LOCALLY NILPOTENT DERIVATIONS AND DANIELEWSKI SURFACES

DANIEL DAIGLE

Introduction

This work started as an attempt to understand the process known as the local slice construction. Introduced by Freudenburg in [3], this is a method for modifying a nonzero locally nilpotent derivation of k[X, Y, Z] so as to obtain another one (where k is a field of characteristic zero). Near the end of the cited paper, Freudenburg defines a graph Γ whose vertices are the kernels of the nonzero locally nilpotent derivations of k[X, Y, Z] and where vertices ker(D) and ker(D′) are joined by an edge whenever D′ can be obtained from D by a local slice construction (in one step).

Over the years, it has become clear that the local slice construction is an interesting idea for studying the locally nilpotent derivations of k[X, Y, Z]. In particular, one would like to know if Γ is connected. Connectedness would mean that every locally nilpotent derivation can be obtained from one of them (say from ∂/∂X) by a finite sequence of local slice constructions. In unpublished work, we have shown that this is indeed the case for derivations which are homogeneous with respect to positive weights.

In the hope of clarifying the local slice construction, we generalize it. Let B be an arbitrary integral domain of characteristic zero. In section 3 of the present paper, we define a graph \underline{klnd} (B) which generalizes Freudenburg’s graph Γ: The vertices of \underline{klnd} (B) are the kernels of the nonzero locally nilpotent derivations of B and the edges, one might say, capture the essence of the local slice construction. Also, the graph \underline{klnd} (B) is an invariant of the ring B and the group of automorphisms of B acts on it in a natural way. In the special case B = k[X, Y, Z], the two graphs Γ and \underline{klnd} (B) have the same vertices and every edge of Γ is an edge of \underline{klnd} (B); we don’t know if every edge of \underline{klnd} (B) is an edge of Γ.

This generalization produces new insight into the local slice construction. In particular, we find that that process is essentially a two-dimensional affair and that it is intimately related to Danielewski surfaces “XY = P (Z)”.

We believe that \underline{klnd} (B) is a suitable tool for studying polynomial rings (B = k^{[n]}). For these rings, the graph \underline{klnd} (B) seems to have just the right amount of edges to be interesting. This is not the case for all rings: One can find examples of rings B for which \underline{klnd} (B) is the empty graph; or \underline{klnd} (B) has only one vertex and no edges; or (see 6.2) \underline{klnd} (B) has infinitely many vertices but no edges.

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 a subsequent paper, we intend to use the methods developed here to investigate the locally nilpotent derivations of k[X, Y, Z].

The material is organized as follows.

Section 1 gives the basic definitions and results that are needed in this paper.

Section 2 gives some algebraic properties of Danielewski surfaces. Note in particular results 2.5, 2.6 and 2.6.2, which characterize Danielewski surfaces in terms of locally nilpotent derivations.

Section 3 defines the graph \underline{klnd} (B), where B is any integral domain of characteristic zero. In addition to \underline{klnd} (B), two other graphs (\underline{klnd}_{∗}(B) and \underline{ℛ}(B)) are defined in that section.

Section 4 describes the graph \underline{klnd} (B) in the case where B is a two-dimensional ring.

Section 5 focuses on the subgraph \underline{klnd}_{∗}(B) of \underline{klnd} (B) obtained by deleting all isolated vertices. If B is a factorial affine domain (of any dimension), Theorem 5.1 states that \underline{klnd}_{∗}(B) is a union of connected subgraphs G_{i} such that: (i) Each G_{i} is isomorphic to \underline{klnd} (B_{i}) for some two-dimensional ring B_{i} (in fact a Danielewski surface); (ii) every edge of \underline{klnd}_{∗}(B) is an edge of exactly one G_{i}; and (iii) if i ≠ j then G_{i} and G_{j} have at most one vertex in common. So the local structure of \underline{klnd}_{∗}(B) is well understood, thanks to the thorough description of the two-dimensional case given in Section 4.

Section 6 gathers some remarks which conclude the paper.

1. Generalities

1.1. Conventions.

• All fields and rings are tacitly assumed to be of characteristic zero.

• Throughout, k denotes an arbitrary field (of characteristic zero).

• The set of units of a ring R is denoted R^{∗}.

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

• If A is a 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.

• By a k-domain of transcendence degree d, we mean an integral domain B containing k and satisfying trdeg_{k}(B) = d.

• If R is a subring of a domain A, then we write A_{R} as an abbreviation for the localized ring S^{−1}A, where S = R \ {0}; in particular, A_{A} = Frac(A); if D : A → A is a derivation, S^{−1}D : S^{−1}A → S^{−1}A is abbreviated D_{R} : A_{R} → A_{R}.

• If α ∈ A then A_{α} = S^{−1}A where S = {1, α, α^{2}, … }.

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

∀_{x,y∈B} xy ∈ R \ {0} ⟹ x, y ∈ R.

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

(1) R^{∗} = B^{∗}

(2) R is algebraically closed in B.

(3) If B is a UFD then so is R.

(4) S^{−1}R is an inert subring of S^{−1}B, for any multiplicative subset S ⊆ R \ {0}.

1.4. A subring R of an integral domain B is inert if and only if B_{R}^{∗} = R_{R}^{∗} and B ∩ R_{R} = R.

1.5. Definitions. Let B be a ring.

(1) A derivation D : B → B is • irreducible if the only principal ideal of B which contains D(B) is B;

• locally nilpotent if ∀_{x∈B} ∃_{k>0} D^{k}(x) = 0.

(2) Notations:

lnd(B) = set of nonzero locally nilpotent derivations D : B → 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) }.

1.6. 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^{∗}, B ∩ Frac A = A and if B is a UFD then so is A. Note, also, that if K is any field contained in B then K^{∗} ⊆ B^{∗} = 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; consequently, B ∩ S^{−1}A = 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.

Note in particular the following consequence of part (5) of 1.6:

1.7. If B is a domain and A ∈ klnd(B) then trdeg_{A} B = 1.

1.8. Rentschler’s Theorem. (See [5].) Let B = k^{[2]}, where k is a field of characteristic zero, and let D : B → B be a nonzero locally nilpotent derivation. Then there exist u, v such that B = k[u, v] and ker D = k[u]. Moreover, given any such u, v we have D = f(u)\frac{∂}{∂v} for some f(u) ∈ k[u].

1.9. Simple derivations. Let B be a k-domain of transcendence degree two.

1.9.1. Definition. A derivation D : B → B is k-simple if it is locally nilpotent, irreducible and satisfies

∃ y ∈ B ker D = k[ Dy].

Note that if this is the case then ker D = k^{[1]}. Consequently:

1.9.2. If B admits a k-simple derivation then B^{∗} = k^{∗}.

1.9.3. Lemma. Suppose that Δ ∈ lnd(B) is k-simple. If D ∈ lnd(B) is irreducible and ker(D) = ker(Δ), then D = λΔ for some λ ∈ k^{∗}. Consequently, D is k-simple.

Proof. Let A = ker D = ker Δ and choose x, y ∈ B such that Δ(y) = x and A = k[x]. Note that Frac B = k(x, y) (by 1.7) and consider the partial derivative ∂ = ∂/∂y : k(x, y) → k(x, y). Extending Δ and D to derivations Δ̃ and D̃ of k(x, y),

Δ̃ = x∂ and D̃ = D(y)∂.

It follows that D(y)Δ = xD. Since D is locally nilpotent and x ∈ ker D, xD is locally nilpotent; so D(y)Δ is locally nilpotent and it follows that D(y) ∈ ker Δ by part (6) of 1.6. Hence, x and D(y) are two elements of the ideal

I = { α ∈ A | α∂(B) ⊆ B }

of A. Observe that 1 ∉ I, for otherwise ∂(B) ⊆ B, so Δ(B) ⊆ xB, so x ∈ B^{∗} (because Δ is irreducible), so x ∈ A^{∗}, but this is false because A = k[x] = k^{[1]}.

Since x is a prime element of A and x ∈ I ≠ A, we have I = xA, so x | D(y) in A. Then D = λΔ where λ = D(y)/x ∈ A. Now D(B) ⊆ λB, so λ ∈ B^{∗} by irreducibility of D. Since B^{∗} = k^{∗}, k[ Dy] = k[ λx] = k[x] and we are done. □

On the number of kernels

Regarding the cardinality of the set klnd(B), we have the following elementary fact:

1.10. Proposition. Let B be a domain of characteristic zero and suppose that k ⊂ B is a field such that trdeg_{k}(B) < ∞. Then the cardinality of klnd(B) is either 0, 1 or |k|.

Proof. As a first step, we show:

(1)

Let B be a ℚ-domain and suppose that A, A′ are distinct elements of klnd(B). Then | klnd(B)| ≥ |A ∩ A′|.

Let A and A′ be distinct elements of klnd(B). Let D, D′ ∈ lnd(B) be such that ker D = A and ker D′ = A′. We first consider the case where:

(2)  D(A′) ⊆ A′ and D′(A) ⊆ A.

Then it follows that

(3)  D ∘ D′ = D′ ∘ D.

Indeed, let δ : B → B denote the derivation D ∘ D′ − D′ ∘ D. Then by assumption (2), we have A ∪ A′ ⊆ ker δ. Since each of A, A′ is algebraically closed in B, and since B has transcendence degree one over each of A, A′, it follows that B is algebraic over ker δ, so δ = 0 and (3) is true.

For each λ ∈ A, let Δ_{λ} : B → B denote the derivation D′+λD. Then (3) immediately implies that Δ_{λ} ∈ lnd(B), so we have a map

(4)

A ⟶ klnd(B)
λ ⟼ ker(Δ_{λ}).

We claim that the map (4) is injective. Indeed, consider distinct elements λ_{1}, λ_{2} of A. Then for each x ∈ ker(Δ_{λ_{1}}) ∩ ker(Δ_{λ_{2}}) we have

D′(x) + λ_{1}D(x) = 0 = D′(x) + λ_{2}D(x),

from which we deduce that D(x) = 0 = D′(x), i.e., x ∈ A∩A′. So ker(Δ_{λ_{1}})∩ker(Δ_{λ_{2}}) ⊆ A ∩ A′ and consequently the transcendence degree of B over ker(Δ_{λ_{1}}) ∩ ker(Δ_{λ_{2}}) is strictly greater than one. It follows that ker(Δ_{λ_{1}}) ≠ ker(Δ_{λ_{2}}), so the map (4) is injective. Thus (1) holds under extra assumption (2).

There remains the case where (2) does not hold; without loss of generality, let us assume that

(5)  D(A′) ⊈ A′.

For each λ ∈ A, let ε_{λ} : B → B be the automorphism of B defined by

ε_{λ}(x) = ∑_{i=0}^{∞} \frac{λ^{i}D^{i}(x)}{i!} (x ∈ B)

i.e., ε_{λ} is the exponential of λD. As is well-known,

(6)  ε_{λ_{1}} ∘ ε_{λ_{2}} = ε_{λ_{1}+λ_{2}} for all λ_{1}, λ_{2} ∈ A.

Since ε_{λ}(A′) = ker (ε_{λ} ∘ D′ ∘ ε_{λ}^{−1}) ∈ klnd(B), the assignment λ ↦ ε_{λ}(A′) is a map from A to klnd(B). We claim that the restriction

(7)  \begin{matrix}A ∩ A′ & ⟶ & klnd(B) \\ λ & ⟼ & ε_{λ}(A′) \end{matrix}

is an injective map. We begin by showing that (5) implies:

(8)  If λ ∈ A ∩ A′ satisfies ε_{λ}(A′) ⊆ A′, then λ = 0.

To see this, consider λ ∈ A ∩ A′ satisfying ε_{λ}(A′) ⊆ A′. By (5), we may pick an x ∈ A′ such that D(x) ∉ A′. Fix such an x and let n be such that D^{i}(x) = 0 for all i > n; consider the polynomial f(T) ∈ B[T] defined by

f(T) = ∑_{i=0}^{∞} D′(\frac{D^{i}(x)}{i!}) T^{i} = ∑_{i=0}^{n} D′(\frac{D^{i}(x)}{i!}) T^{i}

and note that f(T) is not the zero polynomial since the coefficient of T in f(T) is D′(D(x)) ≠ 0. Then for each k ∈ ℕ

f(kλ) = ∑_{i=0}^{n} D′(\frac{D^{i}(x)}{i!}) (kλ)^{i} = D′ (∑_{i=0}^{n} (\frac{D^{i}(x)}{i!}) (kλ)^{i}) = D′(ε_{kλ}(x)) = 0,

where the last equality follows from ε_{kλ}(x) = ε_{λ}^{k}(x) ∈ ε_{λ}^{k}(A′) ⊆ A′. Now f(T) cannot have infinitely many roots, so λ = 0 and (8) is proved.

Now (6) and (8) imply that the map (7) is injective, so (1) holds in this case as well. So (1) is proved.

To complete the proof of the proposition, suppose that k ⊂ B is a field such that trdeg_{k}(B) < ∞. Assuming that | klnd(B)| > 1, we show that | klnd(B)| = |k|.

Consider distinct elements A and A′ of klnd(B). Since k ⊆ A ∩ A′ by part (1) of 1.6, we have | klnd(B)| ≥ |k| by (1).

Consider a finite subset {x_{1}, …, x_{n}} of B such that B is algebraic over k[x_{1}, …, x_{n}]. The map

\begin{matrix}lnd(B) & ⟶ & B^{n} \\ D & ⟼ & (Dx_{1}, …, Dx_{n}) \end{matrix}

is injective, so | lnd(B)| ≤ |B^{n}| = |B| = |k|. Since D ↦ ker D is a surjection from lnd(B) to klnd(B), we have | klnd(B)| ≤ | lnd(B)|, so we are done. □

Remark. It is possible to have | klnd(B)| > |B| if we don’t assume that B has finite transcendence degree over some field. For instance, let k be a field of characteristic zero and let B = k[V] be a polynomial ring, where V is a set of indeterminates satisfying |V | ≥ |k| (thus |V | = |B|). Fix a well-order on the set V. For each subset S of V other than ∅ and V, define a k-derivation D_{S} : B → B by

D_{S}(X) = \begin{cases}0, & if X ∈ S \\ min S, & if X ∉ S. \end{cases}

Then one can verify that D_{S} ∈ lnd(B). Since ker(D_{S}) ∩ V = S, it follows that | klnd(B)| = |2^{B}|.

2. Danielewski surfaces

2.1. Definition. Given a k-algebra B, let Γ_{k}(B) denote the (possibly empty) set of ordered triples (x_{1}, x_{2}, y) ∈ B × B × B satisfying:

The k-homomorphism k[X_{1}, X_{2}, Y] → B defined by

X_{1} ↦ x_{1}, X_{2} ↦ x_{2} and Y ↦ y

is surjective and has kernel equal to (φ − X_{1}X_{2})k[X_{1}, X_{2}, Y] for some nonconstant polynomial in one variable φ ∈ k[Y].

If Γ_{k}(B) ≠ ∅ then we say that (B, k) is a Danielewski surface. If this is the case then B is a k-domain and trdeg_{k}(B) = 2.

Remark. The term “Danielewski surface” usually refers to hypersurfaces of 𝔸^{3} given by an equation of the form xy = φ(z), or sometimes x^{n}y = φ(z), because such surfaces were studied by Danielewski in connection with the cancellation problem (see [2]).

Remarks. Suppose that (B, k) is a Danielewski surface and let (x_{1}, x_{2}, y) ∈ Γ_{k}(B). (1) Any two elements of {x_{1}, x_{2}, y} are algebraically independent over k.

(2) Once (x_{1}, x_{2}, y) ∈ Γ_{k}(B) is chosen, φ ∈ k[Y] \ k is uniquely determined by the condition φ(y) = x_{1}x_{2}.

2.2. Lemma. Let X_{1}, X_{2}, Y be indeterminates over k, let L be a field containing k and let π : k[X_{1}, X_{2}, Y] → L be a k-homomorphism with kernel (φ − X_{1}X_{2})k[X_{1}, X_{2}, Y], where φ is some element of k[Y] \ k. Write x_{1} = π(X_{1}), x_{2} = π(X_{2}) and y = π(Y), then the following hold:

(a) For each element β of the subring k[x_{1}, x_{2}, y] of L, there exists a unique F ∈ k[X_{1}, X_{2}, Y] satisfying F (x_{1}, x_{2}, y) = β and deg_{Y} (F) < deg_{Y} (φ).

(b) k(x_{1})[y] ∩ k(x_{2})[y] = k[x_{1}, x_{2}, y].

Proof. If we view Φ = φ − X_{1}X_{2} as a polynomial in Y with coefficients in k[X_{1}, X_{2}], then the leading coefficient of Φ belongs to k^{∗}. Thus assertion (a) follows from a straightforward application of the division algorithm in k[X_{1}, X_{2}][Y].

To prove (b), it suffices to show that k(x_{1})[y] ∩ k(x_{2})[y] ⊆ k[x_{1}, x_{2}, y]. Let β ∈ k(x_{1})[y] ∩ k(x_{2})[y], then

β = \frac{F (x_{1}, x_{2}, y)}{f(x_{1})} = \frac{G(x_{1}, x_{2}, y)}{g(x_{2})}

for some F, G ∈ k[X_{1}, X_{2}, Y], f ∈ k[X_{1}] \ {0} and g ∈ k[X_{2}] \ {0}. By (a), we may arrange that deg_{Y} (F) < deg_{Y} (φ) and deg_{Y} (G) < deg_{Y} (φ). Then g(x_{2})F (x_{1}, x_{2}, y) = f(x_{1})G(x_{1}, x_{2}, y) and the uniqueness part of (a) imply that gF = fG in k[X_{1}, X_{2}, Y], so f | F in k[X_{1}, X_{2}, Y]. Let Q ∈ k[X_{1}, X_{2}, Y] be such that F = Qf, then β = Q(x_{1}, x_{2}, y) ∈ k[x_{1}, x_{2}, y]. □

The following result gathers the most basic properties of Danielewski surfaces. See 1.9.1 for k-simple derivations.

2.3. Proposition. Let (B, k) be a Danielewski surface, fix an element γ = (x_{1}, x_{2}, y) of Γ_{k}(B) and let φ be the unique element of k[Y] \ k satisfying φ(y) = x_{1}x_{2}.

(a) B is a normal k-domain and B^{∗} = k^{∗}.

(b) B = k^{[2]} ⟺ deg_{Y} (φ) = 1.

(c) B is a UFD ⟺ φ is irreducible in k[Y].

(d) For each i = 1, 2, there exists a unique k-derivation D_{i}^{γ} : B → B satisfying D_{i}^{γ}(x_{i}) = 0 and D_{i}^{γ}(y) = x_{i}. Moreover, ker D_{i}^{γ} = k[x_{i}] and D_{i}^{γ} is a k-simple derivation of B.

Proof. We shall prove assertions (a), (d), (c) and (b), in this order. It is immediate that B is a k-domain and that

(9)  Any two elements of {x_{1}, x_{2}, y} are algebraically independent over k.

By 2.2, there holds

k[x_{1}, x_{2}, y] = k(x_{1})[y] ∩ k(x_{2})[y]

where each k(x_{i})[y] = k(x_{i})^{[1]} is a normal domain, so

(10)  B is normal.

Let us also record that B_{x_{1}} = k[x_{1}, 1/x_{1}, x_{2}, y] = k[x_{1}, 1/x_{1}, y] = k[x_{1}, 1/x_{1}]^{[1]} and similarly for B_{x_{2}}, i.e.,

(11)  For each i = 1, 2, B_{x_{i}} = k[x_{i}, 1/x_{i}, y] = k[x_{i}, 1/x_{i}]^{[1]}.

Suppose that u ∈ B^{∗}. Then u is a unit of each of B_{x_{1}} and B_{x_{2}}, so (11) implies that u ∈ k[x_{1}, 1/x_{1}] ∩ k[x_{2}, 1/x_{2}]. Since x_{1}, x_{2} are algebraically independent over k by (9), we have k(x_{1}) ∩ k(x_{2}) = k and u ∈ k. This shows that B^{∗} = k^{∗}. Together with (10), this proves assertion (a).

We shall now prove assertion (d). Let (i, j) = (1, 2) or (2, 1). Let δ_{i} : k[X_{1}, X_{2}, Y] → k[X_{1}, X_{2}, Y] be the k-derivation given by δ_{i}(X_{i}) = 0, δ_{i}(Y) = X_{i} and δ_{i}(X_{j}) = φ′(Y). Then δ_{i} is triangular, hence locally nilpotent, and clearly δ_{i}(Φ) = 0, where Φ = φ − X_{1}X_{2}. So we may define a locally nilpotent derivation D_{i} : B → B by taking δ_{i} (mod Φ). Then D_{i}(x_{i}) = 0 and D_{i}(y) = x_{i}, thus proving the existence part of assertion (d). If D : B → B is any k-derivation satisfying D(x_{i}) = 0 and D(y) = x_{i}, then x_{i}D(x_{j}) = D(x_{1}x_{2}) = D(φ(y)) = φ′(y)x_{i}, so D(x_{j}) = φ′(y), which proves uniqueness of D_{i}.

It is easy to see that the kernel of the localization B_{x_{1}} → B_{x_{1}} of D_{1} is k[x_{1}, 1/x_{1}], so ker D_{1} = B ∩ k[x_{1}, 1/x_{1}]. Consider an element β of B ∩ k[x_{1}, 1/x_{1}]. By 2.2,

β = F (x_{1}, x_{2}, y), for some F ∈ k[X_{1}, X_{2}, Y] such that deg_{Y} (F) < deg_{Y} (φ).

Since β ∈ k[x_{1}, 1/x_{1}], there exists m > 0 such that x_{1}^{m}F (x_{1}, x_{2}, y) ∈ k[x_{1}], i.e.,

x_{1}^{m}F (x_{1}, x_{2}, y) = f(x_{1}), for some f ∈ k[X_{1}].

Then 2.2 implies that X_{1}^{m}F = f, so X_{1}^{m}F ∈ k[X_{1}], so F ∈ k[X_{1}] and β ∈ k[x_{1}]. This shows that ker D_{1} = k[x_{1}] = k^{[1]} (and by symmetry ker D_{2} = k[x_{2}] = k^{[1]}).

Next, we show that D_{1} is irreducible. Let g ∈ B be such that D(B) ⊆ gB. Since D_{1}(y) = x_{1}, we have g | x_{1} in B, so g ∈ k[x_{1}] because k[x_{1}] = ker D_{1} is inert in B. Hence, g = G(x_{1}) for some G ∈ k[X_{1}]. On the other hand, D_{1}(x_{2}) = φ′(y), so G(x_{1}) | φ′(y) in B and 2.2 allows us to write

G(x_{1})F (x_{1}, x_{2}, y) = φ′(y),

where F ∈ k[X_{1}, X_{2}, Y] and deg_{Y} (F) < deg_{Y} (φ). Now 2.2 implies that GF = φ′ ∈ k[Y] \ {0}, so G ∈ k[X_{1}] ∩ k[Y] = k. Hence D_{1} is irreducible (and so is D_{2} by symmetry). Thus assertion (d) is true.

Next, we prove assertion (c). Since x_{1} is an irreducible element of k[x_{1}] and k[x_{1}] = ker D_{1} is an inert subring of B, x_{1} is an irreducible element of B. On the other hand,

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

≅ k[X_{2}, Y]/(φ) ≅ (k[Y]/(φ))^{[1]}

shows that x_{1} is a prime element of B if and only if φ is a prime element of k[Y]. In particular, if B is a UFD then x_{1} is prime in B, so φ is prime in k[Y].

Conversely, if φ is prime in k[Y] then x_{1} is prime in B and, by (11), B_{x_{1}} is a UFD; so B is a UFD and assertion (c) is true.

For (b), note that if deg_{Y} (φ) = 1 then it is obvious that B = k[x_{1}, x_{2}] = k^{[2]}. Conversely, assume that B = k^{[2]}. By Rentschler’s Theorem 1.8, B = A^{[1]} for any A ∈ klnd(B); in particular B = k[x_{1}]^{[1]}, so B/x_{1}B = k^{[1]}. By (12), deg_{Y} (φ) = 1.

This completes the proof of 2.3. □

We also record the following simple fact:

2.4. Lemma. Suppose that (B, k) is a Danielewski surface and let (x_{1}, x_{2}, y) ∈ Γ_{k}(B). Then x_{1}x_{2} is a generator of the ideal k[y] ∩ x_{1}B of k[y].

Proof. We have x_{1}x_{2} = φ(y) for some nonconstant polynomial φ ∈ k[Y]. Let n = deg_{Y} (φ). Given ξ ∈ k[y] ∩ x_{1}B, we may write ξ = ψ(y), where ψ ∈ k[Y]; by the division algorithm, ψ = qφ + ρ, with q, ρ ∈ k[Y] and deg ρ < n. We have

(13)  ρ(y) = ψ(y) − q(y)φ(y) = ξ − q(y)x_{1}x_{2} ∈ x_{1}B,

so ρ(y) = x_{1}F (x_{1}, x_{2}, y) for some F ∈ k[X_{1}, X_{2}, Y] such that deg_{Y} (F) < n. Then ρ = X_{1}F by 2.2, so X_{1} | ρ in k[X_{1}, X_{2}, Y], which implies that ρ = 0. Then (13) yields ξ = q(y)x_{1}x_{2} ∈ x_{1}x_{2}k[y] and we are done. □

Remark. Applying 2.4 to (x_{2}, x_{1}, y) ∈ Γ_{k}(B) implies that x_{1}x_{2} generates the ideal k[y] ∩ x_{2}B of k[y]. So: The ideals k[y] ∩ x_{1}B and k[y] ∩ x_{2}B of k[y] are equal.

Two characterizations of Danielewski surfaces

Results 2.5, 2.6 and 2.6.2 characterize Danielewski surfaces in terms of locally nilpotent derivations.

2.5. Theorem. Let B be a k-domain, let y ∈ B and let D_{1}, D_{2} : B → B be locally nilpotent derivations. Suppose that (y, D_{1}, D_{2}) satisfies:

(i) ker D_{1} ≠ ker D_{2} (ii) for each i = 1, 2, ker D_{i} = k^{[1]} and D_{i}(y) ∈ ker(D_{i}) \ {0}.

Then (B, k) is a Danielewski surface. Moreover, if D_{1} and D_{2} are irreducible then exactly one of the following holds: (2.5-1) For each i = 1, 2, D_{i}(y) ∈ k^{∗} and B = (ker D_{i})^{[1]} = k^{[2]}.

(2.5-2) Let x_{i} = D_{i}(y) (i = 1, 2), then

ker D_{1} = k[x_{1}], ker D_{2} = k[x_{2}] and (x_{1}, x_{2}, y) ∈ Γ_{k}(B).

For the proof of 2.5 we need the following simple observation, whose proof we leave to the reader:

2.5.1. Let X, Y be indeterminates over the field k and let f ∈ k[Y] \ k. Then

k(X)[Y] ∩ k(X + f)[Y] = k[X, Y],

where the intersection is taken in k(X, Y).

Proof of 2.5. Note that assumption (ii) and 1.7 imply that B has transcendence degree two over k. More precisely, write ker(D_{i}) = k[t_{i}] for each i = 1, 2. Since y is a preslice of D_{i},

(14)  B ⊆ k(t_{i}) ⊗_{k[t_{i}]} B = k(t_{i})[y] = k(t_{i})^{[1]}.

In particular k(y, t_{1}) = Frac B = k(y, t_{2}), so (for each i) t_{i}, y are algebraically independent over k. Since k(y, t_{1}) = k(y, t_{2}),

(15)  t_{2} = \frac{a_{2}t_{1} + b}{ct_{1} + a_{1}}

where a_{1}, a_{2}, b, c ∈ k(y) and a_{1}a_{2} − bc ≠ 0; in fact, we may arrange that

(16)  a_{1}, a_{2}, b, c ∈ k[y], a_{1}a_{2} − bc ≠ 0 and gcd_{k[y]}(a_{1}, a_{2}, b, c) = 1.

Consider the subring R = k[t_{1}, y] of B and note that R = k^{[2]}. We claim:

(17)  a_{2}t_{1} + b and ct_{1} + a_{1} are relatively prime in R.

Indeed, let δ = gcd_{R}(a_{2}t_{1} + b, ct_{1} + a_{1}). If deg_{t_{1}}(δ) > 0 then we easily obtain a contradiction with the condition a_{1}a_{2} − bc ≠ 0. So deg_{t_{1}}(δ) = 0, i.e., δ ∈ k[y]. It follows that δ is a common divisor of a_{1}, a_{2}, b, c, so (17) is a consequence of (16).

Since B ⊆ k(t_{1})[y] by (14), we have t_{2} ∈ k(t_{1})[y] so

t_{2} = f/ζ, where f ∈ R and ζ ∈ k[t_{1}] \ {0}.

This and (15) give:

(a_{2}t_{1} + b)ζ = (ct_{1} + a_{1})f (equation in R),

so (ct_{1} + a_{1}) | (a_{2}t_{1} + b)ζ in R; in view of (17), we obtain (ct_{1} + a_{1}) | ζ in R. Since ζ ∈ k[t_{1}]\{0} and k[t_{1}] is inert in R (because R = k[t_{1}]^{[1]}), it follows that ct_{1}+a_{1} ∈ k[t_{1}]. Hence,

a_{1}, c ∈ k.

Solving (15) for t_{1} gives

(15’)  t_{1} = \frac{−a_{1}t_{2} + b}{ct_{2} − a_{2}}

so, by symmetry, the proof that a_{1}, c ∈ k shows that −a_{2}, c ∈ k. Hence,

(18)  a_{1}, a_{2}, c ∈ k.

Case c = 0. Since a_{1}a_{2} − bc ≠ 0, it follows that a_{1}a_{2} ≠ 0, so a_{1}, a_{2} ∈ k^{∗} by (18). Taking this into account, (15) gives

(19)  t_{2} = αt_{1} + β, where α ∈ k^{∗} and β ∈ k[y].

Note that assumption (i) can be written as k[t_{1}] ≠ k[αt_{1} + β], so β ∉ k. By (14) and (19) we have

k[t_{1}, y] ⊆ B ⊆ k(t_{1})[y] ∩ k(αt_{1} + β)[y],

so 2.5.1 yields B = k[t_{1}, y]. In particular, (B, k) is a Danielewski surface.

Since k[t_{1}, y] = k[t_{2}, y] by (19), we also have B = k[t_{2}, y]. Now D_{i} is a derivation of k[t_{i}, y] with kernel k[t_{i}], so D_{i} = (D_{i}y) ∂/∂y and in particular D_{i}(B) ⊆ (D_{i}y)B. Now if (for each i) D_{i} is assumed to be irreducible, we have D_{i}y ∈ k^{∗} and condition (2.5-1) holds. Case c ≠ 0. Then (18) implies that c ∈ k^{∗} and a_{1}, a_{2} ∈ k.

Define x_{1} = ct_{1} + a_{1} and x_{2} = ct_{2} − a_{2} (so x_{1}, x_{2} are not defined as in the statement). We now show that x_{1}, x_{2} satisfy the following three conditions:

(2.5-2-a) ker D_{1} = k[x_{1}] and ker D_{2} = k[x_{2}]

(2.5-2-b) The k-homomorphism k[X_{1}, X_{2}, Y] → B defined by X_{1} ↦ x_{1}, X_{2} ↦ x_{2}, Y ↦ y is surjective and has kernel (φ − X_{1}X_{2})k[X_{1}, X_{2}, Y] for some nonconstant polynomial φ ∈ k[Y].

(2.5-2-c) If D_{i} is irreducible then D_{i}(y) = λ_{i}x_{i} for some λ_{i} ∈ k^{∗}.

From the definition of x_{1}, x_{2} together with c ∈ k^{∗} and a_{1}, a_{2} ∈ k, we get k[x_{i}] = k[t_{i}], so (2.5-2-a) holds. Clearly, we have x_{1}, x_{2} ∉ k, so

(20)  x_{1}x_{2} ∉ k,

for otherwise x_{1}, x_{2} ∈ B^{∗} = (ker D_{1})^{∗} = k[x_{1}]^{∗} = k^{∗}, which is not the case. Using (15’), we get

x_{1}x_{2} = (ct_{1} + a_{1})(ct_{2} − a_{2}) = [c (\frac{−a_{1}t_{2} + b}{ct_{2} − a_{2}}) + a_{1}] (ct_{2} − a_{2})

= c(−a_{1}t_{2} + b) + a_{1}(ct_{2} − a_{2}) = bc − a_{1}a_{2},

so x_{1}x_{2} ∈ k[y]; thus x_{1}x_{2} ∈ k[y] \ k by (20) and consequently:

(21)  For some nonconstant polynomial φ ∈ k[Y], we have φ(y) = x_{1}x_{2}.

Let π : k[X_{1}, X_{2}, Y] → B be the k-homomorphism defined by π(X_{1}) = x_{1}, π(X_{2}) = x_{2} and π(Y) = y. The image of π is the affine k-domain k[x_{1}, x_{2}, y], whose transcendence degree over k is 2; consequently ker π is a height one prime ideal of k[X_{1}, X_{2}, Y]; since (φ − X_{1}X_{2})k[X_{1}, X_{2}, Y] is a prime ideal and, by (21), is contained in ker π, we have:

(22)  ker π = (φ − X_{1}X_{2})k[X_{1}, X_{2}, Y].

Since we have

k[x_{1}, x_{2}, y] ⊆ B ⊆ k(x_{1})[y] ∩ k(x_{2})[y]

by (14), and since

k(x_{1})[y] ∩ k(x_{2})[y] = k[x_{1}, x_{2}, y]

by 2.2, we obtain:

(23)  B = k[x_{1}, x_{2}, y].

Thus π : k[X_{1}, X_{2}, Y] → B is surjective. Together with (22), this implies that (2.5-2-b) holds.

Hence, (B, k) is a Danielewski surface and (x_{1}, x_{2}, y) ∈ Γ_{k}(B).

Assume that D_{1} and D_{2} are irreducible. Write γ = (x_{1}, x_{2}, y) and consider the D_{i}^{γ} of 2.3. For each i ∈ {1, 2}, applying 1.9.3 to (D_{i}, D_{i}^{γ}) gives D_{i} = λ_{i}D_{i}^{γ} for some λ_{i} ∈ k^{∗}. Thus D_{i}(y) = λ_{i}D_{i}^{γ}(y) = λ_{i}x_{i}, so (2.5-2-c) holds and the condition (2.5-2) of the theorem is satisfied.

In the two cases (c = 0 or c ≠ 0), we proved that (B, k) is a Danielewski surface. Assuming that D_{1} and D_{2} are irreducible, we also proved the two implications c = 0 ⇒ (2.5-1) and c ≠ 0 ⇒ (2.5-2); so exactly one of (2.5-1), (2.5-2) is true and the proof of 2.5 is complete. □

In the special case where B is factorial, we have another characterization of Danielewski surfaces (see also 2.6.2):

2.6. Theorem. Let B be a factorial k-domain of transcendence degree 2. If B admits a k-simple derivation, then (B, k) is a Danielewski surface.

2.6.1. Example. Let B = k[x, y, y^{2}/x, y^{3}/x^{2}] where x, y are indeterminates over k. Then D = x ∂/∂y : B → B is a k-simple derivation but (B, k) is not a Danielewski surface. Note that B is normal but not factorial. (We leave it to the reader to verify that ker D = k[x], that D is irreducible and that B is not a Danielewski surface.)

Proof of 2.6. Consider a k-simple derivation D_{1} : B → B, i.e., an irreducible D_{1} ∈ lnd(B) satisfying k[D_{1}(y)] = ker D_{1} for some y ∈ B. Let x_{1} = D_{1}(y), then

ker D_{1} = k[x_{1}] = k^{[1]}.

In particular, x_{1} is a prime element of ker D_{1}; since B is factorial and ker D_{1} is inert in B,

(24)  x_{1} is a prime element of B.

Observe that D_{1}(y) = x_{1} ∈ ker(D_{1}) \ {0} implies that B_{x_{1}} = (ker D_{1})_{x_{1}}[y] = k[x_{1}, 1/x_{1}, y], so

(25)  B ⊆ k[x_{1}, 1/x_{1}, y].

It follows from (24) that 𝔪 = k[y] ∩ x_{1}B is a prime ideal of k[y]. We claim that 𝔪 is nonzero. To see this, choose β ∈ B such that D_{1}(β) ∉ x_{1}B (this is possible because D_{1} is irreducible). It is clear that D_{1} maps k[x_{1}, y] in x_{1}B, so β ∉ k[x_{1}, y]. In view of (25), we may write

β = \frac{F (x_{1}, y)}{x_{1}^{n}}, for some F ∈ k[X_{1}, Y] and n ≥ 0.

Note that we must have n > 0, because β ∉ k[x_{1}, y]. Assume that n is minimal, i.e., F (0, Y) ≠ 0. Then F (x_{1}, y) = x_{1}^{n}β ∈ x_{1}B, so F (0, y) ∈ x_{1}B and consequently F (0, y) ∈ 𝔪. Since F (0, Y) ≠ 0, we have F (0, y) ≠ 0 (because, by (25), y is transcendental over k), so 𝔪 ≠ {0}. Thus k[y] ∩ x_{1}B is a maximal ideal of k[y] and

(26)  k[y] ∩ x_{1}B = φ(y)k[y] for some irreducible element φ of k[Y].

Let x_{2} = φ(y)/x_{1} ∈ B. Let π : k[X_{1}, X_{2}, Y] → B be the homomorphism of kalgebras defined by π(X_{1}) = x_{1}, π(X_{2}) = x_{2} and π(Y) = y. Since im(π) = k[x_{1}, x_{2}, y] contains k[x_{1}, y], which is birational to B by (25), im(π) has transcendence degree 2 over k. It follows that ker π is a height one prime ideal of k[X_{1}, X_{2}, Y]. It is clear that Φ = φ − X_{1}X_{2} is an irreducible element of k[X_{1}, X_{2}, Y] and that Φ ∈ ker π, so

ker π = Φk[X_{1}, X_{2}, Y].

Let us observe that

Any two elements of {x_{1}, x_{2}, y} are algebraically independent over k.

In fact, (25) implies that x_{1}, y are algebraically independent over k and, from x_{1}x_{2} = φ(y), one easily deduces that each pair, x_{2}, y and x_{1}, x_{2}, is algebraically independent.

Let K = k(x_{2}) and B_{K} = K ⊗_{k[x_{2}]} B. Note that x_{1} ∈ K[y], since x_{2} ∈ K^{∗} and x_{1}x_{2} ∈ k[y] ⊆ K[y]. We claim:

(27)  x_{1} is a prime element of B_{K} and also of K[y].

Begin with the observation that D_{1}(x_{2}) = φ′(y) ∉ φ(y)k[y] = k[y]∩x_{1}B; since D_{1}(x_{2}) ∈ k[y], we get D_{1}(x_{2}) ∉ x_{1}B. So, if \overline{D}_{1} : B/x_{1}B → B/x_{1}B denotes D_{1} (mod x_{1}B), we have x_{2} + x_{1}B ∉ ker(\overline{D}_{1}), so x_{2} + x_{1}B is transcendental over k and consequently k[x_{2}] \ {0} ∩ x_{1}B = ∅. This implies that

(28)  x_{1} ∉ B_{K}^{∗} and x_{1} ∉ K[y]^{∗}.

Since x_{1} is prime in B and x_{1} ∉ B_{K}^{∗}, x_{1} is a prime element of B_{K}.

On the other hand, φ(y) is prime in k[y] ⟹ φ(y) is prime in k[x_{2}, y] = k[y]^{[1]} ⟹ φ(y) is either prime or a unit in k(x_{2})[y] = K[y] ⟹ x_{1} is either prime or a unit in K[y] (because x_{1} and φ(y) are associates in K[y]). By (28), x_{1} is prime in K[y] and (27) is proved.

Next, we show that

(29)  B_{K} = K[y].

In fact, (25) implies that B_{K} ⊆ K[y]_{x_{1}}, so

(30)  K[y] ⊆ B_{K} ⊆ K[y]_{x_{1}}.

By (27), K[y] ∩ x_{1}B_{K} is a prime ideal of K[y] and x_{1}K[y] is a maximal ideal of K[y]; since x_{1}K[y] ⊆ K[y] ∩ x_{1}B_{K}, we have K[y] ∩ x_{1}B_{K} = x_{1}K[y] and by induction on n we deduce:

(31)  ∀ n ∈ ℕ, K[y] ∩ x_{1}^{n}B_{K} = x_{1}^{n}K[y].

Then (29) follows from (30) and (31).

In particular, (29) implies that B ⊆ K[y] = k(x_{2})[y], so (25) gives

k[x_{1}, x_{2}, y] ⊆ B ⊆ k(x_{1})[y] ∩ k(x_{2})[y]

and we obtain B = k[x_{1}, x_{2}, y] by 2.2, i.e., π is surjective. We showed that (B, k) is a Danielewski surface, which completes the proof of 2.6. □

Note the following reformulation of 2.6:

2.6.2. Corollary. Let B be a factorial k-domain and suppose that D ∈ lnd(B) and y ∈ B satisfy:

ker D = k[Dy] = k^{[1]}.

Then (B, k) is a Danielewski surface and the following hold:

(1) If D is irreducible then there exists x ∈ B such that (Dy, x, y) ∈ Γ_{k}(B).

(2) If D is not irreducible then B = k[y, Dy] = k^{[2]}.

Proof. The hypotheses imply that trdeg_{k} B = 2. If D is irreducible then it is ksimple, so the hypothesis of 2.6 is satisfied; then the proof of 2.6 actually shows that (Dy, x_{2}, y) ∈ Γ_{k}(B) for some x_{2} ∈ B, so assertion (1) is true. If D is not irreducible then D = D(y)D_{0} for some D_{0} ∈ lnd(B), because D(y) is a prime element of B; thus D_{0}(y) = 1 and assertion (2) follows from part (3) of 1.6. □

The Transitivity Theorem and some consequences

2.7. Assume that (B, k) is a Danielewski surface, fix an element γ = (x_{1}, x_{2}, y) of Γ_{k}(B) and let φ be the unique element of k[Y] \ k satisfying φ(y) = x_{1}x_{2}. Thus

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

2.7.1. Notations ([1]-2.2).

• 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_{γ} be the subgroup of Aut_{k}(B) generated by {τ} ∪ { Δ_{f} | f ∈ k[x_{1}] }.

• Given f ∈ k[x_{1}], also define δ_{f} = Δ_{f} ∘ τ ∈ G. Note that δ_{0} = τ and that G_{γ} is generated by the set { δ_{f} | f ∈ k[x_{1}] }.

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_{γ} of Aut_{k}(B) defined in 2.7.1. Then the main result of [1] is: 2.7.2. Transitivity Theorem. The action of G_{γ} on klnd(B) is transitive.

Results 2.8, 2.9 and 2.10 are consequences of the Transitivity Theorem.

2.8. Lemma. Suppose that (B, k) is a Danielewski surface and consider an irreducible D ∈ lnd(B). Then D is k-simple, i.e., ∃ (x, y) ∈ B × B such that Dy = x and ker D = k[x]. Moreover, for each such pair (x, y) we have (x, x_{2}, y) ∈ Γ_{k}(B) for some x_{2} ∈ B.

Proof. Case 1: B = k^{[2]}. Rentschler’s Theorem 1.8 gives a pair (x′, s) such that B = k[x′, s], ker D = k[x′] and D = u ∂/∂s for some u ∈ k[x′]. Since D is irreducible, we have u ∈ k^{∗} and in fact we may choose s in such a way that Ds = 1. Then y′ = x′s satisfies D(y′) = x′, showing that D is k-simple.

Now consider any x, y ∈ B such that Dy = x and ker D = k[x]. Since B = k[x′, s] and k[x] = k[x′], B = k[x, s] (where Ds = 1, as before). Then D(y−xs) = 0, so we may write y − xs = a + xf(x) for some a ∈ k and f(x) ∈ k[x]. Define a k-homomorphism π : k[X_{1}, X_{2}, Y] → B by π(X_{1}) = x, π(X_{2}) = s + f(x) and π(Y) = y. Then π is surjective and Y − a − X_{1}X_{2} belongs to ker π, so (x, s + f(x), y) ∈ Γ_{k}(B). Case 2: B ≠ k^{[2]}. Pick any (x_{1}^{(1)}, x_{2}^{(1)}, y^{(1)}) ∈ Γ_{k}(B).

Given (x_{1}^{(j)}, x_{2}^{(j)}, y^{(j)}) ∈ Γ_{k}(B), let D_{1}^{(j)} ∈ lnd(B) be the k-simple derivation given by 2.3, i.e., ker D_{1}^{(j)} = k[x_{1}^{(j)}] and D_{1}^{(j)}(y^{(j)}) = x_{1}^{(j)}.

By the Transitivity Theorem, there exists θ_{1} ∈ Aut_{k}(B) such that θ_{1}(k[x_{1}^{(1)}]) = ker D. Let (x_{1}^{(2)}, x_{2}^{(2)}, y^{(2)}) = (θ_{1}(x_{1}^{(1)}), θ_{1}(x_{2}^{(1)}), θ_{1}(y^{(1)})) ∈ Γ_{k}(B), then ker D = k[x_{1}^{(2)}] = ker D_{1}^{(2)}. By 1.9.3 applied to the pair D_{1}^{(2)}, D,

D is k-simple and D = λ_{2}D_{1}^{(2)} for some λ_{2} ∈ k^{∗}.

For the second assertion, consider x, y ∈ B such that D(y) = x and ker D = k[x]. Then k[x] = k[x_{1}^{(2)}] and consequently x = λ′x_{1}^{(2)} +μ for some λ′ ∈ k^{∗} and μ ∈ k. Define θ_{2} ∈ Aut_{k}(B) by

θ_{2} :  x_{1}^{(2)} ⟼ λ′x_{1}^{(2)},  x_{2}^{(2)} ⟼ (λ′)^{−1}x_{2}^{(2)} and y^{(2)} ⟼ y^{(2)}

and define (x_{1}^{(3)}, x_{2}^{(3)}, y^{(3)}) = (θ_{2}(x_{1}^{(2)}), θ_{2}(x_{2}^{(2)}), θ_{2}(y^{(2)})) ∈ Γ_{k}(B). Then x = x_{1}^{(3)}+μ and D = λD_{1}^{(3)} for some λ ∈ k^{∗}. Let s = λy − y^{(3)}, then D_{1}^{(3)}(s) = (λD_{1}^{(3)})(y) − D_{1}^{(3)}y^{(3)} = x − x_{1}^{(3)} = μ.

We must have μ = 0 for otherwise D_{1}^{(3)}(s) ∈ k^{∗} would imply B = k[x, s] = k^{[2]}, which is not the case. Thus x = x_{1}^{(3)} and, since D_{1}^{(3)}(s) = 0, λy − y^{(3)} = a + xf(x) for some a ∈ k and f(x) ∈ k[x]. As we know, there is an automorphism Δ ∈ Aut_{k}(B) satisfying

Δ : x_{1}^{(3)} ⟼ x_{1}^{(3)} and y^{(3)} ⟼ y^{(3)} + x_{1}^{(3)}f(x_{1}^{(3)}).

Let (x_{1}^{(4)}, x_{2}^{(4)}, y^{(4)}) = (Δ(x_{1}^{(3)}), Δ(x_{2}^{(3)}), Δ(y^{(3)})) ∈ Γ_{k}(B), then

x_{1}^{(4)} = x and y^{(4)} = λy − a.

For each j ∈ {1, 2, 3, 4}, let π_{j} : k[X_{1}, X_{2}, Y] → B be the k-homomorphism defined by π_{j}(X_{1}) = x_{1}^{(j)}, π_{j}(X_{2}) = x_{2}^{(j)} and π_{j}(Y) = y^{(j)}.

Finally, consider Ψ ∈ Aut_{k} (k[X_{1}, X_{2}, Y]) defined by

Ψ :  X_{1} ⟼ X_{1},  X_{2} ⟼ X_{2} and Y ⟼ λY − a

and define π_{5} = π_{4} ∘ Ψ^{−1} : k[X_{1}, X_{2}, Y] → B, i.e., we have constructed a commutative diagram (where we write R = k[X_{1}, X_{2}, Y]):

(32)

R ⟶^{id} R ⟶^{id} R ⟶^{id} R ⟶^{Ψ} R
π_{1}↓  ^{π_{2}}↓  ^{π_{3}}↓  ^{π_{4}}↓  ^{π_{5}}↓
B ⟶_{θ_{1}} B ⟶_{θ_{2}} B ⟶_{Δ} B ⟶_{id} B

Then π_{5} is surjective and ker π_{5} = Ψ( ker π_{4}) is of the required form, i.e., if we define

(x_{1}^{(5)}, x_{2}^{(5)}, y^{(5)}) = (π_{5}(X_{1}), π_{5}(X_{2}), π_{5}(Y))

then (x_{1}^{(5)}, x_{2}^{(5)}, y^{(5)}) ∈ Γ_{k}(B). Since π_{5}(X_{1}) = x and π_{5}(Y) = y, we are done. □

2.9. Lemma. Suppose that (B, k) is a Danielewski surface. If (x_{1}, x_{2}, y), (x′_{1}, x′_{2}, y′) ∈ Γ_{k}(B) then there exists θ ∈ Aut_{k}(B) satisfying:

θ(x′_{1}) = x_{1},  θ(x′_{2}) = cx_{2}, for some c ∈ k^{∗},  and  θ(k[y′]) = k[y].

Proof. If B = k^{[2]} then an element of Γ_{k}(B) is a triple (x_{1}, x_{2}, αx_{1}x_{2} + β) such that B = k[x_{1}, x_{2}], α ∈ k^{∗} and β ∈ k. In this case, the assertion is trivial and we may even arrange c = 1.

The case B ≠ k^{[2]} is in fact a corollary of the proof of 2.8. We know that there exists an irreducible D ∈ lnd(B) such that Dy = x_{1} and ker D = k[x_{1}] (D is the “D_{1}^{γ}” of 2.3, where γ = (x_{1}, x_{2}, y)); so the pair (x_{1}, y) satisfies the hypothesis of 2.8. Start the proof of 2.8 with (x_{1}^{(1)}, x_{2}^{(1)}, y^{(1)}) = (x′_{1}, x′_{2}, y′) instead of picking an arbitrary (x_{1}^{(1)}, x_{2}^{(1)}, y^{(1)}) ∈ Γ_{k}(B). Going through the proof, we obtain the commutative diagram (32).

Now let θ = Δ ∘ θ_{2} ∘ θ_{1}; then θ maps x′_{i} = x_{i}^{(1)} on x_{i}^{(4)} (for i = 1, 2) and y′ = y^{(1)} on y^{(4)}. Recall that the x in the proof of 2.8 corresponds to x_{1} here, so

(33)  θ(x′_{1}) = x_{1}^{(4)} = x = x_{1}.

We also have θ(y′) = y^{(4)} = λy − a (where λ ∈ k^{∗} and a ∈ k), so

(34)  θ(k[y′]) = k[y].

By 2.4, x′_{1}x′_{2} generates the ideal k[y′] ∩ x′_{1}B of k[y′]; applying θ and taking (33) and (34) into account, we obtain:

x_{1}θ(x′_{2}) generates the ideal k[y] ∩ x_{1}B of k[y].

But 2.4 implies that x_{1}x_{2} is another generator of the same ideal of k[y]. Thus x_{1}θ(x′_{2}) and x_{1}x_{2} are associates in B, so θ(x′_{2}) = cx_{2} for some c ∈ k^{∗}. □

2.10. Lemma. Suppose that (B, k) is a Danielewski surface. Then the polynomial φ ∈ k[Y] in a representation

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

is uniquely determined by B, up to a k-automorphism of k[Y] and multiplication by a unit. In particular, the degree of φ is uniquely determined by B.

Proof. Consider (x_{1}, x_{2}, y), (x′_{1}, x′_{2}, y′) ∈ Γ_{k}(B) and the corresponding φ, ψ ∈ k[Y] satisfying x_{1}x_{2} = φ(y) and x′_{1}x′_{2} = ψ(y′). By 2.9, there exists θ ∈ Aut_{k}(B) such that θ(x′_{1}) = x_{1}, θ(x′_{2}) = cx_{2} and θ(y′) = λy − a, for some c, λ ∈ k^{∗} and a ∈ k. Thus

cφ(y) = x_{1}(cx_{2}) = θ(x′_{1}x′_{2}) = θ(ψ(y′)) = ψ(λy − a),

so cφ = ψ(λY − a) and in particular deg_{Y} φ = deg_{Y} ψ, as claimed. □

3. Definition of \underline{klnd} (B) and \underline{ℛ}(B)

Given an arbitrary integral domain B (of characteristic zero), the graphs \underline{klnd} (B) and \underline{ℛ}(B) are defined in 3.3 and 3.8 respectively. These graphs are invariants of the ring B and the group of automorphisms of B acts on each one of them.

See 1.1 for the notations A_{R}, R_{R}, etc. 3.1. Terminology of graphs. By a graph, we mean an undirected graph such that no edge connects a vertex to itself and at most one edge joins any given pair of vertices. In such a graph, the edge joining vertices u and v is represented by the set {u, v}. Two vertices are called neighbors if they are joined by an edge. If u is a vertex in a graph G, the set of neighbors of u in G is denoted N (u) or N_{G}(u). A path in G is a sequence P = (u_{0}, …, u_{k}) of vertices satisfying k ≥ 0 and:

If k ≥ 1 then e_{1} = {u_{0}, u_{1}}, e_{2} = {u_{1}, u_{2}}, …, e_{k} = {u_{k−1}, u_{k}} are edges in G.

If the edges e_{1}, …, e_{k} of P are distinct, we call P a simple path; if P satisfies the weaker condition:

e_{i} ≠ e_{i+1} for 1 ≤ i < k (or equivalently u_{i−1} ≠ u_{i+1} for 1 ≤ i < k),

we say that P is locally simple.

A spanning tree of a graph G is a subgraph of G which is a tree and whose vertex set is equal to that of G.

Let G and H be graphs with vertex sets G and H respectively. By a homomorphism of graphs f : G → H we mean a set map f : G → H satisfying:

for every edge {u, v} of G, {f(u), f(v)} is an edge of H

(note that this condition implies, in particular, that f(u) ≠ f(v)).

3.2. Definitions. Let R ⊂ B be domains such that trdeg_{R}(B) = 2.

3.2.1. If A ∈ klnd_{R}(B), define

Ω_{R}(A) = { y ∈ B | ∃ an irreducible D ∈ lnd_{A}(B) such that A_{R} = R_{R}[Dy] }.

Remarks.

(1) If A ∈ klnd(B), then lnd_{A}(B) is the set of locally nilpotent derivations of B with kernel equal to A. [This is because A is algebraically closed in B and trdeg_{A}(B) = 1, by 1.7.]

(2) If A ∈ klnd_{R}(B) and Ω_{R}(A) ≠ ∅, then A_{R} = (R_{R})^{[1]}.

[Indeed, A_{R} = R_{R}[Dy] for some D and y, and Dy must be transcendental over R_{R} since trdeg_{R}(B) = 2 and (by 1.7) trdeg_{A}(B) = 1.]

3.2.2. Let \underline{klnd}_{R}(B) be the graph with vertex set klnd_{R}(B) and whose edges are defined as follows: Given distinct A, A′ ∈ klnd_{R}(B),

\{A, A′\} is an edge of \underline{klnd}_{R}(B) ⟺ Ω_{R}(A) ∩ Ω_{R}(A′) ≠ ∅.

3.3. Definition. Given an integral domain B, let \underline{klnd} (B) be the graph with vertex set klnd(B) and where distinct A, A′ ∈ klnd(B) are neighbors if:

\{A, A′\} is an edge of \underline{klnd}_{R}(B), for some subring R of B with trdeg_{R}(B) = 2.

We also define:

\underline{klnd}_{∗}(B) = the subgraph of \underline{klnd} (B) obtained by deleting all isolated vertices.

3.4. Lemma. Let R ⊂ B be domains such that trdeg_{R}(B) = 2 and suppose that A and A′ are distinct elements of klnd_{R}(B). If R is inert in B, then R = A ∩ A′.

Proof. Note that one of the inclusions in

(35)  R ⊆ A ∩ A′ ⊆ A

must be an algebraic extension of rings, because trdeg_{R}(A) = 1; since each of R, A, A′, A ∩ A′ is an inert subring of B, and hence is algebraically closed in B, one of the inclusions in (35) must actually be an equality. Now A ≠ A′ and trdeg_{A}(B) = 1 = trdeg_{A′}(B) imply that A ≠ A ∩ A′, so R = A ∩ A′. □

3.5. Lemma. Let B be a domain and let \{A, A′\} be an edge of \underline{klnd} (B). Then there exists a unique inert subring R of B satisfying

trdeg_{R}(B) = 2 and \{A, A′\} is an edge of \underline{klnd}_{R}(B).

Moreover, R = A ∩ A′ and (B_{R}, R_{R}) is a Danielewski surface.

Proof. The assumption implies that the set

Σ = \{ R | R is a subring of B, trdeg_{R}(B) = 2 and \{A, A′\} is an edge of \underline{klnd}_{R}(B) \}

is nonempty. Consider any R_{1} ∈ Σ and define R = B ∩ Frac(R_{1}). Then R_{1} ⊆ R and Frac(R_{1}) = Frac(R); it follows that Ω_{R_{1}}(A) = Ω_{R}(A) and Ω_{R_{1}}(A′) = Ω_{R}(A′), so Ω_{R}(A) ∩ Ω_{R}(A′) ≠ ∅ and R ∈ Σ. Since A is inert in B, A_{R} is inert in B_{R} and consequently (B_{R})^{∗} = (A_{R})^{∗}. On the other hand, the fact that Ω_{R}(A) ≠ ∅ implies that A_{R} = (R_{R})^{[1]}, so (B_{R})^{∗} = (A_{R})^{∗} = (R_{R})^{∗}. Thus the first part of

(B_{R})^{∗} = (R_{R})^{∗} and B ∩ Frac(R) = R

holds, and so does the second part by definition of R. By 1.4, it follows that R is an inert subring of B. This proves that at least one element of Σ is an inert subring of B. Then 3.4 gives:

{ R ∈ Σ | R is an inert subring of B } = {A ∩ A′}.

To complete the proof, we show that if R is any element of Σ then (B_{R}, R_{R}) is a Danielewski surface. Pick y ∈ Ω_{R}(A) ∩ Ω_{R}(A′). Then there exist D, D′ ∈ lnd(B) satisfying ker D = A, ker D′ = A′, A_{R} = R_{R}[Dy] and A′_{R} = R_{R}[D′y]. Then the localized derivations D_{R}, D′_{R} ∈ lnd(B_{R}) satisfy ker(D_{R}) = A_{R} = (R_{R})^{[1]} and ker(D′_{R}) = A′_{R} = (R_{R})^{[1]}. Since D_{R} and D′_{R} are extensions of D and D′ respectively, we have A = B ∩ ker(D_{R}) and A′ = B ∩ ker(D′_{R}); so D_{R} and D′_{R} have distinct kernels. Since y is a preslice of both D_{R} and D′_{R}, 2.5 applied to the triple (y, D_{R}, D′_{R}) gives that (B_{R}, R_{R}) is a Danielewski surface. □

3.6. Definition. Given an integral domain B, let ℛ(B) denote the following set of subrings of B:

ℛ(B) = { R | R is an inert subring of B and (B_{R}, R_{R}) is a Danielewski surface }.

Note that if R ∈ ℛ(B) then trdeg_{R}(B) = 2.

The following result will be improved later (see 5.1). In part (a) of 3.7, “⊆” means “is a subgraph of”.

3.7. Corollary. If B is an integral domain then:

(a) \underline{klnd}_{∗}(B) ⊆ ⋃_{R∈ℛ(B)}\underline{klnd}_{R}(B) ⊆ \underline{klnd} (B)

(b) If R_{1}, R_{2} are distinct elements of ℛ(B), the graphs \underline{klnd}_{R_{1}}(B) and \underline{klnd}_{R_{2}}(B) have at most one vertex in common.

Proof. Assertion (a) follows from 3.5 and (b) from 3.4. □

Result 3.7 suggests a natural way to turn ℛ(B) into a graph:

3.8. Definition. Given an integral domain B, let \underline{ℛ}(B) be the graph with vertex set ℛ(B) and where distinct R_{1}, R_{2} ∈ ℛ(B) are neighbors if and only if klnd_{R_{1}}(B) ∩ klnd_{R_{2}}(B) ≠ ∅.

Equivalently, R_{1}, R_{2} ∈ ℛ(B) are neighbors in \underline{ℛ}(B) if and only if there exists a nonzero locally nilpotent derivation D : B → B satisfying D(R_{1} ∪ R_{2}) = {0}.

The structures of the graphs \underline{klnd} (B) and \underline{ℛ}(B) are closely related and (as can be inferred from 5.1, below) this is particularly true when B is factorial and affine over some field. However, we will not elaborate on this point. Let us simply say that the graphs \underline{klnd} (B) and \underline{ℛ}(B) are two invariants of the ring B, and that \underline{ℛ}(B) should be thought of as a simplified version of \underline{klnd} (B).

3.9. Actions of Aut(B). Let B be an integral domain and θ an automorphism of B. Then the following claims are trivial.

(1) If D ∈ lnd(B) and D′ = θ ∘ D ∘ θ^{−1}, then D′ ∈ lnd(B) and ker D′ = θ(ker D); if D is irreducible then so is D′.

(2) If R ∈ ℛ(B) and A ∈ klnd_{R}(B) then:

θ(R) ∈ ℛ(B), θ(A) ∈ klnd_{θ(R)}(B) and θ(Ω_{R}(A)) = Ω_{θ(R)}(θ(A)).

(3) If R ∈ ℛ(B) and A_{1}, A_{2} are distinct elements of klnd_{R}(B), then:

\{A_{1}, A_{2}\} is an edge of \underline{klnd}_{R}(B) ⇔ \{θ(A_{1}), θ(A_{2})\} is an edge of \underline{klnd}_{θ(R)}(B).

Consequently,

3.9.1. Let Aut(B) denote the group of ring automorphisms of B.

• There is a left-action of Aut(B) on the graph \underline{klnd} (B), given by

(θ, A) ↦ θA = θ(A).

• There is a left-action of Aut(B) on the graph \underline{ℛ}(B), given by

(θ, R) ↦ θR = θ(R).

3.10. The one-dimensional case. Suppose that B is a domain containing a field over which B has transcendence degree one or less.

Then it is well-known that if 0 ≠ D : B → B is a locally nilpotent derivation then B is a polynomial ring in one variable over some field, and this field is in fact the kernel of D. This simple fact can be phrased as follows:

• \underline{klnd} (B) is either the empty graph or the graph with one vertex (and no edge).

• \underline{klnd} (B) is nonempty if and only if B = k^{[1]} for some field k, in which case klnd(B) = {k}.

• \underline{ℛ}(B) is the empty graph [this is because R ∈ ℛ(B) implies trdeg_{R}(B) = 2 and B^{∗} = R^{∗}].

4. Description of the graph \underline{klnd}_{k}(B) in the two-dimensional case

The beginning of this section considers the problem of describing the graph \underline{klnd} (B) where B is an integral domain which has transcendence degree two over some field (of characteristic zero). However 4.3 shows that this problem reduces to the following: Describe the graph \underline{klnd}_{k}(B) where k is a field, B is an integral domain containing k as a subring and B has transcendence degree 2 over k. Solving this reformulated problem then becomes the aim of this section (this viewpoint is adopted in 4.4).

In 4.6 (but see also 4.3) we show that \underline{klnd}_{k}(B) is non-discrete (i.e., has at least one edge) if and only if (B, k) is a Danielewski surface. From 4.7 to the end of the section, we restrict our attention to the case where \underline{klnd}_{k}(B) is non-discrete and give a quite satisfactory description of that graph. In particular, we show that it is connected, we identify in which cases it is a tree and, in all cases, we describe a spanning tree of \underline{klnd}_{k}(B).

Result 5.1, below, is the motivation for giving such a detailed description of \underline{klnd}_{k}(B) in the non-discrete case.

The case where \underline{klnd} (B) is a discrete graph deserves to be investigated, but this is not done in this paper. In particular, one would like to know which two-dimensional rings B are such that \underline{klnd} (B) has many vertices but no edges (see 6.2 for an interesting example).

We begin by showing that the graph \underline{ℛ}(B) has at most one vertex in the twodimensional case:

4.1. Proposition. Let B be an integral domain which has transcendence degree 2 over some field. Then ℛ(B) is the set of fields k contained in B and satisfying: (B, k) is a Danielewski surface. In particular, ℛ(B) has at most one element.

Proof. Consider an arbitrary element k of ℛ(B) (a priori, k is not necessarely a field). Note that B^{∗} = k^{∗}, since k is an inert subring of B. By assumption, there exists a field k ⊂ B such that trdeg_{k}(B) = 2. Then k^{∗} ⊆ B^{∗} = k^{∗}, so k ⊆ k. Since trdeg_{k}(B) = 2 = trdeg_{k}(B), k is integral over k, so k is a field. It follows that k = {0} ∪ k^{∗} = {0} ∪ B^{∗} is uniquely determined by B, so ℛ(B) = {k}. Obviously, k ∈ ℛ(B) implies that (B, k) is a Danielewski surface.

Conversely, suppose that k ⊂ B is a field such that (B, k) is a Danielewski surface. We have B^{∗} = k^{∗} by 2.3, so k is an inert subring of B and k ∈ ℛ(B). □

Next we point out that there are edges in the graph \underline{klnd} (B) of a Danielewski surface:

4.2. Example. Suppose that (B, k) is a Danielewski surface and consider (x_{1}, x_{2}, y) ∈ Γ_{k}(B). Let A_{i} = k[x_{i}] ∈ klnd(B) (i = 1, 2). Then

\{A_{1}, A_{2}\} is an edge in \underline{klnd}_{k}(B).

Proof. For each i ∈ {1, 2}, consider the derivation D_{i}^{γ} : B → B of 2.3, where γ = (x_{1}, x_{2}, y). Then D_{i}^{γ} is an irreducible derivation, belongs to lnd_{A_{i}}(B), and satisfies ker D_{i}^{γ} = k[D_{i}^{γ}(y)]. So

y ∈ Ω_{k}(A_{1}) ∩ Ω_{k}(A_{2})

and \{A_{1}, A_{2}\} is an edge in \underline{klnd}_{k}(B). □

4.3. Corollary. Let B be an integral domain which has transcendence degree 2 over some field. Then the following three conditions are equivalent:

(1) ℛ(B) ≠ ∅

(2) There exists a field k ⊂ B such that (B, k) is a Danielewski surface.

(3) \underline{klnd} (B) has at least one edge. Moreover, we have

(∗) \underline{klnd} (B) = \underline{klnd}_{k}(B) for some field k ⊂ B satisfying trdeg_{k}(B) = 2. More precisely:

(4) If conditions (1–3) hold then the unique element k of ℛ(B) satisfies (∗).

(5) If conditions (1–3) do not hold then (∗) holds for any field k ⊂ B satisfying trdeg_{k}(B) = 2.

Proof. We have (1) ⟺^{4.1} (2) ⟹^{4.2} (3) ⟹^{3.7} (1).

To prove (4), assume that (1–3) hold and consider the unique element k of ℛ(B). Since k is a field contained in B, we have klnd(B) = klnd_{k}(B) by 1.6. So 3.7 and ℛ(B) = \{k\} give \underline{klnd} (B) = \underline{klnd}_{k}(B).

To prove (5), assume that (1–3) do not hold and consider any field k ⊂ B satisfying trdeg_{k}(B) = 2. Again, we have klnd(B) = klnd_{k}(B) by 1.6. This immediately implies that \underline{klnd} (B) = \underline{klnd}_{k}(B), since \underline{klnd} (B) has no edges. □

Result 4.3 implies, in particular, that the study of \underline{klnd} (B) reduces to that of \underline{klnd}_{k}(B). Until the end of this section, our aim is to describe the graph \underline{klnd}_{k}(B) where (B, k) is a pair satisfying: 4.4. Global assumptions. k is a field, B is an integral domain containing k as a subring and B has transcendence degree 2 over k.

4.5. Let (B, k) be a pair satisfying 4.4. Recall the following facts from 3.2: (1) For each A ∈ klnd(B) we define

Ω_{k}(A) = { y ∈ B | ∃ an irreducible D ∈ lnd_{A}(B) such that k[ Dy] = A }.

Regarding the set Ω_{k}(A), note the following.

(i) If y ∈ Ω_{k}(A) then k[ Dy] = A holds for every irreducible D ∈ lnd_{A}(B), by 1.9.3.

(ii) If Ω_{k}(A) ≠ ∅ then A = k^{[1]}.

(iii) If Ω_{k}(A) ≠ ∅ for some A, then B^{∗} = k^{∗} [because B^{∗} = A^{∗} and A = k^{[1]}].

(2) \underline{klnd}_{k}(B) is the graph with vertex set klnd_{k}(B) = klnd(B) and whose edges are defined as follows: Given distinct vertices A, A′ ∈ klnd_{k}(B),

{A, A′} is an edge if and only if Ω_{k}(A) ∩ Ω_{k}(A′) ≠ ∅.

Recall that a graph is non-discrete if it has at least one edge.

4.6. Corollary. Let (B, k) be a pair satisfying 4.4. Then \underline{klnd}_{k}(B) is non-discrete if and only if (B, k) is a Danielewski surface.

Moreover, if \{A_{1}, A_{2}\} is any edge of \underline{klnd}_{k}(B) then there exists (x_{1}, x_{2}, y) ∈ Γ_{k}(B) satisfying A_{1} = k[x_{1}] and A_{2} = k[x_{2}].

Proof. By 4.2, if (B, k) is a Danielewski surface then \underline{klnd}_{k}(B) is non-discrete.

Conversely, suppose that \{A_{1}, A_{2}\} is an edge of \underline{klnd}_{k}(B) (where A_{1}, A_{2} are distinct elements of klnd(B)). Then Ω_{k}(A_{1}) ∩ Ω_{k}(A_{2}) ≠ ∅, so we may pick an element y of that intersection. For each i ∈ {1, 2} we have y ∈ Ω_{k}(A_{i}) and consequently there exists an irreducible D_{i} ∈ lnd_{A_{i}}(B) satisfying A_{i} = k[D_{i}(y)]. Let x_{i} = D_{i}(y), then

ker D_{i} = k[x_{i}] = k^{[1]} (for each i ∈ {1, 2}).

Thus (y, D_{1}, D_{2}) satisfies the hypothesis of 2.5. Since it is clear that (2.5-1) is false, (2.5-2) must hold. So (B, k) is a Danielewski surface and (x_{1}, x_{2}, y) ∈ Γ_{k}(B). □

The graph of a Danielewski surface of degree n

In view of 2.10, the following is well-defined: 4.7. Terminology. Let n be a positive integer. The phrase “(B, k) is a Danielewski surface of degree n” means that (B, k) is a Danielewski surface and that the polynomial φ ∈ k[Y] satisfying B ≅ k[X_{1}, X_{2}, Y]/(φ − X_{1}X_{2}) has degree n.

Until the end of this section, we consider a Danielewski surface (B, k) of degree n and our aim is to describe \underline{klnd}_{k}(B). This is an important problem because of 5.1, below.

4.8. Theorem. If (B, k) is a Danielewski surface then the graph \underline{klnd}_{k}(B) is connected.

Proof. Choose γ = (x_{1}, x_{2}, y) ∈ Γ_{k}(B) and write A_{1} = k[x_{1}] and A_{2} = k[x_{2}]. By 4.2, A_{1} and A_{2} belong to the same connected component C of \underline{klnd}_{k}(B).

Consider the subgroup G = G_{γ} of Aut_{k}(B) generated by the set E = {τ} ∪ { Δ_{f} | f ∈ k[x_{1}] } (see 2.7.1). If g ∈ G then, by 3.9.1, gC is a connected component of \underline{klnd}_{k}(B). It is immediate that if g ∈ E then gA_{1} ∈ \{A_{1}, A_{2}\}, so gA_{1} ∈ C, so gC = C; it follows that gC = C for all g ∈ G. Since G acts transitively on the set klnd(B) (by 2.7.2), we conclude that \underline{klnd}_{k}(B) is connected. □

The main result of this subsection is 4.10.4, but we also point-out:

4.9. Theorem. Suppose that (B, k) is a Danielewski surface of degree n. Then \underline{klnd}_{k}(B) is a tree if and only if n > 2.

The proof of 4.9 consists of 4.9.1, 4.9.2 and part (4) of 4.10.4.

We begin by showing (in 4.9.1 and 4.9.2) that if (B, k) is a Danielewski surface of degree 1 or 2, then \underline{klnd}_{k}(B) is very far from being a tree: Each vertex belongs to a subgraph of \underline{klnd}_{k}(B) isomorphic to the complete graph on the set k.

4.9.1. Example. Suppose that (B, k) is a Danielewski surface of degree 1 (which is equivalent to B = k^{[2]} by 2.3). We first note that:

(36)  The edge set of \underline{klnd}_{k}(B) is \{\{k[u], k[v]\} | B = k[u, v]\}.

Indeed, if u, v ∈ B are such that B = k[u, v], then it is immediate that (u, v, uv) ∈ Γ_{k}(B); so 4.2 implies that {k[u], k[v]} is an edge. Conversely, suppose that {A_{1}, A_{2}} is an edge. Then, by 4.6, there exists (x_{1}, x_{2}, y) ∈ Γ_{k}(B) such that A_{1} = k[x_{1}] and A_{2} = k[x_{2}]. Since φ(y) = x_{1}x_{2} for some φ ∈ k[Y] of degree one, we have y ∈ k[x_{1}, x_{2}], so B = k[x_{1}, x_{2}]. This proves (36).

Let A ∈ klnd(B). By Rentschler’s Theorem 1.8 we may choose x_{1}, x_{2} such that B = k[x_{1}, x_{2}] and A = k[x_{1}]. For each λ = (λ_{1} : λ_{2}) ∈ ℙ_{k}^{1}, let A_{λ} = k[λ_{1}x_{1} + λ_{2}x_{2}]. Then U = { A_{λ} | λ ∈ ℙ_{k}^{1} } is a subset of klnd(B) of cardinality |k|, A ∈ U and, by (36), \{A_{λ}, A_{λ′}\} is an edge of \underline{klnd}_{k}(B) whenever λ, λ′ are distinct elements of ℙ_{k}^{1}. In other words, the complete graph on the set U is a subgraph of \underline{klnd}_{k}(B). □

4.9.2. Example. Suppose that (B, k) is a Danielewski surface of degree 2. Let A ∈ klnd(B). By the Transitivity Theorem (or by 2.8), there exists (x_{1}, x_{2}, y) ∈ Γ_{k}(B) such that A = k[x_{1}]. Consider the polynomial φ ∈ k[Y] which satisfies x_{1}x_{2} = φ(y). Then φ has degree two and depends on our choice of (x_{1}, x_{2}, y). In fact we may choose (x_{1}, x_{2}, y) in Γ_{k}(B) in such a way that A = k[x_{1}] and:

(37)  φ = Y^{2} + c for some c ∈ k.

[To see this, it suffices to observe that if (x_{1}, x_{2}, y) ∈ Γ_{k}(B), ν ∈ k^{∗} and μ ∈ k, then (x_{1}, νx_{2}, y+μ) ∈ Γ_{k}(B).] For each λ = (λ_{1} : λ_{2}) ∈ ℙ_{k}^{1}, let A_{λ} = k[λ_{1}^{2}x_{1}+2λ_{1}λ_{2}y+λ_{2}^{2}x_{2}]. Observe that A = A_{(1:0)}. We claim:

(38) \{A_{λ}, A_{λ′}\} is an edge of \underline{klnd}_{k}(B) whenever λ, λ′ are distinct elements of ℙ_{k}^{1}.

Clearly, if this is true then A belongs to a subgraph of \underline{klnd}_{k}(B) isomorphic to the complete graph on the set k. To prove (38), let α, β ∈ k and consider the element θ = Δ_{α} ∘ τ ∘ Δ_{β} of Aut_{k}(B) (see 2.7.1). Note that, given t ∈ k, Δ_{t}(x_{1}) = x_{1}, Δ_{t}(y) = y + tx_{1} and (taking (37) into account) Δ_{t}(x_{2}) = t^{2}x_{1} + 2ty + x_{2}. It follows that θ(x_{1}) = α^{2}x_{1} + 2αy + x_{2}, so

θ(k[x_{1}]) = A_{(α:1)}.

Also, θ(x_{2}) = (1 + αβ)^{2}x_{1} + 2(1 + αβ)βy + β^{2}x_{2}, so

θ(k[x_{2}]) = A_{(1+αβ:β)}.

Since \{k[x_{1}], k[x_{2}]\} is an edge in \underline{klnd}_{k}(B) by 4.2, so is \{A_{(α:1)}, A_{(1+αβ:β)}\} by 3.9.1. The claim (38) follows from this. □

4.10. Statement of the main result.

Suppose that (B, k) is a Danielewski surface of degree n and fix an element γ = (x_{1}, x_{2}, y) of Γ_{k}(B). Consider the subgroup G = G_{γ} of Aut_{k}(B) and its generating set { δ_{f} | f ∈ k[x_{1}] }, as in 2.7.1. Let A_{1} = k[x_{1}] ∈ klnd(B).

We now define a tree \underline{F}_{γ}, a subtree \underline{F}_{γ}^{∘} of \underline{F}_{γ} and homomorphisms of graphs P_{γ} : \underline{F}_{γ} → \underline{klnd}_{k}(B) and P_{γ}^{∘} : \underline{F}_{γ}^{∘} → \underline{klnd}_{k}(B).

4.10.1. Definition. E_{γ} = \begin{cases}k[x_{1}], & if n > 1 \\ x_{1}k[x_{1}], & if n = 1. \end{cases} 4.10.2. Definition. Let F_{γ} be the set of finite sequences (f_{1}, …, f_{k}) of elements of E_{γ} satisfying :

f_{i} ≠ 0 for all i ≠ 1.

Let F_{γ}^{∘} be the subset of F_{γ} whose elements are the finite sequences (f_{1}, …, f_{k}) in E_{γ} satisfying:

deg_{x_{1}}(f_{i}) ≥ 3 − n for all i ≠ 1.

Note that the empty sequence ∅ is an element of both F_{γ} and F_{γ}^{∘}.

Let \underline{F}_{γ} (resp. \underline{F}_{γ}^{∘}) be the tree with vertex-set F_{γ} (resp. F_{γ}^{∘}) and where the edges are the pairs of the form

{(f_{1}, …, f_{k}), (f_{1}, …, f_{k}, f_{k+1})}.

It is clear that \underline{F}_{γ} is a tree, that \underline{F}_{γ}^{∘} is a subtree of \underline{F}_{γ} and that \underline{F}_{γ}^{∘} = \underline{F}_{γ} whenever n ≥ 3.

4.10.3. Definition. Define a map P_{γ} : F_{γ} → klnd_{k}(B) by declaring that the element (f_{1}, …, f_{k}) of F_{γ} is mapped to the element (δ_{f_{1}} ∘ ⋯ ∘ δ_{f_{k}})(A_{1}) of klnd_{k}(B).

Let P_{γ}^{∘} : F_{γ}^{∘} → klnd_{k}(B) be the restriction of P_{γ} to F_{γ}^{∘}.

4.10.4. Theorem. The maps P_{γ} and P_{γ}^{∘} have the following properties:

(1) P_{γ} : \underline{F}_{γ} → \underline{klnd}_{k}(B) and P_{γ}^{∘} : \underline{F}_{γ}^{∘} → \underline{klnd}_{k}(B) are homomorphisms of graphs (see 3.1 for definition).

(2) P_{γ} is surjective, both as a map of vertices and as a map of edges.

(3) P_{γ}^{∘} is bijective, as a map of vertices. Consequently, P_{γ}^{∘} defines an isomorphism of trees from \underline{F}_{γ}^{∘} to some spanning tree of \underline{klnd}_{k}(B).

(4) If n > 2 then P_{γ} : \underline{F}_{γ} → \underline{klnd}_{k}(B) is an isomorphism and consequently \underline{klnd}_{k}(B) is a tree.

4.11. Preliminaries to the proof of 4.10.4.

Throughout 4.11, we suppose that (B, k) is a Danielewski surface of degree n and we fix an element γ = (x_{1}, x_{2}, y) of Γ_{k}(B). Let φ be the unique element of k[Y] \ k such that x_{1}x_{2} = φ(y). Consider the subgroup G = G_{γ} of Aut_{k}(B) and its elements τ, Δ_{f} and δ_{f} (where f ∈ k[x_{1}]), as in 2.7.1. Let A_{i} = k[x_{i}] ∈ klnd(B) for i ∈ {1, 2}.

Bidegree. Some of the material on bidegree is reproduced from [1], but there are also some additions.

Since γ = (x_{1}, x_{2}, y) ∈ Γ_{k}(B) is fixed, we may embed B in k[x_{1}, x_{1}^{−1}, y]. Each element g of k[x_{1}, x_{1}^{−1}, y] is a sum

g = ∑_{(i,j)∈ℤ×ℕ}g_{ij} x_{1}^{i}y^{j}

where g_{ij} ∈ k for all (i, j) and where the set supp_{γ}(g) = { (i, j) ∈ ℤ × ℕ | g_{ij} ≠ 0 } is finite. As in [1]-2.7, we define the bidegree map determined by γ

\begin{matrix}bideg_{γ} : k[x_{1}, x_{1}^{−1}, y] & ⟶ & ℕ × ℕ \\ g & ⟼ & (u, v) \end{matrix}

26 DANIEL DAIGLE

by declaring that u, v are the following integers:

u = max [{0} ∪ { i ∈ ℕ | (i, 0) ∈ supp_{γ}(g) }]
v = max [{0} ∪ { j ∈ ℕ | j(−1, n) ∈ supp_{γ}(g) }].

Since γ is fixed throughout 4.11, we may simply write supp g and bideg g. 4.11.1. ([1]-2.7.2). Let g ∈ k[x_{1}, 1/x_{1}, y] and (a, b) = bideg g. Then:

a > 0 ⟹ (a, 0) ∈ supp g and b > 0 ⟹ (−b, bn) ∈ supp g. Given g ∈ k[x_{1}, 1/x_{1}, y], (a, b) = bideg g, let C(g) be the unique subset of ℝ^{2} which

is closed, convex and has boundary H_{a} ∪ E ∪ H_{b}, where E is the line segment joining (−b, bn) to (a, 0), H_{a} = { (s, 0) | s ≤ a } and H_{b} = { (s, bn) | s ≤ −b }. 4.11.2. ([1]-2.7.3). Given A ∈ klnd(B) and g ∈ A, supp(g) ⊂ C(g). 4.11.3. ([1]-2.7.6). Given A ∈ klnd(B) and g ∈ A,

bideg g = (a, b) ⟹ bideg τ(g) = (b, a).

As in [1]-3.6.6, let ℕ × ℕ be endowed with the reverse lexicographic order:

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

and define for each A ∈ klnd(B)

bideg_{γ}(A) = min { bideg_{γ} f | f ∈ A \ k } ∈ ℕ × ℕ

[which makes sense because ℕ × ℕ is well-ordered]. So we have a well-defined map

bideg_{γ} : klnd(B) ⟶ ℕ × ℕ.

Recall that A = k^{[1]}; it is a straightforward exercise to prove: 4.11.4. Given A ∈ klnd(B) and f ∈ A, bideg f = bideg(A) ⇔ A = k[f].

So applying 4.11.3 (resp. [1]-3.6.4) to a generator of A yields 4.11.5 (resp. 4.11.6): 4.11.5. Given A ∈ klnd(B), bideg(A) = (a, b) ⟹ bideg(τA) = (b, a). 4.11.6. Let A ∈ klnd(B) and (a, b) = bideg A. Then

a = 0 ⟺ A = k[x_{2}] and b = 0 ⟺ A = k[x_{1}].

Finally we quote: 4.11.7. ([1]-3.9). 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. Remark. The last assertion of 4.11.7 implies, in particular, that u ∈ E_{γ}. To see this, we may assume that n = 1 (otherwise E_{γ} = k[x_{1}]); then s = (a + b)/ gcd(b, a + b) − 1 = (a + b)/ gcd(a, b) − 1, and if this is not positive then a = 0 or b = 0. However, A ≠ A_{1} and 4.11.6 give b ≠ 0, and a ≠ 0 follows from a ≥ b; so s > 0 and u ∈ x_{1}k[x_{1}] = E_{γ}.

We continue to prepare for the proof of 4.10.4. See the beginning of 4.11 for the notation.

4.11.8. Lemma. Suppose that n > 1 (resp. n = 1). Then for f ∈ k[x_{1}] we have:

Δ_{f}(A_{2}) = A_{2} ⟺ f = 0 (resp. f ∈ k).

Proof. Let F ∈ k[X_{1}] be such that F (x_{1}) = f, let φ^{(k)} ∈ k[Y] be the k-th derivative of φ, define

G = X_{2} + ∑_{k=1}^{n} \frac{φ^{(k)}}{k!} X_{1}^{k−1}F (X_{1})^{k} ∈ k[X_{1}, X_{2}, Y]

and note that Δ_{f}(x_{2}) = G(x_{1}, x_{2}, y). Then we have

Δ_{f}(A_{2}) = A_{2} ⟺ ∃\begin{matrix}λ∈k^{∗} \\ μ∈k \end{matrix} Δ_{f}(x_{2}) = λx_{2} + μ

⟺ ∃\begin{matrix}λ∈k^{∗} \\ μ∈k \end{matrix} G(x_{1}, x_{2}, y) = λx_{2} + μ

⟺ ∃\begin{matrix}λ∈k^{∗} \\ μ∈k \end{matrix} G = λX_{2} + μ (equality in k[X_{1}, X_{2}, Y]),

where the last equivalence is a consequence of 2.2 and

deg_{Y} G = \{ \begin{matrix}0, & if f = 0 \\ n − 1, & if f ≠ 0 \end{matrix} \} < n and deg_{Y} (λX_{2} + μ) < n.

The desired result follows. □

In the next result, N (A_{1}) denotes the set of neighbors of the vertex A_{1} in the graph \underline{klnd} (B).

4.11.9. Lemma. Let A_{1} = k[x_{1}] ∈ klnd(B). Then

\begin{matrix}E_{γ} & → & N (A_{1}) \\ f & ↦ & δ_{f}(A_{1}) \end{matrix}

is a well-defined bijection.

Proof. Since A_{1} is a neighbor of A_{2} = k[x_{2}], it follows from 3.9.1 that δ_{f}(A_{1}) is a neighbor of δ_{f}(A_{2}) = A_{1} for every f ∈ k[x_{1}]. Thus η : k[x_{1}] → N (A_{1}), η(f) = δ_{f}(A_{1}), is a well-defined map.

We show that η is surjective. Case n = 1. Let A ∈ N (A_{1}). Define a ∈ k^{∗} by the condition φ = aY + b (for some b ∈ k); note that Δ_{f}(x_{2}) = x_{2} + af, for every f ∈ k[x_{1}].

By (36), A = k[v] for some v satisfying B = k[x_{1}, v]. Then k[x_{1}, x_{2}] = k[x_{1}, v], so λv = x_{2} + af for some λ ∈ k^{∗} and f ∈ k[x_{1}]. Now Δ_{f}(x_{2}) = x_{2} + af = λv, so η(f) = δ_{f}(A_{1}) = Δ_{f}(A_{2}) = k[v] = A. Case n > 1 (i.e., B ≠ k^{[2]}). Let A ∈ N (A_{1}). By 4.6, there exists γ′ = (x′_{1}, x′_{2}, y′) ∈ Γ_{k}(B) such that A_{1} = k[x′_{1}] and A = k[x′_{2}]. Applying 4.2 to γ′ gives y′ ∈ Ω_{k}(A_{1}); thus A_{1} = k[D_{1}(y′)], where D_{1} = D_{1}^{γ} ∈ lnd_{A_{1}}(B) is such that D_{1}(y) = x_{1} (see 2.3 and remark (i) in part (1) of 4.5). So k[D_{1}(y′)] = k[x_{1}], which implies that D_{1}(y′) = λx_{1}+μ for some λ ∈ k^{∗} and μ ∈ k. Since (x′_{1}, x′_{2}, λ^{−1}y′) ∈ Γ_{k}(B), we may in fact arrange that D_{1}(y′) = x_{1} + μ for some μ ∈ k. Since D_{1}(y′ − y) = μ and B ≠ k^{[2]}, μ = 0; so D_{1}(y′) = x_{1} = D_{1}(y).

Note that there is an irreducible D′_{1} ∈ lnd_{A_{1}}(B) such that D′_{1}(y′) = x′_{1} (namely, D′_{1} = D_{1}^{γ′}). By 1.9.3, we have D_{1} = λD′_{1} (some λ ∈ k^{∗}), so x_{1} = D_{1}(y′) = λD′_{1}(y′) = λx′_{1} and consequently (x_{1}, x′_{2}, y′) = (λx′_{1}, x′_{2}, y′) ∈ Γ_{k}(B). To summarize,

(x_{1}, x′_{2}, y′) ∈ Γ_{k}(B), A = k[x′_{2}] and D_{1}(y′) = x_{1} = D_{1}(y).

Since D_{1}(y′) = D_{1}(y), y′ − y ∈ k[x_{1}]. Noting that (x_{1}, x′_{2}, y′ + c) ∈ Γ_{k}(B) for every c ∈ k, we may also arrange that y′ − y ∈ x_{1}k[x_{1}]. Then for some f ∈ k[x_{1}] we have y′ = y + x_{1}f = Δ_{f}(y).

Since (x_{1}, x_{2}, y) ∈ Γ_{k}(B), it follows that

(x_{1}, Δ_{f}(x_{2}), y′) = (Δ_{f}(x_{1}), Δ_{f}(x_{2}), Δ_{f}(y)) ∈ Γ_{k}(B).

Hence, both (x_{1}, x′_{2}, y′) and (x_{1}, Δ_{f}(x_{2}), y′) belong to Γ_{k}(B). By 2.4, each of x_{1}x′_{2} and x_{1}Δ_{f}(x_{2}) generates the ideal k[y′] ∩ x_{1}B of k[y′]. It follows that x′_{2} and Δ_{f}(x_{2}) are associates, so

η(f) = δ_{f}(A_{1}) = Δ_{f}(A_{2}) = k[Δ_{f}(x_{2})] = k[x′_{2}] = A.

So η : k[x_{1}] → N (A_{1}) is a surjective map.

Consider again A ∈ N (A_{1}) and pick f_{0} ∈ k[x_{1}] such that η(f_{0}) = A. Then, for g ∈ k[x_{1}] we have

η(g) = A ⟺ η(g) = η(f_{0}) ⟺ Δ_{g}(A_{2}) = Δ_{f_{0}}(A_{2}) ⟺ Δ_{g−f_{0}}(A_{2}) = A_{2}

and, in view of 4.11.8, the last condition is equivalent to g = f_{0} (resp. g − f_{0} ∈ k) if n > 1 (resp. if n = 1). Thus η is bijective if n > 1; and if n = 1 then exactly one element g of x_{1}k[x_{1}] satisfies η(g) = A. □

4.11.10. Proposition. Let h ∈ B \ k, let f ∈ E_{γ} \ {0} and assume the following: (i) h ∈ A for some A ∈ klnd(B) (ii) a > b, where (a, b) = bideg h.

Then bideg δ_{f}(h) = (ea, a) where e = n[1 + deg_{x_{1}}(f)] − 1. Moreover,

e ≥ 1 and [ e > 1 ⇔ n + deg_{x_{1}}(f) ≥ 3 ].

Proof. Let d = 1 + deg_{x_{1}}(f) ≥ 1, then e = nd − 1 ≥ 0. If e = 0, then nd = 1, so n = 1 and f ∈ k^{∗}, which contradicts the assumption that f ∈ E_{γ}. Hence, e ≥ 1. The equivalence e > 1 ⇔ n + deg_{x_{1}}(f) ≥ 3 is trivial if n ≥ 3, and is easily verified for each n ∈ {1, 2}.

Note that a ≥ 1, because (a, b) ∈ ℕ × ℕ and a > b; also, h ∈ A and 4.11.3 imply

(39)  bideg(τh) = (b, a).

We have B ⊂ R, where R = k[x_{1}, x_{1}^{−1}, y], and observe that Δ_{f} ∈ Aut_{k}(B) extends to Δ_{f} ∈ Aut_{R_{0}}(R), where R_{0} = k[x_{1}, x_{1}^{−1}]. Given (i, j) ∈ ℤ × ℕ, consider

S_{ij} = supp [Δ_{f}(x_{1}^{i}y^{j})] = supp [x_{1}^{i}(y + x_{1}f)^{j}].

Direct calculation shows that (i, j), (i + jd, 0) ∈ S_{ij} ⊂ T_{ij}, where T_{ij} ⊂ ℝ^{2} denotes the triangular region with vertices (i, j), (i + j, 0) and (i + jd, 0).

[figure: i; j; .; .; .; .; .; .; .; .; .; .; .; .; .; .; .; .; .; .; .; .; .; .; .; .; .; .; .; .; .; .; -; this number ≥ nb_{ij}; (i, j); (i + j, 0); (i + jd, 0); (−j/n, j); line ni + j = 0; (−j/n, j)]

[figure: .; .; .; .; .; .; .; .; .; .; .; .; .; .; .; .; .; .; .; .; .; .; .; .; .; .; .; (−a, an); (b, 0); i; j; E; line “i + jd =constant”; C(τh)]
Figure 1  Figure 2

Thus bideg [Δ_{f}(x_{1}^{i}y^{j})] = (i+jd, b_{ij}), for some b_{ij}. By definition of bidegree, b_{ij}(−1, n) ∈ supp[Δ_{f}(x_{1}^{i}y^{j})] ⊂ T_{ij}, so nb_{ij} ≤ j [because any point (i′, j′) of T_{ij} satisfies j′ ≤ j]; we record:

(40)  b_{ij} ≤ j/n,

where equality holds if and only if ni + j = 0 or j = 0 (see Figure 1).

Suppose now that (i, j) ∈ supp(τh). Since τ(h) ∈ τA ∈ klnd(B), we may apply 4.11.2 to τ(h) and conclude that supp(τh) ⊂ C(τh); since bideg(τh) = (b, a) by (39), we have^{1}

(b, 0), (−a, an) ∈ supp(τh) ⊂ C(τh).

In particular, (i, j) ∈ C(τh) implies that j ≤ an, so

(41)  j/n ≤ a.

By (40) and (41) we have b_{ij} ≤ a for all (i, j) ∈ supp(τh). If (i, j) ∈ supp(τh) satisfies b_{ij} = a, then equality must hold in both (40) and (41), so (i, j) = (−a, na). Note that (−a, na) does belong to supp(τh) and b_{−a,na} = a [because if we regard Δ_{f}(x_{1}^{−a}y^{na}) = x_{1}^{−a}(y + x_{1}f)^{na} as a polynomial in y with coefficients in k[x_{1}], then the leading term is x_{1}^{−a}y^{na}, which shows that b_{−a,na} = a]. So the second component of bideg [Δ_{f}(τh)] is a, i.e.,

The second component of bideg [δ_{f}(h)] is a.

Clearly, the slope of a line “i + jd =constant” is equal to −1/d, and the slope of the line segment E joining (b, 0) to (−a, an) is −na/(a + b); thus

(slope of line “i + jd =constant”) − (slope of E) =

\frac{na}{a + b} − \frac{1}{d} = −\frac{nda − a − b}{d(a + b)} = \frac{ea − b}{d(a + b)} > 0

1In our case, C(τh) is the closed and convex subset of ℝ^{2} with boundary H ∪ E ∪ H′, where E is the line segment joining (−a, an) to (b, 0), H = { (s, 0) | s ≤ b } and H′ = { (s, an) | s ≤ −a }.

because e ≥ 1 and a > b. Consequently,

0 > slope of line “i + jd =constant” > slope of E.

Hence, the maximum value of i + jd on supp(τh) is reached at the point (−a, an) and at no other point (see Figure 2). Since bideg [Δ_{f}(x_{1}^{i}y^{j})] = (i + jd, b_{ij}), it follows that the first component of bideg [Δ_{f}(τh)] is −a + and = (−1 + nd)a = ea. So

bideg [δ_{f}(h)] = (ea, a),

as desired. □

4.11.11. Proposition. For each A ∈ klnd(B) \ {A_{1}}, there exists a unique f ∈ E_{γ} satisfying the following condition:

If we define (a, b) = bideg(A), A′ = δ_{f}^{−1}(A) and (a′, b′) = bideg(A′),

then (a′, b′) < (a, b) and a′ > b′.

Moreover, we have deg_{x_{1}}(f) ≥ 3 − n ⟺ a > b.

Proof. We prove the existence of f by induction on bideg(A). Note that bideg(A) ≥ (0, 1), by 4.11.6.

If a < b then f = 0 ∈ E_{γ} satisfies the desired condition, by (4.11.5). In particular, this proves the case bideg(A) = (0, 1), i.e., the base case of induction.

Assume that a ≥ b; by 4.11.7 and the remark following it, there exists u ∈ E_{γ} such that, if we write R = Δ_{u}(A), then bideg(R) = (a_{1}, b) with a_{1} < a, so bideg(R) < bideg(A). Observe that if R = A_{1} then A = Δ_{−u}(A_{1}) = A_{1}, a contradiction; hence R ≠ A_{1}.

Since R ≠ A_{1} and bideg(R) < bideg(A), we may assume by induction that there exists g ∈ E_{γ} such that, if we set A′ = δ_{g}^{−1}(R), then bideg(A′) < bideg(R) and a′ > b′ where (a′, b′) = bideg(A′). Then

A′ = δ_{g}^{−1}(R) = δ_{g}^{−1}Δ_{u}(A) = τΔ_{−g}Δ_{u}(A) = τΔ_{u−g}(A) = δ_{g−u}^{−1}(A).

Note that E_{γ} is closed under addition, so g − u ∈ E_{γ}. Thus f = g − u satisfies the desired condition, which proves the existence of f.

We now prove uniqueness of f. Suppose that f, g ∈ E_{γ} satisfy the conditions a′ > b′ and a′′ > b′′, where:

A′ = δ_{f}^{−1}(A), (a′, b′) = bideg(A′), A′′ = δ_{g}^{−1}(A), (a′′, b′′) = bideg(A′′).

Since δ_{f}(A′) = δ_{g}(A′′), it follows that Δ_{f}τ(A′) = Δ_{g}τ(A′′), so Δ_{f−g}τ(A′) = τ(A′′), i.e.,

(42)  δ_{f−g}(A′) = τA′′.

By (4.11.5), τA′′ has bidegree (b′′, a′′); since b′′ < a′′, the bidegree of the ring δ_{f−g}(A′) = τA′′ cannot be of the form (ea′, a′), where e is a positive integer.

If f−g ≠ 0 then f−g ∈ E_{γ}\{0} (and a′ > b′), so 4.11.10 implies that bideg[Δ_{f−g}τ(A′)] = (ea′, a′) for some positive integer e. This contradicts the preceding paragraph, so f − g = 0, i.e., f is unique.

Finally, we prove the last assertion of 4.11.11. Let e = n[1 + deg_{x_{1}}(f)] − 1.

Suppose that a > b. If f = 0 then A′ = δ_{0}^{−1}(A) = τ(A), so (a′, b′) = bideg(A′) = (b, a) by (4.11.5), and since a′ > b′ we get a < b, a contradiction. Hence, f ∈ E_{γ} \ {0}. Since we also have a′ > b′, 4.11.10 implies that bideg [δ_{f}(A′)] = (ea′, a′), where e ≥ 1. So (a, b) = (ea′, a′); now the assumption a > b implies that e > 1, which gives deg_{x_{1}}(f) ≥ 3 − n.

Conversely, suppose that deg_{x_{1}}(f) ≥ 3 − n. Then f ≠ 0, so f ∈ E_{γ} \ {0}; together with a′ > b′ and 4.11.10, this implies that (a, b) = (ea′, a′) where e ≥ 1. But in fact the condition deg_{x_{1}}(f) ≥ 3 − n implies that e > 1, so a > b. □

4.11.12. Lemma. Let (f_{1}, …, f_{k}) ∈ F_{γ}^{∘} and define

R_{i} = (δ_{f_{i+1}} ⋯ δ_{f_{k}})(A_{1}) and (a_{i}, b_{i}) = bideg(R_{i}) (0 ≤ i ≤ k).

Then (1, 0) = (a_{k}, b_{k}) < ⋯ < (a_{0}, b_{0}) and, for each i > 0, a_{i} > b_{i}. Moreover,

a_{0} > b_{0} ⟺ deg_{x_{1}}(f_{1}) ≥ 3 − n.

Proof. Since R_{k} = A_{1}, (1, 0) = (a_{k}, b_{k}) and a_{k} > b_{k} are clear. Suppose that for some j ∈ {1, …, k} we have

(1, 0) = (a_{k}, b_{k}) < ⋯ < (a_{j}, b_{j}) and, for each i ∈ {j, …, k}, a_{i} > b_{i}.

Proceding by descending induction, it suffices to prove:

(43)  (a_{j}, b_{j}) < (a_{j−1}, b_{j−1}) and [j = 1 or a_{j−1} > b_{j−1}].

We consider two cases. If j > 1 then the definition of F_{γ}^{∘} gives f_{j} ∈ E_{γ} and deg_{x_{1}}(f_{j}) ≥ 3−n; together with a_{j} > b_{j} and 4.11.10, this implies that bideg [δ_{f_{j}}(R_{j})] = (ea_{j}, a_{j}) for some e > 1. Since δ_{f_{j}}(R_{j}) = R_{j−1}, this gives (a_{j−1}, b_{j−1}) = (ea_{j}, a_{j}), so (a_{j}, b_{j}) < (a_{j−1}, b_{j−1}) and a_{j−1} > b_{j−1}, i.e., (43) holds.

If j = 1 then we still have f_{1} ∈ E_{γ} and a_{1} > b_{1}. If f_{1} = 0 then R_{0} = δ_{0}R_{1} = τR_{1} has bidegree (b_{1}, a_{1}) by (4.11.5), so (a_{1}, b_{1}) < (a_{0}, b_{0}); if f_{1} ≠ 0 then 4.11.10 implies that R_{0} = δ_{f_{1}}R_{1} has bidegree (ea_{1}, a_{1}) for some e ≥ 1, so again (a_{1}, b_{1}) < (a_{0}, b_{0}). Hence,

(43) holds in all cases.

To prove that a_{0} > b_{0} ⇔ deg_{x_{1}}(f_{1}) ≥ 3 − n, observe that the conditions

f_{1} ∈ E_{γ}, R_{1} = δ_{f_{1}}^{−1}(R_{0}), (a_{1}, b_{1}) < (a_{0}, b_{0}) and a_{1} > b_{1}

show that f_{1} is the unique element of E_{γ} determined by R_{0} ∈ klnd(B) \ {A_{1}} (see 4.11.11); then the last assertion of 4.11.11 is the desired result. □

Proof of 4.10.4. Consider an edge \{𝔣, 𝔣′\} of \underline{F}_{γ}, where

𝔣 = (f_{1}, …, f_{k}) and 𝔣′ = (f_{1}, …, f_{k}, f_{k+1});

write δ = δ_{f_{1}} ∘ ⋯ ∘ δ_{f_{k}}, A = P_{γ}(𝔣) = δ(A_{1}) and A′ = P_{γ}(𝔣′) = δ ∘ δ_{f_{k+1}}(A_{1}). Since δ^{−1}(A′) = δ_{f_{k+1}}(A_{1}) is a neighbor of δ^{−1}(A) = A_{1} by 4.11.9, it follows that A′ is a neighbor of A. This proves (1).

Observe that the connectedness of \underline{klnd}_{k}(B) (4.8) implies that every vertex of \underline{klnd}_{k}(B) is an endpoint of some edge; so, in order to prove (2), it suffices to prove surjectivity on the edges. Now, again by connectedness of \underline{klnd}_{k}(B), if e is any edge of \underline{klnd}_{k}(B) then there exists a simple path P with initial point A_{1} and which traverses e. So it suffices to prove:

(44)

Suppose that P = (R_{0}, …, R_{k}) is a locally simple path in \underline{klnd}_{k}(B) such that R_{0} = A_{1}. Then there exists 𝔣 = (f_{1}, …, f_{k}) ∈ F_{γ} such that

{(δ_{f_{1}} ⋯ δ_{f_{i}})(A_{1})}_{i=0}^{k} = P.

If k = 0 then 𝔣 = ∅ (empty sequence) satisfies (44). Assume that k > 0 and that (f_{1}, …, f_{k−1}) ∈ F_{γ} is such that

{(δ_{f_{1}} ⋯ δ_{f_{i}})(A_{1})}_{i=0}^{k−1} = (R_{0}, …, R_{k−1}).

Write δ = δ_{f_{1}} ∘ ⋯ ∘ δ_{f_{k−1}}. Then δ^{−1}(R_{k}) is a neighbor of δ^{−1}(R_{k−1}) = A_{1} so, by 4.11.9, there is a unique f_{k} ∈ E_{γ} such that δ_{f_{k}}(A_{1}) = δ^{−1}(R_{k}); this implies that (δ_{f_{1}} ⋯ δ_{f_{k}})(A_{1}) = R_{k} so there remains only to check that (f_{1}, …, f_{k}) ∈ F_{γ}.

Assume that (f_{1}, …, f_{k}) ∉ F_{γ}, then we must have k > 1 and f_{k} = 0; writing δ′ = δ_{f_{1}} ∘ ⋯ ∘ δ_{f_{k−2}}, we have

R_{k} = δ′ ∘ δ_{f_{k−1}} ∘ δ_{0}(A_{1}) = δ′ ∘ Δ_{f_{k−1}} ∘ τ^{2}(A_{1}) = δ′(A_{1}) = R_{k−2},

contradicting the hypothesis that P is locally simple. So (44) is proved and so is assertion (2).

Let A ∈ klnd(B). By induction on (a, b) = bideg(A), we show that A is in the image of P_{γ}^{∘} : F_{γ}^{∘} → klnd_{k}(B). If (a, b) = (1, 0) then A = A_{1} by 4.11.6, so P_{γ}^{∘}(∅) = A.

Suppose that (a, b) > (1, 0); then A ∈ klnd(B) \ {A_{1}}. By 4.11.11, there exists f ∈ E_{γ} such that, if we define

A′ = δ_{f}^{−1}(A) and (a′, b′) = bideg(A′),

then a′ > b′ and (a′, b′) < (a, b). By induction, we may assume that A′ = P_{γ}^{∘}(𝔣′) for some vertex 𝔣′ = (f_{1}, …, f_{k}) of \underline{F}_{γ}^{∘}. We claim that

(45)  𝔣 = (f, f_{1}, …, f_{k}) is a vertex of \underline{F}_{γ}^{∘}.

To see this, it suffices to show that, if 𝔣′ ≠ ∅, then deg_{x_{1}}(f_{1}) ≥ 3 − n. Assume that 𝔣′ ≠ ∅ and apply 4.11.11 to (f_{1}, …, f_{k}); then the last assertion of 4.11.11 reads deg_{x_{1}}(f_{1}) ≥ 3 − n ⇔ a′ > b′. Since a′ > b′ does hold, (45) follows. Clearly, P_{γ}^{∘}(𝔣) = δ_{f}(A′) = A. Thus P_{γ}^{∘} is surjective on vertices.

Notice the following consequence of 4.11.12: If 𝔣 is a vertex of \underline{F}_{γ}^{∘} other than ∅, then P_{γ}^{∘}(𝔣) has bidegree strictly greater than (1, 0); in other words, the only element of P_{γ}^{∘}^{−1}(A_{1}) is the empty sequence.

Suppose that P_{γ}^{∘} is not injective (on vertices). Then we may choose distinct vertices 𝔣, 𝔣′ of \underline{F}_{γ}^{∘} such that P_{γ}^{∘}(𝔣) = P_{γ}^{∘}(𝔣′). Write 𝔣 = (f_{1}, …, f_{k}) and 𝔣′ = (g_{1}, …, g_{m}) and assume that we have chosen 𝔣, 𝔣′ such that k +m is minimal. Write A = P_{γ}^{∘}(𝔣) = P_{γ}^{∘}(𝔣′).

Since 𝔣 ≠ 𝔣′, at least one of 𝔣, 𝔣′ is nonempty, so A ≠ A_{1} by the preceding paragraph, so both 𝔣 and 𝔣′ are nonempty.

Result 4.11.12 implies that each element f of {f_{1}, g_{1}} satisfies:

If we define (a, b) = bideg(A), A′ = δ_{f}^{−1}(A) and (a′, b′) = bideg(A′),

then (a′, b′) < (a, b) and a′ > b′.

So the uniqueness part of 4.11.11 implies that f_{1} = g_{1}.

Notice that 𝔣_{∗} = (f_{2}, …, f_{k}) and 𝔣′_{∗} = (g_{2}, …, g_{m}) belong to F_{γ}^{∘}. Since f_{1} = g_{1}, P_{γ}^{∘}(𝔣_{∗}) = P_{γ}^{∘}(𝔣′_{∗}); by minimality of k + m we obtain 𝔣_{∗} = 𝔣′_{∗}, which implies that 𝔣 = 𝔣′, a contradiction. This proves assertion (3).

If n > 2 then F_{γ} = F_{γ}^{∘} and P_{γ} = P_{γ}^{∘}. By (1–3), P_{γ} is a homomorphism of graphs which is bijective on vertices and surjective on edges; it follows that it is an isomorphism, so (4) is true.

This completes the proof of 4.10.4. □

5. Factorial affine domains

By a factorial affine domain, we mean a UFD which is affine over some field (of characteristic zero, as always in this paper). The main result of this section is 5.1, which improves 3.7.

5.1. Theorem. If B is a factorial affine domain then:

(1) \underline{klnd}_{∗}(B) = ⋃_{R∈ℛ(B)} \underline{klnd}_{R}(B)

(2) For each R ∈ ℛ(B), \underline{klnd}_{R}(B) is isomorphic to \underline{klnd}_{R_{R}}(B_{R}) and (B_{R}, R_{R}) is a Danielewski surface. In particular, \underline{klnd}_{R}(B) is infinite and connected.

(3) If R, R′ are distinct elements of ℛ(B), the graphs \underline{klnd}_{R}(B) and \underline{klnd}_{R′}(B) have at most one vertex in common.

Remarks.

• Assertion 5.1(3) simply repeats 3.7(b).

• In 5.1(2), the fact that (B_{R}, R_{R}) is a Danielewski surface follows from the definition of ℛ(B), and we know by section 4 that the graph of a Danielewski surface is infinite and connected.

The proof of 5.1 requires some preparation. First, we define a set ℛ^{in}(B) of subrings of B which is larger than ℛ(B):

5.2. Definition. Given an integral domain B,

ℛ^{in}(B) = { R | R is an inert subring of B and trdeg_{R}(B) = 2 }.

5.3. Lemma. Let B be a factorial affine domain and R ∈ ℛ^{in}(B). Then:

(a) The map Λ : klnd_{R}(B) → klnd_{R_{R}}(B_{R}), A ↦ A_{R}, is well-defined and bijective. Its inverse is given by A ↦ A ∩ B.

(b) The bijection Λ is an isomorphism of graphs, \underline{klnd}_{R}(B) → \underline{klnd}_{R_{R}}(B_{R}).

Proof. It is clear that B_{R} has transcendence degree two over R_{R}, so it makes sense to consider the graph \underline{klnd}_{R_{R}}(B_{R}). Note that klnd_{R_{R}}(B_{R}) = klnd(B_{R}), by 1.6 and the fact that R_{R} is a field contained in B_{R}.

We prove (a) now, and (b) will be proved after 5.3.3, below.

The fact that Λ : klnd_{R}(B) → klnd_{R_{R}}(B_{R}) is well-defined and injective is a consequence of part (2) of 1.6.

Before proving that Λ is surjective, we first note that B is affine over R. Indeed, we have B^{∗} = R^{∗} because R is an inert subring of B. Let k ⊆ B be a field over which B is affine. Then k^{∗} ⊆ B^{∗} = R^{∗}, so k ⊆ R ⊂ B and it follows that B is affine over R.

To show that Λ is surjective, consider A ∈ klnd_{R_{R}}(B_{R}). Choose D ∈ lnd_{R_{R}}(B_{R}) such that ker D = A. Since B is affine over R, we may consider b_{1}, …, b_{n} such that B = R[b_{1}, …, b_{n}]. For each i ∈ {1, …, n}, we have D(b_{i}) ∈ B_{R}; so there exists r ∈ R \ {0} satisfying

∀_{i} rD(b_{i}) ∈ B.

Since the derivation rD : B_{R} → B_{R} maps R to 0 and maps each b_{i} in B, it maps B into itself; also, rD is locally nilpotent, since r ∈ ker D. Let D : B → B be the restriction of rD, then D ∈ lnd_{R}(B) and ker D = A, where we define A = B ∩ A. Since D has a unique extension to a derivation of B_{R}, we have D_{R} = rD; by 1.6, the kernel of D_{R} is A_{R}, so we obtain A = A_{R} = Λ(A). So Λ is surjective and (a) is proved. □

The next three facts are needed for the proof of 5.3(b). The first one is well-known and easy to prove.

5.3.1. Let B be a UFD and A ∈ klnd(B). Then:

(1) There exists an irreducible D ∈ lnd_{A}(B).

(2) If D_{1}, D_{2} ∈ lnd_{A}(B) are irreducible, then D_{2} = λD_{1} for some λ ∈ B^{∗}.

5.3.2. Lemma. Let B be a UFD, R an inert subring of B and D : B → B an irreducible R-derivation. Then D_{R} : B_{R} → B_{R} is irreducible.

Proof. Assume the contrary; then there exists b ∈ B_{R} \ B_{R}^{∗} such that D_{R}(B_{R}) ⊆ bB_{R}. In fact, such an element b may be chosen in B. Then some prime factor p ∈ B of b satisfies p ∉ B_{R}^{∗}.

Since D is irreducible and p ∉ B^{∗}, we may choose x ∈ B such that Dx ∉ pB. Since D(x) = D_{R}(x) ∈ pB_{R}, there exists r ∈ R \ {0} such that p | rD(x) in B. Then p | r in B; since r ∈ R \ {0} and R is an inert subring of B, p ∈ R \ {0}. Thus p ∈ B_{R}^{∗}, a contradiction. □

5.3.3. Lemma. Let B be a UFD, R ∈ ℛ^{in}(B) and K = R_{R}. Then, for each A ∈ klnd_{R}(B),

Ω_{R}(A) = B ∩ Ω_{K}(A_{R}).

Remark. Since K is a field contained in A_{R}, we have (A_{R})_{K} = A_{R}. So the definition of Ω_{K}(A_{R}) reads:

Ω_{K}(A_{R}) = { ζ ∈ B_{R} | ∃ an irreducible Δ ∈ lnd_{A_{R}}(B_{R}) such that A_{R} = K[Δζ] }.

Proof of 5.3.3. Let y ∈ Ω_{R}(A). Then there exists an irreducible D ∈ lnd_{A}(B) such that A_{R} = K[Dy]. By 1.6, D_{R} : B_{R} → B_{R} belongs to lnd_{A_{R}}(B_{R}); moreover, D_{R} is irreducible by 5.3.2. Since A_{R} = K[Dy] = K[D_{R}(y)], we have y ∈ Ω_{K}(A_{R}). This proves that Ω_{R}(A) ⊆ B ∩ Ω_{K}(A_{R}).

Conversely, suppose that y ∈ B ∩ Ω_{K}(A_{R}). Then there exists an irreducible Δ ∈ lnd_{A_{R}}(B_{R}) such that A_{R} = K[Δy]. On the other hand, 5.3.1 allows us to consider an irreducible D ∈ lnd_{A}(B) and, by 5.3.2, D_{R} is irreducible. Thus D_{R} and Δ are two irreducible derivations belonging to lnd_{A_{R}}(B_{R}); using 5.3.1 again, we get D_{R} = λΔ for some λ ∈ B_{R}^{∗}. Since R is inert in B, K is inert in B_{R}, so B_{R}^{∗} = K^{∗} and λ ∈ K^{∗}. So

K[Dy] = K[D_{R}(y)] = K[λΔ(y)] = K[Δ(y)] = A_{R},

showing that y ∈ Ω_{R}(A). This proves that B ∩ Ω_{K}(A_{R}) ⊆ Ω_{R}(A). □

Proof of 5.3(b). Write K = R_{R}. We have to verify that, given distinct A, A′ ∈ klnd_{R}(B),

(46)  Ω_{R}(A) ∩ Ω_{R}(A′) ≠ ∅ ⟺ Ω_{K}(A_{R}) ∩ Ω_{K}(A′_{R}) ≠ ∅.

By 5.3.3, we have in particular Ω_{R}(A) ⊆ Ω_{K}(A_{R}) and Ω_{R}(A′) ⊆ Ω_{K}(A′_{R}), so “ ⟹ ” holds in (46).

Conversely, suppose that ω ∈ Ω_{K}(A_{R}) ∩ Ω_{K}(A′_{R}). For any λ ∈ K^{∗}, we have λω ∈ Ω_{K}(A_{R}) ∩ Ω_{K}(A′_{R}); choose λ ∈ R \ {0} such that λω ∈ B, then 5.3.3 gives

λω ∈ B ∩ Ω_{K}(A_{R}) ∩ Ω_{K}(A′_{R}) = Ω_{R}(A) ∩ Ω_{R}(A′),

so “⟸” holds in (46). This proves 5.3(b). □

Proof of 5.1. Assertion (3) (of 5.1) is given in 3.7, so only (1) and (2) need proof.

If R ∈ ℛ(B) then (by definition) (B_{R}, R_{R}) is a Danielewski surface; so \underline{klnd}_{R_{R}}(B_{R}) is connected by 4.8, and contains infinitely many vertices by (say) 4.11.9. Now R ∈ ℛ(B) also implies that R ∈ ℛ^{in}(B), so \underline{klnd}_{R}(B) ≅ \underline{klnd}_{R_{R}}(B_{R}) by 5.3; this proves assertion (2).

For each R ∈ ℛ(B), assertion (2) implies that \underline{klnd}_{R}(B) has no isolated vertex; thus ⋃_{R∈ℛ(B)} \underline{klnd}_{R}(B) ⊆ \underline{klnd}_{∗}(B). This and 3.7 imply assertion (1).

This completes the proof of 5.1. □

6. Some philosophical remarks

Given any integral domain B (of characteristic zero) we have defined three graphs, \underline{klnd} (B), \underline{klnd}_{∗}(B) and \underline{ℛ}(B), which are invariants of B up to isomorphism. Moreover, the structures of \underline{klnd}_{∗}(B) and \underline{ℛ}(B) are closely related and \underline{ℛ}(B) should be thought of as a “simplified version” of \underline{klnd}_{∗}(B): If B is factorial and affine, \underline{ℛ}(B) is isomorphic to the graph obtained from \underline{klnd}_{∗}(B) by shrinking each connected subgraph \underline{klnd}_{R}(B) (where R ∈ ℛ(B)) to a single vertex.

To illustrate the claim that \underline{klnd}_{∗}(B) and \underline{ℛ}(B) have closely related structures we mention the following easy consequence of 5.1:

If B is a factorial affine domain then \underline{klnd}_{∗}(B) and \underline{ℛ}(B) have the same number of connected components. In particular,

\underline{klnd}_{∗}(B) is connected ⟺ \underline{ℛ}(B) is connected.

Consider the problem of describing \underline{klnd} (B). In view of 5.1 and of the fact that the graphs \underline{klnd}_{R}(B) ≅ \underline{klnd}_{R_{R}}(B_{R}) are described in section 4, we are justified to state the following: 6.1. Aphorism. Let B be a factorial affine domain. To achieve a satisfactory description of \underline{klnd} (B), it suffices to solve the following problems:

(1) Describe the kernels A ∈ klnd(B) which are isolated vertices of \underline{klnd} (B).

(2) Describe the graph \underline{ℛ}(B).

A particularly interesting factorial affine domain is B = k[X, Y, Z] = k^{[3]}. For this ring, the above problems (1) and (2) are still open but there are some partial results that we intend to give in a subsequent paper. Let us mention that a crucial rôle is played by the polynomials f ∈ k[X, Y, Z] whose generic fiber is a Danielewski surface, i.e.,

the pair (k(f)[X, Y, Z], k(f)) is a Danielewski surface.

In fact, it is not too difficult to show that ℛ(B) is precisely the set of rings k[f] such that f ∈ B is a polynomial whose generic fiber is a Danielewski surface.

It seems to this author that, in order to understand the locally nilpotent derivations (and the automorphisms) of k^{[3]}, it will be necessary to better understand the polynomials whose generic fiber is a Danielewski surface. It may be a good idea to think of those polynomials as generalized variables.

Isolated vertices

This paper made some progress in the understanding of \underline{klnd}_{∗}(B), but essentially nothing has been said about isolated vertices of \underline{klnd} (B). In particular, it would be interesting to classify two-dimensional rings B such that \underline{klnd} (B) is a discrete graph with many vertices. The smooth surfaces X_{m,j} which are studied in [4] give examples of such rings:

6.2. Example. Fix two integers 0 < j < m such that gcd(j, m) = 1. Consider the Danielewski surface B = ℂ[x_{1}, x_{2}, y] defined by x_{1}x_{2} = y^{m} − 1. Let ζ ∈ ℂ be a primitive m-th root of unity and define θ ∈ Aut_{ℂ}(B) by θ(x_{1}) = ζx_{1}, θ(x_{2}) = ζ^{−1}x_{1} and θ(y) = ζ^{j}y. Finally, let B_{m,j} = { b ∈ B | θ(b) = b }. Then Theorem 2.9 of [4] shows, among other things, that the smooth surface X_{m,j} = Spec(B_{m,j}) is a ℚ-homology plane with | Pic(X_{m,j})| = m. We claim:

(47) \underline{klnd} (B_{m,j}) is a discrete graph whose vertex set has the cardinality of ℂ.

Proof of (47). Assume that \underline{klnd} (B_{m,j}) is non-discrete. Then, by 4.3, there exists a field k ⊂ B_{m,j} such that (B_{m,j}, k) is a Danielewski surface; since k must satisfy B_{m,j}^{∗} = k^{∗}, and since it is clear that B_{m,j}^{∗} = ℂ^{∗} (because ℂ ⊂ B_{m,j} ⊆ B and B^{∗} = ℂ^{∗}), we must then have k = ℂ. However, it is known (see 2.8 of [4]) that any smooth Danielewski surface (over ℂ) of degree n has a Picard group isomorphic to ℤ^{n−1}; since X_{m,j} is smooth and has a Picard group of order m, it cannot be a Danielewski surface over ℂ. This contradiction shows that \underline{klnd} (B_{m,j}) is discrete.

Theorem 2.9 of [4] also implies that klnd(B_{m,j}) has at least two elements. In view of 1.10, it follows that | klnd(B_{m,j})| = |ℂ|. □

Local slice construction

As mentioned in the introduction, the present work started as an attempt to understand [3]. In that paper, Freudenburg presents a method for modifying a given kernel A ∈ klnd(B), where B = k^{[3]}, so as to obtain another one, say A′ ∈ klnd(B); in that case he says that A′ is obtained from A by local slice construction.

To conclude this paper, we show that the graph \underline{klnd} (B) can be interpreted as method for modifying kernels, in the same spirit as [3]. This works best when B is a factorial affine domain:

6.3. Proposition. Let B be a factorial affine domain and consider a triple (R, A, y) where R ∈ ℛ(B), A ∈ klnd_{R}(B) and y ∈ Ω_{R}(A). Then there exists exactly one A′ ∈ klnd_{R}(B) such that

y ∈ Ω_{R}(A′) and A′ ≠ A.

6.3.1. Definition. In the situation of 6.3, we say:

A′ is obtained from (R, A, y) by local slice construction.

Remark. There is a method for computing A′ from (R, A, y), similar to the method described in [3], but we leave this aspect to the reader.

The first step in the proof of 6.3 is:

6.3.2. Lemma. Let (B, k) be a Danielewski surface and y ∈ B. Then the set

E = { A ∈ klnd(B) | y ∈ Ω_{k}(A) }

has cardinality zero or two.

Proof. Suppose that A ∈ E. Then y ∈ Ω_{k}(A), so there exists an irreducible D ∈ lnd_{A}(B) satisfying A = k[Dy]. Write x = Dy, then 2.8 implies that (x, x_{2}, y) ∈ Γ_{k}(B) for some x_{2} ∈ B. Write A′ = k[x_{2}], then A′ ∈ klnd(B), A′ ≠ A and 4.2 gives y ∈ Ω_{k}(A) ∩ Ω_{k}(A′); so A′ ∈ E. This shows that if E ≠ ∅ then |E| ≥ 2.

To finish the proof of 6.3.2, it suffices to show that if A_{1}, A_{2}, A_{3} ∈ E satisfy

(48)  A_{1} ≠ A_{2} and A_{1} ≠ A_{3},

then A_{2} = A_{3}.

Suppose that (48) holds. For each i ∈ {1, 2, 3} we have y ∈ Ω_{k}(A_{i}) and consequently there exists an irreducible D_{i} ∈ lnd_{A_{i}}(B) satisfying A_{i} = k[D_{i}(y)]. Let x_{i} = D_{i}(y), then ker D_{i} = A_{i} = k[x_{i}] = k^{[1]} (for each i ∈ {1, 2, 3}).

Let j ∈ {2, 3}. Since A_{1} ≠ A_{j}, (y, D_{1}, D_{j}) satisfies the hypothesis of 2.5; since (2.5-1) is false, (2.5-2) must hold, so (x_{1}, x_{j}, y) ∈ Γ_{k}(B). This and 2.4 imply that k[y] ∩ x_{1}B is the principal ideal of k[y] generated by x_{1}x_{j}.

So x_{1}x_{2} and x_{1}x_{3} are associates in k[y] and consequently x_{3} = λx_{2} for some λ ∈ k^{∗}. So A_{2} = A_{3} and 6.3.2 is proved. □

Proof of 6.3. Let B be a factorial affine domain, let R ∈ ℛ(B) and let y ∈ B. We have to show that the set

E_{(B,R,y)} = { A ∈ klnd(B) | y ∈ Ω_{R}(A) }

has cardinality zero or two. Since R ∈ ℛ(B), the pair (B_{R}, R_{R}) is a Danielewski surface so the set

E = { 𝒜 ∈ klnd(B_{R}) | y ∈ Ω_{R_{R}}(𝒜) }

has cardinality zero or two by 6.3.2. By 5.3,

Λ : klnd_{R}(B) ⟶ klnd(B_{R})
A ⟼ A_{R}

is a well-defined bijection and, in view of 5.3.3, for each A ∈ klnd_{R}(B) we have

A ∈ E_{(B,R,y)} ⇔ y ∈ Ω_{R}(A) ⇔ y ∈ B ∩ Ω_{R_{R}}(A_{R}) ⇔ y ∈ Ω_{R_{R}}(A_{R}) ⇔ A_{R} ∈ E,

i.e., E_{(B,R,y)} = Λ^{−1}(E). So E_{(B,R,y)} has cardinality zero or two and 6.3 is proved. □

References

[1] D. Daigle, On locally nilpotent derivations of k[X_{1}, X_{2}, Y]/(φ(Y) − X_{1}X_{2}), J. Pure and Applied Algebra 181 (2003), 181–208.

[2] W. Danielewski, On the cancellation problem and automorphism groups of affine algebraic varieties, preprint, 1989.

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

[4] K. Masuda and M. Miyanishi, The additive group actions on Q-homology planes, preprint, 2002.

[5] 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