The Nagata automorphism is wild

Abstract

It is proved that the well known Nagata automorphism of the polynomial ring in three variables over a field of characteristic zero is wild, that is, it can not be decomposed into a product of elementary automorphisms.

Let A = F[x₁, x₂,..., x_n] be the polynominal ring in the variables x₁, x₂,..., x_n over a field F, and let Aut A be the group of automorphisms of A as an algebra over F. An automorphism τ ∈ Aut A is called elementary if it has a form

where 0 ≠ α ∈ F, f ∈ F[x₁,..., x_i-1, x_i+1,..., x_n]. The subgroup of Aut A generated by all the elementary automorphisms is called the tame subgroup, and the elements from this subgroup are called tame automorphisms of A.

It is well known (1–5) that the automorphisms of polynomial rings and free associative algebras in two variables are tame. At present, a few new proofs of these results have been found (see refs. 6 and 7). However, in the case of three or more variables the similar question was open and known as “the generation gap problem” (8, 9) or “tame generators problem” (7). The general belief was that the answer is negative, and there were several candidate counterexamples. The most known of them is the following automorphism σ ∈ Aut(F[x, y, z]), constructed by Nagata (10) in 1972:

Observe that the Nagata automorphism is stably tame (11), that is, it becomes tame after adding new variables.

The purpose of the present work is to give a negative answer to the above question. Our main result states that the tame automorphisms of the polynomial ring A = F[x, y, z] over a field F of characteristic 0 are algorithmically recognizable. In particular, the Nagata automorphism σ is “wild,” that is, not tame.

The approach we use is different from the traditional ones. Thenovelty consists of the imbedding of the polynomial ring A into the free Poisson algebra (or the algebra of universal Poisson brackets) on the same set of generatots and of systematical using of brackets as an additional tool.

The crucial role in the proof is played by the description of the structure of two-generated subalgebras of polynomial rings. In particular, we obtain a lower estimate for degrees of elements of these subalgebras, which is essentially used in most of the proofs. Observe that this estimate provides one more proof of the Jung theorem (1) on the tameness of the automorphisms of polynomials in two variables.

Let us give the necessary definitions.

Recall that a vector space B over a field F, endowed with two bilinear operations x·y (a multiplication) and [x, y] (a Poisson bracket), is called a Poisson algebra, if B is a commutative associative algebra under x·y, B is a Lie algebra under [x, y], and B satisfies the Leibuiz identity

[1]

An important class of Poisson algebras is given by the following construction. Let L be a Lie algebra with a linear basis l₁, l₂,..., l_k,.... Denote by P(L) the ring of polynomials on the variables l₁, l₂,..., l_k,.... The operation [x, y] of the algebra L can be uniquely extended to a Poisson bracket [x, y] on the algebra P(L) by means of formula 1, and P(L) becomes a Poisson algebra (12).

Now let L be a free Lie algebra with free generators x₁, x₂,..., x_n. Then P(L) is a free Poisson algebra (12) with the free generators x, x₂,..., x_n. We will denote this algebra by PL(x₁, x₂,..., x_n). If we choose a homogeneous basis

of the algebra L with nondecreasing degrees, then PL〈x₁, x₂,..., x_n〉, as a vector space, coincides with the ring of polynomials on these elements. Evidently, PL〈x₁, x₂,..., x_n〉 is graded by degrees on x₁ as a Poisson algebra, and for every element f ∈ PL(x₁, x₂,..., x_n), the highest homogeneous part f̄ and the degree deg f can be defined in an ordinary way. Note that

[2]

Now, let F be a field of characteristic 0 and let A = F[x₁,..., x_n] be the polynomial ring over F in the variables x₁,..., x_n. We will identify A with the subspace of the algebra PL〈x₁,..., x_n〉 generated by elements

In this way, A is endowed with a Poisson bracket, with the values in PL(x₁,..., x_n). Note that if f, g ∈ A. then

Lemma 1. Elements f, g ∈ A are algebraically dependent if and only if [f, g] = 0.

This implies, in particular, that the inequality for deg[f, g] in Eq. 2 turns to an equality if and only if f̄, ḡ are algebraically independent.

For an element f ∈ A, the set of elements

is called the centralizer of f in A. It follows immediately from identity 1 that C(f) is a subalgebra of A. The next theorem may be considered as an analog of the Bergman theorem (13) on centralizers in free associative algebras.

Theorem 1. For every f ∈ A\F, the centralizer C(f) is a ring of polynomials on single variable.

If f₁, f₂,..., f_k ∈ A, then by 〈 f₁, f₂,..., f_k〉 we denote the subalgebra of A generated by these elements.

Definition 1: A pair of elements f, g of the algebra A is called *-reduced if

1. f, g are algebraically independent;

2. f̄, ḡ are algebraically dependent;

3. f̄ ∈ 〈ḡ〉, ḡ ∈ 〈f̄〉.

Recall that a pair with condition i is called reduced (see ref. 14). We do not consider the case when f, g are algebraically dependent. Concerning this case, recall the well known theorem of S. S. Abhyankar and T.-T. Moh (15), which says that if f, g ∈ F[x] and 〈 f, g〉 = F[x], then f̄ ∈ 〈ḡ〉 or ḡ ∈ 〈f̄〉.

We will mostly consider the pairs that are components of automorphisms, when condition i is automatically satisfied. Condition ii means that we exclude the trivial case when f̄, ḡ are algebraically independent. In this case, by ref. 6, the inclusion h ∈ 〈 f, g〉 always implies that h̄ ∈ 〈f̄, ḡ〉.

Let f, g be a *-reduced pair of elements of A and n = deg f < m = deg g. Put p = n/(n, m), s = m/(n, m), where (n, m) is the greatest common divisor of n, m. Sometimes, we will call the *-reduced pair f, g also a p-reduced pair. In these notations, the following theorem is true.

Theorem 2. Let G(x, y) ∈ F[x, y] and deg_y(G(x, y)) = k = pq + r, 0 ≤ r < p. Then,

[3]

Thus, the low degree of elements in the subalgebra 〈 f, g〉 depends on the degree of the Poisson bracket [f, g]. The estimation of deg[f, g] is a very difficult problem that is related with the famous Jacobian conjecture (16).

One can easily prove that, under the conditions of Theorem 2, the elements f ⁱg ^j with j < p have different degrees for different values of i, j. This implies

Corollary 1. Under the conditions of Theorem 2, for any h ∈ 〈 f, g〉, either h̄ ∈ 〈f̄, ḡ〉 or deg h ≥ pm - m - n + deg[f, g].

Note that we have p ≥ 2 and so pm - m - n + deg[f, g] > deg[f, g] ≥ 2. Therefore, for any *-reduced pair f, g, the subalgebra 〈 f, g〉 does not contain the generators x_i. In particular, we have

Corollary 2. For any *-reduced pair f, g ∈ F[x, y], the endomorphism defined by the condition is not surjective.

Note that this corollary provides one more proof of the Jung theorem (1). In fact, if ϕ: (x y) → (f, g) is an automorphism of F[x, y] then the pair f, g is not *-reduced, and so either f̄, ḡ are algebraically independent, and in this case deg ϕ = 2 since x, y ∈ 〈f̄, ḡ〉, or condition iii of Definition 1 is satisfied and deg ϕ = deg f + deg g can be reduced by an elementary automorphism. The induction on deg ϕ finishes the proof.

Observe that Corollary 2, in turn, follows easily from the Jung theorem.

In what follows, A = F[x₁, x₂, x₃]. A triple Θ = (f₁, f₂, f₃) [or simply (f₁, f₂, f₃)] of elements of the algebra A below always will denote the automorphism Θ of A such that Θ(x_i) = f_i, 1 ≤ i ≤ 3. The number deg Θ = deg f₁ + deg f₂ + deg f₃ we will call the degree of the automorphism Θ. Recall that an elementary transformation of the triple (f₁, f₂, f₃) changes only one coordinate f_i to the element of the form αf_i + g, where 0 ≠ α ∈ F, g ∈ 〈{f_j|j ≠ i}〉. The notation Θ → τ means that the triple τ is obtained from Θ by a single elementary transformation. An automorphism Θ is called elementary reducible or admitting an elementary reduction if there exists τ ∈ Aut A such that Θ → τ and deg τ < deg Θ. The element f_i of Θ which is changed by τ (to an element of lower degree) we will call the reducible element and we will also say that f_i is reduced from Θ by τ.

Lemma 2. The elementary reducibility of automorphisms of the algebra A is algorithmically recognizable.

Now we give an example of tame automorphism, which doesn't admit an elementary reduction.

Example: Put

It is easy to show that (h₁, h₂, h₃) and (g₁, g₂, g₃) are tame automorphisms of the algebra A. Note that deg g₁ = 9, deg g₂ = 6, deg g₃ = 3 and g₁, g₂ form a 2-reduced pair. The element f = g²₁ - g³₂ has degree 8. Hence deg f < deg g₁, and f̄ ∈ 〈ḡ₁, ḡ₂〉. Let us define a tame automorphism (f₁, f₂, f₃) by putting

Then we have deg f₁ = 9, deg f₂ = 6, deg f₃ = 8. It follows easily from Theorem 2 that the automorphism (f₁, f₂, f₃) doesn't admit an elementary reduction.

Thus, the degree of a given tame automorphism not always can be reduced by one elementary automorphism. In general, we need several such automorphisms.

We will now define four types of (nonelementary) reductions for automorphisms of A, all of which are products of at most four elementary automorphisms of special type.

Definition 2: Let Θ = (f₁, f₂, f₃) be an automorphism of A such that deg f₁ = 2n, deg f₂ = ns, s ≥ 3 is an odd number, 2n < deg f₃ ≤ ns, f̄₃ ∈ 〈 f̄₁, f̄₂〉. Suppose that there exists 0 ≠ α ∈ F such that the elements g₁ = f₁, g₂ = f₂ - αf₃ satisfy the conditions:

1. g₁, g₂ is a 2-reduced pair and deg g₁ = deg f₁, deg g₂ = deg f₂;

2. the element f₃ of the automorphism (g₁, g₂, f₃) is reduced by an automorphism (g₁, g₂, g₃) with the condition deg[g₁, g₃] < deg g₂ + deg[g₁, g₂].

Then we will say that Θ admits a reduction (g₁, g₂, g₃) of type I.

Observe that the automorphism from the Example admits a reduction of type I.

Definition 3: Let Θ = (f₁, f₂, f₃) be an automorphism of A such that deg f₁ = 2n, deg f₂ = 3n, 3n/2 < deg f₃ ≤ 2n, and f̄₁, f̄₃ are linearly independent. Suppose that there exist α, β ∈ F, where (α, β) ≠ (0, 0), such that the elements g₁ = f₁ - αf₃, g₂ = f₂ - βf₃ satisfy the conditions i and ii of Definition 2. Then we will say that Θ admits a reduction (g₁, g₂, g₃) of type II.

Definition 4: Let Θ = (f₁, f₂, f₃) be an automorphism of A such that deg f₁ = 2n, and either deg f₂ = 3n, n < deg f₃ ≤ 3n/2, or 5n/2 < deg f₂ ≤ 3n, deg f₃ = 3n/2. Suppose that there exist α, β, γ ∈ F such that the elements g₁ = f₁ - βf₃, g₂ = f₂ - γf₃ - αf ²₃ satisfy the conditions:

1. g₁, g₂ is a 2-reduced pair and deg g₁ = 2n, deg g₂ = 3n;

2. there exists an element g₃ of the form

where 0 ≠ σ ∈ F, g ∈ 〈g₁, g₂〉, such that deg g₃ ≤ 3n/2, deg[g₁, g₃] < 3n + deg[g₁, g₂].

If (α, β, γ) ≠ (0, 0, 0) and deg g₃ < n + deg[g₁, g₂], then we will say that Θ admits a reduction (g₁, g₂, g₃) of type III. On the other hand, if there exists 0 ≠ μ ∈ F such that deg(g₂ - μg²₃) ≤ 2n, then we will say that Θ admits a reduction (g₁, g₂ - μg²₃, g₃) of type IV.

The role of reductions I–IV is justified by the following theorem.

Theorem 3. Let Θ be a tame automorphism of the ring of polynomials A = F[x₁, x₂, x₃] over a field F of characteristic 0. If deg Θ > 3, then Θ admits either an elementary reduction or a reduction of types I–IV.

For the plan of the proof, we follow the method of simple automorphisms developed in ref. 17. Let us call an automorphism Θ ∈ Aut A to be simple if deg Θ = 3 or deg Θ = n > 3 and there exists a simple automorphism of degree < n, which is either an elementary reduction or a reduction of types I–IV of Θ. Evidently, every simple automorphism is tame. To prove the theorem, it suffices to show the contrary, that every tame automorphism of A is simple.

Assume that this is not true. Then there exist tame automorphisms Θ, τ of A such that Θ is simple, τ is not simple, and

In the set of all pairs of automorphisms with this property we choose and fix a pair Θ, τ with the minimal deg Θ.

To obtain a contradiction, it is enough to prove that τ is simple. The proof consists of the analysis of the cases, when Θ admits an elementary reduction or a reduction of types I–IV to a simple automorphism of lower degree. In every case, we show that τ admits a similar reduction as well.

Corollary 3. Under the conditions of Theorem 3, if Θ = (f₁, f₂, f₃) with f₃ = x₃, then Θ admits an elementary reduction.

Proof: In fact, it is easy to see that in this case Θ does not admit reductions of types I–IV.

Corollary 4. The Nagata automorphism σ of the polynomial ring F[x, y, z] over a field F of characteristic 0 is wild.

Proof: Note that the leading terms of the components of σ are mutually algebraically independent, and none of them is contained in the subalgebra, generated by the other two leading terms. Consequently, the automorphism σ does not admit an elementary reduction. By Corollary 3, σ is wild.

In ref. 18, some examples of wild automorphisms of the algebra F[z][x, y] were constructed. The next corollary shows that all those automorphisms are also wild as automorphisms of the algebra F[x, y, z].

Corollary 5. Let F be a field of characteristic 0. An automorphism (f, g) of the F[z]-algebra F[z][x, y] is tame if and only if the automorphism (f, g, z) of the F-algebra F[x, y, z] is tame.

Theorem 4. The tame and the wild automorphisms of the algebra F[x₁, x₂, x₃] of polynomials in three variables over a constructive field F of characteristic 0 are algorithmically recognizable.

Note that a reduction of type I consists of two elementary automorphisms, reductions of type II, III consists of three elementary automorphisms, and a reduction of type IV in general case consists of four elementary automorphisms. Then it follows from Theorem 3 that the degree of any tame automorphism of A of degree > 3 can be reduced by four elementary automorphisms. In this context, the following problem seems very interesting.

Problem. Construct examples of tame automorphisms of the algebra A that admit reductions of types II–IV. In other words, do exist tame automorphisms of A whose degrees can not be reduced by two (or even by three) elementary automorphisms?

The automorphism (f₁, f₂, f₃) from the Example admits a reduction of type I and so its degree can be reduced by two (but not one) elementary automorphisms.