Federal University of Minas Gerais
Department of Mathematics
The Jacobian Conjecture a` la Zp
Wodson Mendson
Submitted in partial fulfillment of the requirements for
the degree of Master of Science in Mathematics at
Federal University of Minas Gerais
Advisor: Israel Vainsencher
Belo Horizonte
2018
Abstract
We survey some classical results about the Jacobian Conjecture and formulate the Invariance Conjecture. We
show that this new conjecture is equivalent, in some sense, to the Jacobian Conjecture. We make a contribution
to the Unimodular Conjecture, cf. [12, Essen-Lipton]. We define a new class of rings: unimodular rings and
the invariant rings. The main objective is to show the close relationship between the Jacobian Conjecture and a
related statement over the p-adic integers Zp.
Resumo
Estudamos alguns resultados cla´ssicos na direc¸a˜o da Conjectura do Jacobiano e formulamos uma nova conjectura:
a Conjectura da Invariaˆncia. Mostramos que tal conjectura e´ equivalente, em um certo sentido, a` Conjectura do
Jacobiano. Apresentamos algumas contribuic¸o˜es a` Conjectura Unimodular, cf. [12, Essen-Lipton], e definimos uma
nova classe de ane´is: ane´is invariantes e ane´is unimodulares. Nosso objetivo principal consiste em mostrar a forte
relac¸a˜o entre a Conjectura do Jacobiano e os inteiros p-a´dicos Zp.
Acknowledgements
Gostaria de agradecer
ao Israel Vainsencher pela competente orientac¸a˜o, desde os tempos de IC, e por sempre estar disposto (e animado)
a discutir matema´tica. Agradec¸o tambe´m pelas sugesto˜es e correc¸o˜es feitas no texto.
aos meus pais pelo apoio e incentivo aos meus estudos.
ao Jeroen van de Graaf por me apresentar ao professor Israel.
a` FAPEMIG pelo apoio financeiro.
Le juge: Accuse´, vous taˆcherez d’eˆtre bref.
L’accuse´: Je taˆcherai d’eˆtre clair.
Introduction
The following theorem is well known in the student world.
Inverse Function Theorem. Let f : Rn −→ Rn be an e´tale map1. Then f is a local homeomorphism.
We remark that “ local ” can not be replaced by “ global ”. Indeed, if we consider
f = (eXcos(Y ), eXsin(Y )) : R2 −→ R2
then det Jf = e2X and so f is an e´tale map. But note that f(0, 0) = f(0, 2pi).
In this dissertation we will study topics related to the global inverse of polynomial maps.
Let k be a field and F : kn −→ kn a polynomial map i.e. F1, . . . , Fn ∈ k[X1, . . . , Xn] where Fi is the i-th
component of F . We say that F is an invertible polynomial map if there exists a polynomial map G : kn −→ kn,
such that F (G(X)) = X and G(F (Y )) = Y .
Linear maps are particular cases where each component is a homogeneous polynomial of degree 1. In this case
it is easy to check that F is an invertible map if and only if F is injective. Furthermore, if Fi =
∑
j AijXj and
M = (Aij) is the associated matrix, we have that F is an invertible map if only if M ∈ GLn(k). Note that M is
the jacobian matrix associated to F .
We can ask whether this is more general. More precisely, let F : kn −→ kn be a polynomial map without restriction
on deg(F ). Denote by JF ∈Mn(k[X1, . . . , Xn]) the associated jacobian matrix.
Question 1. Is F invertible if and only if JF ∈ GLn(k[X1, . . . , Xn])?
Question 2. Is F invertible if and only if F is injective?
In this generality, we can find examples that answer negatively. Indeed, taking k = Fp and F : k −→ k the
polynomial map with x 7→ xp +x we have JF = 1 but F is not invertible. If k = Q, the map F : k −→ k given by
x 7→ x3 is injective but isn’t invertible. Now if we consider k, an algebraically closed field with char(k) = p > 0,
1i.e. f is C∞ and the map in tangent spaces dpf : TpRn −→ Tf(p)Rn are isomophism for all p ∈ Rn.
6
and the map f : kn −→ kn given by (X1, . . . , Xn) 7→ (Xp1 , . . . , Xpn) then f is injective but f is not an isomophism
since the induced map of k-algebras k[Xp1 , . . . , X
p
n] ⊂ k[X1, . . . , Xn] isn’t an isomophism. So, the questions above
has relation with the structure of the field k.
We can suspect that something interesting occurs when k is algebraically closed field and char(k) = 0. Indeed,
formulated initially by Keller in 1939, the first question is an open problem 1 known as Jacobian Conjecture or
Keller Problem. It is a problem well known for its easy formulation in spite of being very difficult. Many results
in positive direction make use of deep theorems of algebraic geometry. For example, in case k is an algebraically
closed field and char(k) = 0, the answer for the 2nd question above is YES. The proof involves Zariski’s Main
Theorem (cf.[8, Theorem 8.45]).
The Keller Problem also is known by the many wrong proofs that can be found on Internet. We quote the paper
“Proof Of Two Dimensional Jacobian Conjecture” by Yucai Su (2005). In 2006, Moh2 published a paper with the
objective of indicating the mistakes in Su’s paper. Most interesting is his remark made at the end:
“ A few words for Mr. Su
The problem of Jacobian Conjecture is very hard. Perhaps it will take human being another 100 years to solve
it. Your attempt is noble, Maybe the Gods of Olympus will smile on you one day. Do not be too disappointed.
B. Sagre has the honor of publishing three wrong proofs and C. Chevalley mistakes a wrong proof for a correct
one in the 1950’s in his Math Review comments, and I.R. Shafarevich uses Jacobian Conjecture (to him it is a
theorem) as a fact. You are in a good company. One only remembers the correct statements from Scientists and
Mathematicians, nobody remembers the wrong ones.”
In the present text, we survey a few aspects of the questions above.
In the first part (chapter 1,2), we are interested in results about reductions. The main point in this part is
to show that for the Jacobian Conjecture is sufficient to consider polynomial maps of type F = X + H where
H = (H1, . . . ,Hn) is homogeneous of degree 3 with jacobian matrix JH nilpotent.
Also, we discuss the 2nd question for the case char(k) = 0 and algebraically closed.
In the second part (chapter 3), we present some reformulations of the Jacobian Conjecture. We show, following
[12, Essen-Lipton], that the Jacobian Conjecture is equivalent to the Unimodular Conjecture. We make some
contribution to the Unimodular Conjecture, and we define two new classes of rings: unimodular rings and invariant
rings. The main objective is to show the close relationship between the Jacobian Conjecture and p-adic integers
Zp. This part contains new results that are denoted by WM. For example, Theorem 3.15 (WM) means that
Theorem 3.15 is a new result.
1This is the 16o problem in Steve Smale list: “Mathematical problems for the next century.”
2https://arxiv.org/pdf/math/0604049.pdf
Contents
Abstract 2
Resumo 3
Acknowledgements 4
Introduction 6
1 Polynomial Maps and the Keller Problem 1
1.1 Polynomial maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 A computational criterion for invertibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.2.1 The criterion of Essen . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.3 Invertible maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.3.1 A refinement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
1.4 Particular cases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
2 Reductions 21
2.1 Classical reduction theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
2.2 Druzkowski maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
2.3 The symmetric case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
3 Jacobian Conjecture via Zp 31
3.1 Completion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
8
3.2 A reformulation of the Jacobian Conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
3.3 The Unimodular Conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
3.3.1 n-dimensional 2-sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
3.4 The Invariance Conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
3.5 Some results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
3.5.1 A refinement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
4 Appendix 50
4.1 k-algebras of finite type . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
4.2 Ka¨hler differentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
4.3 Finite maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
4.4 Normalization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
4.5 Bounds for k-points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
4.6 Ramification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
5 Some problems 63
Bibliography 65
9
Chapter 1
Polynomial Maps and the Keller
Problem
Let R be a domain and F : Rn −→ Rn a polynomial map. In this chapter we will study the following questions:
Question 1. Is F invertible if and only if F is injective?
Question 2. Is F invertible if and only if the jacobian matrix JF is in GLn(R[X1, . . . , Xn])?
We will show that if R = k, an algebraically closed field with char(k) = 0, the answer to first questions is YES.
The second question is false for char(R) 6= 0. In case char(R) = 0 it is an open problem (for n ≥ 2):
Jacobian Conjecture. (or Keller Problem) Let R be a domain with char(R) = 0 and F : Rn −→ Rn a polynomial
map with JF ∈ GLn(R[X1, . . . , Xn]). Then, F is invertible.
1.1 Polynomial maps
In this section R will denote a domain.
We will denote by MPn(R) the collection of polynomial maps F : Rn −→ Rn. Note that we can equip
MPn(R) with a graded structure, via the graded structure of the ring R[X1, . . . , Xn]. More precisely,MPn(R) =⊕
d∈NMPn(R)d where MPn(R)d = {F = (F1, . . . , Fn) ∈ MPn(R) | F1, . . . , Fn are homogeneous of degree d }.
Given F ∈MPn(R) we define deg(F ) := Max{deg(F1), . . . , deg(Fn)} where deg(Fk) is the degree of Fk.
1
1.1. Polynomial maps 2
Given F ∈ MPn(R) we will denote by JF the jacobian matrix associated to F and we say that F is a Keller
map if det JF = 1. We say that F is invertible if there exists a polynomial map G : Rn −→ Rn such that
G(F (X)) = X and F (G(Y )) = Y .
Some notations:
• MPn(R)(0) := {F ∈MPn(R) | F (0) = 0}.
• MPn(R)(1) := {F ∈MPn(R)(0) | J(F )(0) = idn}.
• MPIn(R) := {F ∈MPn(R) | F is invertible}.
Remark 1. To show that F ∈MPn(R) is invertible we can replace it by G ◦F ◦H where H and G are invertible
maps. Furthemore, if F is a Keller map, we can suppose that F is of the form F = X+H where H = (H1, . . . ,Hn)
is a polynomial map with deg(Hi) ≥ 2 for all i.
Definition 1.1. Let F ∈ MPn(R). We say that F is an elementary polynomial map if F = (F1, . . . , Fn)
where Fi = Xi + Gi with Gi ∈ k[X1, . . . , Xˆi, . . . , Xn] (i.e. we omit the variable Xi) for a unique 1 ≤ i ≤ n and
Fj = Xj for i 6= j. We say that F is a regular map if F = X +H where H is homogeneous with deg(F ) = 3. F
is a symmetric regular map if F is regular and JH is a symmetric matrix.
Note that an elementary polynomial map is invertible.
Proposition 1.1. Let F = (F1, . . . , Fn) ∈ MPn(R). Then F is invertible if and only if R[X1, . . . , Xn] =
R[F1, . . . , Fn].
Proof. Follows from the definitions.
Given a domain R we consider A := R[[X1, . . . , Xn]] the ring of formal power series over R. Denote the ideal M =
〈X1, . . . , Xn〉. By definition, every element in A can be written as f = fr + fr+1 + · · · where fr ∈ R[X1, . . . , Xn]
is homogeneous of degree r. If fr 6= 0, we have f ∈Mr and f /∈Mr+1. In particular, we have
⋂
iM
i = 0. 1
Formal Inverse Function Theorem. Let F = (F1, . . . , Fn) be a system of n formal series in n variables without
constant term over a domain R. Suppose that the jacobian JF (0) of F at (0, . . . , 0) ∈ Rn is a unit in the ring
Mn(R). Then F has an inverse system G = (G1, . . . , Gn) which is uniquely determined.
1This is more general in noetherian case: given A noetherian domain for any ideal I ⊂ A we have ⋂i Ii = (0). See [1, Chapter 10].
1.2. A computational criterion for invertibility 3
Proof. Let M := 〈X1, . . . , Xn〉. We will show that there is G1 ∈ R[[X1, . . . , Xn]] such that G1(F1, . . . , Fn) = X1.
Without loss of generality we can assume that Fi = Xi+Hi where Hi has only terms in degree ≥ 2. Let S1 := X1.
Then S1(F1, . . . , Fn) = F1 ≡ X1 mod M2. So S1(F1, . . . , Fn) − X1 = S2 for some S2 ∈ M2. In particular,
S2(F1, . . . , Fn) ≡ S2(X1, . . . , Xn) mod M3. Now note that S1(F1, . . . , Fn) − S2(F1, . . . , Fn) ≡ X1 + S2 − S2 ≡
X1 mod M
3. So S1(F1, . . . , Fn) − S2(F1, . . . , Fn) − X1 = S3 for some S3 ∈ M3. Proceeding in this way, we
obtain S1, . . . , Sk ∈ R[[X1, . . . , Xn]] such that S1(F1, . . . , F2) −
∑k
j≥2 Sj(F1, . . . , Fn) − X1 ∈ Mk+1. Passing to
the M -adic limit, we can determine G1 ∈ R[[X1, . . . , Xn]] such that G1(F1, . . . , Fn) ≡ X1 mod Mk for all k. So
G1(F1, . . . , Fn) = X1. Similarly we can determine G2, . . . , Gn ∈ R[[X1, . . . , Xn]] such that Gj(F1, . . . , Fn) = Xj .
For uniqueness, note that there is H = (H1, . . . ,Hn) such that H ◦G = X. So F = H ◦G◦F = H since G◦F = X.
Thus G is uniquely determined by F .
Now we will give some applications of the theorem above.
Corollary 1.1. Let F ∈MPn(R)(1) with R ⊂ S for some domain S. Then, F ∈MPIn(S)⇐⇒ F ∈MPIn(R).
Proof. Suppose that F ∈ MPIn(S). Let G be the formal inverse of F over R and H the inverse over S. Since
H,G ∈ S[[X1, . . . , Xn]] and the formal inverse of F over S is unique we have G = H. So G ∈ R[[X1, . . . , Xn]] ∩
S[X1, . . . , Xn] ⊆ R[X1, . . . , Xn].
Lemma 1.1. Let R be a Q-algebra of finite type. Then, there exists an immersion φ : R ↪→ C.
Proof. If R is algebraic over Q, by the primitive element theorem, K := Frac(R) = Q(α) for some α ∈ K. So if
m(t) ∈ Q[t] denotes the minimal polynomial, consider the map Q[t] −→ C given by evaluation by α. If R isn’t
algebraic, denote by {α1, . . . , αr} a transcendence basis for K over Q and make use the fact: tr.degQ(C) =∞.
Corollary 1.2. If the Jacobian Conjecture is true over C then it is true over any domain R with char(R) = 0.
Proof. Let F ∈ MPn(R) be a Keller map and suppose that the Jacobian Conjecture is true over C. Let R′ be
the subring of R generated (over Z) by all coefficients wich occur in F . Let φ : R′ ↪→ C be an immersion. By the
corollay above we know that F is invertible over R′ and by the same corollary we conclude that F is invertible
over R.
1.2 A computational criterion for invertibility
In this section we give a computational criterion, due to Arno Essen (cf.[?, poly0]), which determines when a
polynomial map is invertible and, in this case, compute the inverse. Initially we recall some topics about Gro¨bner
1.2. A computational criterion for invertibility 4
Basis. Details can be found in [3].
Let k be a field and consider the ring A := k[X1, . . . , Xn]. We will denote by X
α the monomial Xα11 . . . X
αn
n where
α = (α1, . . . , αn) ∈ Zn≥0. We recall that a monomial order in A is a relation > in the collection of monomials of
A with the following properties:
• > is a total order.
• If Xα > Xβ then XαXγ > XβXγ for all γ = (γ1, . . . , γn) ∈ Zn≥0.
• Any subset S 6= ∅ in the collection of monomials has a minimal element.
Example 1. (Lexicographic Order - lex) Let Xα = Xα11 . . . X
αn
n and X
β = Xβ11 . . . X
βn
n be monomials. Define
>lex by the rule X
α >lex X
β if αi = βi for 1 ≤ i ≤ k and some k and αk > βk. It is possible to show that >lex
defines a monomial order in k[X1, . . . , Xn].
Example 2. (Graded Lex Order- grlex) Let Xα = Xα11 . . . X
αn
n and X
β = Xβ11 . . . X
βn
n monomials in
k[X1, . . . , Xn]. Define >grlex by X
α >grlex X
β if |α| = ∑i αi > |β| = ∑i βi or if |α| = |β| and α >lex β.
It is possible to show that >grlex defines a monomial order in k[X1, . . . , Xn].
Fix > a monomial order and let F ∈ A. Denote by LM(F ) the leading monomial wich occurs in F i.e. LM(F ) > M
for any other monomial that occurs in F . Define LC(F ) the coefficient of LM(F ) and LT (F ) := LC(F )LM(F ).
Let f(X), g(X) ∈ k[X] \ k. By the division algorithm, we know that there exist (unique) q(X), r(X) ∈ k[X] with
r(X) = 0 or deg(r(X)) < deg(g(X)) such that
f(X) = q(X)g(X) + r(X).
For the ring k[X1, . . . , Xn] there is an analogue version
Division Algorithm in k[X1, . . . , Xn]. Fix > a monomial order in k[X1, . . . , Xn] and let G = (g1, . . . , gt) be an
ordered t-tuple of polynomials in k[X1, . . . , Xn]. Then every polynomial f ∈ k[X1, . . . , Xn] can be written as
f = q1g1 + · · ·+ qtgt + r
where r = 0 or r =
∑
i ciMi is a linear combination, with coefficients in k, of monomials Mi such that Mi /∈
〈LT (g1), . . . , LT (gt)〉 for all i. We say that r is the reduction of f by G and write r := rG(f).
Definition 1.2. Let 0 6= I ⊂ k[X1, . . . , Xn] be an ideal. Define LT (I) := 〈cXα | ∃F ∈ I − {0} LT (F ) = cXα〉,
an ideal in k[X1, . . . , Xn].
1.2. A computational criterion for invertibility 5
Let I = 〈F1, . . . , Fm〉 be an ideal in k[X1, . . . , Xn]. By definition it is clear that 〈LT (F1), . . . , LT (Fm)〉 ⊂ LT (I).
The next example shows that, in general, the equality 〈LT (F1), . . . , LT (Fm)〉 = LT (I) is false.
Example 3. Consider k[3] = k[X,Y, Z] and the ideal I = (F1, F2) where F1 = X
3−2XY and F2 = X2Y −2Y 2+X.
Fix the a lexicographic order in k[X,Y, Z]. Then, LT (F1) = X
3 and LT (F2) = X
2Y . Note
X2 = XF2 − Y F1 =⇒ X2 = LT (X2) ∈ LT (I)
Now if X2 ∈ 〈LT (F1), LT (F2)〉 we have X2 ∈ 〈LT (F1)〉 or X2 ∈ 〈LT (F2)〉 and this not occurs. So X2 /∈
〈LT (F1), LT (F2)〉.
Definition 1.3. Fix a monomial order in k[X1, . . . , Xn]. A finite subset G = {g1, . . . , gm} of an ideal 0 6= I ⊂
k[X1, . . . , Xn] it is called a Gro¨bner basis if 〈LT (g1), . . . , LT (gm)〉 = LT (I).
Theorem 1.1. Any ideal 0 6= I ⊂ k[X1, . . . , Xn] has a Gro¨bner basis.
Definition 1.4. Let I ⊂ k[X1, . . . , Xn] be an ideal and G a Gro¨bner basis for I. We say that G is reduced if
(i) LC(g) = 1 for all g ∈ G.
(ii) For all g ∈ G, and M a monomial wich occurs in g, we have M /∈ LT (G− {g}).
Theorem 1.2. Fix > a monomial order in k[X1, . . . , Xn]. Every non-zero ideal I ⊂ k[X1, . . . , Xn] has a reduced
Gro¨bner basis. Furthemore, the reduced Gro¨bner basis is unique.
1.2.1 The criterion of Essen
Let F = (F1, . . . , Fn) ∈MPn(k). So F1, . . . , Fn ∈ k[X1, . . . , Xn]. Introduce new variables Y1, . . . , Yn and consider
the polynomial ring k[X,Y ] := k[X1, . . . , Xn, Y1, . . . , Yn] with monomial order > such that Xn > X2 > · · · >
X1 > Yn > Yn−1 > · · · > Y1.
Let I ⊂ k[X,Y ] be the ideal I = 〈Y1 − F1, . . . , Yn − Fn〉.
Theorem 1.3. (Essen) Let G be the reduced Gro¨bner basis for the ideal I. Then
(i) F is invertible if and only if G = {X1 −G1, . . . , Xn −Gn} for some polynomials Gi ∈ k[Y1, . . . , Yn].
(ii) If this is the case, the inverse is given by G = (G1, . . . , Gn) ∈MPn(k).
Proof. Part (ii) follows immediately from (i).
1.2. A computational criterion for invertibility 6
Suppose that A = {X1−G1, . . . , Xn−Gn} is the reduced Gro¨bner basis for I, for some G1, . . . , Gn ∈ k[Y1, . . . , Yn].
Thus, for each 1 ≤ i ≤ n there is a relation of type
Yi − Fi(X1, . . . , Xn) =
∑
i
Ai(X,Y )(Xi −Gi(Y1, . . . , Yn))
Replacing Xi by Gi(Y1, . . . , Yn) we obtain Yi − Fi(G1(Y ), . . . , Gn(Y )) = 0 =⇒ Y = F (G(Y )). So we conclude
that F is invertible with inverse G.
Now suppose that F is an invertible map with inverse G = (G1, . . . , Gn). We want to show that {X1 −
G1(Y1, . . . , Yn), . . . , Xn −Gn(Y1, . . . , Yn)} is a reduced Gro¨bner basis for I. By definition of I note that
Yi ≡ Fi mod I
So for any g ∈ k[Y1, . . . , Yn] we have g(F1(X), . . . , Fn(X)) ≡ g(Y1, . . . , Yn) mod I. Since F is invertible, we have
that Xi = Gi(F1(X), . . . , Fn(X)). Hence
Xi = Gi(F1(X), . . . , Fn(X)) ≡ Gi(Y1, . . . , Yn) ∈ I.
So Xi = LT (Xi − Gi(Y1, . . . , Yn)) ∈ LT (I) =⇒ Xi = ciLT (bi) for some bi ∈ G and ci ∈ k[X,Y ]. So ci = 1 and
LT (bi) = Xi for all i.
In particular, for i = 1 we have b1 = X1 + h1(Y ) for some h1(Y ) ∈ k[Y1, . . . , Yn] (recall the assumption about the
order).
For i = 2 we have b2 = X2 + h2(X1, Y ) for some h2(X1, Y ) ∈ k[X1, Y1, . . . , Yn]. Now if X1 occurs effectively in
h2 we conclude that b2 contains a monomial that belongs to LT (G− {b2}). Since G is reduced this cannot occur
and so h2 ∈ k[Y1, . . . , Yn]. We can repeat the argument up to i ≤ n and conclude that
bi = Xi + hi(Y ) hi(Y ) ∈ k[Y1, . . . , Yn] ∀ 1 ≤ i ≤ n
So H := {X1 + h1(Y ), . . . , Xn + hn(Y )} ⊆ G. Now note that I = (G) and J = (H) are prime ideals in k[X,Y ]
with height n. Since J ⊂ I we conclude I = J . By uniqueness we have H = G.
Example 4. Let K = k(t) the rational functions field over k and consider the polynomial map F ∈ MP2(K)
defined by
F (X1, X2) = (X1 − t3X31 + 3t2X21X2 − 3tX1X22 +X32 , X2 − t4X31 + 3t3X21X2 − 3t2X1X22 + tX32 )
1.3. Invertible maps 7
Considering the ideal
I = 〈Y1 − (X1 − t3X31 + 3t2X21X2 − 3tX1X22 +X32 ) , Y2 − (X2 − t4X31 + 3t3X21X2 − 3t2X1X22 + tX32 )〉
in the ring k(t)[X1, X2, Y1, Y2] with lexicographic order X2 > X1 > Y2 > Y1, we see (via Singular, cf.[4]) that the
reduced Gro¨bner basis for I is
G = {X1 + Y 32 − 3tY 22 Y1 + 3t2Y2Y 21 − t3Y 31 − Y1 , X2 + tY 32 − 3t2Y 22 Y1 + 3t3Y2Y 21 − Y2 − t4Y 31 }
So by Essen criterion, we have F invertible with inverse map
F−1(X1, X2) = (X1 −X32 + 3tX22X1 − 3t2X2X21 + t3X31 , X2 − tX32 + 3t2X22X1 − 3t3X2X21 + t4X31 )
Remark 2. In the example above, note that F is an invertible regular map of the form F = X + H and the
inverse is given by F−1 = X −H. Also, note that JH2 = 0. Later, we shall see that this fact isn’t arbitrary but a
particular case of a class of invertible maps.
1.3 Invertible maps
In this section we give a version of the following proposition for affine varieties:
Proposition 1.2. Let k be a field and L/k an algebraic extension. If σ : L ↪→ L is an injective morphism over k
then σ is an isomorphism.
Proof. We will reduce to the finite case.
Let α ∈ L. We want to show that there is β ∈ L such that σ(β) = α. Let mα(T ) = Tn+an−1Tn−1+· · ·+a0 ∈ k[T ]
be the minimal polynomial of α over k. Let E = k(α1, . . . αr) (with α1 = α) subfield of L obtained by adjunction
of all roots of mα(T ) that are in L. Note that for each u ∈ E we have σ(u) ∈ E. So σ, by restriction, defines an
injective map σ|E : E ↪→ E over k. Now, E is of finite dimension over k. In particular, σ|E is an isomorphism. So
there is a β ∈ E such that σ(β) = α.
In what follows, k will denote an algebraically closed field. If X = Z(F1, . . . , Fr) ⊂ Ank is an algebraic set, given a
subfield l ⊂ k we will denote by X(l) the set of l-points i.e. X(l) := X ∩ ln.
Theorem 1.4. Let X ⊂ Ank be an affine variety and α : X −→ X an injective endomorphism. Then α is
surjective.
1.3. Invertible maps 8
Proof. Initially we shall write the injectivity, non-surjectivity and membership conditions in terms of equations.
For this let G1, . . . , Gm ∈ k[X1, . . . , Xn] be the equations of X and α1, . . . , αn the components of the map α as
elements of k[X1, . . . , Xn].
(1) Injectivity: If (P,Q) ∈ X ×X are such that α1(P ) = α1(Q), . . . , αn(P ) = αn(Q) then P = Q. Equivalently,
every zero of the ideal
I = 〈G1(X), . . . , Gm(X), G1(Y ), . . . , Gm(Y ), α1(X)− α1(Y ), . . . , αn(X)− αn(Y )〉
is a zero of J = 〈X1 − Y1, . . . , Xn − Yn〉. By Nullstellensatz this is equivalently to: for all i ∈ {1, . . . , n} there is
li ∈ N and polynomials Aik, Bik, Cik such that
(Xk − Yk)lk =
∑
i
Aik(X,Y )Gi(X) +
∑
i
Bik(X,Y )Gi(Y ) +
∑
i
Cik(X,Y )(αi(X)− αi(Y )).
(2) Membership: For all P ∈ X we have α(P ) ∈ X. So if P ∈ Ank is a zero of 〈G1(X), . . . , Gm(X)〉 then P is
a zero of 〈G1(α1, . . . , αn), . . . , Gm(α1, . . . , αn)〉. By Nullstellensatz this is equivalently to: for all u ∈ {1, . . . , n}
there is ti ∈ N and polynomials riu such that
Gk(α1, . . . , αn)
tu =
∑
i
riu(X)Gi.
(3) non-surjectivity: If Q = (q1, . . . , qn) /∈ α(X) this is equivalently to: the ideal
I2 := 〈G1(X), . . . , Gr(X), α1(X)− q1, . . . , αn(X)− qn〉
has no zero. By Nullstellensatz, this is equivalently to I2 = k[X1, . . . , Xn]. So
1 =
∑
l
hl(X)Gl(X) +
∑
k
uk(X)(αk(X)− qk)
for some polynomials hl, uk.
Suppose that α isn’t surjective. Consider all equations wich occur in (1), (2) and in (3).
Case 1: char(k) = p > 0.
This case, Fp is the prime field of k. Consider {a1, . . . , ae} ⊂ k the set of all coefficients that occur in the above
relations: (1), (2) and (3). Let S := Fp[a1, . . . , ae] ⊂ k be the subring generated by all this coefficients. Let
M∈ Specm(S) and consider the ring R = S/M.
1.3. Invertible maps 9
We have R a Fp-algebra of finite type and by algebraic Nullstellensatz we conclude that R is algebraic over Fp, in
particular it is a finite field. Considering X(R) the set obtained from X by reduction of equations mod M and
by α the induced map, we obtain a map
α : X(R) −→ X(R)
injective by relation (1) but non surjective by (3), a contradiction
Caso 2: char(k) = 0.
In this case, consider {a1, . . . , ae} ⊂ k all coefficients that occurs in (1), (2) and (3). Let S := Z[a1, . . . , ae] ⊂ k
be the subring generated by all this coefficient. Let M ∈ Specm(S). If S/M is a finite field we can repeat the
argument of case 1 and conclude the proof. So it is sufficient to show the following
Theorem 1.5. 2 Let R a Z - algebra of finite type and M∈ Spec(R). Then
R/M is a finite field if and only if M∈ Specm(R).
Proof. By the algebraic Nullstellensatz it is sufficient to show that M∩ Z 6= 0. Suppose this false. Then the
natural map Z ↪→ R/M is injective. In this case, we obtain the following commutative diagram
Z 
 // _

R/M
Q
. 
==
By theorem 4.3 we conclude that Q is finitely generated as Z-algebra. A contradiction.
1.3.1 A refinement
Fix k an algebraically closed field with char(k) = 0.
Lemma 1. Let X ⊂ Ank be an affine variety. Denote by (X˜, pi) the normalization of X. Let
S(X) := {f : X −→ k | f ◦ pi ∈ OX˜(X˜)}.
2Another elegant proof: Consider the map of affine schemes: f : Spec(R) −→ Spec(Z). Since Z is noetherian and we have finitude
conditions, by Chevalley theorem (cf.[1]) we know that the map f is constructive i.e. if X ⊂ Spec(R) is a construtive set then f(X)
is constructive. Let M a closed point in Spec(R). In particular, it is constructive. So f({M}) = {M ∩ Z} is constructive set.
Since f({M}) does not contain any open dense, we can applies [1, IV3.5-propositon 3.5] and conclude that f({M}) isn’t dense. In
particular, can’t be the generic point of Spec(Z) i.e. f({M}) 6= {0}. Algebraic Nullstellensatz finishes the proof.
1.3. Invertible maps 10
Then S(X) is a noetherian OX(X)-module.
Proof. By properties of normalization, we know that OX˜(X˜) is noetherian as OX(X)-module (it is of finite type).
Consider the map of OX(X)-modules
L : S(X) −→ OX˜(X˜) f ∈ S(X) 7→ f ◦ pi.
It is easy to check that L is an injective map. So S(X) can be looked at as an OX(X)-submodule of OX˜(X˜) (that
is noetherian). Thus S(X) is noetherian.
Let X be an affine variety and (X˜, pi) its normalization. Let F : X −→ X be a morphism. Consider the following
diagram
X˜
pi //

X
F

X˜
pi // X
By properties of normalization, there is a unique morphism F˜ : X˜ −→ X˜ fitting in the dotted arrow above.
Lemma 2. (i) If F is injective then F˜ is injective.
(ii) If F˜ is surjective then F is surjective.
Proof. (i): Suppose that F is an injective map. By theorem above, we have F is a bijection. Initially, we show
that F˜ is a dominant map. Indeed, suppose this false. Then since X˜ is irreducible, we have dim F˜ (X˜) < dim X˜.
Now, note that pi ◦ F˜ is a dominant map:
pi(F˜ (X˜)) = F (pi(X˜)) = X = X.
Thus,
dimX = dimpi(F˜ (X˜)) = dimpi(F˜ (X˜)) ≤ dim F˜ (X˜) < dim X˜ = dimX
a contradiction.
So F˜ : X˜ −→ X˜ is a dominant map between varieties of same dimension. Furthemore F˜ is quasi-finite map, since
F ◦ pi is too. So by Theorem 4.7, for any point P ∈ X˜ we have #F−1(P ) ≤ deg(F˜ ). Now, 1 = deg(F )deg(pi) =
deg(F ◦ pi) = deg(pi ◦ F˜ ) = deg(F˜ ) =⇒ deg(F˜ ) = 1. In particular F˜ is injective.
1.3. Invertible maps 11
(ii) Suppose that F˜ is surjective. Then
F (X) = F (pi(X˜)) = pi(F˜ (X˜)) = pi(X˜) = X
i.e. F is surjective.
Definition 1.5. In notation as in lemma 1 above, we define Sn(X) := {F : X −→ X | Fi ◦ pi ∈ OX˜(X˜) ∀i =
1, . . . , n}. Note that Sn(X) ∼=
⊕n
i=1 S(X), as OX(X)-module, and so Sn(X) is noetherian as OX(X)-module.
Theorem 1.6. (Cynk-Rusek) Let X ⊂ Ank be an affine variety and F : X −→ X a regular map. The following
conditions are equivalent:
(i) F is injective.
(ii) F is a bijection.
(iii) F is an automorphism.
Proof. The implication (i) =⇒ (ii) follows from the theorem above. We will show (ii)=⇒ (iii). Suppose that
F : X −→ X is a bijection and consider the map F˜ induced in normalization. By the lemma above, we have F˜ an
injective map. Applying the Zariski Main Theorem (see section 4.4) we conclude F˜ is an isomorphism. Let F−1
be the set-theoretic inverse of F . By commutativity of the diagram
X˜
pi //
F˜

X
F

X˜
pi // X
we have pi ◦ F˜−1 = F−1 ◦ pi. So we conclude pi ◦ F˜−d = F−d ◦ pi. So F−d ∈ Sn(X) for all d ∈ N. Consider the
OX(X)-submodules: Sn(X)(e) := OX(X)F−1 + · · ·+OX(X)F−e. The collection {Sn(X)(e)}e∈N is an ascending
chain of submodules
Sn(X)
(1) ⊂ Sn(X)(2) ⊂ Sn(X)(e) · · ·
Since S(X) is noetherian there exists d ∈ N such that F−d−1 ∈ Sn(X)(d). So
F−d−1 =
d∑
j=1
rjF
−j rj ∈ OX(X).
In particular we have F−1 =
∑d
j=1
rjF
d−j = r1F d−1 + r2F d−2 + · · ·+ rdX =⇒ F−1 is a regular map i.e. F is an
1.3. Invertible maps 12
automorphism.
The implication (iii)=⇒(i) is trivial.
Corollary 1.3. (Wang) Let k be an algebraically closed field with char(k) = 0. Let F ∈ MPn(k) be a Keller
map with deg(F ) ≤ 2. Then, F is invertible.
Proof. By 1.6 it is sufficient to show that F is injective. Suppose that this is false. Pick P,Q ∈ kn such that
F (P ) = F (Q) with P 6= Q. Without loss of generality, we can suppose that P = 0 and F (0) = 0 = F (Q). Write
F = F(1) +F(2) where F1 = linear part of F and F2 = quadratic part. Denote by F
(i) the i-th componenent of F .
Then
0 = F (i)(Q) = F
(i)
(1)(Q) + F
(i)
(2)(Q) = F
(i)
(1)(Q) + 2t0F
(i)
(2)(Q), t0 := 1/2.
Now, note that F
(i)
(1)(Q)+2t0F
(i)
(2)(Q) = (tF
(i)
(1)(Q)+t
2F
(i)
(2)(Q))
′|t=t0 = (F (i)(1)(tQ)+F (i)(2)(tQ))′|t=t0 = (F (i)(tQ))′|t=t0 =<
∇F (i)(t0Q), Q > . In terms of matrix we conclude that JF (1/2Q).Q = 0 a contradiction since F is a Keller
map.
The next theorem is fundamental in chapter 3:
Theorem 1.7. (Connell - van den Dries) Let F ∈MPn(C) be a counterexample for the Jacobian Conjecture.
Then there is N ∈ N with N > n and a couterexample G ∈MPN (C) with coefficients in Z such that det(JF ) = 1.
We make use of the following
Lemma 1.2. Let k be a field with char(k) = 0 and K|k a Galois extension of degree m > 1. Let F ∈ MPn(K)
be a Keller map and suppose that F isn’t injective. Then there is a non-injective Keller map G ∈MPnm(k).
Proof. Let {e1, . . . , en} be a k-basis of K and denote σ0, . . . , σm−1 the elements of the group Gal(K/k). Introduce
new variables X11, . . . , X1m, . . . , Xn1, . . . , Xnm and consider the polynomials Gij ∈ k[X11, . . . , Xnm] defined by
relations
Gk1e1 + · · ·+Gkmem = Fk(X11e1 + · · ·+X1mem, . . . , Xn1e1 + · · ·+Xnmem)
Consider the polynomial map G = (G11, . . . , Gnm) : k
nm −→ knm. Since F isn’t injective it follows that G isn’t
injective. We affirm that det JG = 1. For this, consider the polynomial map F˜ : Knm −→ Knm given by P 7→
(Fσ0(P ), . . . , F σm−1(P )). By definition, if Fk =
∑
l αi1...inX
i1
1 · · ·Xinn we set Fσpk =
∑
l α
σp
i1...in
Xi1pn+1 · · ·Xinpn+n ∈
k[Xpn+1, . . . , Xpn+n] for 0 ≤ p ≤ m− 1 and 1 ≤ k ≤ n. Finally, Fσ := (Fσ1 , . . . , F σn ) if σ ∈ Gal(K/k).
1.3. Invertible maps 13
By construction, we have
JF σ =

JF σ0 0 0 . . . 0
... JF σ1
...
. . .
...
...
...
...
. . .
...
...
...
...
. . .
...
0 0 0 0 JF σm−1

So det(F˜ ) =
∏m−1
j=0
det JF σj =
∏m−1
j=0
σj(det JF ) = 1. Now consider L : K
nm −→ Knm the polynomial map
obtained by taking Li(n+j) := σi(e1)Yj1 + · · · + σi(em)Yjm for 0 ≤ i ≤ m − 1 and 1 ≤ j ≤ n. Note that 3
L ◦G = F˜ ◦ L =⇒ G = L−1 ◦ F˜ ◦ L and so detJG = det JF˜ = 1. This finishes the proof.
Proof. (of 1.7) Let F ∈MPn(C) be a counterexample for the Jacobian Conjecture. By 1.6 we know that F isn’t
injective. Thus there are P,Q ∈ Cn such that F (P ) = F (Q). Without loss of generality we can suppose that
P = (0, . . . , 0) and Q = (1, 0, . . . , 0) with F (P ) = 0. In particular we have F = X + F(2) + F(3) + · · · .
Replace each coefficient ai that occurs in terms in F of degree ≥ 2 by a new variable Yi. In this case, we obtain
a map F˜ with coefficients in the ring Z[Y1, . . . , Yl, X1, . . . , Xn] (= Z[X,Y ] for simplicity). Furthemore, we have
det JF˜ = 1 + p(X,Y ) for some p(X,Y ) =
∑k
i=1Di(Y )X
αi ∈ Z[Y,X]. The condition F˜ (1, 0, . . . , 0) = F˜ (0, . . . , 0)
can be expressed by polynomial relations C1, . . . , Cp, in the variables Y1, . . . , Yl. Also, the condition det F˜ = 1 can
be expressed in polynomial relations D1(Y ), . . . , Dk(Y ). Now note that since F satisfies F (1, . . . , 0) = F (0, . . . , 0)
and det JF = 1 and has coefficients in C we know that there is a solution (over C) for the algebraic system:
D1 = D2 = · · · = Dk = C1 = · · · = Cp = 0.
So considering the ideal I = 〈D1, . . . , Dk, C1, . . . , Cp〉 ⊂ Q[Y1, . . . , Yl] we know 1 /∈ I =⇒ Z(I) 6= ∅ (by Nullstel-
lensatz). So there is a conterexample with coefficients in Q. By adjunction of coefficients in Q and taking the
normal closure K we obtain a counterexample F = X + F(2) + F(3) · · · for the Keller Problem with coefficients
in K (Galois extension) with F (1, 0, .., 0) = F (0, . . . , 0). By the lemma above we obtain G : Qnm −→ Qnm a
polynomial map which is not injective and with detJG = 1 where m = [K : Q].
Finally let r ∈ N such that G′i = Xi + rGi1 + rGi2 + · · ·+ ∈ Z[X1, . . . , Xnm] for all 1 ≤ i ≤ n and consider the
map G′ = (G′1, . . . , G
′
nm) : Qnm −→ Qnm.
3we make use of the linear independence of characters: c0σ0+ · · ·+cm−1σm−1 = 0 with c0, . . . , cm−1 ∈ k =⇒ c1 = · · · = cm−1 = 0
1.4. Particular cases 14
We affirm that G′ isn’t injective and satifies det JG′ = 1. Note that G′i(X) = r
−1Gi(rX) ∈ Q[X] =⇒
G(X)′ = r−1G(rX). In particular, the map G′ over Q isn’t injective. We have J(G(X)′) = J(r−1G(rX)) =
r−1rJ(G)(rX) = (JG)(rX). So by condition detJG = 1 we obtain det(J(G(X)′)) = 1 and this finishes the proof.
Corollary 1.4. The Jacobian Conjecture is true over C if and only it is true over Z.
1.4 Particular cases
In this section we study some classes of polynomial maps which satisfy the condition of the Jacobian Conjecture.
We start with some facts about domains of finite type.
Let k be a field and A a domain with k ⊆ A ⊆ k[X1, . . . , Xn] and dimA = 1. We will show that A is a k-algebra
of finite type. For this, we remark that we can assume n = 1. More precisely,
Lemma 1.3. Let A be a domain with dimA = 1 such that k ⊂ A ⊂ k[X1, . . . , Xn]. Then, there is a map of
k-algebras ϕ : k[X1, . . . , Xn] −→ k[Z] that induces an immersion ϕ : A −→ k[Z]. So we can suppose that n = 1.
Proof. Fix the maximal ideal M = (X1, . . . , Xn) ⊂ k[X1, . . . , Xn] and consider P := A ∩ M . If P = 0 then
A ↪→ k[X1, . . . , Xn]/M ∼= k and so A = k a contradiction (since dimA = 1). So P 6= 0. Define Pm :=
〈Xm1 −Xn〉 ∈ Spec(k[X1, . . . , Xn]) for m ∈ N. Fix G ∈ P ∩ M a non zero element and chose m such that
Xm1 −Xn - G. Then, by this choice, and since dimA = 1 we obtain Pm ∩A = 0. So the map ϕ : k[X1, . . . , Xn]
k[X1, . . . , Xn]/Pm ∼= k[X1, . . . , Xn−1] induces an injective map A/Pm ∩ A = A ↪→ k[X1, . . . , Xn−1]. By iteration,
we obtain the result.
Theorem 1.8. Let A a domain with dimA = 1 and k ⊂ A ⊂ k[X1, . . . , Xn] for some field k. Then A is a
k-algebra of finite type and A is a polynomial ring in one variable over k. Here, A is the integral closure of A in
Frac(A).
Proof. By lemma 1.3 we can suppose n = 1 i.e. k ⊂ A ⊂ k[X]. We will show, initially, that A is a k-algebra of
finite type. For this, take g ∈ A \ k and consider the k-algebra k[g]. We have k[g] ⊆ k[X] an integral extension:
indeed, X satisfies p(X) = 0 where p(T ) = g(T ) − g(X) ∈ k[g][T ]. In particular, we have k[X] is a finitely
generated k[g]-module. Since A is a k[g]-submodule we have A a k[g]-module finitely generate. In particular, it is
of finite type over k
1.4. Particular cases 15
Let L := Frac(A). Since k ⊂ L ⊂ k(X) we have L = k(Z), for some rational function Z 4. Let A denote the
integral closure of A in L. By Theorem 4.2, we have
A =
⋂
O∈S(A)
O =
⋂
k[Z]p(Z)
where the intersection is taken over all irreducibles in k[Z] and
S(A) := {O ⊂ L | O is a valuation ring that contains A}.
Now, note that all valuation rings of k(X) contain A, with possible exception of the ring in infinity. This follows
from the fact that A ⊂ k[X] since every valuation ring of “finite distance” of k(X) is of form k[X]p(X) with p(X)
irreducible.
Since every valuation ring of k(Z) is a contraction of some discrete valuation of k(X) 5 we conclude that all
valuation rings of k(Z) contain A, with the exception of those that come from k[1/X]1/X . Now, note that if O is
a DVR of k(Z) that comes from k[1/X]1/X by looking the residues fields we have:
k ↪→ O/P ↪→ k[1/X]1/X
(1/X)
= k
So the possible DVRs that come from k(Z) that do not contain A are of form k[Z]p(Z) for some p(Z) irreducible
linear and k[1/Z]1/Z . In the last case, we have A =
⋂
k[Z]p(Z) = k[Z] and in the first case, we make use of the
fact k(Z) = k(1/Z) and so we obtain A = k[1/Z].
Theorem 1.9. (Formanek) Let F ∈ MPn(k) be a Keller map with k a field with char(k) = 0. Suppose that
there is a Fn+1 ∈ k[X1, . . . , Xn] such that k[F1, . . . , Fn, Fn+1] = k[X1, . . . , Xn]. Then F is an invertible map.
Proof. Let Y1, . . . , Yn, Yn+1 be variables over k. For simplicity, define A := k[Y1, . . . , Yn, Yn+1] and consider the
map of k-algebras α : A k[X1, . . . , Xn] given by Yi 7→ Fi for i = 1, . . . , n+ 1. Let P ∈ Spec(A) be the kernel of
α. Recall that ht(P ) = 1 (apply the formula dimA = dimA/P + ht(P ), true for all domains of finite type over a
field k). So since A is factorial we have P = (F ) for some F ∈ A.
Pick T1, . . . , Tn ∈ A such that α(Tj) = Xj . We have Yj − Fj(T1, . . . , Tn) ∈ Ker(α) = P . So there are
4Recall: Lu¨roth Theorem. Let k(t) be the field of rational functions over a field k and L subfield with k ( L ⊂ k(t). Then,
there is s ∈ k(t) such that L = k(s).
5Let A a DVR with K := Frac(R) and L|K a finite extension. Let R the integral closure of A in L. Suppose that R is a finite
A-module. Then R is a semilocal ring and each DVR over A is obtained by taking RP for some P ∈ Spec(R). In the case in question,
A = k[Z] and by general results we know that R is finite over A.
1.4. Particular cases 16
R1, . . . , Rn+1 ∈ A such that Yj = Fj(T1, . . . , Tn) + RjF . Write, F = HrY rn+1 + · · · + H0 with H0, . . . ,Hr ∈
k[Y1, . . . , Yn]. Note that Hj 6= 0 for some j > 0 (Keller condition (det JF ∈ k∗) implies that F1, . . . , Fn are
algebraically independent over k.)
Applying the operator ∂/∂Yj for j = 1, . . . , n+ 1 we obtain
∂Yi
∂Yj
=
n∑
k=1
∂Fi
∂Xk
(T1, . . . , Tn)
∂Tk
∂Yj
+
∂Ri
∂Yj
F +
∂F
∂Yj
Ri 1 ≤ i, j ≤ n+ 1
In terms of matrices, we have
Idn+1 =

∂F1(T1,...,Tn)
∂X1
...
... ∂F1(T1,...,Tn)∂Xn R1
...
...
...
. . .
...
...
...
...
. . .
...
∂Fn(T1,...,Tn)
∂X1
...
... ∂Fn(T1,...,Tn)∂Xn Rn
∂Fn+1(T1,...,Tn)
∂X1
...
... ∂Fn+1(T1,...,Tn)∂Xn Rn+1


∂T1
∂Y1
...
... ∂T1∂Yn+1
...
...
...
...
...
...
...
...
∂Tn
∂Y1
...
... ∂Tn∂Yn+1
∂F
∂Y1
...
... ∂F∂Yn+1

+ F

∂R1
∂Y1
...
... ∂R1∂Yn+1
...
...
...
...
...
...
...
...
∂Rn
∂Y1
...
... ∂Rn∂Yn+1
∂Rn+1
∂Y1
...
... ∂Rn+1∂Yn+1

Applying α in the above identiy and observing that α(
∂Fj(T1,...,Tn)
∂Xp
) =
∂Fj(X1,...,Xn)
∂Xp
for all 1 ≤ p ≤ n and
1 ≤ i ≤ n+ 1 we obtain the following identity over k[X1, . . . , Xn]:
Idn+1 =

∂F1(X1,...,Xn)
∂X1
...
... ∂F1(X1,...,Xn)∂Xn (∗∗)
...
...
...
. . .
...
...
...
...
. . .
...
∂Fn(X1,...,Xn)
∂X1
...
... ∂Fn(X1,...,Xn)∂Xn (∗∗)
(∗∗) ... ... (∗∗) (∗∗)


(∗∗) ... ... (∗∗)
...
...
...
...
...
...
...
...
(∗∗) ... ... (∗∗)
(∗∗) ... ... α( ∂F∂Yn+1 )

Multiplying by the adjoint matrix and using the fact that (F1, ..., Fn) is a Keller map we obtain α(
∂F
∂Yn+1
) ∈ k∗ =⇒
α( ∂F∂Yn+1 ) = c, for some c ∈ k∗. So ∂F∂Yn+1 − c ∈ P . By comparison of degrees we have ∂F∂Yn+1 − c = 0 =⇒ ∂F∂Yn+1 =
c =⇒ F = cYn+1 + H0(Y1, . . . , Yn). In particular, 0 = cFn+1 + H0(F1, . . . , Fn) =⇒ Fn+1 = −c−1H0(F1, . . . , Fn).
So k[F1, . . . , Fn, Fn+1] = k[F1, . . . , Fn] = k[X1, . . . , Xn] =⇒ (F1, . . . , Fn) is an invertible polynomial map.
Theorem 1.10. Let F = X + H ∈ MPn(k) be a polynomial map with k a Q-algebra. Assume H homogeneous
of degree d. Suppose that JH2 = 0. Then F is invertible with inverse given explicitly by F−1 = X −H.
1.4. Particular cases 17
Remark 3. Let A = ⊕n∈NAd be a graded ring (not necessarily commutative). Let d ∈ A be homogeneous of degree
e > 1. Then
d is nilpotent⇐⇒ 1− d ∈ A∗.
The implication =⇒follows from the formula: (1−d)(1+d+d2+· · ·+dn) = 1−dn+1. Now, suppose that 1−d ∈ A∗
but d isn’t nilpotent. Let α ∈ A be the inverse of 1−d and consider the formula (1−d)(1+d+d2+· · ·+dn) = 1−dn+1.
Multiplying in the left by α we obtain
sn := 1 + d+ d
2 + · · ·+ dn = α− αdn+1.
From α(1 − d) = 1 we obtain α = 1 + αd. Note that deg(α) = 0. Plugging in the above relation we obtain
sn = α− (1 + αd)dn+1 = α− dn+1 − αdn+2. By iteration, and using the fact d isn’t nilpotent we see that sn has
terms of arbitrarily large degree, a contradiction. So d is nilpotent.
Considering A = k[X1, . . . , Xn]
n with its natural graded structure we see that given a map F = X + H with H
homogeneous we have
J(H) is nilpotent if and only if J(F ) ∈ GLn(k[X1, . . . , Xn]).
With notations as above, let H = (H1, . . . ,Hn) with Hi ∈ k[X1, . . . , Xn]. Consider the Euler operator:
4 : k[X1, . . . , Xn] −→ k[X1, . . . , Xn], P 7−→
n∑
i=1
∂P
∂Xi
Hi.
The following properties are easy to check:
• 4 is k-linear.
• 4(PQ) = P4(Q) +Q4(P ) for any P,Q ∈ k[X1, . . . , Xn].
In what follows, 4m := 4 ◦4 ◦ · · · ◦ 4 (n times).
Lemma 1.4. Suppose that 4(H1) = · · · = 4(Hn) = 0. Then for all P ∈ k[X1, . . . , Xn]
4m(P ) =
∑
i1,...,im
∂m(P )
∂Xi1∂Xi2 · · · ∂Xim
Hi1 · · ·Him .
Proof. Induction.
Lemma 1.5. Let F = X +H and G = X −H be polynimial maps (as above). The following are equivalent:
(i) F is ivertible with inverse G.
1.4. Particular cases 18
(ii) Hi(X −H) = Hj for all i = 1, . . . , n.
(iii) 4Hi = 0 for all i = 1, . . . , n.
Proof. (ii) =⇒ (i): We have F (G) = F (X −H) = X −H +H(X −H) = X −H +H = X and so F is invertible
with inverse G.
(iii) =⇒ (ii): By Taylor expansion and by lemma above we have Hi(X −H) = Hi −4Hi −42Hi/2 + · · · . Since
4Hi = 0 for all i we obtain (ii).
(i) =⇒ (iii): We have X = F (X−H) = X−H+H(X−H)⇐⇒ Hi(X−H) = Hi for all i = 1, . . . , n. Considering
the Taylor expansion we obtain
Hi(X −H) = Hi −4Hi −42Hi/2 + · · ·
By hypothesis we know Hi(X − H) = Hi and so 4Hi +42Hi/2 + · · · . By comparison of degrees we see that
4mHi = 0 for all i = 1, . . . , n.
Proof. ( of 1.10) The condition JH2 = 0 implies

∂H1(X1,...,Xn)
∂X1
...
... ∂H1(X1,...,Xn)∂Xn
...
...
...
. . .
∂Hn(X1,...,Xn)
∂X1
...
... ∂Hn(X1,...,Xn)∂Xn


dH1
...
...
dHn

=

d
∑
j
∂H1(X1,...,Xn)
∂Xj
Hj
...
d
∑
j
∂Hn(X1,...,Xn)
∂Xj
Hj
 =

d4H1
...
...
d4Hn

= 0
So d4Hk = 0 =⇒ Hk = 0 for all k = 1, . . . , n. The result follows from the lemma above.
Lemma 1.6. Let k be a field with char(k) = 0 and T ∈ A := k[X1, . . . , Xn]. Denote by Dj := ∂∂Xj the j-th
canonical derivation. Let V be the k-vecor space generated by D1T, . . . ,DnT and r := dimk V . Then there is
a linear change of coordinates Xk 7→ Lk such that if S := T (L) then {D1S, . . . ,DrS} is a basis of V and S is
independent of Xr+1, . . . , Xn. Furthemore, D1S, . . . ,DrS are algebraically independent over k[S].
Proof. If r = dimk V we can suppose, without loss of generality, that {D1T, . . . ,DrT} is a basis for V . So for
r + 1 ≤ j ≤ n: Dr+jT =
∑r
i=1 αijDiT for some constants αij ∈ k. Through such constants, we can find a matrix
A such that the induced map L : kn −→ kn, X 7→ AX is such that if S := T (L) then Dr+jS = 0 for 1 ≤ j ≤ n− r.
So we conclude (char(k) = 0) that S is independent of Xr+1, . . . , Xn. Suppose that
∑r
i=1 Pi(S)DiS = 0 for
1.4. Particular cases 19
some P1(S), . . . , Pr(S) ∈ k[S]. Erasing the common factors we can suppose that gcd(P1(S), . . . , Pr(S)) = 1.
By reduction mod S we obtain 0 =
∑r
i=1 Pi(S)DiS ≡
∑r
i=1 Pi(0)DiS mod S =⇒
∑r
i=1 Pi(0)DiS ∈ (S). By
comparison of degree we obtain
∑r
i=1 Pi(0)DiS = 0 a contradiction.
Theorem 1.11. Let k be a field with char(k) = 0 and F ∈MPn(k) a map of the form F = X+H with H(0) = 0.
Suppose that JH is nilpotent with rank = 1. Then F is an invertible polynomial map.
Proof. Let H = (H1, . . . ,Hn) be the components of H and consider the k-algebra R = k[H1, . . . ,Hn]. Let
L = k(H1, . . . ,Hn) be the fraction field of R and K = k(X1, . . . , Xn). Consider the following exact sequence
ΩL/k ⊗L K α // ΩK/k β // ΩK/L // 0
By definition of α we have α(ΩL/k⊗LK) is a K-subspace of ΩK/k with generators dK/k(H1), . . . , dK/k(Hn). Hence
dimK Im(α) = rankKJ(H) = 1. So
dimK ΩK/L = dimK(ΩK/k)− dim Im(α) = tr.degk(K)− 1
Since dimK ΩK/L = tr.degL(K) (see dimension formula 4.2) we have tr.degk(L) = 1. So dimKrullR = tr.degk(L) =
1. By theorem 1.8 we have that the integral closure, R, of R in Frac(R) is a polynomial ring. Now note
that if P ∈ R then P is integral over k[X1, . . . , Xn] and since that this last ring is integrally closed we obtain
P ∈ k[X1, . . . , Xn]. So R = k[T ] for some T ∈ k[X1, . . . , Xn].
By the lemma above there is a linear change of variables Xi 7→ Li such that if S := T (L) then S is independent
of Xr+1, . . . , Xn and D1S, . . . ,DrS are algebraically linearly independent over k[S]. Write H = (H1, . . . ,Hn) =
(Q1, . . . , Qn) with Q1, . . . , Qn ∈ k[T ]. If G = L−1 ◦ F ◦ L we have
G = X + L−1 ◦H ◦ L = X + L−1 ◦Q(T (L)) = X + L−1(Q(S)).
Define R := (R1, . . . , Rn) with Ri := L
−1 ◦ Q(T ) ∈ k[T ]. Then L−1(Q(S)) = R(S) = (R1(S), . . . , Rn(S)). The
Jacobian matrix of R(S) is nilpotent, of the form

R′1(S)D1S R
′
1(S)D2S
. . .
...
... R′1(S)DnS
R′2(S)D1S R
′
2(S)D2S
. . .
...
... R′2(S)DnS
. . .
...
...
R′n(S)D1S R
′
n(S)D2S
. . .
...
... R′n(S)DnS

In particular, we have 0 = tr(J(R(S))) =
∑n
i=1R
′
i(S)DiS. Since Dr+1S = · · · = DnS = 0 we obtain
1.4. Particular cases 20
∑r
i=1R
′
i(S)DiS = 0 =⇒ R′1(S) = · · · = R′r(S) = 0 =⇒ R1, . . . , Rr ∈ k. Since H(0) = 0 we have R(0) = 0
and so, R1 = · · · = Rr = 0. So G is a map of the form G = (X1, . . . , Xr, Xr+1 + Rr+1(S), . . . , Xn + Rn(S)) with
Rr+j(S) independent of Xr+1, . . . , Xn =⇒ G (and so F ) is invertible.
Chapter 2
Reductions
In this chapter we give some results about reductions. The main objective is to show that the Jacobian Conjecture
(over C) is true if it is so for any regular polynomial map F = X +H with H homogeneous and of degree 3 with
JH nilpotent. This result will be refined: it is sufficient to consider polynomial maps in the form F = X + H
where H is homogeneous, deg(H) = 3 and with JH nilpotent and symmetric.
2.1 Classical reduction theorem
Let R be a ring (as always, commutative with 1) and consider the polynomial ring R[n] := R[X1, . . . , Xn]. Let
M = Xe11 · · ·Xenn be a monomial and define ej(M) := ej for 1 ≤ j ≤ n. If f ∈ R[n] we denote
M˜(f) := {M | M is a monomial wich occurs in f }
and we define
e(f) :=
∑
M∈M˜(f)
n∑
i=1
Max{ej(M)− 1, 0}2.
Definition 2.1. Given g ∈ R[n] we say that g is of linear type if e(g) = 0.
Definition 2.2. Let F ∈MPn(R) with F = (F1, . . . , Fn). The linearity index of F is defined by
e(F ) :=
n∑
i=1
e(Fi).
21
2.1. Classical reduction theorem 22
Let F ∈MPn(R) be a polynomial map and m ∈ Z≥0. Consider the map
αm :MPn(R) −→MPn+m(R), F 7−→ (F,Xn+1, . . . , Xn+m)
For simplicity we denote αm(F ) by F
[m] and we call it the m-expansion of the map F .
It is easy to check the following properties:
(i) detJF = detJF [m].
(ii) F is invertible if and only if F [m] is invertible.
In the sequel MEn(R) will denote the group generated by all elementary n-dimensional maps (via composition)
over R. We also write
ME(0)n (R) :=MEn(R) ∩MP(0)n (R) and ME(1)n (R) :=MEn(R) ∩MP(1)n (R).
Theorem 2.1. (Bass-Connell-Wright)1 Let R be a ring and F ∈ MP(1)n (R) a Keller map. Then there are
m ∈ Z≥0, G,H ∈ME0n+m(R) and E(T ) ∈MPn+m(R[T ]) (T is a variable) such that
(i) E(1) = G ◦ F [m] ◦H. So if E is invertible =⇒ F is invertible.
(ii) Consider the polynomial map gotten from E(T ),
E˜ : Rn+m+1 −→ Rn+m+1
(X,T ) 7−→ (E(T ), T ).
Then E˜ = X +H with H homogeneous of degree 3, JH nilpotent and
E is invertible ⇐⇒ E˜ is invertible.
The proof is in 3 steps:
• reduction of degree.
• reduction to nilpotent case.
• reduction to homogeneous case.
1see [2, (2.1)Theorem]
2.1. Classical reduction theorem 23
Proposition 2.1. Let F ∈ MPn(R). Then there exist m ∈ Z≥0 and maps G,H ∈ ME(1)n+m(R) such that
F ′ = G ◦ F [m] ◦ H has degree ≤ 3. Furthemore, it is possible to take H ∈ ME(0)n+m(R) such that F ′ is of linear
type.
Proof. Let d be the degree of F and denote by g the number of monomials of degree d that occur in F . If d ≤ 3,
finish. Suppose that d > 3. We will show the proposition by induction in the pair (d, g). Let M be a monomial
wich occurs in F with degree d. Without loss of generality we can assume that M occurs in F1. Write M = PQ
where P and Q are monomials with degree ≤ d− 2 and consider the maps G,H ∈ME(1)n+2(R) defined by
G = (X1 −Xn+1Xn+2, X2, . . . , Xn, Xn+1, Xn+2) H = (X1, . . . , Xn, Xn+1 + P,Xn+2 +Q).
We have F ′ := G ◦ F [2] ◦ H = (F1 − Xn+1P − Xn+2Q − Xn+1Xn+2 − PQ,F2, . . . , Fn, Xn+1 + P,Xn+2 + Q).
Consider the pair (d′, g′) associated to F ′. By construction, we have d′ < d or d′ = d. In any case we have
g′ < g and by induction hypothesis there are maps G˜, H˜ ∈ ME(1)n+2+m(R) such that F˜ = G˜ ◦ (F ′)[m] ◦ H˜ =
G˜ ◦G[m] ◦ F [m+2] ◦H [m] ◦ H˜ have degree ≤ 3. This finishes the first part.
Let F ∈ MPn(R) be a polynomial map with degree ≤ 3. Suppose that e(F ) > 0 and let M be a monomial
wich occurs in F1 such that X
2
j |M for some 1 ≤ j ≤ n. Write M = PQ, with Xj |P and Xj |Q. We can
assume that P = Xj and Q = XjN with N ∈ {X1, . . . , Xn} or N ∈ R. By using G and H as above we have
F ′ := G◦F [2] ◦H = (F˜1, . . . , F˜n+2) = (F1−Xn+1P −Xn+2Q−Xn+1Xn+2−PQ,F2, . . . , Fn, Xn+1 +P,Xn+2 +Q)
with G ∈ ME(1)n+2(R) and H ∈ ME(0)n+2(R). Now if N = Xk with k 6= j or N ∈ R we have e(F˜1) = e(F1)− 2 and
e(F˜l) = e(Fl) for 1 ≤ l ≤ n and e(F˜n+1) = e(F˜n+2) = 0. In this case e(F˜ ) < e(F ).
If N = Xj we have P = Xj and Q = X
2
j . So, e(F˜1) = e(F1)− 1 and e(F˜l) = e(Fl) for 1 ≤ l ≤ n, e(F˜n+1) = 0 and
e(F˜n+2) = 1. In this case, e(F˜ ) = e(F ). In order to decrease e(F ) write Q = P
′Q′ with P ′ = Q′ = Xj and consider
the map S = (X1, . . . , Xn+2, Xn+3 + P
′, Xn+4 + Q′) and T = (X1, . . . , Xn+1, Xn+2 − Xn+3Xn+4, Xn+3, Xn+4).
We have
F
′′
:= T ◦ (F ′)[2] ◦ S = T ◦ (F˜1, . . . , F˜n+2, Xn+3 + P ′, Xn+4 +Q′)
= (F˜1, . . . , F˜n+2 −Xn+3Xn+4 − P ′Q′ −Xn+3Q′ −Xn+4P ′, Xn+3 + P ′, Xn+4 +Q′).
It is easy to check that e(F
′′
) < e(F ).
Thus in any case we see that it is possible to reduce the index e(F ). By induction we can find G ∈ME(1)(R) and
H ∈ME(0)n+m such that G ◦ F [m] ◦H has degree ≤ 3 and is of linear type.
Let F ∈MPn(R) be a Keller map. By the result above there are G,H ∈MEn+m(R) with H(0) = G(0) = 0 and
2.1. Classical reduction theorem 24
JG(0) = idn such that F
′ := G ◦F [m] ◦H has maximum degree 3. So to control invertibility we can suppose that
F = F ′. Write F = F(1) + F(2) + F(3) where F(i) is the homogeneous component of F in degree i. Since F is a
Keller map we have, in particular, that det JF (0) = det JF(1) ∈ R∗. So by composition with an automorphism
we can suppose that F = X + F(2) + F(3). Let T be a new variable over R and consider the polynomial map
E(T ) : R[T ]n −→ R[T ]n, X 7−→ X + TF(2)(X) + T 2F(3)(X). Note that
JE(T ) = idn + TJF(2) + T
2JF(3) = JF (TX)
So E(T ) is a Keller map over R[T ].
Consider new variables Y = (Y1, . . . , Yn) and let G(T ), H(T ) ∈MEn+m(R[T ]) be defined by
G(T ) = (X + TY, Y ) H(T ) = (X,Y − TF(3))
We have
E˜(T ) := G(T ) ◦ E(T )[n] ◦H(T ) = G(T ) ◦ (E(T ), Y − TF(3)) = (E(T ) + T (Y − TF(3)), Y − TF(3))
= (E(T ) + TY − T 2F(3), Y − TF(3)).
So E˜(T ) = (X + TF(2) + TY, Y − TF(3)) = (X,Y ) + T (F(2) + Y,−F(3)) = (X,Y ) + TN . By construction E˜(T )
is invertible if and only if E(T ) is invertible. Furthemore, the invertibility of E˜(1) implies the invertibility of the
map F . Also, we have E˜(1) = G(1) ◦ F [m] ◦H(1) and so we obtain the item (i) of theorem above.
Considering the jacobian map J2n we have
J2nE˜(T ) = id2n + TJ2nN ∈ GL2n(R[X1, . . . , Yn, T ])
where
J2nN =
 JnF(2) idn
−JnF(3) 0

Proposition 2.2. J2nN ∈M2n(R[X1, . . . , Yn, T ] is a nilpotent matrix.
Proof. Consider the matrix ring M2n(R[X1, . . . , Xn, T ]) with graded structure by T -degree:
M2n(R[X1, . . . , Xn, T ]) =
⊕
d∈N
M2n(R[X1, . . . , Xn, T ])d
2.1. Classical reduction theorem 25
whereM2n(R[X1, . . . , Xn, T ])d :=M2n(R[X1, . . . , Xn)T d. Note that TJ2nN is a homogeneous element of degree
1. So, the result follows from Remark 3
We can summarize the results above in the following
Proposition 2.3. Let R be a ring. Then the Jacobian Conjecture over R is true if and only if it is so for any
n ∈ N and all regular map F = X +N ∈MPn(R) with JN nilpotent and deg(F ) ≤ 3.
Now let F = X+N be a map as in the proposition above and set N = N(1)+N(2)+N(3) with N(i) the homogeneous
component of degree i. Let T be a variable over R and consider F (T ) = X + T 2N(1) + TN(1) + N(3) = X + N˜ .
Let F˜ : Rn+1 −→ Rn+1 be the map given by (X,T ) 7→ (F (T ), T ) = (X,T ) + (T 2N(2) + TN(1) +N(3), 0). We have
Jn+1F˜ = idn+1 + Jn+1(N˜ , 0) = idn+1 +
T 2Jn(N(2)) + TJn(N(1)) + Jn(N(3)) 2TN(2) +N(1)
0 0

Lemma 2.1. Let A =
⊕
d∈NAd be a graded ring and T a variable over A. Consider a graded structure in A[T ]
by
A[T ] =
⊕
d∈N
A[T ](d)
with A[T ](d) := A0T
d ⊕A1T d−1 ⊕ · · · ⊕Ad. Define A(d) := A0 ⊕A1 ⊕ · · · ⊕Ad.
The map αd : A(d) −→ A[T ](d) given by a0 + · · ·+ ad 7→ a0T d + · · ·+ ad is an isomorphism of groups. Furthemore
a ∈ A(d) is nilpotent if and only if αd(a) is nipotent.
Proof. The first item is easy. Given a ∈ Ad denote αd(a) := a(d) for simplicity. Pick e ∈ N and b ∈ A(e). We
affirm that (ab)(e+d) = a(d)b(e). Indeed, note that α−1d+e(a
(d)b(e)) = (a(d)b(e))(1) = a(d)(1)b(e)(1) = ab and so
(ab)(e+d) = a(d)b(e). In particular taking a = b and n ∈ N we have
(an)(nd) = (a(d))n
and so a is nilpotent ⇐⇒ a(d) is nilpotent.
Proposition 2.4. Jn+1(N˜ , 0) is nilpotent.
Proof. Consider the ring A := Mn(R[X1, . . . , Xn]) with graded structure by X-degree. We have T 2Jn(N(2)) +
TJn(N(1)) + Jn(N(3)) ∈ A[T ](3) and by the lemma above T 2Jn(N(2)) + TJn(N(1)) + Jn(N(3)) is nilpotent if and
2.2. Druzkowski maps 26
only if JnN = Jn(N(2)) + Jn(N(1)) + Jn(N(3)) is nilpotent. Since JnN is nilpotent we get the result.
A consequence of this is the following
Reduction Theorem. Suppose that, for all n ∈ N and all F ∈ MPn(R) of the form F = X + H with H
cubic homogeneous and JN nilpotent, F is invertible. Then for all n ∈ N and for any map F ∈ MPn(R) with
JF ∈ GLn(R[X1, . . . , Xn]), F is invertible.
2.2 Druzkowski maps
Definition 2.3. Let R be a domain and F ∈ MPn(R). We say that F is a Druzkowski map if F = X + H
with H = (H1, . . . ,Hn) cubic homogeneous of the form Hj = (
∑
k aklXk)
3 for some constants aij ∈ R.
We can apply the reduction theorem to show the following
Theorem 2.2. Let R be a Q-algebra. Suppose that for all n ∈ Z≥2 and F ∈MPn(R) all Keller Druzkowski maps
are invertible. Then the Jacobian Conjecture over R is true.
Proof. Let F ∈MPn(R) be a Keller regular map, i.e., F = X+H with JH nilpotent matrix and H homogeneous
of degree 3. We will show that F is invertible.
Given X,Y, Z ∈ R[X1, . . . , Xn] note the following formulas
XY 2 = 1/6(−2X3+(X−Y )3+(X+Y )3) and XY Z = ((X−Y−Z)3−(X+Y−Z)3−(X−Y+Z)3+(X+Y+Z)3)/24.
So we can rewrite Hj ∈ R[X1, . . . , Xn] in the form
Hl =
sj∑
k=1
αlkL
3
lk
where Llk ∈ QX1 + · · ·+QXn ⊂ Q[X1, . . . , Xn]. Let s := s1 + · · ·+ sn be the total quantity of terms of type L3ij .
Introduce new variables Y
(j)
i and consider the map FL ∈MPn+s(R) defined by
FL = (F1, . . . , Fn, Y
(1)
1 + L
3
11, . . . , Y
(1)
s1 + L
3
1s1 , . . . , Y
(n)
1 + L
3
n1, . . . , Y
(n)
sn + L
3
nsn).
By construction we have that F is invertible if FL is invertible. Consider the elementary map
E1 = (X1 −
s1∑
k=1
α1kY
(1)
k , X2, . . . , Xn, Y
(1)
1 , . . . , Y
(1)
r1 , . . . , Y
(n)
1 , . . . , Y
(n)
rn ).
2.3. The symmetric case 27
Note that
E1◦FL = (F1−
s1∑
k=1
α1kY
(1)
k −
r1∑
k=1
α1kL
3
1k, X2, . . . , Xn, Y
(1)
1 +L
3
11, . . . , Y
(1)
s1 +L
3
1s1 , . . . , Y
(n)
1 +L
3
n1, . . . , Y
(n)
sn +L
3
nsn).
Now, considering the elementary map E2 = (X1, X2 −
∑s2
k=1 α2kY
(2)
k , . . . , Xn, Y
(1)
1 , . . . , Y
(1)
r1 , . . . , Y
(n)
1 , . . . , Y
(n)
rn )
and composing with E1 ◦ FL we can erase the term
∑s2
k=1 α2kL
3
2k in the second component.
Proceeding this way we obtain a map of the form
F ′ = (X1 −
s1∑
k=1
α1kY
(1)
k , . . . , Xn −
sn∑
k=1
αnkY
(n)
k , Y
(1)
1 + L
3
11, . . . , Y
(1)
s1 + L
3
1s1 , . . . , Y
(n)
1 + L
3
n1, . . . , Y
(n)
sn + L
3
nsn).
and if F ′ is invertible then so is F . By composition with elementary maps we see that F ′ is invertible if and only
if the map
S = (X1, . . . , Xn, Y
(1)
1 + L˜
3
11, . . . , Y
(1)
s1 + L˜
3
1s1 , . . . , Y
(n)
1 + L˜
3
n1, . . . , Y
(n)
sn + L˜
3
nsn).
for L˜ij ∈ QX1 + · · ·+QXn. This finishes the proof.
2.3 The symmetric case
In this section we will study polynomial maps over C. If X1, . . . , Xn are variables we denote the ring C[X1, . . . , Xn]
by C[X].
Let f ∈ C[X] and consider the hessian matrix
H(f) =

fX1X1 fX1X2
...
...
... fX1Xn
fX2X1 fX2X2
...
...
... fX2Xn
fX3X1 fX3X2
...
...
... fX3Xn
...
...
...
...
...
...
fXnX1 fXnX2
...
...
... fXnXn

where fXiXj :=
∂2f
∂Xi∂Xj
.
Given a polynomial map H = (H1, . . . ,Hn) : Cn −→ Cn introduce new variables Y1, . . . , Yn and define
fH := (−i)H1(X1 + iY1, . . . , Xn + iYn)Y1 + · · ·+ (−i)Hn(X1 + iY1, . . . , Xn + iYi)Yn ∈ C[X,Y ].
2.3. The symmetric case 28
Considering the map S = (X1 − iY1, . . . , Xn − iYn, Y1, . . . , Yn) we have
gH := fH ◦ S = (−i)
∑n
k=1Hk(X)Yk.
Furthemore,
H(gH) =
 (∗∗) −i(JH)t
−iJH 0
 .
Lemma 2.2. Let A ∈Mn(C) and f ∈ C[X]. Define f ◦A := f(AX). Then
H(f ◦A) = AtH(f)|AXA.
Proof. Chain rule.
Lemma 2.3. Let H = (H1, . . . ,Hn) ∈ MPn(C) and suppose that JH is symmetric. Then there exists f ∈ C[X]
such that Hi = fXi for all i = 1, . . . , n.
Proof. Let Hi =
∑
α a
(i)
α Xα. By hypothesis we know
∂Hi
∂Xj
=
∂Hj
∂Xi
. Comparing the coefficients we obtain
αja
(i)
α1···(αi−1)···αn = αia
(j)
α1···(αj−1)···αn . Take f =
∑
α bαX
α where bα = aα1···(αi−1)···αn/αi for αi > 0 and bα = 0
if α = (0, . . . , 0).
Proposition 2.5. Let H = (H1, . . . ,Hn) ∈ MPn(C) be a polynomial map and fH the associated polynomial.
Then
H(fH) is nilpotent ⇐⇒ JH is nilpotent .
Proof. By simplicity denote fH by f and fH ◦ S by g. We know that H(f) is nilpotent if and only if
det(Tid2n −H(f)) = T 2n.
Let Q := 12
∑
k(X
2
k + Y
2
k ). Note that Q ◦ S = 12
∑
k((Xk − iYk)2 + Y 2k ) = 12
∑
kX
2
k − i
∑
kXkYk and so
H(Q ◦ S) =
 idn −i(idn)
−i(idn) 0

Also, we have
H(g) =
 (∗∗) −i(JH)t
−iJH 0

2.3. The symmetric case 29
So, det(H(TQ ◦ S − g)) = det(StH(TQ− f)|S(X,Y )S) = det(H(TQ− f)|S(X,Y )) = det(Tid2n −H(f))|S(X,Y ).
On the other hand, since
H(TQ ◦ S − g) =
 (∗∗) i[Tidn − (JH)t]
i[Tidn − JH] 0

we have det(H(TQ ◦ S − g)) = det(Tidn − JH) det(Tidn − JHt) = det(Tidn − JH)2.
So JH is nilpotent if and only if det(Tidn − JH) = Tn ⇐⇒ det(Tid2n −H(f)) = T 2n ⇐⇒ H(f) is nilpotent.
Now consider the following
Hessian Conjecture. If F = (X1, . . . , Xn) + (fX1 , . . . , fXn) for some f ∈ C[X] with H(f) nilpotent then F is
invertible.
Note that if the Jacobian Conjecture is true for dimension n then so is the Hessian Conjecture since nilpotency of
H(f) implies Keller condition det(F ) = 1. The surprising fact is that the converse holds in the following sense:
Theorem 2.3. Suppose that HP2n(C) is true. Then all regular polynomial map F = X + H with JH nilpotent
in dimension n is invertible, i.e.,
HP2n(C) =⇒ JPn(C).
Proof. Let F = X +H be a regular map in dimension n. By the proposition above we have that JH is nilpotent
if and only if H(f) is nilpotent where
f = (−i)H1(X1 + iY1, . . . , Xn + iYn)Y1 + · · ·+ (−i)Hn(X1 + iY1, . . . , Xn + iYn)Yn.
So since we assume that HP2n(C) is true we conclude that the map
F = (X1 + fX1 , . . . , Xn + fXn , Y1 + fY1 , . . . , Yn + fYn)
is invertible. In particular, F ◦G is invertible where G = (X1 − iY1, . . . , Xn − iYn, Y1, . . . , Yn). Now, we note that
F ◦G = F (X1 − iY1, . . . , Xn − iYn, Y1, . . . , Yn) = (G1, . . . , Gn, Gn+1, . . . , G2n) where
Gp = Xp − iYp + (−i)
∑
k
∂Hk(X1, . . . , Xn)
∂Xp
Yk 1 ≤ p ≤ n.
Gn+l = Yl +
∑
k
∂Hk(X1, . . . , Xn)
∂Xl
Yk − iHl(X1, . . . , Xn) 1 ≤ l ≤ n.
2.3. The symmetric case 30
By composition with G−1 we have that
G−1◦F ◦G = (E1, . . . , E2n) = (G1+iGn+1, . . . , Gn+iG2n, Gn+1, . . . , G2n) = (X1+H1, . . . , Xn+Hn, (∗∗), . . . , (∗∗))
is an invertible map. In particular, F = (X1 +H1, . . . , Xn +Hn) is an invertible polynomial map.
Using the theorem above and the reduction theorem we obtain
Symmetric Reduction Theorem. (Bondt-Essen) It is sufficient to prove the Jacobian Conjecture for all
polynomial maps F : kn −→ kn of the form “ gradient”, i.e., F = X + (fx1 , . . . , fxn) where f is homogeneous
of degree 4 and the hessian matrix H(f) is nilpotent (equivalently, it is sufficient to consider the case where F is
symmetric regular i.e. in the form F = X +H with H homogeneous of degree 3, J(H) nilpotent and symmetric
matrix).
Remark 4. Let k be a field. Recall that k is an ordered field if there is a partition k = P− ∪ {0} ∪ P+ such that
x, y ∈ P+ =⇒ x + y, xy ∈ P+. Note that 1 ∈ P− =⇒ (−1) ∈ P+ =⇒ (−1)2 = 1 ∈ P+ a contradiction. So,
1 ∈ P+ =⇒ 1 + 1 · · ·+ 1 ∈ P+ =⇒ char(k) = 0.
In relation to the theorem above, the following result is interesting:
• Let k be an ordered field. If M ∈Mn(k[X1, . . . , Xn]) is a nilpotent symmetric matrix then M = 0.
Indeed, let M ∈ Mn(k) be nilpotent and symmetric. Suppose that M 6= 0. Then, there is l ∈ N such that
M l = 0 and N := M l−1 6= 0. Since N 6= 0 we have that some row is non zero, say (u11, . . . , u1n). Since
N2 = 0 and N is symmetric we have u211 + · · · + u21n = 0. But this is a contradiction, since k is ordered.
Now let M ∈ Mn(k[X1, . . . , Xn]) be nilpotent and symmetric. By evaluation on (a1, . . . , an) ∈ kn we have
M(a1, . . . , an) = 0. If Mij denote the (i, j)-element in M then Mij is a polynomial that is zero in k
n. Since k is
infinite field (char(k) = 0) we have Mij ≡ 0 for all 1 ≤ i, j ≤ n. So, M = 0.
Chapter 3
Jacobian Conjecture via Zp
In this chapter we describe an approach to the Jacobian Conjecture via p-adic integers and we formulate the
Unimodular Conjecture of [12, Essen-Lipton]. We propose a new conjecture, the Invariance Conjecture, and we
show the equivalences
Jacobian Conjecture over C is true ⇐⇒ Unimodular Conjecture over Zp is true for almost all primes p
Jacobian Conjecture over C is true⇐⇒ Invariance Conjecture over Zp is true for almost all primes p .
In the end of this chapter we make some contributions to the Unimodular Conjecture.
3.1 Completion
Let R be a ring and I ⊂ R an ideal with ∩n∈NIn = 0 1. We can define a topology in R declaring that V ⊂ R is a
neighborhood of a ∈ R iff a+ In ⊂ V for some n ∈ N. Provided with this structure, we obtain a topological ring
(sum/product are continuous). This topology is called the I-adic topology.
Given a sequence s = {an} ⊂ R we say that s is a Cauchy sequence if for any n ∈ N there is m0 such that if
m, r > m0 then am − ar ∈ In. We remark that a convergent sequence is Cauchy, but there are Cauchy sequences
which are not convergent.
1In this chapter, we are interested only in the case where the I-adic topology is Hausdorff. This is equivalent to the condition
∩n∈NIn = 0.
31
3.1. Completion 32
Example 5. Let R = Z with I-adic topology where I = 5Z. We will construct a sequence {an} such that
a2n + 1 ≡ 0 mod 5n an+1 ≡ an mod 5n.
For this, take a1 = 2 and suppose that we have am defined for all m ≤ n. Define an+1 := an + b5n, with b to be
determined. We show that there is a b ∈ Z such that a2n+1 + 1 = (an + 5nb)2 + 1 ≡ 0 mod 5n+1 i.e. d+ 2ban ≡ 0
mod 5, with d = (a2n+1)/5
n ∈ Z. Now, since (an, 5) = 1 we can solve the congruence above and obtain the integer
wanted.
By construction, we have {an} Cauchy. But it isn’t convergent. Indeed, if {an} converge to α then, taking the
limit in the relation a2n + 1 = 0 mod 5
n, by continuity we have α2 + 1 = 0 in Z, a contradiction.
Definition 3.1. Let I be an ideal of R and consider the projective system {Φnm, R/In} where, for m ≥ n,
Φnm : R/I
m  R/In are the natural projections. The completion of R with respect to I-adic topology is by
definition R˜ := lim←−n∈NR/I
n.
Theorem 3.1. Let I ⊂ R with R commutative ring and I an ideal with ∩n∈NIn = 0. Consider the I-adic topology
in R. Then, R˜ is a topological Hausdorff complete ring with topology given by ideal I˜ = IR˜.
Proof. see [1, chapter 10].
Definition 3.2. Let p ∈ Z be a prime and consider the p-adic topology in Z. The p-adic ring is the completion of
Z with respect to p-adic topology, denoted Zp.
Theorem 3.2. Zp is a discrete valuation ring with uniformizer p and residue field Fp.
Proof. The associeted valuation is defined in the following way: If 0 6= a = (an)n∈N ∈ Zp define ordp(a) :=
Min{n ∈ N | an 6= 0}. If a = 0, by convention we define ordp(a) = ∞. We can check that ordp define a discrete
valuation in Zp. For the second affirmation, note that the following sequence
0 // Zp
φ // Zp
pi // Fp // 0
is exact where the map φ is the multiplication by p and pi is the natural projection.
Hensel Lemma. Let (O,M, k) be a complete discrete valuation ring and F1(X1, . . . , Xn), . . . , Fn(X1, . . . , Xn) ∈
O[X1, . . . , Xn]. Choose α = (α1, . . . , αn) ∈ On such that
F1(α1, . . . , αn) ≡ · · · ≡ Fn(α1, . . . , αn) ≡ 0 mod M2m+1
3.1. Completion 33
where m := ordM(det JF (α)) <∞. Then there is unique β = (β1, . . . , βn) ∈ On such that F1(β) = · · · = Fn(β) =
0 and βi ≡ αi mod Mm+1 for all i = 1, . . . , n.
Proof. See [5].
By Hensel lemma we get the following
Proposition 3.1. Let f1, . . . , fn ∈ O[X1, . . . , Xn] satisfying the Keller condition where (O,M, k) is a complete
DVR. If R is an O-algebra denote by X(R) the set of R-points. Then there is a bijection X(O) ∼= X(k).
Proof. As f is Keller we have m = ordM(det JF (α)) = 0 for all α ∈ On. The bijection is natural: given P ∈ On
define ϕ(P ) ∈ X(k) the k-point obtained by reduction mod M. Hensel lemma implies that ϕ : X(R) −→ X(k)
is a bijection: injectivity by uniqueness and surjectivity by lifting.
Proposition 3.2. Let R be a noetherian domain and I ⊂ R an ideal. Then, ∩n∈NIn = 0.
Proof. see [1].
Lemma 3.1. Let Q(T ) = anT
n + · · ·+ a0 ∈ Z[T ] be an irreducible polynomial and E ⊂ Z a finite set of rational
primes. Then there is a prime p ∈ Z \ E and α ∈ Zp such that Q(α) = 0.
Proof. By Hensel lemma, it is sufficient to show that there exists a prime p /∈ E and b ∈ Z such that
Q(b) ≡ 0 mod pZp and Q′(b) 6≡ 0 mod pZp.
For this, denote d the discriminant of Q(T ). We remark that d ∈ Z \ {0} since d := resultant(Q,Q′) ∈ Z and
the separability of Q implies d 6= 0. Define e := 2d∏p∈E p and for l ∈ N denote ml := a0el. We have,
Q(ml) = a0(ana
n−1
0 e
ln + · · ·+ a1el + 1) = a0Q˜(ml) (∗).
Now, since Q˜(t) ∈ {1, 0,−1} occurs for a finite number of t ∈ Z, we can take l ∈ N such that Q˜(ml) 6= 1, 0,−1. Let
p ∈ Z be a prime such that p | Q˜(ml). In particular, we have Q(ml) ≡ 0 mod pZp. Note that p - e. Indeed, if p | e
we have, by (∗), p | 1 a contradiction. In particular, p - d =⇒ Q′(ml) 6≡ 0 mod pZp. So, we get the result.
Proposition 3.3. Let E ⊂ Z be a finite set of rational primes and α ∈ Q. Then there is a prime p ∈ Z \ E and
an injective homomorphism φ : Z[α] ↪→ Zp.
3.1. Completion 34
Proof. Let mα(T ) ∈ Q[T ] be the minimal polynomial of α. We have mα(T ) = c(mα(T ))f(T ) for some primitive
polynomial f(T ) ∈ Z[T ]2. We affirm that
Q[T ]f(T ) ∩ Z[T ] = f(T )Z[T ].
The inclusion ⊃ is trivial. Let h(T ) ∈ Q[T ]f(T ) ∩ Z[T ] and write h(T ) = 1aA(T )f(T ) with A(T ) primitive.
By Gauss lemma we know that the product of primitive polynomials is primitive. So, c(h(T )) = 1/a ∈ Z and
a ∈ {1,−1} =⇒ h(T ) ∈ Z[T ]f(T ).
So
Z[α] ∼= Z[T ]
f(T )Z[T ]
↪→ Q[T ]
f(T )Q[T ]
.
Let p /∈ E be a prime such that f(a) = 0 for some a ∈ Zp and consider the map Φ : Q[T ] −→ Qp, evaluation by a
where Qp := Frac(Zp)
Since f(T ) is irreducible we have ker(Φ) = f(T )Q[T ] and so
Q[T ]
f(T )Q[T ]
↪→ Qp.
By composition of isomorphims we get Z[α] ↪→ Zp.
Immersion Lemma. Let α1, . . . , αn ∈ Q. Then for infinitely many primes p ∈ Z there is an injective homomor-
phism
ϕp : Z[α1, . . . , αn] ↪→ Zp.
Proof. Consider K = Q(α1, . . . , αn), the field obtained by adjunction. By the primite element theorem, we have
K = Q(α) for some α ∈ Q. In particular, we can find d ∈ Z such that dαi ∈ Z[α] for all i ∈ {1, . . . , n}. Consider
the multiplicative system S = {dn | n ∈ N}. We have
Z[α1, . . . , αn] ↪→ S−1Z[α].
Denote E := {q | q prime with q|d}. By the lemma above, we know that there is a prime number p /∈ E and an
injective homomorphism ϕp : Z[α] ↪→ Zp. Since all elements of S are invertible in Zp, the map ϕp induces an
injection S−1Z[α] ↪→ Zp. By composition, we get Z[α1, . . . , αn] ↪→ Zp. Replacing E by E′ = E ∪ {p} and by
2Recall the content of polynomials: Given f(T ) = anTn+an−1Tn−1 + · · ·+a0 ∈ Q[T ] by definition c(f(T )) :=
∏
p p
ordp(f), where
the product take all primes in Z and ordp(f) := Min{ordp(ai) | 0 ≤ i ≤ n}. If f, g ∈ Q[T ], Gauss lemma implies C(fg) = C(f)C(g)
3.2. A reformulation of the Jacobian Conjecture 35
repetion of the argument above we get a prime q /∈ E′ and an injection Z[α1, . . . , αn] ↪→ Zq. So, Z[α1, . . . , αn] ↪→ Zp
for infinite primes.
3.2 A reformulation of the Jacobian Conjecture
We start with some lemmas.
Lemma 3.2. Let R be a domain and F ∈ R[X1, . . . , Xn]n with F (0) = 0. Let P = (p1, . . . , pn), Q = (q1, . . . , qn) ∈
Rn be distinct points.
(i) If F (P ) = F (Q) and det JF (0) ∈ R∗ then the ideal 〈p1, . . . , pn, q1, . . . , qn〉 = R.
(ii) If F (P ) = F (Q) and det JF ∈ R∗ then the ideal 〈p1 − q1, . . . , pn − qn〉 = R.
Proof. Note that we can suppose R is a noetherian ring. Indeed, by adjunction of all elements that occurs in F ,
det JF and the coordinates of P and Q to the prime subring of R we obtain a subring A ⊂ R of finite type over
a noetherian ring. So achieve a noetherian ring.
Suppose that F (P ) = F (Q) and det JF (0) ∈ R∗ but 〈p1, . . . , pn, q1, . . . , qn〉 ⊂ m for some maximal ideal m ∈
Specm(R). Denote R˜ the m-adic completion of R. By injectivity of the natural map R −→ R˜ we can suppose
P,Q ∈ R˜. Let G be the formal inverse of F over R˜. It is easy to see that G(F (P )) and G(F (Q)) are well defined
elements in R˜. In particular, P = G(F (P )) = G(F (Q)) = Q a contradiction.
Given F with F (P ) = F (Q) and det JF ∈ R∗ consider FQ := F (X +Q)−F (Q) and note that that FQ(P −Q) =
F (P )− F (Q) = 0 and FQ(0) = 0. Also, we have det JFQ = det JF (X +Q) ∈ R∗ =⇒ det JFQ(0) ∈ R∗. So by (i)
we obtain 〈p1 − q1, . . . , pn − qn〉 = R.
Proposition 3.4. (Connell - van den Dries) Let R be a domain and F ∈ MPn(R) = R[X1, . . . , Xn]n with
F (0) = 0 and det JF ∈ R∗. Let I ( R be an ideal and consider the induced map F : (R/I)n −→ (R/I)n. If F is
injective then so is F .
Proof. Suppose this false and let P = (p1, . . . , pn), Q = (q1, . . . , qn) ∈ Rn be distinct points such that F (P ) =
F (Q). By Keller condition and the lemma above we have
1 = α1(p1 − q1) + · · ·+ αn(pn − qn) for some α1, . . . , αn ∈ R.
3.2. A reformulation of the Jacobian Conjecture 36
By reduction mod I we have 1 = α1(p1 − q1)+ · · ·+αn(pn − qn). On the other hand, the injectivity of F implies
that pi − qi ∈ I for all i. This implies 1 ∈ I a contradiction.
Lemma 3.3. Let F ∈ MPn(Z) be such that the induced map F ∈ MPn(Zp) is injective for almost all primes
p ∈ Z. Then d := det JF ∈ Z \ {0} and F is invertible over Z[d−1].
Proof. Consider F as a map over Q (by scalar extension). We claim that F ∈MPn(Q) is injective. Suppose this
false and let P = (p1, . . . , pn), Q = (q1, . . . , qn) ∈ Q be distinct points such that F (P ) = F (Q). By the immersion
lemma 3.3, we know that there is an injective homomorphism ϕp : Z[P,Q] ↪→ Zp for infinitely many primes p ∈ Z.
So F (ϕp(P )) = ϕp(F (P )) = ϕp(F (Q)) = F (ϕp(Q)). Since F is injective over Zp (for almost all primes) we have
ϕp(P ) = ϕp(Q) =⇒ P = Q, a contradiction. So F ∈ MPn(Q) is injective. By Cynk-Rusek theorem 1.6 we know
that F is, of course, an isomorphim. In particular, det JF ∈ Q∗ ∩Z[X1, . . . , Xn] = Z \ {0}. By the inverse formal
theorem 1.1 we know that F has a formal inverse over Z[d−1]. Since F is invertible over Q it follows that the
inverse over Z[d−1] is polynomial.
Theorem 3.3. The Jacobian Conjecture is equivalent to the following statement:
• For almost all primes p ∈ Z, n ∈ Z≥2, for any map F ∈ MPn(Z) with det JF = 1 the induced map
F ∈MPn(Fp) is injective.
Proof. Suppose that the Jacobian Conjecture is true and take F ∈MPn(Z) a polynomial map with the condition
det JF = 1. By the inverse formal theorem we know that F is an invertible map over Z if and only if is invertible
over C. By hypothesis we have the Jacobian Conjecture true and so F is an invertible map. Denote G the inverse.
By reduction mod p the relations F ◦ G = X and G ◦ F = Y imply that the induced map F ∈ MPn(Fp) is a
bijection.
Now suppose that for almost all primes p ∈ Z, and all n ∈ Z≥2 any map F ∈MPn(Z) with det JF = 1 is injective.
Let F ∈MPn(C) be a Keller map that isn’t injective, i.e., a counterexemple for the Jacobian Conjecture. We can
assume F ∈ MPn(Z) with detJF = 1 (by Connell-van den Dries theorem, see chapter 1, 3.4). By a translation
we can assume F (0) = 0. The map F is injective over Zp. By the lemma above this implies that F is invertible
over Z[d−1] where d = 1. So F is invertible over Z. Contradiction.
3.3. The Unimodular Conjecture 37
3.3 The Unimodular Conjecture
Let (O,M, k) be a local domain and consider a linear map F : On −→ On i.e. F = (F1, . . . , Fn) with Fi ∈
O[X1, . . . , Xn] homogeneous of degree 1. Suppose that the matrix B := JF is invertible and let A ∈ Matn(O)
be such that AB = BA = idn. The relation BA = idn implies that there exist u1, . . . , un ∈ O such that
F1(u1, . . . , un) = 1. In particular, by reduction mod M we have F : kn −→ kn a non-zero map. The general
case is an open problem, the so called
Unimodular Conjecture. Let (O,M, k) be a local domain with char(O) = 0 and F ∈ MPn(O) (n > 1) a
Keller map. Then the induced map F ∈MPn(k) is a non-zero map.
Given F ∈ MPn(O) over a local domain (O,M, k) we will denote by f ∈ MPn(k) the induced map over the
residue field k.
Definition 3.3. Let (O,M, k) be a local domain. We say that O is a unimodular domain if the Unimodular
Conjecture is true for O. We say that a polynomial map F ∈MPn(O) is unimodular if it satisfies the condition
on Unimodular Conjecture.
Proposition 3.5. (WM)3 Let (O,M, k) be a local domain with k an infinite field. Then O is unimodular.
Proof. Let F ∈ MPn(O) be a Keller map. Let f ∈ MPn(k) be the induced map over the residue field and
suppose that f(α) = 0 for all α ∈ kn. Since k is infinite we have f ≡ 0. So the coefficients that occur in F belong
to the maximal ideal M. In particular, det JF ∈M[X1, .., Xn] a contradiction by Keller condition: det JF = 1.
Theorem 3.4. Suppose that the Jacobian Conjecture over C is true. Then every local domain (O,M, k) with
char(O) = 0 is unimodular.
Proof. Let F ∈ MPn(O) be a Keller map over a local domain O with char(O) = 0. Since we assume that
Jacobian Conjecture is true over C we have F an invertible map over O (see corollary 1.2). So, there is a unique
G ∈ MPn(O) such that F ◦G = X. By reduction mod M we see that the map f ∈ MPn(k) is a bijection, in
particular, non-zero map.
We remark that the Unimodular Conjecture is false for local domains with char(O) = p > 0 and residue field
finite. For example: consider the local domain (Fp[[T ]], TFp[[T ]],Fp) and take the polynomial map F = (X1 −
Xp1 , . . . , Xn − Xpn) ∈ MPn(Fp[[T ]]). Note that F is a Keller map but the induced map over the residue field is
the zero map, since αp = α for all α ∈ Fp.
3This is more general than [12, Theorem 4]
3.3. The Unimodular Conjecture 38
Remark 5. Let (O,M, k) be a local domain. The following table shows the complete set of relations between
char(O) and char(k).
char(O) char(k) #k type
p = 0 q > 0 ∞ unimodular
p = 0 q > 0 <∞ unknown
p = 0 q = 0 ∞ unimodular
p > 0 q = p <∞ non-unimodular
p > 0 q = p ∞ unimodular
Thus the interesting case is (char(O), char(k),#k, type) = (0, p,< ∞, unknown) where p > 0. Indeed, we will
see later (Essen-Lipton theorem) that unknown = unimodular if and only if the Jacobian Conjecture over C is
true.
3.3.1 n-dimensional 2-sets
Definition 3.4. 4 Let R be a domain. We denote by SR(n, 2) the category where the objects are of the form
(P,Q) (ordered) with P 6= Q ∈ Rn and a morphism F : (P1, Q1) −→ (P2, Q2) is restriction of a polynomial
map in MPn(R) with the condition F (P1) = P2 and F (Q1) = Q2. Each object (P,Q) in SR(n, 2) is called an
n-dimensional 2-set.
In what follows, given P = (p1, . . . , pn) and Q = (q1, . . . , qn) in R
n we denote by 〈Q− P 〉 the ideal
〈p1 − q1, . . . , pn − qn〉 ⊂ R.
Theorem 3.5. Let X = (a, b) and Y = (c, d) n-dimensional 2-sets. Then
(i) Hom(X,Y ) 6= ∅⇐⇒ 〈d− c〉 ⊂ 〈b− a〉.
(ii) X ∼= Y ⇐⇒ 〈d− c〉 = 〈b− a〉.
Proof. Suppose that Hom(X,Y ) 6= ∅ and pick F ∈MPn(R) such that F (a) = c and F (b) = d. Let G = (X−c)◦F
so that G(a) = 0 G(b) = d− c. Since Gi(a) = 0 we have
Gi(X1, . . . , Xn) = fi1(X1, . . . , Xn)(X1 − a1) + · · ·+ fin(X1, . . . , Xn)(Xn − an).
So, di − ci = Gi(b1, . . . , bn) ∈ 〈b1 − a1, . . . , bn − an〉 for all i = 1, . . . , n =⇒ 〈d− c〉 ⊂ 〈a− b〉. Reciprocally, if
〈d− c〉 ⊂ 〈b− a〉 then
dj − cj = e1j(b1 − a1) + · · ·+ enj(bn − an)
4see [12, 2-transitivity]
3.3. The Unimodular Conjecture 39
for some eij ∈ R. Define Gj := e1j(X1 − a1) + · · ·+ enj(Xn − an) and take F = (X + c) ◦G.
(ii) follows from (i) trivially.
Theorem 3.6. Let R be a PID. Suppose that 〈a, b〉 ∼= 〈c, d〉. Then, there exists an affine automorphism S ∈
MPn(R) with det JS = 1 such that S(a) = c and S(b) = d.
Proof. Since R is a PID we have 〈b− a〉 = 〈g〉 for some g ∈ R. Write bi − ai = gvi for vi ∈ R with i = 1, . . . , n
and note that 〈v1, . . . , vn〉 = R. In particular, it follows that (v1, . . . , vn) is a unimodular sequence. So there is
B = (bij) ∈ Gln(R) with detB = 1 the first row of which is the n-tuple v = (v1, . . . , vn) 5. Let A = (aij) ∈ Gln(R)
such that AB = BA = idn and define for 1 ≤ k ≤ n
Fk(X1, . . . , Xn) = a1k(X1 − a1) + · · ·+ ank(Xn − an).
Consider the affine automorphism F = (F1, . . . , Fn) ∈ MPn(R). We have Fk(a) = 0 and Fk(b) = a1k(b1 −
a1) + · · · + ank(bn − an) = a1kgv1 + · · · + ankgvn = g(a1kv1 + · · · + ankvn) = gδ1k. In particular, F (a) = 0 and
F (b) = (g, 0, . . . , 0) = ge1. Repeating the argument, change 〈b− a〉 by 〈d− c〉 and take the affine automorphism
G ∈ MPn(R) such that G(c) = 0 and G(d) = ge1. So, define S := G−1 ◦ F . By construction, S is an affine
automorphism with S(a) = c and S(b) = d.
Proposition 3.6. Let R be a local unimodular domain and F ∈ MPn(R) a Keller map. Then for all a ∈ Rn
there is d ∈ Rn such that F (d)− F (a) is unimodular.
Proof. Consider Fa = F (X + a)− F (a). Note that Fa is a Keller map and since R is unimodular there is b ∈ Rn
such that Fa(b) is unimodular. Now take d := b+ a.
Theorem 3.7. Let R be a local unimodular PID and F ∈ MPn(R) Keller map non-injective. Then for each
m ∈ Z≥2 there is a Keller map G ∈MPn(R) and P ∈ Rn with #G−1(P ) ≥ m.
Proof. This will be shown in the next section in more general form.
Proposition 3.7. Let (O,M, k) be a complete discrete valuation ring with finite residue field. Let F ∈MPn(O)
be a Keller map. Then for all Q ∈ On we have #F−1(Q) ≤ #(k)n.
Proof. By a translation we can suppose Q = (0, . . . , 0). So, finding the fiber F−1(Q) is equivalent to solve the
algebraic system over O:
F1 = · · · = Fn = 0.
5This follows from the structure of modules of finite type over a PID.
3.4. The Invariance Conjecture 40
By Hensel lemma we can prove that the natural map
F−1(Q) = {α ∈ On | F1(α) = · · · = Fn(α) = 0} −→ {b ∈ kn | F1(b) = · · · = Fn(b) = 0}
is injective and so #F−1(Q) ≤ #(k)n.
We will now prove the following
Theorem (Essen-Lipton). Zp is a unimodular domain for almost all primes p if and only if the Jacobian
Conjecture is true (over C).
Proof. The implication ⇐= follows from 3.4. Now suppose that Zp is a unimodular domain for almost all primes
p. By a result of Connell-van den Dries (see chapter 1, 1.7) it is sufficient to show that the Jacobian Conjecture
is true over Z. So let F ∈ MPn(Z) be a polynomial map with det JF = 1. Since Z ↪→ Q is sufficient to prove
that the map over the extension F ⊗ Q ∈ MPn(Q) is injective. Indeed, if this occurs then F is an isomorphim
(Cynk-Rusek) over Q and so F−1 is a polynomial map over Z, via the formal inverse function theorem.
Suppose by contradiction that F isn’t injective. Then there are distinct a = (a1, . . . , an), b = (b1, . . . , bn) ∈ Qn
such that F (a) = F (b). By the immersion lemma we know that there exist an injective homomorphism ϕp :
Z[a1, . . . , an, b1, . . . , bn] ↪→ Zp for infinitely many primes p. So the map F ⊗ Zp isn’t injective for infinitely many
primes p. Fix such a prime p for which Zp is a unimodular domain. By theorem 3.7, we can find G ∈ MPn(Zp)
Keller map and Q ∈ Znp such that #G−1(Q) ≥ p2n > pn, a contradiction by proposition 3.7. So F is injective.
3.4 The Invariance Conjecture
Invariance Conjecture. Let (O, M, k) be a local domain with char(O) = 0 and F ∈ MPn(O) a Keller
unimodular map. Let G ∈ Autn(O) be an affine Keller automorphism i.e. G = AX + b where A ∈ Sln(O). Then
F ◦G ◦ F and F − F (a) are unimodular maps for all a ∈ On.
Remark 6. Note that in the above conjecture we ask the unimodular property to be invariant under translation
and composition of a special type. Note also that, as in the unimodular case, if the residue field k is infinite then
the Invariance Conjecture is true for any complete discrete valuation ring (O,M, k) with char(O) = p ≥ 0.
Definition 3.5. Let (O, M, k) be a local domain. Given a map F ∈MPn(O) we say that
• F is an invariant map if F is Keller, unimodular and satisfies the Invariance Conjecture condition.
3.4. The Invariance Conjecture 41
• F is strongly invariant if it is invariant and for all Keller affine automorphisms G1, . . . , Gk ∈ Autn(O)
the map F1 ◦ F2 ◦ F3 ◦ · · · ◦ Fk is invariant where Fj = Gj ◦ F .
The domain O is called an invariant domain if every polynomial map that is Keller and unimodular (in dimen-
sion n > 1) is invariant.
Lemma 3.4. If a map F ∈MPn(O) is strongly invariant then F ◦G◦F is strongly invariant for all Keller affine
automorphism G ∈ Autn(O).
Proof. Indeed, by induction it is sufficient to consider the case k = 2. For this, let G1, G2 ∈ Autn(O) be a Keller
affine automorphisms and remark that G1 ◦ (F ◦G ◦F ) ◦G2 ◦ (F ◦G ◦F ) = (G1 ◦F ) ◦ (G ◦F ) ◦ (G2 ◦F ) ◦ (G ◦F ).
So it is invariant by hypothesis on F .
Proposition 3.8. Let (O, M, k) be a local unimodular domain. Then O is invariant.
Proof. By hypothesis a Keller map F ∈ MPn(O) is unimodular. Since unimodularity is invariant under compo-
sition and translation we have the result.
The condition char(O) = 0 is important.
Example 6. Let F1, . . . , Fn ∈ Fp[[T ]][X1, . . . , Xn] be defined by Fj = 1 −Xpj + Xj and consider the polynomial
map F = (F1, . . . , Fn) ∈MPn(Fp[[T ]]). It is easy to check that det JF = 1 and that F is unimodular. But
F − F (1, . . . , 1) = (−Xp1 +X1, . . . ,−Xpn +Xn).
So, in case (O, M, k) = (Fp[[T ]], TFp[[T ]],Fp) it follows that the property of invariance by translation is false.
Example 7. Let g(X) ∈ Fp[X] be a polynomial wich maps {0, . . . , p − 2} 7→ p − 1 and p − 1 7→ 0. For example,
take p = 5 and consider
g(X) = −1 +X − 1X2 +X3 + 4X4 ∈ F5[X].
It is easy to check that g◦g = 0. Note that g(0) 6= 0. Define the polynomial map F = (F1, . . . , Fn) ∈MPn(Fp[[T ]])
with Fj = Xj −Xpj + g(Xpj ). We have F a Keller map with the induced map over the residue field non-zero. But
by construction we have F ◦ F = 0. Thus, in characteristic p > 0 the invariance by composition is false.
In the next theorem the argument is simitar to the argument given in [12, Theorem 4] with the observation that
is sufficient to require the invariance property.
3.5. Some results 42
Theorem 3.8. (WM) Let (O,M, k) be a complete discrete valuation ring with finite residue field. Let F ∈
MPn(O) be a strongly invariant map. Then F is injective.
Proof. Suppose false and let F be a strongly invariant map over O with F (a1) = · · · = F (am) = c (m > 1)
for some a1, . . . , am ∈ On with ai 6= aj , if i 6= j. We will show that there is a strongly invariant map G with
#G−1(c) > m. By iteration we will get a Keller map G˜ ∈ MPn(O) with #G−1(c) > (#k)n a contradiction by
proposition 3.7 above.
Since F (a1) = F (a2) we have 〈a2 − a1〉 = R (lemma 3.2 above). On the other hand, since F is an invariant
map it is ensured that there exists b ∈ On such that F (b) − F (a1) is unimodular, i.e., 〈F (b)− F (a1)〉 = R. In
particular, 〈a2 − a1〉 = 〈F (b)− F (a1)〉 = 〈F (b)− c〉 = R. So, we have (a2, a1) ∼= (F (b), c). By theorem 3.6 we
know that there is H ∈ MPn(O), Keller affine automorphism such that H(c) = a1 and H(F (b)) = a2. Now
define G = F ◦ H ◦ F . We have G strongly invariant map with G(aj) = F (H(c)) = F (a1) = c for all j and
G(b) = F (H(F (b))) = F (a2) = c. Note that b 6= aj for all j.
Theorem 3.9. (WM) Let (O,M, k) be a complete discrete valuation ring with finite residue field. Suppose that
O is an invariant domain. Then any unimodular Keller polynomial map F ∈MPn(O) is injective.
3.5 Some results
Definition 3.6. Pick d ∈ Z≥1 and let (O,M, k) a local domain. We say that O is a d-unimodular map if any
Keller map F ∈MPn(O) in dimension n > 1 with deg(F ) ≤ d is unimodular.
Note that any local domain O is 1-unimodular and O is a unimodular domain if and only if it is d-unimodular for
all d ∈ N. If O is d-unimodular then it is e-unimodular for all e ≤ d. We will see later that Zp is 3-unimodular
for any prime p > 3. In case char(O) = p > 0 and k finite we have that O isn’t d-unimodular for infinitely many
d ∈ Z. Indeed, for each m ∈ N take d = (#k)m and consider the map F = (X1 −Xd1 , . . . , Xn −Xdn) ∈MPn(O).
Proposition 3.9. (WM) Let F ∈MPn(Z) be a non constant polynomial map. Then for almost all primes p ∈ Z
we have F ⊗ Zp unimodular map over Zp.
Proof. Indeed, suppose F1(X1, . . . , Xn) ∈ Z[X1, . . . , Xn] \ Z. We can choose d ∈ Zn such that F1(d) 6= 0. Note
that F1(d) ∈ Z∗p for all p such that p - F1(d).
3.5. Some results 43
We have seen that to prove the Jacobian Conjecture it is sufficient to consider polynomial maps of Druzkowski
type, i.e., maps in the form F = X + H with Hj = (
∑
k akjXk)
3 and JH nilpotent. We call maps of the form
F = X +H with H =
∑
k akjX
3
k quasi-Druzkowski maps.
Proposition 3.10. (WM) For almost all primes p, the Unimodular Conjecture over Zp is true for quasi-
Druzkowski maps.
Proof. Let F be a quasi-Druzkowski map with H = (H1, . . . ,Hn) where Hj =
∑
k bkjX
3
k . We will show that there
exist u1, . . . , un ∈ Zp, not all null , such that
u1H1(X1, . . . , Xn) + · · ·+ unHn(X1, . . . , Xn) = 0.
Indeed, for this it is sufficient to find a non trivial solution for the homogeneous system:
u1b11 + u2b12 + · · ·+ unb1n = u1b21 + u2b22 + · · ·+ unb2n = · · · = u1bn1 + u2bn2 + · · ·+ unbnn = 0.
Now since JH is nilpotent we have, in particular, det(bij) = 0 and so there is a non-trivial solution (u1, . . . , un) ∈
Qnp for the system above. Without loss of generality we can suppose that u1 ∈ Z∗p and uj ∈ Zp, if j > 1. Now
consider s := u1 + u2p · · ·+ unp ∈ Z∗p. Note that, (1, p, . . . , p) ∈ Znp is such that
〈F1(1, p, . . . , p), . . . , Fn(1, p, . . . , p)〉 = Zp.
Remark 7. The proposition above will be generalized later (see corollary 3.2).
It was seen in the previous section that there are local domains (O,M, k) with char(O) = p > 0 that are not
unimodular domains. On the other hand we know that any local domain with infinite residue field is indeed a
unimodular domain. In particular, if we consider the map F = (X1−Xp1 , . . . , Xn−Xpn) over (Fp[[T ]], TFp[[T ]],Fp)
we have F (α) 6= 0 for some α ∈ Fp ( = algebraically closure of Fp). So, if we take L = the field obtained by
adjunction of α to Fp we see that our F is unimodular over the local domain (L[[T ]], TL[[T ]], L). For the p-adic
case there is an analogue:
Theorem 3.10. (WM) Let F ∈ MPn(Zp) be a Keller map. Then there is a complete discrete valuation ring
(O,M, k) that dominates Zp such that F ⊗O is a unimodular map. Furthermore, O is a free Zp-module with
rankZp(O) = [k : Fp].
Proof. In the proof we use a result of algebraic number theory.
3.5. Some results 44
Consider the map F ∈MPn(Fp) induced over the residue field. By the previous remark we know that there exists
α ∈ Fpn such that F (α) 6= 0. By taking the field k := Fp(α1, . . . , αn) obtained by adjunction we can look at F as
a polynomial map which is non zero over k. Now we recall the following theorem about unramified extensions of
a local field L 6:
Theorem. Let L be a local field with residue field l. There exists a 1-1 correspondence between the following sets
{finite extensions unramified over L } ∼= {finite separable extensions of l}
given by L′ 7→ l′, where l′ is the residue field associated to L′. Furthermore, in this correspondence we have
[L′ : L] = [l′ : l].
Applying the theorem above to L = Qp with l = Fp we see that the extension k|Fp corresponds to a local field
K|Qp such that k is the residue field of K. Denote by (O,M, k) the ring of integers of K. The ring O is the
integral closure of Zp in K and by a general result (cf.[1, proposition 5.17]) we know that O is a free Zp-module
and rankZp(O) = [K : Qp] = [k : Fp]. So F ⊗O ∈MPn(O) is a Keller map with non zero induced map over the
residue field.
The theorem below was gotten in an attempt of the author to show the following
Conjeture. 7 Denote by O the integral closure of Z in Q. Let F : Zn −→ Zn be a Keller map such that F ⊗O
is injective. Then F is an isomorphim.
Lemma 3.5. Let K|Qp be a finite Galois extension with m := [K : Qp] > 1. Let OK be the integral closure of
Z in K. Let F ∈ MPn(OK) be a non-injective Keller unimodular map. Then there exists a non-injective Keller
unimodular map G ∈MPmn(Zp).
Proof. The same argument of lemma 1.2 works. The relevant fact is that OK is a free Zp-module of rankZp(OK) =
m.
By proposition 3.7 we know that if Zp is a unimodular domain then any Keller map over Zp is injective. We can
show a more general result
Theorem 3.11. (WM) Assume that Zp is invariant domain for some prime p. Then for all Keller unimodular
maps F ∈MPn(Zp) and K|Qp finite extension we have F ⊗OK is an injective map.
6See appendix, theorem 4.10
7see mathoverflow “Polynomial maps over Z” for an incomplete argument due to Jason Starr
3.5. Some results 45
Proof. It is sufficient to show that F ⊗ O is an injective map where O denotes the integral closure of Zp in Qp.
Indeed, if α 6= β ∈ O are such that F (α) = F (β) consider the ring R = Zp[α, β] obtained by adjunction and let
K := Frac(R). We have K|Qp a finite extension such that α1, . . . , αn, β1, . . . , βn ∈ K. Note that αi, βi ∈ OK
for all i. Without loss of generality we can suppose that K|Qp is a Galois extension. So, F ⊗ OK is a Keller
unimodular map over OK that isn’t injective. By the lemma above, we obtain G ∈MPN (Zp) a Keller unimodular
map that isn’t injective. Now since we assume that Zp is a invariant domain and G is strongly invariant map, by
theorem 3.8, we know that G is an injective map. A contradiction.
In the direction of the Unimodular Conjecture we have the interesting result:
Theorem 3.12. (WM) Let (O,M, k) be a local domain with q := #k < ∞. Then O is a (q − 1)-unimodular
domain.
Note that there are no restrictions about char(O).
Proof. Let F : On −→ On be a Keller map with deg(F ) ≤ q − 1. Denote by fj the polynomial in k[X1, . . . , Xn]
obtained by reduction of Fj mod M. By Keller condition we have fj is a non zero polynomial, for all indices j.
So we obtain a polynomial map f = (f1, . . . , fn) : k
n −→ kn. We must prove that there exists α ∈ kn such that
f(α) 6= 0.
By passing to algebraic closure consider the algebraic set X ⊂ kn defined by equations f1 = · · · = fn = 0. We
affirm that dimX = 0 8. Indeed, note that X = f−1(0), where f is the map over k defined by tuple f1, . . . , fn. The
Keller condition implies that [k(X1, . . . , Xn) : k(f1, . . . , fn)] <∞ (see [11, proposition 1.1.31]) and by [11, theorem
1.1.32] we know #f−1(Q) ≤ [k(X1, . . . , Xn) : k(f1, . . . , fn)] for all Q ∈ kn. In particular, #X = #f−1(0) < ∞.
So dimX = 0.
Now we make use of the following 9
Bezout Inequality. Let X ⊂ An
k
be an affine algebraic set given by the equations f1 = · · · = fr = 0. Denote by
X(k) the set of k-points. If dimX = 0 then
#X(k) ≤ #X ≤ deg(f1) · · · deg(fr).
8second proof: consider the k-algebra R = k[X1, . . . , Xn]/〈f1, . . . , fn〉 and take M ∈ Specm(R) a maximal ideal. By local
noetherian ring theory we know that dimRm ≤ dimkTPX for all P ∈ X where TPX := Homk(MX,P /M2X,P , k) is the tangent
space. By Jacobian criterion we have dimkTPX = n − rank(JF (P )) = n − n = 0. So we conclude dimRm = 0. In particular,
dimR = 0 and so that R is an artinian k-algebra. In particular, Specm(R) is finite set. By correspondence, we conclude that X is a
finite set. So dimX = 0.
9see appendix.
3.5. Some results 46
Applying the theorem above we have
#X(k) ≤ deg(f1) · · · deg(fn) ≤ deg(F )n < qn
where we use the hypothesis: q > deg(F ). So S = kn \X(k) 6= ∅. Let α ∈ S. By definition fj(α) 6= 0 for some
index j. In particular, the map f : kn −→ kn isn’t zero. So F is a unimodular map.
Corollary 3.1. (WM) For all prime p, Fp[[T ]] and Zp are (p− 1)-unimodular domains .
Note that the bound p− 1 is “maximal” for Fp[[T ]].
Corollary 3.2. (WM) Zp is 3-unimodular domain for all prime p > 3. In particular, for almost all primes p the
Unimodular Conjecture is true for maps of degree ≤ 3 over Zp.
Proof. Since p > 3 we have p− 1 ≥ 3 and so the result follows from theorem above.
In dimension n = 2 and in characteristic 0 the theorem above can be refined. Indeed, we have the following
Theorem 3.13. (WM) Let f = (f1, f2) ∈ MP2(O) be a Keller map over a complete DVR (O,M, k) with
q := #k <∞ and char(O) = 0. If deg(f1) < q2 then f is unimodular.
Remark 8. This is particular for char(O) = 0. Indeed, consider the map
f = (X1 −Xp1 , X2 −Xp2 ) ∈MP2(Fp[[T ]]).
f is a Keller map but is not unimodular. Furthermore deg(f1) = p < p
2.
In order to prove the theorem above we use the following result:
Theorem (Yitang Zhang). 10 Let f = (f1, f2) ∈MP2(K) be a Keller map over an algebraically closed field K
with char(K) = 0. Then [K(X,Y ) : K(f1, f2)] ≤Min{deg(f1), deg(f2)}.
Proof. (of theorem 3.13) Let f = (f1, f2) be a Keller map over O. By proposition 3.1 we have a bijection
S1 := {(u, v) ∈ O2 | f1(u, v) = f2(u, v) = 0} ∼= {(a, b) ∈ k2 | g1(a, b) = g2(a, b) = 0} =: S2
where g1 and g2 are the reductions of f1, f2 mod M.
10see: Zhang, Y.: ” The Jacobian conjecture and the degree of field extension-Thesis, (1991)“.
3.5. Some results 47
By [11, theorem 1.1.32] we know that #S1 ≤ [K(X,Y ) : K(f1, f2)] where K = Frac(O). By Zhang theorem
and hypothesis we have #S1 ≤ Min{deg(f1), deg(f2)} ≤ q2 − 1. In particular there is Q ∈ k2 \ S2. So f is a
unimodular map.
Theorem 3.14. (WM) 11 Let p ∈ Z be a prime. For each d ∈ Z≥1 we can find a finite extension K|Qp such that
the ring of integers OK is a d-unimodular domain.
Proof. Let d ∈ N. If d = 1 take K = Qp. Suppose that d > 1. We know that for any Keller map F ∈ MPn(Zp)
of degree d we have
#X = deg(X) ≤ deg(F )n = dn
where X is the algebraic set in AnFp given by reduction of F mod p. Let n be an integer such that p
n > d and fix
Fpn the unique extension of Fp of degree n in Fp. We have seen in the proof of theorem 3.10 that there is a finite
extension K|Qp such that the residue field of OK is Fpn . By construction, for all Keller map G ∈MPn(OK) with
deg(G) ≤ d we have #{g1 = · · · = gn = 0} ≤ deg(G)n ≤ dn < (pn)n. So G is a unimodular map and OK is a
d-invariant domain.
Proposition 3.11. (WM) Suppose that for all n ∈ N and F = (F1, . . . , Fn) ∈ MPn(Zp) Keller map with
deg(F ) < n is unimodular. Then Zp is a unimodular domain.
Proof. Let F ∈ MPn(Zp) be a Keller map with n ≤ deg(F ). Let m ∈ Z be an integer (to be determined) and
consider the map
F [[m]] = (F1, . . . , Fn, F1, . . . , Fn, . . . , F1, . . . , Fn) ∈MPmn(Zp)
which consists of m-repetitions of the tuple F1, . . . , Fn where in each occurrence of such tuple we introduce n-
distinct variables. By construction we have F [[m]] a Keller map and F is a unimodular map if and only if so is
F [[m]]. We can choose large m such that deg(F ) < mn. Thus, we get the unimodularity of F .
Let R be a domain and f ∈ R[X1, . . . , Xn]. Define d(f) := number of monomials in degree > 3 that occur in f .
If F = (F1, . . . , Fn) ∈MPn(R) we define d(F ) :=
∑
j d(Fj).
Proposition 3.12. (WM) Let p ∈ Z>3 be a prime number and f ∈MPn(Zp) a Keller map. Suppose that
d(f) ≤ log(2)−1log(nlog(p/3)/log(3)) (∗)
11This proposition motivates the following question:
• Let p be a prime and consider the p-adic ring. Is there a finite extension K/Qp such that OK , the ring of integers, is a
unimodular domain?
3.5. Some results 48
where log is the natural logarithm. Then f is unimodular.
Proof. Let f ∈ MPn(Zp) be a Keller map. By the reduction theorem we can find invertible maps G,H ∈
MPn+m(Zp) for some m ∈ N such that g := G ◦ f [m] ◦ H has degree ≤ 3 where f [m] = (f,Xn+1, . . . , Xn+m).
Furthermore, we know that G(0) = H(0) = 0. Denote by Xf (Zp) and Xg(Zp) the set of Zp-points of f and g
respectively. It is easy to check that #Xf (Zp) = #Xg(Zp). Now, since Zp is a 3-unimodular domain (corollary 3.2)
we have #Xg(Zp) < 3n+m. By the proof of reduction theorem we get m = 2d(f). The inequality (∗) implies
3m+n ≤ pn and so we have f a unimodular map.
Theorem 3.15. (WM) Zp is an invariant domain for almost all prime p if and only if the Jacobian Conjecture
(over C) is true.
Proof. The implication⇐= follows from 3.4. Suppose that the Invariance Conjecture is true over Zp for almost all
prime p. By a result of Connel-var den Dries (see chapter 1, 1.7) we know that is sufficient to show the Jacobian
Conjecture over Z. So, suppose some Keller map F ∈ MPn(Z) isn’t invertible. Since F has coefficients in Z
it follows that F is unimodular over Zp for almost all primes p. Also, we know by hypothesis that F ⊗ Q isn’t
injective. By immersion lemma, F isn’t injective over Zp for infinitely many primes p. Fix such a prime p such
that Zp is an invariant domain. So, we obtain F ⊗Zp a Keller map non-injective over the invariant complete local
domain. A contradiction by theorem 3.8.
3.5.1 A refinement
Strong Immersion Lemma. (WM) Let α1, . . . , αm ∈ Q be algebraic numbers. Then there is a finite set E of
rational primes such that for all prime p /∈ E we have an injective homomorphism
Z[α1, . . . , αm] ↪→ OK,p
where OK,p is the ring of integers of some finite K|Qp.
Proof. The proof is similar to proof the immersion lemma. It is sufficient to prove the following
Fact. Let f(T ) ∈ Z[T ] be an irreducible polynomial. Then for almost all prime p there is a finite extension K|Qp
and α ∈ OK,p such that f(α) = 0.
Let d be the discriminant of the polynomial f and E := {p | p is prime with p | d}. Let p ∈ Z \E be a prime and
take f(T ) ∈ Fp[T ], via reduction mod p. Let α ∈ Fp be a root of f(T ) and take Fpk the definition field of α.
3.5. Some results 49
Then
f(α) = 0 and f
′
(α) 6= 0 by condition p /∈ E.
Now we recall that there exists a finite extension K|Qp such that OK,p is a complete discrete valuation with residue
field Fpk . Since OK,p is a complete ring we can use the Hensel lemma to conclude that there is some a ∈ OK,p
such that f(a) = 0.
Theorem 3.16. (WM) Zp is an invariant domain for infinitely many primes p if and only if the Jacobian
Conjecture (over C) is true.
Proof. The implication ⇐= is trivial. Suppose that Zp is invariant domain for infinitely many primes p but the
Jacobian Conjecture is false. Let F ∈ MPN (Z) be a counterexemple with det JF = 1. In particular, F ⊗ Q
isn’t injective. Let α 6= β ∈ QN be such that F (α) = F (β). By the strong immersion lemma we know that
R := Z[α, β] ↪→ OK,p for almost all primes p. Fix a prime p such that R ↪→ OK,p and such that Zp is invariant
domain. So, we obtain F ⊗OK,p a Keller map, not injective, over the domain OK,p. By lemma 3.5 we know that
there exists a Keller map G over Zp that isn’t injective. A contradiction by theorem 3.9.
Theorem 3.17. (WM) There is a finite set of primes E such that for all prime p ∈ Z \ E we have
Zp is an invariant domain⇐⇒ Zp is a unimodular domain.
Proof. The implication ⇐= is easy. Suppose =⇒ false. Then for infinitely many primes p we have Zp is an
invariant non-unimodular domain. Since Zp is invariant for infinitely many primes we have that the Jacobian
Conjecture is true by theorem 3.16. On the other hand since Zp is not unimodular for infinitely many primes we
know, by Essen-Lipton theorem, that the Jacobian Conjecture is false. Contradiction.
Chapter 4
Appendix
In this chapter we include for the reader’s convenience some results in algebraic geometry and commutative algebra
that have been used in the dissertation. Details can be found in the references. In all text the word ring signifies
commutative ring with unity unless otherwise stated.
4.1 k-algebras of finite type
We start with the
Definition 4.1. Let K be a field. A valuation ring of K is a subring O of K such that given x ∈ K \ {0} we have
x ∈ O or 1/x ∈ O.
Proposition 4.1. Let K be a field and O a valuation ring. Then O is an integrally closed local domain.
We say that a valuation ring O is a discrete valuation ring if the maximal ideal is principal.
Theorem 4.1. Let k be a field and K = k(T ) the rational functions field over k. Let O be a valuation ring that
contains k. Then O is a discrete valuation ring and O = k[T ]p(T ) for some irreducible polynomial p(T ) ∈ k[T ] or
O = k[1/T ]1/T , the “infinity” discrete valuation.
Proof. Let O be a valuation ring with O ⊃ k and maximal ideal M. We divide in cases:
(i) T ∈ O: In this case, k[T ] ⊂ O and if P :=M∩ k[T ] = (p(T )) we have k[T ]p(T ) ⊂ O. Since discrete valuation
rings are maximal with respect the domination relation, we have O = k[T ]p(T ).
(ii) If T /∈ O, a similar argument applied to T−1 shows O = k[1/T ]1/T .
50
4.1. k-algebras of finite type 51
The following result has many interesting applications:
Extension Theorem. Let k be an algebraically closed field and R a domain with K = Frac(R). Let φ : R −→ k
be a ring map. Then, there is a valuation ring (O,M) of K containing R and Φ : O −→ k , an extension of φ,
such that Ker(Φ) =M.
Proof. Consider the set S that consists of pairs (A,Φ) where A is a (proper) subring of K containing R and
Φ : A −→ k is an extension of φ. Note that S 6= ∅ since (R,φ) ∈ S. Introduce a partial order in S by
(A1,Φ) ≤ (A2,Φ2) if A1 ⊂ A2 and Φ2 is an extension of Φ1. This implies that S is an inductive set and by Zorn
lemma there exists (O,Φ) a maximal element. I affirm that O is a valuation ring. Note that it is local ring with
maximal ideal Q := Ker(Φ). Indeed, pick x ∈ R with x /∈ Q and consider the ring A1 := O[x−1]. Since x /∈ Q we
have Φ(x) 6= 0. Define ψ : A1 −→ k given by a0 + · · · + anx−n 7→ Φ(a0) + · · · + Φ(an)Φ(x)−n. This map is well
defined and yields a map extending Φ. By maximality we have x−1 ∈ O. So O−O∗ = Q. It is easy to show that
O is a valuation ring.
Algebrically Nullstellensatz. Let k be a field and A a k-algebra of finite type. Let M be a maximal ideal A.
Then, A/M is a finite extension of k.
Proof. Write A′ = A/M = k[α1, . . . , αn] and suppose that d = tr.degk(A′). Without loss of generality we can
suppose that α1, . . . , αd are algebraically independent over k. So R := k[α1, . . . , αd] is a polynomial ring in d
variables. Thus for each d+ 1 ≤ i ≤ n there is a relation (minimal) of type
a0i(α1, . . . , αd)α
ei
i + a1i(α1, . . . , αd)α
ei−1
i + · · ·+ aeii(α1, . . . , αd) = 0.
Since k ( = algebraic closure of k) is infinite there exists P = (a1, . . . , ad) ∈ kn such that a0j(P ) 6= 0 for all
j ∈ {d+ 1, . . . , n}.
Define φ : R −→ k given by αi 7→ ai. By the extension theorem there is a DVR O of k(α1, . . . , αd) wich contains
R and Φ : O −→ k extension of φ with ker(Φ) =MO. We affirm that O contains αd+1, . . . , αn. Indeed, suppose
this false for some d+ 1 ≤ i ≤ n. In particular, α−1i ∈ O −O∗ =M = Ker(Φ).
Then making use of the relation a0i(α1, . . . , αd)α
ei
i + a1i(α1, . . . , αd)α
ei−1
i + · · · + aeii(α1, . . . , αd) = 0 we obtain
a0i(α1, . . . , αd) + a1i(α1, . . . , αd)α
−1
i + · · ·+ aeii(α1, . . . , αd)α−ei = 0.
Applying Φ we obtain a relation of type a0i(a1, . . . , ad) = 0 that is impossible by construction. So, A
′ ⊂ O =⇒
O = A′. Since A′ is a field we should have Ker(Φ) = 0 and thus A′ is a subfield of k. In particular, A′|k is an
algebraic extension.
4.1. k-algebras of finite type 52
Corollary 4.1. Let M be a maximal ideal in k[X1, . . . , Xn] (with k = k). Then there is α1, . . . , αn ∈ k such that
M = (X1 − α1, . . . , Xn − αn).
Corollary 4.2. Let X = Z(I) ⊂ Ank be a closed subset (in Zariski topology) with k = k. Denote I(X) := {f ∈
k[X1, . . . , Xn] | f(P ) = 0 for all P ∈ X}.
I(X) =
√
I.
Theorem 4.2. Let A be a domain with fraction field K. Denote A the integral closure of A in K. Then
A =
⋂
O∈S
O em K
where S := {O | O is a valuation ring of K and A ⊂ O}.
Proof. Since valuation ring is integrally closed we have A ⊂ O provided that O is a valuation ring of K and
A ⊂ O. Now suppose that x ∈ K with x /∈ A. Consider K, the algebraic closure of K, and define the map
φ : A[x−1] −→ K a0 + · · ·+ anx−n 7→ a0.
By extension theorem we know that φ extend to a map Φ : O −→ L, where O is a valuation ring of K and,
furthermore, MO = Ker(Φ). In particular, we have x−1 ∈ Ker(Φ) =⇒ x−1 /∈ O∗ =⇒ x /∈ O.
Theorem 4.3. Suppose that R is a noetherian ring and A an R-algebra of finite type. Let T ⊂ A be an R-algebra
such that A is finite as T -module. Then, T is an R-algebra of finite type.
R //

A
T
/ 
??
Proof. Let A = R[x1, . . . , xn] for generators x1, . . . , xn and A = Tu1 + · · · + Tur as T -module. We have xi =∑r
j=1 tijuj for all i an some tij ∈ T and uluk =
∑
p tlkpup for some elements tikp ∈ T . Consider T0 = R[{tij , tlkp |
1 ≤ i, j, k, l ≤ n}] the subalgebra of T . We affirm that A is finitely generated as T0-module. Assume, this for a
moment. Since T0 is notherian and A is finitely generated as T0-module we have A a noetherian T0-module. So,
since T is a T0-submodule, we have that T is finitely generated as T0-module. Since T0 is of finite type over R we
concluded that T if of finite type over R.
Now, we show that A is finitely generated as T0-module. Let a ∈ A. Then there is p(X1, .., Xn) ∈ R[X1, . . . , Xn]
such that a = p(x1, . . . , xn). Using the relations xi =
∑r
j=1 tijuj and uluk =
∑
p tlkpup we have that we can write
4.2. Ka¨hler differentials 53
a in the form a = v1u1 + · · ·+ vrur with vi ∈ T0. So, A = T0u1 + · · ·+T0ur and so A is module-finite over T0.
4.2 Ka¨hler differentials
Let B be an A-algebra. Given a B-module M an A-derivation consists of an A-linear map D : B −→ M that
satisfies the rule of Leibniz
D(bb′) = bD(b′) + b′D(b) ∀ b, b′ ∈ B.
We denote by DerA(B,M) the set of all A-derivation of B in M .
Theorem 4.4. DerA(B,M) is a B-module and there is a B-module ΩB/A together with a derivation d : B −→
ΩB/A and with the following universal property:
Given M , a B-module, and any A-derivation D : B −→ M there is a unique B-linear map l : ΩB/A −→ M such
that the following diagram is commutative
B
D //
d

M
ΩB/A
l
<<
More precisely, DerA(B,M) ∼= HomB(ΩB/A,M) for all B-module M .
Proof. Consider the map of A-algebras
φ : B ⊗A B −→ B b⊗ b′ 7→ bb′.
Let I := Ker(φ)
Define ΩB/A := I/I
2. Consider ΩB/A as B-module via bb1 ⊗ b2 := bb1 ⊗ b2.
In view of the exact sequence
0 −→ I −→ B ⊗A B −→ B −→ 0
we get 0 −→ I/I2 −→ C −→ B −→ 0 where C := B ⊗ B/I2. Consider the maps β1, β2 : B −→ C given by
b 7→ b ⊗ 1 mod I2 and b 7→ 1 ⊗ b mod I2 respectivaly. We define dB/A : B −→ ΩB/A by γ 7→ β1(γ) − β2(γ). It
can be shown that pair (ΩB/A, dB/A) satisfies the universal property in the statement.
4.3. Finite maps 54
By universal property it follows that (ΩB/A, dB/A) is unique up to isomorphism.
Proposition 4.2. Let A be a k-algebra (k isn’t necessarily a field) and B an A-algebra. Then there is an exact
sequence
ΩA/k ⊗A B α // ΩB/k β // ΩB/A // 0
where α : dA/k(a)⊗ b 7→ bdB/k(a) and β : dB/k(b) 7→ dB/A(b).
Definition 4.2. Let K|k be a finitely generated field extension and d = tr.degk(K). We say that K is separable
over k if exists t1, . . . , td ∈ K such that K|k(t1, . . . , td) is finite and separable.
Theorem 4.5. Let K|k be field extension finitely generated. The following are equivalent
(i) K is separable over k.
(ii) The sequence 0 // Ωk/Z ⊗k K α // ΩK/Z β // ΩK/k // 0 is exact.
(iii) dimK(ΩK/k) = tr.degk(K).
Proof. [7, proposition 5.2-Iversen]
Dimension Formula. Let A be a domain of finite type over a field k with char(k) = 0 or perfect. Let K :=
Frac(A). Then
dimA = tr.degkK = dimK ΩK/k.
Proof. By Noether Normalization lemma we can find X1, . . . , Xn ∈ A, algebraically independent over k, such
that the extension k[X1, . . . , Xn] ↪→ A is integral. In particular, dimA = dim k[X1, . . . , Xn] = n. Denote
L = k(X1, . . . , Xn). Then L|K is an algebraic extension and so tr.degkK = tr.degkL = n. So, dimA = tr.degkK.
The formula tr.degkK = dimK ΩK/k is a consequence of the theorem above.
4.3 Finite maps
In this section, k will denote an algebraically closed field. Given A a domain of finite type over k we will
denote by Specm(A) the subset of Spec(A) that consists of closed points If A = k[X1, . . . , Xn]/J then there is a
homeomorphism Specm(A) ∼= Z(J):
α : Specm(A) −→ Z(J) (X1 − a1, . . . , Xn − an) 7→ (a1, . . . , an).
4.3. Finite maps 55
Proposition 4.3. Let A
f // B be a map of k-algebras of finite type. Let M∈ Specm(B). Then, M∩A is a
maximal ideal of A.
Proof. Let M a maximal ideal of B. By Nullstellensatz we have B/M is a finite extension of k. So, taking
Im(pi) = pi(A), where pi : A −→ B −→ B/M, we have that pi(A) is a k-algebra with dimk pi(A) <∞. So, pi(A) is
a field. Since pi(A) ∼= A/M∩A we have M∩A ∈ Specm(A).
So, given a map of k-algebras A
f // B we obtain a map in closed points f# : Specm(B) −→ Specm(A).
Let X ⊂ Ank and Y ⊂ Amk affine varieties and f : X −→ Y a morphism. Let f# : OY (Y ) −→ OX(X) the map
induced in regular functions given by g ∈ OY (Y ) 7→ g ◦ f ∈ OX(X). So, we obtain a structure of OY (Y ) algebra
in OX(X).
Definition 4.3. We say that the morphism f if finite if OX(X) is integral over OY (Y ).
In particular, if f : X −→ Y is a finite dominant map we have dimX = dimY . Note that, if X is a subvariety of
X then f |X′ : X ′ −→ Y is a finite map.
Theorem 4.6. Let f : X −→ Y be a map of affine varieties. Then
(i) f is finite =⇒ f quasi-finite i.e. f has finite fibers.
(ii) f finite =⇒ f closed.
(iii) f finite and dominant =⇒ f surjective.
Proof. [8, Milne]
Theorem 4.7. Let f : X −→ Y a finite dominant map of affine varieties and assume that Y is normal.
(i) ∀P ∈ Y we have #f−1(P ) ≤ deg(f).
(ii) The set S = {P ∈ Y | #f−1(P ) = deg(f)} is open in Y .
Proof. Let B = OX(X) , A = OY (Y ) and g : A −→ B the induced map in k algebras. Let f−1(P ) = {Q1, . . . , Qr}
and u ∈ OX(X) such that u(Qi) 6= u(Qj) if i 6= j. Let F (T ) ∈ OY (Y )[T ] be the minimal polynomial of u over
k(Y ) (function field of Y ). Since Y is normal we conclude that the coefficients of F (T ) are regular functions in
Y . We have
F (T ) = Tn + an−1Tn−1 + · · ·+ a0
4.4. Normalization 56
for some a0, . . . , an−1 ∈ OY (Y ). Now, note that n ≤ deg(f) and F (u)(Qj) = u(Qj)n + an−1(P )u(Qj)n−1 +
· · · + a0(P ) = 0 and so, u(Q1), . . . , u(Qr) are roots of Tn + an−1(P )Tn−1 + · · · + a0(P ) ∈ k[T ]. In particular,
r ≤ n ≤ deg(f).
Let P ∈ Y such that deg(f) = #f−1(P ) = #{Q1, . . . , Qr}. We want to show that there is an open set U 3 P ,such
that for any P ′ ∈ U we have deg(f) = #f−1(P ′). Pick u ∈ OX(X) a regular function such that u(Qi) 6= u(Qj)
if i 6= j. Denote by F (T ) = Tn + an−1Tn−1 + · · · + a0 ∈ OY (Y )[T ] the minimal polynomial of u. Then
deg(f) = r ≤ deg(F ) ≤ deg(f) =⇒ deg(F ) = deg(f). Let d := disc(F (T )), the discriminant of F (T ). Since
Tn + an−1(P )Tn−1 + · · · + a0(P ) has deg(F ) distinct roots, we have that d ∈ OY (Y ) is such that d(P ) 6= 0.
So, there is an open UP over P such that T
n + an−1(P ′)Tn−1 + · · · + a0(P ′) has distinct roots. Consider the
commutative diagram:
OY (Y )
pi

f# // OX(X)
OY (Y )[T ]/(F (T ))
g
66
Here, g is defined by T 7→ u. This diagram induces, by restriction
Specm(OX(X))
g∗

f˜ // Y = Specm(OY (Y )) ⊃ UP
Specm(OY (Y )[T ]/(F (T )))
pi∗
33
Let P ∈ UP . The fiber with respect to the map pi∗ has deg(f) elements. Furthermore, the map g∗ is surjective.
Indeed, g∗ is finite and dominant since the map in regular functions is injective. In particular, #f−1(P ) ≥ deg(f)
and by (i) we have the equality #f−1(P ) = deg(f).
4.4 Normalization
Let X be an affine variety over a field k (= k) with coordinate ring OX(X). Denote by A the integral closure of
OX(X). It is well know that A is a domain of finite type over k and so it is the coordinate ring of a variety X˜.
The inclusion i : OX(X) ↪→ A induces a finite map pi : X˜ −→ X.
It can be shown that the pair (X˜, pi) is unique, up to isomorphism, and has the following properties:
(i) OX˜(X˜) is an integrally closed domain.
(ii) Suppose that Y is a normal affine variety and g : Y −→ X is a morphism. Then there exists a unique morphism
4.4. Normalization 57
h : Y −→ X˜ such that the following diagram is commutative
Y
h

g // X
X˜
pi
??
The pair (X˜, pi) is called the normalization of X.
Let X be a variety and X
f // Y a finte dominant map of affine varieties. In particular we have f quasi-finite.
Let Q ∈ Y and consider F := f−1(Q) = {P1, . . . , Pn}.
Take the open set UF = X \ F . By restriction, we obtain a map UF f // Y , quasi-finite but non-finite. So,
we obtain the following diagram:
UF
f |UF //
i

Y
X
f
>>
where i is inclusion, in particular an open embedding. The next theorem shows that, conversely, any quasi-finite
map factors through a finite map
Zariski Main Theorem. Every quasi-finite map X
f // Y is factored X
j // Y ′
g // Y where j is an
open immersion and g is a finite map. If Y is normal and f a dominant map then the statement is true with
(Y ′, g) being the normalization of Y .
X
f //
j

Y
Y ′
g
>>
We will not prove the above theorem; details can be found in [10, Mumford] or in [8, Milne].
Example 8. Let X be an affine variety and f : X −→ Y a bijective morphism. It is possible that f isn’t an
isomorphism. Indeed, consider Y = A1k and X = {(x, y) ∈ A2k | y2 − x3 = 0} (k = C). The map f : Y −→ X
given by t 7→ (t2, t3) is a bijective morphism but not isomorphism. Indeed, consider the map in coordinate rings:
f# : OX(X) = k[t2, t3] ↪→ k[t] = OY (Y ). Note that this map isn’t an isomorphism since OY (Y ) is integrally
closed and this is false for OX(X).
The problem is that X is singular over (0, 0):
Corollary 4.3. Let f : Y −→ X be a bijective map of affine varieties with X normal. Then, f is an isomorphism.
4.5. Bounds for k-points 58
Proof. Since f is a bijection, in particular it is dominant. So, by Zariski Main Theorem we have that f is factored
by f = pi ◦ j where pi = id and j = f is an open immersion. In particular, f is an isomorphism.
4.5 Bounds for k-points
In this section k is a field and K is its algebraic closure. The objective is to give a bound for k-points in an algebraic
set X ⊂ PnK (or in AnK). In what follows we will identify AnK as an open subset in PnK via the homeomorphism
AnK −→ D(X0) ⊂ PnK given by (a1, . . . , an) 7→ [1 : a1 : · · · : an].
Definition 4.4. Let X ⊂ Pn be an algebraic set. Let Z1, . . . , Zm be the irreducible components of X. The
cumulative degree of X is defined by degc(X) :=
∑m
l=1 deg(Zl). Here, if Z is a variety deg(Z) means the degree of
Z (via Hilbert polynomial cf.[6, chapter 1]).
Remark 9. We have deg(X) ≤ degc(X). Indeed, by [6, proposition 7.6-chapter 1] we know that deg(X1 ∪X2) =
deg(X1) + deg(X2) if dim(X1 ∩X2) < r where X1 and X2 are algebraic subsets of Pn and r = dimX1 = dimX2.
Furthermore if dimX2 < dimX1 then deg(X1 ∪X2) = deg(X1). Indeed, let S := K[X0, . . . , Xn] be the homogeous
coordinate ring for PnK and let IX1 and IX2 the associated ideals. Then we have an exact sequence of graded rings:
0 // S/IX1 ∩ IX2
f 7→(f,−f)// S/IX1 ⊕ S/IX2
(h,g) 7→h+g// S/IX1 + IX2 // 0
By properties of Hilbert polynomials in exact sequences we have HX1(t)+HX2(t) = HX1∪X2(t)+HX1∩X2(t). Since
dimXi = deg(HXi(t)) we obtain deg(HX1(t)) = deg(HX1∪X2(t)) and so deg(X1) = deg(X1 ∪X2). In particular,
if Z1, . . . , Zm are the components of X we have
deg(X) =
∑
dimZk = dimX
deg(Zk) ≤ degc(X).
Recall that a subset X ⊂ PnK is called a locally closed subset if X = F ∩ U where F is a closed subset in Pn and
U is a open subset. This is equivalent to the equality: X = X ∩U for some open subset in PnK . Examples: closed
subsets in AnK . We define cumulative degree and degree of a locally closed subset X by taking degc(X) := degc(X)
and deg(X) := deg(X). Note that deg(Ad) = deg(Pd) = 1 since that HPd(t) = t
d
d! + terms in degree < d.
Proposition 4.4. Let X ⊂ Pn be a locally closed subset.Then,
• degc(X) = deg(X) if X is a variety.
• If X is finite then degc(X) = #X.
4.5. Bounds for k-points 59
• If X is a variety with dimX > 0 and H ⊂ PnK is a hypersurface with X * H then deg(X ∩ H) =∑
kmkdeg(Ck) where C1, . . . ., Cm are the irreducible components of X ∩H and mk ∈ Z≥1.
Proof. The first item is trivial. For the second item, it is sufficient to compute the degree of the subset Z :=
{Q} ⊂ PnK . But this is trivial since that HZ(t) = 1 ∈ Q[t] and so deg(Z) = 0!(leader coefficient of HZ(t)) = 1.
The last item is a version of Bezout theorem and a proof can be found in [6, Hartshorne].
Lemma 3. Let X ⊂ PnK be a locally closed subset and ∅ 6= U ⊂ X open in X. Then degc(U) ≤ degc(X).
Furthermore, if U is dense in X then degc(U) = degc(X)
Proof. Let U1, . . . , Uk be the irreducible components of U . Let Fi denote the closure of Ui in X. We will show
that F1, . . . , Fk are the distint irreducible components of X. Let X1, . . . , Xm be the irreducible componenets of
X. Suppose that U ∩Xi 6= ∅ for i ∈ {1, . . . , s} and Xi ∩ U = ∅ for i > s. It suffices to show that Xi ∩ U are the
irreducible components of U . For Xi ∩ U = Xi.
Indeed, note that since Xi is irreducible we have that the open subset Xi ∩ U is dense and irreducible. Thus
Xi ∩ U * Xj ∩ U for 1 ≤ i, j ≤ s. Now if U = U1 ∪ · · · ∪ Uk we have degc(U) =
∑s
j=1 deg(U i) =
∑s
j=1 deg(Xj) ≤∑m
j=1 deg(Xj) = degc(X). For the last statement observe that if U is dense in X then Xi ∩U 6= ∅ for all i and so
s = m.
Lemma 4. If X ⊂ PnK is locally closed then deg(X) = deg(X).
Of course, we have X = X ∩U for some open U . So X is an open dense subset in X. By lemma above we obtain
deg(X) = deg(X).
Theorem 4.8. Let X ⊂ Pn a locally closed subset and H1, . . . ,Hm hypersurfaces in PnK . Then
degc(X ∩H1 ∩ · · · ∩Hm) ≤ degc(X)
∏
j
deg(Hj).
Proof. By induction we can suppose that m = 1. We can assume that X is a closed subset. Of course we have
X = X ∩ U for some U open subset of PnK . So X is open dense in X and thus X ∩H1 is open subset of X ∩H1.
By lemma above we have degc(X ∩H1) ≤ degc(X ∩H1).
If X ⊂ H1 then degc(X ∩ H1) = degc(X) ≤ degc(X)deg(H1). Suppose that X * H1 and that X and H1 are
variety. Let C1, . . . , Cl the irreducible components of X ∩H1. Then by proposition above we have
degc(X ∩H1) =
∑
j
deg(Cj) ≤
∑
j
mjdeg(Cj) = deg(X)deg(H1) = degc(X)deg(H1).
4.6. Ramification 60
If X1, · · · , Xj are the irreducible components of X note that each irreducible component of X ∩ H1 is equal to
Xj ∩H1 for some j. In particular
degc(X ∩H1) ≤
∑
l
deg(Xl ∩H1) =
∑
l
deg(Xl)deg(H1) = [
∑
l
deg(Xl)]deg(H1) = degc(X)deg(H1).
The case H1 reducible is similar.
Bezout Inequality. Let X ⊂ AnK an affine algebraic subset defined by equations f1 = · · · = fr = 0. Suppose that
dimX = 0. Then #X(k) ≤ #X(K) ≤ deg(f1) · · · deg(fr).
Proof. By induction is sufficient to consider the case r = 1. Denote h the closed subset described by f1. Let
Z be a locally closed subset in AnK and denote H the closure of h in PnK . We have h = H ∩ AnK and so Z ∩ h
is an open subset of Z ∩ H. By lemma above we have degc(Z ∩ h) ≤ degc(Z ∩ H) and by theorem above
degc(Z ∩ h) ≤ degc(Z)deg(H) = degc(Z)deg(h).
So degc(Z ∩ h1 ∩ · · · ∩ hr) ≤ degc(Z)deg(h1) · · · deg(hr). In particular, if Z = AnK we have
#X(k) ≤ #X(K) ≤ degc(AnK ∩H1 ∩ · · · ∩Hr) ≤ degc(AnK)deg(f1) · · · deg(fr) = deg(f1) · · · deg(fr).
4.6 Ramification
Let K be a field and v a discrete valuation (always normalizated) of K. We say that K is a local field (with
respect to v) if K is complete and kv, the residue field, is finite. In this section, we give a summary of results on
local fields. Details can be found in [9, Milne]. We recall that if v is a discrete valuation in a field K we have an
associated discrete valuation ring Ov := {a ∈ K | v(a) ≥ 0} with maximal ideal Mv := {a ∈ Ov | v(a) > 0}.
Proposition 4.5. Let K be a field with discrete valuation v (not necessarily local field). Let L|K be a finite
separable extension and denote O the integral closure of Ov in L. Then there exists a bijection
ϕ : Specm(O) −→ P(L, v)
where P(L, v) := {w : L∗ −→ Z | w is a discrete valuation normalized in L such that w|v }.
Proof. The map is defined in the following way.
4.6. Ramification 61
LetM∈ Specm(O) and consider the ring OM obtained by localization overM. Note that OM is noetherian ring
(the extension is separable), dimOM = 1 and it is integrally closed. In particular it is a local Dedekind domain
and so a discrete valuation ring in L. Let vM be the associated normalizated discrete valuation. The map is
ϕ :M 7→ vM.
We affirm that ϕ is a bijection. Indeed, let A be a discrete valuation ring of L with maximal ideal P and A ⊃ Ov.
Note that m := P ∩O 6= 0. Indeed, let 0 6= α ∈ P . Since L|K is algebraic we have dα ∈ O for some d ∈ Ov \ {0}.
In particular dα ∈ m.
By localization property we obtain the inclusion dominant A ⊃ Om. Since DVR are maximal subrings with respect
to relation of domination we obtain A = Om. So the map ϕ is sujective. It is easy to check that ϕ is injective.
Let K be a complete field with respect to a valuation v. Let L|K be a separable finite extension and denote A
the integral closure of Ov in L.
Theorem 4.9. Notation as above L is a local field and there exists a unique discrete valuation in L that extends
the discrete valuation of K.
Proof. It is a general result that A is a Dedekind domain. By proposition above we know that Specm(A) ∼= P(L, v).
We want to prove that #P(L, v) = 1. For this supposeM1 andM2 are distinct maximal ideals in A. Take b ∈ A
such that M1 ∩Ov[b] 6=M2 ∩Ov[b]. For example, b ∈M1 \M2. Let mb(T ) ∈ Ov[T ] be the minimal polynomial
of b. By Hensel lemma, (Ov is complete!) we know that f(T ) ∈ kv[T ] is a power of an irreducible polynomial
f(T ) ∈ kv[T ]. In particular, Ov[b]/MvOv[b] ∼= Ov[T ]/(Mv,mb(T )) ∼= kv[T ]/(mb(T )), a local ring with maximal
ideal (f(T ))/(mb(T )). But M1 ∩ Ov[b]/MvOv[b] 6=M2 ∩ Ov[b]/MvOv[b] are maximal ideals in Ov[b]/MvOv[b].
A contradiction.
We have L complete. Of course, let r = [L : K] and take {e1, . . . , er} a K-basis for L. To give a Cauchy sequence
{an} ⊂ L is equivalent to give r-Cauchy sequences {a(1)n }, . . . , {a(r)n } and {an} is convegent if and only if {a(l)n } is
convegent for all l.
By Dedekind ring theory we know that [L : K] = ef where
f = [A/P : Ov/M] and e = ordP (t).
Here, P is the maximal ideal of A = the integral closure of Ov in L and t is a uniformizer for Ov We say that
L|K is an unramified extension if e = 1.
4.6. Ramification 62
Remark 10. Ramification can be studied in terms of discriminants. More precisely, let K|L be a separable
finite extension, where L is a local field with valuation ring Ov, and denote by A the ring of integers of K.
By the primitive element theorem we know that K = L(α) for some α ∈ L integral over Ov. Suppose that
A = O ⊕ Oα ⊕ · · · ⊕ Oαn−1 i.e. {1, α, α2, . . . , αn−1} it is a basis for A. Let mα(T ) ∈ Ov[T ] the associ-
ated minimal polynomial. Let d ∈ Ov the discriminant of mα(T ), i.e., d = resultant(mα(T ),m′α(T )). Then
K|L is unramified if and only if d ∈ O∗v .
Theorem 4.10. Let L be a local field with discrete valuation v and residue field lv. There is a 1-1 correspondence
between
{ unramified finite extension of L} −→ {finite extension of lv}.
given by K 7→ kv.
Proof. (sketch) For surjectivity, let k|lv a finite extension and take α such that k = lv(α). Let m(T ) ∈ Ov[T ] a
monic polynomial such that m(T ) ∈ lv[T ] is a minimal polynomial of α. By Hensel lemma we know that there is
some a ∈ Ov simple root of m(T ) in Ov. Take K := L(a).
Let K|L be an unramified finite extension. By the definition above, we have kv|lv a finite separable extension
with [kv : lv] = [K : L]. We affirm that the map K 7→ kv induces the bijection above. For injectivity, suppose that
K and S are fields that kv = sv. Then, taking the composite (in L) we have K.S an unramified extension with
residue kv, [9, Lemma 6.5], and so [S.K : L] = [kv : lv] = [K : L] =⇒ S ⊂ S.K ⊂ K . By a similar argument, we
conclude that K ⊂ S and so K = S.
Chapter 5
Some problems
Here, we list some problems we wish to find a solution:
Problem 1. Recall the reduction theorem in chapter 2: In order to show the Jacobian Conjecture it is sufficient
to consider the maps of degree ≤ 3. So, we may ask
• Is there an analogue of the reduction theorem for Unimodular Conjecture?
Problem 2. Find a prime p ∈ Z such that Zp is unimodular.
In chapter 3 it was seen that for each integer d ∈ Z there exists a finite extension K|Qp such that OK is d-
unimodular. This motivates the following
Problem 3. Given a prime p ∈ Z find (or show that this is impossible) a finite extension K|Qp such that OK is
a unimodular domain.
The following problem is equivalent to the Jacobian Conjecture 1
Problem 4. Let F ∈MPn(Z) a Keller map. Let O be the integral closure of Z in Q. Then, F ⊗O is injective.
We can compare the problem above with the following
Theorem 5.1. Let F ∈MPn(Z) be a Keller map. Suppose that F ⊗Q is injective. Then F is an isomophism.
The proof of this theorem uses Cynk-Rusek theorem of chapter 1.
Problem 5. In theorem 3.17, what can be said about #E?
1indeed, the theorem of Connell-van den Dries in chapter 1 is more general: Let P ∈ Spec(O) and consider A := OP localization
over P . If the Jacobian Conjecture over C is false then there is a Keller map F ∈ MPn(Z)(0), counterexemple, and such that for all
d ∈ A−A∗ we have a Keller injective dF ∈MPn(A). Here, (dF )l := F1l + dF2l + · · ·+ ddeg(Fl)−1Fdeg(Fl)l where F = (F1, . . . , Fn).
63
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