac{∂R}{∂B}. Assume first that F divides L(A, B). Since R is minimal, it follows that L(A, B) = λR(A, B); λ ∈ K. This implies that L(A, B) is homogeneous in A, B and is a product of linear factors. Then F must divide one of them and we are done. Assume now that F does not divide L(A, B). Then F^{n} must divide [A, B]. Applying D_{A} to R(A, B), we obtain

\frac{∂R}{∂B} [A, B] = D_{A}(U)F^{n} + nUD_{A}(F)F^{n−1}

Therefore F must divide D_{A}(F). Then, by Lemma 2.1, F divides a polynomial of A and, being irreducible, it must divide A − c_{1} for some constant c_{1}. In other words, A − c_{1} = A_{0}F and similarly B − c_{2} = B_{0}F. c_{1} and c_{2} cannot both be 0, since A and B do not have common factors. It is clear that F divides c_{2}A − c_{1}B. This concludes the proof.

Theorem 4.1. Let A ∈ K[x, y] be w.n. and let r, s ∈ ℕ be coprime. Then there are three possible cases:

(1) r = 1, s > 1 and the (r, s)-leading term of A has the form:

Ā_{1,s} = const · x^{n_{1}}(y + ax^{s})^{n_{2}}

a ∈ K, a ≠ 0, n_{1} ≠ n_{2}.

(2) s = 1, r > 1 and the (r, s)-leading term of A has the form:

Ā_{r,1} = const · y^{n_{1}}(x + by^{r})^{n_{2}}

b ∈ K, b ≠ 0, n_{1} ≠ n_{2}.

(3) s = 1, r = 1 and the (r, s)-leading term of A has the form:

Ā_{1,1} = const · (a_{1}x + b_{1}y)^{n_{1}}(a_{2}x + b_{2}y)^{n_{2}}

a_{1}, a_{2}, b_{1}, b_{2} ∈ K, n_{1} ≠ n_{2} and a_{1}b_{2} − a_{2}b_{1} ≠ 0.

Proof. N(A) is a subalgebra of K[x, y] and, since A is w.n., N(A) ≠ N_{0}(A). There exists, therefore, by Lemma 4.1, B ∈ N(A) such that [Ā_{r,s}, B̄_{r,s}] ≠ 0. It is obvious, since A is w.n., that B̄_{r,s} ∈ N(Ā_{r,s}). Thus Dn ; Ā_{r,s}(B̄_{r,s}) = 0

and Dn−1 ; Ā_{r,s}(B̄_{r,s}) ≠ 0 for some n > 1. Let Dn−2 ; Ā_{r,s}(B̄_{r,s}) = H and let G be a primitive (r, s)-homogeneous polynomial such that N_{0}(Ā_{r,s}) = K[G]. Such a polynomial exists by (6) of Proposition 4.1. We now have the following situation: D_{G}(H) ∈ K[G]; D_{G}(H) ≠ 0. Since both G and H are (r, s)-homogeneous, it follows that D_{G}(H) = const · G^{m} for some m ≥ 0.

We can assume without loss of generality that m > 0, since if m = 0, then we can multiply H by G. Our goal is to prove that G^{m−1} divides H. Assume, on the contrary, that G^{m−1} does not divide H. Then there exists an (r, s)-homogeneous, irreducible F such that: G = G_{0}F^{k}; H = H_{0}F^{l}, where k > 0, l ≥ 0, F does not divide G_{0}H_{0} and k(m − 1) > l. Then:

D_{G}(H) = const · G_{0}^{m}F^{km} = F^{k+l−1}(kG_{0}[F, H_{0}] + F [G_{0}, H_{0}] + lH_{0}[G_{0}, F])

Therefore kG_{0}[F, H_{0}] + lH_{0}[G_{0}, F] = UF for some polynomial U. This is equivalent to:

[\frac{G_{0}^{l}}{H_{0}^{k}} , F] = U \frac{G_{0}^{l−1}}{H_{0}^{k+1}} F

It follows now from Lemma 4.2 that there exists a nontrivial pair c_{1}, c_{2} ∈ K such that c_{1}G_{0}^{l} − c_{2}H_{0}^{k} = MF; M ∈ K[x, y]. Moreover, neither c_{1} nor c_{2} can be 0. Indeed, if c_{1} = 0, then c_{2} = 0 since k > 0 and F does not divide H_{0}. If c_{2} = 0, then either c_{1} = 0, which is impossible, or c_{1} ≠ 0 and l = 0 , which would imply that F divides a nonzero constant. Assume for a moment that l deg_{r,s} G_{0} ≠ k deg_{r,s} H_{0}. Then the decomposition of M into the sum of (r, s)-homogeneous polynomials must contain exactly two terms: M_{1} and M_{2}, such that M_{1}F = c_{1}G_{0}^{l} and M_{2}F = c_{2}H_{0}^{k} which is obviously impossibble. Thus G_{0}^{l} and H_{0}^{k} must have the same (r, s)-degree.

Since [G, H] = const · G^{m} ≠ 0 and both G and H are (r, s)-homogeneous, it follows from (4) of Proposition 4.1 that

deg_{r,s}[G, H] = deg_{r,s} G + deg_{r,s} H − (r + s)

Therefore, since [G, H] = const · G^{m},

deg_{r,s} H_{0} = r + s + (m − 1) deg_{r,s} G_{0} + (k(m − 1) − l) deg_{r,s} F

Now we can use the fact that deg_{r,s} G_{0}^{l} = deg_{r,s} H_{0}^{k} and we obtain:

(k(m − 1) − l)(deg_{r,s} G_{0} + k deg_{r,s} F) + k(r + s) = 0

This is impossible if k(m−1) > l. Therefore k(m−1) ≤ l, which contradicts our earlier assumption. Thus G^{m−1} divides H.

Hence L = \frac{H}{G^{m−1}} is an (r, s)-homogeneous polynomial and

[G, L] = const · G ≠ 0

Thus L is w.s. and G is an eigenfunction of L.

deg_{r,s}[G, L] = deg_{r,s} G + deg_{r,s} L − (r + s) by (4) of Proposition 4.1 and, therefore, deg_{r,s} L = r + s. Assume now that L contains a monomial const · x^{i}y^{j} with const ≠ 0. Then r(i − 1) + s(j − 1) = 0. Since r and s are coprime, there are only three alternatives:

(1) i = 0, s = 1, r = j − 1

(2) j = 0, r = 1, s = i − 1

(3) i = 1, j = 1 If L contains only one nontrivial monomial, then alternatives (1) and (2) are obviously impossible since in this case L cannot be w.s. So we can assume L = xy. Thus G is an eigenfunction of xy. Assume for a moment that G contains at least two nontrivial monomials: ax^{i_{1}}y^{j_{1}} and bx^{i_{2}}y^{j_{2}}. Then: r(i_{1} − i_{2}) = s(j_{2} − j_{1}) since G is (r, s)-homogeneous and i_{1} − j_{1} = i_{2} − j_{2} since G is an eigenfunction of xy. i_{1} ≠ i_{2} because the equality would imply that j_{1} = j_{2} and the monomials are distinct. Therefore r = s and, since they are coprime, r = s = 1. This immediately implies that i_{1} = i_{2} and j_{1} = j_{2}. Thus G must be a single monomial: G = const · x^{n_{1}}y^{n_{2}}. n_{1} ≠ n_{2}, since otherwise D_{G}(L) would be 0, and we obtain Case 3 of the theorem since Ā_{r,s} = G^{n}.

If L contains three nontrivial monomials, then the only possible form for L is L = ax^{i} + by^{j} + cxy; abc ≠ 0. In this case both r = 1 and s = 1 and it follows immediately that i = j = 2. Then L = (a_{1}x + b_{1}y)(a_{2}x + b_{2}y) and a_{1}b_{2} − a_{2}b_{1} ≠ 0, since otherwise L = const · (ax + by)^{2} and cannot be w.s. Then there exists an automorphism τ of K[x, y] such that τ(x) = a_{1}x + b_{1}y; τ(y) = a_{2}x + b_{2}y. Thus L = τ(xy) and G = const · (a_{1}x + b_{1}y)^{n_{1}}(a_{2}x + b_{2}y)^{n_{2}} with n_{1} ≠ n_{2} and we obtain Case 3 again.

It is obvious from the previous argument, that L cannot contain more then three nontrivial monomials. So, we have either alternative (1) or alternative (2).

Assume first L = ay^{r+1} + bxy; s = 1. Then L = y(ay^{r} + bx) and there exists an automorphism τ of K[x, y] such that τ(y) = y; τ(x) = bx + ay^{r}. Thus L = τ(xy) and G = (bx + ay^{r})^{n_{1}}y_{2}^{n} with n_{1} ≠ n_{2}. This leads to Case 2 of the theorem.

Similarly, L = ax^{s+1} + bxy leads to Case 1.

Definition 4.2. A ∈ K[x, y] is called quasi-monic if

A = cx^{n_{1}}y^{n_{2}} + ∑_{i≤n_{1},j≤n_{2},i+j<n_{1}+n_{2}} a_{ij}x^{i}y^{j}

where n_{1} > 0, n_{2} > 0, and c is a nonzero constant.

Theorem 4.2. Let A be w.n. Then either A is locally nilpotent and there exists an automorphism τ of K[x, y] such that τ(A) ∈ K[x], or there exists τ such that τ(A) is quasi-monic with deg_{x} τ(A) ≠ deg_{y} τ(A).

Proof. Let d(A) = deg_{x} A + deg_{y} A. Let

d_{0} = min_{τ∈Aut(K[x,y])} d(τ(A))

We can assume, without loss of generality, that d(A) = d_{0}. Assume that A is not quasi-monic. Then either all the points of the Newton polygon of A are located on one of the coordinate axes, or there exist coprime r, s ∈ ℕ such that Ā_{r,s} is not a monomial. In the first case A ∈ K[x] or A ∈ K[y] and is, therefore, locally nilpotent. Thus if A is not locally nilpotent, then the Newton polygon of A must contain an edge with negative slope corresponding to Ā_{r,s}.

Therefore, by Theorem 4.1, there are three possible cases:

(1) s = 1, r > 1 and

A = c(x + ay^{r})^{n_{1}}y^{n_{2}} + ∑_{ri+j<rn_{1}+n_{2}} a_{ij}x^{i}y^{j}

with c ≠ 0, a ≠ 0, n_{1} > 0.

(2) r = 1, s > 1 and

A = c(y + bx^{s})^{n_{1}}x^{n_{2}} + ∑_{i+sj<sn_{1}+n_{2}} a_{ij}x^{i}y^{j}

with c ≠ 0, b ≠ 0, n_{1} > 0.

(3) r = 1, s = 1 and

A = c(a_{1}x + b_{1}y)^{n_{1}}(a_{2}x + b_{2}y)^{n_{2}} + ∑_{i+j<n_{1}+n_{2}} a_{ij}x^{i}y^{j}

where c ≠ 0, a_{1}b_{2} − a_{2}b_{1} ≠ 0 and at most one of a_{1}, b_{1}, a_{2}, b_{2} is 0.

In Cases 1 and 2 there exists a triangular and in Case 3 a linear automorphism τ such that τ(Ā_{r,s}) is a monomial. It is easy to verify that the action of τ on A decreases either deg_{x} A or deg_{y} A or both. This contradicts our choice of d_{0}. Thus A must be quasi-monic and, since A is w.n., it follows from Theorem 4.1 that deg_{x} A ≠ deg_{y}A.

On the other hand, if A is quasi-monic, then its (r, s)-leading term for any r > 0, s > 0 is const · x^{n_{1}}y^{n_{2}} with n_{1} > 0, n_{2} > 0. If n_{1} ≥ n_{2}, then, as can be easily verified, x ∉ N(A) and if n_{1} < n_{2}, then y ∉ N(A), so that a quasi-monic polynomial can never be locally nilpotent.

Corollary. Let S be w.s. Then either S is semisimple and there exists an automorphism τ of K[x, y] such that τ(S) = axy + b; a, b ∈ K, a ≠ 0, or there exists τ such that τ(A) is quasi-monic and deg τ(A) > 2.

Proof. We can assume, without loss of generality, that S is normalized (e.g. E(S) = ℤ). Then, by Theorem 2.2 and its Corollary, there exist two eigenfunctions v_{1} ∈ V_{1}(S) and v_{−1} ∈ V_{−1}(S) such that V (S) = K[S][v_{1}] ⊕ K[S] ⊕ K[S][v_{−1}] and the derivations D_{v_{1}} and D_{v−1} are locally nilpotent on V (S). Applying a suitable automorphism τ, we obtain that either τ(v_{1}) ∈ K[x] or τ(v_{1}) is quasi-monic. Assume first that S is semisimple. Then, of course, τ(S) is also semisimple, V (τ(S)) = K[x, y] and τ(v_{1}) is locally nilpotent. Therefore it cannot be quasi-monic and τ(v_{1}) = R(S) ∈ K[x]. Then

[τ(S), R(x)] = −R′(x)\frac{∂τ(S)}{∂y} = γR(x)

where γ ∈ K, γ ≠ 0.

This implies that R′(x) divides R(x), which is possible only if R(x) = const·(x−c)^{n}; c ∈ K. A straightforward calculation shows that, by applying a triangular automorphism if necessary, τ(S) can be reduced to the form axy + b.

If S is not semisimple, then τ(v_{1}) must be quasi-monic, because otherwise τ(S) can be reduced to the form axy + b, as we have just shown, and this polynomial is semisimple. Thus the (r, s)-leading term of τ(v_{1}) is the same for all r > 0, s > 0 and is const · x^{n_{1}}y^{n_{2}} with n_{1} > 0, n_{2} > 0 and n_{1} ≠ n_{2}. Let V denote this leading term. Let us fix a pair (r, s) of coprime positive integers. Let U denote the (r, s)-leading term of τ(S). Assume first that [U, V] ≠ 0. Then , since [τ(S), τ(v_{1})] = const·τ(v_{1}), we obtain that [U, V] = const · V, const ≠ 0. thus U is w.s., V is its eigenfunction, and

deg_{r,s}[U, V] = deg_{r,s} U + deg_{r,s} V − (r + s) = deg_{r,s} V

Therefore deg_{r,s} U = r +s. Let U contain a nontrivial monomial const·x^{i}y^{j}. Then r(i − 1) + s(j − 1) = 0, so either i = 0 or j = 0 or i = j = 1. Assume first i = 0. Then r = s(j − 1) and, since r and s are coprime, it follows that s = 1, j = r+1. Since V is a monomial and an eigenfunction of U, it must be an eigenfunction of each nontrivial monomial contained in U. V obviously cannot be an eigenfunction of y^{r+1}, so the case i = 0 is impossible, Similarly, j ≠ 0. This leaves i = 1, j = 1 as the only alternative, but in this case τ(S) can be reduced to axy + b, which is impossible since S is not semisimple.

Thus [U, V] = 0. Then there exists a monomial W = x^{i}y^{j} such that V = const · W^{k} and U = const · W^{m}. Since n_{1} > 0, n_{2} > 0, n_{1} ≠ n_{2}, it follows that i > 0, j > 0, i ≠ j and, therefore, U is the (r, s)-leading term for all pairs (r, s) and τ(S) is quasi-monic.

The Newton polygon of a w.n. or a w.s. polynomial may contain vertical or horizontal edges. In fact, it may consist of just one such edge. In such a case the polynomials cannot be ”simplified” by an automorphism.

Example 4.1. Let A = y^{3}(y − 1)^{3}x^{2} and let S = y(y − 1)(2y − 1)x. Then it is easy to verify that D_{S}(A) = A and, therefore, S is w.s. and A is w.n. The Newton polygons of both S and A consist of a single vertical edge.

References

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[3] T. Krasiński and S. Spodzieja, On linear differential operators related to the n-dimensional Jacobian Conjecture, real Algebraic Geometry (M. Coste, L. Mahe, and M.F. Roy, eds.), Lecture Notes in Mathematics, vol. 1524, Springer-Verlag, 1992, 308-315.

[4] T. Moh, On the global Jacobian Conjecture and the configuration of roots, Journal für die reine und Angewandte Mathematik 340 (1983), 140-212.

[5] Y. Stein, On linear differential operators related to the Jacobian Conjecture, Journal of Pure and Applied Algebra 57 (1989), 175-186.

[6] , On the density of image of differential operators generated by polynomials, Journal d’Analyse Mathématique, 52 (1989), 291-300.

[7] , The total reducibility order of a polynomial in two variables, Israel Journal of Mathematics, 68, No. 1, (1989), 109-122.

[8] , Linear differential operators related to the Jacobian Conjecture have a closed image, Journal d’Analyse Mathématique, 54 (1990), 237-245.

MAX-PLANCK-INSTITUT FÜR MATHEMATIK, Vivatsgasse 7, 53111 Bonn, Germany

   E-mail address: ystein@mpim-bonn.mpg.de