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

DANIEL DAIGLE AND GENE FREUDENBURG

Abstract. It is shown that, if k is a field of characteristic zero, then the kernel of any triangular k-derivation of k[X_{1}, X_{2}, X_{3}, X_{4}] is finitely generated as a k-algebra. This is obtained as a corollary of a more general result concerning triangular R-derivations of R[X, Y, Z] for certain rings R.

1. Introduction

Let R be a commutative ring and B = R[X_{1}, …, X_{n}] a polynomial algebra in n variables over R (abbreviated B ≅ R^{[n]}). An R-derivation D : B → B is called triangular if DX_{i} ∈ R[X_{1}, …, X_{i−1}] holds for all i = 1, …, n (in particular, DX_{1} ∈ R). Note that such a derivation is necessarily locally nilpotent, i.e., for each b ∈ B there exists m > 0 satisfying D^{m}(b) = 0. The phrase “A is k-affine” means that A is a finitely generated k-algebra.

Consider a triangular R-derivation 0 ≠ D : R[X, Y, Z] → R[X, Y, Z] where R is a domain containing ℚ. If R is a field, then a well-known theorem of Miyanishi [7] asserts that ker D ≅ R^{[2]} (even in the more general case when D is locally nilpotent). On the other hand, if R ≅ k^{[2]} (k a field) then [1] gives an example where ker D is not finitely generated as an R-algebra. The aim of this paper is to prove: 1.1. Theorem. Let k be a field of characteristic zero and let R be a k-affine Dedekind domain or a localization of such a ring. If D is any triangular R-derivation of R^{[3]}, then the kernel of D is finitely generated as an R-algebra.

Let us immediately derive: 1.2. Corollary. If k is a field of characteristic 0, then the kernel of any triangular kderivation of k^{[4]} is finitely generated as a k-algebra.

Proof. Suppose δ is a triangular k-derivation of B = k[W, X, Y, Z] = k^{[4]} and write δW = λ ∈ k and δX = f(W) ∈ k[W]. Choose F (W) ∈ k[W] such that F′(W) = f(W), noting that δ(λX − F (W)) = 0.

If λ = 0, then δ is a triangular R-derivation of B = R[X, Y, Z] ≅ R^{[3]}, where R = k[W] ≅ k^{[1]}. If λ ≠ 0, then δ is a triangular R-derivation of B = R[W, Y, Z] ≅ R^{[3]}, where R = k[λX − F (W)] ≅ k^{[1]}. In either case, 1.1 implies the desired result. □

Note the following related fact (see [2]): For each positive integer n, there exists a triangular k-derivation Δ_{n} of k^{[4]} whose kernel cannot be generated by n elements. So, in 1.1 and 1.2, there is no a priori upper bound on the number of generators.

2000 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.

We remind the reader that locally nilpotent derivations of k^{[4]} are equivalent to algebraic actions of G_{a} = (k, +) on 𝔸^{4} in such a way that the kernel of a derivation is equal to the ring of invariants of the corresponding G_{a}-action. So 1.2 asserts that the ring of invariants of any triangular G_{a}-action on 𝔸^{4} is finitely generated, whereas the results of [1] (op. cit.) imply that, for each n ≥ 5, there is a triangular G_{a}-action on 𝔸^{n} with non-finitely generated ring of invariants.

Essentially, the proof of 1.1 is based on the well-known algorithm of van den Essen [5] for computing the kernel of a locally nilpotent derivation: We consider a certain increasing sequence of subalgebras of the kernel,

R[f_{0}, f_{1}] ⊆ R[f_{0}, f_{1}, f_{2}] ⊆ R[f_{0}, f_{1}, f_{2}, f_{3}] ⊆ …,

and we have to show that R[f_{0}, …, f_{n}] = R[f_{0}, …, f_{n+1}] for some n. However, in our situation, that integer n can be arbitrarily large and the proof of its existence requires a delicate analysis of the sequence (f_{0}, f_{1}, f_{2}, …). The tools needed for this analysis, simple sequences (2.2) and simple pairs (2.4), are developed in Section 2.

In particular, we need the fact that (k[[t]], t) is a simple pair whenever k is an algebraically closed field of characteristic zero and t is an indeterminate over k. This fact (2.8) is an easy consequence of the following result: 1.3. Theorem (A. Sathaye). Let k be an algebraically closed field of characteristic zero, let R be an affine k-domain with one place at infinity, let t be an indeterminate over R and let f_{0}, f_{1} ∈ R[[t]]. If^{1} f_{1}(0) does not belong to k, then there exists a sequence (u_{i})_{i=0}^{∞} in R[[t]] satisfying:

(1) u_{0} = f_{0} and u_{1} = f_{1};

(2) for each n ≥ 1, there exists a monic polynomial φ_{n}(Y) ∈ k[u_{0}, …, u_{n−1}][Y] such that^{2}

(a) φ̄_{n}(Y) ∈ k[u_{0}(0), …, u_{n−1}(0)][Y] is the minimal polynomial of u_{n}(0) over k(u_{0}(0), …, u_{n−1}(0)),

(b) u_{n+1} = φ_{n}(u_{n})/t. Sathaye obtained this result in [9] as an application of his theory of Generalized Newton-

Puiseux Expansions. We acknowledge that this paper would not exist without Sathaye’s contribution.

2. Simple sequences

2.1. Setup. We consider triples (R, t, S) satisfying:

(1) R is a PID containing ℚ and t a prime element of R;

(2) S is an overdomain of R such that t is prime in S. Given a triple (R, t, S) as above, we use the following notations:

• S̄ = S/tS (an integral domain);

• given s ∈ S, write s̄ = s + tS ∈ S̄; • given a ring A such that R ⊆ A ⊆ S, (i) let Ā be the image of the composite

1Given f = ∑_{i=0}^{∞} a_{i}t^{i} ∈ R[[t]] (where a_{i} ∈ R), we write f(0) = a_{0}.

2If φ_{n}(Y) = ∑_{i} a_{i}Y^{i} (a_{i} ∈ k[u_{0}, …, u_{n−1}]), then we write φ̄_{n}(Y) = ∑_{i} a_{i}(0)Y^{i} ∈ k[u_{0}(0), …, u_{n−1}(0)][Y].

A ↪ S → S̄; (ii) let A^{+} = A[S ∩ \frac{1}{t}A];

• the field R/tR is denoted κ (note that R̄ = κ). 2.2. Definition. Let (R, t, S) be as in 2.1. Let f = (f_{0}, …, f_{r}) be a finite sequence in S and write A_{n} = R[f_{0}, …, f_{n}] and K_{n} = Frac(Ā_{n}) for 0 ≤ n ≤ r. We say that f is a simple sequence of (R, t, S) if the following hold:

(1) r ≥ 1;

(2) f̄_{0} ∈ S̄ is transcendental over κ;

(3) for each n such that 0 < n < r, f̄_{n} is algebraic over K_{n−1} and there exists a monic polynomial φ_{n} ∈ A_{n−1}[T] satisfying: (i) φ̄_{n} ∈ Ā_{n−1}[T] is the minimal polynomial of f̄_{n} over K_{n−1}; and (ii) f_{n+1} = φ_{n}(f_{n})/t.

We distinguish three types of simple sequences:

(i) f is transcendental if f̄_{r} is transcendental over K_{r−1}; (ii) f is extendable if f̄_{r} is algebraic over K_{r−1} and its minimal polynomial is in Ā_{r−1}[T]; (iii) f is obstructed if it is neither transcendental nor extendable, i.e., if f̄_{r} is algebraic

over K_{r−1} but its minimal polynomial fails to have all its coefficients in Ā_{r−1}. Remark. It is easy to see that a simple sequence f = (f_{0}, …, f_{r}) of (R, t, S) is extendable if and only if ∃f_{r+1} ∈ S such that (f_{0}, …, f_{r}, f_{r+1}) is a simple sequence of (R, t, S). 2.3. Lemma. Let (R, t, S) be as in 2.1 and let f = (f_{0}, …, f_{r}) be a simple sequence of (R, t, S). Then the rings A_{0} ⊆ A_{1} ⊆ ⋯ ⊆ A_{r} (where A_{n} = R[f_{0}, …, f_{n}]) have the following properties:

(1) A_{0}^{+} = A_{0};

(2) for all n such that 0 < n < r, A_{n}^{+} = A_{n+1};

(3) if f is of transcendental type, then A_{r}^{+} = A_{r}.

Proof. Assertion (1) easily follows from the fact that f̄_{0} is transcendental over κ. We prove (2) by induction on n. Consider n such that 0 < n < r and note that A_{n−1}^{+} ⊆ A_{n} (by part (1) if n = 1, by inductive hypothesis if n > 1); also, A_{n} = A_{n−1}[f_{n}]. Let φ_{n} ∈ A_{n−1}[T] be the monic polynomial given by 2.2. Since f_{n+1} = φ_{n}(f_{n})/t ∈ S ∩ \frac{1}{t}A_{n}, it is clear that A_{n+1} ⊆ A_{n}^{+}. For the reverse inclusion, it suffices to check that S ∩ \frac{1}{t}A_{n} ⊆ A_{n+1}. Consider x ∈ S ∩ \frac{1}{t}A_{n}; then tx = ψ(f_{n}) for some ψ ∈ A_{n−1}[T]. Since φ_{n} is monic, the division algorithm gives

ψ = qφ_{n} + ρ (q, ρ ∈ A_{n−1}[T], deg(ρ) < deg(φ_{n})).

Now ρ̄ ∈ Ā_{n−1}[T] satisfies ρ̄(f̄_{n}) = 0 and deg(ρ̄) < deg(φ̄_{n}), so ρ̄ = 0 because φ̄_{n} is the minimal polynomial of f̄_{n}. So all coefficients of ρ belong to A_{n−1} ∩ tS ⊆ tA_{n−1}^{+} ⊆ tA_{n}, so \frac{1}{t}ρ ∈ A_{n}[T] and

x = q(f_{n})\frac{φ_{n}(f_{n})}{t} + \frac{ρ(f_{n})}{t} = q(f_{n})f_{n+1} + (\frac{1}{t}ρ)(f_{n}) ∈ A_{n}[f_{n+1}].

This shows that S ∩ \frac{1}{t}A_{n} ⊆ A_{n+1}, so A_{n}^{+} = A_{n+1} and (2) is proved.

We have A_{r−1}^{+} ⊆ A_{r} (by (1) and (2)), A_{r} = A_{r−1}[f_{r}] and (assuming that f is transcendental) f̄_{r} is transcendental over K_{r−1}; it easily follows that A_{r}^{+} = A_{r}, so (3) holds. □

2.4. Definition. Suppose that (R, t) satisfies condition (1) of 2.1. We say that the pair (R, t) is simple if (R, t, R^{[1]}) does not have an obstructed simple sequence (i.e., if every simple sequence of (R, t, R^{[1]}) is extendable). 2.5. Lemma. Let (R, t) be a simple pair and let S = R[X, Y], where X is an indeterminate over R and Y is a family of indeterminates over R[X]. If f = (f_{0}, …, f_{r}) is a simple sequence of (R, t, S) satisfying f̄_{0} ∈ κ[X], then f is not obstructed.

Proof. The hypothesis is that f̄_{0} belongs to the subring κ[X] of S̄ = κ[X, Y]. Assume that f is not transcendental; then

f̄_{n} ∈ κ[X] for all n = 0, …, r,

because each f̄_{n} is algebraic over κ(f̄_{0}) and f̄_{0} ∈ κ[X]. Define a sequence e = (e_{0}, …, e_{r}) in R[X] by e_{n} = f_{n}(X, 0). It is clear that ē_{n} = f̄_{n}(X, 0), so f̄_{n} ∈ κ[X] implies that

ē_{n} = f̄_{n}, for all n = 0, …, r.

We claim that e is a simple sequence of (R, t, R[X]) and that e and f are of the same type. For n = 0, …, r, write E_{n} = R[e_{0}, …, e_{n}], A_{n} = R[f_{0}, …, f_{n}] and K_{n} = Frac(Ā_{n}); clearly, Ē_{n} = Ā_{n} and so Frac(Ē_{n}) = K_{n}. For each n such that 0 < n < r, let φ_{n} ∈ A_{n−1}[T] be a monic polynomial satisfying: (i) φ̄_{n} ∈ Ā_{n−1}[T] is the minimal polynomial of f̄_{n} over K_{n−1}; and (ii) f_{n+1} = φ_{n}(f_{n})/t. Also, choose Φ_{n} ∈ R[T_{0}, …, T_{n−1}, T] monic in T such that Φ_{n}(f_{0}, …, f_{n−1}, T) = φ_{n}.

For each n such that 0 < n < r, define ψ_{n} = Φ_{n}(e_{0}, …, e_{n−1}, T) ∈ E_{n−1}[T]. Then ψ_{n} is monic and ψ̄_{n} = φ̄_{n} is the minimal polynomial of ē_{n} = f̄_{n} over K_{n−1}. Also, setting Y = 0 in the equation tf_{n+1} = Φ_{n}(f_{0}, …, f_{n}) yields te_{n+1} = Φ_{n}(e_{0}, …, e_{n}) = ψ_{n}(e_{n}). Thus e is a simple sequence of (R, t, R[X]); since ē_{r} = f̄_{r} and Ē_{r−1} = Ā_{r−1}, e and f are of the same type, as claimed.

Since (R, t) is simple, e, and therefore f, must be extendable. □

2.6. Lemma. Let (R, t, S) be as in 2.1, with S = R[X, Y, Z] = R^{[3]}. Let f = (f_{0}, …, f_{r}) be a simple sequence of (R, t, S) satisfying f_{0} ∈ R[X, Y], f̄_{0} ∈ κ[X] and f_{1} ∉ R[X, Y]. Let k be the natural number such that f_{1} = w_{0} + w_{1}Zt^{k}, where w_{0} ∈ R[X, Y] and w_{1} ∈ S \ tS. Then r ≤ k + 1. Remark. 2.6 remains true if “Y ” is interpreted as a family of indeterminates.

Proof. Define a map ord : S → ℕ ∪ {∞} by setting ord(b) = ∞ for b ∈ R[X, Y] and ord(b) = s ∈ ℕ for b ∈ S \ R[X, Y], where b = b_{0} + b_{1}Zt^{s} is the unique expression of b satisfying b_{0} ∈ R[X, Y], b_{1} ∈ S \ tS and s ∈ ℕ. For a nonempty subset E of S, define ord(E) = min_{b∈E} ord(b).

Let A_{n} = R[f_{0}, …, f_{n}] and K_{n} = Frac(Ā_{n}) for 0 ≤ n ≤ r. We have ord(f_{1}) = k, so ord(A_{1}) ≤ k; then it suffices to show that {ord(A_{n})}_{n=1}^{r} is a strictly decreasing sequence in ℕ. Fix n such that 0 < n < r; we shall show that ord(A_{n}) is a positive integer and that ord(A_{n}) > ord(A_{n+1}).

Since n ≥ 1, we have A_{1} ⊆ A_{n}, so ord(A_{n}) ≤ ord(A_{1}) ≤ k and consequently ord(A_{n}) ∈ ℕ. Since n < r, K_{n} is algebraic over K_{0} ⊆ κ(X), so Ā_{n} ⊆ κ[X] and consequently ord(A_{n}) > 0. So s = ord(A_{n}) is a positive integer. Choose a ∈ A_{n} such that ord(a) = s and write a = a_{0} + a_{1}Zt^{s} with a_{0} ∈ R[X, Y], a_{1} ∈ S \ tS. The element ā = ā_{0} of κ[X] is algebraic over K_{0}, so we may consider its minimal polynomial φ̄ ∈ K_{0}[T]. Since ā is integral over Ā_{0} and Ā_{0} = κ^{[1]} is a normal ring, φ̄ ∈ Ā_{0}[T] and we may lift φ̄ to a monic polynomial φ ∈ A_{0}[T]. Then φ̄(ā) = 0 implies that φ(a) belongs to tS and hence to tA_{n}^{+} = tA_{n+1} (equality by 2.3). So the Taylor expansion of φ(a) gives

tA_{n+1} ∋ φ(a) = φ(a_{0} + a_{1}Zt^{s}) = φ(a_{0}) + φ′(a_{0})a_{1}Zt^{s} + ϵt^{s+1} (some ϵ ∈ S)

and if we set ã = t^{−1}φ(a) ∈ A_{n+1} then

ã = t^{−1}φ(a_{0}) + φ′(a_{0})a_{1}Zt^{s−1} + ϵt^{s}.

We have t^{−1}φ(a_{0}) ∈ A_{0}[a_{0}]^{+} ⊆ R[X, Y]^{+} = R[X, Y] and (by minimality of φ̄) φ′(a_{0})a_{1} ∈ S \ tS; it follows that ord(ã) = s − 1 and ord(A_{n+1}) < ord(A_{n}). □

In the remainder of the section we prove that certain pairs (R, t) are simple. The main result of this type is 2.11. 2.7. Lemma. Let (R, t, S) and (R, t, S′) be as in 2.1, with S ⊆ S′ and tS′ ∩ S = tS. Let f = (f_{0}, …, f_{r}) and f′ = (f′_{0}, …, f′_{r}) be simple sequences of (R, t, S) and (R, t, S′) respectively, and suppose that f_{0} = f′_{0} and f_{1} = f′_{1}. Then f and f′ are simple sequences of the same type.

Proof. Let A_{n} = R[f_{0}, …, f_{n}] and A′_{n} = R[f′_{0}, …, f′_{n}] for 0 ≤ n ≤ r. By assumption,

(1)  f_{n} − f′_{n} ∈ A_{n−1} and A_{n−1} = A′_{n−1}

holds for n = 1. Consider any n satisfying (1) and 0 < n < r; then it is clear that A_{n} = A′_{n}, and we shall show that f_{n+1} − f′_{n+1} ∈ A_{n}.

Write f_{n} − f′_{n} = a ∈ A_{n−1}. Let φ_{n} ∈ A_{n−1}[T] be a monic polynomial satisfying (i) φ_{n}(f_{n})/t = f_{n+1} and (ii) φ̄_{n} is the minimal polynomial of f̄_{n} over K_{n−1} = Frac(Ā_{n−1}); also, let ψ_{n} ∈ A′_{n−1}[T] = A_{n−1}[T] be a monic polynomial satisfying (i′) ψ_{n}(f′_{n})/t = f′_{n+1} and (ii′) ψ̄_{n} is the minimal polynomial of f̄′_{n} over Frac(Ā′_{n−1}) = K_{n−1}. Then φ̄_{n}(T + ā) is also the minimal polynomial of f̄′_{n} over K_{n−1}, so ψ̄_{n} = φ̄_{n}(T + ā), so the polynomial ψ_{n}(T) − φ_{n}(T + a) has its coefficients in A_{n−1} ∩ tS′ = A_{n−1} ∩ tS ⊆ tA_{n−1}^{+} ⊆ tA_{n}. So ψ_{n}(T) − φ_{n}(T + a) = tζ(T), with ζ(T) ∈ A_{n}[T], and it follows that t(f′_{n+1} − f_{n+1}) = ψ_{n}(f′_{n}) − φ_{n}(f_{n}) = φ_{n}(f′_{n} + a) + tζ(f′_{n}) − φ_{n}(f_{n}) = tζ(f′_{n}), so f′_{n+1} − f_{n+1} = ζ(f′_{n}) ∈ A_{n}.

By induction, this proves that (1) holds for all n such that 0 < n ≤ r. The case n = r implies the desired assertion. □

The last lemma and Sathaye’s Result (1.3) imply the following: 2.8. Corollary. Let k be an algebraically closed field of characteristic zero and t an indeterminate over k. Then (k[[t]], t) is a simple pair.

Proof. Let f = (f_{0}, …, f_{r}) be a simple sequence of (k[[t]], t, k[[t]][x]), where x is an indeterminate over k[[t]]; it suffices to show that f is extendable.

Write S = k[[t]][x] and S′ = k[x][[t]], then S ⊂ S′, so f_{0}, f_{1} ∈ S′. Since k[x] has one place at infinity, we may consider the sequence (u_{i})_{i=0}^{∞} ⊂ S′ given by 1.3. By the properties of (u_{i})_{i=0}^{∞} ⊂ S′ stated in 1.3, it is clear that (u_{0}, …, u_{r+1}) is a simple sequence of (k[[t]], t, S′); thus f′ = (u_{0}, …, u_{r}) is an extendable simple sequence of (k[[t]], t, S′) whose first two terms are f_{0}, f_{1}. Thus f is extendable by 2.7. □

2.9. Definition. Let (R, t) and (R′, t′) be pairs satisfying the condition 1. of 2.1. We use the notation (R, t) ≺ (R′, t′) to indicate that the following conditions hold:

(1) R′ is a DVR with maximal ideal t′R′;

(2) R ⊆ R′ and R ∩ t′R′ = tR;

(3) R′/t′R′ is an algebraic extension of R/tR. 2.10. Lemma. Let (R, t) and (R′, t′) be pairs satisfying the condition 1. of 2.1. If (R, t) ≺ (R′, t′) and (R′, t′) is simple, then (R, t) is simple.

Proof. Suppose that (R, t) ≺ (R′, t′) and that (R′, t′) is simple; let f = (f_{0}, …, f_{r}) be a simple sequence of (R, t, R[X]); we show that f is extendable.

As usual, write κ = R/tR and, for each n such that 0 ≤ n ≤ r, A_{n} = R[f_{0}, …, f_{n}], Ā_{n} = κ[f̄_{0}, …, f̄_{n}], K_{n} = Frac(Ā_{n}) = κ(f̄_{0}, …, f̄_{n}); for each n such that 0 < n < r, let φ_{n}(T) ∈ A_{n−1}[T] be a monic polynomial satisfying (i) φ_{n}(f_{n})/t = f_{n+1} and (ii) φ̄_{n} is the minimal polynomial of f̄_{n} over K_{n−1}. Also, let κ′ = R′/t′R′ and write t = u(t′)^{d} where u is a unit of R′ and d is a positive integer.

Consider the following condition P(n) on an integer n ∈ {1, …, r}: There exists a simple sequence g = (g_{0}, …, g_{m}) of (R′, t′, R′[X]) and an integer k ≥ 0 such that g_{m} = u^{k}f_{n}, R′[g_{0}, …, g_{m−1}] ⊇ A_{n−1} and κ′[ḡ_{0}, …, ḡ_{m−1}] = κ′[Ā_{n−1}].

Note that P(1) is true (with g = (f_{0}, f_{1}) and k = 0). We will use an inductive argument to show that P(r) is true.

Consider any n ∈ {1, …, r} such that P(n) is true and define the following notations: A′_{i} = R′[g_{0}, …, g_{i}], Ā′_{i} = κ′[ḡ_{0}, …, ḡ_{i}] and K′_{i} = Frac(Ā′_{i}) = κ′(ḡ_{0}, …, ḡ_{i}). Note that, for each i, K_{i} is a regular extension of κ (because κ ⊂ K_{i} ⊆ κ(X)). Together with the fact that κ′ is algebraic over κ, this implies that κ′ ⊗_{κ} K_{i} is a field, and hence is equal to κ′(f̄_{0}, …, f̄_{i}). In particular,

κ′ ⊗_{κ} K_{n−1} = κ′(f̄_{0}, …, f̄_{n−1}) =^{P(n)} κ′(ḡ_{0}, …, ḡ_{m−1}) = K′_{m−1}, (2)

κ′ ⊗_{κ} K_{n} = κ′(f̄_{0}, …, f̄_{n−1})(f̄_{n}) =^{(2)} κ′(ḡ_{0}, …, ḡ_{m−1})(f̄_{n}) = κ′(ḡ_{0}, …, ḡ_{m−1})(ḡ_{m}) = K′_{m}

and consequently

(3)  [K′_{m} : K′_{m−1}] = [K_{n} : K_{n−1}].

Assume, now, that n < r; then we may consider the polynomial φ_{n}(T) ∈ A_{n−1}[T]. Let e = deg φ_{n}(T) and define ψ_{m}(T) = u^{ek}φ_{n}(T/u^{k}). Then ψ_{m}(T) is a monic polynomial with coefficients in A_{n−1}[u] ⊆ A′_{m−1} and

ψ_{m}(g_{m}) = u^{ek}φ_{n}(f_{n}) = u^{ek}tf_{n+1} = u^{ek+1}(t′)^{d}f_{n+1}.

This implies that ψ̄_{m}(ḡ_{m}) = 0; since deg ψ̄_{m}(T) = e = [K_{n} : K_{n−1}] = [K′_{m} : K′_{m−1}], ψ̄_{m}(T) is the minimal polynomial of ḡ_{m} over K̄′_{m−1}. So (g_{0}, …, g_{m}, g_{m+1}) is a simple sequence of (R′, t′, R′[X]), where we define

g_{m+1} = \frac{ψ_{m}(g_{m})}{t′} = u^{ek+1}(t′)^{d−1}f_{n+1}.

Consequently, if we set g_{m+i} = u^{ek+1}(t′)^{d−i}f_{n+1} (1 ≤ i ≤ d), then

g′ = (g_{0}, …, g_{m}, g_{m+1}, …, g_{m+d})

is a simple sequence of (R′, t′, R′[X]) and it is easily verified that P(n + 1) holds (with g′ and ek + 1 in place of g and k).

By induction, we obtain that P(r) is true, i.e., there exists a simple sequence g = (g_{0}, …, g_{m}) of (R′, t′, R′[X]) and an integer k ≥ 0 such that

g_{m} = u^{k}f_{r}, A′_{m−1} = R′[g_{0}, …, g_{m−1}] ⊇ A_{r−1} and Ā′_{m−1} = κ′[ḡ_{0}, …, ḡ_{m−1}] = κ′[Ā_{r−1}].

Let P (T) be the minimal polynomial of f̄_{r} over K_{r−1}, let e = deg P (T), define Q(T) = ū^{ke}P (ū^{−k}T) and note that Q(T) is a monic polynomial with coefficients in K′_{m−1}. Since Q(ḡ_{m}) = ū^{ke}P (f̄_{r}) = 0, and since deg Q(T) = e = [K_{r} : K_{r−1}] = [K′_{m} : K′_{m−1}] by (3), Q(T) is the minimal polynomial of ḡ_{m} over K′_{m−1}. Since (R′, t′) is assumed to be simple, g is extendable, so Q(T) has all its coefficients in Ā′_{m−1}; it follows that P (T) has all its coefficients in Ā′_{m−1}, hence in Ā′_{m−1} ∩ K_{r−1}. So, in order to show that f is extendable, it suffices to show that Ā′_{m−1} ∩ K_{r−1} = Ā_{r−1}.

Let B be a basis of κ′ over κ such that 1 ∈ B. Since Ā′_{m−1} = κ′[ḡ_{0}, …, ḡ_{m−1}] = κ′[f̄_{0}, …, f̄_{r−1}] = κ′⊗_{κ}Ā_{r−1}, B is a basis of κ′⊗_{κ}Ā_{r−1} = Ā′_{m−1} as a module over κ⊗_{κ}Ā_{r−1} = Ā_{r−1}. Because 1 ∈ B, this allows us to write Ā′_{m−1} = Ā_{r−1} ⊕ F, where F is a free (hence torsion free) module over Ā_{r−1}. It easily follows that Ā′_{m−1}∩K_{r−1} = Ā_{r−1}, which completes the proof. □

2.11. Corollary. Let R be a DVR containing a field k of characteristic zero and such that Frac(R) is a function field in one variable over k, and let t be a uniformizing parameter of R. Then (R, t) is a simple pair.

Proof. Write K = Frac(R) = k(α_{1}, …, α_{s}) (for some α_{1}, …, α_{s}), let Ω be an algebraically closed field containing K, let k̄ be the algebraic closure of k in Ω and let L = k̄(α_{1}, …, α_{s}); note that L is a function field in one variable over k̄. Choose a DVR R′ of L satisfying R′ ⊇ R and R ∩ t′R′ = tR, where t′ is a uniformizing parameter of R′; then we have (R, t) ≺ (R′, t′). Since R′ is a DVR satisfying k̄ ⊆ R′ and R′/t′R′ = k̄, we also have R′ ⊆ k̄[[t′]], hence (R′, t′) ≺ (k̄[[t′]], t′). Now (k̄[[t′]], t′) is a simple pair by 2.8, so two applications of 2.10 give that (R, t) is simple. □

3. Proof of the main theorem

By combining results 2.14 and 2.20 of [8] with 2.1 of [6], we obtain: 3.1. Let R ⊆ A ⊆ B be domains, where R is noetherian and B is finitely generated as an R-algebra. Then the following are equivalent:

(1) A is finitely generated as an R-algebra;

(2) for every maximal ideal 𝔐 of R, A_{𝔐} is finitely generated as an R_{𝔐}-algebra.

Then it immediately follows: 3.2. Lemma. Let R be a noetherian domain containing ℚ, B a finitely generated overdomain of R and D : B → B a locally nilpotent R-derivation. Then the following conditions are equivalent:

(1) ker D is finitely generated as an R-algebra; (2) for every maximal ideal 𝔐 of R, the kernel of D_{𝔐} is finitely generated as an

R_{𝔐}-algebra (where D_{𝔐} : B_{𝔐} → B_{𝔐} is the R_{𝔐}-derivation obtained by localizing D).

Proof. Let A = ker D and recall that if S ⊆ A\{0} is multiplicatively closed, the derivation S^{−1}D : S^{−1}B → S^{−1}B has kernel S^{−1}A; the result follows from 3.1. □

In view of the above statement, it suffices to prove 1.1 in the special case where R is a DVR. By 2.11, it’s enough to prove the following: 3.3. Proposition. Let R be a DVR containing ℚ, let B = R[X, Y, Z] = R^{[3]} and let D : B → B be an R-derivation such that DX ∈ R, DY ∈ R[X] and DZ ∈ R[X, Y]. Suppose that (R, t) is a simple pair, where t is a uniformizing parameter of R. Then ker D is a finitely generated R-algebra.

Proof. If DX = 0 then D is an R[X]-derivation of B = R[X]^{[2]}, and is locally nilpotent, so ker D = R[X]^{[1]} (by 2.1 of [3], for instance). From now on, we assume that DX ≠ 0. Consider the localization D_{t} : B_{t} → B_{t} of D at {1, t, t^{2}, … }. Then s = X/DX ∈ B_{t} is a “slice” of D_{t}, i.e., D_{t}(s) = 1. In this situation, it is known (see for instance 2.1 of [4]) that the exponential of −sD_{t} is a surjective homomorphism of R_{t}-algebras (which we denote ξ):

ξ : B_{t} → ker(D_{t}) = (ker D)_{t}, w ↦ ∑_{n=0}^{∞} \frac{(−s)^{n}}{n!} D_{t}^{n}(w).

We have (ker D)_{t} = R_{t}[ξ(X), ξ(Y), ξ(Z)] = R_{t}[0, Y + F (X), Z + G(X, Y)], where F (X) ∈ R_{t}[X] and G(X, Y) ∈ R_{t}[X, Y]. Multipliying the generators by suitable powers of t, we obtain polynomials

f_{0} = f_{0}^{∗}(X) + t^{m}Y  (m ∈ ℕ, f_{0}^{∗}(X) ∈ R[X])
f_{1} = f_{1}^{∗}(X, Y) + t^{k}Z  (k ∈ ℕ, f_{1}^{∗}(X, Y) ∈ R[X, Y])

satisfying:

R[f_{0}, f_{1}] ⊆ ker D and R[f_{0}, f_{1}]_{t} = (ker D)_{t},

and it is easy to arrange that the image f̄_{0} of f_{0} in B̄ = κ[X, Y, Z] = κ^{[3]} is nonconstant (where κ = R/tR as usual).

If m = 0, set R′ = R[f_{0}] = R^{[1]}; then D is an R′-derivation of B = R′[X, Z], so ker D = (R′)^{[1]}, again by 2.1 of [3].

From now on, assume that m > 0, so f̄_{0} ∈ κ[X]. Let E be the set of simple sequences f′ = (f′_{0}, …, f′_{r}) of (R, t, B) satisfying f′_{0} = f_{0} and f′_{1} = f_{1}. Note that E is nonempty, since (f_{0}, f_{1}) ∈ E. By 2.6, we have r ≤ k + 1 for all f′ = (f′_{0}, …, f′_{r}) ∈ E; so there exists f = (f_{0}, …, f_{r}) ∈ E with r maximal; this f cannot be extendable and, by 2.5, it cannot be obstructed either. So f is transcendental and, by 2.3,

A^{+} = A, where A = R[f_{0}, …, f_{r}].

Consequently, A ∩ tB = tA. Together with A ⊆ ker D and A_{t} = (ker D)_{t}, this gives ker D = A. □

The proof of Theorem 1.1 is now complete.

Remark. If R is a DVR of characteristic zero obtained by localizing an affine domain at a prime ideal of height one, then R satisfies the hypothesis of Theorem 1.1, i.e., R is the localization of an affine Dedekind domain at a maximal ideal.

Indeed, suppose that R = B_{𝔭}, where B is an affine domain over a field k_{1} of characteristic zero and 𝔭 is a height one prime ideal of B. By Noether’s Normalization Lemma, there exists a subring A = k_{1}[x_{1}, …, x_{d}] ≅ k_{1}^{[d]} of B such that B is integral over A and 𝔭 ∩ A = x_{d}A. Then the field k = k_{1}(x_{1}, …, x_{d−1}) is contained in R and consequently k[B] ⊆ R. The fact that k[B] is k-affine and one-dimensional implies that its integral closure N in Frac(k[B]) is a k-affine Dedekind domain. We have B ⊆ N ⊆ R, so R is the localization of N at 𝔐 ∩ N, where 𝔐 is the maximal ideal of R.

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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