ON LOCALLY NILPOTENT DERIVATIONS OF k[X_{1}, X_{2}, Y ]/(φ(Y ) − X_{1}X_{2})

DANIEL DAIGLE

Abstract. We investigate the locally nilpotent derivations of the k-algebra

B = k[X_{1}, X_{2}, Y]/(φ − X_{1}X_{2}), where k is a field of characteristic zero, X_{1}, X_{2}, Y are indeterminates over k and φ ∈ k[Y] is a non-constant polynomial. We prove that a certain subgroup of Aut_{k}(B) acts transivitely on the set of kernels of nonzero locally nilpotent derivations of B.

Introduction

Given a field k of characteristic zero and a nonconstant polynomial φ ∈ k[Y], consider the k-algebra

B = k[X_{1}, X_{2}, Y]/(φ − X_{1}X_{2}),

where X_{1}, X_{2}, Y are indeterminates over k. The main result of this paper is the Transitivity Theorem (see 2.3), which asserts that a certain subgroup G of Aut_{k}(B) acts transitively on the set of kernels of locally nilpotent derivations 0 ≠ D : B → B.

Note that, in the special case where deg φ = 1, the Transitivity Theorem gives the well-known result [6] of R. Rentschler (see 2.4).

The Transitivity Theorem will be used, in the subsequent paper [2], to describe the locally nilpotent derivations of B and of other rings. In fact this paper is one step in a bigger project: Understanding the locally nilpotent derivations of the polynomial ring k[X, Y, Z]. In [3], Freudenburg described a way to modify a given locally nilpotent derivation of k[X, Y, Z] so as to obtain another one; he called this process the “local slice construction” and asked whether every locally nilpotent derivation D of k[X, Y, Z] could be constructed, starting from ∂/∂X, by finitely many applications of the local slice construction. Note that we show in [1] that this is indeed the case whenever D is w-homogeneous, with w(X), w(Y), w(Z) positive integers. It turns out that describing the locally nilpotent derivations of the two-dimensional ring B improves our understanding of the local slice construction in k[X, Y, Z], and in other rings as well, and this is our motivation for writing the present work. The connection between B and the local slice construction is not treated in this paper (it will be in [2]). Let us only mention that, for that application, we need the results of the present paper to be valid over a ground field k which is not necessarely algebraically closed.

2000 Mathematics Subject Classification. Primary: 14R10. Secondary: 14R20, 13N15.

Key words and phrases. Locally nilpotent derivations, additive group actions, Danielewski surfaces.

Research supported by a grant from NSERC Canada.

In writing this paper, our aim was not to give the shortest possible proof of the Transitivity Theorem, but rather to develop some tools that are needed in [2]. This is the case, in particular, for the bidegree function, whose properties are carefully established in paragraph 2.7.

Section 1 gathers the definitions and basic facts which we need. The Transitivity Theorem is proved in the second section. The purpose of the third section is to prove Lemma 3.1, which is used in the proof of 2.8. Note that section 3 is essentially an exercise in the theory of affine plane curves; that section is called an appendix to reflect the fact that its topic and methods are independent of the rest of the text.

Acknowledgements

Paper [4] gives generators for the group of automorphisms of B and [5] investigates automorphisms and locally nilpotent derivations of k[X_{1}, X_{2}, Y]/(φ(Y)−X_{1}^{k}X_{2}) where k > 1 (but we need the case k = 1). Although our results do not seem to follow from [4] and [5] as corollaries (at least we could not find how to do that), the two papers were still very helpful. In particular, the homogeneization technique used in 2.7.3 is borrowed from [5].

I also wish to mention that I recently received a note, from Makar-Limanov, which sketches a proof of the Transitivity Theorem.

1. Definitions and basic facts

1.1. Conventions.

• Throughout this paper, k denotes an arbitrary field of characteristic zero.

• The word ring means commutative ring with a unity element and of characteristic zero. The same convention applies to the words algebra and field. We stress that all fields and rings are tacitly assumed to be of characteristic zero.

• The group of units of a ring A is denoted A^{∗}.

• If A is a subring of a ring B and r ∈ ℕ, the notation B = A^{[r]} means that B is A-isomorphic to the polynomial ring in r variables over A. (So, B = A^{[r]} and B′ = A^{[r]} do not imply that B = B′.)

• If L/K is a field extension and r ∈ ℕ, the notation L = K^{(r)} means that L is a purely transcendental extension of K, of transcendence degree r.

• If A is an integral domain then Frac A is its field of fractions. If A ⊆ B are domains then trdeg_{A}(B) is the transcendence degree of Frac B over Frac A.

ON LOCALLY NILPOTENT DERIVATIONS OF k[X_{1}, X_{2}, Y]/(φ(Y) − X_{1}X_{2}) 3

1.2. An inert subring of a domain B is a subring A of B satisfying:

∀ x, y ∈ B, xy ∈ A \ {0} ⟹ x, y ∈ A.

If A is an inert subring of B then the following hold.

(1) A^{∗} = B^{∗}.

(2) A is algebraically closed in B.

(3) For each a ∈ A, a is irreducible in A if and only if it is irreducible in B.

(4) If B is a UFD then so is A.

(5) S^{−1}A is an inert subring of S^{−1}B, for any multiplicatively closed set S ⊆ A\{0}. 1.3. A locally nilpotent derivation of a domain B is a derivation D : B → B satisfying ∀_{x∈B} ∃_{e∈ℕ} D^{e}(x) = 0.

If D : B → B is any derivation, we write

ker D = {x ∈ B | D(x) = 0} ,

which is a subring of B. We use the following notations:

lnd(B) = set of nonzero locally nilpotent derivations of B
klnd(B) = {ker D | D ∈ lnd(B)} .

If E is a subset of B,

lnd_{E}(B) = {D ∈ lnd(B) | D(E) = {0}}
klnd_{E}(B) = {ker D | D ∈ lnd_{E}(B)} .

1.4. Basic properties of locally nilpotent derivations. Let B be an integral domain, let D : B → B be a nonzero derivation of B, and let A = ker D. The following facts are well-known.

(1) If D is locally nilpotent then A is an inert subring of B. In particular, B^{∗} = A^{∗} and if B is a UFD then so is A. Note, also, that if K is any field contained in B then K ⊆ A, so D is a K-derivation.

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

(3) Assume that ℚ ⊆ B. If D is locally nilpotent, and if s ∈ B satisfies D(s) ∈ B^{∗}, then B = A[s] = A^{[1]}.

(4) Assume that ℚ ⊆ B. If D is locally nilpotent, choose any s ∈ B such that Ds ≠ 0 and D^{2}s = 0 (such an s exists, and is called a preslice of D), and let S = {1, Ds, (Ds)^{2}, … } ⊂ A. Then S^{−1}D(s) ∈ (S^{−1}B)^{∗} so, by (3), S^{−1}B = (S^{−1}A)[s] = (S^{−1}A)^{[1]}.

(5) If D is locally nilpotent, let S = A \ {0}, then (4) implies S^{−1}B = (Frac A)^{[1]}.

(6) Let a ∈ B \ {0}. The derivation aD : B → B is locally nilpotent if and only if D is locally nilpotent and a ∈ A.

We stress the following consequence of part (5) of 1.4: 1.5. If B is a domain and A ∈ klnd(B) then trdeg_{A} B = 1. 1.6. If B = ⊕_{i∈ℤ} B_{i} is a ℤ-graded domain, a derivation D : B → B is homogeneous if there exists d ∈ ℤ satisfying ∀_{i∈ℕ} D(B_{i}) ⊆ B_{i+d}. If this is the case then ker D is a homogeneous subring of B, i.e., ker D = ⊕_{i∈ℤ}(ker(D) ∩ B_{i}).

2. The Transitivity Theorem

Throughout this section, B = k[X_{1}, X_{2}, Y]/(φ − X_{1}X_{2}) where X_{1}, X_{2}, Y are indeterminates over k and φ ∈ k[Y] is a polynomial of degree n ≥ 1. We also write B = k[x_{1}, x_{2}, y], where x_{1}, x_{2}, y are the images of X_{1}, X_{2}, Y under the canonical epimorphism k[X_{1}, X_{2}, Y] → B. 2.1. It is easy to see that, for each i ∈ {1, 2}, there is a unique locally nilpotent derivation D_{i} : B → B satisfying D_{i}(x_{i}) = 0 and D_{i}(y) = x_{i}. Moreover, ker D_{i} = k[x_{i}], so k[x_{1}] and k[x_{2}] are two elements of klnd(B). 2.2. Notations. (See [4].)

• Define τ ∈ Aut_{k}(B) by τ(x_{1}) = x_{2}, τ(x_{2}) = x_{1} and τ(y) = y.

• For each f ∈ k[x_{1}], define Δ_{f} ∈ Aut_{k}(B) by Δ_{f}(x_{1}) = x_{1} and Δ_{f}(y) = y + x_{1}f. (Then Δ_{f}(x_{2}) = x_{1}^{−1}φ(y + x_{1}f).)

• Let G be the subgroup of Aut_{k}(B) generated by {τ} ∪ {Δ_{f} | f ∈ k[x_{1}]}. (Perhaps one should call G the “tame subgroup” of Aut_{k}(B), see part (2) of 2.4.)

The assignment (α, A) ⟼ α(A), where α ∈ Aut_{k}(B) and A ∈ klnd(B), is a left-action of the group Aut_{k}(B) on the set klnd(B). We restrict this action to the subgroup G of Aut_{k}(B) defined in 2.2. Then: 2.3. Transitivity Theorem. The action of G on klnd(B) is transitive.

2.4. Remarks.

(1) For each f ∈ k[x_{1}], Δ_{f} is an exponential automorphism of B. In fact, Δ_{f} = e^{fD_{1}} where D_{1} is as in 2.1. Also, if f, g ∈ k[x_{1}] then Δ_{f} ∘ Δ_{g} = Δ_{f+g}.

(2) Consider the special case where φ(Y) = Y. Then B = k[x_{1}, x_{2}] = k^{[2]} and G is the tame subgroup of Aut_{k}(B), so the well-known result [6] of Rentschler is a corollary of 2.3.

The proof of the Transitivity Theorem is divided into several parts. Paragraphs 2.5– 2.7 cover some preliminaries; then we prove Proposition 2.8, which is a homogeneous version of the theorem; finally we prove the theorem.

Preliminaries to the proof of 2.3 (p, q)-gradings, filtrations and bidegree

Note that if φ, ψ ∈ k[Y] are associates, then

k[X_{1}, X_{2}, Y]/(φ − X_{1}X_{2}) ≅ k[X_{1}, X_{2}, Y]/(ψ − X_{1}X_{2}).

Thus we will assume throughout that φ ∈ k[Y] is a monic polynomial of degree n ≥ 1.

2.5. It is clear that B is a subalgebra of R = k[x_{1}, 1/x_{1}, y]. Given a pair (p, q) of relatively prime positive integers, we shall now define a graded ring B^{p,q} which is also a subalgebra of R, but which is usually distinct from B.

As a k-algebra, we set B^{p,q} = k[x_{1}, y^{n}/x_{1}, y], i.e.,

B^{p,q} ≅ k[X_{1}, X_{2}, Y]/(Y^{n} − X_{1}X_{2}).

In order to define the grading of B^{p,q}, consider first the ℤ-grading of R determined by the conditions

R = ⊕_{d∈ℤ}R_{d}, k ⊆ R_{0}, x_{1} ∈ R_{p}, y ∈ R_{q}.

Then note that B^{p,q} is a homogeneous subring of R, i.e., if we set

B_{d}^{p,q} = R_{d} ∩ B^{p,q}

for each d ∈ ℤ, then

B^{p,q} = ⊕_{d∈ℤ}B_{d}^{p,q}.

The ring B^{p,q} is always equipped with this grading. Note:

(1) The set {x_{1}^{i}y^{j} | (i, j) ∈ ℤ × ℕ and ni + j ≥ 0} is a basis of B^{p,q} over k.

We also define a map H_{p,q} : B → B^{p,q} as follows. Let f ∈ B. If f = 0, define H_{p,q}(f) = 0. If f ≠ 0, regard f as an element of the graded ring R = ⊕_{d∈ℤ} R_{d}, write

f = ∑_{d∈ℤ}f_{d} (f_{d} ∈ R_{d})

and define H_{p,q}(f) = f_{M} where M = max {d ∈ ℤ | f_{d} ≠ 0}. The fact that

(2)  H_{p,q}(f) ∈ B^{p,q}

can be proved directly, but we will obtain it as a trivial consequence of 2.6, below. Note that H_{p,q}(f) = 0 iff f = 0 and that H_{p,q}(fg) = H_{p,q}(f)H_{p,q}(g) for all f, g ∈ B.

2.6. Associated graded ring. Let B, R and (p, q) be as in 2.5 and let deg : R → ℤ∪ {−∞} denote the degree function associated with the grading ⊕_{d∈ℤ} R_{d} (i.e., deg(x_{1}^{i}y^{j}) = pi + qj). For each d ∈ ℤ, let

B_{(d)} = {f ∈ B | deg(f) ≤ d} ,

then

⋯ ⊆ B_{(d−1)} ⊆ B_{(d)} ⊆ B_{(d+1)} ⊆ ⋯

is a ℤ-filtration of B. Consider the associated graded ring and its “symbol map”

gr(B) = ⊕_{d∈ℤ}B_{(d)}/B_{(d−1)} and B ⟶^{gr} gr(B).

Since the degree function of R satisfies deg(fg) = deg(f) + deg(g) for all f, g ∈ R, it follows that gr(B) is an integral domain and that the symbol map preserves multiplication. Write

x̄_{1} = gr(x_{1}), x̄_{2} = gr(x_{2}), ȳ = gr(y)

and consider the subalgebra k[x̄_{1}, x̄_{2}, ȳ] of gr(B). 2.6.1. The k-homomorphism π : k[X_{1}, X_{2}, Y] → gr(B) defined by

π(X_{1}) = x̄_{1}, π(X_{2}) = x̄_{2}, π(Y) = ȳ

is surjective and has kernel (Y^{n} − X_{1}X_{2})k[X_{1}, X_{2}, Y].

Proof. The proof of surjectivity is straightforward and is left to the reader. Since x_{1}x_{2} = φ(y) holds (in B), and since by assumption φ ∈ k[Y] is monic of degree n, we have deg(x_{1}x_{2}) = deg(y^{n}) > deg(y^{n} − x_{1}x_{2}) and consequently x̄_{1}x̄_{2} = ȳ^{n} in gr(B). Thus the height one prime ideal (Y^{n} − X_{1}X_{2}) of k[X_{1}, X_{2}, Y] is included in ker(π). We know that gr(B) is a domain and (since π is surjective) a finitely generated k-algebra, so we will be done if we can show that x̄_{1}, ȳ are algebraically independent. Consider F ∈ k[X_{1}, Y] \ {0} and write F = ∑_{(i,j)∈S} a_{ij}X_{1}^{i}Y^{j} where S ≠ ∅ is a finite subset of ℕ^{2} and a_{ij} ∈ k^{∗} for all (i, j) ∈ S. It suffices to prove that F (x̄_{1}, ȳ) ≠ 0; moreover, we may assume that F is homogeneous in the sense that pi + qj is the same number d ∈ ℕ for all (i, j) ∈ S. Then by definition of the degree function of R we have

deg ( ∑_{(i,j)∈S} a_{ij}x_{1}^{i}y^{j}) = max {pi + qj | (i, j) ∈ S} = d,

so, in B, ∑_{(i,j)∈S} a_{ij}x_{1}^{i}y^{j} has the same degree as each one of its terms a_{ij}x_{1}^{i}y^{j}. This condition implies that

gr ( ∑_{(i,j)∈S} a_{ij}x_{1}^{i}y^{j}) = ∑_{(i,j)∈S} gr(a_{ij}x_{1}^{i}y^{j}),

where the right hand side is equal to F (x̄_{1}, ȳ) because the symbol map preserves multiplication, and where the left hand side is nonzero because ∑_{(i,j)∈S} a_{ij}x_{1}^{i}y^{j} is a nonzero element of B (x_{1}, y ∈ B are algebraically independent). So we are done. □

The above fact allows us to identify the graded ring gr(B) with the graded ring B^{p,q} defined in 2.5. Moreover, under that identification, the map gr : B → gr(B) corresponds to H_{p,q} : B → B^{p,q} (which proves the claim (2), in 2.5). 2.6.2. Every k-derivation D : B → B satisfies

sup {deg(Df) − deg(f) | f ∈ B \ {0}} < ∞.

Proof. Let D : B → B be a k-derivation. Since R is equal to the localization B_{x_{1}}, D extends to a k-derivation R → R. Since R is a graded domain and a finitely generated k-algebra, it is well-known that every k-derivation ∂ : R → R satisfies

sup {deg(∂f) − deg(f) | f ∈ R \ {0}} < ∞.

The desired result follows. □

Consequently, any nonzero k-derivation D : B → B determines a nonzero homogeneous k-derivation gr(D) : gr(B) → gr(B). Indeed, let

δ = sup {deg(Df) − deg(f) | f ∈ B \ {0}} ∈ ℤ.

Then, for each i ∈ ℤ, D maps B_{(i)} into B_{(i+δ)} and consequently there is an induced linear map D_{i} : B_{(i)}/B_{(i−1)} → B_{(i+δ)}/B_{(i+δ−1)}. These define a homogeneous linear map 0 ≠ gr(D) : gr(B) → gr(B) of degree δ which is easily seen to be a k-derivation. Given f ∈ B \ {0}, we have

gr(D)(gr f) = \begin{cases}gr(Df), & if deg(Df) = δ + deg(f), \\ 0, & if deg(Df) < δ + deg(f) \end{cases}

and it follows that gr(D) is locally nilpotent whenever D is, and that the symbol map B ⟶^{gr} gr(B) maps ker(D) \ {0} into ker(gr D) \ {0}. We may rephrase these facts as follows: 2.6.3. Each nonzero locally nilpotent derivation D : B → B determines a nonzero homogeneous locally nilpotent derivation D_{p,q} : B^{p,q} → B^{p,q}. Moreover, H_{p,q} : B → B^{p,q} maps ker(D) \ {0} into ker(D_{p,q}) \ {0}.

2.7. Bidegree. Recall that B is a subalgebra of R = k[x_{1}, 1/x_{1}, y].

Given f ∈ R, we may write

f = ∑_{(i,j)∈ℤ×ℕ}λ_{i j}x_{1}^{i}y^{j} (λ_{i j} ∈ k)

and define the finite set

supp(f) = {(i, j) ∈ ℤ × ℕ | λ_{i j} ≠ 0} .

Then we define the “bidegree” map

\begin{matrix}bideg : R & ⟶ & ℕ × ℕ \\ f & ⟼ & (a, b) \end{matrix}

by declaring that a and b are the following integers:

a = max [{0} ∪ {i ∈ ℕ | (i, 0) ∈ supp(f)} ]
b = max [{0} ∪ {j ∈ ℕ | j(−1, n) ∈ supp(f)} ].

Warning. If f, g ∈ R, then bideg(fg) is not always equal to bideg f + bideg g: Let f = (x_{1}^{−2} + 1)y^{n} ∈ R, then bideg f = (0, 0) and bideg(f^{2}) = (0, 2). So the properties of the bidegree function require careful proofs.

Note that there is an important case where the formula bideg(f^{N}) = N bideg f is valid: See 2.7.4. 2.7.1. If P ∈ k[T] \ {0} has degree d, then

bideg P (x_{1}) = (d, 0) and bideg P (x_{2}) = (0, d).

Proof. Left to the reader. The second equality requires a little bit of work. □

2.7.2. Let f ∈ R and (a, b) = bideg f. Then:

a > 0 ⟹ (a, 0) ∈ supp f and b > 0 ⟹ (−b, bn) ∈ supp f.

Proof. Trivial. □

Given f ∈ R, let C(f) be the unique subset of ℝ^{2} which is closed, convex and has boundary H_{a} ∪ E ∪ H_{b}, where (a, b) = bideg(f) and:

E = line segment joining (−b, bn) and (a, 0),
H_{a} = horizontal half line {(s, 0) | s ≤ a} ,
H_{b} = horizontal half line {(s, bn) | s ≤ −b} .

Note in particular:

(3)  ∀ (i, j) ∈ supp(f), (j = 0 or ni + j = 0) ⟹ (i, j) ∈ C(f).

A better statement is true in the special case where f ∈ A ∈ klnd(B): 2.7.3. Lemma. If f ∈ A for some A ∈ klnd(B), then supp(f) ⊂ C(f).

Proof. Suppose the contrary and consider a point P = (i_{0}, j_{0}) in supp(f) \ C(f). Let D ∈ lnd(B) be such that ker D = A.

Because supp(f) is a finite set, only finitely many r ∈ ℚ satisfy:

(4)  Some line of slope r contains more than one point of supp(f).

On the other hand, the shape of C(f) and the fact that P is is a point of

{(s, t) ∈ ℝ^{2} | t ≥ 0} \ C(f)

make it clear that there exist infinitely many r ∈ ℚ such that

(5) r < 0 and the line going through P and of slope r is disjoint from C(f).

So we may choose r ∈ ℚ satisfying (5) but not (4). Let p, q be relatively prime positive integers such that r = −p/q and let deg : k[x_{1}, 1/x_{1}, y] → ℤ ∪ {−∞} be the degree function given by deg(x_{1}) = p and deg(y) = q (as in 2.6). Then

(6)  ∀ (i, j) ∈ ℤ^{2} ∩ C(f)  deg(x_{1}^{i}y^{j}) < deg(x_{1}^{i_{0}}y^{j_{0}}).

Consider H_{p,q} : B → B^{p,q} and D_{p,q} : B^{p,q} → B^{p,q} (see 2.5, 2.6) and recall that D_{p,q} is locally nilpotent and nonzero. Let h = H_{p,q}(f), then h ∈ ker(D_{p,q}) \ {0} by 2.6.3. Since r = −p/q does not satisfy (4), supp(h) is a single point, so h = μx_{1}^{i_{1}}y^{j_{1}} for some μ ∈ k^{∗} and (i_{1}, j_{1}) ∈ supp(f). Since deg(x_{1}^{i_{1}}y^{j_{1}}) = deg(h) = deg(f) ≥ deg(x_{1}^{i_{0}}y^{j_{0}}), (6) implies that (i_{1}, j_{1}) ∉ C(f), so (3) gives

j_{1} > 0 and ni_{1} + j_{1} > 0

(note that j_{1} and ni_{1} + j_{1} cannot be negative because of (1) together with h ∈ B^{p,q}). Now this and (1) imply that x_{1}^{i_{1}}y^{j_{1}−1} ∈ B^{p,q}. Since (x_{1}^{i_{1}}y^{j_{1}−1})y ∈ ker(D_{p,q}), it follows that y ∈ ker(D_{p,q}) because ker(D_{p,q}) is an inert subring of B^{p,q} (see 1.4); then x_{1}(y^{n}/x_{1}) = y^{n} ∈ ker(D_{p,q}), so x_{1} ∈ ker(D_{p,q}) for the same reason. But this is impossible because D_{p,q} ≠ 0 and x_{1}, y generate Frac B^{p,q} over k. □

2.7.4. Lemma. If f ∈ A for some A ∈ klnd(B), then bideg(f^{N}) = N bideg(f) holds for all positive integers N.

Proof. By 2.7.3, it suffices to show that the formula is valid for all f ∈ R satisfying

(7)  supp(f) ⊂ C(f).

So consider f ∈ R satisfying (7) and let N be a positive integer. Let (a, b) = bideg(f) and (a′, b′) = bideg(f^{N}). Since

a = max (0, deg_{x_{1}} f(x_{1}, 0)) and a′ = max (0, deg_{x_{1}}(f(x_{1}, 0)^{N})),

it is clear that a′ = Na. So it suffices to show that b′ = Nb.

If b = 0 then f ∈ k[x_{1}, x_{1}^{−1}] by (7), so f^{N} ∈ k[x_{1}, x_{1}^{−1}] and consequently b′ = 0. So b′ = Nb holds true in this case.

So we may assume that b > 0. By 2.7.2 we have (−b, bn) ∈ supp f, so (7) gives

(8)  supp f = {(−b, bn)} ∪ S ∪ T,

where S = {(i, j) ∈ supp f | ni + j < 0 and j = nb} and T = {(i, j) ∈ supp f | j < nb}. Observe that if (α, β) is any element of supp(f^{N}), then (α, β) = u_{1} + ⋯ + u_{N} for some elements u_{1}, …, u_{N} of supp f. So, to complete the proof, it suffices to show:

(9)

If u_{1}, …, u_{N} ∈ supp f satisfy u_{1} + ⋯ + u_{N} = (−k, kn) for some k ≥ Nb, then

u_{ν} = (−b, bn) for all ν.

Write u_{ν} = (i_{ν}, j_{ν}) for all ν. Then j_{ν} ≤ nb for all ν by (8), so the condition k ≥ Nb implies that no u_{ν} can belong to T:

u_{1}, …, u_{N} ∈ {(−b, bn)} ∪ S.

So we have ni_{ν} + j_{ν} ≤ 0 for all ν and

∑_{ν=1}^{N}(ni_{ν} + j_{ν}) = n ∑_{ν=1}^{N} i_{ν} + ∑_{ν=1}^{N} j_{ν} = n(−k) + kn = 0

gives ni_{ν} + j_{ν} = 0 for all ν. So no u_{ν} can belong to S and we are done. □

2.7.5. Lemma. Suppose that f ∈ A for some A ∈ klnd(B) and let (a, b) = bideg f.

(1) a = 0 ⇔ f ∈ k[x_{2}] and b = 0 ⇔ f ∈ k[x_{1}].

(2) Assuming that f ∉ k, a = 0 ⇔ A = k[x_{2}] and b = 0 ⇔ A = k[x_{1}].

Proof. Clearly, it suffices to prove assertion (1). Moreover, the two implications “⇐” are trivial.

If b = 0 then the line segment E joins (0, 0) to (a, 0), so C(f) = (−∞, a] × {0}; by 2.7.3, each (i, j) ∈ supp(f) satisfies j = 0, so f ∈ k[x_{1}, x_{1}^{−1}]. Since k[x_{1}] ∈ klnd(B), k(x_{1}) ∩ B = k[x_{1}]. So f ∈ k[x_{1}].

If a = 0 then the line segment E joins (−b, bn) to the origin, so by 2.7.3

(10)  each (i, j) ∈ supp(f) satisfies ni + j ≤ 0.

Let us argue that f ∈ k[x_{2}].

Given d ∈ ℕ, let H_{d} be the set of elements of B which can be written in the form

(11)  ∑_{i=0}^{d−1} h_{i}^{0}(y)x_{1}^{d−i} + ∑_{j=0}^{e} h_{d}^{j}(y)x_{2}^{j}

for some e ∈ ℕ and for some h_{i}^{j} ∈ k[Y] satisfying deg_{Y} (h_{i}^{j}) ≤ ni, where n = deg_{Y} (φ). Observe that H_{0} = k[x_{2}].

By (10), f = ∑_{i=0}^{r} h_{i}^{0}(y)x_{1}^{−i} for some r ∈ ℕ and for some h_{i}^{0} ∈ k[Y] satisfying deg_{Y} (h_{i}^{0}) ≤ ni. Thus

x_{1}^{r}f = ∑_{i=0}^{r} h_{i}^{0}(y)x_{1}^{r−i} = ∑_{i=0}^{r−1} h_{i}^{0}(y)x_{1}^{r−i} + ∑_{j=0}^{0} h_{r}^{j}(y)x_{2}^{j},

i.e., the condition

(12)  x_{1}^{d}f ∈ H_{d}

is true for d = r. We show by descending induction that (12) is true for all d ∈ {0, 1, …, r}. Assume that d ≥ 0 is such that x_{1}^{d+1}f ∈ H_{d+1}, i.e.,

(13)  x_{1}^{d+1}f = ∑_{i=0}^{d} h_{i}^{0}(y)x_{1}^{d+1−i} + ∑_{j=0}^{e} h_{d+1}^{j}(y)x_{2}^{j}.

Then ∑_{j=0}^{e} h_{d+1}^{j}(y)x_{2}^{j} ∈ x_{1}B, or equivalently:

∑_{j=0}^{e} h_{d+1}^{j}X_{2}^{j} ∈ (X_{1}, φ − X_{1}X_{2})k[X_{1}, X_{2}, Y] = (X_{1}, φ)k[X_{1}, X_{2}, Y],

which implies that φ | h_{d+1}^{j} in k[Y] for all j ∈ {0, 1, …, e}. Write h_{d+1}^{j} = φh_{d}^{j+1} (for 0 ≤ j ≤ e) and notice that deg_{Y} (h_{d+1}^{j}) ≤ (d + 1)n implies deg_{Y} (h_{d}^{j+1}) ≤ dn. We have h_{d+1}^{j}(y)x_{2}^{j} = φ(y)h_{d}^{j+1}(y)x_{2}^{j} = h_{d}^{j+1}(y)x_{1}x_{2}^{j+1}, so, dividing (13) by x_{1} yields

(14) x_{1}^{d}f = ∑_{i=0}^{d} h_{i}^{0}(y)x_{1}^{d−i} + ∑_{j=0}^{e} h_{d}^{j+1}(y)x_{2}^{j+1}

= ∑_{i=0}^{d−1} h_{i}^{0}(y)x_{1}^{d−i} + h_{d}^{0}(y) + ∑_{j=0}^{e} h_{d}^{j+1}(y)x_{2}^{j+1} = ∑_{i=0}^{d−1} h_{i}^{0}(y)x_{1}^{d−i} + ∑_{j=0}^{e+1} h_{d}^{j}(y)x_{2}^{j},

i.e., x_{1}^{d}f ∈ H_{d}. Thus by induction (12) holds for all d ∈ {0, 1, …, r}; in particular, it holds for d = 0, which means that f ∈ H_{0} = k[x_{2}]. □

See 2.2 for the element τ of G. 2.7.6. Lemma. Suppose that f ∈ A for some A ∈ klnd(B) and let (a, b) = bideg(f). Then bideg (τ(f)) = (b, a).

Proof. We may assume that f ≠ 0. If a = 0 then, by 2.7.5 and 2.7.1, f = P (x_{2}) for some P ∈ k[T] with deg_{T} (P) = b. Then τ(f) = P (x_{1}) has bidegree (b, 0) = (b, a) by 2.7.1. Similarly, if b = 0 then the result is trivial. So we may assume that

a, b > 0.

Consider the embeddings B ⊂ R = k[x_{1}, 1/x_{1}, y] ⊂ R′ = K[x_{1}, 1/x_{1}], where K = k((y^{−1})) = Frac k[[y^{−1}]]. Note that τ ∈ Aut_{k}(B) extends to τ ∈ Aut_{K}(R′).

Define an equivalence relation ∼ on R′ \ {0} by declaring that

α ∼ β if and only if α/β ∈ k[[y^{−1}]]^{∗}.

This relation is useful because:

(15) ∀\begin{matrix}β∈R′\\{0\} \\ i,j∈ℤ \end{matrix}  β ∼ x_{1}^{i}y^{j} ⟺ (i, j) ∈ supp(β) ⊂ \{i\} × (−∞, j].

Note that ∼ is a congruence for multiplication, i.e., if α_{1} ∼ α_{2} and β_{1} ∼ β_{2} then α_{1}β_{1} ∼ α_{2}β_{2}. Also, let ξ_{2} = y^{n}/x_{1}; then clearly

x_{2} ∼ ξ_{2}.

Results 2.7.2 and 2.7.3 and the fact that b > 0 imply:

(−b, nb) ∈ supp(f) ⊆ {(−b, nb)} ∪ I,

where I = {(i, j) ∈ ℤ × ℕ | (j = nb and i < −b) or j < nb }; thus we have

(16)  f = λ(x_{1}^{−1}y^{n})^{b} + ∑_{(i,j)∈I} λ_{ij}x_{1}^{i}y^{j},

where λ ∈ k^{∗} and λ_{ij} ∈ k. Similarly,

(a, 0) ∈ supp(f) ⊆ {(a, 0)} ∪ I′,

where I′ = {(i, j) ∈ ℤ × ℕ | i < a and nbi + (a + b)j ≤ nab }; so we may also write f as

(17)  f = μx_{1}^{a} + ∑_{(i,j)∈I′} μ_{ij}x_{1}^{i}y^{j},

where μ ∈ k^{∗} and μ_{ij} ∈ k. Applying τ to (16) and (17) gives

(18) τ(f) = λ(x_{2}^{−1}y^{n})^{b} + ∑_{(i,j)∈I} λ_{ij}x_{2}^{i}y^{j}  and τ(f) = μx_{2}^{a} + ∑_{(i,j)∈I′} μ_{ij}x_{2}^{i}y^{j}.

Note that

(19)  (x_{2}^{−1}y^{n})^{b} ∼ (ξ_{2}^{−1}y^{n})^{b} = x_{1}^{b},  x_{2}^{a} ∼ ξ_{2}^{a} = x_{1}^{−a}y^{na}

and

(20)  x_{2}^{i}y^{j} ∼ (ξ_{2}^{i}y^{j}) = x_{1}^{−i}y^{ni+j}.

Let us now verify that

(21)  ∀ (i, j) ∈ I,  −i < b or ni + j < 0.

Assume that (i, j) ∈ I and that −i ≥ b; we have to show that ni + j < 0. By definition of I, there are two cases to consider: (i) If j = nb and i < −b, then ni+j < −nb+j = 0. (ii) If j < nb, then

−i ≥ b ⇒ ni ≤ −nb ⇒ ni + j ≤ −nb + j < 0.

This proves (21). Similarly, one can prove

(22)  ∀ (i, j) ∈ I′,  ni + j < na.

For each (i, j) ∈ I we have

(s, t) ∈ supp(x_{2}^{i}y^{j}) ⟹ s = −i and t ≤ ni + j  by (20) and (15)
⟹ s < b or t < 0  by (21).

Hence,

(23)  (s, t) ∈ supp ( ∑_{(i,j)∈I} λ_{ij}x_{2}^{i}y^{j}) ⟹ s < b or t < 0.

On the other hand, (19) and (15) imply that (b, 0) ∈ supp [(x_{2}^{−1}y^{n})^{b}]; this, (23) and (18) imply that the first component of bideg(τ(f)) is b.

Using (22) in place of (21), a similar argument shows that

(s, t) ∈ supp ( ∑_{(i,j)∈I′} μ_{ij}x_{2}^{i}y^{j}) ⟹ t < na

and it follows that the second component of bideg(τ(f)) is a. □

2.7.7. To compare bidegrees, let ℕ×ℕ be endowed with the reverse lexicographic order, i.e.,

(a, b) < (a′, b′) ⟺ b < b′ or (b = b′ and a < a′).

Since this is a well-ordering of ℕ × ℕ, given A ∈ klnd(B) it makes sense to define

bideg(A) = min {bideg f | f ∈ A \ k} ∈ ℕ × ℕ.

Homogeneous case

The following result is needed in the proof of 2.3. It is concerned with the special case where φ = Y^{n}. 2.8. Proposition. Let B = k[X_{1}, X_{2}, Y]/(Y^{n} − X_{1}X_{2}), where n ≥ 1, and write B = k[x_{1}, x_{2}, y] where x_{1}, x_{2}, y are the images of X_{1}, X_{2}, Y via the canonical epimorphism k[X_{1}, X_{2}, Y] → B. Let p, q be relatively prime positive integers and let B be endowed with the (p, q)-grading:

B = ⊕_{d∈ℤ}B_{d}, k ⊆ B_{0}, x_{1} ∈ B_{p}, y ∈ B_{q}.

Suppose that D ≠ 0 is a homogeneous locally nilpotent derivation of B whose kernel is neither k[x_{1}] nor k[x_{2}]. Then there exists a homogeneous and irreducible element h of ker D and, given any such h, we have ker D = k[h] and one of the following holds:

(1) For some λ, μ ∈ k^{∗}, h = μx_{1}^{−1}(y − λx_{1}^{q})^{n}

(2) For some λ, μ ∈ k^{∗}, h = μx_{2}^{−1}(y − λx_{2}^{q})^{n}.

2.9. Proof of 2.8. 2.9.1. Set-up. Throughout 2.9, B = k[x_{1}, x_{2}, y] and p, q are as in the statement of 2.8 (so x_{1}x_{2} = y^{n}) and we regard B as a subalgebra of R = k[x_{1}, x_{1}^{−1}, y]. Let R = ⊕_{d∈ℤ} R_{d} be the (p, q)-grading, as in 2.5, and let deg : R → ℤ ∪ {−∞} denote the corresponding degree function (deg(x_{1}) = p and deg(y) = q, so deg(x_{2}) = nq − p). Note that B_{d} = B ∩ R_{d} for all d ∈ ℤ.

2.9.2. We define a valuation

ord : k(x_{1}, y) → ℤ ∪ {∞}

as follows. As a first step, define a map ord : R → ℤ ∪ {∞} by

ord(f) = \begin{cases}min \{ni + j | (i, j) ∈ supp(f)\} , & if f ∈ R \ \{0\} \\ ∞, & if f = 0 \end{cases}

and note that, for all f, g ∈ R, the following conditions hold:

• ord(f) = ∞ ⟺ f = 0

• ord(fg) = ord(f) + ord(g)
• ord(f + g) ≥ min(ord(f), ord(g)).

Then extend ord to a valuation of k(x_{1}, y). Note that ord(x_{1}) = n, ord(y) = 1, ord(x_{2}) = 0 and:

(24)  B = {f ∈ R | ord(f) ≥ 0} .

2.9.3. Lemma. Suppose that f_{1}, …, f_{k} ∈ R satisfy ord(f_{1}) = ⋯ = ord(f_{k}). (1) If ∏_{i=1}^{k} f_{i} belongs to B then f_{1}, …, f_{k} ∈ B.

(2) If ∏_{i=1}^{k} f_{i} belongs to A (for some A ∈ klnd(B)) then f_{1}, …, f_{k} ∈ A.

Proof. Assertion (1) is an immediate consequence of (24). Assertion (2) follows from (1) and the fact that A is an inert subring of B. □

2.9.4. If D ∈ lnd(B) is (p, q)-homogeneous, then there exists a homogeneous and irreducible element h of ker D.

Proof. Let A = ker(D), then A is a homogeneous subring of B and B has transcendence degree one over A. So we can choose a homogeneous element H of A = ker(D) which is neither zero nor a unit. Since B is affine over k, there exists a factorization of H into irreducible elements of B. Let h ∈ B be an irreducible factor of H; then h is homogeneous, and h ∈ A since A is inert in B. Finally, h is irreducible in A because A^{∗} = B^{∗}. □

2.9.5. Reduction. If 2.8 holds in the special case where k is algebraically closed, then it holds in general.

Proof. Suppose that D ≠ 0 is a (p, q)-homogeneous locally nilpotent derivation of B whose kernel is neither k[x_{1}] nor k[x_{2}]. Let k̄ be an algebraic closure of k and define B̄ = k̄ ⊗_{k} B. Note that B̄ ≅ k̄[X_{1}, X_{2}, Y]/(Y^{n} − X_{1}X_{2}) and let D̄ : B̄ → B̄ be the unique extension of D to a k̄-derivation of B̄. Then D̄ is locally nilpotent, (p, q)- homogeneous and such that ker D̄ is neither k̄[x_{1}] nor k̄[x_{2}]. Thus 2.8 can be applied to D̄: There exists a homogeneous and irreducible element h̄ of ker D̄, this h̄ satisfies

ker D̄ = k̄[h̄]

and one of the following holds:

(25)  h̄ = μx_{1}^{−1}(y − λx_{1}^{q})^{n}, for some λ, μ ∈ k̄^{∗},
(26)  h̄ = μx_{2}^{−1}(y − λx_{2}^{q})^{n}, for some λ, μ ∈ k̄^{∗}.

We may assume that μ = 1 in (25) and (26), because ker D̄ = k̄[μ^{−1}h̄]. Then we claim that h̄ ∈ B. To prove this, we first observe that if (25) (resp. (26)) holds then deg(h̄) = (qn − 1) deg(x_{1}) (resp. deg(h̄) = (qn − 1) deg(x_{2})); since h̄ is homogeneous, we also have deg(y) = deg(x_{1}^{q}) (resp. deg(y) = deg(x_{2}^{q})), so deg(x_{1}) = 1 (resp. deg(x_{2}) = 1). Hence, deg(h̄) = nq − 1. We consider two cases.

If deg(h̄) = 0 then n = 1 = q, so (25) (resp. (26)) simplifies to h̄ = x_{2} − λ (resp. h̄ = x_{1} − λ), which is impossible since ker D̄ is neither k̄[x_{1}] nor k̄[x_{2}].

So we must have deg(h̄) > 0. Let h be a homogeneous and irreducible element of ker D (such an h exists by 2.9.4). Since both h and h̄ are homogeneous, the condition deg(h̄) > 0 implies:

h = ϵh̄^{N}

for some ϵ ∈ k̄^{∗} and some positive integer N.

Suppose that (25) holds. Then h ∈ k[x_{1}, x_{1}^{−1}, y] and

h = ϵx_{1}^{−N}(y − λx_{1}^{q})^{nN} = ϵx_{1}^{−N}[y^{nN} − Nnλx_{1}^{q}y^{nN−1} + ∑_{i=2}^{nN} \binom{nN}{i}y^{nN−i}(−λx_{1}^{q})^{i}]

= ϵx_{1}^{−N}y^{nN} − Nnϵλx_{1}^{q−N}y^{nN−1} + ⋯,

so ϵ ∈ k and consequently λ ∈ k. Thus h̄ ∈ B.

If (26) holds then, by a similar argument, h̄ ∈ B.

Finally, it is easy to see that h̄ ∈ B implies that ker D = k[h̄]. This proves 2.9.5. □

2.9.6. From now-on, we assume that k is algebraically closed.

We now proceed to prove 2.8, under assumption 2.9.6.

Consider a (p, q)-homogeneous derivation D ∈ lnd(B), let A = ker D and suppose that A is neither k[x_{1}] nor k[x_{2}]. Let h be a homogeneous irreducible element of A and define

(a, b) = bideg(h).

Then min(a, b) > 0 by 2.7.5 and, by 2.7.2,

(a, 0), (−b, bn) ∈ supp(h).

Since h is (p, q)-homogeneous we must have deg(x_{1}^{a}) = deg(x_{1}^{−b}y^{nb}), so

(27)  p = \frac{nb}{δ} and q = \frac{a + b}{δ}, where δ = gcd(nb, a + b).

Homogeneity of h also implies that the convex hull of supp(h) is the line segment joining (−b, nb) and (a, 0); so the convex hull of supp(x_{1}^{b}h) is the line segment joining (0, nb) and (a + b, 0). Thus x_{1}^{b}h ∈ k[x_{1}, y] and, using the assumption that k is algebraically closed,

x_{1}^{b}h = μ ∏_{i=1}^{δ}(y^{p} − λ_{i}x_{1}^{q})

for some μ, λ_{1}, …, λ_{δ} ∈ k^{∗}. Equivalently,

(28)  h = μx_{1}^{−b} ∏_{i=1}^{δ}(y^{p} − λ_{i}x_{1}^{q}).

Then we have

A ∋ h^{δ} = (μx_{1}^{−b})^{δ} ∏_{i=1}^{δ}(y^{p} − λ_{i}x_{1}^{q})^{δ} = ∏_{i=1}^{δ} μx_{1}^{−b}(y^{p} − λ_{i}x_{1}^{q})^{δ} = ∏_{i=1}^{δ} h_{i},

where h_{i} = μx_{1}^{−b}(y^{p} − λ_{i}x_{1}^{q})^{δ}. Note that ord(h_{i}) = 0 for all i; so 2.9.3 implies that h_{1}, …, h_{δ} all belong to A. We claim:

(29)  h_{1} = ⋯ = h_{δ}.

Indeed, it is easy to see that ord(h_{i} − h_{j}) ≥ 1, so

ord ((h_{i} − h_{j})^{n}) ≥ n ⟹ ord (x_{1}^{−1}(h_{i} − h_{j})^{n}) ≥ 0 ⟹ x_{1}^{−1}(h_{i} − h_{j})^{n} ∈ B,

i.e., x_{1} divides (h_{i} − h_{j})^{n} in B. Assuming that h_{i} ≠ h_{j}, we have the two conditions

(h_{i} − h_{j})^{n} ∈ A \ {0} and x_{1} | (h_{i} − h_{j})^{n} in B,

and these imply that x_{1} ∈ A, contradicting the assumption A ≠ k[x_{1}].

This proves (29); thus (28) simplifies to

(30)  h = μx_{1}^{−b}(y^{p} − λx_{1}^{q})^{δ} for some μ, λ ∈ k^{∗}.

2.9.7. gcd(b, δ) = 1

Proof. Let d = gcd(b, δ) and let h_{0} = x_{1}^{−b/d}(y^{p} −λx_{1}^{q})^{δ/d}. Then h_{0} ∈ R and h_{0}^{d} = μ^{−1}h ∈ A imply (by 2.9.3) that h_{0} ∈ A. Since h is irreducible in A, it follows that d = 1. □

2.9.8. A = k[h] Proof. By (27) and 2.9.7, (b, δ) is the unique pair of relatively prime positive integers satisfying b/δ = p/n. So (b, δ) is independent of our choice of h.

Consequently, if h′ is any homogeneous irreducible element of A then, by (30) applied to h′, we have h′ = νx_{1}^{−b}(y^{p} − ϵx_{1}^{q})^{δ} for some ν, ϵ ∈ k^{∗}. Clearly, the proof of (29) also proves that ν^{−1}h′ = μ^{−1}h. Assertion 2.9.8 follows. □

2.9.9. gcd(a, b) = 1 and b | p.

Proof. gcd(a, b) = gcd(b, a + b) = gcd(b, nb, a + b) = gcd(b, gcd(nb, a + b)) = gcd(b, δ) = 1. Since δp = nb, we have b | δp; so the assertion follows from gcd(b, δ) = 1. □

Define the positive integer k = p/b and rewrite (30) as h = μx_{1}^{−b}(y^{kb} − λx_{1}^{q})^{δ}. In order to prove 2.8, there remains to show that h satisfies one of conditions (1), (2) displayed in the statement of 2.8. For that purpose, we may assume that μ = 1. So we write h as

(31)  h = x_{1}^{−b}(y^{kb} − λx_{1}^{q})^{δ}, for some λ ∈ k^{∗}.

Define

K = Frac A = k(h).

Since A ∈ klnd(B), 1.4 implies that Frac B = (Frac A)^{(1)}, i.e.,

(32)  k(x_{1}, y) = K^{(1)}.

On the other hand, let us prove the following: 2.9.10. k(x_{1}, y) is K-isomorphic to the field of fractions of

K[T, Y]/(Y^{kb} + T^{b}(λ_{1}T^{a} + μ_{1}))

for some μ_{1}, λ_{1} ∈ K^{∗} (where T and Y are indeterminates over K).

Proof. By 2.9.7, there exist u, v ∈ ℤ such that δu + bv = 1. Define

(33)  w = y^{kb} − λx_{1}^{q} and t = w^{v}x_{1}^{u}.

Now (31) gives h = x_{1}^{−b}w^{δ} and the following are easily obtained:

t^{δ} = w^{vδ}x_{1}^{uδ} = w^{vδ}x_{1}^{1−bv} = x_{1}(w^{δ}x_{1}^{−b})^{v} = x_{1}h^{v},
t^{b} = w^{vb}x_{1}^{ub} = w^{1−δu}x_{1}^{ub} = w(w^{−δ}x_{1}^{b})^{u} = wh^{−u},

which we rewrite as

(34)  x_{1} = h^{−v}t^{δ} and w = h^{u}t^{b}.

In particular we have x_{1} ∈ K(t, y), so

(35)  k(x_{1}, y) = K(t, y).

By (33), (34) and the fact that qδ = a + b,

y^{kb} = w + λx_{1}^{q} = (h^{u})t^{b} + (λh^{−vq})t^{a+b} = t^{b}((λh^{−vq})t^{a} + h^{u}),

which gives

(36)  y^{kb} + t^{b}(λ_{1}t^{a} + μ_{1}) = 0,

where λ_{1} = −λh^{−vq} ∈ K^{∗} and μ_{1} = −h^{u} ∈ K^{∗}. Finally, note that Y^{kb} + T^{b}(λ_{1}T^{a} + μ_{1}) is an irreducible polynomial in K[Y, T] (this is proved in the Appendix, see 3.1). So 2.9.10 follows from (35) and (36). □

In view of 2.9.10 and of the fact, observed in (32), that k(x_{1}, y) is rational over K, 3.1 implies that the triple (k, a, b) satisfies one of the following conditions:

(a) k = 1 and min(a, b) = 1

(b) k = 1 and a = 2 = b

(c) k = 2 and a = 1 = b. However, we have gcd(a, b) = 1 by 2.9.9, so (b) cannot hold. We shall now see that (c) cannot hold either and that (a) gives the desired result. 2.9.11. If (k, a, b) satisfies (a), then h satisfies one of conditions (1), (2) displayed in the statement of 2.8.

Proof. Assume that (a) holds and recall that k = p/b = n/δ. Thus k = 1 gives δ = n and, by (31),

(37)  h = x_{1}^{−b}(y^{b} − λx_{1}^{q})^{n}.

We have min(a, b) = 1, so there are two cases to consider. If b = 1 then (37) shows that h satisfies condition (1) of 2.8. Suppose now that a = 1. Keeping in mind that b + 1 = a + b = qδ = qn, we manipulate (37) as follows:

h = x_{1}^{−b}(y^{b} − λx_{1}^{q})^{n} = x_{1}^{−b}y^{−n}(y^{b+1} − λx_{1}^{q}y)^{n} = x_{1}^{−b}y^{−n}(y^{qn} − λx_{1}^{q}y)^{n}

= x_{1}^{−b}y^{−n}((x_{1}x_{2})^{q} − λx_{1}^{q}y)^{n} = x_{1}^{−b}x_{1}^{qn}y^{−n}(x_{2}^{q} − λy)^{n} = x_{1}^{−b}x_{1}^{qn}(x_{1}x_{2})^{−1}(x_{2}^{q} − λy)^{n}

= x_{2}^{−1}(x_{2}^{q} − λy)^{n} = (−λ)^{n}x_{2}^{−1}(y − λ^{−1}x_{2}^{q})^{n},

so h satisfies condition (2) of 2.8. So 2.9.11 is proved. □

Finally we show that (c) cannot hold.

Assume that (c) holds. Then 2 = k = n/δ, so n = 2δ; on the other hand, δ = gcd(nb, a + b) = gcd(n, 2) = gcd(2δ, 2) = 2, so n = 2δ = 4 and q = (a + b)/δ = 1. Hence, in view of (31),

h = x_{1}^{−1}(y^{2} − λx_{1})^{2} and n = 4.

Define ω ∈ Aut_{k}(B) by ω(x_{1}) = −λ^{−1}x_{1}, ω(x_{2}) = −λx_{2} and ω(y) = y; then ω(h) = −λx_{1}^{−1}(y^{2} + x_{1})^{2}. In other words, it suffices to prove the following statement:

2.9.12. Let h = x_{1}^{−1}(y^{2} + x_{1})^{2} and assume that n = 4. Then k[h] is not the kernel of a locally nilpotent derivation of B.

Proof. Suppose that k[h] ∈ klnd(B). Then, by part (4) of 1.4, there exists α ∈ k[h] \ {0} such that B_{α} = (k[h]_{α})^{[1]}. This implies:

(38)  B/(h − μ)B ≅ k^{[1]} for general μ ∈ k.

However, let us compute B/(h − μ)B. Recall that n = 4, so x_{1}x_{2} = y^{4} and

h = x_{1}^{−1}(y^{2} + x_{1})^{2} = x_{1}^{−1}y^{4} + 2y^{2} + x_{1} = x_{2} + 2y^{2} + x_{1}.

Consequently,

B/(h − μ)B ≅ k[X_{1}, X_{2}, Y]/(X_{2} + 2Y^{2} + X_{1} − μ, Y^{4} − X_{1}X_{2})

≅ k[X_{1}, Y]/(Y^{4} + X_{1}(2Y^{2} + X_{1} − μ)) ≅ k[X, Y]/(F),

where we define F = Y^{4} + 2XY^{2} + X^{2} − μX = (Y^{2} + X)^{2} − μX. It is easy to see that k[X, Y]/(F) is not isomorphic to k^{[1]}. One way to see it is to consider θ ∈ Aut_{k} k[X, Y] defined by θ(X) = X −Y^{2} and θ(Y) = Y, and to observe that θ(F) = X^{2} +μY^{2} −μX. Since X^{2} + μY^{2} is not a power of a linear form, we see that k[X, Y]/(F) has two places at infinity, and consequently is not isomorphic to k^{[1]}. This contradicts (38), so 2.9.12 is proved. □

The proof of 2.8 is now complete. □

Proof of the Transitivity Theorem

We return to the general assumptions of this section, i.e.,

B = k[X_{1}, X_{2}, Y]/(φ − X_{1}X_{2}),

where φ ∈ k[Y] is a nonconstant monic polynomial; let n = deg_{Y} (φ) and write B = k[x_{1}, x_{2}, y] as usual.

Recall the map bideg : B → ℕ×ℕ defined in 2.7. From 2.7.7, recall that ℕ×ℕ is wellordered, with the reverse lexicographic order; and that we defined bideg(A) ∈ ℕ × ℕ for each A ∈ klnd(B).

See 2.2 for the definition of Δ_{u}. 2.10. Proposition. Let A ∈ klnd(B)\{k[x_{1}]}, let (a, b) = bideg(A) and suppose that a ≥ b. Then there exists (λ, s) ∈ k^{∗} × ℕ such that if we set u = λx_{1}^{s} then the ring A′ = Δ_{u}(A) satisfies bideg(A′) = (a′, b) and a′ < a. Moreover, s = (a+b)/ gcd(nb, a+b)−1.

Proof. Let A ∈ klnd(B) \ {k[x_{1}]}, let (a, b) = bideg(A) and suppose that a ≥ b. Note that A ≠ k[x_{2}], since a ≥ b. By 2.7.5, we have min(a, b) > 0, so:

a ≥ b > 0.

Let

δ = gcd(bn, a + b),  p = \frac{bn}{δ}  and q = \frac{a + b}{δ},

then p, q are relatively prime positive integers. As in 2.5, consider the (p, q)-graded rings^{1} R = k[x_{1}, 1/x_{1}, y] = ⊕_{d∈ℤ} R_{d} and B^{p,q} = ⊕_{d∈ℤ} B_{d}^{p,q} as well as the corresponding degree function deg : R → ℤ ∪ {−∞} (which satisfies deg(x_{1}) = p and deg(y) = q).

Pick f ∈ A \ k such that bideg(f) = (a, b). By 2.7.2 and 2.7.3,

(a, 0), (−b, bn) ∈ supp(f) ⊂ C(f);

since our choice of p, q is such that deg(x_{1}^{a}) = deg(x_{1}^{−b}y^{bn}), this implies that

(39)  (a, 0), (−b, bn) ∈ supp(h′)

where we define h′ = H_{p,q}(f). Now (39) implies that bideg(h′) = (a, b), which implies that h′ ∉ k[x_{1}] ∪ k[x_{2}] (by 2.7.5, because a, b > 0).

Choose D ∈ lnd(B) such that ker(D) = A and also consider D_{p,q} : B^{p,q} → B^{p,q}. We know from 2.6.3 that h′ = H_{p,q}(f) belongs to ker(D_{p,q}), so ker(D_{p,q}) is neither k[x_{1}] nor k[x_{2}] (because h′ ∉ k[x_{1}] ∪ k[x_{2}]). By 2.8 applied to D_{p,q}, there exists a homogeneous irreducible element h of ker(D_{p,q}), that element satisfies

(40)  ker(D_{p,q}) = k[h]

and one of the following holds:

(41)  h = μx_{1}^{−1}(y − λx_{1}^{q})^{n}, for some λ, μ ∈ k^{∗},
(42)  h = μξ_{2}^{−1}(y − λξ_{2}^{q})^{n}, for some λ, μ ∈ k^{∗},

where we define ξ_{2} = y^{n}/x_{1} (this element cannot be called x_{2}, because x_{2} = φ(y)/x_{1}).

Recall that h′ ∈ ker(D_{p,q}) = k[h] and that deg(h′) > 0 by (39); so deg(h) > 0 as well. Since both h and h′ are homogeneous, we obtain

(43)  h′ = ϵh^{N}

for some ϵ ∈ k^{∗} and some positive integer N.

We claim that if (42) holds then so does (41). Assume that (42) holds; then

h = μξ_{2}^{−1}(y − λξ_{2}^{q})^{n} = μ(y^{n}/x_{1})^{−1}(y − λ(y^{n}/x_{1})^{q})^{n} = μx_{1}^{−(nq−1)}(x_{1}^{q} − λy^{nq−1})^{n},

from which we see that bideg(h) = (1, nq−1). By (43) and 2.7.4 we have N(1, nq−1) = bideg(h′) = (a, b); since we have a ≥ b > 0, we must also have 1 ≥ nq − 1 > 0, so nq = 2 and

h = μx_{1}^{−(nq−1)}(x_{1}^{q} − λy^{nq−1})^{n} = μx_{1}^{−1}(x_{1}^{q} − λy)^{n},

so (41) holds.

From now-on, assume that (41) holds: h = μx_{1}^{−1}(y − λx_{1}^{q})^{n}. Let u = λx_{1}^{q−1} ∈ k[x_{1}] and consider Δ_{u} ∈ G (see 2.2), i.e., Δ_{u}(x_{1}) = x_{1} and Δ_{u}(y) = y + λx_{1}^{q}. Note:

(44)  Δ_{u}(h) = μξ_{2}.

1We remind the reader that B^{p,q} is not necessarely equal to B, see 2.5 for details.

From (41) and the fact that h is homogeneous, we see that Δ_{u}(y) = y + λx_{1}^{q} is homogeneous; so Δ_{u} extends to a homogeneous, degree-preserving automorphism Δ′_{u} : R → R and consequently

H_{p,q}(Δ′_{u}(f)) = Δ′_{u}(H_{p,q}(f)) = Δ′_{u}(h′) = ϵΔ′_{u}(h)^{N} = ϵμ^{N}ξ_{2}^{N}.

So, if we define f′ = Δ_{u}(f), we obtain that supp(H_{p,q}(f′)) = {(−b, bn)}. This shows that bideg(f′) = (a′, b) for some a′ < a, so 2.10 is proved. □

Proof of Theorem 2.3.

It suffices to prove that, given any A ∈ klnd(B) \ {k[x_{1}]},

(45)  There exists θ ∈ G such that bideg(θA) < bideg(A).

Let (a, b) = bideg(A).

If a ≥ b then by 2.10 there exists u ∈ k[x_{1}] such that bideg(Δ_{u}A) = (a′, b), where a′ < a, so bideg(Δ_{u}A) < bideg(A).

If a < b, pick f ∈ A \ k such that bideg(f) = (a, b). Then by 2.7.6 we have bideg[τ(f)] = (b, a) < (a, b) = bideg(A), so bideg [τ(A)] ≤ bideg[τ(f)] < bideg(A).

Hence, (45) is true in all cases. This completes the proof of 2.3. □

2.11. Remark. As a corollary to the Transitivity Theorem, one obtains that each A ∈ klnd(B) satisfies A = k^{[1]} (but of course there are quicker ways to prove this). Clearly, if f is such that A = k[f], then bideg(A) = bideg(f). Consequently, if bideg(A) = (a, b) then bideg (τ(A)) = (b, a) by 2.7.6.

3. APPENDIX: Rationality of certain affine plane curves

The aim of this section is to establish the following fact, which is used in the proof of 2.8 (see between 2.9.10 and 2.9.11). 3.1. Lemma. Let k be a field (of characteristic zero), X and Y indeterminates over k and let F = Y^{kb} + X^{b}(λX^{a} + μ) ∈ k[X, Y], where λ, μ ∈ k^{∗} and a, b, k are positive integers. Then F is an irreducible element of k[X, Y]. Consider the field extension L/k, where L is the field of fractions of k[X, Y]/F k[X, Y]. Then L = k^{(1)} if and only if one of the following holds:

(a) k = 1 and min(a, b) = 1

(b) k = 1 and a = 2 = b

(c) k = 2 and a = 1 = b. 3.2. Lemma. If 3.1 holds in the special case where k is algebraically closed, then it holds in general.

Proof. Assume that 3.1 holds in the special case where k is algebraically closed. Let k be arbitrary and consider

F = Y^{kb} + X^{b}(λX^{a} + μ) ∈ k[X, Y],

where a, b, k ≥ 1 and λ, μ ∈ k^{∗}.

Let k̄ be an algebraic closure of k; the first step is to prove that

(46)  F is irreducible in k̄[X, Y].

Write f = −X^{b}(λX^{a} + μ) ∈ k[X] and K = k̄(X), then F = Y^{kb} − f ∈ K[Y] where f ∈ K. By Gauss Lemma, (46) is equivalent to F being irreducible in K[Y]; and it is well-known that the latter condition is a consequence of

(47)  f^{1/d} ∉ k̄(X) for all d > 1.

The verification of (47) is straightforward, and is left to the reader. So (46) is proved.

Let A = k[X, Y]/F k[X, Y] and Ā = k̄[X, Y]/F k̄[X, Y]. Then (46) implies that Frac A is a subfield of Frac Ā. In fact, it is easy to see that Frac Ā can be obtained by adjoining the elements of k̄ to Frac A; in other words, the function field

(48)  Frac Ā over k̄

is obtained from the function field

(49)  Frac A over k

by “constant field extension”. Since the characteristic is zero, it follows that the function fields (48) and (49) have the same genus. Finally, note that the curve “F = 0” goes through the origin ((X, Y) = (0, 0)); this implies that the function field (49) has at least one place of degree one. Hence, (49) is rational if and only if (48) is rational. □

Until the end of this section, we assume that k is algebraically closed.

3.3. Notations. If G ∈ k[X_{1}, …, X_{n}] then \frac{∂G}{∂X_{i}} is abbreviated G_{X_{i}}. If G ∈ k[X, Y] is a nonconstant polynomial, let V (G) ⊂ 𝔸^{2} be the curve “G = 0”. If H ∈ k[X, Y, Z] is a homogeneous polynomial, let V (H) ⊂ ℙ^{2} be the curve “H = 0”.

3.4. Genus formula. If C ⊂ ℙ^{2} is an irreducible projective curve, its geometric genus g can be computed by the formula

(50)  g = \frac{(deg C − 1)(deg C − 2)}{2} − ∑_{P ∈S} \frac{μ(P)(μ(P) − 1)}{2},

where S is the set of singular points of C—including the infinitely near ones—and, given P ∈ S, μ(P) is the multiplicity of P on (a suitable strict transform of) C.

Given a subset E of ℙ^{2}, it is convenient to write

g′(E) = ∑_{P ∈S_{E}} \frac{μ(P)(μ(P) − 1)}{2},

where S_{E} = {P ∈ S | P ∈ E or P is infinitely near a point of E}. If the set E is a singleton, say E = {P_{0}}, g′({P_{0}}) is abbreviated g′(P_{0}).

For instance, we may partition ℙ^{2} as 𝔸^{2} ∪ L_{∞}, where 𝔸^{2} consists of the points (x : y : 1) and L_{∞} is the “line at infinity”, containing the points (x : y : 0) [we say that the points of 𝔸^{2} are “at finite distance” and that those of L_{∞} are “at infinity”]. Then we may rewrite the genus formula (50) as

(51)  g = \frac{(deg C − 1)(deg C − 2)}{2} − g′(𝔸^{2}) − g′(L_{∞}).

3.5. Lemma. Let P_{0} be a singular point of a curve C ⊂ ℙ^{2} and let F_{0} ∈ k[[U, V]] be a local equation of C at P_{0}. Make the following assumptions:

(1) F_{0}(U, 0) ≠ 0 and F_{0}(0, V) ≠ 0

(2) Let α = ord_{U} F_{0}(U, 0) and β = ord_{V} F_{0}(0, V); then the only elements (i, j) of

supp(F_{0}) satisfying βi + αj ≤ αβ are (α, 0) and (0, β). Then

g′(P_{0}) = \frac{(α − 1)(β − 1) + (δ − 1)}{2},

where δ = gcd(α, β) and where g′(P_{0}) is defined in 3.4.

Proof. Recall that supp(F_{0}) = {(i, j) ∈ ℕ^{2} | a_{ij} ≠ 0}, where F_{0} = ∑_{(i,j)∈ℕ^{2}} a_{ij}U^{i}V^{j} (a_{ij} ∈ k).

If α = β then the assumptions imply that the initial form of F_{0} is a_{α 0}U^{α} + a_{0 β}V^{α} (with a_{α 0}, a_{0 β} ∈ k^{∗} and α > 0). So P_{0} is an ordinary singularity with multiplicity α, in which case it is well-known that g′(P_{0}) = α(α − 1)/2. Since

\frac{(α − 1)(β − 1) + (δ − 1)}{2} = \frac{(α − 1)(α − 1) + (α − 1)}{2} = \frac{α(α − 1)}{2},

we are done in this case.

Assume that α ≠ β and let d = max(α, β) and r_{0} = min(α, β). Consider the euclidean algorithm of the pair (d, r_{0}):

d = q_{0}r_{0} + r_{1}  (q_{0}, r_{1} ∈ ℕ, r_{1} < r_{0})
r_{0} = q_{1}r_{1} + r_{2}  (q_{1}, r_{2} ∈ ℕ, r_{2} < r_{1})
⋮
r_{k−2} = q_{k−1}r_{k−1} + r_{k} (q_{k−1}, r_{k} ∈ ℕ, r_{k} < r_{k−1})
r_{k−1} = q_{k}r_{k} + r_{k+1}  (q_{k}, r_{k+1} ∈ ℕ, r_{k+1} < r_{k}, r_{k+1} = 0),

note that r_{k} = δ and define N = q_{0} + ⋯ + q_{k}. Then we inductively define morphisms

(52)  S_{N} ⟶^{π_{N}} ⋯ ⟶^{π_{1}} S_{0} = ℙ^{2}

as follows:

(a) Let π_{1} : S_{1} → S_{0} = ℙ^{2} be the blowing-up of ℙ^{2} at the given point P_{0}.

(b) If i ∈ {1, …, N} is such that S_{i} ⟶^{π_{i}} ⋯ ⟶^{π_{1}} S_{0} = ℙ^{2} have been defined, set π̂_{i} = π_{1} ∘ ⋯ ∘ π_{i} : S_{i} → S_{0} and let C_{i} ⊂ S_{i} be the strict transform of C with respect to π̂_{i}. If i < N then assumptions (1) and (2) on F_{0} ensure that C_{i} ∩ π̂_{i}^{−1}(P_{0}) is a single point, which we denote P_{i} ∈ S_{i}; let π_{i+1} : S_{i+1} → S_{i} be the blowing-up of S_{i} at P_{i}.

This defines (52). It is known that the multiplicity sequence satisfies:

(c) ( μ(P_{0}), μ(P_{1}), …, μ(P_{N−1}) ) = ( \underbrace{r_{0}, …, r_{0}}_{q_{0}}, \underbrace{r_{1}, …, r_{1}}_{q_{1}}, …, \underbrace{r_{k}, …, r_{k}}_{q_{k}} ).

The fact that C_{i} ∩π̂_{i}^{−1}(P_{0}) is a singleton for all i = 1, …, N −1 (see (b)) implies that the restriction of π̂_{N−1} : S_{N−1} → S_{0} to a map C_{N−1} → C is bijective, but note that the restriction C_{N} ⟶^{π_{N}} C_{N−1} is not necessarely injective. However, our assumptions have the following consequence: Let F_{N−1} ∈ k[[U, V]] be the local equation of C_{N−1} at P_{N−1}, then the initial form of F_{N−1} is a_{α 0}U^{δ} + a_{0 β}V^{δ} (with a_{α 0}, a_{0 β} ∈ k^{∗} and δ > 0). Hence, P_{N−1} is either a smooth point or an ordinary singularity of C_{N−1}. Consequently, every point of C_{N} ∩ π̂_{N}^{−1}(P_{0}) is a smooth point of C_{N}, or in other words:

(d) The N blow-ups (52) completely resolve the singularity of C at P_{0}.

By (d) and the definition of g′(P_{0}),

g′(P_{0}) = ∑_{i=0}^{N−1} \frac{μ(P_{i})(μ(P_{i}) − 1)}{2}.

Finally, it is an exercise of arithmetic to deduce the following formula from (c):

∑_{i=0}^{N−1} \frac{μ(P_{i})(μ(P_{i}) − 1)}{2} = \frac{(d − 1)(r_{0} − 1) + (δ − 1)}{2},

so the proof is complete. □

3.6. Lemma. Let F = Y^{b} + λX^{a} + μ ∈ k[X, Y], where λ, μ ∈ k^{∗} and

min(a, b) ≥ 2 and max(a, b) ≥ 3.

Then the curve V (F) is not rational.

Proof. It is clear that the ideal (F_{X}, F_{Y} , F) contains 1, so the curve is smooth at finite distance, i.e., g′(𝔸^{2}) = 0. Consider the closure V (F^{∗}) of V (F) in ℙ^{2}, where F^{∗} ∈ k[X, Y, Z] is the homogeneization of F.

If a = b then F^{∗} = Y^{a} + λX^{a} + μZ^{a}, so V (F^{∗}) meets the line at infinity L_{∞} in a distinct points, so V (F^{∗}) is a smooth projective curve. Since its degree is a = max(a, b) ≥ 3, it is not rational.

There remains the case a ≠ b; clearly, we may assume that

a < b.

Then F^{∗} = Y^{b} + λX^{a}Z^{b−a} + μZ^{b}, so V (F^{∗}) meets L_{∞} in one point: P_{0} = (1 : 0 : 0). Consequently, the polynomial

F_{0} = F^{∗}(1, Y, Z) = Y^{b} + λZ^{b−a} + μZ^{b}

is a local equation of V (F^{∗}) at P_{0} (and P_{0} is now the origin, (y, z) = (0, 0)). By 3.5,

g′(L_{∞}) = g′(P_{0}) = \frac{(b − 1)(b − a − 1) + (δ − 1)}{2}

where δ = gcd(b, b − a). By the genus formula (51),

g = \frac{(b − 1)(b − 2)}{2} − \frac{(b − 1)(b − a − 1) + (δ − 1)}{2} = \frac{(b − 1)(a − 1)}{2} − \frac{δ − 1}{2}

> \frac{(b − 1)(a − 1)}{2} − \frac{b − a}{2} = \frac{b(a − 2) + 1}{2} > 0,

where we used δ − 1 < b − a. So V (F^{∗}) is not rational. □

3.7. Corollary. Let F = Y^{b} + λX^{a} + μ ∈ k[X, Y], where λ, μ ∈ k^{∗} and a, b ≥ 1. Then the curve V (F) is rational if and only if:

min(a, b) = 1 or a = b = 2.

Proof. Follows immediately from 3.6 □

Proof of 3.1. We may write L = k(x, y) where y^{kb} + x^{b}(λx^{a} + μ) = 0. Set z = y^{k}/x ∈ k(x, y), then

k(x, z) = k(x, y^{k}) ⊆ k(x, y)

and the field k(x, z) is defined by the equation

(53)  z^{b} + λx^{a} + μ = 0.

If (a) or (b) holds, then k = 1 (so k(x, z) = k(x, y)) and 3.7 gives k(x, z) = k^{(1)}, so k(x, y) = k^{(1)}. If (c) holds then V (F) is a curve of degree 2, and so is rational.

Hence, if one of (a), (b), (c) holds then k(x, y) = k^{(1)}.

Conversely, assume that k(x, y) = k^{(1)}. Then k(x, z) = k^{(1)} by Lüroth Theorem and it follows that either min(a, b) = 1 or a = 2 = b, by applying 3.7 to (53). Thus, if k = 1 then (a) or (b) holds. For this reason, we will assume from now-on that k ≥ 2. Let g be the genus of the curve V (F), or equivalently the genus of the function field k(x, y)/k. Since we assumed that k(x, y) = k^{(1)}, we have g = 0. The proof consists in showing that the conditions

k ≥ 2 and g = 0

imply that (c) holds.

We begin by noting that consideration of

F = Y^{kb} + λX^{a+b} + μX^{b} = Y^{kb} + X^{b}(λX^{a} + μ)
F_{Y} = kbY^{kb−1}
F_{X} = λ(a + b)X^{a+b−1} + μbX^{b−1} = X^{b−1}[λaX^{a} + b(λX^{a} + μ)]

shows that the singular locus of V (F) (at finite distance) is

(54)  \begin{cases}\{(0, 0)\} & if b > 1, \\ ∅ & if b = 1. \end{cases}

Now F = Y^{kb} + λX^{a+b} + μX^{b} may be regarded as a local equation of V (F) at (0, 0); so the hypothesis of 3.5 is satisfied, with α = b, β = kb and δ = b; so

(55)  g′(𝔸^{2}) = g′((0, 0)) = \frac{(b − 1)(kb − 1) + (b − 1)}{2} = \frac{kb(b − 1)}{2}.

Observe that this formula is valid in the two cases of (54), i.e., in all cases. In order to compute g′(L_{∞}), we have to consider several cases. Let d = deg(F) = max(kb, a + b).

Case d = kb = a + b. Then F^{∗} = Y^{kb} + λX^{kb} + μX^{b}Z^{(k−1)b}, so V (F^{∗}) is a curve of degree d meeting L_{∞} in d distinct points; so V (F^{∗}) is smooth at infinity, i.e., g′(L_{∞}) = 0. Consequently,

2g = (d − 1)(d − 2) − 2g′(𝔸^{2}) − 2g′(L_{∞}) = (kb − 1)(kb − 2) − kb(b − 1) = kb(kb − b − 2) + 2 = kb(a − 2) + 2.

Since g = 0, we must have a = 1; then kb = a + b implies that b = 1 and that k = 2, i.e., condition (c) holds.

Case kb ≠ a + b. If d = kb > a + b then F^{∗} = Y^{kb} + λX^{a+b}Z^{kb−a−b} + μX^{b}Z^{kb−b}, so V (F^{∗}) ∩ L_{∞} = {P_{0}} where P_{0} = (1 : 0 : 0). Then

F_{0} = F^{∗}(1, Y, Z) = Y^{kb} + λZ^{kb−a−b} + μZ^{kb−b}

is a local equation of V (F^{∗}) at P_{0} and the hypothesis of 3.5 is satisfied, with {α, β} = {kb, kb − a − b} and δ = gcd(kb, a + b). Observe that (d − 1)(d − 2) − (α − 1)(β − 1) = (kb − 1)(kb − 2) − (kb − 1)(kb − a − b − 1), so

(56)  (d − 1)(d − 2) − (α − 1)(β − 1) = (a + b − 1)(kb − 1).

Now

2g = (d − 1)(d − 2) − 2g′(L_{∞}) − 2g′(𝔸^{2}) = (d − 1)(d − 2) − 2g′(P_{0}) − 2g′(𝔸^{2})

= [(d − 1)(d − 2) − (α − 1)(β − 1)] − (δ − 1) − kb(b − 1), so (56) implies

(57)  2g = (a + b − 1)(kb − 1) − kb(b − 1) − (δ − 1).

We claim that (57) is also valid when kb < a + b. Indeed, if kb < a + b then F^{∗} = Y^{kb}Z^{a+b−kb} + λX^{a+b} + μX^{b}Z^{a}, so V (F^{∗}) ∩ L_{∞} = {P_{0}} where P_{0} = (0 : 1 : 0). Then

F_{0} = F^{∗}(X, 1, Z) = Z^{a+b−kb} + λX^{a+b} + μX^{b}Z^{a}

is a local equation of V (F^{∗}) at P_{0}, so the hypothesis of 3.5 is satisfied, with {α, β} = {a + b, a + b − kb} and δ = gcd(kb, a + b). It follows that

(d − 1)(d − 2) − (α − 1)(β − 1) = (a + b − 1)(a + b − 2) − (a + b − 1)(a + b − kb − 1) = (a + b − 1)(kb − 1),

i.e., (56) holds in this case as well. Since (57) follows from (56), (57) is therefore valid whenever kb ≠ a + b. Also,

δ = gcd(kb, a + b) ≤ a + b

holds whenever kb ≠ a + b.

Now an algebraic manipulation of (57) yields

(58)  2g = (k − 2)[ab + (b − 1)^{2}] + 2(b − 1)(a + b − 2) + (a + b − δ),

i.e., 2g is expressed as the sum of three nonnegative integers. Since g = 0, we obtain the three conditions:

k = 2, b = 1 and a + b = δ.

The third condition reads a + 1 = gcd(kb, a + b) = gcd(2, a + 1), so a + 1 | 2. This gives a = 1, so condition (c) holds.

This completes the proof of 3.1. □

References

1. D. Daigle, Affine rulings and homogeneous locally nilpotent derivations, preprint.

2. , Locally nilpotent derivations and Danielewski surfaces, to appear, 2002.

3. G. Freudenburg, Local slice constructions in K[X, Y, Z], Osaka J. Math. 34 (1997), 757–767.

4. L. Makar-Limanov, On groups of automorphisms of a class of surfaces, Israel J. of Math. 69 (1990), 250–256.

5. , On the group of automorphisms of a surface x^{n}y = P (z), preprint, 1996.

6. R. Rentschler, Opérations du groupe additif sur le plan affine, C. R. Acad. Sc. Paris 267 (1968), 384–387.

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

  E-mail address: ddaigle@uottawa.ca