arXiv:1705.01838v1 [math.AG] 4 May 2017

ON PERMUTATIONS INDUCED BY TAME AUTOMORPHISMS
OVER FINITE FIELDS
KEISUKE HAKUTA
Abstract. The present paper deals with permutations induced by tame automorphisms over finite fields. The first main result is a formula for determining
the sign of the permutation induced by a given elementary automorphism over
a finite field. The second main result is a formula for determining the sign of
the permutation induced by a given affine automorphism over a finite field.
We also give a combining method of the above two formulae to determine the
sign of the permutation induced by a given triangular automorphism over a
finite field. As a result, for a given tame automorphism over a finite field, if we
know a decomposition of the tame automorphism into a finite number of affine
automorphisms and elementary automorphisms, then one can easily determine
the sign of the permutation induced by the tame automorphism.

1. Introduction
Let k be a field and let k[X1 , . . . , Xn ] be a polynomial ring in n indeterminates
over k. An n-tuple of polynomials F = (f1 , . . . , fn ) is called a polynomial map,
when fi belongs to k[X1 , . . . , Xn ] for all i, 1 ≤ i ≤ n. A polynomial map F =
(f1 , . . . , fn ) can be viewed as a map from k n to itself by defining
F (a1 , . . . , an ) = (f1 (a1 , . . . , an ), . . . , fn (a1 , . . . , an ))

(1.1)

n

for (a1 , . . . , an ) ∈ k . The set of polynomial maps over k and the set of maps from
k n to itself are denoted by MAn (k) and Maps(k n , k n ), respectively. Then by (1.1),
we see that there exists a natural map
π : MAn (k)

→ Maps(k n , k n ).

MAn (k) (resp. Maps(k n , k n )) is a semi-group with respect to the composition of
polynomial maps (resp. maps). Moreover, these two semi-groups are monoids with
the identity maps as neutral elements, and the map π is a monoid homomorphism.
Let GAn (k) be the subset of invertible elements in MAn (k). An element in
GAn (k) is called a polynomial automorphism. A polynomial automorphism F =
(f1 , . . . , fn ) is said to be affine when deg fi = 1 for i = 1, . . . , n. A polynomial map
Eai of the form
Eai = (X1 , . . . , Xi−1 , Xi + ai , Xi+1 , . . . , Xn ),
ai ∈ k[X1 , . . . , X̂i , . . . , Xn ] = k[X1 , . . . , Xi−1 , Xi+1 , . . . , Xn ],

(1.2)

= E−ai . A polynomial automoris also a polynomial automorphism since Ea−1
i
phism Eai of form (1.2) is said to be elementary. Let Aff n (k) denote the set of
Date: May 4, 2017.
2010 Mathematics Subject Classification. Primary: 14R10; Secondary: 12E20, 20B25.
Key words and phrases. Affine algebraic geometry, polynomial automorphism, tame automorphism, finite field, permutation.
1

all affine automorphisms, and let EAn (k) denote the set of all elementary automorphisms. Aff n (k) and EAn (k) are subgroups of GAn (k). We put TAn (k) :=
hAff n (k), EAn (k)i, where hH1 , H2 i is a subgroup of a group G generated by two
subgroups H1 , H2 ⊂ G. Then TAn (k) is also a subgroup of GAn (k) and is called
the tame subgroup. If τ ∈ TAn (k) then τ is called tame automorphism, and otherwise (i.e., τ ∈ GAn (k) \ TAn (k)) τ is called wild automorphism. A polynomial
automorphism Ja,f of the form
Ja,f = (a1 X1 + f1 (X2 , . . . , Xn ), a2 X2 + f2 (X3 , . . . , Xn ), . . . , an Xn + fn ) ,
ai ∈ k (i = 1, . . . , n) , fi ∈ k[Xi+1 , . . . , Xn ] (i = 1, . . . , n − 1) , fn ∈ k,
is called de Jonquières automorphism (or triangular automorphism). From the
definition, triangular automorphism is trivially polynomial automorphism. The set
of all elements of the form Ja,f is denoted by BAn (k):
BAn (k) := {Ja,f ∈ GAn (k) | ai ∈ k (i = 1, . . . , n) ,
fi ∈ k[Xi+1 , . . . , Xn ] (i = 1, . . . , n − 1) , fn ∈ k}.
BAn (k) is a subgroup of GAn (k). BAn (k) is also a subgroup of TAn (k), and it is
known that TAn (k) = hAff n (k), BAn (k)i (cf. [2, Exercises for § 5.1 – 1, 2]). The
Tame Generators Problem asks whether GAn (k) = TAn (k), and is related to the
famous Jacobian conjecture.
Tame Generators Problem. GAn (k) = TAn (k)?
For any field k, we denote the characteristic of the field k by p = char(k). If k
is a finite field Fq with q elements (p = char(Fq ), q = pm , and m ≥ 1), we use the
symbol πq instead of π:
πq : MAn (Fq ) → Maps(Fnq , Fnq ).
The map πq induces a group homomorphism
πq : GAn (Fq ) → Sym(Fnq ),
where Sym(S) is a symmetric group on a finite set S. Let
sgn : Sym(S) → {±1}
be the sign function. The sign function sgn is a group homomorphism, and Ker(sgn) =
Alt(S), where Alt(S) is the alternating group on S. Recall that for any subgroup
G ⊆ GAn (Fq ), πq (G) is a subgroup of Sym(Fnq ).
In the case where k = Fq , we can consider a slightly different problem from the
Tame Generators Problem, namely, it is natural to investigate the subgroup πq (G)
of Sym(Fnq ). This problem has first been investigated in the case G = TAn (Fq ) by
Maubach [3]. Indeed, Maubach proved the following theorem ([3, Theorem 2.3])
and proposed a following problem ([4, page 3, Problem]):
Theorem 1. ([3, Theorem 2.3]) If n ≥ 2, then πq (TAn (Fq )) = Sym(Fnq ) if q is
odd or q = 2. If q = 2m where m ≥ 2 then πq (TAn (Fq )) = Alt(Fnq ).
Question 1. ([4, page 3, Problem]) For q = 2m and m ≥ 2, do there exist
polynomial automorphisms such that the permutations induced by the polynomial
automorphisms belong to Sym(Fnq ) \ Alt(Fnq )?
2

If there exists F ∈ GAn (F2m ) such that π2m (F ) ∈ Sym(Fn2m ) \ Alt(Fn2m ), then we
must have F ∈ GAn (F2m ) \ TAn (F2m ), namely, F is a wild automorphism. Thus,
Question 1 is a quite important problem for the Tame Generators Problem in positive characteristic. Furthermore, we refer the reader to [5, Section 1.2] for several
questions related to [3, Theorem 2.3]. The present paper deals with permutations
induced by tame automorphisms over finite fields. We address the following questions related to [3, Theorem 2.3] which is a little different from the questions in [5,
Section 1.2].
Question 2. For a given tame automorphism φ ∈ TAn (Fq ), how to determine the
sign of the permutation induced by φ? (how to determine sgn(πq (φ))?)
Question 3. Suppose that G is a subgroup of GAn (Fq ). What are sufficient
conditions on G such that the inclusion relation πq (G) ⊂ Alt(Fnq ) holds?
Question 2 seems to be natural since if q 6= 2m (m ≥ 2), one can not determine
the sign of the permutation induced by a given tame automorphism over a finite
field from [3, Theorem 2.3]. The information about sign of the permutations might
be useful for studying the Tame Generators Problem, Question 1, or other related
questions such as [4, page 5, Conjecture 4.1], [4, page 5, Conjecture 4.2] and so on.
Question 3 also seems to be natural since if q 6= 2m (m ≥ 2) and G 6= TAn (Fq ), one
can not obtain any sufficient condition for the inclusion relation πq (G) ⊂ Alt(Fnq )
from [3, Theorem 2.3].
The contributions of the present paper are as follows. The first main result is a
formula for determining the sign of the permutation induced by a given elementary
automorphism over a finite field (Section 3, Main Theorem 1). Our method to determine the sign of the permutation induced by a given elementary automorphism over
a finite field, is based on group theory. As a consequence of Main Theorem 1, one
can derive a sufficient condition for the inclusion relation πq (EAn (Fq )) ⊂ Alt(Fnq )
(Section 3, Corollary 1). The second main result is a formula for determining the
sign of the permutation induced by a given affine automorphism over a finite field
(Section 4, Main Theorem 2). Our method to determine the sign of the permutation
induced by a given affine automorphism over a finite field, is based on linear algebra.
As a consequence of Main Theorem 2, one can also derive a sufficient condition for
the inclusion relation πq (Aff n (Fq )) ⊂ Alt(Fnq ) (Section 4, Corollary 2). Section 5
gives a combining method of Main Theorem 1 and Main Theorem 2 to determine
the sign of the permutation induced by a given triangular automorphism over a
finite field (Section 5, Corollary 3). As a result, for a given tame automorphism
over a finite field, if we know a decomposition of the tame automorphism into a finite number of affine automorphisms and elementary automorphisms, then one can
easily determine the sign of the permutation induced by the tame automorphism
(Section 5, Corollary 5).
The rest of this paper is organized as follows. In Section 2, we fix our notation.
In Section 3, we give a method to determine the sign of the permutation induced by
a given elementary automorphism over a finite field. In Section 4, we give a method
to determine the sign of the permutation induced by a given affine automorphism
over a finite field. Section 5 deals with how to determine the sign of the permutation
induced by a given triangular automorphism and a given tame automorphism over
a finite field.
3

2. Notation
Throughout this paper, we use the following notation. For any field k, we denote
the characteristic of the field k by p = char(k). We denote the multiplicative
group of a field k by k ∗ = k \ {0}. We use the symbols Z, C, Fq to represent the
rational integer ring, the field of complex numbers, and a finite field with q elements
(p = char(Fq ), q = pm , m ≥ 1). We denote the set of non-negative integers by Z≥0 .
For a group G and g ∈ G, ordG (g) is the order of g, namely, ordG (g) is the smallest
non-negative integer x that holds g x = 1G , where 1G is the identity element in the
group G. We denote by GLn (k) the set of invertible matrices with entries in k.
The polynomial ring n indeterminates over k is denoted by k[X1 , . . . , Xn ]. We
denote the polynomial ring k[X1 , . . . , Xi−1 , Xi+1 , . . . , Xn ] (omit the indeterminate
Xi ) by k[X1 , . . . , X̂i , . . . , Xn ]. Let MAn (k) be the set of polynomial maps over
k. MAn (k) is a monoid with respect to the composition of polynomial maps, and
the neutral elements of the monoid MAn (k) is the identity map. We denote the
subset of invertible elements in MAn (k), the set of all affine automorphisms, the
set of all elementary automorphisms, and the set of all triangular automorphisms
by GAn (k), Aff n (k), EAn (k), and BAn (k), respectively. GAn (k) is a group with
respect to the composition of polynomial automorphisms, and Aff n (k), EAn (k) are
subgroups of GAn (k). Recall that
Aff n (k) ∼
= GLn (k) ⋉ k n .

(2.1)

For a finite set S, we denote the cardinality of S by ♯S, and denote the symmetric
group on S and the alternating group on S by Sym(S) and Alt(S), respectively.
Let S = {s1 , . . . , sn }, ♯S = n, and σ ∈ Sym(S). For σ ∈ Sym(S),


s1
s2
···
sn
σ=
,
sσ(1) sσ(2) · · · sσ(n)
means that σ (si ) = sσ(i) for 1 ≤ i ≤ n. For σ, τ ∈ Sym(S), we denote by τ ◦ σ the
composition


s1
s2
···
sn
τ ◦σ =
.
sτ (σ(1)) sτ (σ(2)) · · · sτ (σ(n))
The permutation on S defined by
S
∈

−→

∈

S
si
si1

7−→ si
7−→ si2
..
.

sir−1
sir

7−→ sir
7−→ si1

if i 6∈ {i1 , . . . , ir }

is called the cycle of length r and is denoted by (i1 i2 . . . ir ). Let δ : R → R be
a function satisfying that δ (x) = 0 when x = 0, and δ (x) = 1 when x 6= 0. Let
χ : F∗q → C∗ be a non-trivial multiplicative character of order ℓ. We extend χ to
Fq by defining χ (0) = 0.
4

3. Sign of permutations induced by elementary automorphisms
In this section, we consider the sign of permutations induced by elementary
automorphisms over finite fields. The main result of this section is as follows.
(q)

Main Theorem 1. (Sign of elementary automorphisms) Suppose that Eai
is an elementary automorphism over a finite field, namely,
= (X1 , . . . , Xi−1 , Xi + ai , Xi+1 , . . . , Xn ) ∈ EAn (Fq ) ,
Ea(q)
i

(3.1)

and ai ∈ Fq [X1 , . . . , X̂i , . . . , Xn ]. If q is odd or q = 2m , m ≥ 2 then we have
(q)
πq Eai ∈ Alt(Fnq ). Namely, if q is odd or q = 2m , m ≥ 2 then
 

sgn πq Ea(q)
= 1.
(3.2)
i

 
(q)
depends only on the number of monomials of the
If q = 2 then sgn πq Eai
e

e

i+1
i−1
· · ·Xnen with c ∈ F∗q and e1 , . . . , en ≥ 1 appearing in the
form cX1e1 · · ·Xi−1
Xi+1

 
(q)
= (−1)Mai , where
polynomial ai . More precisely, if q = 2 then sgn πq Eai

e

e

i+1
i−1
· · ·Xnen with c ∈ F∗q
Xi+1
Mai is the number of monomials of the form cX1e1 · · ·Xi−1
and e1 , . . . , en ≥ 1 appearing in the polynomial ai .

Proof. It suffices to assume that ai ∈ Fq [X1 , . . . , X̂i , . . . , Xn ] is a monomial (Similar
discussion can be found in [3, page 96, Proof of Theorem 5.2.1]). Let ci ∈ Fq , Nî :=
n−1
(e1 , . . . , eˆi , . . . , en ≥
{1, . . . , i − 1, i + 1, . . . , n}, and e := (e1 , . . . , eˆi , . . . , en ) ∈ Z≥0
Q
ej
0). We put ai = ci 1≤j≤n Xj and
j6=i

Ec(q)
i ,e

:=

X1 , . . . , Xi−1 , Xi + ci

Y

e
Xj j , Xi+1 , . . . , Xn

1≤j≤n
j6=i

=

X1 , . . . , Xi−1 , Xi + ci

Y

e
Xj j , Xi+1 , . . . , Xn

j∈Nî

!

!

∈ EAn (Fq ) .

 

(q)
In the following, we determine the value of sgn πq Eci ,e
by decomposing the


(q)
permutation πq Eci ,e as a product of transpositions. Let y1 , . . . , yi−1 , yi+1 , . . . , yn

be elements of Fq . We put y = (y1 , . . . , yi−1 , yi+1 , . . . , yn ) ∈ Fqn−1 . For y ∈ Fqn−1 ,
(q)

we define the map λci ,e,y as follows:
Fnq

Fnq

∈

−→

∈

(q)

λci ,e,y :

(x1 , . . . , xn )
(x1 , . . . , xn )

7−→
(x1 , . . . , xn )
if ∃j ∈ Nî s.t. xj 6= yj ,
(q)
7−→ Eci ,e (x1 , . . . , xn ) xj = yj for all j ∈ Nî .
(q)

It follows from the definition of the map λci ,e,y that
λc(q)
(x1 , . . . , xn ) = (x1 , . . . , xn )
i ,e,y
for each (x1 , . . . , xn ) ∈ Fnq \ {(y1 , . . . , yi−1 , y, yi+1 , . . . , yn ) | y ∈ Fq } and
λc(q)
(x1 , . . . , xn ) = Ec(q)
(x1 , . . . , xn )
i ,e,y
i ,e
5

for each (x1 , . . . , xn ) ∈ {(y1 , . . . , yi−1 , y, yi+1 , . . . , yn ) | y ∈ Fq }. Thus the map
(q)
(q)
(q)
(q)
λci ,e,y is bijective and the inverse of λci ,e,y is λ−ci ,e,y . Remark that λci ,e,y is the
identity map if and only if e = (0, . . . , 0) and ci = 0, or there exists j ∈ Nî such
that ej 6= 0 and yj = 0. Put


n
o
n
(q)
B λc(q)
:=
(x
,
.
.
.
,
x
)
∈
F
λ
(x
,
.
.
.
,
x
)
=
6
(x
,
.
.
.
,
x
)
.
1
n
n
1
n
q
ci ,e,y 1
i ,e,y
Since





(q)
B λc(q)
∩
B
λ
=∅
′
ci ,e,y
i ,e,y

′
′
6 y, the permutation
for any y′ = y1′ , . . . , yi−1
, yi+1
, . . . , yn′ ∈ Fqn−1 satisfying y′ =


(q)
πq Eci ,e can be written as




Y

λc(q)
= πq 
πq Ec(q)
i ,e,y
i ,e
y1 ,...,yi−1 ,yi+1 ,...,yn ∈Fq

Y

=

y1 ,...,yi−1 ,yi+1 ,...,yn ∈Fq



,
πq λc(q)
,e,y
i

(3.3)

which is a composition of disjoint permutations on Fnq . We denote the number
of distinct permutations other than the identity map in the right-hand side of
Equation (3.3) by M1 . It is straightforward to check that M1 satisfies the equation
Y
ℓ
M1 = χ (ci ) ×
(q − δ (ej )) .
(3.4)
j∈Nî



(q)
Next, we decompose each permutation πq λci ,e,y as a composition of disjoint
cycles on Fnq . In order to find such a decomposition, we define an equivalence
relation ∼ on Fq : y ∈ Fq and y ′ ∈ Fq are equivalent
 if and only if there exists
Q
ej
′
l ∈ {0, 1, . . . , p − 1} such that y = y + l ci j∈N yj . Note that the equivalence
î
relation ∼ depends on the choice of y1 , . . . , yi−1 , yi+1 , . . . , yn . Put
Cy := {y ′ ∈ Fq | y ∼ y ′ }.

We now choose a complete system of representatives R for the above equivalence
relation ∼. Since R is a complete system of representatives, it follows that
♯R = q/p = pm /p = pm−1 .
We write
R := {w1 , . . . , wpm−1 } ⊂ Fq .
For any ws ∈ R (1 ≤ s ≤ pm−1 ), we set yws := (y1 , . . . , yi−1 , ws , yi+1 , . . . , yn ). We
(q)
define the bijective map λci ,e,yws as follows:
Fnq

−→

(x1 , . . . , xn )
(x1 , . . . , xn )

∈

Fnq
∈

(q)

λci ,e,yws :

7−→

(x1 , . . . , xn )
if ∃j ∈ Nî s.t. xj 6= yj , or xi 6∈ Cws ,
(q)
7 → Eci ,e (x1 , . . . , xn )
−
xj = yj for all j ∈ Nî and xi ∈ Cws .
6

(q)

Note that the map λci ,e,yws is a cycle of length p, and the standard result from
elementary group theory yields that its sign is (−1)p−1 , namely,

 
p−1
= (−1)
.
(3.5)
sgn πq λc(q)
i ,e,yws
(q)

We also remark that the map λci ,e,yws is the identity map if and only if e =
(0, . . . , 0) and ci = 0, or there exists j ∈ Nî such that ej 6= 0 and yj = 0. If
1 ≤ s1 , s2 ≤ pm−1 and s1 6= s2 , then we have Cws1 ∩ Cws2 = ∅. This yields a
decomposition of the permutation πq (λci ,y ) into a composition of disjoint cycles on
Fnq :
 m−1

m−1
pY
pY




(q)
(q)
.
(3.6)
πq λc(q)
λci ,e,yws  =
πq λci ,e,y = πq 
i ,e,yws
s=1

s=1

We denote the number of disjoint cycles other than the identify map in Equation (3.6) by M2 . By counting the number of disjoint cycles appearing in Equation (3.6), we have
ℓ

M2 = χ (ci ) × pm−1 .



(3.7)

(q)
Eci ,e . By Equation (3.3)

Now we determine the sign of the permutation πq
through Equation (3.7), we obtain

 
= (−1)M ,
sgn πq Ec(q)
i ,e
where

M = M1 × M2 × (p − 1)
ℓ

= χ (ci ) × p

m−1

× (p − 1) ×

Y

j∈Nî



(q − δ (ej )) .

(3.8)

If q is odd (and thus p is odd) or q = 2m , m ≥ 2, we have M ≡ 0 mod 2 and
Q
ℓ
if q = 2 (namely, p = 2 and m = 1), we have M ≡ χ (ci ) × j∈N δ (ej ) mod 2.
î


(q)
Therefore if q is odd or q = 2m , m ≥ 2 then πq Eci ,e ∈ Alt(Fnq ). Hence, we


 

(q)
(q)
= 1. On the other hand, if
have πq Eai ∈ Alt(Fnq ), namely, sgn πq Eai


 
 
ℓ Q
χ(ci ) × j∈N δ(ej )
(q)
(q)
î
= −1
= (−1)
. Thus, sgn πq Eci ,e
q = 2 then sgn πq Eci ,e
 

Q
(q)
if and only if ci 6= 0 and j∈N δ (ej ) = 1, or more generally, sgn πq Eai
= −1
î

e

e

i+1
i−1
· · ·Xnen with
if and only if the number of monomials of the form cX1e1 · · ·Xi−1
Xi+1
∗
c ∈ Fq and e1 , . . . , en ≥ 1 appearing in the polynomial ai , is odd. This completes
the proof.


Corollary 1. If q is odd or q = 2m , m ≥ 2 then we have πq (EAn (Fq )) ⊂ Alt(Fnq ).
Remark 1. Let θ : Fq → C be a map satisfying that θ (c) = 0 when c = 0, and
θ (c) = 1 when c ∈ F∗q . Then one can also prove Main Theorem 1 by replacing
ℓ
χ (ci ) with θ (ci ) in the proof of Main Theorem 1.
Remark 2. Note that Main Theorem 1 is similar to [5, Lemma 6.4], but Main
Theorem 1 is more general result than [5, Lemma 6.4].
7

Example 1. Let ai := αX1 · · · Xi−1 Xi+1 · · · Xn , bi := βX12 X2 · · · Xi−1 Xi+1 · · · Xn ∈
Fq [X1 , . . . , X̂i , . . . , Xn ], and α, β ∈ F∗q . We consider the sign of the permutations
induced by the following elementary automorphisms:
= (X1 , . . . , Xi−1 , Xi + ai , Xi+1 , . . . , Xn ) ∈ EAn (Fq ) ,
Ea(q)
i
(q)

Ebi = (X1 , . . . , Xi−1 , Xi + bi , Xi+1 , . . . , Xn ) ∈ EAn (Fq ) ,
and
(q)

Eai +bi = (X1 , . . . , Xi−1 , Xi + ai + bi , Xi+1 , . . . , Xn ) ∈ EAn (Fq ) .
m
We first assume
q is odd or
, m ≥ 2. Then by Main Theorem 1,
 q = 2 
 that 
(q)
(q)
= 1. By the fact that πq and sgn
= sgn πq Ebi
we have sgn πq Eai


 
 
(q)
(q)
×
= sgn πq Eai
are group homomorphisms, we also have sgn πq Eai +bi

 
(q)
= 1. We next suppose that q = 2. Since α, β ∈ F∗q , we remark that
sgn πq Ebi


 
 
(q)
(q)
=
= sgn πq Ebi
α = β = 1. From Main Theorem 1, we have sgn πq Eai
−1.  Again,
by
the fact that πq and
homomorphisms, we obtain

 sgn are group
(q)

(q)

(q)

sgn πq Eai +bi
= sgn πq Eai
×sgn πq Ebi
= 1. We can directly prove



 
(q)
(q)
= 1 by using the fact that πq Eai +bi = πq ((X1 , . . . , Xn )).
that sgn πq Eai +bi
4. Sign of permutations induced by affine automorphisms

In this section, we consider the sign of permutations induced by affine automorg
(q)
phisms over finite fields. Suppose that Ab is an affine automorphism over a finite
field, where,
!
!
!
n
n
X
X
g
(q)
(4.1)
an,i Xi + bn ∈ Aff n (Fq ) ,
a1,i Xi + b1 , . . . ,
Ab :=
i=1

i=1

and ai,j , bi ∈ Fq for 1 ≤ i, j ≤ n. We also assume that A(q) is the homogeneous
part (linear automorphism) of the affine automorphism (4.1), namely,
!
n
n
X
X
(4.2)
an,i Xi ∈ Aff n (Fq ) .
a1,i Xi , . . . ,
A(q) =
i=1

i=1

By Equation (2.1) and by Main Theorem 1, we obtain

 

 
g
(q)
= sgn πq A(q) .
sgn πq Ab

(4.3)

Thus, it is sufficient to consider the sign of affine automorphisms over finite field of
the form (4.2). We put
Ti,j := (X1 , . . . , Xi−1 , Xj , Xi+1 , . . . , Xj−1 , Xi , Xj+1 , . . . , Xn ) ∈ Aff n (Fq ) ,

(4.4)

Di (c) := (X1 , . . . , Xi−1 , cXi , Xi+1 , . . . , Xn ) ∈ Aff n (Fq ) ,

(4.5)

Ri,j (c) := (X1 , . . . , Xi−1 , Xi + cXj , Xi+1 , . . . , Xn ) ∈ Aff n (Fq ) ,

(4.6)

and
where c ∈ F∗q . It is easy to see that
2
Ti,j
:= Ti,j ◦ Ti,j = (X1 , . . . , Xn ) ,
8

q−1

Di (c)

:= Di (c) ◦ · · · ◦ Di (c) = (X1 , . . . , Xn ) ,
|
{z
}
q − 1 times

and
p

Ri,j (c) := Ri,j (c) ◦ · · · ◦ Ri,j (c) = (X1 , . . . , Xn ) .
{z
}
|
p times

Since each invertible matrix is a product of elementary matrices ([1, Proposition 2.18]), a linear automorphism can be written as a finite composition of linear
automorphisms of form (4.4), (4.5), and (4.6). Namely, there exists ℓA ∈ Z≥0 and
(A)
(A)
linear automorphisms M1 , . . . , MℓA such that
(A)

A(q) = M1

(A)

◦ · · · ◦ M ℓA ,

(4.7)

(A)

where Mi is a linear automorphism of form (4.4), (4.5), or (4.6) for each i (1 ≤
i ≤ ℓA ). We remark that the representation (4.7) is not unique in general. Since
πq and sgn are group homomorphisms, we obtain




(q)

sgn πq A



=

ℓA
Y

i=1


 
(A)
.
sgn πq Mi

(4.8)

Therefore, it is sufficient to consider the sign of the permutation induced by a
linear automorphism of form (4.4), (4.5), and (4.6). We use for the symbols NT (A),
ND (A), and NR (A) to represent the number of linear automorphisms of form (4.4),
(4.5), and (4.6) appearing in (4.7), respectively. Then we have
NT (A) + ND (A) + NR (A) = ℓA .
o
n
(A)
(A)
Furthermore, we suppose that i1 , . . . , iND (A) is the subset of {1, . . . , ℓA } satisfying the following two conditions:


(A)
(A)
(A)
(A)
(i): There exist uj ∈ {1, . . . , n} and cj ∈ F∗q such that M (A) = Du(A) cj
ij

j

for 1 ≤ j ≤ ND (A).

(A)

(A)

(A)

(ii): For i ∈ {1, . . . , ℓA } \ {i1 , . . . , iND (A) }, Mi

is of form (4.4) or (4.6).

In the following, we determine the sign of permutations induced by the above
three types of linear automorphisms of form (4.4), (4.5), and (4.6).
Lemma 1. (Sign of πq (Ti,j )) Suppose that Ti,j ∈ Aff n (Fq ) is a linear automorphism of form (4.4). If n ≥ 2, then

(q = 2m , m ≥ 2 or q = 2, n ≥ 3) ,

1
(q = 2 and n = 2) ,
sgn (πq (Ti,j )) = −1
(4.9)

q−1

(−1) 2
(q is odd) .

Proof. Put

B (Ti,j ) := {(x1 , . . . , xn ) ∈ Fnq | Ti,j ((x1 , . . . , xn )) 6= (x1 , . . . , xn )}
= {(x1 , . . . , xn ) ∈ Fnq | xi 6= xj }.
Since
B (Ti,j ) = Fnq \ {(x1 , . . . , xn ) ∈ Fnq | xi = xj },
9

we have ♯B (Ti,j ) = q n − q n−2 × q = q n−1 (q − 1). Hence one can see that the
permutation πq (Ti,j ) is a compositon of q n−1 (q − 1) /2 transpositions on Fnq . We
write NTi,j = q n−1 (q − 1) /2.
Case 1. q = 2m , m ≥ 2.
Since q n−1 /2 = 2m(n−1)−1 ≥ 2, we have NTi,j ≡ 0 mod 2. Therefore, πq (Ti,j ) is an
even permutation.
Case 2. q = 2.
In this case, we see that NTi,j = 2n−2 . If n ≥ 3, we obtain NTi,j ≡ 0 mod 2.
Otherwise NTi,j = 1.
Case 3. q is odd.
If q is odd then q ≡ 1 mod 4 or q ≡ 3 mod 4. From this fact, we obtain
(
0 (q ≡ 1 mod 4) ,
q−1
≡
NTi,j ≡
2
1 (q ≡ 3 mod 4) .
Thus, Equation (4.9) holds.



Lemma 2. (Sign of πq (Di (c))) Suppose that Di (c) ∈ Aff n (Fq ) is a linear automorphism of form (4.5). If n ≥ 2, then
(
1
(q is even) ,
sgn (πq (Di (c))) =
(4.10)
ordF∗
(c)
q
(−1)
(q is odd) .
Proof. Let us suppose that q is even. We assume that sgn (πq (Di (c))) = −1. Since
sgn and πq are group homomorphisms, we have

 
q−1
q−1
q−1
= sgn (πq (Di (c)))
= (−1)
= −1.
sgn πq Di (c)
q−1

On the other hand, from Di (c)
= (X1 , . . . , Xn ), we must have

 
q−1
= sgn (πq ((X1 , . . . , Xn ))) = 1.
sgn πq Di (c)

This is a contradiction. Therefore, sgn (πq (Di (c))) = 1.
Next suppose that q is odd. We define the map Hc : Fq → Fq as follows:
Hc : F q

∈

Fq

∈

−→

x

7−→ cx.

The map Hc is bijective, and the inverse map is Hc−1 . Since we can regard the
map Hc as a permutation on Fq , it is obious that
sgn (πq (Di (c))) = sgn (Hc ) .
Let g be a generator of the multiplicative group F∗q . We put c = g h , 0 ≤ h =
ordF∗q (c) ≤ q − 2. If h = 0 then sgn (Hc ) = 1. We assume that h 6= 0. If h = 1 then

the map Hc is the length
q − 1 cycle  g 0 g 1 · · · g q−2
as a permutation
on Fq , and



hence Hc = g q−3 g q−2 ◦ g q−4 g q−3 ◦ · · · ◦ g 1 g 2 ◦ g 0 g 1 , the product of q − 2
transpositions as a permutation on Fq . This yields that for 1 ≤ h ≤ q− 2, the map
Hc is the product of h copies of the length q − 1 cycle g 0 g 1 · · · g q−2 , namely, the
product of h × (q − 2) transpositions as a permutation on Fq . Therefore, we obtain
10

h

sgn (Hc ) = (−1)h(q−2) . Since q is odd, it satisfies that (−1)h(q−2) = (−1)q−2 =
h

(−1) . Hence

ordF∗
(c)
q

sgn (Hc ) = (−1)h = (−1)
for c ∈
holds.

F∗q ,

(4.11)

c 6= 1. Equation (4.11) is obviously true for c = 1. Thus the assertion


Lemma 3. (Sign of πq (Ri,j (c))) Suppose that Ri,j (c) ∈ Aff n (Fq ) is a linear
automorphism of form (4.6). If n ≥ 2, then
sgn (πq (Ri,j (c))) = 1.

(4.12)

Proof. By Ri,j (c) ∈ Aff n (Fq ) ∩ EAn (Fq ) ⊂ EAn (Fq ), it follows immediately from
Main Theorem 1.

We are now in the position to prove the main result of this section.
Main Theorem 2. (Sign of affine automorphisms) With notation as above,
the following assertions are hold:
(1) If q = 2m , m ≥ 2 then

 
g
(q)
= 1.
(4.13)
sgn πq Ab
(2) If q = 2 and n = 2 then


 
g
(q)
= (−1)NT (A) .
sgn πq Ab

(4.14)


 
g
(q)
= 1.
sgn πq Ab

(4.15)

(3) If q = 2 and n ≥ 3 then

(4) If q is odd then

 
PN (A)


(A)
D
g
ordF∗
cj
+ q−1
(q)
2 NT (A)
q
= (−1) j=1
sgn πq Ab
.

In particular, if q ≡ 1 mod 4 then

 
P


ND (A)
(A)
g
ordF∗
cj
(q)
q
= (−1) j=1
sgn πq Ab
.

(4.16)

(4.17)

Proof. The assertions (1) through (4) follow immediately from Equation (4.3),
Equation (4.7), and Lemma 1 through Lemma 3. This completes the proof.

Corollary 2. If q = 2m and m ≥ 2, or q = 2 and n ≥ 3 then we have πq (Aff n (Fq )) ⊂
Alt(Fnq ).

Remark 3. One can see that Equation (4.14), (4.15), (4.16), and
 (4.17)
 depend on
g
(q)
is uniquely
the representation (4.7) which is not unique. However, since πq Ab

 
g
(q)
does not depend on the
determined as a permutation on Fnq , sgn πq Ab
representation (4.7).

11

Example 2. Let α, β ∈ F∗q . We consider the sign of the permutation induced
by the affine automorphism A(q) := (X3 , X2 , αX1 + βX3 ) ∈ Aff 3 (Fq ). It is easy
to see that A(q) = (X3 , X2 , X1 ) ◦ (X1 + βX3 , X2 , X3 ) ◦ (αX1 , X2 , X3 ) = T1,3 ◦
R1,3 (β) ◦ D1 (α).We remark that NT (A) = ND (A) = NR (A) = 1, ℓA = 3,
and sgn πq A(q) = sgn (πq (T1,3 )) × sgn (πq (R1,3 (β))) × sgn (πq (D1 (α))). If
p = 2 then by
 Equation (4.13) and by Equation (4.15), one can easily see that
sgn πq A(q) = 1. If q is odd then by Equation (4.16),

 
q−1
ord ∗ (α)
= (−1) Fq
× (−1) 2 .
sgn πq A(q)

ord ∗ (α)
In particular, if q ≡ 1 mod 4 then sgn πq A(q) = (−1) Fq , and if q ≡ 3 mod 4

ord ∗ (α)
ord ∗ (α)+1
= (−1) Fq
.
then sgn πq A(q) = −1 × (−1) Fq
5. Sign of permutations induced by triangular automorphisms and
tame automorphisms

In this section, we consider the sign of permutations induced by triangular automorphisms and tame automorphisms over finite fields. By Main Theorem 1 and
Main Theorem 2, we obtain the following corollary (Corollary 3).
(q)

Corollary 3. (Sign of triangular automorphisms) Suppose that Ja,f is a
triangular automorphism over a finite field, namely,
(q)

Ja,f = (a1 X1 + f1 (X2 , . . . , Xn ), a2 X2 + f2 (X3 , . . . , Xn ), . . . , an Xn + fn ) ∈ BAn (Fq ),
ai ∈ Fq (i = 1, . . . , n) , fi ∈ Fq [Xi+1 , . . . , Xn ] (i = 1, . . . , n − 1) , fn ∈ Fq .
(5.1)


(q)
If q = 2m , m ≥ 2 then πq Ja,f ∈ Alt(Fnq ). Namely, if q = 2m , m ≥ 2 then

 
(q)
= 1.
(5.2)
sgn πq Ja,f

If q is odd then


 
Pn
ord ∗ (a )
(q)
(5.3)
= (−1) i=1 Fq i .
sgn πq Ja,f

 
(q)
depends only on the coefficients
In other words, if q is odd then sgn πq Ja,f
a1 , . . . , an . If q = 2 then

 
(q)
M
= (−1) f1 ,
(5.4)
sgn πq Ja,f

where Mf1 is the number of monomials of the form cX2e2 · · ·Xnen with c ∈ F∗q and
e2 , . . . , en ≥ 1 appearing in the polynomial f1 ∈ Fq [X2 , . . . , Xn ].
Proof. By using the notation of Equation (3.1) and Equation (4.5), we have
(q)

(q)

(q)

Ja,f = Dn (an ) ◦ · · · D1 (a1 ) ◦ Ea−1 f ◦ · · · ◦ Ea−1 f .
1

1

n

n

Hence we obtain
n
n

 
 Y
 
Y
(q)
(q)
.
sgn πq Ea−1 f
sgn (πq (Di (ai ))) ×
=
sgn πq Ja,f
i=1

i=1

i

i

(5.5)

By Equation (5.5), Main Theorem 1, and Main Theorem 2, we obtain the desired
results.

12

(q)

Corollary 4. (Sign of strictly triangular automorphisms) Suppose that Jf
is a strictly triangular automorphism over a finite field, namely,
(q)

Jf

= (X1 + f1 (X2 , . . . , Xn ), X2 + f2 (X3 , . . . , Xn ), . . . , Xn + fn ) ∈ BAn (Fq ),
fi ∈ Fq [Xi+1 , . . . , Xn ] (i = 1, . . . , n − 1) , fn ∈ Fq .
(5.6)


(q)



∈ Alt(Fnq ). Namely, if q is odd or
If q is odd or q = 2m , m ≥ 2 then πq Jf
q = 2m , m ≥ 2 then

 
(q)
= 1.
(5.7)
sgn πq Jf

If q = 2 then


 
(q)
M
= (−1) f1 ,
sgn πq Jf

cX2e2 · · ·Xnen

where Mf1 is the number of monomials of the form
e2 , . . . , en ≥ 1 appearing in the polynomial f1 ∈ Fq [X2 , . . . , Xn ].

(5.8)
with c ∈

Proof. It follows immediately from Corollary 3.

F∗q

and


We recall that for any φ(q) ∈ TAn (Fq ), there exist l ∈ Z≥0 , ǫ1 , ǫ2 ∈ {0, 1} ⊂ Z,
^
(q)
As,b(s) ∈ Aff n (Fq ) (1 ≤ s ≤ l + 1) of the form
!
!
!
n
n
X
X
^
(s)
(s)
(s)
(q)
(s)
,
an,i Xi + bn
a1,i Xi + b1 , . . . ,
As,b(s) =
i=1

i=1

and

(q)
Js,t(s) ,f (s)

such that
(q)

φ

∈ BAn (Fq ) (1 ≤ s ≤ l) of the form


(q)
(s)
(s)
(s)
Js,t(s) ,f (s) = t1 X1 + f1 (X2 , . . . , Xn ), . . . , t(s)
,
n Xn + f n

^
(q)
= A1,b(1)

ǫ1

◦

(q)
J1,t(1) ,f (1)

ǫ2

^
^
(q)
(q)
(q)
◦ · · · ◦ Al,b(l) ◦ Jl,t(l) ,f (l) ◦ Al+1,b(l+1) ,

(5.9)

^
(q)
(q)
As,b(s) 6∈ BAn (Fq ) for 2 ≤ s ≤ l + 1, and Js,t(s) ,f (s) 6∈ Aff n (Fq ) for 1 ≤ s ≤ l (for
example, [2, Lemma 5.1.1]).
(q)
We use the symbol As to denote the homogeneous part (linear automorphism)
^
(q)
of the affine automorphism As,b(s) (as in Equation (4.1) and Equation (4.2)) for
1 ≤ s ≤ l + 1. Furthermore, for each s (1 ≤ s ≤ l), we denote by Mf (s) the number
1
of monomials of the form cX2e2 · · ·Xnen with c ∈ F∗q and e2 , . . . , en ≥ 1 appearing in
(s)

the polynomial f1

∈ Fq [X2 , . . . , Xn ].

The following corollary (Corollary 5) states that if we know Equation (5.9) for
a given φ(q) ∈ TAn (Fq ), then one can easily compute the sign of the permutation
induced by φ(q) ∈ TAn (Fq ).
Corollary 5. (Sign of tame automorphisms) With notation as above, the
following assertions are hold:
(1) If q = 2m , m ≥ 2 then

 
= 1.
(5.10)
sgn πq φ(q)
13

(2) If q = 2 and n = 2 then
P
P

 
ǫ1 NT (A1 )+( ls=2 NT (As ))+ǫ2 NT (Al+1 )+ ℓs=1 M (s)
(q)
f
1
= (−1)
.
sgn πq φ

(5.11)

(3) If q = 2 and n ≥ 3 then

Pℓ
 

s=1 Mf (s)
1
sgn πq φ(q)
.
= (−1)

(5.12)

(4) If q is odd then


 

Pl PND (As )
(A )
ordF∗
cj s
q
sgn πq φ(q)
= (−1) s=2 j=1
ǫ1

PN (A )
1
D

ǫ2

PN (A
D
l+1 )

× (−1)

× (−1)

× (−1)

j=1



(A )
ordF∗
cj 1
q

 (A

)
ordF∗
cj l+1
q

j=1

q−1
2

(ǫ1 NT (A1 )+ǫ2 NT (Al+1 )+



P
(s)
ti
1≤i≤n ordF∗
q

× (−1)

1≤s≤l

Pl

s=2

NT (As ))

.

(5.13)

In particular, if q ≡ 1 mod 4 then



 
Pl PND (As )
(A )
cj s
ordF∗
q
= (−1) s=2 j=1
sgn πq φ(q)
ǫ1

P
ND (A1 )

ǫ2

PN (A
D
l+1 )

× (−1)

× (−1)

j=1

P

× (−1)

j=1

1≤i≤n
1≤s≤l



(A )
ordF∗
cj 1
q

 (A

)
ordF∗
cj l+1
q



(s)
ordF∗
ti
q

.

(5.14)

Proof. By Equation (5.9), we obtain
l
ǫ1
 
 
 
 Y
 
(q)
(q)
× sgn πq A1
sgn πq Ai
=
sgn πq φ(q)
i=2

l

 
ǫ2 Y

 
(q)
(q)
sgn πq Ji,t(i) ,f (i) .
×
× sgn πq Al+1

(5.15)

i=1

By Equation (5.15), Main Theorem 2, and Corollary 3, we obtain the desired
results.

Remark 4. As in Remark 3, one can see that Equation (5.11), (5.12), (5.13), and
(5.14) dependon the representations (4.7) and (5.9) which are not unique. However,

since πq φ(q) is uniquely determined as a permutation on Fnq , sgn πq φ(q) does
not depend on the representations (4.7) and (5.9).
Remark 5. Suppose that n is greater than or equal to two. Corollary 5 tells us
that πq (TAn (Fq )) ⊂ Sym Fnq (this is a trivial inclusion relation) if q is odd or

q = 2, and πq (TAn (Fq )) ⊂ Alt Fnq if q = 2m and m ≥ 2. This indicates that
Corollary 5 is strictly weaker than [3, Theorem 2.3]. However, Main Theorem 1,
Main Theorem 2, Corollary 3, and Corollary 5 are useful for determining the sign
of permutations induced by tame automorphisms over finite fields. The reasons are
as follows. Firstly, for q = 2m and m ≥ 2, we prove that πq (TAn (Fq )) ⊂ Alt(Fnq )
14

by directly showing that πq (EAn (Fq )) ⊂ Alt(Fnq ) and πq (Aff n (Fq )) ⊂ Alt(Fnq ).
Secondly, if q is odd then one can not determine the sign of permutation induced
by an elementary automorphisms over a finite field by using [3, Theorem 2.3]. In
contrast to [3, Theorem 2.3], Main Theorem 1 tells us that if q is odd then each
permutation induced by an elementary automorphism over a finite field is even. In
other words, if q is odd then πq (EAn (Fq )) ⊂ Alt(Fnq ) (Corollary 1). Similarly, in
contrast to [3, Theorem 2.3], Main Theorem 2 tells us that if q = 2 and n ≥ 3 then
each permutation induced by an affine automorphism over a finite field is even.
Namely, if q = 2 and n ≥ 3 then πq (Aff n (Fq )) ⊂ Alt(Fnq ) (Corollary 2). Thus, our
results (Main Theorem 1, Main Theorem 2, Corollary 3, and Corollary 5) and [3,
Theorem 2.3] are complementary to each other.
Acknowledgements
This work was supported by JSPS KAKENHI Grant-in-Aid for Young Scientists
(B) 16K16066.
References
[1] M. Artin, Algebra, Prentice-Hall, 1991.
[2] A. van den Essen, Polynomial Automorphisms and the Jacobian Conjecture, Progress in
Mathematics, Vol.190, Birkhäuser Verlag, Basel-Boston-Berlin, 2000.
[3] S. Maubach, Polynomial automorphisms over finite fields, Serdica Math. J., 27 (2001), no.4,
343–350.
[4] S. Maubach, A problem on polynomial maps over finite fields, arXiv preprint (2008),
arXiv:0802.0630.
[5] S. Maubach and A. Rauf, The profinite polynomial automorphism group, J. Pure Appl.
Algebra, 219 (2015), no.10, 4708–4727.
Interdisciplinary Graduate School of Science and Engineering, Shimane University,
1060 Nishikawatsu-cho, Matsue, Shimane 690-8504, Japan
E-mail address: hakuta(at)cis.shimane-u.ac.jp

15

