Table Of ContentAutomorphism groups of polynomial
rings
Bachelor Thesis
Ajith Urundolil Kumaran
Advisor
Prof. Richard Pink
Fall Semester 2018
ETH Zürich
Departement of Mathematics
Contents
1 Introduction 1
2 First definitions and results 2
3 Basic notions on weighted degrees 3
4 Key lemmas 9
5 The proof of Theorem 2.8 15
6 Introduction to amalgamated products 19
7 Group actions on trees 21
8 Proof of the decomposition 25
9 Case n ≥ 3 26
1 Introduction
Polynomial rings in n variables over a field K appear everywhere in mathematics. Since
these rings are so fundamental we believe it is very natural to study the structure of
their automorphism groups. This would be the way an algebraist would think about
this problem, whereas an algebraic geometer would regard this automorphism group
as the automorphism group of affine n-space, a geometric object. There are many
problems surrounding this group which have not been solved for general n. An example
is the linearization problem, which asks if every element of finite order in the group is
conjugate to an affine automorphism. A counterexample to this conjecture is given in
[1] by Asanuma over a field of positive characteristic, but it is still an open problem
for fields of characteristic 0. A more famous problem on polynomial automorphisms is
the Jacobian conjecture which states that a K-algebra homomorphism whose Jacobian
determinant is a nonzero constant is an automorphism. Again the conjecture is wrong
for fields of positive characteristic but open for fields of characteristic 0. The kind
of problem we will be focusing on in this thesis is the problem of giving elementary
generators for the group.
The aim of this thesis is to study the cases n = 1 and n = 2 and give a summary of
the progress in the case n ≥ 3. The case n = 2 will be the main part of this paper. In
this case we will show that the group can be generated by two kinds of automorphisms
which have a simple description.
I would like to thank my advisor, Professor Richard Pink for his support, and en-
couragement to pursue this subject.
1
2 First definitions and results
We will now introduce some notation which will be used throughout the thesis: Fix
R := R(cid:48)[X ,...,X ] the polynomial ring in n variables over a commutative ring R(cid:48). The
1 n
automorphism group of R we will denote by Aut (R). We notice that an element φ ∈
R(cid:48)
Aut (R) is uniquely determined by the tuple (φ(X ),...,φ(X )). Therefore we will
R(cid:48) 1 n
often write (f ,..,f ) ∈ Rn for the unique automorphism that sends X to f for all i.
1 n i i
We write X for the set {X ,..,X }.
1 n
Definition 2.1. We call φ ∈ Aut (R) affine if for all i = 1,...,n : φ(X ) is a linear
R(cid:48) i
polynomial.
Definition 2.2. Let i ∈ {1,...,n}. We call φ ∈ Aut (R) an X \{X }-based shear if
R(cid:48) i
for all j ∈ {1,...,n}\{i} : φ(X ) = X and φ(X ) = X +f for some fixed polynomial
j j i i
f in the variables X \{X }
i
An easy computation shows that the inverse of a shear is again a shear where the
polynomial f in Definition 2.2 is replaced by −f. More generally we have the following
definition:
Definition 2.3. Let i ∈ {1,...,n}. We call φ ∈ Aut (R) an X \ {X }-based de
R(cid:48) i
Jonquières automorphism if φ(R(cid:48)[X \{X }]) = R(cid:48)[X \{X }].
i i
Example 2.4. An X \{X }-based shear is an X \{X }-based de Jonqières automor-
i i
phism.
Definition 2.5. We call φ ∈ Aut (R) tame if it can be written as a composition of
R(cid:48)
affine and shear automorphisms. The group of all tame automorphisms will be denoted
as T (R(cid:48)).
n
Definition 2.6. We call φ ∈ Aut (R) triangular if φ(X ) = X +f (X ,...,X ) for
R(cid:48) i i i i+1 n
f in K[X ,...,X ] and i = 1,..,n.
i i+1 n
Remark 2.7. We notice that all triangular automorphisms are tame.
As mentioned before, the main part of this thesis will be about the case n = 2 and
R(cid:48) a field. Our aim will be to prove the following theorem:
Theorem 2.8. (Jung,Van der Kulk) All automorphisms of K[X ,X ] are tame. Fur-
1 2
thermore Aut (K[X ,X ]) = A ∗ B the amalgamated product of A and B over C,
K 1 2 C
where A is the group of affine automorphisms of K[X ,X ] and B the group of X -based
1 2 2
de Jonquières automorphism and C = A∩B.
This theorem was first proven by Jung [2] in 1942 in the case where K = C and
without the amalgamated product decomposition. Van der Kulk [3] then proved the
general case over an arbitrary field with the product decomposition in 1953. In later
years many different kind of proofs have been proposed, some (algebro)-geometric in
nature such as in [4] and others purely algebraic like the proof by Makar-Limanov [5].
Our interest in this paper lies in this purely algebraic approach. Dicks [6] simplified
2
the proof of Makar-Limanov and we will be closely following this proof of Dicks as it is
explained in [7]. But first we will study the case n = 1:
Theorem 2.9. Every φ ∈ Aut (K[X]) is affine.
K
Proof. Since φ is surjective there exists f ∈ K[X] such that
(1) φ(f) = X,
If we write f(X) = (cid:80)n a Xi with n ∈ Z≥0 and a in K for all i and a nonzero, the
i=0 i i n
equation (1) becomes (cid:80)n a φ(X)i = X by the homomorphism property and φ being
i=0 i
the identity function on K. This tells us that the degree of φ(X) cannot be greater
than 1 or else
n
(cid:88)
(2) 1 = deg(X) = deg( a φ(X)i) = deg(φ(X)n) = ndeg(φ(X)) (cid:54)= 1,
i
i=0
which is a contradiction. So we have deg(φ(X)) ≤ 1 but φ(X) cannot be in K, because
then X = φ(f) would be in K. Therefore deg(φ(X)) = 1.
This theorem also tells us that we can embed Aut (K[X]) into a subgroup of
K
GL (K): Since every φ ∈ Aut (K[X]) is affine there exist unique a ∈ K× and b ∈ K
2 K
such that φ(X) = aX +b. This implies that
(cid:18) (cid:19)
a b
(3) φ (cid:55)→
0 1
defines a well-defined map, and an easy computation shows that this is a group homo-
morphism. So this means that in the case n = 1 we can think of the automorphism
group as a finite dimensional object over K. This makes the case n = 1 quite easy
to study compared to the cases where n ≥ 2. In the cases where n ≥ 2 the shear
automorphisms allow us to embed K[X ], which is a infinite-dimensional vector space
1
over K, into the automorphism group Aut (K[X ,...,X ]). A way we can do this is by
K 1 n
sending an f ∈ K[X ] to the X \{X }-based shear automorphism defined by f(X ).
1 2 1
3 Basic notions on weighted degrees
InthissectionwewillproveTheorem2.8. Foranautomorphism(g ,g ) ∈ Aut (K[X ,X ]),
1 2 K 1 2
let (f ,f ) be its inverse. The basic idea of the proof is to introduce a weighted degree
1 2
on K[X ,X ] using the (1,0)-degree d and (0,1)-degree d of f (see Definition 3.1).
1 2 1 2 1
With this weighted degree we can prove that d divides d or the other way around.
1 2
This is a key step that also appears in all other proofs of Theorem 2.8. Fix (n ,n ) ∈ Z2
1 2
with (n ,n ) (cid:54)= (0,0).
1 2
3
Definition 3.1. We define a function d : K[X ,X ] → Z∪{−∞} by setting
1 2
d(f) := sup{n i+n j | a (cid:54)= 0}
1 2 ij
(cid:88)
where f = a XiXj. It is called the (n ,n )-degree of f.
ij 1 2 1 2
i,j≥0
The (1,1)-degree is the usual total degree and instead of d(f) we denote this by
deg(f). We call the (1,0)-degree the X -degree. Similarly we call the (0,1)-degree the
1
X -degree.
2
Definition 3.2. For (a,b) and (a(cid:48),b(cid:48)) in Z2 we say (a,b) ≤ (a(cid:48),b(cid:48)) if an + bn <
1 2
a(cid:48)n +b(cid:48)n or an +bn = a(cid:48)n +b(cid:48)n with a(cid:48)n ≥ an and bn ≥ b(cid:48)n .
1 2 1 2 1 2 2 2 1 1
Lemma 3.3. (Z2,≤) is a totally ordered abelian group
Proof. We first show that ≤ defines a total order on Z2:
Totality: Let (a,b) and (a(cid:48),b(cid:48)) be elements in Z2 such that (a,b) (cid:2) (a(cid:48),b(cid:48)). By Defini-
tion3.2thisisequivalenttosayingthatn a+n b > n a(cid:48)+n b(cid:48) orn a+n b = n a(cid:48)+n b(cid:48)
1 2 1 2 1 2 1 2
with a(cid:48)n < an or bn < b(cid:48)n . If n a+n b > n a(cid:48) +n b(cid:48) we are finished since then by
2 2 1 1 1 2 1 2
Definition 3.2 we have (a(cid:48),b(cid:48)) ≤ (a,b). Therefore we assume that n a+n b = n a(cid:48)+n b(cid:48)
1 2 1 2
with a(cid:48)n < an or bn < b(cid:48)n . Suppose that a(cid:48)n < an . This implies that n is
2 2 1 1 2 2 2
nonzero. We multiply the equation n a+n b = n a(cid:48)+n b(cid:48) by n and get the equation
1 2 1 2 2
n n a + n2b = n n a(cid:48) + n2b(cid:48). Let us assume n ≤ 0. From a(cid:48)n < an we get that
1 2 2 1 2 2 1 2 2
n n a ≤ n n a(cid:48) which implies that n2b ≥ n2b(cid:48). Since n2 > 0 we get that b ≥ b(cid:48). If
1 2 1 2 2 2 2
instead we assume n > 0 we would get b ≤ b(cid:48). But in both cases we get bn ≤ b(cid:48)n .
1 1 1
By Definition 3.2 this means that (a(cid:48),b(cid:48)) ≤ (a,b).
Transitivity: Let (a,b),(a(cid:48),b(cid:48)) and (a(cid:48)(cid:48),b(cid:48)(cid:48)) be elements in Z such that (a,b) ≤ (a(cid:48),b(cid:48))
and (a(cid:48),b(cid:48)) ≤ (a(cid:48)(cid:48),b(cid:48)(cid:48)). If n a + n b < n a(cid:48) + n b(cid:48) or n a(cid:48) + n b(cid:48) < n a(cid:48)(cid:48) + n b(cid:48)(cid:48), the
1 2 1 2 1 2 1 2
inequality n a+n b < n a(cid:48)(cid:48)+n b(cid:48)(cid:48) follows via the transitivity of ≤ on Z. This implies
1 2 1 2
(a,b) ≤ (a(cid:48)(cid:48),b(cid:48)(cid:48)) by Definition 3.2. So assume that n a+n b = n a(cid:48)+n b(cid:48) = n a(cid:48)(cid:48)+n b(cid:48)(cid:48).
1 2 1 2 1 2
By Definition 3.2 we know have the inequalities a(cid:48)n ≥ an , bn ≥ b(cid:48)n , a(cid:48)(cid:48)n ≥ a(cid:48)n ,
2 2 1 1 2 2
b(cid:48)n ≥ b(cid:48)(cid:48)n . The transitivity of ≤ on Z implies that a(cid:48)(cid:48)n ≥ an and bn ≥ b(cid:48)(cid:48)n . The
1 1 2 2 1 1
case where bn < b(cid:48)n works analogously. By Definition 3.2 this implies (a,b) ≤ (a(cid:48)(cid:48),b(cid:48)(cid:48)).
1 1
Antisymmetry: Let (a,b) and (a(cid:48),b(cid:48)) be elements in Z2 such that (a,b) ≤ (a(cid:48),b(cid:48))
and (a(cid:48),b(cid:48)) ≤ (a,b). This implies an +bn = a(cid:48)n +b(cid:48)n and a(cid:48)n = an and bn = b(cid:48)n
1 2 1 2 2 2 1 1
by the antisymmetry of the usual total order on Z. Since (n ,n ) (cid:54)= (0,0) we know
1 2
n (cid:54)= 0 or n (cid:54)= 0. Let us assume n (cid:54)= 0. From bn = b(cid:48)n , we get b = b(cid:48). But
1 2 1 1 1
this implies a = a(cid:48) via the equation an + bn = a(cid:48)n + b(cid:48)n . The other case works
1 2 1 2
analogously.
Compatibility: The only thing left to show is the compatibility of the total order
with the additive structure of Z2. This means for all (a,b),(a(cid:48),b(cid:48)),(c,d) in Z2 such that
(a,b) ≤ (a(cid:48),b(cid:48)) it follows that (a,b)+(c,d) ≤ (a(cid:48),b(cid:48))+(c,d). This immediately follows
from Definition 3.2 and the fact that (Z,≤) is a totally ordered abelian group.
4
Definition 3.4. We define a function D : K[X ,X ] → Z2 ∪{−∞} by setting
1 2
D(f) := sup{(i,j) | a (cid:54)= 0}
ij
where f = (cid:80) a XiXj and the total order on Z2 as in Definition 3.1.2.
i,j≥0 ij 1 2
Remark 3.5. In all the equations and inequalities involving elements of G∪{−∞} we
use the convention that the symbol −∞ satisfies the following: ∀g ∈ G : −∞ < g and
g +−∞ = −∞.
Definition 3.6. Let (G,≤,+) be a totally ordered abelian group. Let R be a commu-
tative ring. We call a function δ : R → G∪{−∞} a degree function if it satisfies the
following properties:
For all x,y ∈ R :
(i) δ(xy) = δ(x)+δ(y)
(ii) δ(x+y) ≤ max{δ(x),δ(y)}
(iii) δ(x) = −∞ ⇔ x = 0
Remark 3.7. Letδ : R → G∪{−∞}beadegreefunction. Sinceδ(1) = δ(1·1) = 2δ(1)
we get δ(1) = 0. Now we want to compute δ(−1). We have 0 = δ(1) = δ((−1)·(−1)) =
2δ(−1) which implies δ(−1) = 0 since a totally ordered abelian group has no torsion
elements.
Lemma 3.8. For a degree function δ : R → G∪{−∞} we have the following property:
For all x,y ∈ R :
if δ(x) (cid:54)= δ(y) it follows that δ(x+y) = max{δ(x),δ(y)}.
Proof. Assume δ(x) > δ(y). We leave the case where y = 0 to the reader. Therefore
we assume that y is nonzero. we know that
δ(x) = δ((x+y)+(−y)) ≤ max{δ(x+y),δ(−y)}.
If δ(x + y) were strictly smaller than max{δ(x),δ(y)} = δ(x) it would imply δ(x) ≤
δ(−y) = δ(−1)+δ(y) = δ(y)whichisacontradiction. Thereforeδ(x+y) = max{δ(x),δ(y)}.
Lemma 3.9. Let R be an integral domain and δ a degree-function on R. We can extend
δ to δ˜: Quot(R) → G∪{−∞} by setting δ˜(a) = δ(a)−δ(b) for an a ∈ Quot(R). Then
b b
˜
δ is well-defined and is again a degree-function on Quot(R).
Proof. We leave the proof to the reader.
Definition 3.10. Let G,G(cid:48) be totally ordered abelian groups. We call a function
φ : G∪{−∞} → G(cid:48) ∪{−∞} a homomorphism of totally ordered abelian groups if it
satisfies the following properties:
For all g,h ∈ G∪{−∞}:
(i) g ≤ h ⇒ φ(g) ≤ φ(h)
(ii) φ(g) = −∞ ⇔ g = −∞
(iii)φ(g +h) = φ(g)+φ(h)
5
Lemma 3.11. Let φ : G∪{−∞} → G(cid:48)∪{−∞} be a homomorphism of totally ordered
abelian groups. Let S be a finite subset of G ∪ {−∞}. Then we have the following
equation:
(4) sup{φ(S)} = φ(sup{S})
Proof. This immediately follows by Definition 3.10 property (i) and if S is empty prop-
erty (ii).
Lemma 3.12. Let δ : R → G ∪ {−∞} be a degree function. Let φ : G ∪ {−∞} →
G(cid:48)∪{−∞} be a homomorphism of totally ordered abelian groups. Then the composition
φ◦δ is again a degree-function on R.
Proof. We leave the proof to the reader.
Lemma 3.13. Both d and D are degree functions on K[X ,X ].
1 2
Proof. We first define the function φ : Z2 ∪{−∞} → Z∪{−∞} by setting φ((a,b)) =
n a + n b and φ(−∞) = −∞ and notice this is a homomorphism of totally ordered
1 2
abelian groups. Then by Lemma 3.11 it follows that φ ◦ D = d. Then Lemma 3.12
tells us that it is enough to prove that D is a degree function on K[X ,X ]. We leave
1 2
the proof that D satisfies property (ii) and (iii) in Definition 3.6 to the reader and only
prove property (i): Let f,g be elements in K[X ,X ]. The case where fg = 0 is clear
1 2
since then f or g will be zero, so we assume that f and g are both nonzero. We write
f and g in the following form:
(cid:88) (cid:88)
(5) f = a XiXj, g = b XiXj.
ij 1 2 ij 1 2
i,j≥0 i,j≥0
We write D(f) = (i ,j ) and D(g) = (i(cid:48),j(cid:48)). The product fg can be written in the
0 0 0 0
following form:
(6) (cid:88)((cid:88)a b )Xi(cid:48)(cid:48)Xj(cid:48)(cid:48)
ij i(cid:48)j(cid:48) 1 2
i(cid:48)(cid:48),j(cid:48)(cid:48) (i,j)+(i(cid:48),j(cid:48))
≥0 =(i(cid:48)(cid:48),j(cid:48)(cid:48))
If we have a (i(cid:48)(cid:48),j(cid:48)(cid:48)) such that the inner sum is nonzero then we know that a term a b
ij i(cid:48)j(cid:48)
which is nonzero must appear in the inner sum. For this term to be nonzero both a
ij
and b have to be nonzero. This implies (i,j) ≤ D(f) and (i(cid:48),j(cid:48)) ≤ D(g). And from
i(cid:48)j(cid:48)
these two inequalities we get that (i(cid:48)(cid:48),j(cid:48)(cid:48)) = (i,j)+(i(cid:48),j(cid:48)) ≤ D(f)+D(g). This gives
us the inequality D(fg) ≤ D(f)+D(g).
Claim 3.14. Let (i,j),(i(cid:48),j(cid:48)) be elements in (Z≥0)2 such that (i,j)+(i(cid:48),j(cid:48)) = D(f)+
D(g) and the term a b is nonzero. Then (i,j) = (i ,j ) and (i(cid:48),j(cid:48)) = (i(cid:48),j(cid:48)).
ij i(cid:48)j(cid:48) 0 0 0 0
Proof. Consider (i,j,i(cid:48),j(cid:48)) ∈ (Z≥0)4 which differs from (i ,j ,i(cid:48),j(cid:48)) and which satisfies
0 0 0 0
the equation (i,j) + (i(cid:48),j(cid:48)) = D(f) + D(g). Let us assume (i,j) (cid:54)= (i ,j ) or else we
0 0
work with (i(cid:48),j(cid:48)) (cid:54)= (i(cid:48),j(cid:48)). Then we must have (i,j) < (i ,j ) or the other way around.
0 0 0 0
Suppose(i,j) < (i ,j ). Thistogetherwiththeequation(i,j)+(i(cid:48),j(cid:48)) = (i ,j )+(i(cid:48),j(cid:48))
0 0 0 0 0 0
implies (i(cid:48),j(cid:48)) > (i(cid:48),j(cid:48)). This means that b is zero so the term a b is also zero.
0 0 i(cid:48)j(cid:48) ij i(cid:48)j(cid:48)
6
This shows us that the inner sum for (i(cid:48)(cid:48),j(cid:48)(cid:48)) = D(f)+D(g) consist only of the term
a b which is nonzero. This implies the inequality D(fg) ≥ D(f)+D(g). Since we
i0j0 i(cid:48)0j0(cid:48)
have both inequalities the equality D(fg) = D(f)+D(g) follows.
Lemma 3.9 tells us that we can extend d and D to K(X ,X ). We will still denote
1 2
these extensions by the same letters d and D. The equation φ◦D = d in the beginning
of the proof of Lemma 3.13 tells us that D is a kind of refinement of d. The next lemma
tells us this equation still holds for the extensions of d and D to K(X ,X )
1 2
Lemma 3.15. The equation φ◦D = d holds on K(X ,X ).
1 2
Proof. We know the equation holds for polynomials. Let q = f be a rational function
g
where f,g ∈ K[X ,X ] and g nonzero. We now have φ(D(q)) = φ(D(f) − D(g)) =
1 2
φ(D(f))−φ(D(g)) = d(f)−d(g) = d(f).
g
Definition 3.16. We define the function |·| : K[X ,X ] → K[X ,X ] by setting
1 2 1 2
(cid:88)
(7) |f| = a XiXj.
ij 1 2
i,j≥0
n1i+n2j=d(f)
It is called the leading term of f.
Definition 3.17. Let R,R(cid:48) be a commutative rings. We call a function ψ : R → R(cid:48)
multiplicative if for all x,y ∈ R : ψ(xy) = ψ(x)ψ(y).
Lemma 3.18. If R and R(cid:48) are integral domains and ψ a multiplicative function from
R to R(cid:48) such that ψ(x) (cid:54)= 0 for all x (cid:54)= 0, we can extend ψ to ψ˜ : Quot(R) → Quot(R(cid:48))
by setting ψ˜(a) := ψ(a) where a ∈ Quot(R). The extension ψ˜ is well defined and is
b ψ(b) b
again a multiplicative function.
Proof. We leave the proof to the reader.
Lemma 3.19. The leading term function is multiplicative.
Proof. Let f,g be elements in K[X ,X ]. We write f and g in the following form:
1 2
(cid:88) (cid:88)
(8) f = a XiXj, g = b XiXj.
ij 1 2 ij 1 2
i,j≥0 i,j≥0
Both |fg| and |f||g| are polynomials where the monomials Xi(cid:48)(cid:48)Xj(cid:48)(cid:48) with nonzero co-
1 2
efficients satisfy n i(cid:48)(cid:48) + n j(cid:48)(cid:48) = d(fg) = d(f) + d(g). Therefore to show the equality
1 2
|fg| = |f||g| it is enough to show that for any (i(cid:48)(cid:48),j(cid:48)(cid:48)) ∈ (Z≥0)2 such that n i(cid:48)(cid:48) +
1
n j(cid:48)(cid:48) = d(f) + d(g), the corresponding coefficients of the monomial Xi(cid:48)(cid:48)Xj(cid:48)(cid:48) are the
2 1 2
same. In |fg| this corresponding coefficient is the sum over all a b such that
ij i(cid:48)j(cid:48)
(i,j) + (i(cid:48),j(cid:48)) = (i(cid:48)(cid:48),j(cid:48)(cid:48)). In |f||g| this coefficient is the sum over all a b such
ij i(cid:48)j(cid:48)
that n i + n j = d(f) and n i(cid:48) + n j(cid:48) = d(g) and (i,j) + (i(cid:48),j(cid:48)) = (i(cid:48)(cid:48),j(cid:48)(cid:48)). These
1 2 1 2
sums are the same since for any (i,j,i(cid:48),j(cid:48)) ∈ (Z≥0)4 such that a b is nonzero and
ij i(cid:48)j(cid:48)
(i,j) + (i(cid:48),j(cid:48)) = (i(cid:48)(cid:48),j(cid:48)(cid:48)), we know that n i + n j ≤ d(f) and n i(cid:48) + n j(cid:48) ≤ d(g).
1 2 1 2
Since (n i + n j) + (n i(cid:48) + n j(cid:48)) = n i(cid:48)(cid:48) + n j(cid:48)(cid:48) = d(f) + d(g) we can conclude that
1 2 1 2 1 2
n i+n j = d(f) and n i(cid:48) +n j(cid:48) = d(g).
1 2 1 2
7
UsingLemma3.18weextendtheleadingtermfunctiontoK(X ,X ), againdenoted
1 2
by |·|.
Definition 3.20. We call an element f in K(X ,X ) (n ,n )-homogeneous if |f| = f.
1 2 1 2
Example 3.21. The notion of a (1,1)-homogeneous polynomial coincides with the
usual notion of a homogeneous polynomial.
Lemma 3.22. Let f ,f be in K(X ,X ) such that d(f ) = d(f ) = c ∈ Z. Then we
1 2 1 2 1 2
have
(9) d(f +f ) < c ⇔ |f |+|f | = 0.
1 2 1 2
Proof. In the case where f and f are polynomials the lemma is clear since
1 2
d(f +f ) < c if and only if the highest terms of f and f cancel each other out, in other
1 2 1 2
words d(f +f ) < c if and only if |f |+|f | = 0. For the general case we write f = p1
1 2 1 2 1 q1
and f = p2 with p ,q ,p ,q polynomials where q and q are nonzero. The inequality
2 q2 1 1 2 2 1 2
d(f +f ) < cisequivalenttod(p q +p q ) < c+d(q )+d(q ) = d(p q ) = d(p q ). This
1 2 1 2 2 1 1 2 1 2 2 1
is equivalent to |p q |+|p q | = 0 by applying the lemma in the case of polynomials.
1 2 2 1
This equation is then equivalent to |f |+|f | = 0, using Lemma 3.19.
1 2
Remark 3.23. Lemma3.22canbeeasilyextendedtoasumofmorethan2polynomials
of the same (n ,n )-degree.
1 2
Lemma 3.24. Let f ,f be in K(X ,X ) such that d(f ) > d(f ). Then we have
1 2 1 2 1 2
(10) |f +f | = |f |.
1 2 1
Proof. We notice d(f +f ) = d(f ) due to Lemma 3.8. Since d((f +f )+(−f )) =
1 2 1 1 2 1
d(f ) < d(f ), Lemma 3.22 tells us that |f +f |+|−f | = 0. This implies (10).
2 1 1 2 1
Lemma 3.25. Let f ,f be in K(X ,X ) such that |f | and |f | are algebraically inde-
1 2 1 2 1 2
pendent over K. Then we have
(11) |K[f ,f ]| ⊆ K[|f |,|f |].
1 2 1 2
Proof. A nonzero element g in |K[f ,f ]| is of the form |p(f ,f )| with p a nonzero
1 2 1 2
polynomial in two variables. Let us write p(Y ,Y ) in the following form:
1 2
(cid:88)
(12) p(Y ,Y ) = a YiYj.
1 2 ij 1 2
i,j≥0
We now isolate the monomials of highest degree. If (d(f ),d(f )) = (0,0) we set
1 2
p := p and µ := 0. Notice that the cases d(f ) = −∞ or d(f ) = −∞ do not
0 1 2
occur since then |f | and |f | would satisfy the equation |f ||f | = 0 which would
1 2 1 2
contradict the assumption that |f | and |f | are algebraically independent over K. If
1 2
(d(f ),d(f )) (cid:54)= (0,0) we define p to be equal to the leading term of p with respect to
1 2 0
the (d(f ),d(f ))-degree function. We also define µ to be the (d(f ),d(f ))-degree of p.
1 2 1 2
Claim 3.26. d(p (f ,f )) = µ.
0 1 2
8
Proof. The element p (f ,f ) is a sum of monomial expressions in f and f where each
0 1 2 1 2
of those expressions has (n ,n )-degree µ. If d(p (f ,f )) were strictly smaller than µ,
1 2 0 1 2
Lemma 3.22 with Remark 3.23 would imply p (|f |,|f |) = 0. But this is not possible
0 1 2
since |f | and |f | are algebraically independent over K.
1 2
By Lemma 3.24 we know that |p(f ,f )| = |p (f ,f )+(terms of lower degree)| =
1 2 0 1 2
|p (f ,f )|. By Lemma 3.22 with Remark 3.23 and Claim 3.26 we know that
0 1 2
d(p (f ,f )−p (|f |,|f |)) < µ = d(p (f ,f )) = d(p (|f |,|f |)).
0 1 2 0 1 2 0 1 2 0 1 2
This tells us that
|p (f ,f )| = |p (|f |,|f |)+p (f ,f )−p (|f |,|f |)| = |p (|f |,|f |)|
0 1 2 0 1 2 0 1 2 0 1 2 0 1 2
byusingLemma3.24. Sincep (|f |,|f |)isalreadyhomogeneous(itisthesumofhomo-
0 1 2
geneous elements of the same degree), we have |p(f ,f )| = |p (|f |,|f |)| = p (|f |,|f |)
1 2 0 1 2 0 1 2
which lies in K[|f |,|f |].
1 2
4 Key lemmas
In this section we will prove some rather technical lemmas which play an important
role in the proof of Theorem 2.0.2.
Lemma 4.1. If f and f are non-zero (n ,n )-homogeneous elements of K(X ,X )
1 2 1 2 1 2
that are algebraically dependent over K, then there exists λ ∈ K× such that
(13) fd(f2) = λfd(f1).
1 2
Proof. Since f and f are algebraically dependent over K, we have p(f ,f ) = 0 for
1 2 1 2
some nonzero polynomial p. Since the case where (d(f ),d(f )) = (0,0) is clear we
1 2
assume (d(f ),d(f )) (cid:54)= (0,0). We can assume p to be (d(f ),d(f ))-homogeneous by
1 2 1 2
possibly replacing p by its (d(f ),d(f ))-leading term. Let (d(f ),d(f )) = (sd ,sd )
1 2 1 2 1 2
with s = gcd(d(f ),d(f )) then p is also (d ,d )-homogeneous. We write p in the
1 2 1 2
following form:
(cid:88)
(14) p = a YiYj.
ij 1 2
i,j≥0
Since p is (d ,d )-homogeneous we know that if a is nonzero, then id + jd is a
1 2 ij 1 2
constant independent of i and j. We fix an i and j such that a is nonzero. For
0 0 i0j0
any other pair (i(cid:48),j(cid:48)) such that a is nonzero we have the following equation:
i(cid:48)j(cid:48)
(15) i d +j d = i(cid:48)d +j(cid:48)d .
0 1 0 2 1 2
Using the fact that d and d are coprime this equation implies that
1 2
(16) (i(cid:48),j(cid:48)) = (i −kd ,j +kd )
0 2 0 1
for some k in Z. If we divide out fi0fj0 from the equation p(f ,f ) = 0 and possibly
1 2 1 2
multiply by an appropriate power of f−d2fd1 we get a nonzero polynomial equation for
1 2
f−d2fd1 with coefficients in K. Since K is relatively algebraically closed in K(X ,X )
1 2 1 2
we get that f−d2fd1 lies in K×. This implies the lemma.
1 2
9
