Copyright © 2004, The National Academy of Sciences Mathematics The strong Nagata conjecture § To whom correspondence should be addressed. E-mail: yujt/at/hkucc.hku.hk. Communicated by Efim I. Zelmanov, University of California at San Diego, La Jolla, CA, December 9, 2003 Received August 5, 2003. This article has been cited by other articles in PMC. | |||||||||||||||||||||||||||||||||||||||||||||||
Abstract It is proved that there exist wild coordinates in the polynomial algebra in three variables over a field of characteristic zero. This result implies the famous Nagata conjecture. | |||||||||||||||||||||||||||||||||||||||||||||||
Let Pn = K[x1, x2,..., xn] be the polynomial algebra in n variables x1, x2,..., xn over a field K of characteristic 0. A K-automorphism AutKPn is called elementary if preserves all but one xi for i = 1,..., n. In other words, a K-automorphism AutKPn is elementary if it can be expressed as | |||||||||||||||||||||||||||||||||||||||||||||||
Strong Nagata Conjecture There exist wild coordinates of P3. The history of the strong Nagata conjecture naturally goes back to 1972 when the Nagata conjecture was formulated. However, it was formulated formally in June 2002 by J.-T.Y. in an algebra seminar held at the University of Hong Kong. Obviously the strong Nagata conjecture implies the Nagata conjecture but not vice versa, since it is possible that an image f of x under a wild automorphism :(x, y, z) → (f, g, h) of P3 = K[x, y, z] can be taken to x by a tame automorphism. In this article, we settle the strong Nagata conjecture affirmatively. Our result is the following. | |||||||||||||||||||||||||||||||||||||||||||||||
Main Theorem There exist wild coordinates of P3. In particular, all wild coordinates of K[z][x, y] (see refs. 1 and 2) are also wild coordinates of P3 = K[x, y, z]. Moreover, the image of x, y, z under a wild automorphism in AutK(P3) must contain at least two wild coordinates. Before proving the main theorem in the next section, let us describe the main idea of this article introduced by J.-T.Y. when he formulated the strong Nagata conjecture in the same algebra seminar [this idea was motivated by a similar idea used by Shpilrain and Yu (12) to effectively classify parametrized curves; see also ref. 13]: a given coordinate p = p(x, y, z) P3: = K[x, y, z] can be faithfully parametrized as a triple , where p(x(u, v), y(u, v), z(u, v)) = 0 and K[x(u, v), y(u, v), z(u, v)] = K[u, v]. It can be seen that p = p(x, y, z) is a tame coordinate of P3 if and only if there exists a sequence of elementary automorphisms of P3 that takes the triple (x(u, v), y(u, v), z(u, v)) to (0, u, v). This condition can be effectively determined by an algorithm motivated by ideas of Shestakov and Umirbaev (9–11). By the algorithm, we are able to prove that all wild coordinates of K[z][x, y] (1, 2) are also wild coordinates of P3 = K[x, y, z], hence we obtain many wild coordinates of P3. Note that a tame automorphism takes a wild coordinate to a wild coordinate; this way we also obtain many wild coordinates. In the sequel we use the following conventions: (i) K is a field of characteristic zero; (ii) all automorphisms are K-automorphisms unless specified otherwise, and AutPn always means AutKPn; (iii) algebraic (linear) independency (dependency) always means K-algebraic (K-linear) independency (dependency); (iv) an automorphism of Pn with (xi) = pi is denoted (p1,..., pn) sometimes, and each pi is called a coordinate of ; and (v) denotes the highest homogeneous part of a polynomial p Pn. | |||||||||||||||||||||||||||||||||||||||||||||||
Proof of the Main Theorem We start with some preliminaries. In ref. 14, the Poisson bracket of elements of the algebra Pn was defined. If f, g Pn, then
The next lemma was proved in ref. 14. Lemma 1. Two polynomials f, g Pn are algebraically dependent if and only if [f, g] = 0. Lemma 2. Elements f, g Pn are algebraically dependent if and only if there exists t Pn such that f, g K[t]. Proof: It was proved in ref. 14 that the centralizer C(f) = {h Pn|[f, h] = 0} of any element f Pn\K is a polynomial algebra in one indeterminate, i.e., there exists t Pn such that C(f) = K[t]. By Lemma 1, g C(f) if and only if f and g are algebraically dependent. Two generated subalgebras of the algebra Pn were studied in ref. 14. The lower degree bound of an element in such a subalgebra obtained there plays a major role in the study of the automorphisms in AutP3 in refs. 9–11 as well as in the study of coordinates of P3 in this article. Recall that a pair of elements (f, g) of the algebra Pn is called reduced (15), if K[] and K[]. According to ref. 14, an algebraically independent reduced pair of elements (f, g) of the algebra Pn is called *-reduced if and are algebraically dependent. Let f, g be a *-reduced pair of elements of Pn and k = deg f < m = deg g. Put
A polynomial f1 in the triple (f1, f2, f3) of P2 is reducible, if there exists g K[f2, f3] such that 1 = , otherwise it is irreducible. Put , where 0 ≠ c K, then deg() < deg(f) and deg(, f2, f3) < deg θ. In this situation we say f1 is reduced in θ by the triple (, f2, f3). We may define the reducibility of f2 and f3 similarly. θ is elementarily reducible (or θ admits an elementary reduction) if one of fi is reducible. In the sequel the next lemma is useful. Lemma 3. Let f, g, h P2 and let K[f, g, h] = P2. If f, g are algebraically dependent, then there exists at P2 such that K[f, g] = K[t] and P2 = K[t, h]. Proof: By Lemma 2 there exists a t P2 such that f, g K[t]. So P2 = K[t, h]. If K[f, g] is a proper subring of K[t], then K[f, g, h] is a proper subring of K[t, h] = P2. Therefore K[f, g] = K[t]. Lemma 3 implies the following. Lemma 4. Let f, g, h P2 and let K[f, g, h] = P2. If f, g are algebraically dependent, then there exists a coordinate t P2, and a sequence of elementary transformations taking (f, g) to (0, t). Proof: By Lemma 3 and the famous Abhyankar-Moh-Suzuki Theorem (13) we obtain a proof. Theorem 1. Let f, g, h P2 such that K[f, g, h] = P2. Then the elementary reducibility of the triple (f, g, h) is algorithmically recognizable.¶ Proof: We only need to prove the reducibility of an element in the triple is algorithmically recognizable. By Lemma 4, we may assume that f, g, h are pairwisely algebraically independent. Now we are going to recognize the reducibility of an element, say, h of the triple. If and are algebraically independent, then h is reducible if and only if K[, ]. Since and are homogeneous, this can be effectively determined. If K[] and = ck, where c is a nonzero element of K, then h is reducible in (f, g, h) if and only if it is reducible in (f, g-cfk, h). Since deg (f, g-cfk, h) < deg (f, g, h), we may apply induction in this case. Now we may assume (f, g) is a -reduced pair and deg (f) < deg (g). Suppose that there exists *G(x, y) P2 such that = the highest homogeneous part of G(f, g). Then 2 and 3 give a upper bound k for degx (G) and degy (G). Hence G(f, g) is in the vector space generated by figj, i, j < k. The highest homogeneous part of elements in this space can be effectively determined by triangulation. Now we are going to define four types of (nonelementary) reductions for triples of P2; all of these reductions are compositions of at most four elementary transformations of a special type. These definitions are motivated by the definitions of the four types of nonelementary reductions for automorphisms in AutP3 in refs. 9–11. Definition 1: Let θ = (f1, f2, f3) be a triple of P2 such that deg f1 = 2k, deg f2 = ks, s ≥ 3 is an odd number, 2k < deg f3 ≤ ks, 3 K[1, 2]. Suppose that there exists 0 ≠ α K such that the elements g1 = f1, g2 = f2 – αf3 satisfy the conditions:
Definition 2: Let θ = (f1, f2, f3) be a triple of P2 such that deg f1 = 2k, deg f2 = 3k, 3k/2 < deg f3 ≤ 2k, and 1, 3 are linearly independent. Suppose that there exist α, β K, where (α, β) ≠ (0, 0), such that the elements g1 = f1 – αf3, g2 = f2 – βf3 satisfy the conditions i and ii in Definition 1. Then we say that θ admits a reduction (g1, g2, g3) of type II. Definition 3: Let θ = (f1, f2, f3) be a triple of P2 such that deg f1 = 2k, and either deg f2 = 3k, k < deg f3 ≤ 3k/2, or 5k/2 < deg , deg f3 = 3k/2. Suppose that there exist α, β, K such that the elements g1 = f1 – βf3, g2 = f2 – βf3 – αf3 satisfy the conditions:
If (α, β, β) ≠ (0, 0, 0) and deg g3 < k + deg[g1, g2], then we say that θ admits a reduction (g1, g2, g3) of type III. On the other hand, if there exists 0 ≠ μ K such that deg(g2 – ) ≤ 2k, then we say that θ admits a reduction (g1, g2 – g3) of type IV. Definition 4: A simple triple in P2 is defined inductively as follows: A triple is a simple triple if deg (f, g, h) = 2 and K[f, g, h] = P2. A triple β of deg β = k > 2 is a simple triple, if there exists an elementary reduction or a reduction of one of the types I–IV to β, and after such a reduction, the resulting triple β of deg β < k is simple. The above definition of a simple triple of P2 is motivated by the definition of a simple automorphism in AutP3 in refs. 9–11. Theorem 2. A triple is tame if and only if it is simple. Proof: It is obvious that a simple triple is a tame one according to Definition 4. To prove the converse, first note that by Lemma 2, we may assume that f, g, h are pairwisely algebraically independent for a tame triple (f, g, h) of P2. Then the proof of theorem 1 in ref. 9 shows that (f, g, h) is either elementarily reducible, or admits a reduction of one of the types I–IV. Theorem 3. Let f, g, h P2 such that K[f, g, h] = P2. Then the tameness of the triple (f, g, h) is algorithmically recognizable. Proof: By Theorem 2, we only need to prove that the simple reducibility of (f, g, h) is algorithmically recognizable. The elementary reducibility of (f, g, h) is algorithmically recognizable by Theorem 1, if (f, g, h) admits a reduction of one of types I–IV, then the proof is similar to the proof of theorem 3 in ref. 9. Remark 1: If p P3 is irreducible and p(f, g, h) = 0, then we do not know whether the hypothesis of the above theorem implies that p is a coordinate of P3. The famous embedding conjecture of Abhyankar-Sathaye claims that in this case p is a coordinate. See, for instance, ref. 16. On the other hand, Lemma 5 below shows that p is a tame coordinate if and only if (f, g, h) is a tame triple. Lemma 5. Let be a tame triple with deg(fi) ≤ 1 for some i {1, 2, 3}. Then θ admits an elementary reduction. Proof: It is easy to see from Definitions 1–3 that in this case θ does not admit a reduction of types I–IV. Now we consider coordinates of P3 = K[x, y, z]. If one of the coordinates of θ = (f, g, h) AutP3 is wild, then obviously θ is a wild automorphism. But even if θ is a wild automorphism, it does not necessarily imply that all coordinates of the θ are wild. For example, the Nagata automorphism σ: = (x – 2wy – w2z, y + wz, z) of P3, where w = y2 + xz, is wild (9–11), but the coordinate z of σ is obviously tame.
Lemma 6. Let p ε P3 be a coordinate of P3 and let (x(u, v), y(u, v), z(u, v)) be a faithful parametrization of the p. Then the coordinate p is tame if and only if the triple (x(u, v), y(u, v), z(u, v)) is tame. Proof: Put = x(u, v), ŷ = y(u, v), = z(u, v). Assume that p is a tame coordinate and let (p, q, r) be a tame automorphism. Then Recall that the well-known Nagata automorphism σ AutP3 is defined as follows (see ref. 8): Theorem 4. The coordinates f and g of the Nagata automorphism σ are wild. Lemma 7. Let p be a coordinate of the algebra Pn and let be a tame automorphism in AutPn. Then p is a tame coordinate if and only if (p) is a tame coordinate. Proof: The conclusion is obvious by the definition of a tame coordinate. Theorem 5. Let θ = (f, g, z) AutP3. If θ is a wild automorphism, then both coordinates f and g are wild. Proof: Put θ–1 = (p, q, z). Then the homomorphism (see 4) defined by Remark 2: As a consequence, a wild coordinate of K[z][x, y] is always a wild coordinate of P3 = K[x, y, z], hence by Drensky and Yu (1, 2) (where all wild coordinates of K[z][x, y] are effectively classified), we obtain many wild coordinates of P3. Theorem 6. Let θ = (f, g, h) be a wild automorphism in AutP3. Then at least two coordinates among the coordinates f, g, h of θ are wild coordinates of P3. Proof: Suppose that h is a tame coordinate and let AutP3 be a tame automorphism with (h) = z. Then θ1 = θ = ((f), (g), z) is also a wild automorphism. By Theorem 5, the coordinates (f) and (g) are wild. Now the wildness of both coordinates f and g follows from Lemma 7. The conclusion of the main theorem follows from Theorems 4–6. | |||||||||||||||||||||||||||||||||||||||||||||||
Some Open Problems Recall that an element q of the free associative algebra An:= Kx1,..., xn is called a primitive element if there exists an automorphism AutAn such that (q) = x1. Tame and wild automorphisms and primitive elements of An can also be defined naturally. Problem 1. Let p P3 be a coordinate. Can p be lifted to a primitive element p′ of the free associative algebra A3? In other words, does there exist a primitive element p′ A3 such that the natural homomorphism (the abelianization) from A3 to P3 takes p′ to p? Remark 3: If the answer to Problem 1 is yes for some wild coordinate p of P3, then obviously p′ is a wild primitive element of A3, hence there exist wild primitive elements and wild automorphisms of A3 that would settle a famous conjecture of Cohn (17) affirmatively. On the other hand, if the answer is no for some coordinate q of P3, then we would obtain a new proof of the Nagata conjecture without using the previous results of Shestakov and Umirbaev (9–11). Suppose that f1,..., fk Pn and K[f1,..., fk] = Pn. Then obviously n ≤ k. The following two problems are naturally raised. Problem 2. Does there exist an automorphism AutPk such that
Problem 3. Does there exist a tame automorphism AutPk such that
If k = n, then Problem 2 becomes the problem of recognizing automorphisms of Pn and was solved by van den Essen (16) by the Gröbner basis method; and Problem 3 becomes the Nagata conjecture for Pn. The case k = 2, n = 1 of both Problems 2 and 3 have positive solutions if Char(K) = 0 (that is the well-known Abhyankar-Moh Theorem) and a negative solution if Char(K) > 0 (13). The case k ≥ 2n + 2of Problem 2 was solved positively in refs. 18–21 (see also refs. 12 and 22–24). In fact, the automorphisms obtained in refs. 12 and 18–24 are all tame, hence also give the positive solutions of the case k ≥ 2n + 2 of Problem 3. The case n = k = 3 of Problem 3 was solved negatively by Shestakov and Umirbaev (9–11). Note that some other examples considered in refs. 1, 2, and 25 are also wild by refs. 9–11. Also Theorems 4 and 5 give the answer to the case n = 3, k = 2 of Problem 3 negatively. It is natural to raise the following. Problem 4. Let p P3 be a given coordinate. Does there exist an algorithm to determine whether p is tame or wild? This problem is closely related to the following question: for a given polynomial p P3, how do we effectively determine whether p is a coordinate? Moreover, if p is a coordinate, how do we effectively construct an automorphism AutP3 such that (p) = x? Theorem 7. Let p be a given coordinate of P3 and let θ = (f, g, h) be a given automorphism in AutP3 taking p to x. Then there exists an algorithm to determine whether the coordinate p is tame or wild. Proof: Suppose θ = (f, g, h). Then the homomorphism (see 4) defined by Finally, in view of Theorems 5 and 6, we may propose the following. Problem 5. Is AutK[x, y, z] essentially generated by ? Or, more precisely, is it true that any automorphism in AutK[x, y, z] can be decomposed as a product of automorphisms fixing z and linear automorphisms? | |||||||||||||||||||||||||||||||||||||||||||||||
Acknowledgments We thank Leonid Makar-Limanov, Vladimir Shpilrain, and Efim Zelmanov for stimulating discussion and helpful suggestions. This work was partially supported by Hong Kong Research Grants Council Competitive E-marked Research Grant Grants 10203186 and 10203669. | |||||||||||||||||||||||||||||||||||||||||||||||
Footnotes ¶In formulation of algorithmic results, we always assume that the ground field K is constructive. | |||||||||||||||||||||||||||||||||||||||||||||||
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