This paper provides formulas to determine the sign of permutations induced by tame automorphisms over finite fields, enabling easier analysis of their permutation properties based on their decompositions.
Contribution
It introduces explicit formulas for the permutation sign induced by elementary and affine automorphisms over finite fields, and combines these to analyze tame automorphisms.
Findings
01
Formulas for signs of elementary automorphisms
02
Formulas for signs of affine automorphisms
03
Method to determine signs of tame automorphisms
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.
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Full text
On permutations induced by tame automorphisms over finite fields
Keisuke Hakuta
Interdisciplinary Graduate School of Science and Engineering,
Shimane University,
1060 Nishikawatsu-cho, Matsue, Shimane 690-8504, Japan
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.
Key words and phrases:
Affine algebraic geometry,
polynomial automorphism,
tame automorphism,
finite field,
permutation.
2010 Mathematics Subject Classification:
Primary:
14R10; Secondary:
12E20, 20B25.
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 kn to itself
by defining
[TABLE]
for (a1,…,an)∈kn.
The set of polynomial maps over k and the set of maps from kn to itself
are denoted by MAn(k) and Maps(kn,kn), respectively.
Then by (1.1), we see that there exists a natural map
[TABLE]
MAn(k) (resp. Maps(kn,kn)) 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 degfi=1 for i=1,…,n.
A polynomial map Eai of the form
[TABLE]
is also a polynomial automorphism since Eai−1=E−ai.
A polynomial automorphism Eai of form
(1.2) is said to be elementary.
Let Affn(k) denote the set of all affine automorphisms,
and let EAn(k) denote the set of all elementary automorphisms.
Affn(k) and EAn(k) are subgroups of GAn(k).
We put TAn(k):=⟨Affn(k),EAn(k)⟩,
where ⟨H1,H2⟩ 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
[TABLE]
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):
[TABLE]
BAn(k) is a subgroup of GAn(k).
BAn(k) is also a subgroup of TAn(k), and it is known that
TAn(k)=⟨Affn(k),BAn(k)⟩ (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:
[TABLE]
The map πq induces a group homomorphism
[TABLE]
where Sym(S) is a symmetric group on a finite set S.
Let
[TABLE]
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(Fqn).
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(Fqn).
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(Fqn)
if q is odd or q=2.
If q=2m where m≥2 then
πq(TAn(Fq))=Alt(Fqn).*
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(Fqn)∖Alt(Fqn)?
**
If there exists F∈GAn(F2m)
such that π2m(F)∈Sym(F2mn)∖Alt(F2mn),
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(Fqn) holds?
**
Question 2 seems to be natural
since if q=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=2m (m≥2) and G=TAn(Fq),
one can not obtain any sufficient condition for the inclusion relation
πq(G)⊂Alt(Fqn)
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(Fqn)
(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(Affn(Fq))⊂Alt(Fqn)
(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.
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 gx=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,…,Xi^,…,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), Affn(k), EAn(k),
and BAn(k), respectively.
GAn(k) is a group with respect to the composition of polynomial automorphisms,
and Affn(k), EAn(k) are subgroups of GAn(k).
Recall that
[TABLE]
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),
[TABLE]
means that σ(si)=sσ(i) for 1≤i≤n.
For σ,τ∈Sym(S), we denote by τ∘σ the composition
[TABLE]
The permutation on S defined by
[TABLE]
is called the cycle of length r and is denoted
by (i1i2…ir).
Let δ:R→R be a function
satisfying that δ(x)=0 when x=0,
and δ(x)=1 when x=0.
Let χ:Fq∗→C∗ be a non-trivial multiplicative character of order ℓ.
We extend χ to Fq by defining χ(0)=0.
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.
Main Theorem 1**.**
(Sign of elementary automorphisms) Suppose that Eai(q)
is an elementary automorphism over a finite field, namely,
[TABLE]
and ai∈Fq[X1,…,Xi^,…,Xn].
If q is odd or q=2m, m≥2 then we have πq(Eai(q))∈Alt(Fqn).
Namely, if q is odd or q=2m, m≥2 then
[TABLE]
If q=2 then sgn(πq(Eai(q)))
depends only on the number of monomials of the form
cX1e1⋯Xi−1ei−1Xi+1ei+1⋯Xnen
with c∈Fq∗ and e1,…,en≥1 appearing in the polynomial ai.
More precisely, if q=2 then
sgn(πq(Eai(q)))=(−1)Mai,
where Mai is the number of monomials of the form
cX1e1⋯Xi−1ei−1Xi+1ei+1⋯Xnen
with c∈Fq∗ and e1,…,en≥1 appearing in the polynomial ai.
Proof.
It suffices to assume that ai∈Fq[X1,…,Xi^,…,Xn] is a monomial
(Similar discussion can be found in [3, page 96, Proof of Theorem 5.2.1]).
Let ci∈Fq, Ni^:={1,…,i−1,i+1,…,n}, and
e:=(e1,…,ei^,…,en)∈Z≥0n−1 (e1,…,ei^,…,en≥0).
We put ai=ci∏1≤j≤nj=iXjej and
[TABLE]
In the following, we determine the value of sgn(πq(Eci,e(q)))
by decomposing the permutation πq(Eci,e(q))
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,
we define the map λci,e,y(q) as follows:
[TABLE]
It follows from the definition of the map λci,e,y(q) that
[TABLE]
for each
(x1,…,xn)∈Fqn∖{(y1,…,yi−1,y,yi+1,…,yn)∣y∈Fq}
and
[TABLE]
for each
(x1,…,xn)∈{(y1,…,yi−1,y,yi+1,…,yn)∣y∈Fq}.
Thus the map λci,e,y(q) is bijective
and the inverse of λci,e,y(q) is λ−ci,e,y(q).
Remark that
λci,e,y(q) is the identity map
if and only if e=(0,…,0) and ci=0,
or there exists j∈Ni^ such that ej=0 and yj=0.
Put
[TABLE]
Since
[TABLE]
for any y′=(y1′,…,yi−1′,yi+1′,…,yn′)∈Fqn−1 satisfying y′=y,
the permutation πq(Eci,e(q)) can be written as
[TABLE]
which is a composition of disjoint permutations on Fqn.
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
[TABLE]
Next, we decompose each permutation πq(λci,e,y(q))
as a composition of disjoint cycles on Fqn.
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 l∈{0,1,…,p−1} such that
y′=y+l(ci∏j∈Ni^yjej).
Note that the equivalence relation ∼ depends on the choice of y1,…,yi−1,yi+1,…,yn.
Put
[TABLE]
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
[TABLE]
We write
[TABLE]
For any ws∈R (1≤s≤pm−1), we set
yws:=(y1,…,yi−1,ws,yi+1,…,yn).
We define the bijective map λci,e,yws(q) as follows:
[TABLE]
Note that the map λci,e,yws(q) is a cycle of length p,
and the standard result from elementary group theory yields that its sign is (−1)p−1, namely,
[TABLE]
We also remark that the map λci,e,yws(q) is the identity map if and only if
e=(0,…,0) and ci=0,
or there exists j∈Ni^ such that ej=0 and yj=0.
If 1≤s1,s2≤pm−1 and s1=s2,
then we have Cws1∩Cws2=∅.
This yields a decomposition of the permutation πq(λci,y) into a composition of disjoint cycles on Fqn:
[TABLE]
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
[TABLE]
Now we determine the sign of the permutation πq(Eci,e(q)).
By Equation (3.3)
through Equation (3.7),
we obtain
[TABLE]
where
[TABLE]
If q is odd (and thus p is odd) or q=2m, m≥2, we have
M≡0mod2 and if q=2 (namely, p=2 and m=1),
we have M≡χ(ci)ℓ×∏j∈Ni^δ(ej)mod2.
Therefore if q is odd or q=2m, m≥2 then
πq(Eci,e(q))∈Alt(Fqn).
Hence, we have πq(Eai(q))∈Alt(Fqn),
namely, sgn(πq(Eai(q)))=1.
On the other hand, if q=2 then sgn(πq(Eci,e(q)))=(−1)χ(ci)ℓ×∏j∈Ni^δ(ej).
Thus, sgn(πq(Eci,e(q)))=−1
if and only if ci=0 and ∏j∈Ni^δ(ej)=1,
or more generally, sgn(πq(Eai(q)))=−1
if and only if the number of monomials of the form
cX1e1⋯Xi−1ei−1Xi+1ei+1⋯Xnen
with 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(Fqn).
Remark 1**.**
Let θ:Fq→C be a map satisfying that
θ(c)=0 when c=0,
and θ(c)=1 when c∈Fq∗.
Then one can also prove Main Theorem 1
by replacing χ(ci)ℓ 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].
**
Example 1**.**
Let ai:=αX1⋯Xi−1Xi+1⋯Xn,bi:=βX12X2⋯Xi−1Xi+1⋯Xn∈Fq[X1,…,Xi^,…,Xn],
and α,β∈Fq∗.
We consider the sign of the permutations induced by the following elementary automorphisms:
[TABLE]
[TABLE]
and
[TABLE]
We first assume that q is odd or q=2m, m≥2.
Then by Main Theorem 1,
we have sgn(πq(Eai(q)))=sgn(πq(Ebi(q)))=1.
By the fact that πq and sgn are group homomorphisms,
we also have sgn(πq(Eai+bi(q)))=sgn(πq(Eai(q)))×sgn(πq(Ebi(q)))=1.
We next suppose that q=2.
Since α,β∈Fq∗, we remark that α=β=1.
From Main Theorem 1,
we have sgn(πq(Eai(q)))=sgn(πq(Ebi(q)))=−1.
Again, by the fact that πq and sgn are group homomorphisms,
we obtain sgn(πq(Eai+bi(q)))=sgn(πq(Eai(q)))×sgn(πq(Ebi(q)))=1.
We can directly prove that sgn(πq(Eai+bi(q)))=1
by using the fact that πq(Eai+bi(q))=πq((X1,…,Xn)).
**
4. Sign of permutations induced by affine automorphisms
In this section, we consider the sign of permutations induced by affine automorphisms over finite fields.
Suppose that Ab(q)
is an affine automorphism over a finite field, where,
[TABLE]
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,
[TABLE]
By Equation (2.1)
and by Main Theorem 1,
we obtain
[TABLE]
Thus, it is sufficient to consider the sign of affine automorphisms over finite field of the form (4.2).
We put
[TABLE]
[TABLE]
and
[TABLE]
where c∈Fq∗.
It is easy to see that
[TABLE]
[TABLE]
and
[TABLE]
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 linear automorphisms M1(A),…,MℓA(A) such that
[TABLE]
where Mi(A) 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
[TABLE]
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
[TABLE]
Furthermore, we suppose that {i1(A),…,iND(A)(A)}
is the subset of {1,…,ℓA} satisfying the following two conditions:
(i):
There exist uj(A)∈{1,…,n} and cj(A)∈Fq∗ such that Mij(A)(A)=Duj(A)(cj(A)) for 1≤j≤ND(A).
(ii):
For i∈{1,…,ℓA}∖{i1(A),…,iND(A)(A)},
Mi(A) 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∈Affn(Fq)
is a linear automorphism of form (4.4).
If n≥2, then*
[TABLE]
Proof.
Put
[TABLE]
Since
[TABLE]
we have ♯B(Ti,j)=qn−qn−2×q=qn−1(q−1).
Hence one can see that the permutation πq(Ti,j) is a compositon of qn−1(q−1)/2 transpositions on Fqn.
We write NTi,j=qn−1(q−1)/2.
**Case 1. ** q=2m,m≥2.
Since qn−1/2=2m(n−1)−1≥2, we have NTi,j≡0mod2.
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≡0mod2.
Otherwise NTi,j=1.
**Case 3. ** q is odd.
If q is odd then q≡1mod4 or q≡3mod4.
From this fact, we obtain
(Sign of πq(Di(c)))* Suppose that Di(c)∈Affn(Fq)
is a linear automorphism of form (4.5).
If n≥2, then*
[TABLE]
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
[TABLE]
On the other hand, from Di(c)q−1=(X1,…,Xn),
we must have
[TABLE]
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:
[TABLE]
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
[TABLE]
Let g be a generator of the multiplicative group Fq∗.
We put c=gh, 0≤h=ordFq∗(c)≤q−2.
If h=0 then sgn(Hc)=1.
We assume that h=0.
If h=1 then the map Hc is the length q−1 cycle
(g0g1⋯gq−2) as a permutation on Fq,
and hence Hc=(gq−3gq−2)∘(gq−4gq−3)∘⋯∘(g1g2)∘(g0g1),
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 (g0g1⋯gq−2),
namely, the product of h×(q−2) transpositions as a permutation on Fq.
Therefore, we obtain sgn(Hc)=(−1)h(q−2).
Since q is odd, it satisfies that (−1)h(q−2)=((−1)q−2)h=(−1)h.
Hence
[TABLE]
for c∈Fq∗, c=1.
Equation (4.11) is obviously true for c=1.
Thus the assertion holds.
∎
Lemma 3**.**
(Sign of πq(Ri,j(c)))* Suppose that Ri,j(c)∈Affn(Fq)
is a linear automorphism of form (4.6).
If n≥2, then*
[TABLE]
Proof.
By Ri,j(c)∈Affn(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*
[TABLE]
(2)* If q=2 and n=2 then*
[TABLE]
(3)* If q=2 and n≥3 then*
[TABLE]
(4)* If q is odd then*
[TABLE]
In particular, if q≡1mod4 then
[TABLE]
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(Affn(Fq))⊂Alt(Fqn).
Remark 3**.**
One can see that
Equation (4.14),
(4.15),
(4.16),
and (4.17)
depend on the representation (4.7) which is not unique.
However, since πq(Ab(q)) is uniquely determined as a permutation on Fqn,
sgn(πq(Ab(q))) does not depend on
the representation (4.7).
**
Example 2**.**
Let α,β∈Fq∗.
We consider the sign of the permutation induced by the affine automorphism
A(q):=(X3,X2,αX1+βX3)∈Aff3(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),
[TABLE]
In particular, if q≡1mod4 then
sgn(πq(A(q)))=(−1)ordFq∗(α),
and if q≡3mod4 then
sgn(πq(A(q)))=−1×(−1)ordFq∗(α)=(−1)ordFq∗(α)+1.
**
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).
Corollary 3**.**
(Sign of triangular automorphisms) Suppose that Ja,f(q)
is a triangular automorphism over a finite field, namely,
[TABLE]
If q=2m, m≥2 then πq(Ja,f(q))∈Alt(Fqn).
Namely, if q=2m, m≥2 then
[TABLE]
If q is odd then
[TABLE]
In other words, if q is odd then sgn(πq(Ja,f(q)))
depends only on the coefficients a1,…,an.
If q=2 then
[TABLE]
where Mf1 is the number of monomials of the form
cX2e2⋯Xnen
with c∈Fq∗ 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
[TABLE]
Hence we obtain
[TABLE]
By Equation (5.5),
Main Theorem 1,
and Main Theorem 2,
we obtain the desired results.
∎
Corollary 4**.**
(Sign of strictly triangular automorphisms) Suppose that Jf(q)
is a strictly triangular automorphism over a finite field, namely,
[TABLE]
If q is odd or q=2m, m≥2 then πq(Jf(q))∈Alt(Fqn).
Namely, if q is odd or q=2m, m≥2 then
[TABLE]
If q=2 then
[TABLE]
where Mf1 is the number of monomials of the form
cX2e2⋯Xnen
with c∈Fq∗ and e2,…,en≥1
appearing in the polynomial f1∈Fq[X2,…,Xn].
We recall that for any ϕ(q)∈TAn(Fq),
there exist l∈Z≥0, ϵ1,ϵ2∈{0,1}⊂Z,
As,b(s)(q)∈Affn(Fq) (1≤s≤l+1) of the form
[TABLE]
and
Js,t(s),f(s)(q)∈BAn(Fq) (1≤s≤l) of the form
[TABLE]
such that
[TABLE]
As,b(s)(q)∈BAn(Fq) for 2≤s≤l+1,
and Js,t(s),f(s)(q)∈Affn(Fq) for 1≤s≤l
(for example, [2, Lemma 5.1.1]).
We use the symbol As(q) to denote the homogeneous part (linear automorphism)
of the affine automorphism As,b(s)(q)
(as in Equation (4.1) and Equation (4.2)) for 1≤s≤l+1.
Furthermore, for each s (1≤s≤l),
we denote by Mf1(s) the number of monomials of the form
cX2e2⋯Xnen
with c∈Fq∗ and e2,…,en≥1
appearing in the polynomial f1(s)∈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:
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)
depend on the representations (4.7)
and (5.9) which are not unique.
However, since πq(ϕ(q)) is uniquely determined as a permutation on Fqn,
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(Fqn)
(this is a trivial inclusion relation) if q is odd or q=2, and
πq(TAn(Fq))⊂Alt(Fqn)
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(Fqn)
by directly showing that πq(EAn(Fq))⊂Alt(Fqn)
and πq(Affn(Fq))⊂Alt(Fqn).
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(Fqn)
(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(Affn(Fq))⊂Alt(Fqn)
(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.
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