A NOTE ON TRIANGULAR DERIVATIONS OF k[X_{1}, X_{2}, X_{3}, X_{4}]

DANIEL DAIGLE AND GENE FREUDENBURG

Abstract. For a field k of characteristic zero, and for each integer n ≥ 4, we construct a triangular derivation of k[X_{1}, X_{2}, X_{3}, X_{4}] whose ring of constants, though finitely generated over k, cannot be generated by fewer than n elements.

1. Introduction

Let k be a field of characteristic zero. If R is a finitely generated k-algebra, we write #(R) = s to indicate that R can be generated by s elements but not by s − 1. The purpose of this note is to show: Theorem. Given any integer n ≥ 3, there exists a triangular derivation Δ of the polynomial ring k[X_{1}, X_{2}, X_{3}, X_{4}] whose kernel satisfies n ≤ #(ker Δ) ≤ n + 1. Equivalently, the theorem asserts that, given n ≥ 3, there exists a triangular action of G_{a} = (k, +) on 𝔸^{4} whose ring of invariants satisfies n ≤ #O(𝔸^{4})^{G_{a}} ≤ n + 1. The theorem is proved by constructing Δ explicitly for n ≥ 4 (for n = 3, just use a partial derivative).

In contrast to our present result, the well-known theorem of Miyanishi [2] states that, for any locally nilpotent k-derivation D of k[X_{1}, X_{2}, X_{3}], #(ker D) = 2. At the other extreme, the authors recently found a triangular derivation of k[X_{1}, X_{2}, X_{3}, X_{4}, X_{5}] whose kernel is not finitely generated as a k-algebra [1]. It is not known whether such kernels in dimension four are always finitely generated, even for triangular derivations.

2. Preliminaries

A triangular derivation of k[X_{1}, …, X_{n}] is a k-derivation Δ : k[X_{1}, …, X_{n}] → k[X_{1}, …, X_{n}] satisfying Δ(X_{i}) ∈ k[X_{1}, …, X_{i−1}] for all i = 1, …, n.

An element of a submonoid Γ of (ℕ, +) is primitive if it is positive and cannot be written as the sum of two positive elements of Γ. It is easy to see that the set of primitive elements in Γ is a finite set which generates Γ, and which is contained in every generating set.

The support of an element f = ∑_{i=0}^{∞} a_{i}X^{i} of the power series ring k[[X]] is Supp(f) = {i ∈ ℕ | a_{i} ≠ 0}. Given a submonoid Γ of (ℕ, +), the elements f of k[[X]] satisfying Supp(f) ⊆ Γ form a subalgebra of k[[X]] which we denote k[[Γ]]. We observe:

1991 Mathematics Subject Classification. Primary: 14R10. Secondary: 14R20, 13N15.

Key words and phrases. Derivations, Hilbert fourteenth problem, additive group actions, invariants.

Research of the first author supported by NSERC Canada.

(1)

If g_{1}, …, g_{r} ∈ k[[Γ]] and P ∈ k[[T_{1}, …, T_{r}]] satisfy ord(g_{i}) ≥ 1 for all i and ord(P) ≥ 2, then no primitive element of Γ belongs to the support of P (g_{1}, …, g_{r}).

Indeed, let γ ∈ Supp P (g_{1}, …, g_{r}); then γ must be in the support of some monomial g_{1}^{i_{1}} … g_{r}^{i_{r}} with i_{1}+⋯+i_{r} ≥ 2, so γ is the sum of i_{1}+⋯+i_{r} elements of ∪_{i=1}^{r} Supp(g_{i}) ⊆ Γ \ {0} and hence is not primitive. Lemma 1. Let Γ be a submonoid of (ℕ, +), let e_{1} < ⋯ < e_{h} be the primitive elements of Γ, let R = k[X^{e_{1}}, …, X^{e_{h}}] and let T be an indeterminate over R. Then:

#(R) = h and #(R[T]) = h + 1.

Proof. Given f ∈ R[T], let f(0) ∈ k[X] be the result of evaluating f at T = 0; and let ord(f) ∈ ℕ ∪ {∞} be the X-order of f(0), i.e., the largest s ≥ 0 such that X^{s} divides f(0) in k[X]. Note that #(R) = h is a consequence of #(R[T]) = h + 1, so it suffices to prove the latter.

Assume that #(R[T]) ≠ h + 1; then R[T] can be generated by h elements, say R[T] = k[f_{1}, …, f_{h}]. We begin by showing that, replacing if necessary the generating set {f_{1}, …, f_{h}} by another one with the same cardinality h, we may arrange that ord(f_{j}) = e_{j} for all j = 1, …, h. To see this, consider an integer i satisfying 1 ≤ i ≤ h and:

(2)  ord(f_{j}) = e_{j},  for all j < i

(this certainly holds for i = 1). Observe that every element of Γ strictly less than e_{i} belongs to the monoid generated by {e_{1}, …, e_{i−1}}; hence, replacing each f_{j} (with j ≥ i) by f_{j} plus a suitable polynomial in (f_{1}, …, f_{i−1}), we may arrange that ord(f_{j}) ≥ e_{i} for all j ≥ i. After relabelling, we obtain that f_{i}, …, f_{h} satisfy:

e_{i} ≤ ord(f_{i}) ≤ ord(f_{i+1}) ≤ ⋯ ≤ ord(f_{h}).

Since X^{e_{i}} ∈ R[T], we may write

X^{e_{i}} = λ_{1}f_{1} + ⋯ + λ_{h}f_{h} + P (f_{1}, …, f_{h}),

where λ_{j} ∈ k, P ∈ k[T_{1}, …, T_{h}] (the T_{j} are indeterminates) and where every monomial occuring in P (T_{1}, …, T_{h}) has degree at least two. Now P (f_{1}, …, f_{h})_{T =0} = P (f_{1}(0), …, f_{n}(0)) = ∑_{j} μ_{j}X^{γ_{j}} where μ_{j} ∈ k and γ_{j} ∈ Γ, but none of these γ_{j} can be a primitive element of Γ by (1). It follows that λ_{j} = 0 for all j < i; also, ord(f_{i}) = e_{i}, so we arranged that (2) holds for a larger value of i. Thus we may arrange that

ord(f_{j}) = e_{j} for all j = 1, …, h.

Since T ∈ R[T], we may write

(3)  T = λ′_{1}f_{1} + ⋯ + λ′_{h}f_{h} + P′(f_{1}, …, f_{h}),

where λ′_{j} ∈ k, P′ ∈ k[T_{1}, …, T_{h}], and where every monomial occuring in P′(T_{1}, …, T_{h}) has degree at least two. Evaluating (3) at T = 0 shows that λ′_{j} = 0 for all j (as before, P′(f_{1}(0), …, f_{h}(0)) can’t produce a term X^{e_{j}}, by (1)). On the other hand, each f_{j} evaluated at X = 0 is an element of T k[T]. Thus, evaluating the equation T = P′(f_{1}, …, f_{h}) at X = 0 yields T = T^{2}Q(T) for some Q(T) ∈ k[T]. This is a contradiction, so #(R[T]) = h + 1 cannot be false. □

Lemma 2. Let h, p, q be positive integers. If gcd(p, q) = 1, then the ideal

(T_{0}^{q} − T_{1}^{p}, T_{1}^{q} − T_{2}^{p}, …, T_{h−1}^{q} − T_{h}^{p})

of k[T_{0}, …, T_{h}] is prime.

Proof. Consider the ideals 𝔭 = (T_{0}^{q} − T_{1}^{p}, …, T_{h−1}^{q} − T_{h}^{p}) of k[T_{0}, …, T_{h}] and 𝔭′ = (T_{0}^{q} − T_{1}^{p}, …, T_{h−2}^{q} − T_{h−1}^{p}) of k[T_{0}, …, T_{h−1}]. By induction, we may assume that 𝔭′ is prime. This allows us to identify R′ = k[T_{0}, …, T_{h−1}]/𝔭′ with k[X^{e_{0}}, …, X^{e_{h−1}}], where X is an indeterminate and e_{j} = p^{h−j}q^{j}. Let K′ be the field of fractions of R′ and note that K′ = k(X^{p}). Since k[T_{0}, …, T_{h}]/𝔭 ≅ R′[T_{h}]/(T_{h}^{p}−θ^{q}), where θ = T_{h−1}+𝔭′ ∈ R′, it suffices to show that T_{h}^{p} − θ^{q} is an irreducible element of K′[T_{h}]; for this, it’s enough to verify that (θ^{q})^{i/p} ∉ K′ for all i = 1, …, p−1. But θ = X^{e_{h−1}}, so (θ^{q})^{i/p} = X^{iq^{h}} ∉ k(X^{p}) for all i = 1, …, p − 1. □

The following is a well-known fact about extracting roots in a power series ring. Lemma 3. Let q be a positive integer, R a domain containing ℚ, W an indeterminate over R and σ an element of R[[W]] with constant term equal to 1 (i.e., σ = 1 + s_{1}W + s_{2}W^{2} + ⋯ where s_{i} ∈ R). Then there exists a unique ρ ∈ R[[W]] satisfying ρ^{q} = σ and having constant term equal to 1. Lemma 4. Let h ≥ 2 be an integer and p, q prime numbers such that p^{2} < q. Then there exist f_{0}, …, f_{h} ∈ k[W, X] satisfying:

(i) f_{j}(0, X) = X^{p^{h−j}q^{j}} for all j such that 0 ≤ j ≤ h; (ii) f_{j+1} ≡ \frac{f_{j−1}^{q}−f_{j}^{p}}{W} (mod W^{h−j}) for all j such that 0 < j < h; (iii) f_{h−1}^{q} − f_{h}^{p} = 0.

Proof. Define f_{h} = X^{q^{h}} and f_{h−1} = X^{pq^{h−1}}. Suppose that f_{h}, …, f_{i} ∈ k[W, X] have been defined (where 0 < i < h) and satisfy (i–iii) and

(4)  X^{p^{h−j}q^{j}} | f_{j} (i ≤ j ≤ h).

Note that the assumption p^{2} < q implies that f_{i}^{p} + W f_{i+1} is divisible by X^{p^{h−i+1}q^{i}}; define

σ = \frac{f_{i}^{p} + W f_{i+1}}{X^{p^{h−i+1}q^{i}}} ∈ k[W, X] ⊂ k[X][[W]]

and note that σ has the form σ = 1+s_{1}W +s_{2}W^{2}+⋯ (with s_{j} ∈ k[X]). By Lemma 3, we may consider ρ = 1 + r_{1}W + r_{2}W^{2} + ⋯ ∈ k[X][[W]] (with r_{j} ∈ k[X]) such that ρ^{q} = σ. Then f̃_{i−1} := X^{p^{h−i+1}q^{i−1}}ρ ∈ k[X][[W]] satisfies

\frac{f̃_{i−1}^{q} − f_{i}^{p}}{W} = f_{i+1} and X^{p^{h−i+1}q^{i−1}} | f̃_{i−1}

so, if f_{i−1} ∈ k[W, X] is a suitable truncation of f̃_{i−1}, then f_{h}, …, f_{i−1} satisfy (i–iii) and (4). So we are done by induction. □

3. The examples

Given an integer h ≥ 2, we construct a triangular derivation Δ : k[W, X, Y, Z] → k[W, X, Y, Z] whose kernel satisfies h + 2 ≤ #(ker Δ) ≤ h + 3.

Choose prime numbers p, q satisfying p^{2} < q; consider f_{0}, …, f_{h} ∈ k[W, X] as in Lemma 4 and define F_{0} = f_{0} + Y W^{h+1}, F_{1} = f_{1} + ZW^{h} and

F_{i+1} = \frac{F_{i−1}^{q} − F_{i}^{p}}{W} (1 ≤ i ≤ h).

Let A = k[W, F_{0}, …, F_{h+1}]. We have to prove the following two claims:

h + 2 ≤ #(A) ≤ h + 3; (5)

(6) A is the kernel of some triangular derivation Δ : k[W, X, Y, Z] → k[W, X, Y, Z].

We begin by showing that

(7)  F_{j} = f_{j} + b_{j}W^{h+1−j}  (0 ≤ j ≤ h + 1),

where b_{j} ∈ k[W, X, Y, Z] and b_{j}(0, X, Y, Z) ∉ k[X], and where we define f_{h+1} = 0. We proceed by induction and note that the assertion is clear for j ≤ 1. Assume that (7) holds for 0 ≤ j ≤ i, for some i such that 1 ≤ i ≤ h. Then F_{i} = f_{i} + b_{i}W^{h+1−i} and F_{i−1} = f_{i−1} + b_{i−1}W^{h+2−i}, so

F_{i−1}^{q} − F_{i}^{p} = (f_{i−1} + b_{i−1}W^{h+2−i})^{q} − (f_{i} + b_{i}W^{h+1−i})^{p}
= (f_{i−1}^{q} − f_{i}^{p}) − pf_{i}^{p−1}b_{i}W^{h+1−i} + ε_{1}W^{h+2−i}

for some ε_{1} ∈ k[W, X, Y, Z]. Write \frac{f_{i−1}^{q}−f_{i}^{p}}{W} = f_{i+1} + ε_{2}W^{h−i}, with ε_{2} ∈ k[W, X]; then dividing (F_{i−1}^{q} − F_{i}^{p}) by W gives

F_{i+1} = (f_{i+1} + ε_{2}W^{h−i}) − pf_{i}^{p−1}b_{i}W^{h−i} + ε_{1}W^{h+1−i}
= f_{i+1} + (ε_{2} − pf_{i}^{p−1}b_{i} + ε_{1}W)W^{h−i},

which proves (7). (In particular, the F_{j}’s are polynomials.)

Let π : k[W, X, Y, Z] → k[X, Y, Z] be the surjective k-homomorphism defined by W ↦ 0, X ↦ X, Y ↦ Y, Z ↦ Z. Then (7) implies that π(A) = k[X^{e_{0}}, …, X^{e_{h}}, τ], where e_{i} = p^{h−i}q^{i} and where τ is transcendental over k(X). Let R = k[X^{e_{0}}, …, X^{e_{h}}]; since R[τ] is a homomorphic image of A, #(A) ≥ #(R[τ]); since #(R[τ]) = h + 2 by Lemma 1, (5) holds. Define a derivation Δ : k[W, X, Y, Z] → k[W, X, Y, Z] by

(8)  Δ = |\begin{matrix}\frac{∂}{∂X} & \frac{∂}{∂Y} & \frac{∂}{∂Z} \\ \frac{∂F_{0}}{∂X} & \frac{∂F_{0}}{∂Y} & \frac{∂F_{0}}{∂Z} \\ \frac{∂F_{1}}{∂X} & \frac{∂F_{1}}{∂Y} & \frac{∂F_{1}}{∂Z} \end{matrix} | .

Then ΔZ = −W^{h+1} \frac{∂f_{1}}{∂X}, ΔY = −W^{h} \frac{∂f_{0}}{∂X}, ΔX = W^{2h+1} and ΔW = 0, so Δ is a triangular derivation of k[W, X, Y, Z]. It is clear that k[W, F_{0}, F_{1}] ⊆ ker Δ, so A ⊆ ker Δ; let us now argue that ker Δ ⊆ A_{W}. Write B = k[W, X, Y, Z]; since

B_{W} ⊇ A_{W} [X] ⊇ k[W, W^{−1}, X, F_{0}, F_{1}] = k[W, W^{−1}, X, Y, Z] = B_{W} ,

B_{W} is a polynomial ring over A_{W}. On the other hand, (ker Δ)_{W} contains A_{W} and is the kernel of the nonzero derivation Δ_{W} : B_{W} → B_{W}; so (ker Δ)_{W} = A_{W} and we have shown that A ⊆ ker Δ ⊂ A_{W}. So, in order to prove (6), there remains only to prove:

(9)  A ∩ W B = W A.

It is easy to see that the proof of (9) reduces to that of the following: if T_{0}, …, T_{h+1} are indeterminates and ψ ∈ k[T_{0}, …, T_{h+1}], then ψ(F_{0}, …, F_{h+1}) ∈ W B implies ψ(F_{0}, …, F_{h+1}) ∈ W A. Write ψ = ∑_{n≥0} ψ_{n}T_{h+1}^{n} with ψ_{n} ∈ k[T_{0}, …, T_{h}]; then

0 = π(ψ(F_{0}, …, F_{h+1})) = ∑_{n≥0} ψ_{n}(X^{e_{0}}, …, X^{e_{h}})τ^{n},

where τ = π(F_{h+1}) is transcendental over k(X), and consequently ψ_{n} ∈ ker φ for all n, where φ : k[T_{0}, …, T_{h}] → k[X] is the k-homomorphism which maps T_{i} to X^{e_{i}}. By Lemma 2, ker φ = (T_{0}^{q} − T_{1}^{p}, …, T_{h−1}^{q} − T_{h}^{p}), so ψ_{n} = ∑_{j=1}^{h} α_{j}(T_{j−1}^{q} − T_{j}^{p}) for some α_{j} ∈ k[T_{0}, …, T_{h}]. Then

ψ_{n}(F_{0}, …, F_{h}) = ∑_{j=1}^{h} α_{j}(F_{0}, …, F_{h})(F_{j−1}^{q} − F_{j}^{p}) = ∑_{j=1}^{h} α_{j}(F_{0}, …, F_{h})W F_{j+1} ∈ W A.

So (9) holds and, consequently, ker(Δ) = A. So (5) and (6) are proved. Example. We exhibit a triangular derivation Δ of k[W, X, Y, Z] whose kernel cannot be generated by five elements over k. Let p = 2, q = 5 and h = 4 and, following the proof of Lemma 4, successively define f_{4}, f_{3}, f_{2}, f_{1}, f_{0} by:^{1}

f_{4} = X^{625}, f_{3} = X^{250}, f_{2} = X^{100} + \frac{1}{5} X^{225}W,
f_{1} = X^{40} + (\frac{2}{25}X^{165} + \frac{1}{5}X^{90}) W + (−\frac{3}{625}X^{290} − \frac{8}{125}X^{215} − \frac{2}{25}X^{140}) W^{2}

1Note that the f_{j}’s are not unique.

and

f_{0} = X^{16} + (\frac{2}{25}X^{66} + \frac{1}{5}X^{36} + \frac{4}{125}X^{141}) W

+ (−\frac{92}{3125}X^{191} − \frac{23}{625}X^{116} − \frac{42}{15625}X^{266} + \frac{9}{625}X^{161} − \frac{8}{125}X^{86} − \frac{2}{25}X^{56}) W^{2}
+ ( \frac{408}{78125}X^{241} + \frac{68}{15625}X^{166} + \frac{666}{390625}X^{316} + \frac{328}{15625}X^{211} + \frac{132}{3125}X^{136}

+ \frac{36}{625}X^{106} − \frac{72}{78125}X^{286} − \frac{28}{3125}X^{181} + \frac{6}{125}X^{76} + \frac{244}{1953125}X^{391} ) W^{3}.

Define Δ as in (8) or, equivalently, by

ΔW = 0, ΔX = W^{9}, ΔY = −W^{4} \frac{∂f_{0}}{∂X} and ΔZ = −W^{5} \frac{∂f_{1}}{∂X} .

Then, by (5) and (6), we have 6 ≤ #(ker Δ) ≤ 7.

References

1. D. Daigle, G. Freudenburg, A counterexample to Hilbert’s Fourteenth Problem in dimension five, ppt 1999 (9 pages)

2. M. Miyanishi, Normal affine subalgebras of a polynomial ring, in: Algebraic and Topological Theories – to the Memory of Dr. Takehiko Miyata, Kinokuniya, Tokyo (1985) 37-51

Department of Mathematics and Statistics, University of Ottawa, Ottawa, Canada K1N 6N5

  E-mail address: daniel@mathstat.uottawa.ca

Department of Mathematics, University of Southern Indiana, Evansville, IN 47712, U.S.A.

  E-mail address: freudenb@usi.edu