locally nilpotent derivations on polynomial rings in …€¦ · locally nilpotent derivations are...

LOCALLY NILPOTENT DERIVATIONS ON

POLYNOMIAL RINGS IN TWO VARIABLES OVER A

FIELD OF CHARACTERISTIC ZERO

By

Nyobe Likeng Samuel Aristide, B.Sc., M.S.

March 2017

A Thesis

submitted to the School of Graduate Studies and Research

in partial ful�llment of the requirements

for the degree of

Master of Science in Mathematics1

c© Samuel Aristide Nyobe Likeng, Ottawa, Canada, 2017

1The M.Sc. Program is a joint program with Carleton University, administered by the Ottawa-

Carleton Institute of Mathematics and Statistics

Abstract

The main goal of this thesis is to present the theory of Locally Nilpotent Derivations

and to show how it can be used to investigate the structure of the polynomial ring

in two variables k[X, Y ] over a �eld k of characteristic zero. The thesis gives a com-

plete proof of Rentschler's Theorem, which describes all locally nilpotent derivations

of k[X, Y ]. Then we present Rentschler's proof of Jung's Theorem, which partially

describes the group of automorphisms of k[X, Y ]. Finally, we present the proof of the

Structure Theorem for the group of automorphisms of k[X, Y ].

Key Words: Locally nilpotent derivation, Rentschler's Theorem, Tame automor-

phisms, Wild automorphisms.

ii

Acknowledgements

I would like to thank my supervisor Professor Daniel Daigle, who has always been

there for me when I was facing some di�culties in my courses and who patiently

allowed me to understand the theory developed in this thesis. He also helped with

editing of my English in this thesis; and all over who believed in me when I was facing

some hard moment at the beginning of my program.

iii

Dedication

I dedicate this work to my parents.

iv

Contents

Abstract ii

Acknowledgements iii

Dedication iv

1 Preliminaries 6

1.1 Basic de�nitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.2 Basic properties of polynomial rings . . . . . . . . . . . . . . . . . . . 11

1.3 Degree functions and factorially closed subrings . . . . . . . . . . . . 19

1.4 Z-graded rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

1.5 Rings of fractions and transcendence degree . . . . . . . . . . . . . . 28

2 Locally Nilpotent Derivations 36

2.1 Derivations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

2.2 Locally nilpotent derivations . . . . . . . . . . . . . . . . . . . . . . . 46

2.3 Slices and preslices . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

2.4 Degree and Homogenization of Derivations . . . . . . . . . . . . . . . 59

3 Derivations and automorphisms of k [X ,Y ] 66

3.1 Some facts on polynomial rings . . . . . . . . . . . . . . . . . . . . . 66

3.2 Preliminaries on automorphisms of k[X, Y ] . . . . . . . . . . . . . . . 68

3.3 Rentschler's Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

3.4 Automorphism theorem . . . . . . . . . . . . . . . . . . . . . . . . . 78

v

4 Polynomial rings in three variables 89

4.1 Locally nilpotent derivations . . . . . . . . . . . . . . . . . . . . . . . 89

4.2 Automorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

Bibliography 94

vi

Introduction

Throughout this thesis the word ring means a commutative ring with a unity.

A derivation of a ring B is a map D : B −→ B satisfying D(a+ b) = D(a) +D(b)

and D(ab) = D(a)b+ aD(b) for all a, b ∈ B.A derivation D of a ring B is said to be locally nilpotent if:

for each b ∈ B, there exists n ∈ N such that Dn(b) = 0.

Let k be a �eld. Given the ring B = k[X, Y ], the polynomial ring in two variables

over k, two problems may be stated:

(1) Describe the structure of the group of automorphisms of the k-algebra k[X, Y ].

(2) Assuming that k has characteristic zero, describe all locally nilpotent derivations

of k[X, Y ].

Problem (1) has been solved by Jung in 1942 and by van der Kulk in 1953. Problem

(2) was solved by Rentschler in 1968. The objective of this thesis is to present these

solutions, using the book of Freudenburg [6] as our main reference.

Let k be a �eld and n ≥ 1, and let k[X1, . . . , Xn] be the polynomial ring in n

variables. The automorphism group of the k-algebra k[X1, . . . , Xn] is denoted GAn(k),

and is called the general a�ne group in dimension n. Let Afn(k) be the set of

φ ∈ GAn(k) given by

φ(Xj) = a1jX1 + · · ·+ anjXn + bj for 1 ≤ j ≤ n,

where b1, . . . , bn ∈ k and (aij) ∈ GLn(k) and BAn(k) be the set of φ ∈ GAn(k) given

by

φ(Xi) = aiXi + fi for 1 ≤ i ≤ n,

1

2

where a1, . . . , an ∈ k×, f1 ∈ k, and fi ∈ k[X1, . . . , Xi−1] for all i such that 1 < i ≤ n.

Then Afn(k) and BAn(k) are subgroups of GAn(k). The elements of Afn(k) are called

the a�ne automorphisms and those of BAn(k) the triangular automorphisms ; Afn(k)

and BAn(k) are sometimes called the a�ne subgroup and the triangular subgroup

of GAn(k). The subgroup of GAn(k) generated by Afn(k) ∪ BAn(k) is called the

tame subgroup of GAn(k), and the elements of that subgroup are called the tame

automorphisms of k[X1, . . . , Xn]. An automorphism that is not tame is said to be

wild. It is natural to ask:

Is it the case that all automorphisms of k[X1, . . . , Xn] are tame?

In 1942 Jung showed in his paper [9] that GA2(k) = 〈Af2(k) ∪ BA2(k)〉 in the

particular case where k is of characteristic zero; in other words, he showed that all

automorphisms of k[X, Y ] are tame. In 1953 van der Kulk showed in [10] that this

result is also true when k is of positive characteristic. The same article of van der

Kulk also contains a Structure Theorem for GA2(k) (Theorem 3.4.4 of this thesis),

which asserts that each element of this group has a unique factorization of a certain

kind. This result of van der Kulk completely describes the structure of the group

GA2(k).

It is interesting to note that the results of Jung and van der Kulk cannot be

extended in higher dimension. As we will see in the last chapter, there exist wild

automorphisms of k[X, Y, Z] and it is an open problem to describe the structure of

GA3(k).

The following theorem, proved by Rentschler in [18] in 1968, gives a complete

solution to the Problem (2) stated above:

Rentschler's Theorem. Let B = k[X, Y ] where k is a �eld of charac-

teristic zero, and let D : B → B be a locally nilpotent derivation. Then

there exist a tame automorphism α of B and a polynomial f(X) ∈ k[X]

such that α ◦D ◦ α−1 = f(X) ∂∂Y

.

Rentschler also showed that his Theorem implies Jung's Theorem as a corollary,

and this will be our approach for proving Jung's Theorem in this thesis.

3

Let us discuss locally nilpotent derivations in general, to give some perspective

on Problem (2). We write LND(B) for the set of locally nilpotent derivations of a

ring B.

It was known before Rentschler's article that locally nilpotent derivations are

closely related to group actions on algebraic varieties. More precisely, if X ⊆ Cn

is an a�ne algebraic variety and A(X) is its a�ne coordinate ring (i.e., A(X) =

C[X1, . . . , Xn]/I(X) where I(X) ⊂ C[X1, . . . , Xn] is the ideal of X), then studying

the locally nilpotent derivations of the ring A(X) is equivalent to studying the actions

of the algebraic group Ga = (C,+) on the algebraic variety X. In particular, studying

the locally nilpotent derivations of C[X, Y ] is equivalent to studying the actions of

Ga on the algebraic variety C2. This is actually the reason why Rentschler proved

the above Theorem: his article [18] gives a complete description of all actions of Ga

on C2, and he obtained that result as a consequence of his description of the locally

nilpotent derivations of C[X, Y ].

Given an algebraB over a �eld k of characteristic zero, locally nilpotent derivations

may be used for investigating the automorphisms of B. Indeed, it can be shown that

for each D ∈ LND(B) the map

exp(D) : B → B

b 7→∑∞

n=01n!Dn(b)

is an automorphism of B as a k-algebra. In the case where B = k[X1, . . . , Xn], it is

conjectured that GAn(k) =⟨

Afn(k) ∪ {exp(D) | D ∈ LND(B)}⟩, and it is believed

that in order to understand GAn(k) it will be necessary to �rst understand LND(B).

Locally nilpotent derivations are also very useful for de�ning invariants of rings.

For instance, if B is a ring then ML(B) =⋂D∈LND(B)

ker(D) is a subring of B which

is called the Makar-Limanov invariant of B. The ring invariants de�ned in terms of

locally nilpotent derivations are currently the subject of much research activity.

These are some of the reasons why locally nilpotent derivations are now regarded

as an important tool for investigating the structure of commutative rings. The aim of

the thesis is to present the basic theory of locally nilpotent derivations, and to show

how that theory can be used to investigate the ring k[X, Y ].

4

This thesis is organised as follows:

Chapter 1 gathers some preliminaries on several topics: basic algebra, polynomial

rings and systems of variables, transcendence degree, degree functions and graded

rings.

Chapter 2 is about Locally Nilpotent Derivations. We �rst give some basic facts

from this theory, and prove the Slice Theorem. At the end of this chapter we present

the technique of homogenization of derivations, which has a particular importance

for the proof of Rentschler's Theorem.

Chapter 3 focuses on the polynomial ring k[X, Y ], and in particular on the above

problems (1) and (2). We give a complete proof of Rentschler's Theorem, and deduce

Jung's Theorem as a corollary. Our proof of Jung's Theorem is based on the one given

in [6], which is essentially the one given by Rentschler. As we already mentioned, van

der Kulk showed that Jung's Theorem is valid over an arbitrary �eld. However our

proof is only valid in characteristic zero.

We also give a complete proof of the Structure Theorem for GA2(k) (our proof

is for the case char k = 0). This, however, turned out to be more complicated than

expected. We were planning to follow the proof given in [6], but there is a step

in that argument that we could not understand. Namely, on page 90 of [6], one

considers an arbitrary κ ∈ GA2(k) and a factorization κ = ch1 · · ·hn satisfying certainrequirements, and one has to show that such a factorization of κ is unique. On line

−7 of that page, it is claimed that it is enough to prove the special case �κ = identity�

of that statement; no justi�cation is given for that claim, and we don't see why it is

true. The case �κ = identity� is correctly proved in [6], but as far as we can see, this

does not prove the Theorem. Resolving this issue added several pages to the proof,

so �nally our proof of the Structure Theorem is quite di�erent from that given in [6].

Chapter 4 gives some remarks on the locally nilpotent derivations and the auto-

morphisms of the polynomial ring in three variables. More precisely we observe that

Rentschler's Theorem is not true in dimension 3 and that some elements of GA3(k)

are wild.

References. Our main references are Freudenburg's book [6], Daigle's lecture notes

5

[3], and the articles of Jung [9], van der Kulk [10] and Rentschler [18].

Since this thesis is of an expository nature, all results contained in it are known,

and in principle each result should come with a reference. We do that whenever

possible, but there remains a certain number of results for which we are unable to

�nd suitable references. Typically, these are simple results known to the experts, but

whose proofs are not published or hard to locate. An example of that is the sequence

of results (1.2.7, 1.2.9, 1.2.19, 1.2.20 and 1.2.22) on polynomial rings.

Chapter 1

Preliminaries

In this chapter, we recall some basic notions which will be used later in this thesis;

most of these results can be found in any textbook of commutative algebra.

1.1 Basic de�nitions

In this thesis, the word ring has the following meaning:

De�nition 1.1.1. A ring is a set B with two binary operations (addition and mul-

tiplication) satisfying:

1. (B,+) is an abelian group.

2. Multiplication is commutative, associative and distributive over addition:

xy = yx, for all x, y ∈ B

(xy)z = x(yz), for all x, y, z ∈ B

x(y + z) = xy + xz, for all x, y, z ∈ B.

3. Multiplication has an identity element, denoted 1 or 1B:

x1 = x = 1x, for all x ∈ B.

6

CHAPTER 1. PRELIMINARIES 7

De�nition 1.1.2. A unit of a ring B is an element x ∈ B satisfying xy = 1 for some

y ∈ B. The symbol B× denotes the set of all units of B.

A ring which has only one element is called a zero ring. It is well known that a

ring B is a zero ring if and only if 1B = 0B.

De�nition 1.1.3. 1. An integral domain (or simply a domain) is a nonzero ring

B such that, for all x, y ∈ B, the condition xy = 0 implies x = 0 or y = 0.

2. A �eld is a nonzero ring F such that each element of F \ {0} is a unit.

De�nition 1.1.4. A subring of a ring B is a subgroup of (B,+) which is closed under

the multiplication of B and which contains 1B, the identity of B.

De�nition 1.1.5. Let A and B be rings. A ring homomorphism from A to B is a

set map ϕ : A −→ B satisfying ϕ(1A) = 1B and:

ϕ(a1 + a2) = ϕ(a1) + ϕ(a2) and ϕ(a1a2) = ϕ(a1)ϕ(a2) for all a1, a2 ∈ A.

For any ring B there exists exactly one ring homomorphism from Z to B. This

unique homomorphism η : Z −→ B is de�ned by:

For each n ∈ Z, η(n) = n1B =

1 + 1 + · · ·+ 1︸︷︷︸n

if n > 0

−(1 + 1 + · · ·+ 1︸︷︷︸|n|

) if n < 0

0 if n = 0.

Note that ker η is an ideal of Z. So there exists a unique nonnegative integer p ≥ 0

such that ker η = (p).

De�nition 1.1.6. The nonnegative integer p de�ned above is called the characteris-

tic of B.

It is easy to see that if B is a domain, its characteristic is either zero or a prime

number.


De�nition 1.1.7. Let B be a domain.

1. An element p of B is said to be irreducible if p /∈ B× ∪ {0} and if the condition

p = xy (with x, y ∈ B) implies that {x, y} ∩B× 6= ∅.

2. An element p of B is said to be prime if p /∈ B× ∪{0} and if the condition p|xy(with x, y ∈ B) implies that p|x or p|y.

3. Two elements a and b are said to be associates in B, if a = ub for some u ∈ B×.

The following result is Proposition 10 of the Section 8.3 of [4].

Proposition 1.1.8. In a domain B, every prime element is irreducible.

Proof. Let p be prime such that p = xy in B. Then p divides x or y (since p is prime).

Assume without loss of generality that p divides x, then x = µp for some µ ∈ B. Thisimplies that x = µp = µxy; so x(1 − yµ) = 0. Since B is a domain and x 6= 0, we

thus have yµ = 1, that is y ∈ B×. Thus p is irreducible.

Remark 1.1.9. The converse is not necessarily true.

De�nition 1.1.10. A unique factorization domain (UFD) is a domain B such that

each element of B \ (B× ∪ {0}) is a product of prime elements.

Example 1.1.11. 1. The ring Z of integers is a UFD.

2. A �eld is trivially a UFD.

3. If A is a UFD, then the polynomial ring A[X1, · · · , Xn] is a UFD.

The following well known result is Proposition 12 of Section 8.3 of [4] but the

approach here is quite di�erent.

Proposition 1.1.12. If B is a UFD, then an element of B is irreducible if and only

if it is prime.

Proof. We have already shown in Proposition 1.1.8 that every prime element is ir-

reducible in a domain. Now assume that p is an irreducible element in B, then

p ∈ B \ (B× ∪ {0}) and since B is a UFD p can be written as p = q1q2 · · · qr; whereall the qi for i = 1, · · · , r are prime. But p is irreducible, this forces r = 1. Hence

p = q1 and then p is prime.


De�nition 1.1.13. Let A be a ring. An A-module is a a pair (M,µ), where M is an

abelian group (written additively) and µ is a mapping of A×M into M such that, if

we write ax for µ(a, x) (a ∈ A, x ∈M), the following axioms are satis�ed:

1. a(x+ y) = ax+ ay,

2. (a+ b)x = ax+ bx,

3. (ab)x = a(bx),

4. 1x = x (a, b ∈ A; x, y ∈M).

A submodule of an A-module M is a subgroup N of M satisfying ax ∈ N for all

a ∈ A, x ∈ N .

We shall assume that the reader is familiar with the basic concepts of the theory

of modules.

De�nition 1.1.14. Let A be a ring. An A-algebra is a pair (B, f), where B is a ring

and f : A −→ B is a ring homomorphism.

Example 1.1.15. 1. Suppose A is a subring of a ring B. Then the inclusion map

i : A −→ B is a ring homomorphism, so (B, i) is an A-algebra. One says B is

an A-algebra without mentioning i.

2. Let A be a ring. The zero ring is an A-algebra.

3. Let A be a ring and n ∈ Z+. Then the polynomial ring A[X1, · · · , Xn] is an

A-algebra with the inclusion homomorphism i : A −→ A[X1, · · · , Xn].

Remark 1.1.16. 1. If k is a �eld and (B, f) is a k-algebra such that B 6= 0, then

f is injective. So k can be canonically identi�ed with its image in B. Thus

a nonzero k-algebra (where k is a �eld) is e�ectively a ring containing k as a

subring.

2. For any ring A, there exists a unique ring homomorphism Z −→ A. Thus every

ring is automatically a Z-algebra. Moreover, if A is of characteristic 0, then A

is a ring containing Z as a subring.


3. Let (B, f) be an A-algebra. Given a ∈ A and b ∈ B, we can de�ne a product

ab = f(a)b. This product confers to B a structure of A-module.

De�nition 1.1.17. Let (B, f) and (C, g) be two A-algebras. An A-algebra homo-

morphism (or simply an A-homomorphism) h : B −→ C is a ring homomorphism

which satis�es h ◦ f = g. If this is the case, then h : B −→ C is a homomorphism of

A-modules.

Remark 1.1.18. Let A be a ring.

1. Let C be a subring of two rings B and D. Then it is easy to check that a ring

homomorphism ϕ : B −→ D is a C-algebra homomorphism if and only if it

satis�es ϕ(c) = c for all c ∈ C.

2. If B is an A-algebra, then the identity map B −→ B is an A-homomorphism.

3. A composition of A-homomorphisms is an A-homomorphism.

4. If ϕ : B −→ C is a bijective A-homomorphism, then ϕ−1 : C −→ B is an

A-homomorphism. In that case we say that ϕ is an isomorphism of A-algebras,

or simply an A-isomorphism.

De�nition 1.1.19. Let A be a ring and (B, f) an A-algebra. A subalgebra R of B

is a subring R of B satisfying f(A) ⊆ R.

Remark 1.1.20. Let A be a ring.

1. If R is a subalgebra of an A-algebra B, then R is also an A-algebra and the

inclusion R −→ B is an A-homomorphism.

2. If B is an A-algebra then an arbitrary intersection of subalgebras of B is a

subalgebra of B.

3. Given (B, f) and (C, g) two A-algebras and an A-homomorphism h : B −→ C,

we have h(B) is a subalgebra of C.


De�nition 1.1.21. Let A be a ring, (B, f) an A-algebra and S a subset of B. The

subalgebra of B generated by S is de�ned to be the intersection of all subalgebras R

of B satisfying S ⊆ R. It is denoted by A[S].

Remark 1.1.22. Let A be a ring, (B, f) an A-algebra and S a subset of B.

1. The subalgebra A[S] of (B, f) is the intersection of all the subrings R of B

satisfying f(A) ∪ S ⊆ R. In the case where A is a subring of B, A[S] is the

smallest subring of B containing A ∪ S.

2. The subalgebra A[S] of (B, f) is the smallest subalgebra of B that contains S.

3. If S is a �nite set, say S = {f1, · · · , fn}, we write A[f1, · · · , fn] instead of

A[{f1, · · · , fn}].

4. If S = ∅, then A[∅] = f(A).

1.2 Basic properties of polynomial rings

In this section we recall some basic results about polynomial rings which will be very

useful later in this thesis.

De�nition 1.2.1. Given a ring A and n ∈ Z+, we write A[X1, · · · , Xn] for the

polynomial ring in n variables over A. We assume that the reader is familiar with

this ring. Note the following:

1. Each element P of A[X1, · · · , Xn] is a formal expression

P =∑

(j1,··· ,jn)∈Nn

aj1,··· ,jnXj11 · · ·Xjn

n , where aj1,··· ,jn ∈ A for all (j1, · · · , jn) ∈ Nn

and aj1,··· ,jn = 0 for all (j1, · · · , jn) except possibly �nitely many of them. Note

that a polynomial is a formal sum, not a function.

2. The element aj1,··· ,jnXj11 · · ·Xjn

n of this formal sum is called a monomial term of

P .


3. If aj1,··· ,jn 6= 0, then the exponent ji of Xi is called the degree in Xi of the

monomial aj1,··· ,jnXj11 · · ·Xjn

n . The sum j = j1 + j2 + · · ·+ jn is called the degree

of this monomial. A polynomial is called homogeneous if all its monomials with

nonzero coe�cients aj1,··· ,jn have the same degree.

4. The degree of a nonzero polynomial is the largest degree of any of its nonzero

monomial terms. The degree of the zero polynomial is −∞.

5. If P is a polynomial in n variables, the sum of all the monomials in P of degree

m is called the homogeneous component of P of degree m (m ∈ N).

Remark 1.2.2. • If P 6= 0 is of degree n, then P may be written as the sum

P0 + P1 + · · ·+ Pn where Pm is the homogeneous component of P of degree m,

for 0 ≤ m ≤ n (where some Pm may be zero).

• Note that A[X1, · · · , Xn] can be de�ned inductively by

A[X1, · · · , Xn] = A[X1, · · · , Xn−1][Xn].

This means that we can consider polynomials in n variables with coe�cients in

A, simply as polynomials in one variable (say Xn), but now with coe�cients

that are themselves polynomials in n− 1 variables.

• The degree of a polynomial (as de�ned in part (4) of De�nition 1.2.1) is an

element of N ∪ {−∞}.

• When A is a domain we have A[X1, · · · .Xn]× = A×.

De�nition 1.2.3. Let A be a ring and B an A-algebra.

a. Given P =∑

i1,··· ,in ai1,··· ,inXi11 · · ·X in

n ∈ A[X1, · · · , Xn] and (b1, · · · , bn) ∈ Bn,

we de�ne P (b1, · · · , bn) ∈ B by:

P (b1, · · · , bn) =∑i1,··· ,in

ai1,··· ,inbi11 · · · binn .


b. Given b = (b1, · · · , bn) ∈ Bn, we de�ne the map

evb : A[X1, · · · , Xn] −→ B

P (X1, · · · , Xn) 7−→ P (b1, · · · , bn),

which we call �evaluation at b�.

The following is the universal property of the polynomial algebra. It implies,

among other things, that evb is a homomorphism of A-algebras. The proof of the

universal property can be found in standard algebra textbooks.

Theorem 1.2.4. Let A be a ring, n ≥ 1 and A[X1, · · · , Xn] the polynomial ring in n

variables over A. Then for any choice of an A-algebra (B, f) and of an n-tuple b =

(b1, · · · , bn) ∈ Bn, there exists a unique A-homomorphism ϕ : A[X1, · · · , Xn] −→ B

satisfying ϕ(Xi) = bi for all i = 1, · · · , n.

A[X1, · · · , Xn]∃!ϕ

Xi 7→bi// B

A?�i

OO

f

99

Moreover, the following hold:

a. ϕ = evb

b. Imϕ = A[b1, · · · , bn] = {P (b1, · · · , bn) | P ∈ A[X1, · · · , Xn]}.

De�nition 1.2.5. Let A be a subring of a ring B. An element b ∈ B is said to be

algebraic over A if there exists a nonzero polynomial P ∈ A[T ] such that P (b) = 0.

Otherwise b is transcendental over A. If P can be chosen to be monic over A then b

is said to be integral over A.

We say that B is algebraic over A if each element of B is algebraic over A.

We say that A is algebraically closed in B if each element of B \ A is transcendental

over A.

We say that A is integrally closed in B if no element of B \ A is integral over A.


De�nition 1.2.6. Let B be an algebra over a ring A. An element f = (f1, · · · , fn) ∈Bn is said to be algebraically dependent over A if there exists P ∈ A[X1, · · · , Xn]\{0}such that P (f1, · · · , fn) = 0. Otherwise f = (f1, · · · , fn) is said to be algebraically

independent over A.

Notation. Let B be an algebra over a ring A and let n ∈ N. We write

B = A[n]

as an abbreviation for the sentence:

B is A-isomorphic to the polynomial algebra A[X1, . . . , Xn].

Note that the conditions B1 = A[n] and B2 = A[n] do NOT imply that B1 = B2.

Corollary 1.2.7. Let B be an algebra over a ring A. If f = (f1, · · · , fn) ∈ Bn is

algebraically independent over A then A[f1, · · · , fn] = A[n].

Proof. The universal property 1.2.4 implies that evf : A[X1, · · · , Xn] −→ A[f1, · · · , fn]

is a surjective A-homomorphism. If P ∈ ker(evf ) then P (f1, . . . , fn) = evf (P ) = 0,

so P = 0 since f is algebraically independent over A. Therefore ker(evf ) = 0 and so

evf is an A-isomorphism.

The converse of Corollary 1.2.7 is true, but is more di�cult to prove. We will

prove it in Proposition 1.2.22.

Example 1.2.8. 1. Let B = C[X, Y, Z]; the elementX2Y 3+1 of B is transcendental

over C, so Corollary 1.2.7 implies that C[X2Y 3 + 1] = C[1].

2. Let B = C[X, Y, Z] and (X2, Y 3, Z2) ∈ B3. Since (X2, Y 3, Z2) is algebraically

independent over C, Corollary 1.2.7 implies that C[X2, Y 3, Z2] = C[3].

Proposition 1.2.9. For an algebra B over a ring A, the following conditions are

equivalent:

a. B = A[n]


b. There exist f1, · · · , fn ∈ B such that (f1, · · · , fn) is algebraically independent

over A and A[f1, · · · , fn] = B.

Proof. b)⇒ a) is Corollary 1.2.7.

a) ⇒ b) The condition B = A[n] implies that there exists an A-isomorphism φ :

A[X1, · · · , Xn] −→ B. Set f1 = φ(X1), · · · , fn = φ(Xn). Since φ is anA-homomorphism

that sends Xi to fi for all i = 1, . . . , n, we must have Im(φ) = A[f1, · · · , fn] by part

(b) of Theorem 1.2.4. As φ is surjective, B = Im(φ) = A[f1, · · · , fn].

Note that φ = evf by part (a) of Theorem 1.2.4. If P ∈ A[X1, · · · , Xn] is such

that P (f1, · · · , fn) = 0 then φ(P ) = evf (P ) = P (f1, · · · , fn) = 0, so P = 0 since φ is

injective. Thus (f1, · · · , fn) is algebraically independent over A.

De�nition 1.2.10. A ring B is Noetherian if it satis�es the following equivalent

conditions:

(i) Every in�nite ascending sequence I0 ⊆ I1 ⊆ I2 ⊆ · · · of ideals of B stabilizes;

that is there exists a nonnegative integer n such that In = In+1 = In+2 = · · · .

(ii) Every non empty collection of ideals of B has a maximal element.

(iii) Every ideal of B is �nitely generated.

Example 1.2.11. Every �eld is a Noetherian ring.

The following basic results on Noetherian rings (1.2.12�1.2.15) can be found in

Chapter 7 of [1].

Theorem 1.2.12 (Hilbert's Basis Theorem). If A is a Noetherian ring then so is the

polynomial ring A[X].

Corollary 1.2.13. If A is a Noetherian ring, then so is the polynomial ring in n

variables A[X1, · · · , Xn], for every n ≥ 1.

Lemma 1.2.14. If A is a Noetherian ring and A→ B is a surjective ring homomor-

phism then B is a Noetherian ring.

Lemma 1.2.15. Any �nitely generated algebra over a Noetherian ring is itself a

Noetherian ring.


The following result is Exercise 10, Section 2, Chapter 1 of [11].

Lemma 1.2.16. Let B be a Noetherian ring and ϕ : B −→ B a ring homomorphism.

If ϕ is surjective then ϕ is bijective.

De�nition 1.2.17. Let A be a ring, n ∈ Z+ and B = A[X1, · · · , Xn] a polynomial

ring in n variables over A. A system of variables in B is an n-tuple (f1, · · · , fn) ∈ Bn

satisfying B = A[f1, · · · , fn].

Example 1.2.18. Let B = C[X, Y, Z].

1. (X, Y, Z) is a system of variables in B.

2. Since C[X, Y, Z +X2Y ] = B, (X, Y, Z +X2Y ) is a system of variables in B.

3. We have C[X, Y, Z,XY + Z2] = B but (X, Y, Z,XY + Z2) is not a system of

variables in B (because such a system must be a triple by de�nition).

Proposition 1.2.19. Let A be a ring and B = A[X1, · · · , Xn]. Let (f1, · · · , fn) ∈ Bn

be a system of variables in B. Then:

i. (f1, · · · , fn) is algebraically independent over A.

ii. There exists an A-automorphism Φ : B −→ B such that Φ(Xi) = fi for all

i = 1, . . . , n.

Proof. We have A[f1, · · · , fn] = B since f = (f1, · · · , fn) is a system of variables in

B. Consider the A-homomorphism evf : A[X1, · · · , Xn] −→ A[X1, · · · , Xn]. Then

Im(evf ) = A[f1, · · · , fn] = B, so evf is surjective.

To prove the Proposition, it's enough to prove that condition (i) is true. Indeed,

if (f1, . . . , fn) is algebraically independent over A then ker(evf ) = 0, so evf is an

A-automorphism, so Φ = evf satis�es condition (ii).

So let us prove that f = (f1, . . . , fn) is algebraically independent over A. We �rst

prove this in the case where A is Noetherian. Then A[X1, · · · , Xn] is Noetherian by

Corollary 1.2.13 and evf : A[X1, · · · , Xn] → A[X1, · · · , Xn] is surjective, so Lemma

1.2.16 implies that evf is an A-automorphism. Therefore ker(evf ) = {0} and so f is

algebraically independent over A.


We now drop the assumption that A is Noetherian. By contradiction, assume that

(f1, . . . , fn) is algebraically dependent over A. Then there exists Q ∈ A[T1, . . . , Tn] \{0} such that Q(f1, . . . , fn) = 0 (where A[T1, . . . , Tn] is a polynomial ring in n vari-

ables over A). Since A[f1, . . . , fn] = B, there also exist P1, . . . , Pn ∈ A[T1, . . . , Tn]

such that Xi = Pi(f1, . . . , fn) for i = 1, . . . , n. Choose a �nite subset S of A such

that all coe�cients of Q,P1, . . . , Pn, f1, . . . , fn belong to S.

Consider the minimal subring R of A. Since R is isomorphic to Z or to Z/mZ for

some m > 0, R is a Noetherian ring. Since S is �nite, the subring A0 = R[S] of A is

a �nitely generated algebra over R, so A0 is Noetherian by Lemma 1.2.15. Note that

f1, . . . , fn ∈ A0[X1, . . . , Xn] and Q,P1, . . . , Pn ∈ A0[T1, . . . , Tn]. For each i = 1, . . . , n

we have

Xi = Pi(f1, . . . , fn) ∈ A0[f1, . . . , fn].

It follows that A0[f1, . . . , fn] = A0[X1, . . . , Xn], so (f1, . . . , fn) is a system of variables

of A0[X1, . . . , Xn]. Since A0 is Noetherian, the �rst part of the proof implies that

(f1, . . . , fn) is algebraically independent over A0. But Q ∈ A0[T1, . . . , Tn]\{0} satis�esQ(f1, . . . , fn) = 0, so we have a contradiction.

Corollary 1.2.20. Let A be a ring and let (f1, · · · , fn) be a system of variables of

A[X1, · · · , Xn]. Then for each A-algebra B and each (b1, · · · , bn) ∈ Bn, there exists

a unique A-homomorphism Φ : A[X1, · · · , Xn] → B such that Φ(fi) = bi for all

i = 1, . . . , n.

Proof. Write f = (f1, . . . , fn). By Theorem 1.2.4, there exist uniqueA-homomorphisms

Ψ : A[X1, · · · , Xn] −→ B and φ : A[X1, · · · , Xn] −→ A[X1, · · · , Xn]

such that Ψ(Xi) = bi and φ(Xi) = fi for i = 1, · · · , n. By the same result we have

φ = evf and Im(φ) = A[f1, . . . , fn], so φ is surjective (note that A[X1, · · · , Xn] =

A[f1, · · · , fn] since f is a system of variables of A[X1, · · · , Xn]). By Proposition 1.2.19,

f is algebraically independent over A, and this implies that ker(φ) = ker(evf ) = 0, so

φ is an A-automorphism of A[X1, . . . , Xn]. Thus we obtain the following commutative


diagram.

A[X1, · · · , Xn] Ψ //

φ

��

B

A[X1, · · · , Xn]Ψ◦φ−1

66

Set Φ = Ψ ◦ φ−1, then clearly Φ(fi) = bi for all i = 1, · · · , n.If Φ′ : A[X1, · · · , Xn] → B is any A-homomorphism satisfying Φ′(fi) = bi for

all i, then Φ′ ◦ φ sends Xi to bi for all i, so Φ′ ◦ φ = Ψ by uniqueness of Ψ. Then

Φ′ = Ψ ◦ φ−1 = Φ, proving uniqueness.

Example 1.2.21. The triple (X, Y, Z +X2Y ) is a system of variables in Z[X, Y, Z], Ris a Z-algebra and (

√2,

1

7,√

3) ∈ R3. So Corollary 1.2.20 implies that there exists a

unique Z-homomorphism Φ : Z[X, Y, Z] −→ R such that Φ(X) =√

2, Φ(Y ) =1

7and

Φ(Z +X2Y ) =√

3.

Proposition 1.2.22. Let B be an algebra over a ring A, let (f1, · · · , fn) ∈ Bn and

consider the subalgebra A[f1, · · · , fn] of B. The following are equivalent:

1. A[f1, · · · , fn] = A[n]

2. (f1, · · · , fn) is algebraically independent over A.

Proof. 2)⇒ 1) is Corollary 1.2.7.

1)⇒ 2). Since A[f1, · · · , fn] = A[n], there exists an A-isomorphism

Φ : A[f1, · · · , fn] −→ A[X1, · · · , Xn].

Set gi = Φ(fi), i = 1, · · · , n. We have gi ∈ A[X1, · · · , Xn] for all i, so A[g1, · · · , gn] ⊆A[X1, · · · , Xn]; let us prove that equality holds. Let Q ∈ A[X1, · · · , Xn]. Since

Φ is surjective there exists ξ ∈ A[f1, · · · , fn] such that Φ(ξ) = Q. We have ξ =

P (f1, · · · , fn) for some P ∈ A[X1, · · · , Xn], so

Q = Φ(ξ) = Φ(P (f1, · · · , fn)) = P (g1, · · · , gn) ∈ A[g1, · · · , gn].

So A[g1, · · · , gn] = A[X1, · · · , Xn], hence g = (g1, · · · , gn) is a system of variables. By

Proposition 1.2.19, (g1, · · · , gn) is algebraically independent over A.


If H ∈ A[X1, · · · , Xn] is such that H(f1, · · · , fn) = 0, then

H(g1, . . . , gn) = H(Φ(f1), . . . ,Φ(fn)) = Φ(H(f1, · · · , fn)) = 0,

so H = 0 because (g1, · · · , gn) is algebraically independent over A. This shows that

(f1, · · · , fn) is algebraically independent over A, so (1)⇒ (2).

We generalize the notion of a system of variables (compare with De�nition 1.2.17):

De�nition 1.2.23. Let A be a ring and B = A[n]. By a system of variables in B, we

mean an n-tuple (f1, . . . , fn) ∈ Bn satisfying B = A[f1, . . . , fn].

Remark 1.2.24. Let A be a ring and B = A[n].

1. There exists a system of variables in B. Indeed, there exists an A-isomorphism

φ : A[X1, . . . , Xn]→ B; then (φ(X1), . . . , φ(Xn)) is a system of variables in B.

2. If (f1, . . . , fn) is a system of variables in B, then (f1, . . . , fn) is algebraically

independent over A. Indeed, this follows from Proposition 1.2.22.

1.3 Degree functions and factorially closed subrings

De�nition 1.3.1. Let B be a ring. A Z-valued degree function on B is a set map

deg : B −→ Z ∪ {−∞}

satisfying the following conditions for all f, g ∈ B:

1. deg(f) = −∞ if and only if f = 0;

2. deg(fg) = deg(f) + deg(g);

3. deg(f + g) ≤ max{deg f, deg g}.

An N-valued degree function on B is a set map deg : B −→ N ∪ {−∞} that satis�esthe above conditions 1− 3 for all f, g ∈ B.

By a degree function, we mean either a Z-valued or an N-valued degree function.


Example 1.3.2. Let B = A[X1, · · · , Xn] be the polynomial ring in n variables over

a ring A. Given f ∈ B, let deg(f) denote the usual degree of f as a polynomial

in X1, · · · , Xn (as de�ned in De�nition 1.2.1), and recall that deg(0) = −∞ by

convention. This de�nes a set map deg : B −→ N ∪ {−∞}, f 7−→ deg(f). It is

well known that if A is an integral domain then this map deg : B −→ N∪ {−∞} is adegree function in the sense of De�nition 1.3.1. If A is not an integral domain then

we have deg(fg) ≤ deg(f) + deg(g) for all f, g ∈ B, but equality does not necessarily

hold, so the map deg is not necessarily a degree function.

The following lemma is Exercise 4.2 of [3].

Lemma 1.3.3. If B is a nonzero ring that admits a degree function, then B is an

integral domain.

Proof. Let deg : B −→M ∪ {−∞} be a degree function (where M is either Z or N).Let f, g ∈ B \ {0}. Then deg(f), deg(g) ∈ M and deg(fg) = deg(f) + deg(g) ∈ M ,

so deg(fg) 6= −∞, so fg 6= 0.

Example 1.3.4. Let B = A[X, Y ] = A[2] be a polynomial ring in two variables over a

domain A. Consider the two systems of variables f1 = (X, Y ), f2 = (X + Y 2, Y ) in

B and the element P = X + Y 2 of B. If we denote by degf1 and degf2 the degree

functions associated to the systems of variables f1 and f2 respectively, then we have

degf1(P ) = 2 and degf2(P ) = 1.

The following result is Exercise 4.5 of [3].

Lemma 1.3.5. Let ϕ : B −→ B′ be an injective ring homomorphism. If d : B′ −→N∪{−∞} is a degree function, then d◦ϕ : B −→ N∪{−∞} is also a degree function.

Proof. Straightforward.

Lemma 1.3.6. Let A be a domain and B = A[n]. Then each system of variables

f = (f1, . . . , fn) in B determines an N-valued degree function

degf : B → N ∪ {−∞}

that has the following property:


(∗) for each b ∈ B, degf (b) is equal to the degree of the unique polynomial

P (X1, . . . , Xn) ∈ A[X1, . . . , Xn] that satis�es b = P (f1, . . . , fn).

Proof. Let f = (f1, . . . , fn) be a system of variables in B (i.e., B = A[f1, . . . , fn]). Let

A[X1, . . . , Xn] be the polynomial ring in n variables and consider theA-homomorphism

evf : A[X1, . . . , Xn]→ B. Since B = A[f1, . . . , fn], evf is surjective. Since (f1, . . . , fn)

is algebraically independent over A, evf is injective. So we may consider the A-

isomorphism φ = (evf )−1 : B → A[X1, . . . , Xn]. Let d : A[X1, . . . , Xn] → N ∪ {−∞}

be the standard degree function on A[X1, . . . , Xn] (i.e., d(g) = degree of g as a poly-

nomial in X1, . . . , Xn) and de�ne

degf = d ◦ φ : B → N ∪ {−∞}.

By Lemma 1.3.5, degf is a degree function on B. For each b ∈ B, φ(b) is equal to the

unique polynomial P (X1, . . . , Xn) ∈ A[X1, . . . , Xn] that satis�es b = P (f1, . . . , fn).

So degf satis�es condition (∗).

De�nition 1.3.7. Let A ⊆ B be domains. We say that A is factorially closed in B

if:

∀x, y ∈ B \ {0}, xy ∈ A =⇒ x, y ∈ A.

Example 1.3.8. Let B[X] be the polynomial ring in one variable over an integral

domain B. Then B is factorially closed in B[X].

The next result gives an important property of N-valued degree functions. Note

that Z-valued degree functions do not have the analogous property.

The following is Lemma 5.2 of [3].

Lemma 1.3.9. Let B be a domain with a degree function deg : B −→ N ∪ {−∞}.Then the set {x ∈ B | deg x ≤ 0} is a factorially closed subring of B.

Proof. Clearly this set is a subring of B and for x, y ∈ B \{0} such that deg(xy) ≤ 0,

we have deg(x) ≤ 0 and deg(y) ≤ 0, since deg(xy) = deg(x) + deg(y) and the degree

function has values in N ∪ {−∞}.


Lemma 1.3.10. If A and B are two domains such that B = A[n], then A is factorially

closed in B.

Proof. Let X = (X1, · · · , Xn) be a system of variables in B. By Lemma 1.3.6, X

determine a degree function deg : B −→ N ∪ {−∞} (in the sense of De�nition 1.2.1)

such that {P ∈ B | deg(P ) ≤ 0} = A. By Lemma 1.3.9 we have that A is factorially

closed in B.

Example 1.3.11. Let B = k[X, Y ] = k[2] where k is a �eld and A = k[X2] be a

subring of B. Then we have B 6= A[1]. In fact assume that B = A[1], then A must be

factorially closed in B. But X ∈ B \ {0} and X /∈ A while XX = X2 ∈ A, so A is

not factorially closed in B. Hence B 6= A[1].

The following proposition can be found in [3] as Exercise 5.7.

Proposition 1.3.12. Given A and B two domains such that A ⊆ B, we have:

A is factorially closed in B =⇒ A is algebraically closed in B

=⇒ A is integrally closed in B.

Proof. Assume that A is factorially closed in B and consider b ∈ B algebraic over A.

Let f ∈ A[T ] \ {0} of minimal degree such that f(b) = 0. If b = 0 then b ∈ A. Nowassume that b 6= 0 and f(T ) =

∑ni=0 aiT

i ∈ A[T ], an 6= 0 (n ≥ 1).

f(b) = 0 =⇒ a0 + a1b+ a2b2 + · · ·+ anb

n = 0

=⇒ a1b+ a2b2 + · · ·+ anb

n = −a0

=⇒ bg(b) ∈ A,

where g(T ) = a1 + a2T + · · · + anTn−1 ∈ A[T ] is nonzero because an 6= 0. Since

deg(g) < deg(f), we have g(b) 6= 0 and hence b ∈ A (because A is factorially closed

in B and bg(b) ∈ A). Hence A is algebraically closed in B.

Assume now that A is algebraically closed in B, and let C be the set of integral

elements of B over A. We show that A = C. For all a ∈ A, we have f(X) = X − a ∈A[X] \ {0}, and f(a) = 0, so A ⊆ C. Moreover, if c ∈ C then f(c) = 0 for some

nonzero monic f ∈ A[T ] so c is algebraic over A and hence c ∈ A, showing that

C ⊆ A. Thus A = C and so A is integrally closed in B.


Lemma 1.3.13. Let A be a domain and f ∈ B = A[X1, · · · , Xn] = A[n]. If deg(f) =

d ≥ 1 then f is transcendental over A.

Proof. By Lemma 1.3.10, A is factorially closed in B. So A is algebraically closed in

B by Proposition 1.3.12. Since f ∈ B \ A, f is transcendental over A.

De�nition 1.3.14. A �eld L is said to be algebraically closed if every nonconstant

polynomial in L[X] splits in L[X].

Proposition 1.3.15. Let B = k[n], where k is an algebraically closed �eld. Given

f ∈ B \ k and A = k[f ] a subring of B, the following are equivalent:

(i) A is factorially closed in B.

(ii) For all λ ∈ k, f − λ is irreducible in B.

Proof. (i) =⇒ (ii) Note that f is transcendental over k and hence A = k[f ] = k[1]

by Corollary 1.2.7. So f is a system of variables in A and, by Lemma 1.3.6, we may

consider the degree function degf : A −→ N ∪ {−∞} associated to f (that is, if

ω ∈ A then degf (ω) is the degree of ω as a polynomial in f). Assume that f−λ = gh

where g, h ∈ B \ {0}. Since f − λ ∈ A, then we must have g, h ∈ A since A is

factorially closed in B; but f − λ = gh =⇒ degf (f − λ) = degf (g) + degf (h) that is

1 = degf (g) + degf (h), so degf (g) = 0 or degf (h) = 0 and thus g ∈ k× or h ∈ k×.Therefore f − λ is irreducible in B.

(ii) =⇒ (i) Let g, h ∈ B \ {0} such that gh ∈ A = k[f ]. Then there is P (t) ∈ k[t]

such that gh = P (f).

If degt(P ) < 1, then we have P (t) ∈ k× and so gh = P (f) ∈ k×. Thus g, h ∈ k×

since by Lemma 1.3.10, k is factorially closed in B; therefore g, h ∈ A.Now if m = degt(P ) ≥ 1, then P (t) = λΠm

i=1(t− λi) where λ ∈ k× and λi ∈ k, i ∈{1, · · · ,m}, since k is a algebraically closed �eld. Thus gh = P (f) = λΠm

i=1(f − λi);since f −λi is irreducible in B for all i, g and h are subproducts of λΠm

i=1(f −λi) andhence g, h ∈ A.

Example 1.3.16. 1. Let B = k[X, Y ] = k[2] where k is an algebraically closed �eld.

We have f = XY + 1 ∈ B \ k but f − 1 is not irreducible in B. Thus k[f ] is

not factorially closed in B.


2. B = C[X, Y ] = C[2] and f = X2 +Y 2 ∈ B \{0}. We have C[f ] is not factorially

closed since f is not irreducible in B.

3. Let B = k[X, Y ] = k[2] where k is an algebraically closed �eld; we have k[X2 +

Y 3] is factorially closed in B since for all λ ∈ k, X2 − Y 3 − λ is irreducible in

B.

The following lemma can be found in [3] as Exercise 5.4.

Lemma 1.3.17. Let A be factorially closed subring in a domain B. The following

hold:

1. A× = B×.

2. An element of A is irreducible in A if and only if it is irreducible in B.

3. If an element of A is prime in B then it is prime in A.

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

Proof. 1. Since A is a subring of B we have A× ⊆ B×. Now let x ∈ B×, we havex 6= 0 and there is y ∈ B \ {0} such that xy = 1B. Morever 1B ∈ A since A is

a subring of B. Thus we have x, y ∈ A, and so x ∈ A×. Hence B× ⊆ A× and

thus A× = B×.

2. =⇒) Let p ∈ A be an irreducible element in A. We have p /∈ B× ∪{0} (by part

(1)). Moreover assume that p = ab in B, then a, b ∈ A (since A is factorially

closed in B and p ∈ A), thus a ∈ A× = B× or b ∈ A× = B× (since p is

irreducible in A). Therefore p is irreducible in B.

⇐=) Now assume that p ∈ A is irreducible in B, we have p /∈ A× ∪ {0} (by 1).

Let a, b ∈ A such that p = ab we have a ∈ B× = A× or b ∈ B× = A× (since p

is irreducible in B). Hence p is irreducible in A.

3. Let p ∈ A such that p is prime in B. Then p ∈ A \ (A× ∪ {0}) (by part (1)).

Given a, b ∈ A such that p|ab in A, we have p|a or p|b in B (since p is prime in

B). We may assume that p|a in B. Then there exists x ∈ B such that px = a.


Since A is factorially closed in B it follows that x ∈ A, so p|a in A. This shows

that p is prime in A.

4. Let a ∈ A \ (A× ∪ {0}). Then a ∈ B \ (B× ∪ {0}), so we have a = p1p2 · · · pmwhere pi, i = 1, · · · ,m are prime elements in B. But all pi are in A (since A is

factorially closed) and so they must be prime in A (by part (3)). Therefore A

is a UFD.

1.4 Z-graded rings

De�nition 1.4.1. A Z-graded ring is a ring B together with a direct sum decomposi-

tion B = ⊕n∈ZBn, where all the Bn are subgroups of (B,+) satisfying BnBm ⊆ Bm+n

for all n,m ∈ Z. The decomposition B = ⊕n∈ZBn is called a Z-grading of B.

Remark 1.4.2. Let B = ⊕n∈ZBn be a Z-graded ring.

1. An element of B is homogeneous if it belongs to ∪n∈ZBn.

2. If f ∈ B is homogeneous and f 6= 0, then there is a unique d ∈ Z such that

f ∈ Bd. In this situation we say that f is homogeneous of degree d.

3. The element 0 is homogeneous of degree −∞.

4. An element f of B is of the form f =∑

n∈Z fn, where fn ∈ Bn for all n ∈ Z,and fn = 0 for all n except possibly �nitely many of them. So f is a �nite sum

of homogeneous elements and this decomposition is unique.

Example 1.4.3. Take B = k[X] where k is a �eld.

We de�ne

Bn =

{spank{Xn} = {aXn | a ∈ k} if n ≥ 0

0 if n < 0.

Clearly B = ⊕n∈ZBn, so k[X] is a Z-graded ring.


Example 1.4.4. Let k be a �eld and consider the polynomial ring B = k[X1, · · · , Xn].

Let ω = (ω1, · · · , ωn) ∈ Zn, and for each d ∈ Z de�ne

Bd = spank {Xe11 X

e22 · · ·Xen

n | (e1, · · · , en) ∈ Nn and e1ω1 + · · ·+ enωn = d} .

In fact B is a k-vector space and each Bd is a subspace of B. Clearly B =

k[X1, · · · , Xn] = ⊕d∈ZBd and Bd1Bd2 ⊆ Bd1+d2 for all d1, d2 ∈ Z. So the n-tuple

ω determines a Z-grading of B. In this grading, if an element f is homogeneous of

degree d, then we write degω(f) = d.

The Z-grading de�ned above is such that degω(Xi) = ωi, i = 1, · · · , n and is

called an ω-grading of k[X1, · · · , Xn]. When ω = (1, · · · , 1), the grading is called

the standard Z-grading of B. A polynomial that is homogeneous with respect to the

standard Z-grading is called a standard homogeneous polynomial.

Example 1.4.5. Let B = k[X, Y ] where k is a �eld, and let ω = (2,−3). Then we have

the ω-grading B = ⊕n∈ZBn, where Bn = spank {X iY j | 2i− 3j = n}. The elements

X2Y and X2Y +X5Y 3 are homogeneous of degree 1; the element XY is homogeneous

of degree −1.

Lemma 1.4.6. Let B = ⊕n∈ZBn be a Z-graded ring. The following hold:

(i) 1B ∈ B0

(ii) The subgroup B0 is a subring of B.

Proof. We have 1B ∈ B, then 1B =∑

n∈Z fn, where fn ∈ Bn. Thus for every gm ∈ Bm

(m ∈ Z), we have gm = gm.1B =∑

n∈Z fngm and fngm ∈ Bn+m. Consequently

fngm = 0 for all n 6= 0. It follows that if n 6= 0 then fng = 0 for all g ∈ B. In

particular for g = 1B, we have fn = 0 for any n 6= 0. Hence 1B = f0 ∈ B0. Moreover,

B0 is a subgroup of (B,+), 1B ∈ B0 and B0B0 ⊆ B0, so B0 is a subring and then (ii)

is shown.

De�nition 1.4.7. Let B = ⊕d∈ZBd be a Z-graded ring.

a) Let f =∑

d∈Z fd be an element of B (where fd ∈ Bd for all d ∈ Z). The set

supp(f) = {d ∈ Z | fd 6= 0} is �nite and is called the support of f .


b) We de�ne the set map

deg : B −→ Z ∪ {−∞}

f 7−→

{max(supp(f)) if f 6= 0

−∞ if f = 0.

Remark 1.4.8. Let B be a Z-graded ring.

a) The map deg : B −→ Z ∪ {−∞} de�ned in De�nition 1.4.7 satis�es:

(i) deg(f) = −∞⇐⇒ f = 0

(ii) deg(fg) ≤ deg(f) + deg(g) ∀ f, g ∈ B

(iii) deg(f + g) ≤ max{deg f, deg g} ∀ f, g ∈ B

(iv) If deg(f) 6= deg(g), then deg(f + g) = max{deg(f), deg(g)}.

b) Assume that B is a domain. Then equality holds in a-ii) and consequently the

map deg : B −→ Z∪{−∞} is a degree function in the sense of De�nition 1.3.1.

We call it the degree function determined by the grading.

Proposition 1.4.9. Let B be a Z-graded domain and let f, g ∈ B \ {0}. If fg is

homogeneous, then f and g are homogeneous.

Proof. For f ∈ B \ {0}, set m(f) = min(supp(f)) and de�ne the map:

L : B \ {0} −→ N

f 7−→ deg(f)−m(f).

Since B is domain, we have m(fg) = m(f) +m(g) and consequently L(fg) = L(f) +

L(g). Now observe that an element f ∈ B\{0} is homogeneous if and only if L(f) = 0.

Thus since fg is homogeneous, we have:

L(fg) = 0 =⇒ L(f) + L(g) = 0

=⇒ L(f) = L(g) = 0.

Therefore f and g are homogeneous.


1.5 Rings of fractions and transcendence degree

De�nition 1.5.1. Let A be a ring. A subset S of A is said to be multiplicatively

closed if 1A ∈ S and S is closed under multiplication.

Example 1.5.2. • Given α ∈ A, the subset S = {αn}n≥0 of A is multiplicatively

closed.

• If p is a prime ideal of a ring A, the subset S = A \ p of A is multiplicatively

closed.

If S is a multiplicatively closed subset of a ring A then we can form the ring S−1A,

called the ring of fractions of A with respect to S. If M is an A-module then we

can form the S−1A-module S−1M , called a module of fractions. We assume that the

reader is familiar with the basic properties of S−1A and S−1M .

Notation. Let A be a ring and S a multiplicatively closed subset of A.

• If α ∈ A and S = {αn}n≥0, then we write Aα for S−1A.

• If p is a prime ideal of A and S = A \ p, we write Ap for S−1A.

Remark 1.5.3. Given a ring A and a multiplicatively closed subset S of A, the

following hold:

(a) S−1A is the zero ring if and only if 0 ∈ S.

(b) Let A,B be two rings such that A ⊆ B, and let S a multiplicatively closed

subset of A. Then we have S−1A ⊆ S−1B. Moreover, if M ⊆ N are A-modules

then S−1M ⊆ S−1N .

The following proposition is the �rst Remark of the Chapter 3 of [1].

Proposition 1.5.4. If A is a domain and S = A \ {0}, then S−1A is a �eld.

Proof. Clearly by Remark 1.5.3, S−1A 6= 0. Furthermore for all a/s ∈ S−1A \ {0} wehave a 6= 0 so s/a ∈ S−1A. Thus (a/s)(s/a) = 1/1 = 1S−1A and S−1A is a �eld.


De�nition 1.5.5. Let A be a domain. The �eld S−1A de�ned in the previous propo-

sition is called the �eld of fractions of A. It is denoted by FracA.

Remark 1.5.6. If A is a domain of characteristic zero, then Q ⊆ FracA.

Theorem 1.5.7. Let A be a domain and k a �eld. If ϕ : A −→ k is a ring monomor-

phism, then there is a unique homomorphism ϕ : Frac(A) −→ k such that ϕ(t) = ϕ(t)

for all t ∈ A ⊆ Frac(A).

Aϕ //

��

k

Frac(A)∃!ϕ

;;

Proof. This is Theorem 1.17 of [2].

Remark 1.5.8. Let A be a domain and k a �eld containing A. Then we have

Frac(A) ⊆ k. The �eld Frac(A) is the smallest sub�eld of k containing A.

Lemma 1.5.9. Let A,B and C be domains, and let ϕ : A −→ B be a ring monomor-

phism. There exists a unique ring homomorphism ϕ∗ : FracA −→ FracB which

satis�es ϕ∗(t) = ϕ(t) whenever t ∈ A ⊆ FracA. That is the following diagram

commutes:

Aϕ //� _

��

B� _

��FracA ϕ∗

// FracB

Moreover this construction has the following properties:

• If ϕ : A −→ B and θ : B −→ C are monomorphisms between integral domains,

then (θ ◦ ϕ)∗ = θ∗ ◦ ϕ∗.

• For any integral domain B, the identity homomorphism id : B −→ B induces

the identity homomorphism id∗ : FracB −→ FracB.

Aϕ //� _

��

Bθ //� _

��

C� _

��FracA ϕ∗

// FracBθ∗// FracC

Bid //� _

��

B� _

��FracB

id∗// FracB


Proof. This follows from Theorem 1.17 of [2].

Notation. If k is a �eld, k(n) = Frac(k[n]) = �eld of fractions of k[n].

De�nition 1.5.10. Let A be a ring. A sequence (Mi, fi)i where Mi is an A-module

and fi an A-linear map, de�ned as:

· · · //Mi−1fi //Mi

fi+1 //Mi+1// · · ·

is said to be exact at Mi if Im(fi) = ker(fi+1). Such a sequence is said to be exact if

it is exact at each Mi.

In particular the sequence:

0 //M ′ f //Mg //M ′′ // 0

is exact if and only if it is exact at M ′, M and M ′′, if and only if f is injective, g is

surjective and Im(f) = ker(g). The last sequence, when it is exact, is called a short

exact sequence.

The following result is Proposition 3.3 of Chapter 3 of [1].

Proposition 1.5.11. Let A be a ring, f : M ′ −→ M and g : M −→ M ′′ two

A-module homomorphisms. Given S a multiplicatively closed subset of A, we have:

(a) The map S−1f : S−1M ′ −→ S−1M, m/s 7−→ f(m)/s is a well de�ned S−1A-

module homomorphism. Moreover, S−1(g ◦ f) = (S−1g) ◦ (S−1f).

(b) The operation S−1 is exact: If M ′ f //Mg //M ′′ is exact at M , then

S−1M ′ S−1f // S−1M

S−1g // S−1M ′′ is exact at S−1M .

Proof. a) is straightforward.

Now by assumption in b) we have g ◦ f = 0, hence S−1g ◦ S−1f = S−1(0) = 0; thus

Im(S−1f) ⊆ ker(S−1g). Let m/s ∈ ker(S−1g), we have g(m)/s = 0 in S−1M ; hence

there exists t ∈ S such that tg(m) = 0 in M ′′. But tg(m) = g(tm), since g is an A-

module homomorphism. Thus tm ∈ ker(g) = Im(f), therefore there exists m′ ∈ M ′

such that tm = f(m′).

Hence in S−1M , we have m/s = f(m′)/st = (S−1f)(m′/st) ∈ Im(S−1f).

Thus ker(S−1g) ⊆ Im(S−1f). Therefore b) is shown.


De�nition 1.5.12. Let L,K be two �elds such that K ⊆ L. Then we say that K is

a sub�eld of L or L is an extension (�eld) of K.

1. A subset {s1, · · · , sn} of L (where n ≥ 1 and s1, · · · , sn are distinct) is alge-

braically independent over K if there is no P ∈ K[X1, · · · , Xn] \ {0} such that

P (s1, · · · , sn) = 0. An arbitrary subset S of L is called algebraically indepen-

dent if every �nite subset of S is algebraically independent. A subset S of L is

algebraically dependent over K if it is not algebraically independent.

2. A �eld L is said to be an algebraic closure of a sub�eldK when it is algebraically

closed and algebraic over K.

3. If S is a subset of L, we de�ne K(S) to be the intersection of all sub�elds of L

that contain K ∪ S. Then K(S) is a �eld such that K ⊆ K(S) ⊆ L. The �eld

K(S) is the smallest sub�eld of L containing K and the subset S.

If S = {s1, · · · , sn} ⊆ L, then it can be shown that K(s1, · · · , sn) is the set

of elements of the formf(s1, · · · , sn)

g(s1, · · · , sn)where f(X1, · · · , Xn), g(X1, · · · , Xn) ∈

K[X1, · · · , Xn] and g(s1, · · · , sn) 6= 0.

Remark 1.5.13. Let K ⊆ L be a �eld extension.

(a) The empty set is algebraically independent over K.

(b) If b ∈ L, then {b} is algebraically independent over K if and only if b is tran-

scendental over K.

Theorem 1.5.14. Let F ⊆ K ⊆ L be �eld extensions. If K is algebraic over F and

L is algebraic over K, then L is algebraic over F .

Proof. This is Theorem 20 in chapter 13 of [4].

Notation. Let K ⊆ L be a �eld extension. We denote by K the set of all elements of

L which are algebraic over K.

Proposition 1.5.15. Let K ⊆ L be a �eld extension. The subset K of L is a sub�eld

of L containing K and the extension K ⊆ K is algebraic.


Proof. This is corollary 19 in chapter 13 of [4].

De�nition 1.5.16. Let K ⊆ L be a �eld extension. A transcendence basis of K ⊆ L

is a subset S ⊆ L satisfying:

• S is algebraically independent over K;

• L is algebraic over K(S).

Lemma 1.5.17. Let K ⊆ L be a �eld extension.

• A transcendence basis of K ⊆ L is a maximal subset (with respect to inclusion)

of L which is algebraically independent over K.

• If S is a subset of L such that L is algebraic over K(S) and S is minimal among

subsets of L with this property, then S is a transcendence basis of L over K.

Proof. These are Proposition 9.12 and Lemma 9.9 of [13].

Theorem 1.5.18. Let K ⊆ L be a �eld extension.

1. There exists at least one transcendence basis of K ⊆ L.

2. Any two transcendence bases of K ⊆ L have the same cardinality.

Proof. These are shown in Theorem 19.14 and 19.15 of [14].

De�nition 1.5.19. The transcendence degree of K ⊆ L is the cardinality of a tran-

scendence basis of K ⊆ L; it is denoted by trdegK(L).

Proposition 1.5.20. Let K ⊆ L ⊆M be �eld extensions. Then we have

trdegK(M) = trdegK(L) + trdegL(M).

Proof. This is Proposition 19.18 of [14].

Corollary 1.5.21. Let K ⊆ L ⊆M be �eld extensions. Then we have

trdegK(M) <∞ if and only if trdegK(L) <∞ and trdegL(M) <∞.

Proof. This follows from Proposition 19.18 of [14].


Example 1.5.22. (a) A �eld extension K ⊆ L is algebraic if and only if trdegK(L) =

0 (the empty set is a transcendence basis).

(b) Let K[X1, · · · , Xn] be a polynomial ring over a �eld K, and consider L =

K(X1, · · · , Xn) the �eld of fractions ofK[X1, · · · , Xn]. The subset {X1, · · · , Xn} ⊂L is a transcendence basis of the �eld extensionK ⊆ L, hence trdegK K(X1, · · · , Xn) =

n. Furthermore, K(X1, · · · , Xn) = Frac(K[X1, · · · , Xn]) = Frac(K [n]) = K(n),

hence trdegK K(n) = n.

Lemma 1.5.23. Consider the polynomial ring B = A[X1, · · · , Xn] where A is a

domain. Then FracB = (FracA)(X1, · · · , Xn).

Proof.

We have B = A[X1, · · · , Xn] ⊆ (FracA)[X1, · · · , Xn]

⊆ Frac((FracA)[X1, · · · , Xn]) = (FracA)(X1, · · · , Xn).

Since (FracA)(X1, · · · , Xn) is a �eld that contains B, we have

B ⊆ FracB ⊆ (FracA)(X1, · · · , Xn).

Since A ⊆ B we have FracA ⊆ FracB. Furthermore, since FracA ⊆ FracB and

X1, · · · , Xn ∈ FracB, we have (FracA)(X1, · · · , Xn) ⊆ FracB.

So FracB = (FracA)(X1, · · · , Xn).

Notation. Given A,B two domains such that A ⊆ B, we de�ne trdegA(B) to be equal

to the transcendence degree of FracB over FracA.

The following proposition can be found in [3] as Exercise 6.3.

Proposition 1.5.24. 1. Let A ⊆ B be domains. Then we have trdegA(B) = 0 if

and only if B is algebraic over A.

2. Given a domain A and B = A[n], we have trdegA(B) = n.

3. Given three domains A,B and C such that A ⊆ B ⊆ C, we have

trdegA(C) < ∞ if and only if both trdegA(B) < ∞ and trdegB(C) < ∞.

Moreover if trdegA(C) <∞, then

trdegA(C) = trdegA(B) + trdegB(C).


Proof. 1. ⇒) Assume that trdegAB = 0, that is trdegFracA FracB = 0. Then

FracB is algebraic over FracA. Now since B ⊆ FracB, then for all b ∈ B,

there exists f ∈ (FracA)[X] \ {0} such that f(b) = 0. Set g(X) = sf(X),

where s = s0s1 · · · sn, n = deg(f) and the si ∈ A \ {0} are the denominators

of the coe�cients of f . Thus g(X) ∈ A[X] \ {0}, and g(b) = 0; therefore B is

algebraic over A.

⇐) Let b/s ∈ FracB and consider the �eld extension FracA ⊆ FracB. Since

B is algebraic over A, we have b, s are algebraic over A and thus over FracA.

So b, s ∈ FracA and since FracA is a �eld by Proposition 1.5.15 we obtain that

b/s ∈ FracA. Thus b/s is algebraic over FracA, then FracB is algebraic over

FracA. Therefore trdegFracA FracB = 0; that is trdegAB = 0.

2. Let us writeB = A[X1, · · · , Xn]. By Lemma 1.5.23, FracB = (FracA)(X1, · · · , Xn).

So by Example 1.5.22-b), we get trdegFracA(FracB) = n and hence

trdegA(B) = n.

3. Since A ⊆ B ⊆ C, we have FracA ⊆ FracB ⊆ FracC by Remark 1.5.3-b); and

the result follows from Proposition 1.5.20.

Lemma 1.5.25. Let A, B and C be three domains such that A ⊆ B ⊆ C. If B is

algebraic over A and C is algebraic over B, then C is algebraic over A.

Proof. It is a consequence of Proposition 1.5.24.

The next corollary is Exercise 6.4 of [3].

Corollary 1.5.26. Let A ⊆ B be domains, such that A is algebraically closed in

B and trdegA(B) < ∞. If A′ is a ring such that A ⊆ A′ ⊆ B and trdegA(B) =

trdegA′(B), then A = A′.

Proof. By Proposition 1.5.24-3), we get trdegAA′ = 0 and so A′ is algebraic over A.

Hence A′ = A since A is algebraically closed in B.

The following two corollaries will be very useful later on in the text.


Corollary 1.5.27. Given domains A ⊆ B, de�ne A = {b ∈ B | b is algebraic over A}.Then A is a subring of B that contains A and A is algebraic over A.

Proof. By Proposition 1.5.15 the set FracA = {x ∈ FracB | x is algebraic over FracA}is a sub�eld of FracB that contains FracA. We have the inclusions A ⊆ FracA ⊆FracA ⊆ FracB ⊇ B. Since B and FracA are subrings of FracB, it follows that

B ∩ FracA is a subring of FracB. It is easy to see that A = B ∩ FracA, so A is

a subring of FracB; since A ⊆ B, it is in fact a subring of B. By de�nition, every

element of A is algebraic over A, so A is algebraic over A.

Corollary 1.5.28. Let A ⊆ B be domains and suppose that b1, · · · , bn ∈ B are such

that B = A[b1, · · · , bn]. If each bi is algebraic over A, then B is algebraic over A.

Proof. De�ne A = {b ∈ B | b is algebraic over A}, then Corollary 1.5.27 implies that

A is a subring of B, A ⊆ A ⊆ B and A is algebraic over A. Since A ⊆ A and

b1, · · · , bn ∈ A, we have A[b1, · · · , bn] ⊆ A, so B = A and hence B is algebraic over

A.

Chapter 2

Locally Nilpotent Derivations

In this chapter we introduce the theory of locally nilpotent derivations and we present

some results which we will use in the next chapter.

2.1 Derivations

De�nition 2.1.1. Let B be a ring. A derivation of a ring B is a map D : B −→ B

satisfying: For all f, g ∈ B

D(f + g) = D(f) +D(g)

D(fg) = D(f)g + fD(g).

Notation. The set of all derivations of a ring B is denoted by DerB.

Example 2.1.2. Let B be a ring.

1. The zero map B −→ B, x 7−→ 0, is a derivation, called the the zero derivation.

2. Let B[t] be the polynomial ring in one variable over B. The map D : B[t] −→B[t] de�ned by:

D

(n∑i=0

aiti

)=

n∑i=1

iaiti−1, where

n∑i=0

aiti ∈ B[t],

is a derivation. It is denoted byd

dt: B[t] −→ B[t].

36

CHAPTER 2. LOCALLY NILPOTENT DERIVATIONS 37

The next proposition is Exercise 1.7 of [3].

Proposition 2.1.3. Let B be a ring, D ∈ DerB and f1, · · · , fn ∈ B where n is a

positive integer. The following hold:

1. For all f ∈ B, the map fD : B −→ B, x 7−→ fD(x) is a derivation.

2. D(f1 + · · ·+ fn) = D(f1) + · · ·+D(fn)

3. For all f, g, h ∈ B, D(fgh) = ghD(f) + fhD(g) + fgD(h). More generally,

D(f1 · · · fn) = f2 · · · fnD(f1) + f1f3 · · · fnD(f2) + · · ·

+ f1 · · · fn−2fnD(fn−1) + f1 · · · fn−1D(fn).

4. For f ∈ B and k > 0, D(fk) = kfk−1D(f).

Proof. Straightforward.

Corollary 2.1.4. The set DerB of all derivations of a ring B is a B-module.

Proof. It is easy to show that (DerB,+) is an abelian group. Moreover, by Proposi-

tion 2.1.3 we can de�ne the map B×DerB −→ DerB, (f,D) 7−→ fD which satis�es

the axioms of De�nition 1.1.13. Hence DerB is a B-module.

De�nition 2.1.5. Given, D ∈ DerB, the set kerD = {b ∈ B | D(b) = 0} is calledthe kernel of D. It can also be called the ring of constants of D and be denoted BD.

The proposition below can be found in [3] as Exercise 1.5.

Proposition 2.1.6. Let B be a ring and D ∈ DerB. Then kerD is a subring of B.

Proof. We have D(1) = D(12) = 2D(1) =⇒ D(1) = 0. Moreover D(1− 1) = D(1) +

D(−1), but D(0) = D(0 + 0) = 2D(0) =⇒ D(0) = 0; hence D(−1) = −D(1) = 0.

Thus 1 ∈ kerD and −1 ∈ kerD. Now given x, y ∈ kerD it is easy to see that

x+ y ∈ kerD and xy ∈ kerD. Therefore kerD is a subring of B.

Example 2.1.7. Let k be a �eld. Consider the polynomial ring k[t] and the derivation

D =d

dt. When k is of positive characteristic p we have kerD = k[tp], otherwise

kerD = k.


Remark 2.1.8. The only derivation D : Z −→ Z is the zero derivation; this is

because kerD is a subring of Z and the only subring of Z is Z.

De�nition 2.1.9. Let A,B be two rings such that A ⊆ B. An A-derivation of B is

a derivation D : B −→ B satisfying D(a) = 0 for all a ∈ A.The set of all A-derivations of a ring B is denoted by DerA(B). It is clear that

DerA(B) is a B-submodule of Der(B).

The lemma below is Exercise 3.1 of [3].

Lemma 2.1.10. Let B be a Q-algebra. Then we have Der(B) = DerQ(B).

Proof. If B = 0 the result is trivial. Assume that B 6= 0 and hence that Q ⊆ B. Let

D ∈ Der(B). Since kerD is a subring of B we have Z ⊆ kerD. If n ∈ Z and n > 0

then

0 = D(1) = D(1

n+ · · ·+ 1

n︸︷︷︸n times

) = nD

(1

n

)=⇒ D

(1

n

)= 0.

Since Z ⊆ kerD and 1n∈ kerD for all n > 0, we have Q ⊆ kerD.

The next result is Exercise 1.10 of [3].

Proposition 2.1.11. Let A be a ring and B = A[t] = A[1]. If D1, D2 ∈ DerA(B)

satisfy D1(t) = D2(t), then D1 = D2.

Proof. Since DerA(B) is a B-submodule of DerB and D1, D2 ∈ DerA(B), we have

D1−D2 ∈ DerA(B) and so ker(D1−D2) is a subring of B. Moreover A ⊆ ker(D1−D2)

and t ∈ ker(D1−D2). Hence A[t] ⊆ ker(D1−D2), which implies thatD1−D2 = 0.

Example 2.1.12. Let A be a ring and B = A[X1, · · · , Xn] = A[n]. For all j ∈{1, · · · , n}, de�ne ∂

∂Xj: B −→ B by

∂

∂Xj

( ∑i1,··· ,in

ai1···inXi11 · · ·X in

n

)=∑i1,··· ,in

ijai1···inXi11 · · ·X

ij−1

j−1 Xij−1j X

ij+1

j+1 · · ·X inn ,


where∑

i1,··· ,in ai1···inXi11 · · ·X in

n ∈ B.Then ∂

∂Xj∈ DerA(B). This derivation is called the partial derivative with respect to

Xj. Also by de�nition of ∂∂Xj

we have:

∂

∂Xj

(Xi) =

1, if i = j

0, if i 6= j.

The following is Lemma 1.12 of [3].

Lemma 2.1.13. Let A be a ring and B = A[X1, · · · , Xn].

1. Given any f1, · · · , fn ∈ B there exists a unique D ∈ DerA(B) satisfying D(Xi) =

fi for all i = 1, · · · , n.

2. The set DerA(B) is a free B-module with basis { ∂∂X1

, · · · , ∂∂Xn}.

Proof. 1. Since DerA(B) is a B-submodule of Der(B) and { ∂∂X1

, · · · , ∂∂Xn} are el-

ements of DerA(B) it follows that∑n

i=1 fi∂∂Xi∈ DerA(B). Set D =

∑ni=1 fi

∂∂Xi

,

then clearly D(Xi) = fi for all i ∈ {1, · · · , n}. Let D′ ∈ DerA(B) be such that

D′(Xi) = fi for all i. Set D0 = D − D′, then D0 ∈ DerA(B), hence kerD0

is a subring of B, Xi ∈ kerD0 for all i ∈ {1, · · · , n} and A ⊆ ker(D0). So

B ⊆ ker(D0), hence D0 = 0, that is D = D′.

2. Note that { ∂∂X1

, · · · , ∂∂Xn} is a �nite subset of DerA(B). Let D ∈ DerA(B) and

set ai = D(Xi) ∈ B; by part (1), D =∑n

i=1 ai∂∂Xi

. Hence { ∂∂X1

, · · · , ∂∂Xn} is a

generating set for the module DerA(B).

Suppose that a1, · · · , an ∈ B are such that∑n

i=1 ai∂∂Xi

= 0. Evaluating at

each Xj, j = 1, · · · , n, we get aj = 0 for all j. So { ∂∂X1

, · · · , ∂∂Xn} is linearly

independent over B and hence is a basis of DerA(B). So DerA(B) is free.

Remark 2.1.14. • Given rings A ⊆ B, the B-module DerA(B) is not necessarily

free. The previous result tell us that it is free when B is a polynomial ring over

A.


• When B = A[n] where A is a ring, each system of variables γ = (U1, · · · , Un) of

B determines a system of A-derivation ∂∂U1

, · · · , ∂∂Un

. The derivation ∂∂Uj

is the

unique element of DerA(B) satisfying ∂Ui

∂Uj= δij. For instance let B = A[2] and

consider the two systems of variables γ = (X, Y ) and γ′ = (X,X2−Y ) = (X, V ).

Relatively to the �rst system of variables, we have ∂Y∂X

= 0 and relatively to the

second we have ∂Y∂X

= ∂(X2−V )∂X

= 2X. So ∂γX 6= ∂γ′

X , and therefore distinct

systems of variables may induce distinct bases (in Lemma 2.1.13 (2)).

The following useful result is Exercise 1.14 of [3].

Lemma 2.1.15. Let B be a ring and D ∈ DerB.

1. Consider a polynomial f(T ) =∑n

i=0 aiTi ∈ B[T ] (ai ∈ B) and b ∈ B. Then

D(f(b)) = f (D)(b) + f ′(b)D(b)

where f (D)(T ) =∑n

i=0D(ai)Ti ∈ B[T ] and f ′(T ) =

∑ni=1 iaiT

i−1.

2. Consider a polynomial f(T1, · · · , Tn) ∈ B[T1, · · · , Tn] and b1, · · · , bn ∈ B. Then

D(f(b1, · · · , bn)) = f (D)(b1, · · · , bn) +n∑i=1

fTi(b1, · · · , bn)D(bi),

where fTi =∂f

∂Ti∈ B[T1, · · · , Tn].

Proof. 1. We have

D(f(b)) = D(n∑i=0

aibi)

= D(a0 + a1b+ a2b2 + · · ·+ anb

n)

= D(a0) +D(a1)b+ · · ·+D(an)bn + a1D(b) + a2D(b2) + · · ·+ anD(bn)

=n∑i=0

D(ai)bi + a1D(b) + 2a2bD(b) + · · ·+ nanb

n−1D(b)

= f (D)(b) + f ′(b)D(b).


2. Let us �rst consider a monomial g(T1, · · · , Tn) = aT i11 · · ·T inn ∈ B[T1, · · · , Tn]

(where a ∈ B and i1, · · · , in ∈ N). Then

D(g(b1, · · · , bn)) = D(abi11 · · · binn )

= D(a)bi11 · · · binn +n∑i

abi11 · · · bij−1j−1 D(b

ijj )b

ij+1j+1 · · · binn

= D(a)bi11 · · · binn +n∑i

ijabi11 · · · b

ij−1j−1 b

ij−1j b

ij+1j+1 · · · binn D(bj)

= g(D)(b1, · · · , bn) +n∑j=1

gTj(b1, · · · , bn)D(bj).

Since a general polynomial f(T1, · · · , Tn) ∈ B[T1, · · · , Tn] is a �nite sum of

monomials such as g(T1, · · · , Tn), it follows thatD(f(b1, · · · , bn)) = f (D)(b1, · · · , bn)+∑ni=1 fTi(b1, · · · , bn)D(bi).

Corollary 2.1.16. Let A ⊆ B be rings, let D ∈ DerA(B) and let s ∈ B be such that

D(s) = 1. Then for any f(T ) ∈ A[T ] and j ∈ N,

Dj(f(s)) = f (j)(s)

where Dj = D ◦ · · · ◦D︸︷︷︸j

and f (j)(T ) = ( ∂∂T

)j(f(T )) ∈ A[T ].

Proof. By induction on j and using Lemma 2.1.15-1).

Example 2.1.17. Let B = Z[X, Y ] = Z[2] and D =∂

∂Y+Y

∂

∂X∈ DerB. Let us prove

that ker(D) = Z[2X−Y 2]. In fact let D : Q[X, Y ] −→ Q[X, Y ] be the extension of D

toQ[X, Y ], we �rst show that ker(D) = Q[2X−Y 2]. ClearlyQ[X, Y ] = Q[2X−Y 2, Y ]

and Q[2X − Y 2] ⊆ kerD. Let h ∈ ker(D) \ {0}, then h ∈ Q[X, Y ] = Q[2X − Y 2, Y ]

and D(h) = 0. Hence there is a nonzero polynomial F ∈ Q[U, V ] such that h =

F (2X − Y 2, Y ). Furthermore,

D(h) = 0⇒ D(F (2X − Y 2, Y )) = 0

⇒ FU(2X − Y 2, Y )D(2X − Y 2) + FV (2X − Y 2, Y )D(Y ) = 0 by Lemma 2.1.15.

⇒ FV (2X − Y 2, Y ) = 0, because D(2X − Y 2) = 0 and D(Y ) = 1.


Since (2X − Y 2, Y ) is algebraically independent over Q, it follows that FV (U, V ) = 0

and then F (U, V ) = T (U), T (U) ∈ Q[U ]. So h = T (2X − Y 2) ∈ Q[2X − Y 2] and

ker(D) = Q[2X − Y 2]. Moreover, kerD = Z[X, Y ]∩ ker(D) = Z[X, Y ]∩Q[2X − Y 2].

Clearly Z[2X − Y 2] ⊆ Z[X, Y ] ∩Q[2X − Y 2]. Given P ∈ Z[X, Y ] ∩Q[2X − Y 2]

we have P (X, Y ) = a0 + a1(2X − Y 2) + · · ·+ an(2X − Y 2)n, where ai ∈ Q.Since P (X, Y ) ∈ Z[X, Y ], then

P (0, Y ) ∈ Z[Y ] =⇒ a0 − a1Y2 + a2Y

4 + · · ·+ (−1)nanY2n ∈ Z[Y ].

Hence ai ∈ Z and so P (X, Y ) ∈ Z[2X−Y 2]; thus Z[X, Y ]∩Q[2X−Y 2] ⊆ Z[2X−Y 2].

Therefore Z[2X − Y 2] = Z[X, Y ] ∩Q[2X − Y 2], we thus have kerD = Z[2X − Y 2].

The following useful result is Lemma 1.17 of [3].

Proposition 2.1.18. If B is a domain of characteristic zero and D ∈ Der(B), then

kerD is algebraically closed in B.

Proof. Consider b ∈ B such that b is algebraic over kerD. Set A = kerD and let

f ∈ A[T ] be a nonzero polynomial of minimal degree such that f(b) = 0. Note that

deg(f) ≥ 1. Then we have,

0 = D(f(b)) = f (D)(b) + f ′(b)D(b) = f ′(b)D(b).

Since deg(f) ≥ 1 and B is a domain of characteristic zero, we have f ′ 6= 0. As f ′ is

a nonzero polynomial and deg(f ′) < deg(f), then f ′(b) 6= 0 by minimality of deg(f),

and so D(b) = 0 (since B is a domain). Therefore b algebraic over kerD implies that

b ∈ kerD; that is kerD is algebraically closed in B.

Remark 2.1.19. If B is a ring of characteristic p > 1, it is easy to observe that

for D ∈ Der(B) a nonzero derivation, kerD is not algebraically closed in B. In fact

for all x ∈ B, we have D(xp) = pxp−1D(x) = 0 so xp ∈ kerD. Set A = kerD and

take z ∈ B \A, consider the polynomial P (X) = Xp − zp; P (X) ∈ A[X] \ {0} (sincezp ∈ A) and P (z) = 0. Hence z is algebraic over A = kerD. Therefore kerD is not

algebraically closed in B.


Lemma 2.1.20. Let B = A[X1, · · · , Xn] where A is a ring and f = (f1, · · · , fn) a

system of variables in B. Then

Mf =

∂f1

∂X1

· · · ∂f1

∂Xn...

...∂fn∂X1

· · · ∂fn∂Xn

∈ Mn(B) is invertible.

Proof. We have A[f ] = A[X1, · · · , Xn], so Xi ∈ A[f ] for each i ∈ {1, · · · , n}. Hencethere is Pi(U1, · · · , Un) ∈ A[U1, · · · , Un] such that Xi = Pi(f) = Pi(f1, · · · , fn) for

each i ∈ {1, · · · , n}. Moreover, δij =∂Xi

∂Xj

=∂Pi(f1, · · · , fn)

∂Xj

=∑n

k=1

∂Pi∂Uk

(f1, · · · , fn)∂fk∂Xj

(by Lemma 2.1.15). SetQik =∂Pi∂Uk

(f1, · · · , fn) ∈ B, thus δij = (Qi1, · · · , Qin)

∂f1

∂Xj...∂fn∂Xj

which is the (i, j)-element of the matrix

Q11 · · · Q1n

... · · · ...

Qn1 · · · Qnn

∂f1

∂X1

· · · ∂f1

∂Xn... · · · ...∂fn∂X1

· · · ∂fn∂Xn

So this matrix product is equal to In. Thus det(Mf ) ∈ B×, and soMf is invertible.

Remark 2.1.21. When A is domain, det(Mf ) ∈ A× = B×.


Proposition 2.1.22. Let B be a ring, D ∈ Der(B) and let S be a multiplicative

subset of B and S−1B the ring of fractions of B. The map S−1D : S−1B −→ S−1B

de�ned by

(S−1D)(b/s) =sD(b)− bD(s)

s2for all b ∈ B and s ∈ S.

is a derivation of S−1B.


Proof. First we need to show that the map S−1D is well de�ned. Let b1, b2 ∈ B and

s1, s2 ∈ S be such thatb1

s1

=b2

s2

in S−1B, we have to show that S−1D(b1

s1

) = S−1D(b2

s2

)

in S−1B.

Recall that:b1

s1

=b2

s2

in S−1B =⇒ (b1s2 − b2s1)u = 0 for some u ∈ S.

We have

(s1D(b1)− b1D(s1))s22 − (s2D(b2)− b2D(s2))s2

1

= s1s22D(b1)− s2

2b1D(s1)− s21s2D(b2) + s2

1b2D(s2)

= s1s22D(b1)− s2

2b1D(s1)− s21s2D(b2) + s2

1b2D(s2) + s1s2b2D(s1)− s1s2b2D(s1) + s1s2b1D(s2)

− s1s2b1D(s2)

= s1s2(s2D(b1) + b1D(s2)− s1D(b2)− b2D(s1))− b1s2(s1D(s2) + s2D(s1))

+ b2s1(s1D(s2) + s2D(s1))

= s1s2(D(b1s2)−D(s1b2))− b1s2(D(s1s2)) + b2s1(D(s1s2))

= s1s2(D(b1s2)−D(s1b2)) +D(s1s2)(b2s1 − b1s2)

= s1s2D(b1s2 − s1b2) +D(s1s2)(b2s1 − b1s2).

Now u ∈ S =⇒ u2 ∈ S and we have:

((s1D(b1)−b1D(s1))s22−(s2D(b2)−b2D(s2))s2

1)u2 = s1s2D(b1s2−s1b2)u2+D(s1s2)(b2s1−b1s2)u2.

But (b2s1 − b1s2)u = 0 =⇒ (b2s1 − b1s2)u2 = 0 and

(b2s1 − b1s2)u = 0 =⇒ D((b2s1 − b1s2)u) = 0

=⇒ D(s2b1 − b2s1)u+ (b2s1 − b2s1)D(u) = 0

=⇒ D(s2b1 − b2s1)u2 = 0 (by multiplying by u both terms of the equality).

Therefore ((s1D(b1)− b1D(s1))s22 − (s2D(b2)− b2D(s2))s2

1)u2 = 0. That is

s1D(b1)− b1D(s1)

s21

=s2D(b2)− b2D(s2)

s22

in S−1B.

So S−1D is well de�ned, now we show that it is a derivation. Let x1, x2 ∈ S−1B.


Then there exist b1, b2 ∈ B and s ∈ S such that x1 = b1sand x2 = b2

s. Then

S−1D(x1) + S−1D(x2) = S−1D

(b1

s

)+ S−1D

(b2

s

)=D(b1)s− b1Ds

s2+D(b2)s− b2Ds

s2

=D(b1)s− b1Ds+D(b2)s− b2Ds

s2=D(b1 + b2)s− (b1 + b2)Ds

s2

= S−1D

(b1 + b2

s

)= S−1D(x1 + x2),

and

S−1D(x1) · x2 + x1S−1D(x2) = S−1D

(b1

s

)·(b2

s

)+

(b1

s

)S−1D

(b2

s

)=D(b1)s− b1Ds

s2· b2

s+b1

s· D(b2)s− b2Ds

s2=

(D(b1)s− b1Ds)b2s

s4+b1s(D(b2)s− b2Ds)

s4

=(D(b1)b2 + b1D(b2))s2 − 2b1b2sDs

s4=D(b1b2)s2 − b1b2D(s2)

s4= S−1D

(b1b2

s2

)= S−1D

(b1

s· b2

s

)= S−1D(x1x2).

Remark 2.1.23. Let B be a ring, D ∈ Der(B), S be a multiplicative subset of B

and S−1B the ring of fractions of B.

1. Proposition 2.1.22 allows us to de�ne a map

Der(B) −→ Der(S−1B), D 7−→ S−1D.

By this we say that every derivation of B can be extended to a derivation of

S−1B.

2. If B is a domain and 0 /∈ S, then the canonical homomorphism B −→ S−1B

is injective and we may consider that B ⊆ S−1B. Then we have ker(D) =

ker(S−1D) ∩B.

3. In the special case where S is included in kerD, the de�nition of S−1D simpli�es

as follows:

S−1D(b/s) =D(b)

sfor all b ∈ B and s ∈ S.


Example 2.1.24. Let B = C[X, Y ] = C[2] and S = B \ {0}. Then S−1B = C(X, Y )

is the �eld of rational functions in two variables over C. Consider the C-derivation∂

∂X: C[X, Y ] −→ C[X, Y ] and its extension to a derivation of C(X, Y ),

S−1 ∂

∂X: C(X, Y ) −→ C(X, Y ).

Then forXY

X2 + Y 2∈ C(X, Y ) we have S−1 ∂

∂X

(XY

X2 + Y 2

)=

Y 3 −X2Y

(X2 + Y 2)2.

De�nition 2.1.25. Let B = A[n] where A is a ring and γ = (X1, · · · , Xn) a system

of variables of B. Given f = (f1, · · · , fn−1) ∈ Bn−1, we de�ne the map ∆γf : B −→ B

by ∆γf (g) = det

(∂(f1, · · · , fn−1, g)

∂(X1, · · · , Xn)

), for all g ∈ B. By using some basic properties

of the determinant function it is not di�cult to show that ∆γf ∈ DerA(B) and that

A[f1, · · · , fn−1] ⊆ ker(∆γf ).

We say that ∆γf is a Jacobian derivation.

Remark 2.1.26. Let B = A[2] and γ = (X, Y ) a system of variables of B, we have

∆γX =

∂

∂Yand ∆γ

Y = − ∂

∂X.

2.2 Locally nilpotent derivations

Given a ring B, D ∈ Der(B) and n > 0, we denote by Dn : B −→ B the composition

of D with itself n times; also we de�ne D0 : B −→ B to be the identity map (even in

the case where D = 0). Note that Dn is usually not a derivation when n 6= 1.

De�nition 2.2.1. Given a ring B and D ∈ DerB, we de�ne

Nil(D) = {x ∈ B | ∃n ∈ N such that Dn(x) = 0} .

Clearly kerD ⊆ Nil(D) ⊆ B.

The following Lemma is known as the Leibnitz Rule and can be found in [3] as

Exercise 1.19.


Lemma 2.2.2. Let B be a ring and D ∈ Der(B). For all x, y ∈ B and n ∈ N we

have:

Dn(xy) =n∑i=0

(n

i

)Dn−i(x)Di(y).

Proof. By induction on n; for n = 0 the result is trivial. Now we consider n ∈ N such

that Dn(xy) =∑n

i=0

(ni

)Dn−i(x)Di(y) is true and we show that

Dn+1(xy) =n+1∑i=0

(n+ 1

i

)Dn+1−i(x)Di(y).

We have:

Dn+1(xy) = D(Dn(xy)) = D

( n∑i=0

(n

i

)Dn−i(x)Di(y)

)=

n∑i=0

(n

i

)Dn+1−i(x)Di(y) +

n∑i=0

(n

i

)Dn−i(x)Di+1(y)

=n∑i=0

(n

i

)Dn+1−i(x)Di(y) +

n−1∑i=0

(n

i

)Dn−i(x)Di+1(y) + xDn+1(y)

=n∑i=0

(n+ 1

i

)Dn+1−i(x)Di(y)−

n∑i=1

(n

i− 1

)Dn+1−i(x)Di(y)

+n−1∑i=0

(n

i

)Dn−i(x)Di+1(y) + xDn+1(y)

=n∑i=0

(n+ 1

i

)Dn+1−i(x)Di(y) + xDn+1(y) =

n+1∑i=0

(n+ 1

i

)Dn+1−i(x)Di(y)

where the second equality is true by assumption and the �fth follows from(n+1i

)=(

ni−1

)+(ni

).


Proposition 2.2.3. If B is a ring and D ∈ Der(B), then Nil(D) is a subring of B.

Proof. Clearly 1 belongs to Nil(D). Now let x, y ∈ Nil(D); there exist m,n ∈ N such

that Dn(x) = 0 and Dm(y) = 0. Take l = m+n+ 1, it is easy to check by the lemma

above that Dl(xy) = 0 and Dl(x− y) = 0. Hence Nil(D) is a subring of B.


De�nition 2.2.4. Let B be any ring. A derivation D : B −→ B is locally nilpotent

if it satis�es Nil(D) = B, that is if ∀b ∈ B ∃n ∈ N such that Dn(b) = 0.

The set of all locally nilpotent derivations of a ring B is denoted by LND(B).

Example 2.2.5. • Let A be a ring and B = A[X1, · · · , Xn] = A[n]. Then∂

∂Xi

∈LND(B) for each i = 1, · · · , n.

• Let B = C[T ] = C[1], we know that d/dT ∈ DerC(B) and for each P (T ) ∈ B,D = P (T )

d

dT∈ DerC(B). In particular in the case where P (T ) ∈ C we have

D ∈ LND(B).

De�nition 2.2.6. Let A be a ring andB = A[X1, · · · , Xn]. A derivationD : B −→ B

is triangular if D(A) = {0} and:

∀i D(Xi) ∈ A[X1, · · · , Xi−1]; in particular D(X1) ∈ A.

Example 2.2.7. Let B = C[X, Y, Z] = C[3], the element D = X2 ∂

∂Y+ (X2 + Y 3)

∂

∂Zof DerC(B) is a triangular derivation.

Remark 2.2.8. Let D be a derivation and f ∈ B, if D(f) ∈ Nil(D) then f ∈ Nil(D).

Moreover, if D is triangular it is in particular an A-derivation, thus A ⊆ ker(D) and

so A ⊆ Nil(D).

The next result is Lemma 2.6 of [3].

Lemma 2.2.9. Let A be a ring and B = A[X1, · · · , Xn] = A[n]. Then every triangular

derivation of B is a locally nilpotent derivation.

Proof. Let D : B −→ B be a triangular derivation. We show by induction on i that

A[X1, · · · , Xi] ⊆ Nil(D), for all i = 1, · · · , n.

Since D is triangular we have D(X1) ∈ A ⊆ Nil(D), by the previous remark we have

X1 ∈ Nil(D). By Proposition 2.2.3 Nil(D) is a subring of B and A ⊆ Nil(D) imply

A[X1] ⊆ Nil(D).

Suppose now that i < n is such that A[X1, · · · , Xi] ⊆ Nil(D); we have to show that


A[X1, · · · , Xi+1] ⊆ Nil(D). Since D(Xi+1) ∈ A[X1, · · · , Xi] ⊆ Nil(D) it follows that

Xi+1 ∈ Nil(D). Using the fact that Nil(D) is a subring we get A[X1, · · · , Xi+1] ⊆Nil(D). Therefore A[X1, · · · , Xi] ⊆ Nil(D), for all i = 1, · · · , n. In particular for

i = n, we have B ⊆ Nil(D) that is Nil(D) = B. Thus D is locally nilpotent.

Example 2.2.10. Let B = C[X, Y ] = C[Y,X] = C[2] and let D1 = Y∂

∂X, D2 = X

∂

∂Ybe two derivations of B such that D1 : C[Y,X] −→ C[Y,X] and D2 : C[X, Y ] −→C[X, Y ]. The derivations D1 and D2 are triangular, hence D1, D2 ∈ LND(B). How-

ever (D1 + D2)2(X) = X, hence D1 + D2 /∈ LND(B). Also,∂

∂X∈ LND(B), but

X∂

∂X/∈ LND(B). This shows that the set LND(B) does not have any algebraic

structure; it is just a set.

The following Lemma is Exercise 2.9 of [3].

Lemma 2.2.11. Let B be a ring, D ∈ LND(B) and A = kerD. The following hold:

1. If a ∈ A, then (aD)n = anDn for all n ∈ N.

2. If a ∈ A, then aD ∈ LND(B).

3. Let S ⊆ A be a multiplicatively closed subset of A. Then the map

S−1D : S−1B −→ S−1B, x/s 7−→ D(x)/s,

is a locally nilpotent derivation of S−1B. Furthermore, ker(S−1D) = S−1A.

Proof. 1. By induction on n.

2. It is a consequence of 1.

3. By Proposition 2.1.22 and part (3) of Remark 2.1.23, S−1D is well de�ned and

is an element of Der(S−1B). Let x/s ∈ S−1B (x ∈ B, s ∈ S); since Nil(D) = B

there exists m ∈ N such that Dm(x) = 0. Now we have

(S−1D)m(x/s) = S−1D ◦ S−1D ◦ · · · ◦ S−1D︸︷︷︸m

(x/s) = Dm(x)/s = 0/s = 0 in S−1B.


Thus S−1D ∈ LND(S−1B). Observe that if i : A −→ B is the inclusion map

then 0 −→ Ai−→ B

D−→ B is an exact sequence of A-modules and A-linear

maps. By Proposition 1.5.11 the sequence

0 −→ S−1AS−1i−→ S−1B

S−1D−→ S−1B

is an exact sequence of S−1A-modules. Hence ker(S−1D) = Im(S−1i) = S−1A.

De�nition 2.2.12. Let B be a ring. We de�ne the set

KLND(B) = {kerD | D ∈ LND(B), D 6= 0} .

Example 2.2.13. LetB = C[X, Y ]. We have∂

∂X∈ LND(B) and

∂

∂X6= 0, ker

(∂

∂X

)∈

KLND(B). Since ker

(∂

∂X

)= C[Y ], we have C[Y ] ∈ KLND(B).

Similarly we have C[X] ∈ KLND(B). More generally, if (U, V ) is a system of variables

of B then C[U ],C[V ] ∈ KLND(B).

De�nition 2.2.14. Given a Q-algebra B and D ∈ LND(B), we de�ne the map

ξD : B −→ B[T ], b 7−→∑n∈N

1

n!Dn(b)T n.

The map ξD is called the exponential map associated to D.

The following theorem can be found in [3] as Theorem 3.3.

Theorem 2.2.15. Let B be a Q-algebra and D ∈ LND(B). Then the map ξD : B −→B[T ] is an injective homomorphism of A-algebras, where A = ker(D).

Proof. Let ev0 : B[T ] −→ B, f 7−→ f(0). Then we have ev0 ◦ξD = idB, hence ξD is

injective. It is clear that ξD preserves addition and restricts to the identity map on

A, hence by Remark 1.1.18 it su�ces to verify that ξD(x)ξD(y) = ξD(xy). By Lemma


2.2.2 we have:

ξD(x)ξD(y) =

(∑i∈N

1

i!Di(x)T i

)(∑j∈N

1

j!Dj(y)T j

)=∑n∈N

( ∑i+j=n

1

i!j!Di(x)Dj(y)

)T n

=∑n∈N

1

n!

( ∑i+j=n

n!

i!j!Di(x)Dj(y)

)T n

=∑n∈N

1

n!Dn(xy)T n

= ξD(xy).

Therefore ξD is an injective A-homomorphism.

Remark 2.2.16. The A-homomorphism ξD is the inclusion map if and only if D = 0.

Example 2.2.17. Let B = C[X, Y, Z] and B[T ] = C[X, Y, Z, T ], and consider D =

X∂

∂Y+ Y

∂

∂Z, D ∈ LND(B). The map ξD is an homomorphism of C-algebras and

we have: ξD(X) = X, ξD(Y ) = Y +XT and ξD(Z) = Z + Y T +1

2XT 2.

De�nition 2.2.18. Let B be a ring. Each element D ∈ LND(B) determines a map

degD : B −→ N ∪ {−∞} de�ned as follows:

degD(x) =

max {n ∈ N | Dnx 6= 0} if x 6= 0

−∞ if x = 0.

Note that kerD = {x ∈ B | degD(x) ≤ 0}.

Example 2.2.19. Let A a domain of characteristic zero and B = A[t]. The derivative

D =d

dt: B −→ B is an element of the set LND(B); so D determines the map

degD : A[t] −→ N ∪ {−∞}. The map degD in this case is the usual t-degree.

The following proposition can be found in [3] as Proposition 4.8.

Proposition 2.2.20. If B be a domain of characteristic zero and D ∈ LND(B), then

degD is a degree function.


Proof. We begin by proving the special case where Q ⊆ B. In this case we may

consider the map ξD : B −→ B[T ], which is injective by Theorem 2.2.15, and the

function degT : B[T ] −→ N ∪ {−∞}, which is a degree function (since B[T ] is a

domain). We have degT ◦ξD = degD so by Lemma 1.3.5, degD is a degree function.

In the general case we have Z ⊆ B and so Z ⊆ kerD. Let S = Z \ {0} and consider

S−1D : S−1B −→ S−1B, S−1D ∈ LND(S−1B) by Lemma 2.2.11. As Q ⊆ S−1B, the

�rst part of the proof implies that degS−1D : S−1B −→ N∪{∞} is a degree function.

Now consider the following commutative diagrams:

S−1BS−1D //

S−1B

BD //

?�

OO

B?�

OO S−1BdegS−1D // N ∪ {−∞}

B?�

OO

degD

66

Note that the map B −→ S−1B, x 7−→ x

1is injective since B is a domain and 0 /∈ S.

So D is the restriction of S−1D and consequently degD is the restriction of degS−1D;

it follows that degD is a degree function.

The next result is Corollary 5.3 of [3].

Proposition 2.2.21. Let B be a domain of characteristic zero and D ∈ LND(B).

Then ker(D) is a factorially closed subring of B.

Proof. By Proposition 2.2.20 degD is a degree function and kerD = {x ∈ B | degD x ≤ 0};since {x ∈ B | degD x ≤ 0} is factorially closed by Lemma 1.3.9, the result follows.

The next corollary is Corollary 5.5 of [3].

Corollary 2.2.22. Let B be a domain of characteristic zero, D ∈ LND(B) and

A = ker(D). The following hold:

1. A× = B×.

2. If k is any �eld included in B, then D is a k-derivation.

3. If B is a UFD then so is A.

Proof. By applying Proposition 2.2.21 and Lemma 1.3.17 we obtain the result.


Corollary 2.2.23. Let k be a �eld of characteristic zero and let B be a domain such

that k ⊆ B. Then we have k ⊆ A for each element A of KLND(B).

Proof. This follows from part (2) of Corollary 2.2.22.

Example 2.2.24. Let B = C[X, Y ] = C[2] and f = XY ∈ B. If A is a factorially

closed subring of B satisfying f ∈ A, then we have X, Y ∈ A. Thus C[X, Y ] ⊆ A and

so B = A. Therefore, if D ∈ LND(B) and f ∈ ker(D) then we have D = 0.

Consider ∆f de�ned in De�nition 2.1.25. Then ∆f (X) = −X, so ∆f is not locally

nilpotent. Since f ∈ ker(∆f ) and ker(∆f ) 6= B, the above paragraph implies that

ker(∆f ) is not factorially closed in B. Note, however, that ker(∆f ) is algebraically

closed in B, by Proposition 2.1.18.

The next result can be found in [3] as Exercise 4.9.

Corollary 2.2.25. Let B be a domain of characteristic zero. If D ∈ Der(B) is such

that Dn = 0 for some n > 0, then D = 0.

Proof. By contradiction, assume that D 6= 0. Since Dn = 0 for some n > 0, then

D ∈ LND(B). By Proposition 2.2.20, degD exists and is a degree function. Now since

D 6= 0 there exists some x ∈ B such that degD(x) ≥ 1. Furthermore, degD(xn) =

n degD(x) (because degD is a degree function) and Dn = 0 imply Dn(xn) = 0. Thus

max{k ∈ N : Dk(xn) 6= 0} < n; that is degD(xn) < n. Hence we have n degD(x) < n,

a contradiction (since degD(x) ≥ 1). Therefore we must have D = 0.

The next useful result can be found in [6] as Principle 5.

Proposition 2.2.26. Let B be a domain of characteristic zero, D ∈ LND(B) and

f, g ∈ B. If Df ∈ gB and Dg ∈ fB, then either f ∈ ker(D) or g ∈ ker(D).

Proof. Assume that Df 6= 0 and Dg 6= 0. Then degD(f) ≥ 1 and Df = gb for some

b ∈ B \ {0}. Thus degD(b) ≥ 0 and degD(Df) = degD(f)− 1 = degD(g) + degD(b) ≥degD(g). By the same argument, we show that degD(g) − 1 ≥ degD(f). Adding

this two inequalities, we obtain that degD(f) + degD(g)− 2 ≥ degD(f) + degD(g), a

contradiction. Therefore either f ∈ ker(D) or g ∈ ker(D).


Corollary 2.2.27. Let B be a domain of characteristic zero, D ∈ LND(B) and

f ∈ B. If Df ∈ fB, then f ∈ ker(D).

Proof. Take f = g in Proposition 2.2.26.

Lemma 2.2.28. Let K/k be a �eld extension of characteristic zero and let B =

k[X, Y ] and B = K[X, Y ]. For each D ∈ LND(B), there exists a unique δ ∈ LND(B)

that is an extension of D.

Proof. SinceD : B −→ B is locally nilpotent, it is a k-derivation. So there exist f, g ∈B such that D = f ∂

∂X+ g ∂

∂Y. Now we may de�ne δ ∈ DerK(B) by δ = f ∂

∂X+ g ∂

∂Y.

Then δ is an extension of D and we have δn(X) = Dn(X) and δn(Y ) = Dn(Y ) for all

n ∈ N, so X, Y ∈ Nil(δ). Since δ is a K-derivation we have K ⊆ Nil(δ), so Nil(δ) = B

and δ ∈ LND(B). So there exists at least one δ ∈ LND(B) that extends D.

Suppose that δ1, δ2 ∈ LND(B) are extensions of D. Since δ1, δ2 are locally nilpo-

tent, they areK-derivations; so δ1−δ2 ∈ DerK(B). Moreover, δ1(X) = D(X) = δ2(X)

and δ1(Y ) = D(Y ) = δ2(Y ), so X, Y ∈ ker(δ1 − δ2) and hence δ1 − δ2 = 0.

2.3 Slices and preslices

De�nition 2.3.1. Let B be a ring and D ∈ LND(B). A slice of D is an element

s ∈ B satisfying D(s) = 1.

Example 2.3.2. Let B = C[X, Y, Z] = C[3].

1. Clearly, X is a slice of∂

∂X∈ LND(B).

2. De�ne D ∈ LND(B) by D(Z) = Y , D(Y ) = X and D(X) = 0, that is D =

X∂

∂Y+ Y

∂

∂Z. Then given f ∈ B, we have D(f) = fYX + fZY . Thus D(B) ⊆

(X, Y )B and so D does not have a slice. (Here, (X, Y )B denotes the ideal of B

generated by X, Y .)

When a slice exists, the situation is very special. (The following result is Theorem

7.3 of [3].)


Theorem 2.3.3. Let B be a Q-algebra, D ∈ LND(B) and A = ker(D). If s ∈ B is

a slice of D, then B = A[s] = A[1] and D =d

ds: A[s] −→ A[s].

Proof. Consider f(T ) =∑n

i=0 aiTi ∈ A[T ] \ {0} where n ≥ 0, ai ∈ A and an 6= 0. By

Corollary 2.1.16 we have Dj(f(s)) = f (j)(s) for all j ≥ 0, where f (j) denotes the j-th

derivative of f ; so Dn(f(s)) = n!an 6= 0 (because n! ∈ Q× ⊆ B×) and in particular

f(s) 6= 0. So s is transcendental over A, that is A[s] = A[1].

We show that B = A[s]. Consider the homomorphism of A-algebras ξ : B −→ B

de�ned by ξ = ev−s ◦ξD where ξD : B −→ B[T ] is de�ned in De�nition 2.2.14 and

ev−s : B[T ] −→ B is de�ned in De�nition 1.2.3. Hence ξ(x) =∑∞

j=0

Dj(x)

j!(−s)j.

For each x ∈ B,

D(ξ(x)) =∞∑j=0

Dj+1(x)

j!(−s)j +

∞∑j=1

Dj(x)

j!j(−s)j−1(−1) = 0,

so ξ(B) ⊆ A; moreover for x ∈ A ⊆ B we have x = ξ(x) ∈ ξ(B) hence A ⊆ ξ(B) and

ξ(B) = A.

By induction on degD(x) we show that ∀x ∈ B, x ∈ A[s].

If degD(x) ≤ 0 it is obvious. Now assume that degD(x) ≥ 1 and that y ∈ A[s] is

true for all y ∈ B of degree degD(y) < degD(x).

Since x = ξ(x) + (x− ξ(x)) where ξ(x) ∈ A and x− ξ(x) ∈ sB, we have

x = a+ x′s, for some a ∈ A and x′ ∈ B. (1)

This implies that D(x) = D(x′)s+ x′ and it easily follows by induction that

∀ m ≥ 1 Dm(x) = Dm(x′)s+mDm−1(x′). (2)

Since degD(x) ≥ 1 we have x /∈ A and x′ 6= 0. Now the fact that D ∈ LND(B)

implies that there is m ≥ 1 such that Dm−1(x′) 6= 0 and Dm(x′) = 0, so degD(x′) =

m− 1. Then (2) gives Dm(x) = mDm−1(x′) 6= 0 and Dm+1(x) = 0, so degD(x) = m

and so degD(x′) = degD(x)− 1. By inductive hypothesis, we have x′ ∈ A[s]; then (1)

gives x ∈ A[s]. So B = A[s] = A[1].


Remark 2.3.4. Let B = Z[Y,X] = Z[2] and D =∂

∂Y+ Y

∂

∂X∈ Der(B). Note that

D is a nonzero triangular derivation (hence locally nilpotent derivation). We have

D(Y ) = 1 and we have shown in Example 2.1.17 that A = ker(D) = Z[2X−Y 2]. We

claim that B is not a polynomial ring over A. In fact assume that B = A[1], then there

is a g ∈ B (deg(g) > 0) such that B = Z[2X−Y 2, g]. Hence (2X−Y 2, g) is a system

of variables of Z[X, Y ]; thus by Lemma 2.1.20 we must have 2(gY + Y gX) ∈ {1,−1},a contradiction. Therefore B 6= A[1]. Thus the hypothesis that B is a Q-algebra in

Theorem 2.3.3 is not super�uous.

The following three corollaries are Corollary 7.4, 7.5 and 7.12 of [3].

Corollary 2.3.5. Let B be a Q-algebra, D ∈ LND(B) and A = ker(D). If s ∈ Bsatis�es Ds ∈ A×, then B = A[s] = A[1].

Proof. Let a ∈ A be such that aD(s) = 1. Then as is a slice of D; so by Theorem

2.3.3 B = A[as] = A[s] and s is transcendental over A (since as is and a ∈ A×).

De�nition 2.3.6. Let B be a ring and D ∈ LND(B). A preslice of D is an element

s of B satisfying D(s) 6= 0 and D2(s) = 0 (that is degD(s) = 1).

Lemma 2.3.7. Let B be a ring and D ∈ LND(B) \ {0}. Then there exists a preslice

of D.

Proof. Since D 6= 0, there exists b ∈ B such that D(b) 6= 0. Set m = degD(b) and

s = Dm−1(b). Then D(s) 6= 0 and D2(s) = 0, so s is a preslice of D.

Corollary 2.3.8. Let B be a domain of characteristic zero and A ∈ KLND(B). Then

S−1B = (FracA)[1], where S = A \ {0}. In particular trdegA(B) = 1.

Proof. Let A ∈ KLND(B), then we have A = ker(D) where D ∈ LND(B) \ {0}. ByLemma 2.3.7 there is x ∈ B such thatD(x) 6= 0 andD2(x) = 0. Set a = D(x), we have

a ∈ S, S−1D(x/a) = 1 and Q ⊆ S−1B (since B is of characteristic zero). Moreover,

by Lemma 2.2.11 we have S−1D ∈ LND(S−1B) and ker(S−1D) = S−1A = Frac(A).

We thus have by Theorem 2.3.3 S−1B = (FracA)[1]. Furthermore, Frac(S−1B) =

Frac(B), hence 1 = trdegFrac(A) S−1B = trdegFrac(A) Frac(B) = trdegA(B).


Corollary 2.3.9. Let B be a Q-algebra, D ∈ LND(B) and A = ker(D). If s ∈ Bsatis�es D(s) 6= 0 and D2s = 0, then Bα = Aα[s] = (Aα)[1] where α = D(s) ∈ A\{0}.

Proof. Let S = {1, α, α2, · · · }, then Bα = S−1B and Aα = S−1A. Consider

S−1D : Bα −→ Bα, then S−1D ∈ LND(Bα) and ker(S−1D) = Aα (by Lemma 2.2.11).

As S−1D(s) = α ∈ (Aα)×, the result follows from Corollary 2.3.5.

The next lemma is Exercise 7.6 of [3].

Lemma 2.3.10. Let B be a domain of characteristic zero. If A,A′ ∈ KLND(B) and

A ⊆ A′, then A = A′.

Proof. We have A ⊆ A′ ⊆ B and A is algebraically closed in B by Proposition 2.1.18.

Corollary 2.3.8 implies trdegAB = 1 = trdegA′ B, so by Corollary 1.5.26 we have

A = A′.

The following lemma will be very helpful in the next chapter.

Lemma 2.3.11. Let k be a �eld of characteristic zero, B = k[X, Y ] = k[2] and

D ∈ LND(B). Then some nonconstant polynomial belongs to ker(D).

Proof. If D = 0 then ker(D) = B, so the claim is true.

If D 6= 0 then trdegker(D)(B) = 1 by Corollary 2.3.8, so trdegk(kerD) = 1, so again

the claim is true.

Example 2.3.12. Let k and B be like in Lemma 2.3.11. Consider D = X ∂∂X

+ Y ∂∂Y

,

it can be shown that ker(D) = k, hence D is not locally nilpotent.

De�nition 2.3.13. Let B be a domain of characteristic zero. We say that B is rigid

if it satis�es the following equivalent conditions:

(i) LND(B) = {0}

(ii) KLND(B) = ∅.

The lemma below can be found in [3] as Exercise 7.8.


Lemma 2.3.14. Let B be a domain of characteristic zero that is not rigid. If k0 is

a �eld such that k0 ⊆ B and trdegk0 B = 1, then B = k[1] for some �eld k such that

k0 ⊆ k ⊆ B.

Proof. Since B is not rigid, there is D ∈ LND(B) \ {0}. Set k = kerD, then k is

factorially closed in B and since k0 is a �eld, we have k0 ⊆ k ⊆ B (by Corollary

2.2.22-(2)). Furthermore Corollary 2.3.8 implies that trdegk B = 1, so trdegk0 k = 0,

that is k is algebraic over k0 and thus k is a �eld (indeed, k is a domain and an

integral extension of the �eld k0, so k is a �eld by Proposition 5.7 of [1]). Since

D ∈ LND \{0}, Lemma 2.3.7 implies that there exists s ∈ B satisfying D(s) 6= 0 and

D2(s) = 0. Then we have D(s) ∈ k×, so Corollary 2.3.5 implies that B = k[1].

Example 2.3.15. Let us prove that the subring B = C[T 2, T 3] of C[T ] = C[1] is

rigid. We have B 6= C[1]. In fact T 2 is an irreducible element of B, however it is

not prime in B because T 2|(T 3 · T 3) in B but T 2 - T 3 in B. Hence by Proposition

1.1.12 B is not an UFD, therefore B 6= C[1]. Furthermore, C ⊆ B ⊆ C[T ] hence

1 = trdegC(C[T ]) = trdegB(C[T ]) + trdegC(B). Since T 2 ∈ B, then T is algebraic

over B and by Corollary 1.5.28 C[T ] is algebraic over B, so trdegB(C[T ]) = 0. Hence

trdegC(B) = 1. If B is not rigid, then by Lemma 2.3.14 there exists a �eld K such

that C ⊆ K ⊆ B and B = K [1], thus trdegCK = 0; since C is algebraically closed we

have K = C, a contradiction (because B 6= C[1]).

The following is an improved version of Lemma 2.2.11.

Corollary 2.3.16. Let B be a domain of characteristic zero, D ∈ Der(B) \ {0} andb ∈ B \ {0}, and let S be a multiplicatively closed subset of B such that 0 /∈ S.

1. We have bD ∈ LND(B) if and only if D ∈ LND(D) and b ∈ ker(D).

2. We have S−1D ∈ LND(S−1B) if and only if D ∈ LND(B) and S ⊆ ker(D).

Proof. 1. ⇐=) Lemma 2.2.11.

=⇒) We know that ker(bD) = ker(D) is factorially closed in B (since bD ∈LND(B)\{0}). Moreover there is x ∈ B such that (bD)(x) 6= 0 and (bD)2(x) =

0 (bD has a preslice). Hence bD(x) = (bD)(x) ∈ ker(bD) \ {0} and so b,D(x) ∈


ker(D) since ker(D) is factorially closed. Moreover, b ∈ ker(D) implies that

(bD)n = bnDn for all n ≥ 1, thus bD ∈ LND(B) implies that D ∈ LND(B).

2. ⇐=) Lemma 2.2.11.

=⇒) The canonical homomorphism B −→ S−1B is injective, so we may con-

sider that B ⊆ S−1B. Since D is the restriction of S−1D, we have Dn(b) =

(S−1D)n(b) for all b ∈ B and n ∈ N, so S−1D ∈ LND(S−1B) implies D ∈LND(B). Since S−1D ∈ LND(S−1B), part (1) of Corollary 2.2.22 implies that

(S−1B)× = ker(S−1D)×. Since S ⊆ (S−1B)×, we have S ⊆ ker(S−1D)× and

hence S ⊆ B ∩ ker(S−1D) = ker(D).

Example 2.3.17. Let B = C[X, Y ] and D =∂

∂X∈ LND(B); consider S = B \ {0},

S−1B = C(X, Y ) and S−1D ∈ Der(S−1B). Since S * ker(D), Corollary 2.3.16

implies that S−1D is not locally nilpotent. We can see this directly by noting that1X∈ S−1B and (S−1D)n( 1

X) 6= 0 for all n ≥ 1.

Notation. If A ⊆ B are rings, de�ne LNDA(B) = LND(B) ∩DerA(B).

The following proposition can be found in Principle 8 of [6].

Proposition 2.3.18. Given a domain A of characteristic zero and B = A[T ] = A[1],

we have LNDA(B) = A · ddT.

Proof. By Example 2.2.5 ddT∈ LND(B), and it is easy to show that ker( d

dT) = A

(since we are in characteristic zero). Therefore by Corollary 2.3.16 part (1) we have

A · ddT⊆ LNDA(B). Conversely for all D ∈ LNDA(B) we have LNDA(B) 3 D =

D(T ) ddT

(by the proof of Lemma 2.1.13), thus by part (1) of Corollary 2.3.16 we have

D(T ) ∈ kerD = A. Therefore LNDA(B) ⊆ A · ddT.

2.4 Degree and Homogenization of Derivations

Throughout this section, let B = ⊕n∈ZBn be a Z-graded domain and let deg : B −→Z ∪ {−∞} be the degree function determined by the grading (see Remark 1.4.8).


De�nition 2.4.1. Let D ∈ Der(B) and consider the nonempty subset

U = {deg(D(f))− deg(f) | f ∈ B \ {0}}

of Z ∪ {−∞}. If U has a greatest element, then we say that the degree of D exists

and we de�ne deg(D) to be that element.

Note that deg(D) = −∞ ⇐⇒ D = 0. Furthermore if deg(D) exists and D 6= 0 then

there is f ∈ B \ ker(D) such that deg(D) = deg(D(f))− deg(f).

De�nition 2.4.2. Given D ∈ Der(B), the defect function of D is the map

defD : B −→ Z ∪ {−∞}

f 7−→

{deg(D(f))− deg(f) if f 6= 0

−∞ if f = 0.

The proofs of results from 2.4.3-2.4.15 follow the reasoning given in section 2.6 of

[17].

Lemma 2.4.3. Let D ∈ Der(B).

(i) defD(fg) ≤ max{defD(f), defD(g)} for all f, g ∈ B.

(ii) If f1, . . . , fm ∈ B are such that deg(∑m

i=1 fi) = max1≤i≤m deg(fi),

then defD(∑m

i=1 fi) ≤ max1≤i≤m defD(fi).

Proof. (i) If f = 0 or g = 0 the result is clear. Assume that f 6= 0 and g 6= 0. We

have

defD(fg) = deg(D(fg))− deg(fg)

= deg(fD(g) + gD(f))− (deg(f) + deg(g))

≤ max{deg(fD(g)), deg(gD(f))} − (deg(f) + deg(g))

= max{deg(f) + deg(D(g)), deg(g) + deg(D(f))} − (deg(f) + deg(g))

In the case that

max{deg(f) + deg(D(g)), deg(g) + deg(D(f))} = deg(f) + deg(D(g)),


we have defD(fg) ≤ deg(D(g))− deg(g) = defD(g).

Otherwise,

max{deg(f) + deg(D(g)), deg(g) + deg(D(f))} = deg(g) + deg(D(f)),

so defD(fg) ≤ deg(D(f))− deg(f) = defD(f).

Therefore defD(fg) ≤ max{defD(f), defD(g)}.

(ii) Observe that

defD(n∑i=1

fi) = deg(D(m∑i=1

fi))− deg(m∑i=1

fi)

= deg(m∑i=1

D(fi))−max {deg(fi) | 1 ≤ i ≤ m}

≤ max {deg(D(fi)) | 1 ≤ i ≤ m} −max {deg(fi) | 1 ≤ i ≤ m}

≤ max {deg(D(fi))− deg(fi) | 1 ≤ i ≤ m}

= max {defD(fi) | 1 ≤ i ≤ m} .

Proposition 2.4.4. Let k be a �eld of characteristic zero. Assume that the Z-graded domain B is a �nitely generated k-algebra. Then deg(D) exists for every

D ∈ Derk(B).

More precisely, if h1, h2, · · · , hn are homogeneous elements of B which generate B as

a k-algebra, then

deg(D) = max {deg(D(hi))− deg(hi) | 1 ≤ i ≤ n} .

Proof. It is clear that there exist homogeneous elements h1, · · · , hn ∈ B such that

B = k[h1, · · · , hn]. Let def = defD : B −→ Z ∪ {−∞} be the defect function of D

and K = max{def(h1), · · · , def(hn)}. To prove the proposition, we have to show that

def(f) ≤ K for all f ∈ B \ {0}.From Lemma 2.4.3 (i) we know that if f1, · · · , fs ∈ B then def(f1 · · · fs) ≤ max1≤i≤s def(fi).

In particular, it is easy to show that def(he11 · · ·henn ) ≤ K for any e1, · · · , en ∈ N.


Furthermore def(λhe11 · · ·henn ) ≤ K for any λ ∈ k× and e1, · · · , en ∈ N. In fact since

D is a k-derivation and λ ∈ k× we have def(λ) = −∞. Thus

def(λhe11 · · ·henn ) ≤ max{def(λ), def(he11 · · ·henn )} ≤ K. (∗)

Claim: If h ∈ B is homogeneous, then def(h) ≤ K.

Proof: If h = 0, then def(h) = −∞ ≤ K, so suppose h 6= 0 and h is homogeneous of

degree r. We can write h =∑m

i=1 fi where fi = λihei11 · · ·heinn and

∑nj=1 eij deg(hj) = r

for all 1 ≤ i ≤ m. So deg(fi) = deg(h) for all 1 ≤ i ≤ m and h =∑m

j=1 fi satis�es

the condition in Lemma 2.4.3 (ii). Then def(h) ≤ max{def(f1), · · · , def(fm)}. But

def(fi) ≤ K for all 1 ≤ i ≤ m by (∗), so def(h) ≤ K and the claim is proved.

Now let f ∈ B \ {0}, then f = f1 + · · ·+ fm where fi are homogeneous elements

of distinct degrees. Then deg(f) = deg(∑m

i=1 fi) = max {deg(fi) | 1 ≤ i ≤ m} so by

Lemma 2.4.3 (ii) def(f) ≤ max{def(f1), · · · , def(fn)}. But by the claim above we

have def(fi) ≤ K for all 1 ≤ i ≤ m. So def(f) ≤ K.

Example 2.4.5. Regard B = k[X, Y, Z] as a graded ring (with standard Z-grading)and let D = X ∂

∂Y+ Y 3 ∂

∂Z. We have D(X) = 0, D(Y ) = X and D(Z) = Y 3. Thus

deg(D) = max{deg(DX)− deg(X), deg(DY )− deg(Y ), deg(DZ)− deg(Z)}

= max{−∞, deg(X)− deg(Y ), deg(Y 3)− deg(Z)}

= max{−∞, 1− 1, 3− 1}

= 2.

De�nition 2.4.6. Let D ∈ Der(B) and D 6= 0. If there exists d ∈ Z such that

D(Bi) ⊆ Bi+d for all i ∈ Z, then d is unique and we say that D is homogeneous of

degree d. We adopt the convention that the zero derivation is homogeneous of degree

−∞.

Example 2.4.7. Let B = k[X] = ⊕n∈NBn, where the Bn are de�ned as in Example

1.4.3. The derivation D = ddX

, de�ned in Example 2.1.2, is homogeneous of degree

−1.


Lemma 2.4.8. Let D ∈ Der(B) be such that D is homogeneous of degree d, then

deg(D) exists and deg(D) = d.

Proof. Trivial.

De�nition 2.4.9. Let D ∈ Der(B) be such that deg(D) exists. De�ne the homoge-

nization of D, D : B −→ B, as follows:

If D = 0, then D = 0.

If D 6= 0, let d = deg(D) ∈ Z, and for all i ∈ Z, de�ne:

Di : Bi −→ Bi+d

fi 7−→ Pi+d(D(fi))

where Pj : B −→ Bj is the canonical projection for all j ∈ Z.Now given f ∈ B, f =

∑i∈Z fi (where all the fi ∈ Bi are zero expect for �nitely

many of them), we de�ne

D(f) =∑i∈Z

Di(fi).

Remark 2.4.10. If D ∈ Der(B) is homogeneous, then D = D.

Proposition 2.4.11. Let D ∈ Der(B). If deg(D) exists, then

(i) D : B −→ B is a homogeneous derivation of degree deg(D),

(ii) D = 0⇐⇒ D = 0,

(iii) deg(D) = deg(D).

Proof. Assertion (i) is clear if D = 0, so assume that D 6= 0 and let d = deg(D) ∈ Z.We leave it to the reader to verify that D : B −→ B is a derivation. It is clear

from the de�nition of D that D(Bi) ⊆ Bi+d for all i ∈ Z, so if D 6= 0 then D is

homogeneous of degree deg(D). So the proof of (i) will be complete once we show

that D 6= 0 (we show it in the proof of (ii)).


(ii) If D = 0, then D = 0. Now assume that D 6= 0 and let us prove that D 6= 0.

Let d = deg(D) ∈ Z. Then there is f ∈ B \ {0} such that d = deg(D(f)) − deg(f).

Set deg(f) = n and write f = fn +∑

i<n fi; we have

deg(D(f)) ≤ max{deg(D(fn)), deg(∑i<n

D(fi))}.

If max{deg(D(fn)), deg(∑

i<nD(fi))} = deg(∑

i<nD(fi)), then there is an i0 < n

such that deg(D(fi0)) ≥ deg(D(f)) = n+ d, so

d ≤ deg(D(fi0))− n < deg(D(fi0))− i0 = deg(D(fi0))− deg(fi0), a contradiction.

Thus max{deg(D(fn)), deg(∑

i<n fi)} = deg(D(fn)), which implies that deg(D(fn)) ≥deg(D(f)) = n+ d. As deg(D(fn)) ≤ n+ d is clear, we have deg(D(fn)) = d+ n and

Pn+d(D(fn)) 6= 0.

Therefore D(fn) = Dn(fn) = Pn+d(D(fn)) 6= 0 and so D 6= 0. This proved (ii), and

also completes the proof of (i).

(iii) Follows from (i) and Lemma 2.4.8.

Example 2.4.12. Let B = k[X, Y, Z] and D = X∂

∂Y+Y 3 ∂

∂Zbe like in Example 2.4.5.

We have D1(X) = P1+2(DX) = P3(0) = 0, D1(Y ) = P1+2(DY ) = P3(X) = 0 and

D1(Z) = P1+2(DZ) = P3(Y 3) = Y 3. So D : B −→ B is the derivation Y 3 ∂∂Z.

Lemma 2.4.13. Let D ∈ Der(B)\{0} be such that deg(D) exists. Set d = deg(D) ∈Z.

(i) D(Pj+nd(Dn(f))) = Pj+(n+1)d(D

n+1(f)) for all f ∈ B, n ∈ N and all j ≥deg(f).

(ii) If h ∈ Bj for some j, then for all n ∈ N we have Dn(h) = Pj+nd(Dn(h)).

Proof. (i) For f ∈ B, write f as f =∑

i≤j fi =∑

i<j fi + fj. For n = 0 we have

D(Pj(D0(f))) = D(Pj(f)) = D(fj) (since D

0 = idB),

= Dj(fj)

= Pj+d(D(fj))

= Pj+d(D(f)) (since for i < j we have deg(D(fi)) < j + d).


Now given n ∈ N, set g = Dn(f) and i = j + nd. Observe that

deg(Dn(f)) ≤ deg(f) +nd ≤ j+nd. Then i ≥ deg(g) and by the previous argument,

we get

D(Pj+nd(Dn(f))) = D(Pi(g))

= Pi+d(D(g))

= Pj+(n+1)d(Dn+1(f)).

(ii) By induction on n. Let h ∈ Bj for some j ∈ Z.For n = 0, Dn(h) = D0(h) = h and Pj+nd(D

n(h)) = Pj(h) = h.

Suppose Dn(h) = Pj+nd(Dn(h)) for some n ∈ N. Then

Dn+1(h) = D(Dn(h)) = D(Pj+nd(Dn(h)))

= Pj+(n+1)d(Dn+1(h))

where the last equality follows from (i). Thus Dn(h) = Pj+nd(Dn(h)) for all n ∈

N.

Proposition 2.4.14. Let D ∈ Der(B) be such that deg(D) exists. If D ∈ LND(B)

then D ∈ LND(B).

Proof. If D ∈ LND(B) \ {0} then Lemma 2.4.13 (ii) implies that all homogeneous

elements of B belong to Nil(D), so Nil(D) = B.

Proposition 2.4.15. Let D ∈ Der(B) be such that deg(D) exists. If f =∑

i≤n fi ∈ker(D), then fn ∈ ker(D).

Proof. We may assume that D 6= 0. Let d = deg(D) then Lemma 2.4.13 (i) implies

that D(fn) = D(Pn(f)) = Pn+d(D(f)) = 0.

Chapter 3

Locally Nilpotent Derivations and

Automorphisms of k [X ,Y ]

In this chapter we present how the theory of locally nilpotent derivations can be used

to describe the structure of the polynomial ring k[X, Y ]. We begin by giving some

useful results about polynomial rings over a �eld which will help us later in our work.

3.1 Some facts on polynomial rings

Lemma 3.1.1. Let B = k[n] where k is a �eld. Then we have B× = k×. Furthermore,

if K is a �eld which is a subring of B, then K ⊆ k.

Proof. By Lemma 1.3.10, k is factorially closed in B. So B× = k× follows from

Lemma 1.3.17. Moreover, if K ⊆ B then K× ⊆ B× = k× so K ⊆ k.

Proposition 3.1.2. Let A be a domain and U ∈ A[X]. The following are equivalent:

(i) A[X] = A[U ],

(ii) U = aX + b, a ∈ A×, b ∈ A.

Proof. ii) =⇒ i) Trivial.

i) =⇒ ii) We have X ∈ A[U ]; so there exists P =∑n

i=0 aiTi ∈ A[T ] \ {0} such that

66

CHAPTER 3. DERIVATIONS AND AUTOMORPHISMS OF k[X, Y ] 67

X = P (U). Let n = degT (P ), m = degX(U) and observe that n > 0 and m > 0. It

is easy to see that degX(P (U)) = mn = degX(U) degT (P ). We thus have

X = P (U)⇒ degT (P ) degX(U) = 1⇒ degX(U) = 1 = degT (P ).

Therefore U = aX + b, a ∈ A \ {0}, b ∈ A. Furthermore, degT (P ) = 1 then

P (T ) = cT + d, c ∈ A \ {0}, d ∈ A; so X = P (U) = cU + d = caX + bc + d.

Therefore ca = 1 and we thus have a ∈ A×.

Corollary 3.1.3. Let B = k[X, Y ] and F be an element of B such that k[X, Y ] =

k[X,F ]. Then we have F = aY + b, a ∈ k \ {0} and b ∈ k[X].

Proof. Apply Proposition 3.1.2 to B = A[Y ] = A[F ], A = k[X].

Lemma 3.1.4. Let k be an algebraically closed �eld and B = k[X, Y ]. If g(X, Y )

is a nonconstant standard homogeneous polynomial, then g = L1L2 · · ·LN , where

Li = aiX + biY, ai, bi ∈ k, (ai, bi) 6= (0, 0) and N = deg(g).

Proof. Since g is standard homogeneous, it can be written as:

g(X, Y ) =N∑i=0

biXiY N−i = Y NP (

X

Y), where P (t) ∈ k[t].

Let M = deg(P (t)) then we have M ≤ N and P (t) =∑M

i=0 biti. Moreover, k is

algebraically closed, so the polynomial P (t) ∈ k[t] can be factored as

P (t) = c∏M

i=1(t−ci), for some c ∈ k×, ci ∈ k and so g(X, Y ) = Y Nc∏M

i=1(X

Y−ci) =

cY N−MΠMi=1(X − ciY ) = L1L2 · · ·LN , with Li = aiX + biY ai, bi ∈ k, (ai, bi) 6=

(0, 0).

Lemma 3.1.5. Let B = k[X, Y ] = k[2] where k is a �eld. Then there are in�nitely

many subrings A of B such that B = A[1].

Proof. In fact let m,n be two elements of N \ {0} such that n 6= m; assume without

loss of generality that 0 < m < n and consider the subrings An = k[X + Y n] and

Am = k[X + Y m] of B. Clearly we have that An[Y ] = B = Am[Y ].

Let us prove that An 6= Am. If An = Am, then X + Y m ∈ k[X + Y n] that is

X+Y m = g(X+Y n) where g(T ) =∑s

i=0 aiTi ∈ k[T ] is a nonconstant polynomial and


as 6= 0. So ev(0,Y )(X+Y m) = ev(0,Y ) g(X+Y n) that is Y m = a0 +a1Yn+ · · ·+asY

ns,

a contradiction (since 0 < m < n). Hence An 6= Am for all n 6= m.

We show that Y is transcendental over An for all n ∈ N. By contradiction assume

Y is algebraic over An. Then, by Corollary 1.5.28, B = An[Y ] is algebraic over An

and consequently trdegAn(B) = 0. Since X + Y n is transcendental over k by Lemma

1.3.13, we have An = k[X + Y n] = k[1] and consequently trdegk(An) = 1. So

2 = trdegk(B) = trdegk(An) + trdegAn(B) = 1 + 0 = 1,

a contradiction. So Y is transcendental over An.

Therefore B = An[Y ] = A[1]n for all n ∈ N; and so there is an in�nite number of

subrings A of B such that B = A[1].

Lemma 3.1.6. Let B = k[n], where k is a �eld and let K be a �eld which is a subring

of B. If there is m ∈ N such that B = K [m], then K = k and m = n.

Proof. Since K is a subring of B, by Lemma 3.1.1 we have K ⊆ k; and k is also a

subring of B. Since B = K [m] we must also have k ⊆ K; hence K = k. Moreover

k[n] = B = K [m] = k[m] =⇒ trdegk k[n] = trdegk k

[m] ⇒ n = m.

3.2 Preliminaries on automorphisms of k[X, Y ]

The aim of this section is to present some particular subgroups of the group of k-

automorphism of k[X, Y ] which we will use regularly throughout this chapter, namely

in Theorems 3.3.8, 3.4.1 and 3.4.4.

In this section k is an arbitrary �eld.

Let B = k[X, Y ] and let Autk(B) be the set of all k-automorphisms of B. It is

well known that Autk(B) is a group.

De�nition 3.2.1. The group of k-automorphisms of B is called the general a�ne

group of dimension 2 and is denoted GA2(k).


Remark 3.2.2. By Theorem 1.2.4, each element ϕ ∈ GA2(k) is completely de-

termined by the pair (ϕ(X), ϕ(Y )) ∈ B × B. It is customary to identify ϕ with

(ϕ(X), ϕ(Y )) and to write simply ϕ = (ϕ(X), ϕ(Y )). Then the rule for composing

elements of GA2(k) is:

(ϕ(X), ϕ(Y )) ◦ (ψ(X), ψ(Y )) = ((ϕ ◦ ψ)(X), (ϕ ◦ ψ)(Y )).

For instance, consider the automorphisms ϕ = (Y,X + Y 2) and ψ = (Y,X). This

means that ϕ : B −→ B is the k-automorphism that satis�es ϕ(X) = Y and ϕ(Y ) =

X + Y 2, and that ψ : B −→ B is the k-automorphism that satis�es ψ(X) = Y and

ψ(Y ) = X. Then (ϕ ◦ ψ)(X) = ϕ(Y ) = X + Y 2 and (ϕ ◦ ψ)(Y ) = ϕ(X) = Y , so

(Y,X + Y 2) ◦ (Y,X) = (X + Y 2, Y ),

and the reader can verify that

(Y,X) ◦ (Y,X + Y 2) = (X, Y +X2).

Lemma 3.2.3. Let ϕ : B −→ B be a k-homomorphism and let u = ϕ(X) and

v = ϕ(Y ). Then we have ϕ ∈ GA2(k) if and only if (u, v) is a system of variables of

B.

Proof. =⇒) Let α = (u, v) ∈ GA2(k), then the automorphism α : B −→ B satis�es

α(X) = u and α(Y ) = v. By Theorem 1.2.4, Im(α) = k[u, v]. Since α is surjective,

B = k[u, v] so (u, v) is a system of variables of B.

⇐=) Assume that (u, v) is a system of variables of B. We know that there is an

unique k-homomorphism such that α(X) = u and α(Y ) = v. Then Im(α) = k[u, v]

by Theorem 1.2.4 and k[u, v] = B by hypothesis, so α is surjective and since B is

Noetherian α is an automorphism.

A�ne linear subgroup

Given A =

(a b

c d

)∈ GL2(k) there exists a unique k-homomorphism ϕA : B −→ B

such that ϕA(X) = aX + cY and ϕA(Y ) = bX + dY . It is easy to verify that if


I ∈ GL2(k) is the identity matrix then ϕI is the identity map of B, and that if

A,B ∈ GL2(k) then ϕA ◦ ϕB = ϕAB. It follows that A 7−→ ϕA is a group ho-

momorphism GL2(k) −→ GA2(k). The image of this homomorphism is denoted

GL2(k) = {ϕA | A ∈ GL2(k)}.Given (a, b) ∈ k2, de�ne the k-homomorphism ϕ(a,b) : B −→ B by ϕ(a,b)(X) =

X + a and ϕ(a,b)(Y ) = Y + b. It is clear that ϕ(0,0) is the identity map of B and

that if (a, b), (c, d) ∈ k2 then ϕ(a,b) ◦ ϕ(c,d) = ϕ(a+c,b+d). So (a, b) 7−→ ϕ(a,b) is a group

homomorphism k2 −→ GA2(k). Its image is denoted T ={ϕ(a,b)

∣∣ (a, b) ∈ k2}.

De�nition 3.2.4. • The subgroup GL2(k) of GA2(k) is called the subgroup of

linear automorphisms.

• The subgroup 〈GL2(k)∪T 〉 of GA2(k) generated by GL2(k) and T is called the

a�ne linear subgroup. It is denoted by Af2(k).

Remark 3.2.5. 1. The elements of Af2(k) are of the form ϕ1 ◦ϕ2 ◦ · · · ◦ϕs, whereϕi ∈ GL2(k) ∪ T for i = 1, 2, · · · , s. ThereforeAf2(k) = {(a1X + b1Y + c1, a2X + b2Y + c2) | ai, bi, ci ∈ k; a1b2 − a2b1 6= 0}.

2. There exists a semi-commuting relation among elements of GL2(k) and T , givenby:

(X+c1, Y +c2)(a1X+b1Y, a2X+b2Y ) = (a1X+b1Y, a2X+b2Y )(X+c, Y +c′),

where c = a1c1 + b1c2 and c′ = a2c1 + b2c2.

Subgroup of triangular automorphisms

Let a, c ∈ k×, b ∈ k and f(X) ∈ k[X] ⊆ k[X, Y ]; like above, de�ne a k-homomorphism

τ(a,b,c,f(X)) such that τ(a,b,c,f(X))(X) = aX + b and τ(a,b,c,f(X))(Y ) = cY + f(X). Given

a′, c′ ∈ k×, b′ ∈ k and f ′(X) ∈ k[X] we have

τ(a′,b′,c′,f ′(X)) ◦ τ(a,b,c,f(X))(X) = aa′X + ab′ + b

and

τ(a′,b′,c′,f ′(X)) ◦ τ(a,b,c,f(X))(Y ) = cc′Y + cf ′(X) + f(a′X + b′),


so τ(a,b,c,f(X)) ◦ τ(a′,b′,c′,f ′(X)) = τ(aa′,ab′+b,cc′,cf ′(X)+f(a′X+b′)).

Hence τ(a,b,c,f(X)) has τ(a−1,−a−1b,c−1,−c−1f(a−1X−a−1b)) as inverse, thus τ(a,b,c,f(X)) is

bijective and τ(a,b,c,f(X)) ∈ GA2(k).

Therefore, the set

BA2(k) ={τ(a,b,c,f(X)) = (aX + b, cY + f(X))

∣∣ a, c ∈ k×, b ∈ k, f ∈ k[X]}

is a subgroup of GA2(k).

De�nition 3.2.6. The subgroup of GA2(k) de�ned above is called the subgroup of

triangular automorphisms. It is denoted by BA2(k).

Remark 3.2.7. Observe that T ⊆ BA2(k) and that consequently the subgroup of

GA2(k) generated by GL2(k) and BA2(k) is equal to the subgroup generated by Af2(k)

and BA2(k).

Subgroup of tame automorphisms

De�nition 3.2.8. The tame subgroup of GA2(k) is the subgroup 〈GL2(k)∪BA2(k)〉generated by the subgroups of linear and triangular automorphisms. An automor-

phism is tame if it is of the form ϕ1 ◦ ϕ2 ◦ · · · ◦ ϕs, where ϕi ∈ GL2(k) ∪ BA2(k) for

i = 1, 2, · · · , s for some s.

3.3 Rentschler's Theorem

From now-on, k denotes a �eld of characteristic zero.

Proposition 3.3.1. Let α : A −→ A be an automorphism of a ring A, let D ∈LND(A), and consider the map δ = α ◦D ◦ α−1 : A −→ A. Then δ ∈ LND(A) and

ker(δ) = α(kerD).

Proof. One can verify directly that δ is a derivation of A. Since D is locally nilpotent

and δn = α ◦ Dn ◦ α−1 for all n ∈ N, it follows that δ is locally nilpotent. For any

x ∈ A we have

δ(x) = 0⇔ α(D(α−1(x))) = 0⇔ D(α−1(x)) = 0⇔ α−1(x) ∈ kerD ⇔ x ∈ α(kerD),


so ker(δ) = α(ker(D)).

De�nition 3.3.2. Let B = A[X1, · · · , Xn] = A[n] where A is a ring. An element

f ∈ B is called a variable of B over A if B = A[f ][n−1]. (In other words, f is a

variable of B over A if there exist f2, · · · , fn ∈ B such that B = k[f, f2, · · · , fn].)

Proposition 3.3.3. Let k be a �eld of characteristic zero and B = k[X, Y ]. If

D ∈ LND(B) \ {0} and X ∈ ker(D), then there exists f ∈ k[X] \ {0} such that

D = f(X)∂

∂Y.

Proof. From Corollary 2.2.23, D is a k-derivation and by Lemma 2.1.13 we have

D = h(X, Y )∂

∂X+ g(X, Y )

∂

∂Y,

where h(X, Y ), g(X, Y ) ∈ B. Since D(X) = 0 we have h(X, Y ) = 0, hence D =

g(X, Y )∂

∂Yand g(X, Y ) 6= 0 (since D 6= 0). It follows that kerD = k[X]. Then

Corollary 2.3.16 (1) implies that g(X, Y ) ∈ kerD, so g(X, Y ) = f(X) ∈ k[X] and we

are done.

Corollary 3.3.4. Let B = k[X, Y ] where k is a �eld of characteristic zero, and

D ∈ LND(B) \ {0}. If there exists f(X) ∈ k[X] \ k (a nonconstant polynomial) such

that f(X) ∈ ker(D), then D = h(X)∂

∂Y, where h ∈ k[X] \ {0}.

Proof. Since f is a non constant polynomial we have f(X)−f(0) 6= 0 and by Corollary

2.2.23D(α) = 0, for all α ∈ k; henceD(f(X)−f(0)) = D(f(X))−D(f(0)) = 0. Thus

f(X)− f(0) = Xg(X) ∈ ker(D), with g(X) 6= 0. The fact that ker(D) is factorially

closed implies that X ∈ ker(D) and by Proposition 3.3.3, we have D = h(X)∂

∂Y.

Remark 3.3.5. The above results (Proposition 3.3.3 and Corollary 3.3.4) remain

valid with X and Y interchanged.

The lemma below is a weak version of the Lemma 4.6 of [6].

Lemma 3.3.6. Let B = k[X, Y ] and ω = (a, b) where k is a �eld of characteristic

zero, and a, b are relatively prime positive integers, and regard B as being endowed with

ω-grading (see Example 1.4.4). Then if f ∈ B is ω-homogeneous with degω f = d and

ab divides d, then there exists a standard homogeneous polynomial g(S, T ) ∈ k[S, T ]

such that f = g(Xb, Y a).


Proof. If a = b = 1, there is nothing to prove. Assume that ab > 1 and let f =∑(i,j)∈N×N ci,jX

iY j with ia + jb = d. Since ab divides d we have ib

+ ja∈ Z, that

isia+ bj

ab∈ Z, so ia = k1b, jb = k2a (k1, k2 ∈ N). Thus b divides i and a divides

j (because a and b are relatively prime). Set l =i

band m =

j

a, l,m ∈ N, f =∑

bl,ma cbl,maXlbY ma = g(Xb, Y a), with g(S, T ) =

∑l,m el,mS

lTm and l + m =d

ab;

therefore g is a standard homogeneous polynomial with degreed

ab.

The next result is an important part of the proof of Theorem 4.1 in [6].

Lemma 3.3.7. Let B = k[X, Y ] and let D ∈ LND(B) \ {0} be such that D(X) 6= 0

and D(Y ) 6= 0. If f is a nonconstant polynomial such that f ∈ ker(D), then there

exists an N-grading of B relative to which the highest degree homogeneous component

f of f has the form f = d(X+cY r)s or f = d′(Y +c′X t)s (c, c′, d, d′ ∈ k× and r, s, t ∈Z+).

Proof. Assume without loss of generality that f(0, 0) = 0, if it is not the case we

replace f by f(X, Y ) − f(0, 0). Thus we can write f as f(X, Y ) = P (X) + Q(Y ) +

XY F (X, Y ), where F ∈ B, P (X) ∈ Xk[X] and Q(Y ) ∈ Y k[Y ].

If P = 0, then f = Y T (X, Y ); since f 6= 0 and ker(D) is factorially closed, we

must have D(Y ) = 0, a contradiction; hence P 6= 0 and by the same argument we

obtain that Q 6= 0.

Now let m = deg(P (X)) and n = deg(Q(Y )), and note that m ≥ 1 and n ≥ 1.

Set e = gcd(m,n), a =n

eand b =

m

e. Let ω = (a, b) and let B be endowed

with the ω-grading. Then we have degω(P (X)) = am = bn = degω(Q(Y )) and so

degω(f) ≥ am.

Let f and F denote the highest degree homogeneous components of f and F

respectively and D the homogenization of D. By Proposition 2.4.14, D ∈ LND(B) \{0} and by Proposition 2.4.15 D(f) = 0. Note that D 6= 0 by Proposition 2.4.11.

Assume that degω(f) > am; then we must have f = XY F , F 6= 0. Since F 6= 0

and ker(D) is factorially closed we must have D(X) = 0 = D(Y ) and so D = 0, a

contradiction. Thus degω(f) ≤ am and so degω(f) = am. Therefore f can be written


as f = uXm + vY n + XY F , for some u, v ∈ k. Furthermore, u 6= 0 and v 6= 0

since they are the highest degree coe�cients of P and Q respectively. Assume that

D(X) = 0, then D(f −uXm) = 0 and so (vY n+XY F ) = Y (vY n−1 +XF ) ∈ ker(D),

thus D(Y ) = 0 (since vY n−1 + XF 6= 0). Therefore D = 0, a contradiction. Thus

D(X) 6= 0 and by the same argument one can show that D(Y ) 6= 0. Since ab divides

am = degω(f) and gcd(a, b) = 1, by Lemma 3.3.6, f(X, Y ) = g(Xb, Y a) where g is a

nonconstant standard homogeneous polynomial with deg(g) =am

ab= e.

Let K denote the algebraic closure of k, then by Lemma 3.1.4 g(X, Y ) factors in

K[X, Y ] as a product of linear polynomials; hence we have f =∏e

i=1(ciXb + diY

a)

where all the ci, di ∈ K, (ci, di) 6= (0, 0). In fact ci, di ∈ K×; if any ci is zero we will

have f = Y H(X, Y ), H ∈ B, thus u must be zero, a contradiction; hence all the ci

are nonzero. By the same argument we can show that di 6= 0.

Set B = K[X, Y ] and let δ ∈ LND(B) be the unique extension of D (see Lemma

2.2.28). Now δ(f) = D(f) = 0 implies that f ∈ ker(δ); f can also be written as f =

C∏e

i=1(Xb + LiYa) where C = c1c2 · · · ce ∈ K×, Li ∈ K×. Hence Xb + LiY

a ∈ ker δ

for all i = 1, · · · , e (since ker δ is factorially closed and each term of the product is

nonzero).

We claim that all the Li are equal. Indeed, assume that Li 6= Lj for some i 6= j,

then (Li−Lj)Y a ∈ ker δ which implies that 0 = δ(Y ) = D(Y ), a contradiction. So all

the Li are equal to, say L. We thus have f = C(Xb + LY a)e with C,L ∈ K×. Sincef(X, Y ) ∈ k[X, Y ] by hypothesis, we have that f(X, Y ) = C

∑el=0

(el

)(Xb)e−l(LY a)l ∈

k[X, Y ]. Recall that k is of characteristic zero, so Q ⊆ k. Thus for l = 0, we obtain

that C ∈ k× and for l = 1 we have that CeL ∈ k× which implies that L ∈ k×. Thusf = C(Xb + LY a)e with C,L ∈ k×; and f ∈ ker(D) implies Xb + LY a ∈ ker(D).

Now assume that a > 1 and b > 1; then

D(Xb + LY a) = 0 =⇒ bXb−1D(X) = −aLY a−1D(Y ).

Hence D(X) ∈ Y B and D(Y ) ∈ XB, so Proposition 2.2.26 implies that D(X) = 0

or D(Y ) = 0, a contradiction. Thus a = 1 or b = 1.

Therefore f = C(X + LY a)e or f = C ′(Y + L′Xb)e where C,C ′, L and L′ ∈ k×

a, b, e ∈ Z+.


We now present Rentschler's Theorem, which is an important result in the theory

of polynomial rings. Our proof follows the reasoning given in Theorem 4.1 of [6], but

our use of the integer ‖D‖ clari�es the argument.

Theorem 3.3.8 (Rentschler's Theorem). Let k be a �eld of characteristic zero. If

D ∈ LND(k[X, Y ]), then there exists f ∈ k[X] and a tame automorphism α ∈ GA2(k)

such that αDα−1 = f(X)∂Y .

Proof. The result is trivial if D = 0, so assume throughout that D 6= 0. If DX = 0,

the result follows from Proposition 3.3.3.

If DY = 0, considering the tame automorphism α = (Y,X), we get α ◦ D ◦α−1(X) = 0 and the result follows from Proposition 3.3.3.

De�ne an equivalence relation ∼ on the set LND \{0} as follows: given D,D′ ∈LND \{0}, we declare that D ∼ D′ if there exists a tame automorphism α of B

satisfying D′ = α ◦D ◦ α−1.

For each D ∈ LND(B)\{0}, note that ker(D)\k 6= ∅ by Lemma 2.3.11 and de�ne

the positive integer ‖D‖ = min {degX(f) + degY (f) | f ∈ ker(D) \ k} .Consider D ∈ LND(B) \ {0} such that D(X) 6= 0 and D(Y ) 6= 0. We may choose

f ∈ ker(D) \ k such that ‖D‖ = degX(f) + degY (f) and f(0, 0) = 0.

By Lemma 3.3.7 there exists an N-grading of B such that we have f(X, Y ) =

d(Y + cXb)e or f(X, Y ) = d(X + cY a)e where f is the highest degree homogeneous

component of f .

Suppose f(X, Y ) = d(Y + cXb)e, note that degX(f) = degX(f) and degY (f) =

degY (f). De�ne the triangular (hence tame) automorphism α = (X, Y−cXb). Clearly

α(f)(0, 0) = 0. We claim:

degX(α(f)) < degX(f) and degY (α(f)) = degY (f). (3)

To prove this, we write f = f + f , where f =∑

i+bj<be aijXiY j (aij ∈ k) and f is as

before. Since α(f) = d(α(Y ) + c(Xb))e = dY e, we have

degX(α(f)) = 0 < be and degY (α(f)) = e. (4)


For each term aijXiY j of f we have α(aijX

iY j) = aijXi(Y−cXb)j, so degX(α(aijX

iY j)) =

i+ bj < be and degY (α(aijXiY j)) = j < e; so

degX(α(f)) < be and degY (α(f)) < e. (5)

Combining (4) and (5), we obtain degX α(f) < be and degY (α(f)) = e, so (3) is

proved.

Set D′ = α ◦D ◦ α−1; we have D′ ∈ LND(B) \ {0} by Proposition 3.3.1, so D ∼ D′.

Again by Proposition 3.3.1 we have α(f) ∈ ker(D′), so (3) implies that ‖D′‖ < ‖D‖.Similarly if f = d(X+ cY a)e, consider the tame automorphism β = (X− cY a, Y ).

Note that α is tame, because β = (Y,X) ◦ (X, Y − cXa) ◦ (Y,X); and clearly

β(f)(0, 0) = 0. Arguing as in the �rst case, we obtain

degY (β(f)) < degY (f) and degX(β(f)) = degX(f).

Set D′ = β ◦D ◦ β−1; then (as in the �rst case) D′ ∼ D and ‖D′‖ < ‖D‖.We have shown the following:

If D ∈ LND(B) \ {0} satis�es D(X) 6= 0 and D(Y ) 6= 0, then there exists

D′ ∈ LND(B) \ {0} such that D′ ∼ D and ‖D′‖ < ‖D‖.

Since this process of lowering ‖D‖ cannot continue inde�nitely, there must exist

D′′ ∈ LND(B) \ {0} such that D′′ ∼ D and either D′′(X) = 0 or D′′(Y ) = 0. If

D′′(Y ) = 0, then (Y,X) ◦D′′ ◦ (Y,X)−1 is equivalent to D′′ and maps X to 0, so in

fact there exists D′′ ∈ LND(B) \ {0} such that D′′ ∼ D and D′′(X) = 0. Proposition

3.3.3 gives D′′ = h(X)∂Y for some h(X) ∈ k[X], so we are done.

Corollary 3.3.9. Given B = k[2] and D ∈ LND(B), there exist X, Y ∈ B such that

B = k[X, Y ] and D = f(X)∂

∂Yfor some f(X) ∈ k[X].

Proof. The case D = 0 is trivial.

We may assume that B = k[Z, T ]; by Rentschler's theorem there is α ∈ GA2(k) such

that α ◦D ◦α−1 = f(Z)∂

∂T, f(Z) ∈ k[Z]\{0}. De�ne X = α−1(Z) and Y = α−1(T ),

and note that B = k[X, Y ] by Lemma 3.2.3. Now α ◦ D ◦ α−1(Z) = 0 implies that


α(D(X)) = 0, so D(X) = 0 (since α is an automorphism). Therefore by Proposition

3.3.3 we have D = h(X)∂

∂Yfor some h(X) ∈ k[X] \ {0}.

Remark 3.3.10. Let B = k[2], where k is a �eld of characteristic zero.

(i) If D ∈ LND(B) \ {0} and A = kerD, then Rentschler's Theorem gives us that

A = k[1] and B = A[1].

(ii) Rentschler's Theorem implies also that if D is a locally nilpotent derivation of

B then D(v) = 0 for some variable v of B.

Corollary 3.3.11. Let B = k[X, Y ] where k is a �eld of characteristic zero. If f is

a variable of B, then f satis�es one of the following conditions:

(a) f = aX + b, where a ∈ k× and b ∈ k.

(b) f = cY + d, where c ∈ k× and d ∈ k.

(c) There exists an N-grading of B relative to which the highest degree homogeneous

component f of f has the form f = d(X+cY r)s or f = d′(Y +c′X t)s (c, c′, d, d′ ∈k× and r, s, t ∈ Z+).

Proof. Since f is variable of B, there exists g ∈ B such that B = k[f, g]. Clearly

there exists D ∈ LND(B) \ {0} such that D(f) = 0.

If D(X) = 0 then ker(D) = k[X] (by Proposition 3.3.3), so f ∈ k[X]. Since

(f, g) is a system of variables of B, Lemma 2.1.20 implies that f ′(X)∂g

∂Y∈ k×, hence

f ′(X) ∈ k× (since k is factorially closed in B). Therefore f = aX + b where a ∈ k×

and b ∈ k.If D(Y ) = 0 then using Remark 3.3.5 and arguing as above, we obtain that

f = cY + d where c ∈ k× and d ∈ k.Assume that D(X) 6= 0 and D(Y ) 6= 0. Since f is a nonconstant element of B

(because f is a variable) and f ∈ ker(D) the result follows by Lemma 3.3.7.

De�nition 3.3.12. Let B = k[X, Y ] = k[2] where k is a �eld of characteristic zero

and f =∑

i,j aijXiY j ∈ B, (aij ∈ k). The support of f is the set: Supp(f) =

{(i, j) ∈ N× N | aij 6= 0}.


Remark 3.3.13. Let k be a �eld of characteristic zero, B = k[X, Y ] and f a variable

of B. Set m = degX(f) and n = degY (f). Then Corollary 3.3.11 implies that the

following hold:

1. n|m or m|n.

2. If f is not of the form aX + b or aY + b with a ∈ k× and b ∈ k, then the convex

hull in R2 of the set Supp(f)∪{(0, 0)} is the triangle with vertices (0, 0), (m, 0)

and (0, n).

•

•

•

Supp(f)

(0, n)

(m, 0)(0, 0)

3.4 Automorphism theorem

Observe �rst that the group GA1(k) is well understood. Indeed, every k-automorphism

φ of k[X] = k[1] is of the form φ(X) = aX + b, a ∈ k×, b ∈ k. This follows from

Proposition 3.1.2 and Remark 3.2.2.

The �rst step in understanding the group GA2(k) was accomplished by Jung in

1942 (see [9]), when he proved that if k is a �eld of characteristic zero then GA2(k)

is generated by its subgroups GL2(k) and BA2(k). In 1953, van der Kulk proved that

this is in fact true for k an arbitrary �eld. The proof that we present here is due to

Rentschler, and is valid only in characteristic zero. In fact we obtain Jung's Theorem

as a corollary to Rentschler's Theorem. Our proof is adapted from that of Theorem

4.8 of [6].

Theorem 3.4.1 (Jung's Theorem). Let k be a �eld of characteristic zero. The

group GA2(k) of k-automorphisms of B = k[X, Y ] is generated by its linear and


triangular subgroups, GL2(k) and BA2(k). In other words, every automorphism of B

is tame.

Proof. Let φ ∈ GA2(k). Let u = φ(X) and v = φ(Y ). Since B = k[u, v] = k[2], the

partial derivatives ∂u = ∂∂u

and ∂v = ∂∂v

are de�ned and we have ∂v ∈ LND(B) and

ker(∂v) = k[u]. Applying Rentschler's Theorem to ∂v, we obtain that there exists a

tame automorphism α of B such that α ◦ ∂v ◦ α−1 = f(X)∂Y for some f(X) ∈ k[X]

(where f(X) 6= 0 since ∂v 6= 0). Moreover,

k[X] = ker(∂Y ) = ker(f(X)∂Y ) = ker(α ◦ ∂v ◦ α−1) = α(ker(∂v)) = α(k[u]),

so k[X] = k[α(u)] and consequently α(u) = aX + b for some a ∈ k×, and b ∈ k, byProposition 3.1.2. Now

k[X, Y ] = k[α(u), α(v)] = k[aX + b, α(v)] = k[X,α(v)]

and the fact that k[X,α(v)] = k[X, Y ] implies that α(v) = cY +g(X) for some c ∈ k×

and g(X) ∈ k[X], by Proposition 3.1.2. Now the automorphism θ = (aX + b, cY +

g(X)) is triangular, hence tame, so α−1 ◦ θ is tame. Now

(α−1 ◦ θ)(X) = α−1(aX + b) = α−1(α(u)) = u = φ(X),

(α−1 ◦ θ)(Y ) = α−1(cY + g(X)) = α−1(α(v)) = v = φ(Y ),

so α−1 ◦ θ = φ, so φ is tame.

As we already mentioned, van der Kulk proved in 1953 that Jung's Theorem is in

fact true over an arbitrary �eld (see [10]). Let us state this as follows (where the last

equality is Remark 3.2.7):

Theorem 3.4.2 (van der Kulk). If k is a �eld then

GA2(k) = 〈GL2(k) ∪ BA2(k)〉 = 〈Af2(k) ∪ BA2(k)〉.

In the same article, van der Kulk also proves a unique factorization theorem for

elements of GA2(k). Theorem 3.4.4, below, is equivalent to van der Kulk's result.


De�nition 3.4.3. Let H be a subgroup of a group G such that H 6= G. By a system

of nontrivial left coset representatives of H in G, we mean a subset G ⊆ G satisfying:

1. G ∩H = ∅;

2. for each left coset gH (where g ∈ G) such that gH 6= H, we have |G ∩ gH| = 1.

Theorem 3.4.4 (Structure theorem). Let k be a �eld and consider the subgroups

A = Af2(k), B = BA2(k) and C = A ∩B of GA2(k), that is

A = {(a1X + b1Y + c1, a2X + b2Y + c2) | ai, bi, ci ∈ k; a1b2 − a2b1 6= 0}

B ={

(aX + b, cY + f(X)) | a, c ∈ k×, b ∈ k, f ∈ k[X]}

C = A ∩B ={

(aX + b, cX + dY + e) | a, d ∈ k×, b, c, e ∈ k}.

De�ne the sets A ⊆ A and B ⊆ B by

A = {(tX + Y,X) | t ∈ k} and B ={

(X, Y +X2f(X))∣∣ f(X) ∈ k[X], f 6= 0

}.

Then A (resp. B) is a system of non trivial left coset representatives of C in A (resp.

of C in B). Moreover, A∩B = ∅ and each φ ∈ GA2(k) has a unique factorization of

the form

φ = x1 · · ·xnc (6)

with n ≥ 0, c ∈ C and xi ∈ A ∪ B for all i, where the xi alternate between A and B.

We now proceed to prove the Structure Theorem. Our proof makes use of Theorem

3.4.2, even though we only proved it for �elds of characteristic zero. In fact the proof

shows that if k is any �eld for which GA2(k) = 〈Af2(k)∪BA2(k)〉, then the Structure

Theorem is true over k.

Our initial plan was to reproduce the proof given in Freudenburg's book [6], but

there is a step in that proof that we could not understand (see the introduction).

Resolving this issue resulted in the proof that we give below, and that is quite di�erent

from that of [6].

There are several steps in the proof. We begin with:


Lemma 3.4.5. Let A and B be subgroups of a group G satisfying G = 〈A ∪ B〉,A * B and B * A. Let C = A ∩ B, and let A ⊂ A (resp. B ⊂ B) be a system of

nontrivial left coset representatives of C in A (resp. of C in B). Then A ∩ B = ∅and each g ∈ G has a (not necessarily unique) factorization of the form

g = x1 · · · xnc (7)

with n ≥ 0, c ∈ C and xi ∈ A ∪ B for all i, where the xi alternate between A and B.

Proof. Note that C 6= A and C 6= B, since A * B and B * A. By De�nition 3.4.3

we have A ⊆ A and B ⊆ B, thus A ∩ B ⊆ A ∩ B = C. Assume that A ∩ B 6= ∅.Let g ∈ A ∩ B ⊆ C, then we have g ∈ C. De�nition 3.4.3-1) implies that g /∈ C, acontradiction. Thus we must have A ∩ B = ∅.

Let g ∈ G. If g ∈ C then it is trivial that g factors as in (7) (take n = 0). Assume

now that g ∈ G \ C. We shall now prove that g can be factored as g = g1g2 · · · gnwith n ≥ 1, gi ∈ A ∪ B and gi /∈ C for all i, and where the gi alternate between A

and B.

Since g ∈ G \ C and G = 〈A ∪B〉, g can be factored as

g = g1 · · · gn, with gi ∈ A ∪B for all i. (8)

Among all these factorizations (8) of g, choose one that minimizes n. Then we cannot

have consecutive factors gi−1 and gi both in A or both in B, because then we could

replace gi−1 and gi by their product and obtain a shorter factorization (8) of g, which

would violate minimality of n. Similarly, if gi ∈ C and i > 1 (resp. i = 1) then

replacing gi−1 and gi (resp. gi and gi+1) by their product would produce a shorter

factorization (8) of g, which is impossible. So we obtain that the gi alternate between

A and B and that gi /∈ C for all i.

Furthermore, for every element a ∈ A \ C we have a = xc for some x ∈ A and

c ∈ C. Also for every element b ∈ B \ C we have b = xc for some x ∈ B and c ∈ C.Observe that g1 belongs to A\C or to B\C. Assume now without loss of generality

that g1 ∈ A \ C. Then g1 = x1c1 with x1 ∈ A and c1 ∈ C. So

g = g1g2 · · · gn = (x1c1)g2 · · · gn = x1(c1g2)g3 · · · gn = x1g′2g3 · · · gn


where g′2 = c1g2 ∈ B \ C. Then g′2 = x2c2, where x2 ∈ B and c2 ∈ C. Therefore

g = x1x2g′3 · · · gn with g′3 ∈ A \ C. We can repeat inductively this process to get at

the end that g = x1x2 · · · xnc, where the xi ∈ A ∪ B alternate between A and B and

c ∈ C.

In what follows, k is any �eld and A, B are de�ned as in the Structure Theorem,

i.e.,

A = {(tX + Y,X) | t ∈ k} and B ={

(X, Y +X2f(X))∣∣ f(X) ∈ k[X], f 6= 0

}.

We also de�ne

U ={

(X + tY, Y ) | t ∈ k×}, V = {(Y,X − tY ) | t ∈ k} and A = A ∪ U ∪ V .

De�nition 3.4.6. Let φ = (F,G) ∈ GA2(k).

• We de�ne deg φ = max{deg(F ), deg(G)}.

• If deg(F ) ≥ deg(G), we say that φ is of type 1.

• If deg(F ) < deg(G), we say that φ is of type 2.

Note that each element of GA2(k) is either of type 1 or of type 2. All elements of Aare of type 1 and all elements of B are of type 2.

Lemma 3.4.7. Let φ ∈ GA2(k) and h ∈ A∪B. If φ and h are of distinct types, then

1. the type of φ ◦ h is the same as that of h;

2. deg(φ ◦ h) = deg φ deg h.

Proof. Observe �rst that A ∩ B = ∅ and the sets A, U and V are pairwise disjoint.

Since φ and h are of distinct types, it su�ces to show the Lemma in the following

cases:

Case 1: φ = (F,G) is of type 1 and h = (X, Y + g(X)) ∈ B. It is easy to show that

φ ◦ h = (F,G+ g(F )) = (P,Q).

Since deg(F ) ≥ deg(G) and deg(g) > 1, we have deg(Q) = deg(g(F )) =

deg(g) deg(F ) > deg(F ) = deg(P ), i.e., φ ◦ h is of type 2 (the type of h)

and deg(φ ◦ h) = deg(Q) = deg(F ) deg(g) = deg φ deg h.


Case 2: φ = (F,G) is of type 2 (deg(φ) = deg(G)) and h = (tX + Y,X) ∈ A. We have

φ ◦ h = (tF +G,F ) = (P,Q),

so deg(φ ◦ h) = deg(G) = deg(P ) (since deg(G) > deg(F )). Thus φ ◦ h is of the

same type as h and deg(φ ◦ h) = deg(φ) deg h.

Case 3: φ = (F,G) is of type 2 and h = (X + tY, Y ) ∈ U . Likewise

φ ◦ h = (F + tG,G) = (P,Q),

since t 6= 0 and deg(G) > deg(F ), we have deg(P ) = deg(Q). Hence φ ◦ h is of

the same type as h and deg(φ ◦ h) = deg(G) = deg(φ) = deg(φ) deg(h).

Case 4: φ = (F,G) is of type 2 and h = (Y,X − tY ) ∈ V . Thus

φ ◦ h = (G,F − tG) = (P,Q)

and so degP ≥ degQ (since degG > degF ). Thus φ ◦ h is of the same type as

h and deg(φ ◦ h) = deg(P ) = deg(G) = deg(φ) = deg(φ) deg(h).

Lemma 3.4.8. Suppose that φ ∈ GA2(k) satis�es

φ = h1 ◦ · · · ◦ hn

where n ≥ 0 and hi ∈ A ∪ B for all i, where the hi alternate between A and B. Thendeg φ =

∏ni=1 deg hi, and if n > 0 then φ has the same type as hn.

Proof. If n = 0 then �φ = h1◦· · ·◦hn� should be interpreted as φ = id, and the empty

product∏n

i=1 deg hi is by convention equal to 1. Since deg(id) = 1, the Lemma is

true when n = 0. From now on we assume that n ≥ 1. Let us prove by induction on

j that the following statement P (j) is true for all j = 1, . . . , n:

P (j) : φjdef

= h1 ◦ · · · ◦ hj has the same type as hj, and deg φj =∏j

i=1 deg hi.


It is obvious that P (1) is true. Suppose that 1 < j ≤ n and that P (j − 1) is true;

let us prove that P (j) is true. By P (j − 1), φj−1 has the same type as hj−1 and

deg φj−1 =∏j−1

i=1 deg hi. Since hj−1 and hj have distinct types, it follows that φj−1

and hj have distinct types; this together with hj ∈ A ∪ B implies (by Lemma 3.4.7)

that φj = φj−1 ◦ hj has the same type as hj and that

deg φj = deg φj−1 deg hj =∏j

i=1 deg hi.

So P (j) is true. By induction, this shows that P (1), . . . , P (n) are true.

Since P (n) is true, the Lemma is proved.

Remark 3.4.9.

1. If h ∈ A then h−1 ∈ V ⊆ A.

2. If h, h′ are distinct elements of A then (h′)−1 ◦ h ∈ U ⊆ A.

3. If h ∈ B then h−1 ∈ B.

4. If h, h′ are distinct elements of B then (h′)−1 ◦ h ∈ B.

Lemma 3.4.10. If h1h2 · · ·hnc = h′1h′2 · · ·h′mc′, where n,m ∈ N, the hi and h′i alter-

nate between A and B and c, c′ are elements of C, then we have n = m, hi = h′i and

c = c′.

Proof. We show this by induction on l = min(n,m) ≥ 0. Let P (l) be the following

statement:

P (l) : If min(m,n) = l then the condition h1 · · ·hnc = h′1 · · ·h′mc′ implies

that n = m, hi = h′i for all i, and c = c′.

We begin by showing that P (0) and P (1) are true.

Case 1: l = 0.

• If n = 0 and m > 0, then we have c = h′1 · · ·h′mc′, so

cc′−1 = h′1 · · ·h′m.


If m = 1, then we must have h′1 ∈ C, a contradiction (since A∩C = ∅ andB ∩ C = ∅).

If m > 1, then by Lemma 3.4.8 we have 1 = Πmi=1 deg(h′i). Hence h′i ∈ A

for all i, a contradiction (since h′i alternate between A and B).

• If n > 0 and m = 0 by applying the same argument above we obtain a

contradiction.

Hence we must have n = m = 0 and therefore c = c′, so P (0) is true.

Case 2: l = 1.

If h1 = h′1 then h2 · · ·hnc = h′2 · · ·h′mc′, so the fact that P (0) is true implies

that n = m = 1 and c = c′. This shows that P (1) is true whenever h1 = h′1.

We now assume that h1 6= h′1, and we show that this leads to contradictions in

all possible cases. This will complete the proof that P (1) is true.

• If n = 1 and m > 1, then we have h1c = h′1 · · ·h′mc′, so

cc′−1 = h−11 h′1 · · ·h′m.

If h1 ∈ A and h′1 ∈ B, then h−11 , h′1, h

′2, · · · , h′m alternate between A and B.

Thus we can apply the Lemma 3.4.8 and get that 1 = deg(h−11 )Πm

i=1 deg(h′i).

Hence we have deg(h′i) = 1 for all i and so h′i ∈ A for all i = 1, 2, · · · ,m,

a contradiction.

If h1 ∈ B and h′1 ∈ A, then h−11 , h′1, h

′2, · · · , h′m alternate between A and B.

Thus we can apply the Lemma 3.4.8, and obtain the same contradiction.

Hence h1 and h′1 must both belong to the same set A or B.

If {h1, h′1} ⊆ A, then since h1 6= h′1 by assumption, we have h−1

1 h′1 ∈ Aby Remark 3.4.9; so (h−1

1 h′1), h′2, · · · , h′m alternate between A and B. ThusLemma 3.4.8 implies 1 = deg(h−1

1 h′1)Πmi=2 deg(hi), and so deg(h′2) = 1, a

contradiction.

Assume that {h1, h′1} ⊆ B. Since h1 6= h′1 by assumption, we have h−1

1 h′1 ∈B by Remark 3.4.9. Then (h−1

1 h′1), h′2, · · · , h′m alternate between A and B.


Lemma 3.4.8 implies 1 = deg(h−11 h′1)Πm

i=2 deg(h′i), and so 1 = deg(h−11 h′1),

a contradiction.

• If n > 1 and m = 1 by applying the same argument above we obtain a

contradiction.

Hence we must have n = m = 1. We thus have h1c1 = h′1c′1, so

cc′−1 = h−11 h′1.

If h1 and h′1 alternate between A and B, then h−1

1 and h′1 alternate between

A and B. So by Lemma 3.4.8 we have 1 = deg(h−11 ) deg(h′1) and thus h1

and h′1 are elements ofA, a contradiction. Therefore h1 and h′1 must belong

to the same set A or B.

Since h1 and h′1 are distinct, we have h

−11 h′1 ∈ U ∪B by Remark 3.4.9 hence

cc′−1 ∈ U ∪ B which is impossible since (U ∪ B) ∩ C = ∅.

These contradictions show that P (1) is true.

Let d > 1. Assume that P (l) is true for 0 ≤ l < d. We show that P (d) is true. So

assume that h1 · · ·hnc = h′1 · · ·h′mc′ with min(n,m) = d.

If h1 = h′1 then h2 · · ·hnc = h′2 · · ·h′mc′, so the fact that P (d − 1) is true implies

that n = m, c = c′ and hi = h′i for all i. This shows that P (d) is true whenever

h1 = h′1. We now assume that h1 6= h′1, and we show that this leads to contradictions

in all possible cases. This will complete the proof of the Lemma.

We have

cc′−1 = h−1n · · ·h−1

2 h−11 h′1 · · ·h′m. (9)

Consider the case where (h1 ∈ B and h′1 ∈ A) or (h1 ∈ A and h′1 ∈ B).Then the h−1

i and h′i (in the right hand side of (9)) alternate between A and B so we

can apply Lemma 3.4.8. Equation (9) implies that

1 =

(n∏i=1

deg(h−1i )

)(m∏i=1

deg(h′i)

),

so h1, h′1 ∈ A, a contradiction. Hence h1 and h′1 must both belong to one of the sets

A or B.


Assume that {h1, h′1} ⊆ A. Since h1 and h′1 are distinct then h−1

1 h′1 ∈ U ⊆ A (by

Remark 3.4.9). So

h−1n , . . . , h−1

2 , (h−11 h′1), h′2, . . . , h

′m alternate between A and B. (10)

Hence we can apply Lemma 3.4.8 and obtain

1 =

(n∏i=2

deg(h−1i )

)deg(h−1

1 h′1)

(m∏i=2

deg(h′i)

). (11)

Hence h2, h′2 ∈ A, a contradiction. (Note that h2 and h′2 exist since d ≥ 2.)

Assume that {h1, h′1} ⊆ B. Since h1 and h′1 are distinct h−1

1 h′1 ∈ B (by Remark

3.4.9). Hence condition (10) is satis�ed and we can apply Lemma 3.4.8 and obtain

that (11) is true in this case too. Hence deg(h−11 h′1) = 1, a contradiction.

These contradictions show that P (d) is true and complete the proof of the Lemma.

Proof of the Structure Theorem. We begin by verifying that the hypothesis of

Lemma 3.4.5 is satis�ed. It is clear that A * B and B * A, and we have GA2(k) =

〈A ∪B〉 by Theorem 3.4.2.

Let us show that A (resp. B) is a system of nontrivial left coset representatives of

C in A (resp. of C in B). Clearly A ⊆ A, B ⊆ B, A ∩ C = ∅ and B ∩ C = ∅. Givenα = (a1X + b1Y + c1, a2X + b2Y + c2) ∈ A \C and β = (aX + b, cY + f(X)) ∈ B \C,we have b1 6= 0, c 6= 0 and f(X) = r + sX + X2g(X), for some r, s ∈ k and

g(X) ∈ k[X] \ {0}. It is not di�cult to check that

α = (a1b−11 X + Y,X)(b1X + c1, b2X + (a2b1 − b2a1)b−1

1 Y + c2) (12)

and

β = (X, Y + c−1X2g(X))(aX + b, cY + r + sX). (13)

We can now show that |A∩αC| = 1. Set α′ = (a1b−11 X+Y,X), and note that α′ ∈ A.

We have A∩ αC = A∩ α′C (by Equation (12)). Clearly α′ ∈ A∩ α′C. Assume that

there is some c ∈ C, c 6= 1 such that α′c ∈ A, hence we must have c ∈ U (de�ned just

before De�nition 3.4.6), a contradiction. Therefore we have A∩αC = A∩α′C = {α′}and so |A ∩ αC| = 1.


Likewise, using the same argument as above we can show that |B ∩ βC| = 1.

So A (resp. B) is a system of nontrivial left coset representatives of C in A (resp.

of C in B).

Therefore by Lemma 3.4.5, every g ∈ GA2(k) has a factorization of the form

g = x1 · · ·xnc,

with n ≥ 0, c ∈ C and xi ∈ A ∪ B for all i, where the xi alternate between A and B.By Lemma 3.4.10, this factorization is unique. So we are done.

Remark 3.4.11. In group theory there is a notion of free product with amalgamation,

which we will not de�ne here. It turns out that the above Structure Theorem is

equivalent to the statement that GA2(k) is the free product of Af2(k) and BA2(k)

with amalgamation along C = Af2(k) ∩ BA2(k), which is written as follows:

GA2(k) = Af2(k) ∗C BA2(k).

According to [6], page 84, Nagata (in 1972) seems to be the �rst to have stated

and proved the Structure Theorem in terms of amalgamated free product, but the

statement had appeared without proof in a 1966 paper by Shafarevich.

Chapter 4

Some remarks on polynomial rings in

three variables

The preceding chapter gives a complete description of the locally nilpotent derivations

and automorphisms of k[X, Y ]. In this very short chapter we make a few comments

on the same problems for k[X, Y, Z].

4.1 Locally nilpotent derivations

In this section, k is a �eld of characteristic zero.

Given any n ≥ 1, one can state:

Problem. Describe all locally nilpotent derivations of k[n].

The case n = 1 of this problem is trivial, and Rentschler's Theorem completely

solves the case n = 2. However, the problem is open for all n ≥ 3.

Our aim, in this section, is to explain why Rentschler's Theorem cannot be ex-

tended to n = 3. More precisely, we will see that a certain consequence of Rentschler's

Theorem is false in the case n = 3.

De�nition 4.1.1. Let B = k[n] and D ∈ Derk(B).

1. The corank of D is the maximum integer i such that there exists a system of

variables (X1, · · · , Xn) of B satisfying {X1, · · · , Xi} ⊆ ker(D).

89

CHAPTER 4. POLYNOMIAL RINGS IN THREE VARIABLES 90

2. The rank of D is de�ned by rank(D) = n− corank(D).

Observe in particular that, if B = k[n] and D ∈ Derk(B),

rank(D) < n⇐⇒ some variable of B belongs to ker(D).

As noted in Remark 3.3.10 (ii), Rentschler's Theorem implies that every locally

nilpotent derivation of k[2] has a variable in its kernel. So the following is a conse-

quence of Rentschler's Theorem:

Corollary 4.1.2. Every locally nilpotent derivation of k[2] has rank < 2.

It is interesting to ask if it the case that every locally nilpotent derivation of k[n]

has rank strictly less than n. For n ≥ 3, the answer was not known until, in 1995,

Freudenburg produced the following example on B = k[X, Y, Z] having rank 3.

Theorem 4.1.3. Let B = k[X, Y, Z] = k[3], F = XZ−Y 2 and G = ZF 2 +2X2Y F +

X5. De�ne the k-derivation ∆ : B −→ B, by ∆ = ∆(F,G). Then ∆ ∈ LND(B),

ker ∆ = k[F,G] and rank(∆) = 3.

Proof. This is Theorem 5.19 of [6].

Remark 4.1.4. This explains the claim that we made at the beginning of this section:

Rentschler's Theorem cannot be extended to n = 3, since one of its consequences

(corollary 4.1.2) is false when n = 3.

4.2 Automorphisms

De�nition 4.2.1. Let B = k[X1, · · · , Xn] = k[n]. The group of k-automorphisms of

B is called the general a�ne group of dimension n. It is denoted by GAn(k).

According to Theorem 1.2.4, Remark 3.2.2 of Section 3.2 can be extended for the

case n > 2. Thus every element of ϕ ∈ GAn(k) is completely determined by the

n-tuple (ϕ(X1), · · · , ϕ(Xn)).

As in Section 3.2, we de�ne some subgroups of GAn(k).


De�nition 4.2.2. Let A = (aij) ∈ GLn(k) and a = (a1, · · · , an) ∈ kn.

1. The element of ϕ ∈ GAn(k) de�ned by ϕ(Xi) = a1iX1 + · · ·+aniXn (1 ≤ i ≤ n)

is denoted by ϕA.

The set

GLn(k) = {ϕA | A ∈ GLn(k)} ,

is a subgroup of GAn(k) and is called the subgroup of linear automorphisms of

dimension n.

2. The element ϕ ∈ GAn(k) de�ned by ϕ(Xi) = Xi + ai (1 ≤ i ≤ n) is denoted by

ϕa.

The set

T = {ϕa | a ∈ kn} ,

is a subgroup of GAn(k).

3. The subgroup 〈GLn(k) ∪ T 〉 of GAn(k), generated by GLn(k) and T , is calledthe a�ne linear subgroup of dimension n. It is denoted by Afn(k).

De�nition 4.2.3. Let BAn(k) be the subgroup of GAn(k) whose elements are the

automorphisms φ ∈ GAn(k) of the form

φ =(a1X1 + f1, . . . , anXn + fn

),

where a1, . . . , an ∈ k×, f1 ∈ k, and fi ∈ k[X1, · · · , Xi−1] for 1 < i ≤ n.

De�nition 4.2.4. Let ϕ ∈ GAn(k). One says that ϕ is a tame automorphism if ϕ

belongs to the subgroup 〈GLn(k) ∪ BAn(k)〉 = 〈Afn(k) ∪ BAn(k)〉 of GAn(k). One

says that ϕ is wild if it is not tame.

Therefore the following can be stated:

Problem. Is GAn(k) = 〈GLn(k) ∪ BAn(k)〉? In other words is every element of

GAn(k) tame?


Jung's Theorem (Theorem 3.4.1), proved in 1942, tells us that the answer to this

question is yes if n = 2. The case n = 3 remained open until 2004.

In 1972, Nagata de�ned the following automorphism σ of k[X, Y, Z] = k[3]:

σ(X) = X − 2(XZ + Y 2)Y − (XZ + Y 2)2Z,

σ(Y ) = Y + (XZ + Y 2)Z,

σ(Z) = Z

and conjectured:

Conjecture (Nagata, [15]). σ is not tame!

In 2004, Shestakov and Umirbaev showed in [21] that Nagata's conjecture was

true (assuming that k has characteristic zero). This also proved that

GA3(k) 6= 〈Af3(k) ∪ BA3(k)〉.

Therefore, not all automorphisms of k[3] are tame. It is still an open problem to

describe the structure of the group GA3(k).

Bibliography

[1] M. F. Atiyah and I. G. MacDonald, Introduction to Commutative Algebra,

Univ. Oxford, (1969).

[2] A. Baker, An Introduction to Galois Theory, Lecture Notes, Univ. Glasgow,

(2013).

[3] D. Daigle, Introduction to locally nilpotent derivations, Lecture notes (2010).

Available on the website http://aix1.uottawa.ca/∼ddaigle/

[4] David S. Dummit and Richard M. Foote, Abstract Algebra, Third edition

Wiley-Interscience, New York (1957).

[5] van den Essen, Polynomial Automorphisms and the Jacobian Conjecture,

Birkhauser 190 (2000), 1-11.

[6] G. Freudenburg, Algebraic Theory of Locally Nilpotent Derivations, Berlin

Heidelberg, Springer-Verlag 136 (2006).

[7] M. Hall, Jr., The Theory of Groups, Macmillan Company, (1959).

[8] A. Hatcher, Algebraic Topology, Cambridge University Press 2 (2002) ISBN

0-521-79160-0.

[9] H. W. E. Jung, Über ganze birationale Transformationen der Ebene, J. Reine

Angew. Math. 184 (1942), 161-174.

[10] W. van der Kulk, On polynomial rings in two variables, Nieuw Arch. Wisk. 1

(1953), 33-41.

93

BIBLIOGRAPHY 94

[11] E. Kunz, Introduction to Commutative Algebra and Algebraic Geometry, Mod-

ern Birkhauser Classics (1985).

[12] C. F. Miller, Combinatorial Group Theory, Lecture Notes, Univ. Melbourne

(2004).

[13] J. S. Milne, Fields and Galois Theory, Lecture Notes. Available at

www.jmilne.org/math/, 4.51 (2015).

[14] P. Morandi, Field and Galois Theory, Graduate Texts in Mathematics,

Springer 167 (1996).

[15] N. Masayoshi, On automorphism group of k[x, y], Department of Mathemat-

ics, Kyoto University, Lectures in Mathematics, No. 5, Kinokuniya Book-Store

Co., Ltd., Tokyo, v+53 (1972).

[16] H. Neumann, Generalized free products with amalgamated subgroups I, Amer.

J. Math. 70 (1948), 590-625.

[17] A. Nur, Locally nilpotent derivations and the cancellation problem in a�ne

algebraic geometry, M.Sc. thesis, Univ. Ottawa (2011).

[18] R. Rentschler, OpÃ c©rations du groupe additif sur le plan a�ne, C.R. Acad.

Sc. Paris, 267 (1968).

[19] J. Rotman, The Theory of Groups: An Introduction, Allyn and Bacon Series

in Advanced Mathematics (1965).

[20] R. Y. Sharp, Steps in Commutative Algebra, Second edition, London Mathe-

matical Society, Student Texts 51, 2000.

[21] P. Shestakov and U. Umirbaev, The tame and the wild automorphisms of

polynomial rings in three variables, American Mathematical Society, 17, 197-

227 (2004).

locally nilpotent derivations on polynomial rings in …€¦ · locally nilpotent derivations are...

Documents