A COUNTEREXAMPLE TO HILBERT’S FOURTEENTH PROBLEM IN DIMENSION FIVE

DANIEL DAIGLE AND GENE FREUDENBURG

Abstract. The kernel of a certain triangular derivation of the polynomial ring k[x_{1}, x_{2}, x_{3}, x_{4}, x_{5}] is shown to be non-finitely generated over k (a field of charactersitic zero), thus giving a new counterexample to Hilbert’s Fourteenth Problem, in the lowest dimension to date. (Math. Subject Class. 13A50, 13E15)

1. Introduction

In a recent paper of the second author [2], a counterexample to Hilbert’s Fourteenth Problem is given in dimension six, in the form of the kernel of a triangular derivation of k^{[6]}, the polynomial ring in six variables over a field k of characteristic zero. This dimension-six example has its roots in Roberts’ well-known counterexample in dimension seven [5], which can be realized as the kernel of a triangular derivation of k^{[7]}. Roberts’ example was generalized in both [1] and [3] to produce triangular counterexamples in every dimension n ≥ 7.

In the present work, we use a quotient of the dimension-six example to produce a remarkably simple counterexample in dimension five. Like its predecessors, this new example is the kernel of a triangular derivation of a polynomial ring, namely, D : k[a, b, x, y, z] → k[a, b, x, y, z],

D = a^{2} \frac{∂}{∂x} + (ax + b)\frac{∂}{∂y} + y \frac{∂}{∂z}

(see Section 3). Thus, the question of finite generation remains open only in dimension four, since it is known that no counterexample can be found in dimension less than four (see [6]).

Preliminaries. In what follows, k is a field of characteristic zero. The term derivation will mean a k-derivation of the form D : B → B, where B is an integral k-domain. D is locally nilpotent if, to each f ∈ B, there is an integer n ≥ 0 such that D^{n}f = 0.

For an integral k-domain R, the symbol R^{[n]} will denote the polynomial ring in n variables over R. If B = R[X_{1}, ..., X_{n}] = R^{[n]}, an R-derivation T of B is called triangular (with respect to X_{1}, ..., X_{n}) if T X_{1} ∈ R, and T X_{i} ∈ R[X_{1}, ..., X_{i−1}] for i = 2, ..., n. Note that triangular derivations are locally nilpotent.

Research of the first author supported by a grant from NSERC Canada.

The proof of Proposition 2.5 uses the following well-known fact: Suppose that D : B → B is a locally nilpotent derivation and that σ ∈ B satisfies Dσ ≠ 0 and D^{2}σ = 0. Then b ↦ ∑_{n=0}^{∞} \frac{D^{n}b}{n!} (−\frac{σ}{Dσ})^{n} defines a ring homomorphism B → (ker D)_{Dσ}, where (ker D)_{Dσ} is the localization of ker D with respect to {1, Dσ, (Dσ)^{2}, … }.

2. Kernel Sequences

Let k be a field of characteristic zero. Consider a pair (B, D), where B = ⊕_{i∈N}B_{i} is an N-graded k-domain and D : B → B is a homogeneous locally nilpotent derivation. Let A = ker D.

We are interested in elements α ∈ A such that:

(NFG) ker(D̃) is not finitely generated as a k-algebra, where D̃ : B[T] → B[T] is the unique derivation extending D which maps T on α (T is an indeterminate over B).

When this is the case, we say that “(B, D, α) satisfies (NFG) over k”. Lemma 2.1. Suppose that the grading of B satisfies B_{0} = k and let M[T] be the extension to B[T] of the maximal ideal M = ⊕_{i>0}B_{i} of B. If α ∈ A \ D(B) is a homogeneous element and D̃ : B[T] → B[T] is the unique derivation extending D and satisfying D̃(T) = α, then

(1) ker D̃ ⊆ k + M[T].

(2) Let {ϕ_{n}} be an infinite sequence in ker D̃ and let b_{n} ∈ B be the leading coefficient of ϕ_{n}. If the sequence {deg(b_{n})} is bounded above but {deg_{T} (ϕ_{n})} is not, then the k-algebra ker D̃ is not finitely generated.

Proof. Assertion (2) is an immediate consequence of (1), so we prove only (1).

Let δ = deg(α) − deg(D) and, for each d ∈ Z, define

B[T]_{d} = ∑_{i∈N} B_{d−δi}T^{i}.

Then B[T] = ⊕_{d∈Z}B[T]_{d} is a Z-grading of B[T] satisfying:

• B_{i} ⊆ B[T]_{i}, for all i ∈ N.

• If ϕ = ∑_{i} ϕ_{i}T^{i} is a homogeneous element of B[T] (with ϕ_{i} ∈ B) then, for all i ∈ N, ϕ_{i} is a homogeneous element of B.

• D̃ is homogeneous, so ker D̃ is a homogeneous subring of B[T].

Let ϕ be a homogeneous element of ker D̃. Write ϕ = ∑_{i∈N} ϕ_{i}T^{i}, ϕ_{i} ∈ B, ϕ_{i} = 0 for i ≫ 0. Since 0 = D̃(ϕ) = ϕ^{(D)}(T) + αϕ′(T),

(1)  Dϕ_{i−1} = −iαϕ_{i}, for all i > 0.

Thus ϕ_{i} ∉ k^{∗} for all i > 0, for otherwise equation (1) would give α ∈ D(B), a contradiction; since ϕ_{i} is a homogeneous element of B, it must belong to M. So

ϕ_{0} ∈ k + M and ϕ_{i} ∈ M for all i > 0.

This means that ϕ ∈ k + M[T], so (1) is proved. □

Definition 2.2. By a kernel-sequence of (B, D), we mean a sequence a = {a_{n}}_{n=1}^{∞} of nonzero homogeneous elements of A for which there exists an integer p > 0 satisfying

a_{n+p} = a_{n}, for all n ≥ 1.

Such an integer p is called a period of a. Clearly, if p is the minimal period then the set of all periods of a is {p, 2p, 3p, … }. If p is a period and α ∈ A satisfies a_{1} ⋯ a_{p} = α^{p}, we call α a p-average of a. The phrase “α is an average of a” means that α is a p-average of a for some period p. Remark 2.3. Assume that k contains all roots of unity and let a = {a_{n}}_{n=1}^{∞} be a kernel-sequence of (B, D).

(1) If α is an average of a then so is ζα for every root of unity ζ.

(2) If α is an average of a then there exists a root of unity ζ such that, for every period p, ζα is a p-average of a.

(3) If α_{1} and α_{2} are averages of a, then α_{1} = ζα_{2} for some root of

unity ζ. Definition 2.4. A kernel-sequence a = {a_{n}}_{n=1}^{∞} of (B, D) is said to be effective if the following equivalent conditions are satisfied:

(1) There exists a sequence b = {b_{n}}_{n=0}^{∞} in B, other than the zero sequence, satisfying Db_{n} = a_{n}b_{n−1} for all n > 0.

(2) There exists a sequence b satisfying condition (1) and such that

every term b_{n} is nonzero and homogeneous. If a is an effective kernel-sequence of (B, D), then a sequence b satisfying condition (2) is called an integral sequence of (B, D) belonging to a.

A kernel-sequence a of (B, D) is strictly effective if: (i) a is effective; (ii) a has an average α; and (iii) α ∉ D(B). Proposition 2.5. Suppose that B_{0} = k and that (B, D) admits a strictly effective kernel-sequence a = {a_{n}}_{n=1}^{∞}. Let α ∈ A be an average of a. Then (B, D, α) satisfies (NFG) over k.

Proof. Let p be a period of a = {a_{n}}_{n=1}^{∞} such that α is a p-average of a. Let b = {b_{n}}_{n=0}^{∞} be an integral sequence belonging to a and let ν ≥ 0 be the unique integer such that D^{ν}(b_{0}) ≠ 0 and D^{ν+1}(b_{0}) = 0. Define the integer e = ν + p − 1 ≥ 0.

Let D̃ : B[T] → B[T] be the extension of D satisfying D̃(T) = α. Consider the ring homomorphism ξ : B[T] → (ker D̃)_{α} defined by

ξ(f) = ∑_{j∈N} \frac{D̃^{j}(f)}{j!} (− \frac{T}{α} )^{j} (f ∈ B[T]),

where (ker D̃)_{α} is the localization of ker D̃ with respect to {1, α, α^{2}, … }. Let ϕ_{n} = ξ(α^{e}b_{n}) for n ≥ 0. In view of Lemma 2.1, it suffices to show that the following two claims hold for all n ≥ 0:

(1) ϕ_{n} ∈ ker D̃;

(2) the leading term of ϕ_{n} is \frac{(−1)^{n+ν}}{(n + ν)!} γ_{n} T^{n+ν}, where \{γ_{n}\}_{n=0}^{∞} is a

sequence in A \ {0} with finite image. We have

D^{j}(b_{n}) = a_{n} … a_{n−j+1}b_{n−j} (0 ≤ j ≤ n)

and

D^{n+i}(b_{n}) = a_{n} … a_{1}D^{i}(b_{0}) (i ≥ 0),

so

ϕ_{n} = ∑_{j∈N} \frac{(−1)^{j}}{j!} \frac{α^{e}D^{j}(b_{n})}{α^{j}}T^{j}

= ∑_{j=0}^{n} \frac{(−1)^{j}}{j!} (\frac{α^{p−1}a_{n} … a_{n−j+1}}{α^{j}} ) α^{ν}b_{n−j}T^{j}

+ ∑_{i=1}^{ν} \frac{(−1)^{n+i}}{(n + i)!} (\frac{α^{p−1}a_{n} … a_{1}}{α^{n}} ) α^{ν−i}D^{i}(b_{0})T^{n+i}.

Since the product of any p consecutive terms of the sequence {a_{n}}_{n=1}^{∞} gives α^{p},

S = \{\frac{α^{p−1}a_{n} … a_{n−j+1}}{α^{j}} | (j, n) ∈ N^{2} and j ≤ n\}

is a finite subset of A \ {0}. In particular, S ⊂ A implies that ϕ_{n} ∈ B[T] ∩ (ker D̃)_{α} = ker D̃, so (1) holds. Since the leading term of ϕ_{n} is

\frac{(−1)^{n+ν}}{(n + ν)!} (\frac{α^{p−1}a_{n} … a_{1}}{α^{n}} ) D^{ν}(b_{0})T^{n+ν},

and S is a finite subset of A, (2) holds. □

Corollary 2.6. Suppose that B is noetherian or factorial, and that (B, D) admits a strictly effective kernel-sequence a = {a_{n}}_{n=1}^{∞} with average α ∈ A. Suppose that q is a homogeneous prime element of B satisfying: q ∈ A, and

(1) q |̸ a_{n} for all n ≥ 1;

(2) α ∉ D(B) + qB. Then \overline{a} = \{\overline{a}_{n}\} is a strictly effective kernel-sequence of (\overline{B}, \overline{D}) with average \overline{α}, where \overline{B} = B/qB, \overline{D} : \overline{B} → \overline{B} is the induced derivation, \overline{a}_{n} = a_{n} + qB and \overline{α} = α + qB. If, moreover,

(3) qB ∩ B_{0} is a maximal ideal of B_{0}, then (\overline{B}, \overline{D}, \overline{α}) satisfies (NFG) over \overline{B}_{0} = B_{0}/(qB ∩ B_{0}).

Proof. It is immediate that \overline{a} is a kernel-sequence of (\overline{B}, \overline{D}) with average \overline{α}.

Consider an integral sequence b = {b_{n}}_{n=0}^{∞} of (B, D) belonging to a. Observe that, if q | b_{n} for all n ≥ 0, then \{\frac{b_{n}}{q} \}_{n=0}^{∞} is another integral sequence belonging to a; since B is noetherian or factorial we have ⋂_{n=1}^{∞} q^{n}B = {0}, so we may assume that some b_{n} is not divisible by q. Then \overline{b} = \{b_{n} + qB\}_{n=0}^{∞} satisfies condition (1) of Definition 2.4, so \overline{a} is effective. Since we assumed that α ∉ D(B) + qB, we have \overline{α} ∉ \overline{D}(\overline{B}), i.e., \overline{a} is strictly effective. The last assertion follows from Proposition 2.5 because \overline{B}_{0} is a field. □

3. A Non-Finitely Generated Kernel in Dimension Five

The following six-dimensional counterexample to Hilbert’s Fourteenth Problem was given in [2]. We interpret it in terms of the concepts of Section 2. Example 3.1. Let R = k[a, b, s, t, u] = k^{[5]}, with grading R = ⊕_{i∈N}R_{i} determined by R_{0} = k, a, b, s ∈ R_{3}, t ∈ R_{6} and u ∈ R_{9}. Define d : R → R by

d = a\frac{∂}{∂ s} + bs\frac{∂}{∂ t} + bt\frac{∂}{∂ u}.

Then d is locally nilpotent and homogeneous of degree 0. Consider the kernel-sequence a = {a_{n}}_{n=1}^{∞} = {a, b, ab, a, b, ab, … } of (R, d). Section 2 of [2] proves that a is effective (the proof of this is non-trivial!). However, a is not strictly effective because it does not have an average ((a^{2}b^{2})^{1/3} ∉ R).

Consider the overring B = k[x, y, s, t, u] = k^{[5]} of R, where x^{3} = a and y^{3} = b, with grading B = ⊕_{i∈N}B_{i} determined by B_{0} = k, x, y ∈ B_{1}, s ∈ B_{3}, t ∈ B_{6} and u ∈ B_{9}. Define D : B → B by

D = x^{3} \frac{∂}{∂ s} + y^{3}s\frac{∂}{∂ t} + y^{3}t\frac{∂}{∂ u}.

Then D is locally nilpotent and homogeneous of degree 0. Since D extends d, a is an effective kernel-sequence of (B, D). Clearly, α = x^{2}y^{2} is an average of a; since D(B) is contained in the ideal (x^{3}, y^{3}) of B, α ∉ D(B), i.e., a is a strictly effective kernel-sequence of (B, D). Consequently, if R = B[v] = B^{[1]} = k^{[6]} and D : R → R is the unique extension of D satisfying D(v) = x^{2}y^{2}, then Proposition 2.5 implies that ker D is not finitely generated as a k-algebra.

Using Corollary 2.6, we deduce a counterexample in dimension five: Example 3.2. Let B, D, a and α be as in Example 3.1 and consider a different grading B = ⊕_{i∈N}B_{i}, where B_{0} = k[y], x ∈ B_{1}, s, t, u ∈ B_{3}. Then D is homogeneous of degree 0 and a is a strictly effective kernel sequence of (B, D) with average α. Note that q = y − 1 is a homogeneous prime element of B satisfying Dq = 0 and q |̸ a_{n} for all n. Let \overline{B} = B/qB = k[x, s, t, u] = k^{[4]} then the induced derivation \overline{D} : \overline{B} → \overline{B} is:

\overline{D} = x^{3} \frac{∂}{∂ s} + s\frac{∂}{∂ t} + t\frac{∂}{∂ u}.

Since \overline{α} = x^{2} does not belong to the ideal (x^{3}, s, t) of \overline{B}, we have \overline{α} ∉ \overline{D}(\overline{B}) (or equivalently condition (2) of Corollary 2.6 holds). By 2.6, the kernel of T : \overline{B}[v] → \overline{B}[v] is not finitely generated as a kalgebra, where T is the extension of \overline{D} which maps v to α. Explicitly, T : k[x, s, t, u, v] → k[x, s, t, u, v] is given by:

(2)  T = x^{3} \frac{∂}{∂ s} + s\frac{∂}{∂ t} + t\frac{∂}{∂ u} + x^{2} \frac{∂}{∂ v} .

Observe that T (s−xv) = 0 and that k[x, s, t, u, v] = k[x, s−xv, t, u, v] (s − xv is a variable). Let σ be the polynomial automorphism of k[x, s, t, u, v] fixing x, t, u, v and mapping s ↦ (s + xv). Then

σT σ^{−1} = x^{2} \frac{∂}{∂v} + (xv + s)\frac{∂}{∂t} + t\frac{∂}{∂u} ,

which gives the following: Theorem 3.3. Let D be the triangular derivation of k[a, b, x, y, z] = k^{[5]} defined by

(3)  D = a^{2} \frac{∂}{∂x} + (ax + b)\frac{∂}{∂y} + y \frac{∂}{∂z} .

Then the kernel of D is not finitely generated as a k-algebra. Remarks (1) By (2), there exists a triangular monomial derivation of k^{[5]}

whose kernel is not a finitely generated k-algebra. (See [4].)

(2) Let R = k^{[2]} and let X, Y, Z be indeterminates over R. By (3), there exists a triangular R-derivation of R[X, Y, Z] whose kernel is not a finitely generated R-algebra.

References

[1] A. van den Essen, T. Janssen, Kernels of elementary derivations, Univ. Nijmegen Report No. 9548 (Nijmegen, The Netherlands) 1995

[2] G. Freudenburg, A counterexample to Hilbert’s Fourteenth Problem in dimension six, Transformation Groups (to appear)

[3] H. Kojima, M. Miyanishi, On Roberts’ counterexample to the fourteenth problem of Hilbert, J. Pure Appl. Algebra 122 (1997) 277-292

[4] S. Maubach, Triangular monomial derivations on k[X_{1}, X_{2}, X_{3}, X_{4}] have kernel generated by at most four elements, Univ. Nijmegen Report No. 9822 (Nijmegen, The Netherlands) 1998

[5] P. Roberts, An infinitely generated symbolic blow-up in a power series ring and a new counterexample to Hilbert’s Fourteenth Problem, J. Algebra 132 (1990) 461-473

[6] O. Zariski, Interpretations algebrico-geometriques du quatorzieme probleme de Hilbert, Bull. Sci. Math. 78 (1954) 155-168

DEPARTMENT OF MATHEMATICS, UNIVERSITY OF OTTAWA, OTTAWA, ONTARIO

K1N 6N5 CANADA

E-mail: ddaigle@uottawa.ca

DEPARTMENT OF MATHEMATICS, UNIVERSITY OF SOUTHERN INDIANA, EVANS- VILLE, INDIANA 47712 USA

E-mail: freudenb@usi.edu