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arXiv:2305.20008v1 [math.NT] 31 May 2023
NUMBER OF EQUIVALENCE CLASSES OF RATIONAL
FUNCTIONS OVER FINITE FIELDS
XIANG-DONG HOU
Abstract. Two rational functions f, g ∈Fq(X) are said to be equivalent if
there exist φ, ψ ∈Fq(X) of degree one such that g=φ◦f◦ψ. We give an
explicit formula for the number of equivalence classes of rational functions of
a given degree in Fq(X). This result should provide guidance for the current
and future work on classifications of low degree rational functions over finite
fields. We also determine the number of equivalence classes of polynomials of
a given degree in Fq[X].
1. Introduction
For a nonconstant rational function f(X) over a field F, written in the form
f(X) = P(X)/Q(X), where P, Q ∈F[X], Q6= 0, and gcd(P, Q) = 1, we define
deg f= max{deg P, deg Q}. Then [F(X) : F(f)] = deg f. By L¨uroth theorem,
every subfield E⊂F(X) with [F(X) : E] = dis of the form F(f) for some f∈F(X)
with deg f=d. Let
(1.1) G(F) = {φ∈F(X) : deg φ= 1}.
The group (G(F),◦) is isomorphic to the projective linear group PGL(2,F) and
the Galois group Aut(F(X)/F) of F(X) over F. For A=a b
c d ∈PGL(2,F), its
corresponding element in G(F), denoted by φA, is φA= (aX +b)/(cX +b). For
φ∈G(F), its corresponding element in Aut(F(X)/F), denoted by σφ, is the F-
automorphism of F(X) defined by σφ(X) = φ(X).
Two rational functions f , g ∈F(X)\Fare said to be equivalent, denoted as
f∼g, if there exist φ, ψ ∈G(F) such that g=φ◦f◦ψ. This happens if and only
if F(g) = σ(F(f)) for some σ∈Aut(F(X)/F).
The set F(X)\Fequipped with composition ◦is a monoid and G(F) is the group
of units of (F(X)\F,◦). In a parallel setting, one replaces F(X) with F[X] and
G(F) with the affine linear group AGL(1,F) = {φ∈F[X] : deg φ= 1}. Then
(F[X]\F,◦) is a submonoid of (F(X)\F,◦) and AGL(1,F) is its group of units. If
two polynomials f, g ∈F[X]\Fare equivalent as rational functions, i.e., g=φ◦f◦ψ
for some φ, ψ ∈G(F), then there are α, β ∈AGL(1,F) such that g=α◦f◦β; see
Lemma 8.1. Factorizations in the monoids (F(X)\F,◦) and (F[X]\F,◦) are difficult
questions that have attracted much attention [1, 2, 3, 9, 10, 18, 19]. Factorizations
in (F(X)\F,◦) are determined by the lattice L(F) of the subfields of F(X) and vice
versa. The Galois group Aut(F(X)/F) is an automorphism group of L(F) and the
Aut(F(X)/F)-orbits in L(F) correspond to the equivalence classes in F(X)\F.
2020 Mathematics Subject Classification. 05E18, 11T06, 12E20, 12F20, 20G40.
Key words and phrases. finite field, general linear group, projective linear group, rational
function.
1
2 XIANG-DONG HOU
Many intrinsic properties of rational functions are preserved under equivalence.
The degree of a rational function in F(X)\Fis invariant under equivalence. Equiv-
alent rational functions in F(X)\Fhave isomorphic arithmetic monodromy groups.
The number of ramification points and their ramification indices of a rational func-
tion are preserved under equivalence [16]. When F=Fq, the finite field with q
elements, there is another important invariant: |f(P1(Fq))|, the number of values
of f∈Fq(X) on the projective line P1(Fq). In the theory and applications of finite
fields, an important question is to understand the polynomials that permute Fq
and the rational functions that permute P1(Fq) under the aforementioned equiva-
lence. For classifications of low degree permutation polynomials of finite fields, see
[4, 6, 7, 14, 17]. Permutation rational functions of P1(Fq) of degree 3 and 4 were
classified recently [5, 8, 13]. Equivalence of rational functions over finite fields also
arises in other circumstances. There is a construction of irreducible polynomials
over Fqusing a rational function R(X)∈Fq[X]; the number of irreducible polyno-
mials produced by the construction depends only on the equivalence class of R(X)
[15]. It is known that the equivalence classes of rational functions f∈Fq(X)\Fq
such that Fq(X)/Fq(f) is Galois are in one-to-one correspondence with the classes
of conjugate subgroups of PGL(2,Fq); see [12].
When F=Fq, there are only finitely many equivalence classes of rational func-
tions in Fq(X)\Fqwith a given degree n. We shall denote this number by N(q, n).
Despite its obvious significance, this number was not known previously. The main
contribution of the present paper is the determination of N(q, n) for all qand n
(Theorem 6.1). For example, when n= 3, we have
N(q, 3) =
2(q+ 1) if q≡1,4 (mod 6),
2qif q≡2,5 (mod 6),
2q+ 1 if q≡3 (mod 6).
The classification of rational functions of degree n≤2 over Fqis straightforward;
see Sections 7.1 and 7.2. When n= 3 and qis even, the classification was obtained
recently by Mattarei and Pizzato [16] using the fact that such rational functions
have at most two ramification points. The case n= 3 and qodd is still unsolved.
(In this case, it was shown in [16] that N(q, 3) ≤4q.) A complete classification of
rational functions over Fqappears to be out of reach. However, the determination
of N(q, n) is an important step towards understanding the equivalence classes of
rational functions over finite fields, especially those with low degree.
Here is the outline of our approach. There is an action of GL(2,Fq) on the set
of subfields F⊂Fq(X) with [Fq(X) : F] = n, and N(q, n) is the number of orbits
of this action. To compute N(q, n) by Burnside’s lemma, it suffices to determine
the number of such subfields of Fq(X) fixed by each member Aof GL(2,Fq). From
there on, the computation becomes quite technical and depends on the canonical
form of A.
The paper is organized as follows: In Section 2, we include some preliminaries and
lay out the plan for computing N(q , n). The ingredients of the formula for N(q, n)
are computed in Sections 3 – 5 and the explicit formula for N(q, n) is presented in
Section 6. A discussion of low degree rational functions over Fqensued in Section 7.
The last section is devoted to equivalence classes of polynomials over finite fields.
The situation is much simpler compared with the case of rational functions. The
number of equivalence classes are computed and, as concrete examples, polynomials
NUMBER OF EQUIVALENCE CLASSES OF RATIONAL FUNCTIONS 3
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.
Fq
Fq(X)
F1F2Fq2(n−1)
···
n n n
Figure 1. Subfields of Fq(X) of degree n
of degree up to 5 are classified. Several counting lemmas used in the paper are
gathered in the appendix.
2. Preliminaries
2.1. Rational functions and subfields.
Let
(2.1) Rq,n ={f∈Fq(X) : deg f=n}.
By Lemma A2,
|Rq,n|=(q−1 if n= 0,
q2n−1(q2−1) if n > 0.
For f1, f2∈Fq(X)\Fq, we define f1∼f2if f2=φ◦f1◦ψfor some φ, ψ ∈G(Fq)
and we define f1
L
∼f2if there exists φ∈G(Fq) such that f2=φ◦f1. It is clear
that
f1
L
∼f2⇔Fq(f1) = Fq(f2)
and
f1∼f2⇔Fq(f2) = σ(Fq(f1)) for some σ∈Aut(Fq(X)/Fq).
Recall that N(q, n) denotes the number of ∼equivalence classes in Rq,n ; this number
is the main subject of our investigation.
For f=P /Q ∈Fq(X)\Fq, where P, Q ∈Fq[X], gcd(P, Q) = 1, let
S(f) = hP, QiFq={aP +bQ :a, b ∈Fq},
the Fq-span of {P, Q}. (Throughout this paper, an Fq-span is denoted by h iFq.)
Then f1
L
∼f2⇔ S(f1) = S(f2). By L¨uroth theorem, every subfield F⊂Fq(X)
with [Fq(X) : F] = n < ∞is of the form F=Fq(f), where f∈Fq(X) is of degree
n. The number of such Fis
|Rq,n|
|G(Fq)|=q2n−1(q2−1)
q(q2−1) =q2(n−1).
Denote the set of these fields by Fn={F1,...,Fq2(n−1) }(Figure 1) and let Aut(Fq(X)/Fq)
act on Fn. Then N(q, n) is precisely the number of orbits of this action.
4 XIANG-DONG HOU
2.2. Conjugacy classes of GL(2,Fq).
Let
Aa=a0
0a, a ∈F∗
q,
A{a,b}=a0
0b, a, b ∈F∗
q,
A{α,αq}=α+αq−α1+q
1 0 , α ∈Fq2\Fq,
Ba=a a
0a, a ∈F∗
q.
Then
C:= {Aa:a∈F∗
q} ∪ {A{a,b}:a, b ∈F∗
q, a 6=b}(2.2)
∪ {A{α,αq}:α∈F2
q\Fq} ∪ {Ba:a∈F∗
q}
forms a set of representatives of the conjugacy classes of GL(2,Fq). Additional
information about these representatives is given in Table 1, where cent(A) denotes
the centralizer of Ain GL(2,Fq); see [11, §6.3].
Table 1. Conjugacy classes of GL(2,Fq)
A∈ C elementary divisors |cent(A)|
Aa, a ∈F∗
qX−a, X −a q(q−1)2(q+ 1)
A{a,b}, a, b ∈F∗
q, a 6=b X −a, X −b(q−1)2
A{α,αq}, α ∈F2
q\Fq(X−α)(X−αq)q2−1
Ba, a ∈F∗
q(X−a)2q(q−1)
2.3. Burnside’s lemma.
Let GL(n, Fq) act on Fnas follows: For A=a b
c d ∈GL(n, Fq) and Fq(f)∈ Fn,
where f∈Fq(X) is of degree n,A(Fq(f)) = Fq(f◦φA), where φA= (aX +b)/(cX +
d). By Burnside’s lemma,
N(q, n) = X
A∈C
Fix(A)
|cent(A)|
(2.3)
=1
q(q−1)2(q+ 1) X
a∈F∗
q
Fix(Aa) + 1
(q−1)2X
{a,b}⊂F∗
q, a6=b
Fix(A{a,b})
+1
q2−1X
{α,αq}⊂Fq2\Fq
Fix(A{α,αq}) + 1
q(q−1) X
a∈F∗
q
Fix(Ba),
where
Fix(A) = |{F∈ Fn:A(F) = F}|.
Obviously,
(2.4) Fix(Aa) = |Fn|=q2(n−1).
We will determine Fix(A{a,b}), Fix(A{α,αq}), and Fix(Ba) in the subsequent sec-
tions; in doing so, we will need a number of counting lemmas which are given in
the appendix.
NUMBER OF EQUIVALENCE CLASSES OF RATIONAL FUNCTIONS 5
3. Determination of Fix(A{a,b})
Let a, b ∈F∗
q,a6=band c=a/b. Then φA{a,b}=cX. Therefore, a field Fq(f),
where f∈Fq(X)\Fq, is fixed by A{a,b}if and only if Fq(f(X)) = Fq(f(cX)).
Lemma 3.1. Let f∈Fq(X)with deg f=n > 0and 16=c∈F∗
qwith o(c) = d,
where o(c)denotes the multiplicative order of c. Then Fq(f(X)) = Fq(f(cX)) if
and only if
S(f) = hXr1P1(Xd), Xr2Q1(Xd)iFq,
where P1, Q1∈Fq[X]are monic, 0≤r1, r2< d,deg(Xr2Q1(Xd)) <deg(Xr1P1(Xd)) =
n, and gcd(Xr1P1, Xr2Q1) = 1.
Proof. (⇐) Obvious.
(⇒) We may assume that f=P/Q, where P, Q ∈Fq[X] are monic, deg P=n,
deg Q=m < n, gcd(P, Q) = 1, and the coefficient of Xmin Pis 0. Let n≡r1
(mod d) and m≡r2(mod d), where 0 ≤r1, r2< d. Such a pair (P, Q) is uniquely
determined by S(f). Since
hP(X), Q(X)iFq=S(f) = S(f(cX)) = hc−nP(cX), c−mQ(cX)iFq,
we have
P(X) = c−nP(cX), Q(X) = c−mQ(cX ).
Thus the coefficient of Xiin P(X) is 0 for all iwith i6≡ n(mod d), whence
P(X) = xr1P1(Xd). In the same way, Q(X) = Xr2Q1(Xd). Since gcd(P, Q) = 1,
we have gcd(Xr1P1, Xr2Q1) = 1.
In Lemma 3.1, let m= deg(Xr2Q1(Xd)). Note that gcd(Xr1P1, Xr2Q1) = 1 if
and only if gcd(P1, Q1) = 1 plus one of the following: (i) r1=r2= 0; (ii) r1= 0,
r2>0, P1(0) 6= 0; (iii) r1>0, r2= 0, Q1(0) 6= 0. When r1=r2= 0, i.e.,
n≡m≡0 (mod d), the number of the fields Fq(f) in Lemma 3.1 fixed by A{a,b}
is q−1αm/d,n/d, where
αi,j =|{(f, g ) : f, g ∈Fq[X] monic,deg f=i, deg g=j, gcd(f , g) = 1}|.
When r1= 0 and r2>0, i.e., n≡0 (mod d) but m6≡ 0 (mod d), the number of
Fq(f) fixed by A{a,b}is βn/d,⌊m/d⌋, where
βi,j =|{(f, g ) : f, g ∈Fq[X] monic,deg f=i, deg g=j, f (0) 6= 0,gcd(f, g ) = 1}|.
When r1>0 and r2= 0, i.e., m≡0 (mod d) but n6≡ 0 (mod d), the number of
Fq(f) fixed by A{a,b}is βm/d,⌊n/d⌋.
Define
αj=|{(f, g ) : f, g ∈Fq[X] monic,deg f < j, deg g=j, gcd(f , g) = 1}| =X
0≤i≤j
αi,j .
The numbers αi,j ,αjand βi,j are determined in Appendix, Lemmas A1 and A3.
Theorem 3.2. Let a, b ∈F∗
q,a6=b, and d=o(a/b). Then
Fix(A{a,b}) =
q2n/d−2+(d−1)(q2n/d −1)
q+ 1 if n≡0 (mod d),
q2⌊n/d⌋+1 + 1
q+ 1 if n6≡ 0 (mod d).
6 XIANG-DONG HOU
Proof. If n≡0 (mod d), using Lemmas A1 and A3, we have
Fix(A{a,b}) = X
0≤m<n
m≡0 (mod d)
q−1αm/d,n/d +X
0≤m<n
m6≡0 (mod d)
βn/d,⌊m/d⌋
=q−1X
0≤i<n/d
αi,n/d +X
0≤i<n/d
(d−1)βn/d,i
=q−1αn/d + (d−1) X
0≤i<n/d
qn/d−i−1(q−1)q2i+1 + 1
q+ 1
=q2n/d−2+(d−1)(q−1)
q+ 1 X
0≤i<n/d
(qn/d ·qi+qn/d−1−i)
=q2n/d−2+(d−1)(q−1)
q+ 1 qn/d qn/d −1
q−1+qn/d −1
q−1
=q2n/d−2+(d−1)(q2n/d −1)
q+ 1 .
If n6≡ 0 (mod d), we have
Fix(A{a,b}) = X
0≤m<n
m≡0 (mod d)
βm/d,⌊n/d⌋=X
0≤i≤⌊n/d⌋
βi,⌊n/d⌋
=q⌊n/d⌋+X
1≤i≤⌊n/d⌋
q⌊n/d⌋−i(q−1)q2i−1
q+ 1 (by Lemma A3)
=q⌊n/d⌋+q−1
q+ 1 X
1≤i≤⌊n/d⌋
(q⌊n/d⌋+1 ·qi−1−q⌊n/d⌋−i)
=q⌊n/d⌋+q−1
q+ 1 q⌊n/d⌋+1 q⌊n/d⌋−1
q−1−q⌊n/d⌋−1
q−1
=q⌊n/d⌋+(q⌊n/d⌋−1)(q⌊n/d⌋+1 −1)
q+ 1
=q2⌊n/d⌋+1 + 1
q+ 1 .
4. Determination of Fix(A{α,αq})
Let
A=A{α,αq}=α+αq−α1+q
1 0 , α ∈Fq2\Fq.
We have
(4.1) BAB−1=D,
where
D=αq0
0α, B =1−α
1−αq∈GL(2,Fq2).
Note that φD=αq−1X∈G(Fq2).
NUMBER OF EQUIVALENCE CLASSES OF RATIONAL FUNCTIONS 7
Lemma 4.1. Let f∈Fq(X)\Fqand g=f◦φ−1
B∈Fq2(X). Then Fq(f)is fixed
by Aif and only if Fq2(g)is fixed by D.
Proof. We have
Fq2(g) is fixed by D
⇔g◦φD=ψ◦gfor some ψ∈G(Fq2)
⇔f◦φA=ψ◦ffor some ψ∈G(Fq2) (by (4.1))
⇔f◦φA=ψ◦ffor some ψ∈G(Fq) (by Lemma 4.2)
⇔Fq(f) is fixed by A.
Lemma 4.2. Let f1, f2∈Fq(X)\Fqbe such that there exists ψ∈G(F), where F
is an extension of Fq, such that f2=ψ◦f1. Then there exists θ∈G(Fq)such that
f1=θ◦f2.
Proof. Let fi=Pi/Qi, where Pi, Qi∈Fq[X] and gcd(Pi, Qi) = 1. It suffices to
show that there exist a0, b0, c0, d0∈Fqsuch that
(4.2) a0b0
c0d0P1
Q1=P2
Q2.
By assumption, there exist a, b, c, d ∈Fsuch that
a b
c dP1
Q1=P2
Q2.
Write F=Fq⊕Vas a direct sum of Fq-subspaces, and write a=a0+a1,b=b0+b1,
c=c0+c1,d=d0+d1, where a0, b0, c0, d0∈Fqand a1, b1, c1, d1∈V. Then
a0b0
c0d0P1
Q1+a1b1
c1d1P1
Q1=P2
Q2.
Comparing the coefficients in the above gives (4.2).
Lemma 4.3. For g∈Fq2(X),g◦φB∈Fq(X)if and only if ¯g(X) = g(X−1), where
¯gdenotes the rational function obtained by applying ( )qto the coefficients of g.
Proof. Recall that φB(X) = (X−α)/(X−αq). Since ¯
φB=X−1◦φB, we have
g◦φB∈Fq(X)⇔g◦φB=g◦φB
⇔¯g◦X−1◦φB=g◦φB
⇔¯g=g◦X−1.
Lemmas 4.1 and 4.3 suggest the following strategy (which we will follow) to
determine Fix(A{α,αq}):
Step 1. Determine all g∈Fq2(X) of degree nsuch that Fq2(g(αq−1X)) = Fq2(g(X)).
Step 2. Among all g’s in Step 1, determine those such that ¯g(X) = g(X−1).
Step 3. Conclude that Fix(A{α,αq}) = |G(Fq)|−1·(the number of g’s in Step 2).
8 XIANG-DONG HOU
We now carry out these steps in detail.
Step 1. Determine all g∈Fq2(X) of degree nsuch that Fq2(g(αq−1X)) =
Fq2(g(X)).
Let d=o(αq−1). By Lemma 3.1, for g∈Fq2(X) with deg g=n,Fq2(g(αq−1X)) =
Fq2(g(X)) if and only if
(4.3) S(g) = hXr1P1(Xd), Xr2Q1(Xd)iFq2,
where 0 ≤r1, r2< d,P1, Q1∈Fq2[X] are monic, deg(Xr2Q1(Xd)) <deg(Xr1P1(Xd)) =
n, gcd(Xr1P1, Xr2Q1) = 1, and h iFq2is the Fq2-span.
In (4.3), let m= deg(Xr2Q1(Xd)). Note that n≡r1(mod d), m≡r2(mod d),
gcd(P1, Q1) = 1, and one of the following holds: (i) r1=r2= 0; (ii) r1= 0, r2>0,
P1(0) 6= 0; (iii) r1>0, r2= 0, Q1(0) 6= 0. Let g∈Fq2(X) satisfy (4.3), i.e.,
(4.4) g=sXr1P1(Xd) + tXr2Q1(Xd)
uXr1P1(Xd) + vX r2Q1(Xd),
where [ s t
u v ]∈GL(2,Fq2).
Step 2. Among all g’s in Step 1, determine those such that ¯g(X) = g(X−1).
For fixed r1and r2, let
N(r1, r2) = the number of gsatisfying (4.3) and ¯g(X) = g(X−1).
Case (i) Assume r1=r2= 0. In this case, we may write (4.4) as
g(X) = ǫP(Xd)
Q(Xd),
where ǫ∈F∗
q2,P, Q ∈Fq2[X] are monic, deg P=l1, deg Q=l2, max{l1, l2}=
n/d =: k, and gcd(P, Q) = 1. Then
g(X−1) = ǫXdkP(X−d)
XdkQ(X−d),
so ¯g(X) = g(X−1) if and only if
¯ǫP(X) = cǫXkP(X−1),(4.5)
Q(X) = cXkQ(X−1)(4.6)
for some c∈F∗
q2.
First, assume that l1=k, and l2≤kis fixed. Then (4.6) is equivalent to
Q(X) = Xk−l2Q1(X),
where Q1∈Fq2[X], deg Q1= 2l2−k(thus k/2≤l2≤k), and
(4.6′)Q1(X) = cX2l2−kQ1(X−1).
We call a polynomial f∈Fq2[X]\ {0}self-dual if Xdeg f¯
f(X−1) = cf (X) for some
c∈F∗
q2. Thus, if gsatisfies (4.3) and ¯g(X) = g(X−1), then both Pand Q1are
self-dual. On the other hand, if both Pand Q1are self-dual, then the cin (4.6)
belongs to µq+1 := {x∈Fq2:xq+1 = 1}and cis uniquely determined by Q1.
Subsequently, in (4.5), ǫq−1is uniquely determined and there are q−1 choices for
ǫ. Therefore, in this case, the number of gsatisfying (4.3) and ¯g(X) = g(X−1) is
(q−1) |{(P, Q1) : P, Q1∈Fq2[X] are monic and self-dual,(4.7)
NUMBER OF EQUIVALENCE CLASSES OF RATIONAL FUNCTIONS 9
deg P=k, deg Q1= 2l2−k, gcd(P, Q1) = 1}|
= (q−1)Γk,2l2−k,
where
Γi,j ={(f1, f2) : f1, f2∈Fq2[X] are monic and self-dual,
deg f1=i, deg f2=j, gcd(f1, f2) = 1}|.
The number Γi,j is determined in Appendix, Lemma A5.
Next, assume that l2=kand l1< k is fixed. By the same argument, the number
of gsatisfying (4.3) and ¯g(X) = g(X−1) is (q−1)Γk,2l1−k.
Therefore, the total number of gsatisfying (4.3) and ¯g(X) = g(X−1) in Case (i)
is
N(0,0) = (q−1) X
k/2≤l2≤k
Γk,2l2−k+ (q−1) X
k/2≤l1<k
Γk,2l1−k
= (q−1)2X
0≤i<k
i≡k(mod 2)
Γi,k + Γk,k.
If k= 2k1,
N(0,0) = (q−1)2X
0≤i<k1
Γ2i,2k1+ Γ2k1,2k1
= (q−1)h2q2k1−1(q+ 1) + X
1≤i<k1
q2(k1−i)−1(q+ 1)(q2−1)
q2+ 1 (q4i−1)
+q(q+ 1)
q2+ 1 (q4k1−q4k1−2−2)i(by Lemma A5)
= (q−1)h2q2k1−1(q+ 1) + 2 (q+ 1)(q2−1)q2k1−1
q2+ 1 X
1≤i<k1
(q2i−q−2i)
+q(q+ 1)
q2+ 1 (q4k1−q4k1−2−2)i
= (q2−1)h2q2k1−1+2(q2−1)q2k1−1
q2+ 1 q21−q2(k1−1)
1−q2−(q−21−q−2(k1−1)
1−q−2
+q
q2+ 1(q4k1−q4k1−2−2)i
= (q2−1)q4k1−1.
If k= 2k1+ 1,
N(0,0) = (q−1)2X
0≤i<k1
Γ2i+1,2k1+1 + Γ2k1+1,2k1+1
= (q−1)h2X
0≤i<k1
q2(k1−i)−1(q+ 1)(q2−1)
q2+ 1 (q4i+2 + 1)
+q(q+ 1)
q2+ 1 (q4k1+2 −q4k1+ 2)i(by Lemma A5)
= (q−1)h2(q+ 1)(q2−1)q2k1−1
q2+ 1 X
0≤i<k1
(q2i+2 +q−2i)
10 XIANG-DONG HOU
+q(q+ 1)
q2+ 1 (q4k1+2 −q4k1+ 2)i
= (q2−1)h2(q2−1)q2k1−1
q2+ 1 q21−q2k1
1−q2+1−q−2k1
1−q−2
+q
q2+ 1(q4k1+2 −q4k1+ 2)i
= (q2−1)q4k1+1.
Therefore, we always have
(4.8) N(0,0) = (q2−1)q2k−1.
Case (ii) Assume r1= 0, r2>0 and P1(0) 6= 0. By (4.4),
g(X−1) = sXnP1(X−d) + tXn−r2Q1(X−d)
uXnP1(X−d) + vX n−r2Q1(X−d).
Hence ¯g(X) = g(X−1) if and only if
(¯sP1(Xd) + ¯
tXr2Q1(Xd) = csXnP1(X−d) + tXn−r2Q1(X−d),
¯uP1(Xd) + ¯vXr2Q1(Xd) = cuXnP1(X−d) + vX n−r2Q1(X−d)
for some c∈F∗
q2, which is equivalent to
(4.9)
¯sP1(Xd) = csXnP1(X−d),
¯uP1(Xd) = cuXnP1(X−d),
¯
tXr2Q1(Xd) = ctXn−r2Q1(X−d),
¯vXr2Q1(Xd) = cvX n−r2Q1(X−d).
Let k=n/d and l= (m−r2)/d. The above equations imply that P1(X) self-
dual and Q1(Xd) = δX n−2r2Q1(X−d) for some δ∈F∗
q2. It is necessary that
n−2r2≡0 (mod d), i.e., dis even and r2=d/2. Hence Q1(X) = δXk−1Q1(X−1).
It follows that Q1(X) = Xk−l−1Q2(X), where Q2(X) is monic and self-dual of
degree 2l−k+ 1. (So (k−1)/2≤l≤k−1.)
On the other hand, let P1, Q2∈Fq2[X] be monic and self-dual with deg P1=k
and deg Q2= 2l−k+ 1 ((k−1)/2≤l≤k−1). Then P1(X) = ǫX kP1(X−1) and
Q2(X) = δX2l−k+1 Q2(X−1) for some ǫ, δ ∈µq+1 . Let Q1(X) = Xk−l−1Q2(X).
Then (4.9) is satisfied if and only if
(4.10)
¯sǫ =cs,
¯uǫ =cu,
¯
tδ =ct,
¯vδ =cv.
Under the assumption that det [ s t
u v ]6= 0, (4.10) implies that c∈µq+1. Write
ǫ=ǫq−1
0,δ=δq−1
0and c=cq−1
0, where ǫ0, δ0, c0∈F∗
q2. Then (4.10) is satisfied if
and only if s t
u v=s1c0/ǫ0t1c0/δ0
u1c0/ǫ0v1c0/δ0,
where s1, t1, u1, v1∈Fq. Therefore, the number of [ s t
u v ] satisfying (4.9) is
(q+ 1) |GL(2,Fq)|=q(q2−1)2.
NUMBER OF EQUIVALENCE CLASSES OF RATIONAL FUNCTIONS 11
To recap, when dis even, r2=d/2 and l((k−1)/2≤l≤k−1) is fixed, the
number of gsatisfying (4.3) and ¯g(X) = g(X−1) is
1
q2−1q(q2−1)2Γk,2l−k+1 =q(q2−1)Γk,2l−k+1.
Hence, when dis even,
N(0, r2) = q(q2−1) X
(k−1)/2≤l≤k−1
Γk,2l−k+1 =q(q2−1) X
0≤i≤k−1
i≡k−1 (mod 2)
Γi,k.
In the above, if k= 2k1,
N(0, r2) = q(q2−1) X
1≤i≤k1
Γ2i−1,2k1
=q(q2−1) X
1≤i≤k1
q2k1−(2i−1)−1(q+ 1)(q2−1)
q2+ 1 (q4i−2+ 1)
(by Lemma A5)
=q(q+ 1)(q2−1)2
q2+ 1 q2k1X
1≤i≤k1
(q2i−2+q−2i)
=(q+ 1)(q2−1)2q2k1+1
q2+ 1 1−q2k1
1−q2+q−21−q−2k1
1−q−2
=(q+ 1)(q2−1)2q2k1+1
q2+ 1 ·q−2k1(q4k1−1)
q2−1
=q(q+ 1)(q2−1)(q4k1−1)
q2+ 1 .
If k= 2k1+ 1,
N(0, r2) = q(q2−1) X
0≤i≤k1
Γ2i,2k1+1
=q(q2−1)hq2k1(q+ 1) + X
1≤i≤k1
q2k1+1−2i−1(q+ 1)(q2−1)
q2+ 1 (q4i−1)i
(by Lemma A5)
=q(q2−1)(q+ 1)hq2k1+(q2−1)q2k1
q2+ 1 X
1≤i≤k1
(q2i−q−2i)i
=q(q2−1)(q+ 1)hq2k1+(q2−1)q2k1
q2+ 1 q21−q2k1
1−q2−q−21−q−2k1
1−q−2i
=q(q2−1)(q+ 1)1 + q4k1+2
1 + q2
=q(q+ 1)(q2−1)(q4k1+2 + 1)
q2+ 1 .
To summarize, we have
(4.11) N(0, r2) =
q(q+ 1)(q2−1)(q2k−(−1)k)
q2+ 1 if dis even,
0 if dis odd.
12 XIANG-DONG HOU
Case (iii) Assume r1>0, r2= 0 and Q1(0) 6= 0. By (4.4),
g(X−1) = sXn−r1P1(X−d) + tXnQ1(X−d)
uXn−r1P1(X−d) + vX nQ1(X−d).
Hence ¯g(X) = g(X−1) if and only if
(¯sXr1P1(Xd) + ¯
t Q1(Xd) = csXn−r1P1(X−d) + tXnQ1(X−d),
¯uXr1P1(Xd) + ¯vQ1(Xd) = cuXn−r1P1(X−d) + vXnQ1(X−d)
for some c∈F∗
q2, which is equivalent to
(4.12)
¯sXr1P1(Xd) = ctX nQ1(X−d),
¯uXr1P1(Xd) = cvXnQ1(X−d),
¯
t Q1(Xd) = csXn−r1P1(X−d),
¯vQ1(Xd) = cuXn−r1P1(X−d).
Under the assumption that det [ s t
u v ]6= 0, (4.12) implies that s, t, u, v 6= 0 and
c∈µq+1. Without loss of generality, assume s= 1. Then (4.12) becomes
(4.13)
P1(X) = ctXkQ1(X−1),
c∈µq+1,
v= ¯ut,
where k= (n−r1)/d. Moreover,
det 1t
u v= det 1t
u¯ut=t(¯u−u),
which is nonzero if and only if t∈F∗
q2and u∈Fq2\Fq.
Condition (4.13) implies that
e
P1=XkP1(X−1) = ctQ1(X),
where gcd(P1,e
P1) = gcd(P1, Q1) = 1.
On the other hand, to satisfy (4.13) with u∈Fq2\Fq, we first choose monic
P1(X)∈Fq2[X] of degree ksuch that gcd(P1,f
P1) = 1; the number of choices
of such P1, denoted by Θk, is determined in Appendix, Lemma A4. Next, let
Q1(X) = ǫXkP1(X−1), where ǫ∈F∗
q2is such that Q1is monic. Afterwards, choose
c∈µq+1 and u∈Fq2\Fqarbitrarily, and let tand vbe uniquely determined by
(4.13). Hence the total number of gsatisfying (4.3) and ¯g(X) = g(X−1) in Case
(iii) is
N(r1,0) = (q+ 1)(q2−q)Θk
(4.14)
=q(q2−1)
1 + q2(−1)k(1 + q) + q2k+1(q−1)(by Lemma A4).
Step 3. We have
Fix(A{α,αq}) = 1
|G(Fq)|(the number of g’s in Step 2).
NUMBER OF EQUIVALENCE CLASSES OF RATIONAL FUNCTIONS 13
Theorem 4.4. Let α∈Fq2\Fqwith o(αq−1) = d. Then
Fix(A{α,αq}) =
q2n/d−2+(q+ 1)(q2n/d −(−1)n/d)
q2+ 1 if d|nand dis even,
q2n/d−2if d|nand dis odd,
1
1 + q2(−1)⌊n/d⌋(1 + q) + q2⌊n/d⌋+1(q+ 1)if d∤n.
Proof. 1◦Assume that d|nand dis even. By (4.8) and (4.11),
Fix(A{α,αq}) = 1
q(q2−1)h(q2−1)q2n/d−1+q(q+ 1)(q2−1)(q2n/d −(−1)n/d)
q2+ 1 i
=q2n/d−2+(q+ 1)(q2n/d −(−1)n/d)
q2+ 1 .
2◦Assume that d|nand dis odd. By (4.8) and (4.11),
Fix(A{α,αq}) = 1
q(q2−1)q2n/d−1=q2n/d−2.
3◦Assume that d∤n. By (4.14),
Fix(A{α,αq}) = 1
q(q2−1) ·q(q2−1)
1 + q2(−1)k(1 + q) + q2k+1(q−1)
=1
1 + q2(−1)k(1 + q) + q2k+1(q−1).
5. Determination of Fix(Ba)
5.1. A useful lemma.
Let p= char Fq. Every f(X)∈Fq[X] has a representation
(5.1) f(X) = gp−1(Xp−X)Xp−1+gp−2(Xp−X)Xp−2+···+g0(Xp−X),
where gi∈Fq[X]. Define ∆f=f(X+ 1) −f(X). Then ∆pf= 0, and for
0≤i≤p−1,
∆if=gi(Xp−X)i! +
p−1
X
j=i+1
gj(Xp−X)∆iXj.
It follows that giin (5.1) are uniquely determined by f.
Lemma 5.1. Let 0≤i≤p−1. Then ∆if= 0 if and only if gj= 0 for all
i≤j≤p−1in (5.1).
Proof. (⇐) Obvious.
(⇒) Assume the contrary. Let j0be the largest jsuch that gj6= 0. Then
i≤j0≤p−1. We have
∆if=gj0(Xp−X)∆iXj0+X
j<j0
gj(Xp−X)∆iXj
=gj0(Xp−X)j0
iXj0−i+X
j<j0−i
hj(Xp−X)Xj(hj∈Fq[X])
14 XIANG-DONG HOU
6= 0,
which is a contradiction.
5.2. Determination of Fix(Ba).
Recall that Ba= [ a a
0a], a∈F∗
q, so φBa=X+ 1. Let F=Fq(P /Q), where
P, Q ∈Fq[X] are monic, deg P=n > deg Q, and gcd(P, Q) = 1. Then Ba(F) =
Fq(P(X+ 1)/Q(X+ 1)). Hence Ba(F) = Fif and only if
(5.2) (Q(X+ 1) = Q(X),
P(X+ 1) = P(X) + cQ(X) for some c∈Fq.
Case 1. Assume c= 0. Then (5.2) holds if and only if P(X) = P1(Xp−X),
Q(X) = Q1(Xp−X), where P1, Q1∈Fq[X] are such that deg P1=n/p > deg Q1
(must have p|n) and gcd(P1, Q1) = 1. The number of such (P, Q) is αn/p.
Case 2. Assume c6= 0. Then (5.2) holds if and only if
(5.3)
Q=c−1∆P,
∆2P= 0,
gcd(P(X), P (X+ 1)) = 1.
Condition (5.3) is equivalent to
(5.4) (∆2P= 0,∆P6= 0,gcd(P(X), P (X+ 1)) = 1,
Q=c−1∆P, where cis uniquely determined by P .
By Lemma 5.1, the P(X) in (5.4) has the form
P(X) = P1(Xp−X)X+P0(Xp−X),
where P16= 0. Since P(X+ 1) −P(X) = P1(Xp−X), gcd(P(X), P (X+ 1)) = 1
if and only if gcd(P0, P1) = 1. Also note that
deg P= max{pdeg P1+ 1, p deg P0}.
Hence the number of (P, Q) satisfying (5.4) is
(q−1)αn/p if n≡0 (mod p),
qif n= 1,
(q−1)(α(n−1)/p +α(n−1)/p,(n−1)/p) if n≡1 (mod p), n > 1,
0 otherwise.
Therefore,
Fix(Ba) =
1
q(αn/p + (q−1)αn/p) if n≡0 (mod p),
1 if n= 1,
q−1
q(α(n−1)/p +α(n−1)/p,(n−1)/p) if n≡1 (mod p), n > 1,
0 otherwise.
Recall that αiand αi,j are given by Lemma A1. When n≡0 (mod p),
Fix(Ba) = αn/p =q2n/p−1.
NUMBER OF EQUIVALENCE CLASSES OF RATIONAL FUNCTIONS 15
When n≡1 (mod p) and n > 1,
Fix(Ba) = q−1
q(q2(n−1)/p−1+q2(n−1)/p(1 −q−1)) = q2(n−1)/p−1(q−1).
To summarise,
(5.5) Fix(Ba) =
q2n/p−1if n≡0 (mod p),
1 if n= 1,
q2(n−1)/p−1(q−1) if n≡1 (mod p), n > 1,
0 otherwise.
6. The Main Theorem
Theorem 6.1. For n≥1, we have
(6.1) N(q, n) = q2n−3
q2−1+1
2(q−1) A(q, n) + 1
2(q+ 1)B(q, n) + 1
qC(q, n),
where
(6.2)
A(q, n) = X
1<d |q−1
d|n
φ(d)q2n/d−2+(d−1)(q2n/d −1)
q+ 1 +X
1<d |q−1
d∤n
φ(d)q2⌊n/d⌋+1 + 1
q+ 1 ,
B(q, n) = X
deven
d|gcd(q+1,n)
φ(d)q2n/d−2+(q+ 1)(q2n/d −(−1)n/d)
q2+ 1
(6.3)
+X
dodd
1<d |gcd(q+1,n)
φ(d)q2n/d−2
+1
(q+ 1)(q2+ 1) X
d|q+1
d∤n
φ(d)1 + (−1)⌊n/d⌋
2(1 + q)2+q(q2⌊n/d⌋+2 −1)
+1
q2+ 1 X
d|q+1
d∤n
φ(d)(−1)⌊n/d⌋(1 + q) + q2⌊n/d⌋+1(q−1),
(6.4) C(q, n) =
q2n/p−1if n≡0 (mod p),
1if n= 1,
q2(n−1)/p−1(q−1) if n≡1 (mod p), n > 1,
0otherwise.
In (6.2) and (6.3),φis the Euler function.
Proof. We have
N(q, n) = 1
q(q−1)2(q+ 1) X
a∈F∗
q
Fix(Aa) + 1
(q−1)2X
{a,b}⊂F∗
q
a6=b
Fix(A{a,b})
16 XIANG-DONG HOU
+1
q2−1X
{α,αq}⊂Fq2\Fq
Fix(A{α,αq}) + 1
q(q−1) X
a∈F∗
q
Fix(Ba).
We now compute the four sums in the above.
1◦We have X
a∈F∗
q
Fix(Aa) = (q−1)q2(n−1).
2◦We have
X
{a,b}⊂F∗
q
a6=b
Fix(A{a,b}) = 1
2X
a∈F∗
qX
b∈F∗
q\{1}
Fix(A{ab,a}) = q−1
2X
b∈F∗
q\{1}
Fix(A{b,1})
=q−1
2hX
1<d |q−1
d|n
φ(d)q2n/d−2+(d−1)(q2n/d −1)
q+ 1 +X
1<d |q−1
d∤n
φ(d)(q2⌊n/d⌋+ 1)
q+ 1 i
(by Theorem 3.2)
=q−1
2A(q, n).
3◦By Theorem 4.4,
X
{α,αq}⊂Fq2\Fq
Fix(A{α,αq})
=q−1
2hX
deven
d|gcd(q+1,n)
φ(d)q2n/d−2+(q+ 1)(q2n/d −(−1)n/d)
q2+ 1
+X
dodd
1<d |gcd(q+1,n)
φ(d)q2n/d−2
+1
(q+ 1)(q2+ 1) X
d|q+1
d∤n
φ(d)1 + (−1)⌊n/d⌋
2(1 + q)2+q(q2⌊n/d⌋+2 −1)
+1
q2+ 1 X
d|q+1
d∤n
φ(d)(−1)⌊n/d⌋(1 + q) + q2⌊n/d⌋+1(q−1)i
=q−1
2B(q, n).
4◦By (5.5), X
a∈F∗
q
Fix(Ba) = (q−1)C(q, n).
NUMBER OF EQUIVALENCE CLASSES OF RATIONAL FUNCTIONS 17
7. N(q, n)for Small n
7.1. n= 1.
We have
A(q, 1) = X
1<d |q−1
φ(d) = q−2,
B(q, 1) = 1
q2+ 1 X
1<d |q+1
φ(d)(1 + q) + q(q−1)=X
1<d |q+1
φ(d) = q+ 1 −1 = q,
C(q, 1) = 1.
Hence
N(q, 1) = q−1
q2−1+1
2(q−1) (q−2) + 1
2(q+ 1) q+1
q= 1,
as expected.
7.2. n= 2.
Case 1. Assume qis even. We have
A(q, 2) = X
1<d |q−1
φ(d) = q−2,
B(q, 2) = 1
q2+ 1 X
1<d |q+1
d∤2
φ(d)(1 + q) + q(q−1)=X
1<d |q+1
φ(d) = q+ 1 −1 = q,
C(q, 2) = q.
Hence
N(q, 2) = q
q2−1+1
2(q−1) (q−2) + 1
2(q+ 1) q+1
qq= 2.
Since X2and X2+Xare nonequivalent (X2is a permutation of P1(Fq) but
X2+Xis not),
X2, X2+X
is a list of representatives of the equivalence classes of rational functions of degree
2 over Fq.
Case 2. Assume qis odd. We have
A(q, 2) = φ(2)1 + q2−1
q+ 1 +X
2<d |q−1
φ(d) = q+q−1−2 = 2q−3,
B(q, 2) = φ(2)1 + (q+ 1)(q2+ 1)
q2+ 1 +1
q2+ 1 X
d|q+1
d∤2
φ(d)(1 + q) + q(q−1)
=q+ 2 + X
2<d |q+1
φ(d) = q+ 2 + q+ 2 −2 = 2q+ 1,
C(q, 2) = 0.
Hence
N(q, 2) = q
q2−1+1
2(q−1) (2q−3) + 1
2(q+ 1)(2q+ 1) = 2.
18 XIANG-DONG HOU
In this case, a list of representatives of the equivalence classes of rational func-
tions of degree 2 over Fqis given by
X2,X2+b
X,
where bis any fixed nonsquare of Fq.
Proof. It suffices to show that every f∈Fq(X) of degree 2 is equivalent to one of
the above two rational functions.
If fis a polynomial, then f∼X2.
If fis not a polynomial, then f∼(X2+aX +b)/X , where b∈F∗
q. Thus
f∼(X2+b)/X. If b=c2for some c∈F∗
q, then
f∼X2+ 2cX +c2
X=(X+c)2
X∼X
(X+c)2∼X−c
X2=1
X−c1
X2
∼X−cX2∼X2.
7.3. n= 3.
1◦Computing A(q, 3).
First assume qis even.
If q−1≡0 (mod 3),
A(q, 3) = φ(3)1 + 2(q2−1)
q+ 1 +X
1<d |q−1
d∤3
φ(d)
= 2(1 + 2(q−1)) + q−1−φ(1) −φ(3)
= 2(2q−1) + q−1−3 = 5q−6.
If q−16≡ 0 (mod 3),
A(q, 3) = X
1<d |q−1
d∤3
φ(d) = q−1−φ(1) = q−2.
Next, assume qis odd.
If q−1≡0 (mod 3),
A(q, 3) = φ(3)1 + 2(q2−1)
q+ 1 +φ(2) q3+ 1
q+ 1 +X
3<d |q−1
φ(d)
= 2(1 + 2(q−1)) + q2−q+ 1 + q−1−φ(1) −φ(2) −φ(3)
= 2(2q−1) + q2−4 = q2+ 4q−6.
If q−16≡ 0 (mod 3),
A(q, 3) = φ(2) q3+ 1
q+ 1 +X
3<d |q−1
φ(d) = q2−q+ 1 + q−1−φ(1) −φ(2) = q2−2.
NUMBER OF EQUIVALENCE CLASSES OF RATIONAL FUNCTIONS 19
To summarize,
A(q, 3) =
5q−6 if q≡4 (mod 6),
q−2 if q≡2 (mod 6),
q2+ 4q−6 if q≡1 (mod 6),
q2−2 if q≡3,5 (mod 6).
2◦Computing B(q, 3).
First assume qis even.
If q+ 1 ≡0 (mod 3),
B(q, 3) = φ(3) + 1
q2+ 1 X
d|q+1
d∤3
φ(d)(1 + q) + q(q−1)
= 2 + X
d|q+1
d∤3
φ(d) = 2 + q+ 1 −φ(1) −φ(3) = q.
If q+ 1 6≡ 0 (mod 3),
B(q, 3) = 1
q2+ 1 X
d|q+1
d∤3
φ(d)(1 + q) + q(q−1)=X
d|q+1
d∤3
φ(d) = q+ 1 −φ(1) = q.
Next, assume qis odd.
If q+ 1 ≡0 (mod 3),
B(q, 3) = φ(3) + 1
q2+ 1 hφ(2)(−(1 + q) + q3(q−1)) + X
3<d |q+1
φ(d)(1 + q+q(q−1))i
= 2 + 1
q2+ 1 hq4−q3−q−1 + (q2+ 1) X
3<d |q+1
φ(d)i
= 2 + 1
q2+ 1 (q2+ 1)(q2−q−1) + (q2+ 1)(q+ 1 −φ(1) −φ(2) −φ(3))
= 2 + q2−q−1 + q+ 1 −4 = q2−2.
If q+ 1 6≡ 0 (mod 3),
B(q, 3) = 1
q2+ 1 hφ(2)(−(1 + q) + q3(q−1)) + X
3<d |q+1
φ(d)((1 + q) + q(q−1))i
=1
q2+ 1 (q2+ 1)(q2−q−1) + (q2+ 1)(q+ 1 −φ(1) −φ(2))
=q2−q−1 + q+ 1 −2 = q2−2.
To summarize,
B(q, 3) = (qif qis even,
q2−2 if qis odd.
20 XIANG-DONG HOU
3◦Computing C(q, 3). We have
C(q, 3) =
q(q−1) if p= 2,
qif p= 3,
0 otherwise.
4◦Computing N(q, 3).
If q≡1 (mod 6),
N(q, 3) = q3
q2−1+1
2(q−1) (q2+ 4q−6) + 1
2(q+ 1) (q2−2) = 2(q+ 1).
If q≡2 (mod 6),
N(q, 3) = q3
q2−1+1
2(q−1) (q−2) + 1
2(q+ 1) q+1
qq(q−1) = 2q.
If q≡3 (mod 6), i.e., p= 3,
N(q, 3) = q3
q2−1+1
2(q−1) (q2−2) + 1
2(q+ 1)(q2−2) + 1
qq= 2q+ 1.
If q≡4 (mod 6),
N(q, 3) = q3
q2−1+1
2(q−1) (5q−6) + 1
2(q+ 1) q+1
qq(q−1) = 2(q+ 1).
If q≡5 (mod 6),
N(q, 3) = q3
q2−1+1
2(q−1) (q2−2) + 1
2(q+ 1)(q2−2) = 2q.
To summarize,
N(q, 3) =
2(q+ 1) if q≡1,4 (mod 6),
2qif q≡2,5 (mod 6),
2q+ 1 if q≡3 (mod 6).
As mentioned in Section 1, rational functions of degree 3 in Fq(X) have been
classified for even n[16]; for odd q, the question is still open.
7.4. n= 4.
We include the formulas for A(q, 4), B(q, 4), C(q, 4) and N(q, 4) but omit the
details of the computations.
A(q, 4) =
−2−q+ 2q2if q≡4,10 (mod 12),
−2 + qif q≡2,8 (mod 12),
−10 + 6q+ 2q2+q3if q≡1 (mod 12),
−10 + 8q+q3if q≡5,9 (mod 12),
−4 + 2q2+q3if q≡7 (mod 12),
−4 + 2q+q3if q≡3,11 (mod 12).
NUMBER OF EQUIVALENCE CLASSES OF RATIONAL FUNCTIONS 21
B(q, 4) =
−2 + q2if q≡2,8 (mod 12),
qif q≡4 (mod 12),
−4 + 4q2+q3if q≡11 (mod 12),
2q+ 2q2+q3if q≡3,7 (mod 12),
−6−2q+ 4q2+q3if q≡5 (mod 12),
−2 + 2q2+q3if q≡1,9 (mod 12).
C(q, 4) =
q3if p= 2,
q(q−1) if p= 3,
0 otherwise.
N(q, 4) =
4 + 3q+q2+q3if q≡1 (mod 12),
3
2q+q2+q3if q≡2,8 (mod 12),
1 + 3q+q2+q3if q≡3 (mod 12),
1 + 2q+q2+q3if q≡4 (mod 12),
2 + 3q+q2+q3if q≡5,7 (mod 12),
3 + 3q+q2+q3if q≡9 (mod 12),
3q+q2+q3if q≡11 (mod 12).
8. Equivalence Classes of Polynomials
Lemma 8.1. Let f, g ∈Fq[X]\Fq. Then g=φ◦f◦ψfor some φ, ψ ∈G(Fq)if
and only if g=α◦f◦βfor some α, β ∈AGL(1,Fq).
Proof. (⇒) Let ψ(X) = A(X)/B(X).
Case 1. Assume that B(X) = 1. Then ψ=A∈AGL(1,Fq). Since f◦A=
f(A(X)) ∈Fq[X] and φ◦f◦A∈Fq[X], it follows that φ∈AGL(1,Fq).
Case 2. Assume that B(X)/∈Fq. Let B(X) = X+dand A(X) = aX +b. Let
f(X) = Xn+an−1Xn−1+···+a0. Then
f(φ(X)) = A(X)n+an−1A(X)n−1B(X) + ···+a0B(X)n
B(X)n.
Let φ(X) = (sX +t)/(uX +v). Then
(8.1) uA(X)n+an−1A(X)n−1B(X) + ···+a0B(X)n+vB(X)n= 1
and
g(X) = sA(X)n+an−1A(X)n−1B(X) + ···+a0B(X)n+tB(X)n.
By (8.1), u6= 0 and
g(X) = su−1(1 −vB(X)n) + tB(X)n=su−1+ (t−su−1v)B(X)n.
Hence we may assume g(X) = Xn. By (8.1) again,
ufA(X)
B(X)+v=1
B(X)n=1
X+dn=AX +b
X+d−an(b−ad)−n
=A(X)
B(X)−an(b−ad)−n.
22 XIANG-DONG HOU
So f(X) = u−1(b−ad)−n(X−a)n−u−1v. Hence we may assume f(X) = Xn.
Then f=g.
Because of Lemma 8.1, we define two polynomials f, g ∈Fq[X]\Fqto be equiv-
alent if there exist α, β ∈AGL(1,Fq) such that g=α◦f◦β; the meaning of
equivalence between fand gis the same whether they are treated as polynomials
or as rational functions.
Let
Pq,n ={f∈Fq[X] : deg f=n}
and let M(q, n) denote the number of equivalence classes in Pq,n. Compared with
N(q, n), M(q, n) is much easier to determine.
For f , g ∈Fq[X]\Fq, define fL
∼gif there exists α∈AGL(1,Fq) such that
g=α◦f. Let [f] denote the L
∼equivalence class of f. Each L
∼equivalence class
has a unique representative Xn+an−1Xn−1+···+a1X. Let AGL(1,Fq) act on
the set of L
∼equivalence classes in Fq[X]\Fqas follows: For f∈Fq[X]\Fqand
α∈AGL(1,Fq), [f]α= [f◦α]. Then M(q, n) is the number of AGL(1,Fq)-orbits in
Ωn:= {[f] : f∈ Pq,n}. The information about the conjugacy classes of AGL(1,Fq)
is given in Table 2. For α∈AGL(1,Fq), let Fix(α) be the number of elements in
Ωnfixed by α. All we have to do is to determine Fix(α) for each representative α
in Table 2.
Table 2. Conjugacy classes of AGL(1,Fq)
representative size of the centralizer
X q(q−1)
aX, a ∈F∗
q, a 6= 1 q−1
X+ 1 q
Clearly,
(8.2) Fix(X) = qn−1.
Next, we compute Fix(aX), where a∈F∗
q,a6= 1. Let o(a) = d. Then [f]∈Ωn
is fixed by aX if and only if
fL
∼Xrh(Xd),
where 0 ≤r < d,n≡r(mod d), h∈Fq[X] is monic of degree (n−r)/d, and
h(0) = 0 if r= 0. Thus
Fix(aX) = (qn/d−1if d|n
q⌊n/d⌋if d∤n
(8.3)
=q⌈n/d⌉−1.
Now we comput Fix(X+ 1). For [f]∈Ωn,
[f] is fixed by X+ 1
⇔f(X+ 1) = f(X) + a, where a∈Fq
⇔f(X) = g(X) + aX, where a∈Fq, g ∈Fq[X],∆g= 0
⇔f(X) = h(Xp−X) + aX, where a∈Fq, h ∈Fq[X], p = char Fq.
NUMBER OF EQUIVALENCE CLASSES OF RATIONAL FUNCTIONS 23
In the above, we may assume that fis monic and f(0) = 0. Therefore, when p|n,
his of degree n/p with h(0) = 0; when p∤n,h= 0, n= 1 and a= 1. So,
(8.4) Fix(X+ 1) =
qn/p−1·q=qn/p if p|n,
1 if n= 1,
0 if p∤nand n > 1.
Theorem 8.2. Let p= char Fq. We have
M(q, n) = qn−2
q−1+1
q−1X
1<d |q−1
φ(d)q⌈n/d⌉−1+
qn/p−1if p|n,
q−1if n= 1,
0if p∤nand n > 1.
Proof. By Burnside’s lemma and (8.2) – (8.4),
M(q, n) = 1
q(q−1)Fix(X) + 1
q−1X
a∈F∗
q\{1}
Fix(aX) + 1
qFix(X+ 1)
=qn−2
q−1+1
q−1X
1<d |q−1
φ(d)q⌈n/d⌉−1+1
qFix(X+ 1),
where
1
qFix(X+ 1) =
qn/p−1if p|n,
q−1if n= 1,
0 if p∤nand n > 1.
In Theorem 8.2, we can write
qn−2
q−1+1
q−1X
1<d |q−1
φ(d)q⌈n/d⌉−1
=qn−2
q−1+1
q−1X
d|q−1
φ(d)q⌈n/d⌉−1−qn−1
=1
q−1X
d|q−1
φ(d)q⌈n/d⌉−1+qn−2−qn−1
q−1
=1
q−1X
d|q−1
φ(d)(q⌈n/d⌉−1−1) + X
d|q−1
φ(d)−qn−2
=1
q−1X
d|q−1
d<n
φ(d)(q⌈n/d⌉−1−1) + 1 −qn−2.
Hence
M(q, n) = 1
q−1X
d|q−1
d<n
φ(d)(q⌈n/d⌉−1−1)+
1−qn−2+qn/p−1if p|n,
1 if n= 1,
1−qn−2if p∤nand n > 1.
24 XIANG-DONG HOU
In the above, the sum
1
q−1X
d|q−1
d<n
φ(d)(q⌈n/d⌉−1−1)
can be made more explicit as follows: Write
lcm{1,2,...,n−1}=Y
rprime
rνr, νr=⌊logr(n−1)⌋,
and
gcd(lcm{1,2,...,n−1}, q −1) = Y
rprime
rur.
Then
1
q−1X
d|q−1
d<n
φ(d)(q⌈n/d⌉−1−1)
=X
er≤ur
Qrrer≤n−1
φY
r
rer(q⌈n/ Qrrer⌉−1−1)
=X
er≤ur
Qrrer≤n−1Y
r
rer Y
r:er>0
(1 −r−1)(q⌈n/ Qrrer⌉−1−1).
As concrete examples, we include the formulas for M(q , n) with 1 ≤n≤5.
M(q, 1) = 1.
M(q, 2) = (2 if p= 2,
1 if p > 2.
M(q, 3) =
2 if p= 2,
4 if p= 3,
3 if p > 3.
M(q, 4) =
q+ 5 if q≡1 (mod 6),
2q+ 2 if q≡2 (mod 6),
q+ 3 if q≡3,5 (mod 6),
2q+ 4 if q≡4 (mod 6).
M(q, 5) =
q2+ 2q+ 8 if q≡1 (mod 12) and p= 5,
q2+ 2q+ 7 if q≡1 (mod 12) and p6= 5,
q2+q+ 2 if q≡2,8 (mod 12),
q2+ 2q+ 3 if q≡3,11 (mod 12),
q2+q+ 4 if q≡4 (mod 12),
q2+ 2q+ 6 if q≡5 (mod 12) and p= 5,
q2+ 2q+ 5 if q≡5,7,9 (mod 12) and p6= 5.
With M(q, n) known, it is not difficult to classify polynomials of low degree
over Fq. Tables 3 – 7 give the representatives of the equivalence classes in Pq,n
for 1 ≤n≤5. In each of these cases, it is easy to verify that every f∈ Pq,n
NUMBER OF EQUIVALENCE CLASSES OF RATIONAL FUNCTIONS 25
is equivalent to one of the representatives, and since their total number equals
M(q, n), the representatives are pairwise nonequivalent. In these tables, Cidenotes
a system of representatives of the cosets of {xi:x∈F∗
q}in F∗
q.
Table 3. Equivalence classes of Pq,1
representative number
X1
1
Table 4. Equivalence classes of Pq,2
qrepresentative number
even X2+X1
X21
2
odd X21
1
Table 5. Equivalence classes of Pq,3
qrepresentative number
p= 2 X3+X1
X31
2
p= 3 X3+X21
X3+aX, a ∈ C22
X31
4
p > 3X3+aX, a ∈ C22
X31
3
Appendix: Counting Lemmas
For m, n ≥0, let
αm,n =|{(f, g ) : f, g ∈Fq[X] monic,deg f=m, deg g=n, gcd(f , g) = 1}|,
αn=|{(f, g ) : f, g ∈Fq[X] monic,deg f < n, deg g=n, gcd(f , g) = 1}|.
Lemma A1. We have
(A1) αm,n =(qnif m= 0,
qm+n(1 −q−1)if m, n > 0,
26 XIANG-DONG HOU
Table 6. Equivalence classes of Pq,4
qrepresentative number
q≡1 (mod 6) X4+a(X2+X), a ∈F∗
qq−1
X4+aX2, a ∈ C22
X4+aX, a ∈ C33
X41
q+ 5
q≡2 (mod 6) X4+X3+aX, a ∈Fqq
X4+X2+aX, a ∈Fqq
X4+X1
X41
2q+ 2
q≡3,5 (mod 6) X4+a(X2+X), a ∈F∗
qq−1
X4+aX2, a ∈ C22
X4+X1
X41
q+ 3
q≡4 (mod 6) X4+X3+aX, a ∈Fqq
X4+X2+aX, a ∈Fqq
X4+aX, a ∈ C33
X41
2q+ 4
and
(A2) αn=q2n−1, n ≥1.
Proof. For (A1), we may assume that n−m=d≥0, and it suffices to show that
(A3) αm,m+d=(qdif m= 0,
q2m+d(1 −q−1) if m > 0,
The pairs (f, g ), where f, g ∈Fq[X] are monic, deg f=mand deg g=m+d,
are of the form (hf1, hg1), where h, f1, g1∈Fq[X] are monic, deg f1=m−deg h,
deg g1=m+d−deg h, and gcd(f1, g1) = 1. Hence
q2m+d=X
i≥0
qiαm−i,m+d−i,
whence X
m≥0
q2m+dXm=X
i≥0
qiXiX
j≥0
αj,j+dXj.
NUMBER OF EQUIVALENCE CLASSES OF RATIONAL FUNCTIONS 27
Table 7. Equivalence classes of Pq,5
qrepresentative number
q≡1 (mod 12) X5+X4+aX 2+bX, a, b ∈Fqq2
p= 5 X5+aX3+bX, a ∈ C2, b ∈Fq2q
X5+aX2, a ∈ C33
X5+aX, a ∈ C44
X51
q2+ 2q+ 8
q≡1 (mod 12) X5+a(X3+X2) + bX, a ∈F∗
q, b ∈Fqq2−q
p6= 5 X5+aX3+bX, a ∈ C2, b ∈Fq2q
X5+a(X2+X), a ∈F∗
qq−1
X5+aX2, a ∈ C33
X5+aX, a ∈ C44
X51
q2+ 2q+ 7
q≡2,8 (mod 12) X5+a(X3+X2) + bX, a ∈F∗
q, b ∈Fqq2−q
X5+X3+aX, a ∈Fqq
X5+X2+aX, a ∈Fqq
X5+X1
X51
q2+q+ 2
q≡3,11 (mod 12) X5+a(X3+X2) + bX, a ∈F∗
q, b ∈Fqq2−q
X5+aX3+bX, a ∈ C2, b ∈Fq2q
X5+X2+aX, a ∈Fqq
X5+aX, a ∈ C22
X51
q2+ 2q+ 3
q≡4 (mod 12) X5+a(X3+X2) + bX, a ∈F∗
q, b ∈Fqq2−q
X5+X3+aX, a ∈Fqq
X5+a(X2+X), a ∈F∗
qq−1
X5+aX2, a ∈ C33
X5+X1
X51
q2+q+ 4
Therefore,
X
j≥0
αj,j+dXj= (1 −qX)X
m≥0
q2m+dXm=qdX
m≥0
q2mXm−X
m≥0
q2m+1Xm+1
28 XIANG-DONG HOU
Table 7. continued
qrepresentative number
q≡5 (mod 12) X5+X4+aX2+bX, a, b ∈Fqq2
p= 5 X5+aX3+bX, a ∈ C2, b ∈Fq2q
X5+X21
X5+aX, a ∈ C44
X51
q2+ 2q+ 6
q≡5,9 (mod 12) X5+a(X3+X2) + bX, a ∈F∗
q, b ∈Fqq2−q
p6= 5 X5+aX3+bX, a ∈ C2, b ∈Fq2q
X5+X2+aX, a ∈Fqq
X5+aX, a ∈ C44
X51
q2+ 2q+ 5
q≡7 (mod 12) X5+a(X3+X2) + bX, a ∈F∗
q, b ∈Fqq2−q
X5+aX3+bX, a ∈ C2, b ∈Fq2q
X5+a(X2+X), a ∈F∗
qq−1
X5+aX2, a ∈ C33
X5+aX, a ∈ C22
X51
q2+ 2q+ 5
=qd1 + X
m≥1
(q2m−q2m−1)Xm=qd1 + X
m≥1
q2m(1 −q−1)Xm,
which is (A3) (with jin place of m).
For (A2), we have
αn=
n−1
X
m=0
αm,n =qn+
n−1
X
m=1
qm+n(1 −q−1)
=qn+qn(q−1)
n−2
X
m=0
qm=qn+qn(qn−1−1)
=q2n−1.
Lemma A2. Let Rq,n ={f∈Fq[X] : deg f=n}. Then
|Rq,n|=(q−1if n= 0,
q2n−1(q2−1) if n > 0.
Proof. For n > 0, we have
|Rq,n|= (q−1)(2αn+αn,n) = (q−1)(2q2n−1+q2n(1 −q−1)) = q2n−1(q2−1).
NUMBER OF EQUIVALENCE CLASSES OF RATIONAL FUNCTIONS 29
For m, n ≥0, let
βm,n =
|{(f, g ) : f, g ∈Fq[X] monic,deg f=m, deg g=n, f (0) 6= 0,gcd(f, g ) = 1}|.
Lemma A3. We have
βm,n =
qm−n−1(q−1)q2n+1 + 1
q+ 1 if m > n ≥0,
qnif m= 0,
qn−m(q−1)q2m−1
q+ 1 if 1≤m≤n.
Proof. We have
αm,n =βm,n +βn,m−1.
Therefore,
βm,n =αm,n −βn,m−1=αm,n −(αn,m−1−βm−1,n−1)(A4)
=αm,n −αm−1,n +βm−1,n−1=cm,n +βm−1,n−1,
where
cm,n =αm,n −αm−1,n
=
qnif m= 0,
qm−1(q−1) if m > 0, n = 0,
qn(q−2) if m= 1, n > 0,
qm+n−2(q−1)2if m > 1, n > 0.
By (A4),
βm,n =X
i≥0
cm−i,n−i.
When m > n,
βm,n =cm,n +cm−1,n−1+···+cm−n,0
=cm,n +cm−1,n−1+···+cm−n+1,1+qm−n−1(q−1)
=
n
X
i=1
qm−n+2i−2(q−1)2+qm−n−1(q−1)
=qm−n(q−1)2q2n−1
q2−1+qm−n−1(q−1)
=qm−n−1(q−1)q2n+1 −1
q+ 1 .
When m≤n,
βm,n =cm,n +cm−1,n−1+···+c0,n−m
=cm,n +cm−1,n−1+···+c1,n−m+1 +qn−m.
30 XIANG-DONG HOU
In the above, if m= 0,
β0,n =qn;
if m≥1,
βm,n =
m
X
i=2
qn−m+2i−2(q−1)2+qn−m+1(q−2) + qn−m
=qn−m+2(q−1)2q2(m−1) −1
q2−1+qn−m+1(q−2) + qn−m
=qn−m(q−1)q2m−1
q+ 1 .
Let ( ) = ( )qbe the Frobenius of Fq2over Fq, and for g=Pn
i=0 aiXi∈Fq2[X],
define ¯g=Pn
i=0 ¯aiXi. For 0 6=g∈Fq2[X], define ˜g=Xdeg g¯g(X−1); that is, for
g=amXm+am−1Xm−1+···+a0∈Fq2[X], am6= 0,
˜g= ¯a0Xm+ ¯a1Xm−1+···+ ¯am.
Clearly, gg1g2= ˜g1˜g2,g
Xm= 1, and ˜
˜g=gif g(0) 6= 0. We say the gis self-dual
if ˜g=cg for some c∈F∗
q2. In this case, (¯a0,¯am) = c(am, a0), which implies that
a0/am∈µq+1 and c= ¯a0/am∈µq+1 .
Define
Λi=|{g∈Fq2[X] : gis monic, self-dual, deg g=i}|,
Θi=|{g∈Fq2[X] : gis monic, deg g=i, gcd(g, ˜g) = 1}|,
Γi,j =|{(g, h) : g, h ∈Fq2[X] monic, self-dual, gcd(g, h) = 1}|.
Lemma A4. We have
(A5) Λi=(1if i= 0,
(q+ 1)qi−1if i > 0,
Θi=1
1 + q2(−1)i(1 + q) + q2i+1(q−1).
Proof. Every monic g∈Fq2[X] has a unique representation g=g1h, where h=
gcd(g, ˜g), which is monic and self-dual, and g1∈Fq2[X] is monic such that gcd(g1,˜g1) =
1. Therefore,
l
X
i=0
ΛiΘl−i=|{g∈Fq2[X] monic of degree l}| =q2l,
that is,
(A6) ∞
X
i=0
ΛiXi∞
X
j=0
ΘjXj=
∞
X
l=0
q2lXl=1
1−q2X.
Clearly Λ0= 1. Assume l≥1. Let g(X) = Xl+al−1Xl−1+···+a0∈Fq2[X], so
˜g(X) = ¯a0Xl+ ¯a1Xl−1+···+ 1. Then gis self-dual if and only if
(A7) ¯a0(a0a1. . . al−1)
= ( 1 ¯al−1... ¯a1).
NUMBER OF EQUIVALENCE CLASSES OF RATIONAL FUNCTIONS 31
If l−1 is even, to satisfy
¯a0(a0a1. . . a(l−1)/2a(l+1)/2. . . al−1)
= ( 1 ¯al−1. . . a(l+1)/2a(l−1)/2... ¯a1),
we can choose a0∈µq+1, choose a1,...,a(l−1)/2∈Fq2arbitrarily and let ai=
al−i/¯a0for (l+ 1)/2≤i≤l−1. Hence Λl= (q+ 1)(q2)(l−1)/2= (q+ 1)ql−1. If
l−1 is odd, to satisfy
¯a0(a0a1. . . al/2−1al/2al/2+1 . . . al−1)
= ( 1 ¯al−1... al/2+1 al/2al/2−1... ¯a1),
we can choose a0∈µq+1, choose a1,...,al/2−1∈Fq2arbitrarily, choose al/2∈Fq2
such that ¯a0al/2=al/2and let ai=al−i/¯a0for l/2 +1 ≤i≤l−1. Since a0∈µq+1,
the number of choices for al/2is q. Thus we also have Λl= (q+ 1)q(q2)l/2−1=
(q+ 1)ql−1. Therefore,
Λl=(1 if l= 0,
(q+ 1)ql−1if l > 0.
We then have
∞
X
i=0
ΛiXi= 1 +
∞
X
i=1
(q+ 1)qi−1Xi=
∞
X
i=0
(q+ 1)qi−1Xi+ 1 −(q+ 1)q−1
(A8)
=q+ 1
q
1
1−qX −1
q=1 + X
1−qX .
By (A6) and (A8),
∞
X
j=0
ΘjXj=1−qX
1 + X·1
1−q2X=1 + q
1 + q2
1
1 + X+q(q−1)
1 + q2
1
1−q2X
=1 + q
1 + q2
∞
X
j=0
(−1)jXj+q(q−1)
1 + q2
∞
X
j=0
q2jXj
=1
1 + q2
∞
X
j=0(−1)j(1 + q) + q2j+1 (q−1)Xj.
Hence
Θj=1
1 + q2(−1)j(1 + q) + q2j+1(q−1).
Lemma A5. For i, j ≥0, we have
Γi,i+j=
1if i=j= 0,
qj−1(q+ 1) if i= 0, j > 0,
q(q+ 1)
q2+ 1 (q2i−q2i−2−(−1)i2) if i > 0, j = 0,
qj−1(q+ 1)(q2−1)
q2+ 1 (q2i−(−1)i)if i > 0, j > 0.
32 XIANG-DONG HOU
Proof. Each ordered pair (f , g), where f, g ∈Fq2[X] are monic and self-dual with
deg f=iand deg g=i+j, has a unique representation (f, g) = (f1h, g1h), where
f1, g1, h ∈Fq2[X] are monic and self-dual and gcd(f1, g1) = 1. Thus
X
k
ΛkΓi−k,i+j−k= ΛiΛi+j.
Therefore,
(A9) X
k≥0
ΛkXkX
l≥0
Γl,l+jXl=X
i≥0
ΛiΛi+jXi.
When j= 0, by (A5),
X
i≥0
ΛiΛiXi= 1 + X
i≥1
(q+ 1)2q2(i−1)Xi
(A10)
=X
i≥0
(q+ 1)2q2(i−1)Xi+ 1 −(q+ 1)2q−2
= (q+ 1)2q−21
1−q2X+ 1 −(q+ 1)2q−2
=1 + (2q+ 1)X
1−q2X.
Combining (A9), (A8) and (A10) gives
X
l≥0
Γl,lXl=1−qX
1 + X·1 + (2q+ 1)X
1−q2X
=2q+ 1
q−2q(q+ 1)
q2+ 1 ·1
1 + X+(q−1)(q+ 1)2
q(q2+ 1) ·1
1−q2X
=2q+ 1
q−2q(q+ 1)
q2+ 1 X
l≥0
(−1)lXl+(q−1)(q+ 1)2
q(q2+ 1) X
l≥0
q2lXl.
Hence
Γl,l =
1 if l= 0,
q(q+ 1)
q2+ 1 (q2l−q2l−2−(−1)l2) if l > 0.
When j > 0, by (A5),
X
i≥0
ΛiΛi+jXi= (q+ 1)qj−1+X
i≥1
(q+ 1)2q2i+j−2Xi
(A11)
=X
i≥0
(q+ 1)2q2i+j−2Xi+ (q+ 1)qj−1−(q+ 1)2qj−2
= (q+ 1)2qj−21
1−q2X−(q+ 1)qj−2
=qj−1(q+ 1) 1 + qX
1−q2X.
Combining (A9), (A8) and (A11) gives
X
l≥0
Γl,l+jXl=qj−1(q+ 1) 1−qX
1 + X·1 + qX
1−q2X
NUMBER OF EQUIVALENCE CLASSES OF RATIONAL FUNCTIONS 33
=qj−1(q+ 1)1 + 1−q2
1 + q2·1
1 + X−1−q2
1 + q2·1
1−q2X
=qj−1(q+ 1) + qj−1(q+ 1)(1 −q2)
1 + q2X
l≥0
(−1)lXl−X
l≥0
q2lXl.
Hence
Γl,l+j=
qj−1(q+ 1) if l= 0,
qj−1(q+ 1)(q2−1)
q2+ 1 (q2l−(−1)l) if l > 0.
.
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Department of Mathematics and Statistics, University of South Florida, Tampa, FL
33620
Email address:xhou@usf.edu
