HOMOGENEOUS LOCALLY NILPOTENT DERIVATIONS OF k[X, Y, Z]

D. DAIGLE

Abstract. Let k be a field of characteristic zero and let B = k[X, Y, Z] be a polynomial ring in three variables. A derivation D : B → B is said to be locally nilpotent if for each b ∈ B we have D^{n}(b) = 0 for n ≫ 0. This paper describes the class of homogeneous locally nilpotent derivations of B, where homogeneity is relative to any N-grading of B which is obtained by assigning weights to the variables, w(X) = a, w(Y) = b and w(Z) = c, with a, b, c ∈ N and gcd(a, b, c) = 1. It is known that the kernel of a derivation from this class is a subalgebra k[f, g] of B = k[X, Y, Z], where f and g are w-homogeneous and algebraically independent, and it is also known that such a derivation is essentially determined by its kernel. The main results of this paper give explicit conditions on f, g ∈ k[X, Y, Z] which are equivalent to k[f, g] being the kernel of a homogeneous locally nilpotent derivation of k[X, Y, Z].

Let B be a ring and let D be a derivation of B (by this, we mean a Z-derivation going from B to B). We say that D is locally nilpotent if for each b ∈ B we have D^{n}(b) = 0 for n ≫ 0.

Several researchers have investigated locally nilpotent derivations of k[X_{1}, …, X_{n}], where k is a field of characteristic zero. These derivations are fully understood for n < 3 (see [14]), partially understood when n = 3, and very little is known when n > 3. Instead of counting the number n of variables, one may consider the “rank” of derivations (Definition 1.9); then one can say that locally nilpotent derivations D of k[X_{1}, …, X_{n}] are well understood when rank D < 3 (see [4]), and very mysterious when rank D ≥ 3. In view of Freudenburg’s recent discovery [6] that there exist rank 3 locally nilpotent derivations of k[X, Y, Z], it is now clear that our understanding of the derivations of k[X, Y, Z] is very incomplete.

It would be very nice to classify all locally nilpotent derivations of k[X, Y, Z], but that problem seems to be quite difficult. This paper addresses the less ambitious problem of describing the class of homogeneous locally nilpotent derivations of k[X, Y, Z]; to be precise, we consider homogeneity with respect to any N-grading of k[X, Y, Z] which is obtained by assigning weights to the variables, say w(X) = a, w(Y) = b and w(Z) = c, where a, b, c ∈ N and gcd(a, b, c) = 1.

1991 Mathematics Subject Classification. Primary: 14L30.

Supported by a grant from NSERC Canada.

It is known that the kernel of such a derivation is a subalgebra k[f, g] of k[X, Y, Z], where f and g are w-homogeneous and algebraically independent, and it is also known that such a derivation is essentially determined by its kernel. Hence our task is to understand which f, g ∈ k[X, Y, Z] are such that k[f, g] is a kernel. One should note that some ring-theoretical characterizations of the kernels are relatively easy to obtain (1.3 for instance), but these are unsatisfactory—they don’t say anything explicit about f and g.

We now state the solution to our problem; for the purpose of this introduction, we restrict ourselves to the special case where k is algebraically closed and a = b = c = 1, i.e., the grading is the standard one (the statement we give here is a corollary of Theorem 3.5 and Corollary 3.1): Theorem . Let k be an algebraically closed field of characteristic zero and let f, g ∈ B = k[X, Y, Z] be homogeneous polynomials. Then the following conditions are equivalent:

(1) k[f, g] is the kernel of some homogeneous locally nilpotent derivation of B;

(2) f, g are irreducible and the algebraic surface P^{2} \ V (fg) is isomorphic to P^{2} minus two lines.

In this statement, P^{2} is the projective plane over k and V (fg) denotes the union of the two curves “ f = 0 ” and “ g = 0 ” in P^{2}.

The theorem enables us to construct infinite families of rank 3 homogeneous locally nilpotent derivations of k[X, Y, Z]. Indeed, if f, g satisfy the equivalent conditions of the theorem, then Δ_{(f,g)} (see 1.6 and 3.4) is a homogeneous locally nilpotent derivation of k[X, Y, Z] with kernel k[f, g]; moreover, Δ_{(f,g)} has rank 3 if and only if deg f > 1 and deg g > 1. Obviously, this raises the following: Question . Which pairs of irreducible curves C_{1}, C_{2} in P^{2} have the property that P^{2} \ (C_{1} ∪ C_{2}) is isomorphic to P^{2} minus two lines?

There is no doubt that this question is interesting in itself, and that there exists relevant literature about it ([9], [13], [16], [18], …), but it is not entirely clear (to this author) to what extent it has been satisfactorily answered. In any case, we stress that we do not address the above question in this paper, except for the following: Example . We answer the above question in the special case deg C_{1} = 2 ≤ deg C_{2}. (In view of the above discussion, this gives rise to an infinite family of examples of rank three homogeneous locally nilpotent derivations of k[X, Y, Z].)

(1) Given a finite sequence n = (n_{1}, …, n_{ℓ}) of positive integers (where ℓ ≥ 1),

define an infinite sequence μ_{n} as follows: • Let i_{1}, …, i_{ℓ} be the integers given by i_{ℓ} = 2 and i_{ν−1} = (4n_{ν} + 1)i_{ν};

• let S_{1}, …, S_{ℓ} be the sequences given by

S_{ν} = (n_{ν}i_{ν}, n_{ν}i_{ν}, n_{ν}i_{ν}, n_{ν}i_{ν}, \underbrace{i_{ν}, …, i_{ν}}_{2n_{ν}});

• let μ_{n} be the concatenation of S_{1}, …, S_{ℓ} and of the infinite constant

sequence (1, 1, …), i.e., μ_{n} = (n_{1}i_{1}, …, i_{ℓ}, 1, 1, …). Observe that n can be recovered from μ_{n}, so that n ↦ μ_{n} is an injective map.

(2) Fix an irreducible curve C_{1} in P^{2}, of degree 2, and let P ∈ C_{1}. One can show that, for each choice of a sequence n, there exists at least one irreducible curve C_{2}^{(n)} (in P^{2}) which satisfies:

• C_{1} ∩ C_{2}^{(n)} = {P } and C_{2}^{(n)} has a cusp at P;

• deg C_{2}^{(n)} = ∏_{ν=1}^{ℓ}(4n_{ν} + 1);

• the multiplicity sequence^{1} of C_{2}^{(n)} at P is μ_{n}. Moreover, one can see that any such C_{2}^{(n)} satisfies: P^{2} \ (C_{1} ∪ C_{2}^{(n)}) ≅ P^{2} minus two lines.

(3) Conversely, if C_{1} and C_{2} are irreducible curves in P^{2} such that deg C_{1} = 2 ≤ deg C_{2} and P^{2} \ (C_{1} ∪ C_{2}) ≅ P^{2} minus two lines, then one can prove that C_{1} and C_{2} meet in one point P, C_{2} has a cusp at P and there exists a unique sequence n = (n_{1}, …, n_{ℓ}) such that μ_{n} is the multiplicity sequence of C_{2} at P and deg C_{2} = ∏_{ν=1}^{ℓ}(4n_{ν} + 1).

Remark . We think that the above family of examples should be mentioned in this paper but, unfortunately, we have to omit the proof. A satisfactory proof would be rather long and would use techniques completely different from those of this paper. Remark . In the notation of the above example, Freudenburg’s example [6] corresponds to the sequence n = (1) (i.e., ℓ = 1 and n_{1} = 1).

1. Preliminaries

Throughout this paper, all rings are commutative and have an identity element. If B is a ring then B^{∗} denotes the group of units of B; if B is an integral domain then qt B denotes the field of fractions of B.

If B is an A-algebra and n > 0 is an integer then the notation B = A^{[n]} means that B is A-isomorphic to the polynomial algebra in n variables over A. Suppose that k is a field and B = k^{[n]}; by a variable of B we mean an element b ∈ B which satisfies B = k[b, b_{2}, …, b_{n}] for some b_{2}, …, b_{n} ∈ B; by a coordinate system of B we mean an ordered n-tuple (b_{1}, …, b_{n}) of elements of B such that B = k[b_{1}, …, b_{n}].

A subring A of a domain B is said to be inert in B (or factorially closed in B) if for all x, y ∈ B we have xy ∈ A \ {0} ⇒ x, y ∈ A.

A subring A of a domain B is said to be algebraically closed in B if, for every b ∈ B and every nonzero polynomial f(T) ∈ A[T], we have f(b) = 0 ⇒ b ∈ A.

1We mean the sequence (μ(C_{2}^{(n)}, P_{i}))_{i=0}^{∞}, where P_{0} = P, P_{i} is the i-th infinitely near point over P and μ(C_{2}^{(n)}, P_{i}) is the multiplicity of P_{i} on the suitable strict transform of C_{2}^{(n)}.

The following fact will be used in the proof of Theorem 3.5. It can be derived from the proof of Theorem 2.3.1 of [15].

1.1. Let A ⊂ B be integral domains, where B is finitely generated as an A-algebra. Suppose that there exists a multiplicatively closed subset S of A such that S^{−1}B = (S^{−1}A)^{[1]}, and that every element of S \ A^{∗} is a finite product of elements π of A satisfying:

(1) π is a prime element of B

(2) πB ∩ A = πA

(3) A/πA is algebraically closed in B/πB.

Then B = A^{[1]}.

1.2. Basic properties of locally nilpotent derivations. Let B be an integral domain of characteristic zero, let D : B → B be a nonzero derivation of B, and let A = ker D. Then the following hold:

(1) If D is locally nilpotent then A is an inert subring of B. In particular, if D is locally nilpotent and B is a UFD then A is a UFD.

(2) Let S be a multiplicatively closed subset of B\{0}, and consider the derivation S^{−1}D of S^{−1}B. Then

(a) S^{−1}D is locally nilpotent if and only if D is locally nilpotent and S ⊆ A.

(b) If S ⊆ A then ker S^{−1}D = S^{−1}A and S^{−1}A ∩ B = A.

(3) A is algebraically closed in B; in particular, B ∩ qt A = A.

(4) Assume that Q ⊆ B and that D is locally nilpotent.

(a) If D(b) = 1 for some b ∈ B, then B = A[b] = A^{[1]}.

(b) More generally, let b be any element of B such that Db ≠ 0 and D^{2}b = 0,

and write a = Db ∈ A \ {0}. Then B_{a} = A_{a}[b] = A_{a}^{[1]}.

(5) Let b ∈ B \ {0}. The derivation bD (of B) is locally nilpotent if and only if D is locally nilpotent and b ∈ A.

References. (1) is given in 2o of [14]; for (2), see the proposition in [8]; (3) is wellknown; (4 a) is Proposition 2.1 of [17]; (4 b) is an easy consequence of (4 a), and is pointed out in [5]; the “if” part of (5) is trivial, and the “only if” part can be derived from the proof of the Claim in [8].

1.3. ([3], Proposition 1.4) Let B be an integral domain of characteristic zero and let A ≠ B be a subring of B such that B is finitely generated as an A-algebra. Then the following are equivalent:

(1) A is the kernel of some locally nilpotent derivation D : B → B;

(2) there exists a multiplicatively closed subset S of A \ {0} such that S^{−1}B = (S^{−1}A)^{[1]} and B ∩ (S^{−1}A) = A;

(3) S^{−1}B = (S^{−1}A)^{[1]} and B ∩ (S^{−1}A) = A, where S = A \ {0}.

The next result will be used in the proof of 3.10. Recall that a locally nilpotent derivation D : B → B is said to be irreducible if the only principal ideal of B which contains D(B) is B itself. Lemma 1.4. Let D : B → B be an irreducible locally nilpotent derivation, where B is a UFD of characteristic zero, let A = ker D and let S ⊆ A \ {0} be a multiplicatively closed set such that S^{−1}B = (S^{−1}A)^{[1]}. Then each v ∈ B satisfying S^{−1}B = (S^{−1}A)[v] also satisfies: Dv ∈ A \ {0} and, for some a ∈ A \ {0}, a Dv ∈ S.

Proof. Let v ∈ B be such that S^{−1}B = (S^{−1}A)[v] and let ∂ : (S^{−1}A)[v] → (S^{−1}A)[v] denote the derivative with respect to v. Being an (S^{−1}A)-derivation of (S^{−1}A)[v], S^{−1}D is a multiple of ∂; more precisely, S^{−1}D = β∂ with β = (S^{−1}D)(v) = Dv ∈ B.

Since D is assumed to be irreducible, it is in particular nonzero, so S^{−1}D ≠ 0 and consequently β ≠ 0. Let p ∈ B be a prime factor of β. Since D is irreducible, there exists b ∈ B such that Db ∉ pB; on the other hand, Db = β∂b (where ∂b ∈ S^{−1}B) so there exists s ∈ S such that β | sDb (in B). Thus p | sDb and, since p |̸ Db, we have p | s. Hence, each prime factor of β = Dv divides some element of S and, consequently, Dv divides some element of S. Then the conclusion follows from the fact that A is an inert subring of B (see 1.2). □

The following theorem was proved by Miyanishi in [12], in the special case where k is algebraically closed; the general case is then a straightforward consequence of [10]. 1.5. Let k be a field of characteristic zero and let D be a nonzero locally nilpotent derivation of k^{[3]}. Then ker D = k^{[2]}.

It should be noted that, in the homogeneous case, V.D. Zurkowski [19] gave a different proof of 1.5. 1.6. ([3], Corollary 2.6) Let k be a field of characteristic zero, B = k[X, Y, Z] = k^{[3]}, let D : B → B be a nonzero locally nilpotent derivation and let f, g ∈ B be such that ker D = k[f, g]. If Δ_{(f,g)} : B → B denotes the k-derivation defined by the jacobian determinant

Δ_{(f,g)}(b) = |\frac{∂(f, g, b)}{∂(X, Y, Z)}| (b ∈ B),

then D = aΔ_{(f,g)} for some a ∈ ker D \ {0}. Moreover, Δ_{(f,g)} is locally nilpotent and irreducible.

We prove two facts which follow easily from 1.6: Corollary 1.7. Let k be a field of characteristic zero and let f, g ∈ k[X, Y, Z]. Then the following are equivalent:

(1) k[f, g] is the kernel of some locally nilpotent derivation of k[X, Y, Z] ;

(2) for all field extensions K/k, K[f, g] is the kernel of some locally nilpotent derivation of K[X, Y, Z]; (3) for some field extension K/k, K[f, g] is the kernel of a locally nilpotent deriva-

tion of K[X, Y, Z]. Remark . In the above statement, by K[X, Y, Z] and K[f, g] we mean K⊗_{k}k[X, Y, Z] and K ⊗_{k} k[f, g] respectively.

Proof. If D is a locally nilpotent derivation of k[X, Y, Z] with kernel k[f, g], and if K/k is any field extension, then K ⊗_{k}D is a locally nilpotent derivation of K[X, Y, Z] with kernel K[f, g]. Hence (1) implies (2).

Obviously, (2) implies (3). Now assume that (3) holds and let Δ_{(f,g)} (resp. \overline{Δ}_{(f,g)}) be the derivation of k[X, Y, Z] (resp. of K[X, Y, Z]) defined by means of a jacobian determinant, as in 1.6. Since (3) holds, 1.6 implies that \overline{Δ}_{(f,g)} is locally nilpotent and has kernel K[f, g]. It follows immediately that Δ_{(f,g)} is locally nilpotent and has kernel K[f, g] ∩ k[X, Y, Z] = k[f, g]. Hence (3) implies (1). □

Corollary 1.8. Let k be a field of characteristic zero and let f, g ∈ k[X, Y, Z] be such that k[f, g] is the kernel of some locally nilpotent derivation of k[X, Y, Z]. Then f and g are irreducible elements of \overline{k}[X, Y, Z], where \overline{k} denotes the algebraic closure of k.

Proof. By 1.2 and 1.7, \overline{k}[f, g] is an inert subring of \overline{k}[X, Y, Z]; since f and g are irreducible elements of \overline{k}[f, g], it follows that they are irreducible in \overline{k}[X, Y, Z]. □

Definition 1.9. ([7], section 3) Let k be a field of characteristic zero and D a kderivation of B = k^{[n]}. The rank of D is the least integer r ≥ 0 for which there exists a coordinate system (X_{1}, …, X_{n}) of B satisfying DX_{i} = 0 for all i > r.

Let the notation be as in the above definition. Obviously, rank D = 0 ⇔ D = 0, and it is easy to see that rank D = 1 is equivalent to the two conditions B = (ker D)^{[1]} and ker D = k^{[n−1]}. Locally nilpotent derivations of rank at most two are studied in [4]. About derivations of high rank, observe that rank D < n if and only if ker D contains some variable of B. Note that, for every integer n > 0 other than 2, there exists a locally nilpotent derivation of k^{[n]} of rank n (see [6]). Definition 1.10. ([2], Definition 2.3) Let B be a domain of characteristic zero, D : B → B a locally nilpotent derivation of B, A = ker D, and let π : Spec B → Spec A be the morphism determined by the inclusion map A ↪ B. Given a prime ideal α ∈ Spec A, let κ(α) denote the field A_{α}/αA_{α}. Then the fibre π^{−1}(α) is, in a natural way, a scheme over κ(α). We define

R(D) = {α ∈ Spec A | π^{−1}(α) ≠ A_{κ(α)}^{1}}
= {α ∈ Spec A | κ(α) ⊗_{A} B ≠ κ(α)^{[1]}}.

We will also consider the closure \overline{R}(D) of R(D) in Spec A. 1.11. ([2], Lemma 2.4) If D is a locally nilpotent derivation of B = k^{[3]} of rank 2, then (1) dim \overline{R}(D) = 1.

(2) Let ξ be an irreducible element of A = ker D such that V (ξ) is an irreducible component of \overline{R}(D), and let (X, Y, Z) be a coordinate system of B which satisfies DX = 0. Then k[X] is the integral closure of k[ξ] in B.

1.12. Conventions on graded rings. Every graded ring considered in this paper is either Z-graded or N-graded. Let B = ⨁_{i} B_{i} be a graded domain.

(1) The following notation is convenient: d(B) = gcd{i | B_{i} ≠ 0}.

(2) If b ∈ B_{i} \ {0} then we write deg b = i, or sometimes w(b) = i.

(3) If S is a multiplicatively closed subset of ⋃_{i}(B_{i} \ {0}) then the localized ring S^{−1}B inherits a Z-grading in a natural way: If b ∈ B_{i} \ {0} and s ∈ S then b/s is declared to be homogeneous of degree deg b − deg s. Note that d(S^{−1}B) = d(B).

(4) A subring A of B is said to be homogeneous if A = ⨁_{i}(A ∩ B_{i}). If this is the case then we set A_{i} = A ∩ B_{i} and regard A = ⨁_{i} A_{i} as a graded ring.

(5) If D : B → B is a derivation, we say that D is homogeneous if there exists an integer n such that D(B_{i}) ⊆ B_{i+n} holds for all i. Note that if D is homogeneous then the subring ker D of B is homogeneous.

(6) Let D : B → B be a homogeneous derivation and let A = ker D. Then d(B) divides d(A) and we define d(D) ∈ N by

d(D) = \begin{cases}d(A)/d(B), & if d(B) ≠ 0, \\ 1, & if d(B) = 0. \end{cases}

(Note that d(D) = 1 ⟺ d(A) = d(B).)

1.13. In this paper, we are particularly interested in the following situation. Let B = k[X, Y, Z] = k^{[3]}, where k is a field of characteristic zero, and let w = (a, b, c) ∈ N^{3} be such that gcd(a, b, c) = 1. By the symbol (B, w) we mean B regarded as an N-graded ring, where the grading is determined (in the usual way) by the weights w(X) = a, w(Y) = b and w(Z) = c. Then we want to study locally nilpotent derivations of B which are w-homogeneous (i.e., homogeneous with respect to the grading determined by w).

If w = (1, 1, 1) then we obtain the standard grading of B. In this case, whomogeneous derivations are simply called homogeneous derivations.

We close this section by recalling that every algebraic action of G_{m}(k) on the affine plane Ak^{2} is linearizable ([1], [11]). This can be stated as follows:^{2}

1.14. Let k be a field, A = k^{[2]}, and let A = ⨁_{i} A_{i} be a Z-grading. Then there exist homogeneous elements f, g of A such that A = k[f, g].

2The given references assume that k is algebraically closed, but the result is known to be true for arbitrary k.

2. Derivations of graded rings

If B is a graded ring and D : B → B is a homogeneous locally nilpotent derivation, then the integer d(D) (see 1.12) may or may not be equal to 1. The first result of this section gives some information about the case where d(D) ≠ 1; its two corollaries are concerned with the special case where B = k^{[n]}. Proposition 2.1. Let B = ⨁_{i} B_{i} be an N-graded UFD containing Q and D : B → B a homogeneous locally nilpotent derivation such that d(D) ≠ 1. Let A = ker D, d = d(A) and

H = {H ∈ B \ {0} | H is homogeneous and prime, and deg H ≢ 0 (mod d)}.

Also, let R be the homogeneous subring of B defined by R_{n} = B_{n} if n ≡ 0 (mod d), R_{n} = 0 if n ≢ 0 (mod d). Then:

(1) H ≠ ∅ and any two elements of H are associates in B.

(2) For each H ∈ H, and for each homogeneous element v ∈ B such that Dv ≠ 0 and D^{2}v = 0, we have v = aH for some a ∈ A \ {0}. In particular, each H ∈ H satisfies DH ≠ 0 and D^{2}H = 0.

(3) For some integer n > 0, B_{0} ≅ B_{n} as B_{0}-modules.

(4) B = R[H], for every H ∈ H.

(5) Let δ = d(D) = d/d(B), assume that δ ≠ 0 and let H ∈ H. Then the restriction of H^{1−δ}D to R is a homogeneous locally nilpotent derivation Δ : R → R satisfying ker Δ = A and consequently d(Δ) = 1. Moreover, R is a UFD, H^{δ} is a prime element of R, Δ(H^{δ}) ≠ 0 and Δ^{2}(H^{δ}) = 0.

Remark . In a sense, part (5) of Proposition 2.1 is a reduction to the case where d(D) = 1. More precisely, if d(D) ≠ 1 then we may study the derivation Δ : R → R, which is closely related to D and which satisfies d(Δ) = 1.

Proof. Observe that the assumption d(D) ≠ 1 (or d ≠ d(B)) implies, in particular, that d(B) ≠ 0 and D ≠ 0.

Since B is a UFD and d ≠ d(B), H is nonempty. Consider H ∈ H and let n = deg H. Note that H ∉ A, because otherwise we would have H ∈ A_{n}, so A_{n} ≠ 0, and this is impossible because n ≢ 0 (mod d). By 1.2 we have qt A ∩ B = A, so we obtain

H ∉ qt A.  (1)

Since D is homogeneous, locally nilpotent and nonzero, we can find a homogeneous element v ∈ B such that Dv ≠ 0 and D^{2}v = 0. Let α = Dv ∈ A \ {0}, then by part (4) of 1.2 we have

B_{α} = A_{α}[v] = A_{α}^{[1]} .

In particular there exists an integer m ≥ 0 such that α^{m}H ∈ A[v]. So we may write α^{m}H = ∑_{i∈I} a_{i}v^{i}, where I is a nonempty subset of N and, for each i ∈ I, a_{i} ∈ A \ {0} is homogeneous and deg(a_{i}v^{i}) = deg(α^{m}H). Since deg a_{i} ≡ 0 and deg α ≡ 0 (mod d), we obtain i deg v ≡ deg H ≢ 0 (mod d), so in particular i > 0 for each i ∈ I. It follows that

α^{m}H = bv,  (2)

for some b ∈ B \ {0}. We claim that H | v (in B). Indeed, if H |̸ v then H | b (because H is a prime element of B), so equation (2) gives v | α^{m}, so v ∈ A by inertness of A in B. This is absurd, because Dv ≠ 0. Hence H | v and, if we write v = aH (with a ∈ B), then equation (2) gives ab = α^{m}. Since A is inert in B it follows that a, b ∈ A, so

v = aH, for some a ∈ A \ {0}.  (3)

Now if H′ ∈ H we have v = a′H′ (a′ ∈ A \ {0}), so

aH = a′H′.

If H | a′ then H ∈ A (because A is inert in B) and this contradicts (1); so H |̸ a′ and consequently H | H′. Similarly we obtain that H′ | H, i.e., we proved part (1) of the proposition. Part (2) follows immediately from equation (3).

Consider the ring R and, again, let H ∈ H. If β is a nonzero homogeneous element of B, we may write β = β_{0}H^{s}, where β_{0} ∈ B, s ∈ N and H |̸ β_{0}. Since H |̸ β_{0}, no prime factor of β_{0} belongs to H (by part (1) of the proposition), so every prime factor of β_{0} has degree divisible by d; consequently, β_{0} ∈ R. Hence every nonzero homogeneous element β of B has the form

β = β_{0}H^{s}, where β_{0} ∈ R, s ∈ N and H |̸ β_{0}.  (4)

In particular, β ∈ R[H]. This shows that B = R[H], which is part (4) of the proposition.

From equation (4), we also deduce that the submodule B_{0}H of the B_{0}-module B_{n} is actually equal to B_{n}, where n = deg H. Indeed, if deg β = n then deg β ≢ 0 (mod d), so β ∉ R and consequently s > 0 in (4). Since

deg H = deg β = deg β_{0} + s deg H

and deg β_{0} ≥ 0, we must have s = 1 and deg β_{0} = 0, i.e., β ∈ B_{0}H. This shows that B_{n} = B_{0}H ≅ B_{0}, which proves part (3) of the proposition.

At this point, we assume that δ = d(D) is nonzero (i.e., d ≠ 0). Note the following property of δ:

For all i ∈ Z, if i deg H ≡ 0 (mod d) then δ | i.  (5)

Indeed, the fact that B = R[H] implies that gcd(d, deg H) = d(B), and this easily implies that (5) holds. Let us now check that

D(R) ⊆ RH^{δ−1}.  (6)

Since DH ≠ 0 and D^{2}H = 0, we have

B_{α} = A_{α}[H] = A_{α}^{[1]},

where α = DH ∈ A\{0}. Thus, if r ∈ R\{0} is a homogeneous element, α^{m}r ∈ A[H] for some integer m ≥ 0. So we may write α^{m}r = ∑_{i∈I} a_{i}H^{i}, where I is a nonempty subset of N and, for each i ∈ I, a_{i} ∈ A \ {0} is homogeneous and deg(a_{i}H^{i}) = deg(α^{m}r). Since deg a_{i} ≡ deg α ≡ deg r ≡ 0 (mod d), we obtain i deg H ≡ 0 (mod d) and, by observation (5), this implies that δ | i (for each i ∈ I). Hence α^{m}r = f(H^{δ}) for some f(T) ∈ A[T], so

α^{m}Dr = Df(H^{δ}) = f′(H^{δ}) δ H^{δ−1} α,

and since H^{δ} ∈ R we have α^{m}Dr ∈ RH^{δ−1}. Since H |̸ α (by (1) and inertness of A in B), we have H^{δ−1} | Dr. If we write ρ = (Dr)/H^{δ−1} then ρ ∈ B and α^{m}ρ ∈ R; thus deg ρ ≡ 0 (mod d), i.e., ρ ∈ R, which proves (6).

It follows that Δ : R → R is a well-defined derivation with kernel A, and it is clear that Δ is homogeneous. To see that Δ is locally nilpotent, observe that the equation α^{m}r = f(H^{δ}), which was obtained in the preceding paragraph, actually implies that R_{α} = A_{α}[H^{δ}] = A_{α}^{[1]}. Since Δ(H^{δ}) = δα, the localization Δ_{α} : R_{α} → R_{α} of Δ is simply the partial derivative ∂/∂H^{δ} multiplied by the element δα of A_{α}. Hence Δ_{α} is locally nilpotent and so is Δ. Also, we see that Δ(H^{δ}) ≠ 0 and Δ^{2}(H^{δ}) = 0.

Since A is a UFD and R_{α} = A_{α}^{[1]}, it is clear that R_{α} is a UFD; so, in order to prove that R is a UFD, it suffices to show that α is a product of prime elements of R. Since α is homogeneous, it is enough to verify that every homogeneous prime element p of A is a prime element of R. So, let r_{1}, r_{2} be homogeneous elements of R such that p | r_{1}r_{2} (in R). Since p is irreducible in A and A is inert in B, it follows that p is irreducible in B, hence p is a prime element of B and consequently we may assume that r_{1} = bp, for some b ∈ B. Then b is homogeneous and 0 ≡ deg r_{1} = deg(bp) ≡ deg b (mod d), so b ∈ R. This shows that p is a prime element of R and hence that R is a UFD.

Finally, we show that H^{δ} is a prime element of R. Let Q be a prime factor of H^{δ} in R; factoring Q in B gives Q = uH^{i}, where u ∈ B^{∗} and 0 < i ≤ δ. Then u is homogeneous of degree zero, so i deg H = deg Q ≡ 0 (mod d). By observation (5), this implies that δ | i, so i = δ and H^{δ} is a prime element of R. □

Observe that if D ≠ 0 is a locally nilpotent derivation of k[X] = k^{[1]} which is homogeneous with respect to the standard grading, then d(D) = 0. In the case of a polynomial ring in several variables, we have: Corollary 2.2. Let D be a locally nilpotent derivation of B = k[X_{1}, …, X_{n}] = k^{[n]}, where n > 1 and where k is a field of characteristic zero. If D is homogeneous with respect to the standard grading, then d(D) = d(ker D) = 1.

Proof. Observe that B_{0} ≅ B̸_{r} for each r > 0; by part (3) of Proposition 2.1, it follows that d(D) = 1, so d(ker D) = d(B) = 1. □

In the following statement, observe that d(B) = 1, and consequently d(D) = d(ker D).

Corollary 2.3. Let k be a field of characteristic zero and consider the graded ring (B, w), where B = k[X_{1}, …, X_{n}] = k^{[n]}, w ∈ N^{n} and gcd(w) = 1. Let D : B → B be a w-homogeneous locally nilpotent derivation, let d = d(D) = d(ker D) and assume that d ≠ 1. Then d |̸ w(X_{j}) holds for exactly one value of j, and this j also satisfies:

(1) DX_{j} ≠ 0, D^{2}X_{j} = 0;

(2) assume that d ≠ 0 and define R as in Proposition 2.1, then

R = k[X_{1}, …, X_{j−1}, X_{j}^{d}, X_{j+1}, …, X_{n}].

Proof. Define H as in Proposition 2.1. Since d ≠ 1 = gcd(w(X_{1}), …, w(X_{n})), d |̸ w(X_{j}) holds for at least one value of j; since X_{j} ∈ H for any such j, the uniqueness of j and the assertion (1) follow respectively from parts (1) and (2) of Proposition 2.1.

Write R′ = k[X_{1}, …, X_{j−1}, X_{j}^{d}, X_{j+1}, …, X_{n}], then R′ ⊆ R is obvious. For the other inclusion, let r ∈ R \ {0} be a homogeneous element and consider a monomial X_{1}^{s_{1}} ⋯ X_{n}^{s_{n}} which occurs in r. Observe that B = R[X_{j}] and d(B) = 1 imply that gcd(d, w(X_{j})) = 1. Thus,

0 ≡ w(r) = ∑_{i=1}^{n} s_{i}w(X_{i}) ≡ s_{j}w(X_{j}) ⟹ s_{j} ≡ 0 (mod d),

i.e., X_{1}^{s_{1}} ⋯ X_{n}^{s_{n}} ∈ R′. This shows that r ∈ R′ and consequently that R = R′. □

The second part of this section, which consists of the results or paragraphs labeled from 2.4 to 2.7, is concerned with the case d(D) = 1, and in particular with the following type of question: Let B be a graded ring and A a homogeneous subring of B such that d(A) = d(B). Find conditions on A which are equivalent to the existence of a homogeneous locally nilpotent derivation D : B → B with kernel A.

We begin with two lemmas: Lemma 2.4. Let R = ⨁ R_{n} be a Z-graded UFD satisfying:

For all n ∈ Z, if R_{n} ≠ 0 then R_{−n} ∩ R^{∗} ≠ ∅.

Then R_{0} is a UFD.

Proof. If p is a homogeneous prime of R, say p ∈ R_{n}, then choose u ∈ R_{−n} ∩ R^{∗} and define p_{0} = up. Then p_{0} ∈ R_{0} and p_{0} is a prime element of R; moreover, it is easy to see that p_{0} is a prime element of R_{0}.

These remarks imply that every element of R_{0} (which is neither zero nor a unit) is a product of prime elements of R_{0}. □

Lemma 2.5. Let R = ⨁ R_{n} be a Z-graded domain and Q a homogeneous subring of R satisfying:

For all n ∈ Z, if R_{n} ≠ 0 then Q_{−n} ∩ Q^{∗} ≠ ∅.

Then the following are equivalent:

(1) There exists a homogeneous element v of R such that R = Q[v] = Q^{[1]};

(2) R_{0} = (Q_{0})^{[1]}.

Proof. Suppose that (1) holds, with v ∈ R_{n} \ {0}. Since R_{n} ≠ 0, we can pick θ ∈ Q_{−n} ∩ Q^{∗} and define v_{0} = θv ∈ R_{0}. Then Q[v_{0}] = R = Q^{[1]}, so v_{0} is algebraically independent over Q and in particular it is algebraically independent over Q_{0}. It is clear that R_{0} = Q_{0}[v_{0}], so (2) holds.

Conversely, assume that (2) holds and consider v ∈ R_{0} such that R_{0} = Q_{0}[v] = (Q_{0})^{[1]}. If r ≠ 0 is any homogeneous element of R (say r ∈ R_{n}) then, since R_{n} ≠ 0, we may pick θ ∈ Q_{−n}∩Q^{∗} and define r_{0} = θr ∈ R_{0}; then r_{0} ∈ Q_{0}[v], whence r ∈ Q[v]. This shows that R = Q[v]. In order to prove that v is algebraically independent over Q, let us assume the contrary and consider a polynomial 0 ≠ f(T) ∈ Q[T] such that f(v) = 0. Write f(T) = ∑_{i} f_{i}(T) with f_{i}(T) ∈ Q_{i}[T], choose m such that f_{m}(T) ≠ 0 and define g(T) = θ′f_{m}(T) ∈ Q_{0}[T], where θ′ ∈ Q_{−m} ∩ Q^{∗}. Then g(v) = 0, which contradicts the fact that v is algebraically independent over Q_{0}. Hence (1) holds. □

2.6. We consider pairs (A, B) of N-graded rings satisfying:

(1) B is an integral domain containing the field Q and A is a homogeneous subring of B;

(2) A ≠ B and B is finitely generated as an A-algebra;

(3) d(A) = d(B). Given such a pair (A, B), consider the multiplicatively closed subset S = ⋃_{i}(A_{i}\{0}) of A and denote the localized rings S^{−1}A and S^{−1}B by A = ⨁_{i∈Z} A_{i} and B = ⨁_{i∈Z} B_{i} respectively. Note that A is a homogeneous subring of B, that A_{0} is a field and that B_{0} is a finitely generated A_{0}-domain. Moreover, observe that if we write d = d(A) = d(B) then for all n ∈ Z we have: A_{n} ≠ 0 ⇔ n ∈ dZ ⇔ B_{n} ≠ 0, and A_{n} \ {0} ⊆ A^{∗}.

The following will be used in the proof of 3.5 : Lemma 2.7. Let A, B, A and B be as in 2.6. Then the following conditions are equivalent:

(1) There exists a nonzero homogeneous locally nilpotent derivation D of B such that A ⊆ ker D ⊆ qt A;

(2) B_{0} = A_{0}^{[1]}.

Proof. Suppose (1) holds. We claim that

ker D ⊆ A.  (7)

Since ker D is a homogeneous subring of B, the proof of (7) reduces to that of:

If b ∈ B \ {0} is homogeneous and Db = 0, then b ∈ A.

Let b ∈ B \ {0}, homogeneous and such that Db = 0; then b ∈ qt A, so we may write ab = a′ for some a, a′ ∈ A \ {0}. Moreover, we may arrange that a and a′ be homogeneous, and this shows that b ∈ A. Hence (7) holds.

If we define S as in 2.6 then (7) implies

S^{−1} ker D = A.  (8)

On the other hand, we may consider a homogeneous element v of B such that Dv ≠ 0 and D^{2}v = 0. Write α = Dv, then B_{α} = (ker D)_{α}[v] = (ker D)_{α}^{[1]} by 1.2. This, together with (8) and the fact that α ∈ S, implies that B = A[v] = A^{[1]}. Then (2) follows from Lemma 2.5.

Conversely, assume that (2) holds; then Lemma 2.5 implies that

B = A[v] = A^{[1]},

for some homogeneous element v of B. Thus we may consider the locally nilpotent derivation ∂ : B → B, where ∂ denotes the derivative with respect to v. Observe that if b ∈ B then ∂b ∈ B = S^{−1}B; since B is assumed to be finitely generated as an A-algebra we see that, for some s ∈ S, the derivation s∂ maps B into itself. Let D : B → B be the restriction of s∂; then D is locally nilpotent (by part (5) of 1.2) and ker D = B ∩ ker ∂ = B ∩ A. Thus A ⊆ ker D ⊆ qt A and (1) holds. □

We conclude this section with a fact concerning subrings of a graded k^{[2]}; it will be used in the proof of Proposition 3.4. Lemma 2.8. Let u and v be indeterminates over a field k and let A = k[u, v]. Let m, n ∈ N be such that gcd(m, n) = 1 and regard A as an N-graded ring, the grading being given by the weights w(u) = m and w(v) = n. Let f and g be elements of A satisfying:

(i) f and g are w-homogeneous and gcd(w(f), w(g)) = 1 ; (ii) f and g are irreducible in \overline{k}[u, v], where \overline{k} is the algebraic closure of k ; (iii) f, g are not associates.

Then k[f, g] = k[u, v].

Proof. Consider the inclusion map ν : k[f, g] → k[u, v] and the corresponding

\overline{k} ⊗_{k} ν : \overline{k}[f, g] → \overline{k}[u, v],

and recall that if \overline{k} ⊗_{k} ν is surjective then so is ν. Hence we may assume that k is algebraically closed.

Let “x ∼ y ” denote that x and y are associates in A. It is an easy exercise (left to the reader) to show that if P is an irreducible homogeneous element of k[u, v] then

P ∼ u or P ∼ v or P ∼ u^{n} + λv^{m} (some λ ∈ k^{∗}).

To verify that k[f, g] = k[u, v], we consider four cases. Case 1: f ∼ u and g ∼ v. The claim is obvious in this case. Case 2: f ∼ u and g ∼ u^{n} + λv^{m}, with λ ∈ k^{∗}. We have

1 = gcd(w(f), w(g)) = gcd(m, mn) = m,

so m = 1, so g ∼ u^{n} + λv. Hence k[f, g] = k[u, u^{n} + λv] = k[u, v].

Case 3: f ∼ v and g ∼ u^{n} + λv^{m}, with λ ∈ k^{∗}. Similar to case 2. Case 4: f ∼ u^{n} + λv^{m} and g ∼ u^{n} + μv^{m}, with λ, μ ∈ k^{∗}. We have

1 = gcd(w(f), w(g)) = gcd(mn, mn) = mn,

so m = 1 = n, so f ∼ u + λv and g ∼ u + μv. By assumption (iii), λ ≠ μ, so k[f, g] = k[u + λv, u + μv] = k[u, v].

This proves the lemma. □

3. Homogeneous derivations of k[X, Y, Z]

Let k be a field of characteristic zero and B = k[X, Y, Z] = k^{[3]}. We consider the problem of describing the set of nonzero, w-homogeneous locally nilpotent derivations of B (see 1.13 for definition). By 1.6, such a derivation is determined by its kernel, up to multiplication by a w-homogeneous element of that kernel; so our basic goal is to characterize the kernels of these derivations. Note that 1.5, 1.14 and corollaries 1.8 and 2.2 imply the following: Corollary 3.1. Let k be a field of characteristic zero and consider the N-graded ring (B, w) (see 1.13), where B = k[X, Y, Z] = k^{[3]} and where w ∈ N^{3} satisfies gcd(w) = 1. Let D ≠ 0 be a w-homogeneous locally nilpotent derivation of B. Then ker D = k[f, g] = k^{[2]}, for some w-homogeneous polynomials f, g, and for any such f, g we have:

(1) f and g are irreducible in \overline{k}[X, Y, Z], where \overline{k} is the algebraic closure of k ;

(2) d(D) = gcd(w(f), w(g)) ;

(3) if w = (1, 1, 1) then d(D) = gcd(deg f, deg g) = 1.

In view of this fact, we can phrase our question as follows: Problem 3.2. Let f and g be w-homogeneous polynomials in B = k[X, Y, Z]. Give conditions, on the pair (f, g), which are necessary and sufficient for the existence of a w-homogeneous locally nilpotent derivation of B with kernel k[f, g].

This investigation splits naturally into two cases, according to the value of the natural number d(D) defined in 1.12. Perhaps surprisingly, it turns out that “ d(D) = 1 ” is the generic case and that “ d(D) ≠ 1 ” is very special. The reader should note that Theorem 3.7 implies, in particular, that all derivations satisfying d(D) ≠ 1 have rank strictly less than 3.

The case d(D) = 1

3.3. Let k be a field of characteristic zero and consider the N-graded ring (B, w), where B = k[X, Y, Z] = k^{[3]} and where w ∈ N^{3} satisfies gcd(w) = 1. Let f, g ∈ B be w-homogeneous polynomials and consider the following conditions on f and g:

(i) gcd(w(f), w(g)) = 1 ; (ii) f and g are irreducible in \overline{k}[X, Y, Z], where \overline{k} is the algebraic closure of k ; (iii) f, g are not associates in k[X, Y, Z].

Note that any f, g satisfying these conditions are algebraically independent over k. Note, also, that these conditions are necessary for the existence of a w-homogeneous locally nilpotent derivation D : B → B with kernel k[f, g] and such that d(D) = 1.

The following result clarifies the meaning of Problem 3.2:

Proposition 3.4. Let k be a field of characteristic zero and consider the N-graded ring (B, w), where B = k[X, Y, Z] = k^{[3]} and where w ∈ N^{3} satisfies gcd(w) = 1. Suppose that f, g ∈ B are w-homogeneous and satisfy the conditions of 3.3. Then the following hold:

(1) If D ≠ 0 is any locally nilpotent derivation of B which satisfies Df = 0 = Dg, then ker D = k[f, g].

(2) For the pair (f, g), the following conditions are equivalent:

(a) There exists a w-homogeneous locally nilpotent derivation D of B such that ker D = k[f, g];

(b) the derivation Δ_{(f,g)} of B (see 1.6) is locally nilpotent;

(c) there exists a nonzero locally nilpotent derivation D of B such that ker D ⊇ k[f, g].

Proof. To prove (1), let D ≠ 0 be a locally nilpotent derivation of B such that Df = 0 = Dg, and write A = ker D. By 1.5, A = k[u, v] for some polynomials u, v ∈ B. By 1.6, we have D = αΔ_{(u,v)} for some α ∈ A \ {0}; note that Δ_{(u,v)} is locally nilpotent and has kernel A.

On the other hand, we may consider the derivation Δ_{(f,g)} of B. Since f and g are polynomials in u and v, the chain rule implies that Δ_{(f,g)} = aΔ_{(u,v)} for some a ∈ k[u, v]. Since f and g are algebraically independent over k, we have Δ_{(f,g)} ≠ 0 and consequently a ≠ 0. Now Δ_{(f,g)} is a w-homogeneous derivation; thus aΔ_{(u,v)} is w-homogeneous, and this clearly implies that Δ_{(u,v)} is w-homogeneous. By Corollary 3.1, A = k[u′, v′] for some w-homogeneous polynomials u′ and v′, and these polynomials satisfy gcd(w(u′), w(v′)) = 1 (because f, g ∈ k[u′, v′]). Also, by condition (ii) of 3.3, f and g are irreducible in \overline{k}[u′, v′]. Thus statement (1) follows from Lemma 2.8.

Assertion (2) is now easy: (a)⇒(b) follows from 1.6; (b)⇒(c) is obvious, since Δ_{(f,g)} ≠ 0 was noted in the proof of (1). To prove (c)⇒(a), assume that (c) holds; then, by assertion (1), ker D = k[f, g] and consequently (by 1.6) D = αΔ_{(f,g)} (some α ∈ ker D \ {0}) and Δ_{(f,g)} is locally nilpotent. Since ker Δ_{(f,g)} = ker D = k[f, g], (a) holds. □

The next result is the solution of Problem 3.2 in the case d(D) = 1. In the statement of the theorem, V (fg) is the closed subset “ fg = 0 ” of the weighted projective space Proj \overline{B}. In most cases, it is correct to say that the surface Proj \overline{B} \ V (fg) is obtained by removing the two curves V (f), V (g) from Proj \overline{B}, but one should be aware that, in some cases, one of those two “curves” is empty! For example, let w = (0, 0, 1), f = X and g = Z; then V (f) is a line in Proj \overline{B} ≅ A_{\overline{k}}^{2} and V (g) is empty, so Proj \overline{B} \ V (fg) is the affine plane minus a line. Theorem 3.5. Let k be a field of characteristic zero, with algebraic closure \overline{k}, and consider the N-graded ring (B, w), where B = k[X, Y, Z] = k^{[3]} and where w ∈ N^{3} satisfies gcd(w) = 1. Suppose that f, g ∈ B are w-homogeneous and satisfy gcd(w(f), w(g)) = 1. Then the following are equivalent:

(1) There exists a w-homogeneous locally nilpotent derivation D of k[X, Y, Z] such that ker D = k[f, g];

(2) f and g are irreducible elements of \overline{B} = \overline{k}[X, Y, Z] and the surface Proj \overline{B} \ V (fg) is isomorphic to P_{\overline{k}}^{2} minus two lines.

Remark . The conditions (1) and (2) of the theorem are also equivalent to:

(1\frac{1}{2}) B_{(fg)} = (A_{(fg)})^{[1]}. Indeed, (1) ⇒ (1\frac{1}{2}) follows from Corollary 3.10 and we shall now argue that (1\frac{1}{2}) ⇒ (2). Assume that (1\frac{1}{2}) holds; then tensoring by \overline{k} yields \overline{B}_{(fg)} = (\overline{A}_{(fg)})^{[1]}, where \overline{B} = \overline{k}[X, Y, Z] and \overline{A} = \overline{k}[f, g], so we may assume that k = \overline{k}. In this case, the proof of the theorem shows that (1\frac{1}{2}) does imply (2) (see statement (15) in the proof, below).

Proof of Theorem 3.5. In view of Corollary 1.7, it is clear that we may assume that k is algebraically closed. Hence \overline{k} = k and \overline{B} = B. Also, it is convenient to assume, throughout the proof, that

at least one of the conditions (3.5.1), (3.5.2) is satisfied.

Let A = k[f, g], B = k[X, Y, Z], p = w(f) and q = w(g). Since it is assumed that gcd(p, q) = 1, we clearly have:

(A, B) satisfies the conditions of 2.6

and we may consider the graded rings A and B defined in 2.6. On the other hand, let A_{(fg)} and B_{(fg)} denote the homogeneous localizations at fg, i.e., the degree zero components of the graded rings A_{fg} and B_{fg} respectively.

Observe that, since B is a UFD, so is the Z-graded ring B_{fg}. Since gcd(p, q) = 1, if n ∈ Z then there exist i, j ∈ Z such that pi + qj = −n, so f^{i}g^{j} ∈ (B_{fg})_{−n} ∩ (B_{fg})^{∗}. Thus 2.4 implies that

B_{(fg)} is a UFD.  (9)

We will now verify that

(f, g) satisfies the conditions of 3.3.  (10)

Condition (i) of 3.3 holds, since it is one of the assumptions of this theorem. Condition (ii) of 3.3 follows from Corollary 3.1 if (3.5.1) holds, and holds by assumption if (3.5.2) holds. Condition (iii) is obvious if (3.5.1) holds because, say, ker D = k^{[2]} by 1.5; there remains to check that (3.5.2) implies that f and g are not associates in B. To see this, observe that Proj B \ V (fg) ≅ Spec B_{(fg)} and that P_{k}^{2} minus two lines is isomorphic to Spec k[W, W^{−1}]^{[1]}, so (3.5.2) implies that B_{(fg)} ≅ k[W, W^{−1}]^{[1]} (W is an indeterminate); in particular, k^{∗} is a proper subset of B_{(fg)}^{∗}. On the other hand, pick h ∈ {f, g} such that w(h) > 0; using only the fact that h is a homogeneous prime element of B and w(h) > 0, it is easy to see that B_{(h)}^{∗} = k^{∗}, hence B_{(h)} is not isomorphic to B_{(fg)}. Hence f, g cannot be associates, and (10) is true.

Let ξ = f^{q}/g^{p}. It is easy to see that every element of A_{n} \ {0} (where n ∈ N) has the form f^{i}g^{s}H(f^{q}, g^{p}), for some i, s ∈ N and some homogeneous polynomial in two variables H ∈ k[T_{1}, T_{2}] (that is, homogeneous with respect to the standard grading of k[T_{1}, T_{2}]). This can be rewritten as f^{i}g^{j}H(ξ, 1) for some j ∈ N, so we conclude that every a ∈ A_{n} \ {0} has the form

a = f^{i}g^{j}θ, for some i, j ∈ N and θ ∈ k[ξ].  (11)

From this, one easily deduces

A_{(fg)} = k[ξ, ξ^{−1}],  (12)
B_{(fg)}^{∗} = ⋃_{n∈Z} k^{∗}ξ^{n},  (13)
A_{0} = T^{−1}A_{(fg)} and B_{0} = T^{−1}B_{(fg)},  (14)

where T = A_{(fg)} \ {0}.

We now consider the equivalence of conditions (3.5.1) and (3.5.2). First, we prove:

The condition (3.5.2) is equivalent to B_{(fg)} = (A_{(fg)})^{[1]}.  (15)

Recall that Spec B_{(fg)} ≅ Proj B \V (fg) and that, by (12), Spec (A_{(fg)}^{[1]}) is isomorphic to P_{k}^{2} minus two lines. So it is obvious that B_{(fg)} = (A_{(fg)})^{[1]} implies (3.5.2). Conversely, suppose that (3.5.2) holds. Then, for some element ζ of B_{(fg)}, we have B_{(fg)} = k[ζ, ζ^{−1}]^{[1]}. That element ζ must then satisfy

B_{(fg)}^{∗} = ⋃_{n∈Z} k^{∗}ζ^{n},

and this, together with (13), implies that ζ = λξ^{±1} (for some λ ∈ k^{∗}) and consequently that k[ζ, ζ^{−1}] = k[ξ, ξ^{−1}] = A_{(fg)}. Hence (3.5.2) implies that B_{(fg)} = (A_{(fg)})^{[1]}, and (15) is proved.

There remains to prove:

The condition (3.5.1) is equivalent to B_{(fg)} = (A_{(fg)})^{[1]}.  (16)

One way is easy: If B_{(fg)} = (A_{(fg)})^{[1]} then (by (14)) B_{0} = A_{0}^{[1]}, so (3.5.1) holds by Lemma 2.7 and Proposition 3.4.

Conversely, assume that (3.5.1) holds. We begin by observing that A_{(fg)} is an inert subring of B_{(fg)}. To see this, consider a locally nilpotent derivation D : B → B with kernel A. Then, by 1.2, localization at fg yields a locally nilpotent derivation, D_{fg} : B_{fg} → B_{fg}, with kernel A_{fg}. Thus (again by 1.2) A_{fg} is an inert subring of B_{fg}, and it follows immediately that A_{(fg)} is an inert subring of B_{(fg)}.

We also point out that Lemma 2.7 and (14) imply that

T^{−1}B_{(fg)} = (T^{−1}A_{(fg)})^{[1]}.  (17)

We will now verify that, for each a ∈ k^{∗}, the following hold (where we write π = ξ − a):

(i) π is a prime element of B_{(fg)}, (ii) A_{(fg)} ∩ πB_{(fg)} = πA_{(fg)}, (iii) A_{(fg)}/πA_{(fg)} is algebraically closed in B_{(fg)}/πB_{(fg)}. Since π is an irreducible element of A_{(fg)} and A_{(fg)} is an inert subring of B_{(fg)}, π

is an irreducible element of B_{(fg)}. Since B_{(fg)} is a UFD by (9), π is prime in B_{(fg)} and (i) holds.

If β ∈ B_{(fg)} \ {0} satisfies πβ ∈ A_{(fg)} then, again by inertness of A_{(fg)}, we have β ∈ A_{(fg)}; so (ii) holds.

Condition (iii) is obvious, since k is assumed to be algebraically closed.

In view of 1.1, conditions (17), (i), (ii) and (iii) imply that B_{(fg)} = (A_{(fg)})^{[1]}, which means that (16) is proved. This completes the proof of the theorem. □

The case d(D) ≠ 1

Our first comment is that this case does occur. To obtain examples, it suffices to make sure that gcd(w(X), w(Y)) ≠ 1 in the following: Example 3.6. Let k be a field of characteristic zero and consider (B, w), where B = k[X, Y, Z] and w ∈ N^{3} satisfies gcd w = 1. Let φ ∈ k[X] and ψ ∈ k[X, Z] be w-homogeneous polynomials such that

(1) φ is not the zero polynomial,

(2) φ and ∂ψ/∂Z are relatively prime in B,

(3) P is w-homogeneous, where P = φ(X)Y + ψ(X, Z). Then Δ_{(X,P)} : B → B is a w-homogeneous locally nilpotent derivation, ker Δ_{(X,P)} = k[X, P] and d(Δ_{(X,P)}) = gcd(w(X), w(Y)).

(The verification of these claims is left to the reader. This is very easy if one uses the third section of [4].)

The aim of this sub-section is to prove that Example 3.6 is in fact the only way that the case d(D) ≠ 1 can occur. More precisely, Theorem 3.7. Let k be a field of characteristic zero and consider (B, w), where B = k[X′, Y′, Z′] and w ∈ N^{3} satisfies gcd w = 1. Suppose that D : B → B is a w-homogeneous locally nilpotent derivation such that d(D) ≠ 1. Then there exists a w-homogeneous coordinate system^{3} (X, Y, Z) of B, and polynomials φ ∈ k[X] and ψ ∈ k[X, Z] satisfying the three conditions of Example 3.6, such that ker D = k[X, P]. Consequently, D = aΔ_{(X,P)} for some nonzero element a of k[X, P].

3We mean that X, Y and Z are w-homogeneous elements of B satisfying B = k[X, Y, Z].

In order to prove Theorem 3.7, we will establish two lemmas. Lemma 3.8. Let D : B → B be a locally nilpotent derivation, where B = k[X, Y, Z]. Assume that DZ ∈ k[f] \ {0} for some variable f of ker D. Then:

(1) B = k[f, Z]^{[1]}.

(2) If D and f are w-homogeneous (for some w ∈ N^{3}), then B = k[f, h, Z] for some w-homogeneous element h of B.

Proof. We leave it to the reader to verify the following statement:

Let B be an N-graded domain and A a homogeneous subring of B such that B = A^{[1]}. Then B = A[h], for some homogeneous element h of B.

Hence the second part of the lemma follows from the first. We prove part (1) of the lemma.

Since DZ ≠ 0 we have D ≠ 0 so, by 1.5, ker D = k[f, g] for some g ∈ B. Write DZ = φ ∈ k[f] \ {0}, then B_{φ} = k[f, g]_{φ}[Z] = k[f, Z]_{φ}[g], so

B_{φ} = (A_{φ})^{[1]},  (18)

where A = k[f, Z]. For future use, we also note that

f ∉ k[Z].  (19)

This follows from equation (18) or, alternatively, from the fact that ker D is algebraically closed in B, together with Df = 0 and DZ ≠ 0.

The first part of the proof consists in proving

A is the kernel of some locally nilpotent derivation of B.  (20)

We first prove this claim under the assumption that k is algebraically closed. In view of 1.3 and equation (18), it suffices to prove that A_{φ} ∩ B = A.

First we show that, for each λ ∈ k, if we write π = f − λ ∈ k[f] then

A ∩ πB = πA.  (21)

Since π is prime in k[f], it is prime in ker D = k[f]^{[1]}; since ker D is an inert subring of B, π is prime in B. Hence p = A ∩ πB is a prime ideal of A and πA ⊆ p. Suppose that p ≠ πA; then p is a maximal ideal of A, so p = (f −λ, Z −μ) (where μ ∈ k) and consequently Z − μ ∈ πB, which contradicts (19). Hence p = πA and equality (21) holds.

To see that A_{φ} ∩ B = A, consider b ∈ A_{φ} ∩ B. Then there exist prime factors π_{1}, …, π_{n} of φ in k[f] such that π_{1} ⋯ π_{n}b ∈ A. If n > 0 then π_{1} ⋯ π_{n}b ∈ A ∩ π_{n}B = π_{n}A by equality (21), so π_{1} ⋯ π_{n−1}b ∈ A. Clearly, this argument shows that b ∈ A and consequently that A_{φ} ∩ B = A. It follows that the claim (20) holds (if k is algebraically closed).

Actually, (20) holds without assuming that k is algebraically closed. Indeed, consider the algebraic closure \overline{k} of k, the rings \overline{B} = \overline{k}[X, Y, Z] and \overline{A} = \overline{k}[f, Z] and the locally nilpotent derivation \overline{D} = \overline{k} ⊗_{k} D : \overline{B} → \overline{B}. Since ker \overline{D} = \overline{k}[f, g], \overline{D} satisfies all hypotheses of this lemma (\overline{D}Z ∈ \overline{k}[f] \ \{0\}, where f is a variable of ker \overline{D}). Since (20) holds in the special case where k is algebraically closed, \overline{A} is the kernel of some locally nilpotent derivation of \overline{B} and, by Corollary 1.7, it follows that (20) holds.

Consider a locally nilpotent derivation Δ : B → B with kernel A. Note that rank Δ < 3, because A contains the variable Z of B. In order to prove this lemma, there remains to show that rank Δ ≠ 2 (then rank Δ = 1 and consequently B = A^{[1]}).

Assume that rank Δ = 2. We have dim \overline{R}(D) = 1 by 1.11, so we may consider a prime element ξ of A such that V (ξ) ⊆ \overline{R}(D); then, again by 1.11, k[Z] is the integral closure of k[ξ] in B. On the other hand, equality (18) implies that \overline{R}(D) is disjoint from the open subset Spec A_{φ} of Spec A. Thus V (ξ) ⊆ V (φ), and consequently ξ | φ in A, whence ξ ∈ k[f]. We obtain that f is integral over k[ξ], so f ∈ k[Z], which contradicts statement (19). We conclude that Δ cannot have rank two, which completes the proof. □

Lemma 3.9. Let k be a field of characteristic zero and consider (B, w), where B = k[X, Y, Z] and w ∈ N^{3}. Let D : B → B be a w-homogeneous locally nilpotent derivation. If w(Z) > 0, DZ ≠ 0 and D^{2}Z = 0, then B = k[u, v, Z] for some w-homogeneous variable u of ker D and some w-homogeneous element v of B.

Proof. Let A = ker D. Since DZ ≠ 0, D ≠ 0 and consequently ker D = k[f, g] for some w-homogeneous elements f, g ∈ B (by 3.1); note that

f, g ∉ k[Z],  (22)

because Df = 0 = Dg, DZ ≠ 0 and A is algebraically closed in B. Note, also, that we may assume that D = Δ_{(f,g)} (see 1.6). Write p = w(f), q = w(g) and w = (a, b, c).

If p = 0 = q then A ⊆ B_{0}, so B_{0} has transcendence degree 2 over k. Consequently, B_{0} = k[X, Y]. Since A is algebraically closed in B, A ⊆ k[X, Y] implies A = k[X, Y], thus the assertion is satisfied with u = X, v = Y.

We may therefore assume that q ≠ 0. Then we claim that a + b > 0. Indeed, if it is not the case then a = 0 = b, so B_{0} = k[X, Y], and since q > 0 we have g = αZ^{n} for some n > 0 and α ∈ k[X, Y]. Thus Z | g and, since g is irreducible by 3.1, we obtain that g and Z are associates, which contradicts statement (22).

Since we assumed that D = Δ_{(f,g)}, we have

w(DZ) = w ( | \begin{matrix}f_{X} & f_{Y} \\ g_{X} & g_{Y} \end{matrix} | ) = p + q − a − b < p + q.

If the monomial f^{i}g^{j} occurs in the polynomial DZ ∈ k[f, g] then

p + q > w(DZ) = w(f^{i}g^{j}) = pi + qj,

so i = 0 or j = 0. In other words, DZ ∈ k[f] + k[g]; since DZ is w-homogeneous and q > 0, we have

DZ = φ(f) + μ g^{j}  (for some φ(T) ∈ k[T], μ ∈ k and j ∈ N).  (23)

In view of Lemma 3.8, it suffices to show that

DZ ∈ k[u], for some w-homogeneous variable u of A.  (24)

If p = 0 then w(DZ) < p + q = q, so j = 0 in equation (23), thus DZ ∈ k[f] and (24) holds.

Assume that p > 0 then, since DZ is w-homogeneous, (23) reads

DZ = λf^{i} + μ g^{j} (for some λ, μ ∈ k).

It is clear that (24) holds if one of the following conditions is satisfied:

λ = 0, μ = 0, i < 2, or j < 2.

Now we claim that if none of these conditions is satisfied then we have a contradiction. Indeed, if p ≤ q then

p + q > w(DZ) = qj ≥ 2q ≥ p + q

is absurd, and the argument is similar if p > q. This completes the proof of the lemma. □

Proof of Theorem 3.7. Let d = d(D). Since d ≠ 1, Corollary 2.3 implies that a unique element Z of {X′, Y′, Z′} satisfies d |̸ w(Z), and that Z also satisfies DZ ≠ 0 and D^{2}Z = 0. Thus, by Lemma 3.9, there exist a w-homogeneous variable X of ker D and a w-homogeneous element Y of B such that B = k[X, Y, Z]. Since ker D = k^{[2]} and X is a w-homogeneous variable of ker D, we have ker D = k[X, P] for some w-homogeneous polynomial P ∈ k[X, Y, Z]. By result 1.6, D = aΔ_{(X,P)} for some a ∈ k[X, P] \ {0}.

Write Δ = Δ_{(X,P)}, then ΔZ ≠ 0 and Δ^{2}Z = 0; since ΔZ = P_{Y} (partial derivative with respect to Y), we obtain P_{Y} ∈ k[X, P] \ {0}. Write

P_{Y} = H(X, P),  (25)

where H(T_{1}, T_{2}) ∈ k[T_{1}, T_{2}] is a nonzero polynomial in two variables, and let n = deg_{T_{2}} H (the degree of H as a polynomial in T_{2} with coefficients in k[T_{1}]). Then equation (25) gives

deg_{Y} P > deg_{Y} (P_{Y}) = n deg_{Y} P,

so n = 0 and P_{Y} ∈ k[X]. Thus P = φY + ψ, where φ ∈ k[X] \ {0} and ψ ∈ k[X, Z].

There remains to verify that φ and ∂ψ/∂Z are relatively prime. Let h be the gcd of these two polynomials; then ψ − ψ(X, 0) is divisible by h, and consequently P = hQ + ψ(X, 0) for some Q ∈ B. It follows that k[X, hQ] = k[X, P] = ker D; then Corollary 3.1 implies that hQ is an irreducible element of B, so h ∈ B^{∗}. □

Some consequences

Our last result gathers some consequences of Theorem 3.5 and of the second section. Corollary 3.10. Let k be a field of characteristic zero and consider the N-graded ring (B, w), where B = k[X, Y, Z], w ∈ N^{3} and gcd w = 1. Let 0 ≠ D : B → B be a w-homogeneous locally nilpotent derivation and let A = k[f, g] = ker D, where f and g are w-homogeneous. Then the following hold.

If d(D) ≥ 1 then:

(1) B_{(fg)} = (A_{(fg)})^{[1]};

(2) Proj B \ V (fg) is isomorphic to P_{k}^{2} minus two lines.

If d(D) ≤ 1 then:

(3) B_{fg} = A_{fg}[v] = (A_{fg})^{[1]}, for some w-homogeneous element v of B;

(4) rank D < 3 ⇔ f or g is a variable of B;

(5) Δ_{(f,g)}(v) = f^{i}g^{j}, for some i, j ∈ N and some w-homogeneous v ∈ B. Remark . If d(D) < 1 (i.e., d(D) = 0) then assertions (1) and (2) do not hold, because we have A = B_{0} and consequently B_{(fg)} = A_{(fg)}.

Assertions (3–5) do not always hold when d(D) > 1. For instance, consider the derivation D = Δ_{(X,P)} as in Example 3.6, with w = (n, 0, 1) (where n > 1), φ = X and ψ = Z^{n}. Let f = X + P = X + XY + Z^{n} and g = P = XY + Z^{n}. Then (3–5) do not hold.

Proof. The first part of the proof consists in showing that condition (5) holds if d(D) = 1. By assumption, the condition (1) of Theorem 3.5 is satisfied; if k is algebraically closed then the proof of that result implies that assertion (1) (of this corollary) holds, so assertion (3) holds by Lemma 2.5 and assertion (5) follows from Lemma 1.4. Hence, if we drop the assumption that k is algebraically closed, we obtain that there exists a w-homogeneous element \overline{v} ∈ \overline{B} = \overline{k}[X, Y, Z] which satisfies \overline{Δ}(\overline{v}) = f^{i}g^{j} for some i, j ∈ N (here, \overline{k} is the algebraic closure of k and \overline{Δ} : \overline{B} → \overline{B} is the extension of Δ_{(f,g)} : B → B). It follows from linear algebra that Δ_{(f,g)}(v) = f^{i}g^{j} for some w-homogeneous element v of B. Indeed, let ω = w(\overline{v}) and let E be a finite set of monomials X^{α}Y^{β}Z^{γ} of weight ω such that \overline{v} ∈ Span\overline{k}(E); let V = Span_{k}(E), W = Δ_{(f,g)}(V) and restrict Δ_{(f,g)} to a linear map of finite dimensional vector spaces over k, δ : V → W. Then f^{i}g^{j} is in the image of the \overline{k}-linear map \overline{k} ⊗_{k} δ : \overline{k} ⊗_{k} V → \overline{k} ⊗_{k} W, and it follows that f^{i}g^{j} is in the image of δ. In other words, (5) holds.

Next, we show that if d(D) = 1 then assertions (1–5) hold. Since we have shown that (5) holds, part (4) of 1.2 implies that assertion (3) holds, which implies that (1) holds (by Lemma 2.5); clearly, (2) follows from (1). As to (4), the only nontrivial part is “⇒” when rank D = 2. Suppose that rank D = 2; then dim \overline{R}(D) = 1 by 1.11, so we may consider an irreducible element ξ of A such that V (ξ) is an irreducible component of \overline{R}(D). Since B_{fg} = (A_{fg})^{[1]}, we have \overline{R}(D) ⊆ V (f) ∪ V (g) in Spec A, so V (ξ) ⊆ V (f) or V (ξ) ⊆ V (g). Suppose, for instance, that V (ξ) ⊆ V (f); then ξ and f are associates, so k[ξ] = k[f], so 1.11 implies that the integral closure of k[f] in B is k[T] for some variable T of B. Since A = k[f]^{[1]} is algebraically closed in B, k[f] is integrally closed in B; we therefore obtain that f is a variable of B and that (4) holds. Hence, (1–5) hold if d(D) = 1.

If d(D) > 1 then consider the homogeneous subring R of B and the homogeneous locally nilpotent derivation Δ : R → R which are defined in Proposition 2.1. Because of the obvious equality B_{(fg)} = R_{(fg)}, in order to show that (1) holds it suffices to show that

R_{(fg)} = A_{(fg)}.  (26)

By Corollary 2.3, we have R = k[X′, Y′, Z′] = k^{[3]} and the grading of R is determined by some w′ ∈ N^{3}. Since ker Δ = A and d(Δ) = 1, the preceding paragraph implies that assertions (1–5) hold for Δ; in particular assertion (1) holds for Δ, i.e., (26) holds. (To be precise, consider the graded ring (R, w′′) where w′′ = w′/ gcd(w′), then gcd(w′′) = 1 so the preceding paragraph implies that assertions (1–5) hold for the w′′-homogeneous derivation Δ. Note that the rings R_{(fg)} and A_{(fg)} are not affected by that change of grading.) Hence (1) holds—and (2) follows from (1).

If d(D) < 1 then d(A) = d(D) = 0, so A ⊆ B_{0} and it follows that A is one of the rings k[X, Y], k[X, Z], or k[Y, Z]. Thus it is clear that assertions (3–5) hold in this case. This completes the proof. □

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

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