On algebraic automorphisms and their rational invariants
On algebraic automorphisms and
their rational invariants
Philippe Bonnet
Mathematisches Institut, Universität Basel
Rheinsprung 21, 4051 Basel, Switzerland
e-mail: Philippe.bonnet@unibas.ch
Abstract
Let X be an affine irreducible variety over an algebraically closed field k of char-
acteristic zero. Given an automorphism Φ, we denote by k(X)Φ its field of invariants,
i.e the set of rational functions f on X such that f ◦Φ = f . Let n(Φ) be the transcen-
dence degree of k(X)Φ over k. In this paper, we study the class of automorphisms Φ
of X for which n(Φ) = dimX − 1. More precisely, we show that under some condi-
tions on X, every such automorphism is of the form Φ = ϕg, where ϕ is an algebraic
action of a linear algebraic group G of dimension 1 on X, and where g belongs to
G. As an application, we determine the conjugacy classes of automorphisms of the
plane for which n(Φ) = 1.
1 Introduction
Let k be an algebraically closed field of characteristic zero. Let X be an affine irreducible
variety of dimension n over k. We denote by O(X) its ring of regular functions, and by
k(X) its field of rational functions. Given an algebraic automorphism Φ of X , denote by
Φ∗ the field automorphism induced by Φ on k(X), i.e. Φ∗(f) = f ◦Φ for any f ∈ k(X). An
element f of k(X) is invariant for Φ (or simply invariant) if Φ∗(f) = f . Invariant rational
functions form a field denoted k(X)Φ, and we set:
n(Φ) = trdegk k(X)
In this paper, we are going to study the class of automorphisms of X for which n(Φ) =
n − 1. There are natural candidates for such automorphisms, such as exponentials of
locally nilpotent derivations (see [M] or [Da]). More generally, one can construct such
automorphisms by means of algebraic group actions as follows. Let G be a linear algebraic
group over k. An algebraic action of G on X is a regular map:
ϕ : G×X −→ X
http://arxiv.org/abs/0704.1588v1
of affine varieties, such that ϕ(g.g′, x) = ϕ(g, ϕ(g′, x)) for any (g, g′, x) in G × G × X .
Given an element g of G, denote by ϕg the map x 7→ ϕ(g, x). Then ϕg clearly defines an
automorphism of X . Let k(X)G be the field of invariants of G, i.e. the set of rational
functions f on X such that f ◦ ϕg = f for any g ∈ G. If G is an algebraic group of
dimension 1, acting faithfully on X , and if g is an element of G of infinite order, then one
can prove by Rosenlicht’s Theorem (see [Ro]) that:
n(ϕg) = trdegk k(X)
G = n− 1
We are going to see that, under some mild conditions on X , there are no other automor-
phisms with n(Φ) = n−1 than those constructed above. In what follows, denote by O(X)ν
the normalization of O(X), and by G(X) the group of invertible elements of O(X)ν .
Theorem 1.1 Let X be an affine irreducible variety of dimension n over k, such that
char(k) = 0 and G(X)∗ = k∗. Let Φ be an algebraic automorphism of X such that
n(Φ) = n − 1. Then there exist an abelian linear algebraic group G of dimension 1, and
an algebraic action ϕ of G on X such that Φ = ϕg for some g ∈ G of infinite order.
Note that the structure of G is fairly simple. Since every connected linear algebraic group
of dimension 1 is either isomorphic to Ga(k) = (k,+) or Gm(k) = (k
∗,×) (see [Hum], p.
131), there exists a finite abelian group H such that G is either equal to H × Ga(k) or
H × Gm(k). Moreover, the assumption on the group G(X) is essential. Indeed, consider
the automorphism Φ of k∗ × k given by Φ(x, y) = (x, xy). Obviously, its field of invariants
is equal to k(x). However, it is easy to check that Φ cannot have the form given in the
conclusion of Theorem 1.1.
This theorem is analogous to a result given by Van den Essen and Peretz (see [V-P]).
More precisely, they establish a criterion to decide if an automorphism Φ is the exponential
of a locally nilpotent derivation, based on the invariants and on the form of Φ. A similar
result has been developed by Daigle (see [Da]).
We apply these results to the group of automorphisms of the plane. First, we obtain
a classification of the automorphisms Φ of k2 for which n(Φ) = 1. Second, we derive a
criterion on automorphisms of k2 to have no nonconstant rational invariants.
Corollary 1.2 Let Φ be an algebraic automorphism of k2. If n(Φ) = 1, then Φ is conjugate
to one of the following forms:
• Φ1(x, y) = (a
nx, amby), where (n,m) 6= (0, 0), a, b ∈ k, b is a root of unity but a is
• Φ2(x, y) = (ax, by + P (x)), where P belongs to k[t]− {0}, a, b ∈ k are roots of unity.
Corollary 1.3 Let Φ be an algebraic automorphism of k2. Assume that Φ has a unique
fixpoint p and that dΦp is unipotent. Then n(Φ) = 0.
We then apply Corollary 1.3 to an automorphism of C3 recently discovered by Pierre-Marie
Poloni and Lucy Moser (see [M-P]).
We may wonder whether Theorem 1.1 still holds if the ground field k is not algebraically
closed or has positive characteristic. The answer is not known for the moment. In fact,
two obstructions appear in the proof of Theorem 1.1 when k is arbitrary. First, the group
Gm(k) needs to be divisible (see Lemma 4.2), which is not always the case if k is not
algebraically closed. Second, the proof uses the fact that every Ga(k)-action on X can be
reconstructed from a locally nilpotent derivation on O(X) (see subsection 4.1), which is
no longer true if k has positive characteristic. This phenomenom is due to the existence
of differents forms for the affine line (see [Ru]). Note that, in case Theorem 1.1 holds and
k is not algebraically closed, the algebraic group G needs not be isomorphic to H ×Ga(k)
or H ×Gm(k), where H is finite. Indeed consider the unit circle X in the plane R
2, given
by the equation x2 + y2 = 1. Let Φ be a rotation in R2 with center at the origin and
angle θ 6∈ 2πQ. Then Φ defines an algebraic automorphism of X with n(Φ) = 0, and the
subgroup spanned by Φ is dense in SO2(R). But SO2(R) is not isomorphic to either Ga(R)
or Gm(R), even though it is a connected linear algebraic group of dimension 1.
We may also wonder what happens to the automorphisms Φ of X for which n(Φ) =
dimX − 2. More precisely, does there exist an action ϕ of a linear algebraic group G
on X , of dimension 2, such that Φ = ϕg for a given g ∈ G? The answer is no. Indeed
consider the automorphism Φ of k2 given by Φ = f ◦ g, where f(x, y) = (x + y2, y) and
g(x, y) = (x, y + x2). Let d(n) denote the maximum of the homogeneous degrees of the
coordinate functions of the iterate Φn. If there existed an action ϕ of a linear algebraic
group G such that Φ = ϕg, then the function d would be bounded, which is impossible
since d(n) = 4n. A similar argument on the length of the iterates also yields the result. But
if we restrict to some specific varieties X , for instance X = k3, one may ask the following
question: If n(Φ) = 1, is Φ birationally conjugate to an automorphism that leaves the first
coordinate of k3 invariant? The answer is still unknown.
2 Reduction to an affine curve C
Let X be an affine irreducible variety of dimension n over k. Let Φ be an algebraic
automorphism of X such that n(Φ) = n − 1. In this section, we are going to construct
an irreducible affine curve on which Φ acts naturally. This will allow us to use some
well-known results on automorphisms of curves. We set:
K = {f ∈ k(X)|∃m > 0, f ◦ Φm = f ◦ Φ ◦ ... ◦ Φ = f}
It is straightforward that K is a subfield of k(X) containing both k and k(X)Φ. We begin
with some properties of this field.
Lemma 2.1 K has transcendence degree (n − 1) over k, and is algebraically closed in
k(X). In particular, the automorphism Φ of X has infinite order.
Proof: First we show that K has transcendence degree (n − 1) over k. Since K contains
the field k(X)Φ, whose transcendence degree is (n − 1), we only need to show that the
extension K/k(X)Φ is algebraic, or in other words that every element of K is algebraic
over k(X)Φ. Let f be any element of K. By definition, there exists an integer m > 0 such
that f ◦ Φm = f . Let P (t) be the polynomial of k(X)[t] defined as:
P (t) =
(t− f ◦ Φi)
By construction, the coefficients of this polynomial are all invariant for Φ, and P (t) belongs
to k(X)Φ[t]. Moreover P (f) = 0, f is algebraic over k(X)Φ and the first assertion follows.
Second we show that K is algebraically closed in k(X). Let f be an element of k(X)
that is algebraic over K. We need to prove that f belongs to K. By the first assertion
of the lemma, f is algebraic over k(X)Φ. Let P (t) = a0 + a1t + ... + apt
p be a nonzero
minimal polynomial of f over k(X)Φ. Since P (f) = 0 and all ai are invariant, we have
P (f ◦ Φ) = P (f) ◦ Φ = 0. In particular, all elements of the form f ◦ Φi, with i ∈ N, are
roots of P . Since P has finitely many roots, there exist two distinct integers m′ < m′′ such
that f ◦ Φm
= f ◦ Φm
. In particular, f ◦ Φm
′′−m′ = f and f belongs to K.
Now if Φ were an automorphism of finite order, then K would be equal to k(X). But
this is impossible since K and k(X) have different transcendence degrees.
Lemma 2.2 There exists an integer m > 0 such that K = k(X)Φ
Proof: By definition, k(X) is a field of finite type over k. Since K is contained in k(X),
K has also finite type over k. Let f1, ..., fr be some elements of k(X) such that K =
k(f1, ..., fr). Let m1, ..., mr be some positive integers such that fi ◦ Φ
mi = fi, and set
m = m1...mr. By construction, all fi are invariant for Φ
m. In particular, K is invariant
for Φm and K ⊆ k(X)Φ
. Since k(X)Φ
⊆ K, the result follows.
Let L be the algebraic closure of k(X), and let A be the K-subalgebra of L spanned by
O(X). By construction, A is an integral K-algebra of finite type of dimension 1. Let m
be an integer satisfying the conditions of lemma 2.2. The automorphism Ψ∗ = (Φm)∗ of
O(X) stabilizes A, hence it defines a K-automorphism of A, of infinite order (see lemma
2.1). Let B be the integral closure of A. Then B is also an integral K-algebra of finite
type, of dimension 1, and the K-automorphism Ψ∗ extends uniquely to B. If K stands for
the algebraic closure of K, we set:
C = B ⊗K K
By construction, C = Spec(C) is an affine curve over the algebraically closed field K.
Moreover the automorphism Ψ∗ acts on C via the operation:
Ψ∗ : C −→ C, x⊗ y 7−→ Ψ∗(x)⊗ y
This makes sense since Ψ∗ fixes the field K. Therefore Ψ∗ induces an algebraic automor-
phism of the curve C. Since K is algebraically closed in k(X) by lemma 2.1, C is integral
(see [Z-S], Chap. VII, §11, Theorem 38). But by construction, B and K are normal rings.
Since C is a domain and char(K) = 0, C is also integrally closed by a result of Bourbaki
(see [Bou], p. 29). So C is a normal domain and C is a smooth irreducible curve.
Lemma 2.3 Let C be the K-algebra constructed above. Then either C = K[t] or C =
K[t, 1/t].
Proof: By lemma 2.1, the automorphism Φ of X has infinite order. Since the fraction field
of B is equal to k(X), Ψ∗ has infinite order on B. But B ⊗ 1 ⊂ C, so Ψ∗ has infinite
order on C. In particular, Ψ acts like an automorphism of infinite order on C. Since C is
affine, it has genus zero (see [Ro2]). Since K is algebraically closed, the curve C is rational
(see [Che], p. 23 ). Since C is smooth, it is isomorphic to P1(K) − E, where E is a finite
set. Moreover, Ψ acts like an automorphism of P1(K) that stabilizes P1(K) − E. Up to
replacing Ψ by one of its iterates, we may assume that Ψ fixes every point of E. But an
automorphism of P1(K) that fixes at least three points is the identity, which is impossible.
Therefore E consists of at most two points, and C is either isomorphic to K or to K
particular, either C = K[t] or C = K[t, 1/t].
3 Normal forms for the automorphism Ψ
Let C and Ψ∗ be the K-algebra and the K-automorphism constructed in the previous
section. In this section, we are going to give normal forms for the couple (C,Ψ∗), in case
the group G(X) is trivial, i.e. G(X) = k∗. We begin with a few lemmas.
Lemma 3.1 Let X be an irreducible affine variety over k. Let Ψ be an automorphism of
X. Let α, f be some elements of k(X)∗ such that (Ψ∗)n(f) = αnf for any n ∈ Z. Then α
belongs to G(X).
Proof: Given an element h of k(X)∗ and a prime divisor D on the normalization Xν , we
consider h as a rational function on Xν , and denote by ordD(h) the multiplicity of h along
D. This makes sense since the variety Xν is normal. Fix any prime divisor D on X . Since
(Ψ∗)n(f) = αnf for any n ∈ Z, we obtain:
ordD((Ψ
∗)n(f)) = nordD(α) + ordD(f)
Since Ψ is an algebraic automorphism of X , it extends uniquely to an algebraic automor-
phism of Xν , which is still denoted Ψ. Moreover, this extension maps every prime divisor
to another prime divisor, does not change the multiplicity and maps distinct prime divisors
into distinct ones. If div(f) =
i niDi, where all Di are prime, then we have:
div((Ψ∗)n(f)) =
∗)n(Di)
where all (Ψ∗)n(Di) are prime and distinct. So the multiplicity of (Ψ
∗)n(f) along D is
equal to zero if D is none of the (Ψ∗)n(Di), and equal to ni if D = (Ψ
∗)n(Di). In all cases,
if R = max{|ni|}, then we find that |ordD((Ψ
∗)n(f))| ≤ R and |ordD(f)| ≤ R, and this
implies for any integer n:
|nordD(α)| ≤ 2R
In particular we find ordD(α) = 0. Since this holds for any prime divisor D, the support
of div(α) in Xν is empty and div(α) = 0. Since Xν is normal, α is an invertible element
of O(X)ν , hence it belongs to G(X).
Lemma 3.2 Let K be a field of characteristic zero and K its algebraic closure. Let C be
either equal to K[t] or to K[t, 1/t]. Let Ψ∗ be a K-automorphism of C such that Ψ∗(t) = at,
where a belongs to K. Let σ1 be a K-automorphism of C, commuting with Ψ
∗, such that
σ1(K) = K. Then σ1(a) is either equal to a or to 1/a.
Proof: We distinguish two cases depending on the ring C. First assume that C = K[t].
Since σ1 is a K-automorphism of C that maps K to itself, we have K[t] = K[σ1(t)]. In
particular σ1(t) = λt + µ, where λ, µ belong to K and λ 6= 0. Since Ψ
∗ and σ1 commute,
we obtain:
Ψ∗ ◦ σ1(t) = λat + µ = σ1 ◦Ψ
∗(t) = σ1(a)(λt + µ)
In particular, we have σ1(a) = a and the lemma follows in this case. Second assume that
C = K[t, 1/t]. Since σ1 is a K-automorphism of C, we find:
σ1(t)σ1(1/t) = σ1(t.1/t) = σ1(1) = 1
Therefore σ1(t) is an invertible element of C, and has the form σ1(t) = a1t
n1 , where
a1 ∈ K
and n1 is an integer. Since σ1 is a K-automorphism of C that maps K to K,
we have K[t, 1/t] = K[σ1(t), 1/σ1(t)]. In particular |n1| = 1 and either σ1(t) = a1t or
σ1(t) = a1/t. If σ1(t) = a1t, the relation Ψ
∗ ◦ σ1(t) = σ1 ◦ Ψ
∗(t) yields σ(a) = a. If
σ1(t) = a1/t, then the same relation yields σ(a) = 1/a.
Lemma 3.3 Let X be an irreducible affine variety of dimension n over k, such that
G(X) = k∗. Let Φ be an automorphism of X such that n(Φ) = (n− 1). Let Ψ∗ be the au-
tomorphism of C constructed in the previous section. If either C = K[t] or C = K[t, 1/t],
and if Ψ∗(t) = at, then a belongs to k∗.
Proof: We are going to prove by contradiction that a belongs to k∗. So assume that a 6∈ k∗.
Let σ be any element of Gal(K/K), and denote by σ1 the K-automorphism of C defined
as follows:
∀(x, y) ∈ B ×K, σ1(x⊗ y) = x⊗ σ1(y)
Since Ψ∗ ◦σ1(x⊗ y) = Ψ
∗(x)⊗σ1(y) = σ1 ◦Ψ
∗(x⊗ y) for any element x⊗ y of B⊗KK, Ψ
and σ1 commute. Moreover if we identify K with 1⊗K, then σ1(K) = K by construction.
By lemma 3.2, we obtain:
∀σ ∈ Gal(K/K), σ(a) = a or σ(a) =
In particular, the element (ai + a−i) is invariant under the action of Gal(K/K) for any i,
and so it belongs to K because char(K) = 0. Now let f be an element of B −K. Since f
belongs to C, we can express f as follows:
Choose an f ∈ B−K such that the difference (s−r) is minimal. We claim that (s−r) = 0,
i.e. f = fst
s. Indeed, assume that s > r. Since f is an element of B, the following
expressions:
Ψ∗(f) + (Ψ∗)−1(f)− (as + a−s)f =
i=r fi(a
i + a−i − as − a−s)ti
Ψ∗(f) + (Ψ∗)−1(f)− (ar + a−r)f =
i=r+1
i + a−i − ar − a−r)ti
also belong to B. By minimality of (s− r), these expressions belong to K. In other words,
i+ a−i− as− a−s) = 0 (resp. fi(a
i+ a−i− ar− a−r) = 0) for any i 6= 0, s (resp. for any
i 6= 0, r). Since k is algebraically closed and a 6∈ k∗ by assumption, (ai + a−i − as − a−s)
(resp. (ai+a−i−ar−a−r)) is nonzero for any i 6= s (resp. for any i 6= r). Therefore fi = 0
for any i 6= 0, and f belongs to K, a contradiction. Therefore s = r and f = fst
s. Since
f belongs to B, it also belongs to k(X). Since Ψ is an automorphism of X , the element
as = Ψ∗(f)/f belongs to k(X). Moreover (Ψ∗)n(f) = ansf for any n ∈ Z. By lemma
3.1, as belongs to G(X) = k∗. Since k is algebraically closed, a belongs to k∗, hence a
contradiction, and the result follows.
Proposition 3.4 Let X be an irreducible affine variety of dimension n over k, such that
G(X) = k∗. Let Φ be an automorphism of X such that n(Φ) = (n− 1). Let C and Ψ∗ be
the K-algebra and the K-automorphism constructed in the previous section. Then up to
conjugation, one of the following three cases occurs:
• C = K[t] and Ψ∗(t) = t + 1,
• C = K[t] and Ψ∗(t) = at, where a ∈ k∗ is not a root of unity,
• C = K[t, 1/t] and Ψ∗(t) = at, where a ∈ k∗ is not a root of unity.
Proof: By lemma 2.3, we know that either C = K[t] or C = K[t, 1/t]. We are going to
study both cases.
First case: C = K[t].
The automorphism Ψ∗ maps t to at + b, where a ∈ K
and b ∈ K. If a = 1, then
b 6= 0 and up to replacing t with t/b, we may assume that Ψ∗(t) = t+ 1. If a 6= 1, then up
to replacing t with t− c for a suitable c, we may assume that Ψ∗(t) = at. But then lemma
3.3 implies that a belongs to k∗. Since Ψ∗ has infinite order, a cannot be a root of unity.
Second case: C = K[t, 1/t].
Since Ψ∗(t)Ψ∗(1/t) = Ψ∗(1) = 1, Ψ∗(t) is an invertible element of C. So Ψ∗(t) = atn,
where a ∈ K
and n 6= 0. Since Ψ∗ is an automorphism, n is either equal to 1 or to
−1. But if n were equal to −1, then a simple computation shows that (Ψ∗)2 would be the
identity, which is impossible. So Ψ∗(t) = at, where a ∈ K
. By lemma 3.3, a belongs to
k∗. As before, a cannot be a root of unity.
4 Proof of the main theorem
In this section, we are going to establish Theorem 1.1. We will split its proof in two steps
depending on the form of the automorphism Ψ∗ given in Proposition 3.4. But before, we
begin with a few lemmas.
Lemma 4.1 Let Φ be an automorphism of an affine irreducible variety X. Let G be a
linear algebraic group and ψ be an algebraic G-action on X. Let h be an element of G such
that the group <h> spanned by h is Zariski dense in G. If Φ and ψh commute, then Φ
and ψg commute for any g in G.
Proof: It suffices to check that Φ∗ and ψ∗g commute for any g ∈ G. For any k-algebra
automorphisms α, β of O(X), denote by [α, β] their commutator, i.e. [α, β] = α ◦β ◦α−1 ◦
β−1. For any f ∈ O(X), set:
λ(g, f)(x) = [Φ∗, ψ∗g ](f)(x)− f(x)
Since G is a linear algebraic group acting algebraically on the affine variety X , λ(g, f)(x)
is a regular function on G×X . Since Φ∗ and ψ∗h commute, the automorphisms Φ
∗ and
ψ∗hn commute for any integer n. So the regular function λ(g, f)(x) vanishes on <h> ×X .
Since < h > is dense in G by assumption, < h > ×X is dense in G × X and λ(g, f)(x)
vanishes identically on G × X . In particular, [Φ∗, ψ∗g ](f) = f for any g ∈ G. Since this
holds for any element f of O(X), the bracket [Φ∗, ψ∗g ] coincides with the identity on O(X)
for any g ∈ G, and the result follows.
Lemma 4.2 Let Φ be an automorphism of an affine irreducible variety X. Let G be a
linear algebraic group and ψ be an algebraic G-action on X. Let h be an element of G such
that the group <h> spanned by h is Zariski dense in G. Assume there exists a nonzero
integer r such that Φr = ψh, and that G is divisible. Then there exists an algebraic action
ϕ of G′ = Z/rZ×G such that Φ = ϕg′ for some g
′ in G′.
Proof: Fix an element b in G such that br = h, and set ∆ = Φ◦ψb−1 . This is possible since
G is divisible. By construction, ∆ is an automorphism of X . Since Φr = ψh, Φ and ψh
commute. By lemma 4.1, Φ and ψg commute for any g ∈ G. In particular, we have:
∆r = (Φr) ◦ ψb−r = (Φ
r) ◦ ψh−1 = Id
So ∆ is finite, Φ = ∆ ◦ ψb and ∆ commutes with ψg for any g ∈ G. The group G
′ then
acts on X via the map ϕ defined by:
ϕ(i,g)(x) = ∆
i ◦ ψg(x)
Moreover we have Φ = ϕg′ for g
′ = (1, b).
The proof of Theorem 1.1 will then go as follows. In the following subsections, we are going
to exhibit an algebraic action ψ of Ga(k) (resp. Gm(k)) on X , such that Ψ = Φ
m = ψh for
some h. In both cases, the group G we will consider will be linear algebraic of dimension 1,
and divisible. Moreover the element h will span a Zariski dense set because h 6= 0 (resp. h
is not a root of unity). With these conditions, Theorem 1.1 will become a direct application
of Lemma 4.2.
4.1 The case Ψ∗(t) = t+ 1
Assume that C = K[t] and Ψ∗(t) = t+1. We are going to construct a nontrivial algebraic
Ga(k)-action ψ on X such that Ψ = ψ1. Since O(X) ⊂ C, every element f of O(X) can
be written as f = P (t), where P belongs to K[t]. We set r = degt P (t). Since Ψ
∗ stabilizes
O(X), the expression:
(Ψi)∗(f) = P (t+ i) =
P (j)(t)
belongs to O(X) for any integer i. Since the matrix M = (ij/j!)0≤i,j≤r is invertible in
Mr+1(Q), the polynomial P
(j)(t) belongs to O(X) for any j ≤ r. So the K-derivation
D = ∂/∂t on C stabilizes the k-algebra O(X). Since Dr+1(f) = 0, the operator D,
considered as a k-derivation on O(X), is locally nilpotent (see [Van]). Therefore the
exponential map:
exp uD : O(X) −→ O(X)[u], f 7−→
Dj(f)
is a well-defined k-algebra morphism. But exp uD defines also a K-algebra morphism from
C to C[u]. Since exp uD(t) = t + u, expD coincides with Ψ∗ on C. Since C contains the
ring O(X), we have expD = Ψ∗ on O(X). So the exponential map induces an algebraic
Ga(k)-action ψ on X such that Ψ = ψ1 (see [Van]).
4.2 The case Ψ∗(t) = at
Assume that Ψ∗(t) = at and a is not a root of unity. We are going to construct a nontrivial
algebraic Gm(k)-action ψ on X such that Ψ = ψa. First note that either C = K[t] or
C = K[t, 1/t]. Let f be any element of O(X). Since O(X) ⊂ C, we can write f as:
f = P (t) =
where the fit
i belong a priori to C. Since Ψ∗ stabilizes O(X), the expression:
(Ψj)∗(f) = P (ajt) =
ajifit
belongs to O(X) for any integer j. Since a belongs to k∗ and is not a root of unity, the
Vandermonde matrix M = (aij)0≤i,j≤s−r is invertible in Ms−r+1(k). So the elements fit
all belong to O(X) for any integer i. Consider the map:
ψ∗ : O(X) −→ O(X)[v, 1/v], f 7−→
Then ψ∗ is a well-defined k-algebra morphism, which induces a regular map ψ from k∗×X
to X . Moreover we have ψv ◦ ψv′ = ψvv′ on X for any v, v
′ ∈ k∗. So ψ defines an algebraic
Gm(k)-action on X such that Ψ = ψa.
5 Proof of Corollary 1.2
Let Φ be an automorphism of the affine plane k2, such that n(Φ) = 1. By Theorem 1.1,
there exists an algebraic action ϕ of an abelian linear algebraic group G of dimension 1
such that Φ = ϕg. We will distinguish the cases G = Z/rZ×Gm(k) and G = Z/rZ×Ga(k).
First case: G = Z/rZ×Gm(k).
Then G is linearly reductive and ϕ is conjugate to a representation in Gl2(k) (see [Ka]
or [Kr]). Since G consists solely of semisimple elements, ϕ is even diagonalizable. In par-
ticular, there exists a system (x, y) of polynomial coordinates, some integers n,m and some
r-roots of unity a, b such that:
ϕ(i,u)(x, y) = (a
iunx, biumy)
Note that, since the action is faithful, the couple (n,m) is distinct from (0, 0). Since k is
algebraically closed, we can even reduce Φ = ϕg to the first form given in Corollary 1.2.
Second case: G = Z/rZ×Ga(k).
Let ψ and ∆ be respectively the Ga(k)-action and finite automorphism constructed in
Lemma 4.2. By Rentschler’s theorem (see [Re]), there exists a system (x, y) of polynomial
coordinates and an element P of k[t] such that:
ψu(x, y) = (x, y + uP (x))
For any f ∈ k[x, y], set degψ(f) = degu exp uD(f). It is well-known that this defines a
degree function on k[x, y] (see [Da]). Since ψ and ∆ commute, ∆∗ preserves the space En
of polynomials of degree ≤ n with respect to degψ. In particular, ∆
∗ preserves E0 = k[x].
So ∆∗ induces a finite automorphism of k[x], hence ∆∗(x) = ax + b, where a is a root of
unity. Since ∆ is finite, either a 6= 1 or a = 1 and b = 0. In any case, up to replacing
x by x − µ for a suitable constant µ, we may assume that ∆∗(x) = ax. Moreover ∆∗
preserves the space E1 = k[x]{1, y}. With the same arguments as before, we obtain that
∆∗(y) = cy+ d(x), where c is a root of unity and d(x) belongs to k[x]. Composing ∆ with
ψ1/m then yields the second form given in Corollary 1.2.
6 Proof of Corollary 1.3
Let Φ be an algebraic automorphism of k2. We assume that Φ has a unique fixpoint p and
that dΦp is unipotent. We are going to prove that n(Φ) = 0.
First we check that n(Φ) cannot be equal to 2. Assume that n(Φ) = 2. Then k(x, y)Φ
has transcendence degree 2, and the extension k(x, y)/k(x, y)Φ is algebraic, hence finite.
Moreover Φ∗ acts like an element of the Galois group of this extension. In particular, Φ∗ is
finite. By a result of Kambayashi (see [Ka]), Φ can be written as h◦A◦h−1, where A is an
element of Gl2(k) of finite order and h belongs to Aut(k
2). Since Φ has a unique fixpoint p,
we have h(0, 0) = p. In particular, dΦp is conjugate to A in Gl2(k). Since dΦp is unipotent
and A is finite, A is the identity. Therefore Φ is also the identity, which contradicts the
fact that it has a unique fixpoint.
Second we check that n(Φ) cannot be equal to 1. Assume that n(Φ) = 1. By the
previous corollary, up to conjugacy, we may assume that Φ has one of the following forms:
• Φ1(x, y) = (a
nx, amby), where (n,m) 6= (0, 0), b is a root of unity but a is not,
• Φ2(x, y) = (ax, by + P (x)), where P belongs to k[t]− {0} and a, b are roots of unity.
Assume that Φ is an automorphism of type Φ1. Then dΦp is a diagonal matrix of Gl2(k),
distinct from the identity. But this is impossible since dΦp is unipotent. So assume that Φ
is an automorphism of type Φ2. Then dΦp is a linear map of the form (u, v) 7→ (au, bv+du),
with d ∈ k. Since dΦp is unipotent, we have a = b = 1. So (α, β) is a fixpoint if and only
if P (α) = 0. In particular, the set of fixpoints is either empty or a finite union of parallel
lines. But this is impossible since there is only one fixpoint by assumption. Therefore
n(Φ) = 0.
7 An application of Corollary 1.3
In this section, we are going to see how Corollary 1.3 can be applied to the determination
of invariants for automorphisms of C3. Set Q(x, y, z) = x2y − z2 − xz3 and consider the
following automorphism (see [M-P]):
Φ : C3 −→ C3,
 7−→
y(1− xz) + Q
z − Q
We are going to show that:
C(x, y, z)Φ = C(x) and C[x, y, z]Φ = C[x]
Let k be the algebraic closure of C(x). Since Φ∗(x) = x, the morphism Φ∗ induces an
automorphism of k[y, z], which we denote by Ψ∗. The automorphism Ψ has clearly (0, 0)
as a fixpoint, and its differential at this point is unipotent, distinct from the identity.
Indeed, it is given by the matrix:
dΨ(0,0) =
1 −x3/2
Moreover, the set of fixpoints of Ψ is reduced to the origin. Indeed, if (α, β) is a point of
k2 fixed by Ψ, then xQ = 0 and 4β4 − 4xαβ +Q2 = 0. Since x belongs to k∗, we have:
Q = x2α− β2 − xβ3 = 0 and β4 − xαβ = 0
If β = 0, then α = 0 and we find the origin. If β 6= 0, then dividing by β and multiplying
by −x yields the relation:
x2α− xβ3 = 0
This implies β2 = 0 and β = 0, hence a contradiction. By Corollary 1.3, the field of invari-
ants of Ψ has transcendence degree zero. So the field of invariants of Φ has transcendence
degree ≤ 1 over C. Since this field contains C(x) and that C(x) is algebraically closed in
C(x, y, z), we obtain that C(x, y, z)Φ = C(x). As a consequence, the ring of invariants of
Φ is equal to C[x].
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	Introduction
	Reduction to an affine curve C
	Normal forms for the automorphism 
	Proof of the main theorem
	The case *(t)=t+1
	The case *(t)=at
	Proof of Corollary ??
	Proof of Corollary ??
	An application of Corollary ??
