Internat. J. Math. & Math. Sci.
Vol. 8 No. 2 (1985) 373-388
373
ON FUNDAMENTAL SETS OVER A FINITE FIELD
YOUSEF ABBAS and JOSEPH J. LIANG
Department of Mathematics
University of South Florida
Tampa, Florida
33620
U.S.A.
(Received March 3, 1983)
A partition over finite field is defined and each equivalence class is
constructed and represented by a set called the fundamental set. If a primitive
ABSTRACT.
element is used to construct the addition table over these fundamental sets then
The number of partitions is given
all additions over the field can be computed.
for some finite fields.
KEY WORDS AND PHRASES. Finite field, primitive polynomial.
19B0 MATHEmaTICS SUBJECT CLASSIFICATION CODE. 12C99.
i.
INTRODUCTION
Throughout p,q will be fixed but arbitrary primes.
n > 1 and define A
A
E GF(p n)
{a, a+l, +2,
then
B E
Define
(p-l)A such that
1,2,...,p- I, thus
Let
A,
A A.
As As.
DEFINITION
Note that if 8 6
DEFINITION.
,
Ae,
Aa
p=
p/x
{x
A
P+ (p-l)} Aap
the values of
,
1
,
then
(
Since
pn
=i
AS"
AC
for
{a p3
A
n-l||
+bla Z*p
b
EXa,
Z
.
p
* (A
a pj
+
-1
or
A A8"
b.
we have Apn
A
+I,
Therefore
’
This implies
8C
P
,Ap
and j E Z }
n
X A--B.
LEMMA i.i.
If 8
then
PROOF. If 8 6 A
then there exist a 6 Z
8
P ,P
so, A(
GF(p n)
for every
DEFINITION.
X
p
0,i 2
,
Ap
Aap
a
Since
If
A}
in the definition can be limited to 0,1,2,...,n-i.
DEFINITION.
thus
{x/x
P-IU=
DA
*A
=+ (p-l)}.
A
2A, 3A
P
b 6 Z
(Zp,+,.)
P
and
J
Z
n
such that
is a field and a
0 both
a
Y. ABBAS AND J. J. LIANG
374
Since
pn
A
such that A
will be called a Fundamental Set.
A
DEFINITION.
E GF(p n), there
A for every
a
pm
aA
for some a
Let u,8 E
DEFINITION.
6 Z*
GF(pn).
From Lemma i.i.
in GF(p
n)
as
Aa AB
is an equivalence relation.
the relation
will partition the field GF(p n) into equivalence
This equivalence relation
classes A
n
*A
We define the relation
8 iff
<_
p
It follows that A
a
THEOREM i.I.
exists a least positive integer m
m-i
and each class is represented by a fundamental set.
A will be called a Fundamental Class.
DEFINITION.
pn_ i
Let 8 be a primitive element in
,
GF(pn).
8p
Since,
-I
is primitive in Z
P
where i < k < p- i. So, k can be determined
there exists k
a
a
a
p
set A are expressed as powers of 8, then
fundamental
of
the
elements
easily. If the
can be expressed from A as powers of 8. So to calculate the addition table of
Z
then for every a
Aa
it is sufficient to have the addition table of
GF(pn),
are all the fundamental classes in
addition tables over A
GF(pn).
,
Z
P
B
integer for every
as
Z
and b
a
Therefore, if
AI,Aa2,...,A=s,
it will be enough to tabulate only the
with respect to 8 to do all the calculations over
,...,A
GF(pn),
If for some a
for some a
i
A.
apm
m is the least positive integer such that
a+
b
then it is true that m will be the smallest positive
P
Bpm
such that
aB+ b’ where b’E Z
P
This will be shown
below.
aa+ b for
GF(pn),
Let a
DEFINITION.
ap m
some a
nd
coefficient of
Z*
P
we say
b
a has index 0 with coefficient a
if m is the least positive integer such that
Z
P
and a is the
then m is called the index of
If
has an index m with coefficient a.
Z
P
we say
i.
A has the
If a has index m with coefficient a, then every 8
LEMMA 1.2.
=
same index m and the same coefficient a.
6
Z and
Z
There exist E
p
P
P
m
pj
pj
pm
p
+ 6
b)
+
(aa
+ 6)
(a
+ 6. This implies 8
Z such that 8
n
Hence
A
Therefore s _< m. But from Lemma i.i. we have
+ c where c
8.
a.
coefficient
with
m
index
has
<
s, which implies 8
A. Therefore m
PROOF
j
a
a
with index s and
Let B
B Z
PJ
As
Zp.
DEFINITION
The fundamental set A
has index m with coefficient a if and only
pm
if m is the least positive integer such that A
THEOREM 1.2.
If
aA.
has index m with coefficient a
coefficient a and each fundamental set A C
8
Aa
then A
has index m with
has the same index m with the same
coefficient a.
PROOF.
Follows from Lemma 12
From the above theorem we can define the index A
to be the index of any element
or any fundamental set included in A
Now, we want to discuss some properties of the index and the coefficient of
the fundamental set.
375
FUNDAMENTAL SETS OVER A FINITE FILED
In GF(p n), if A has index m and coefficient a then m dfvldes
THEOREM 1.3.
n/m
n and a
I.
mpm
m+
Let n
PROOF
a(m)
+ l)b
A
pr
aA which
A
So
bA, b
r
Hence
ml
for some bE Z
m
r
a
r
n/m
.
a -m
+ (a
we have
p
-i
+ a -2 +
i.
Let
km
am+ 6 and mp
pm
m
such that
If
+ r,
0
_<
b+ 6’.
But
Therefore
This implies r > m.
6".
r < m.
Since all finite fields of the same order are
Note:
b.
and a
i E GFI(P n)
isomorphic, if
am+ b,
(mp )p
A, we have m <
r
, r P
(ako + 6 )P + 6’ akm p +
r
(P )P
0.
n
and a
then m divides
P
Since m is the index of
km
b+ 6’
m
0 and a
implies r
A, then there exist 6, 6’ Z
If a
p
Let A be a fundamental set with index m and coefficient a.
/m b.
E Z*
PROOF.
mp
+ (a -I + a -2 +...+ l)b and m
THEOREM 1.4.
p
Since m
r, 0 < r < m.
having index m with coefficient a, is the image of
a. There2 GF2(pn) under an isomorphism o, then 2 has index m with coefficient
fore; GFl(pn) and GF2(pn) have the same partitions with respect to the equivalence
relation
GF(52)
EXAMPLE: Let F
the primitive polynomial
two equivalence classes, Z
8
3
817
+ 4
2A
8
3A
4A
and A
2.
4A
8
5
0
8
8228/ 815
,
Z
P
GF(pm)
in
GF(pn).
822
i
Since
8
815
+ 2
86
3
82
812
and
+
8
2, then 4
87 + 2 8 4 87 + 4 =821 87 + 1 88}
{019 019 + 3 016 019 + i 0 I0 019 + 4 821 019 +
{813,813 + 4 810,813 + 3 83, 813 + 2 814, 813 +
8
(a)
+
82 817.
SOLUTIONS OF EQUAl IONS OF THE FORM Ap
solutions of:
-
8
{8 7
To study the fundamental sets in
6
{8,
A8
and
So, we have 8
Therefore,
818.
5
and 8 be a primitive element in F such that 8 satisfies
2
x + 4x + 2. The field F has
polynomial], P(x)
indexing
pm
x
Note:
ax
+ 6;
a
2
I
06
85
m
aA.
GF(p n)
with index m < n, we have to study the
Z*
# i, 6 E Z
a
P
P
and (b)
x
pm
x
+ 6;
0 in (b), then all the elements of the subfield
If 6
will satisfy (b).
We will now consider the solutions of
x
where
,
a 6 Zp,
a
LEMMA 2 i.
n’m
n and a
# 1 and
6
p
ax
+
(2.1)
6
Zp
Equation (2 i) has a solution e
Z
p
with index m only if m divides
i.
The proof is a direct application of Theorem 1.3 and Theorem m1.4.
ay.
we have
x +
LEMMA 2.2. In equation (2.1), for y
a-1
PROOF
yP
m
)pm=xpm + a-i
6
ax + 6 + a--- a(y- a_l
yP
(x +
ay.
a
+ a-i
376
Y. ABBAS AND J. J. LIANG
Therefore, to study the solutions of (2.1), it is sufficient to study the solutions of
ypm
ay.
LEMMA 2.3.
-a
x
is a solution of equation
If 0 is a primitive element in
GF(pn),
(2.1).
then the following statements are true:
p -i
P-I
(a) 0
(b)
Z
Z * there exists an integer ka such that a
p
If a
i, ka # 0.
For every a
where 0 < k
< p-i.
a
LEMMA 2.4.
n
p -I k
p
For every m such that m divides n and
n
m
0
n
p-i, then
+ I.
n
m
Let
0(mod k).
LEMMA 2.5.
If a
k
n
PROOF.
n/m
n
a
_p -i
1
m
n/m
p -i
n
p -i
p
Pn-! -ka .nm
i
p -I
Therefore, p-i k a
m
p
Given x
ax
0(a)
i and
where m n, a
If
P
r
Z
(a)
If x
(b)
Since x
n
k
1
a
p-1
m
(c)
Or
ax.
x
ax and a
# i,
is a solution of
then x
n
p
1
m
p -i
It follows from Lemma 2.5
a
p-i
+
(pro) k-2
is an integer.
pn-l_.ka
implies
So p-i k
a
Z
(mod
P
is primitive in
GF(pn),
pn -!).
m
p -I
P
k
a
p-i
O.
_GF(pm).
(2.2) then
(or) pm-I
=- pnp-ii ka
pn_ 1).
(mod
r(pm- i)
r
k
0 mod(pn-l)
Therefore x
m
p
n/m
m
then xp
P
I
n
m
(2.2)
0 is the only solution of (2.2) in Z
PROOF.
which implies
n
m
in
p -I
x
i
Therefore,
i
0 p-
Also, a
GF(p n) and
r
O
O
a solution of (2 2) where a
is
i
then
Equation (2.2) has solution
If
I
p
i.
THEOREM 2.1.
(c)
p-1
is an integer.
n/m
i
1,2
n
-i
then p
m
p -i
Im_(n/m) ka
n
n
p- i
0
(a)
(b)
m
l(mod k) for j
1 and
i, a
p-i
But, from Lemma 2.4
(pro)k- 1
pm -i
a
pn_ i
pm 1
n
m
Hence
j=O
n/m
a
pJ
l(mod k), we have
k-i
(pm)j
(pro) k -I
m
p -i
-=
Since p
pn -1
l(mod k) and
So, p
k.
Pm
p
is an integer,
PROOF.
1
p
pn-i
=a= 8
(mod
pn
p-i
k
a
I)
m
p -i
that the above congruence is meaningful.
Thus,
then
FUNDAMENTAL SETS OVER A FINITE FIELD
a
P
Let r
PROOF
n
pn_ i
-k
o p-I
i
I
a
(Or) p
So,
a.
O
m
0,1,2,..
j
n
(8p-l)j
a
pn_ i)
r
is a solution of
p -i
a
p-I + j pm -I
-i
m
p -i
m_l
pn
k
(mod
I
p
p -I
(2.2).
(Sr)P
k
n
p -i
m
For every r
THEOREM 2.2.
377
then
m
aO.
Now consider the solutions of
x
where a
# i,
Z
a
we have
Theorem 2.2,
m
+
ax
n
m
and 0(a)
(2.3)
6
From Lemma 2.2, Theorem 2.1 and
is a solution of
(2.3),
where
k
n
p -I
r
Zp, mln
6
r
O + -a
6
p
p
a
"p-i (mod
m
p -i
is the only solution in Z
and
m
p -i
P
Let A be a fundamental set in GF(pn) with index m and coefficient
n
and
If A is not included in any proper subfield of GF(p n), then O(a)
m
THEOREM 2.3.
i
a
n
l(P- i).
m
Let 0(a)
PROOF.
p
Thus, s
coefficient a.
dm
d
p
a
+
a s
GF(pdm)
for some 6 6 Z
+ 1)6
+
dl
and a has an index m with
Hence
P
Therefore s 6GF(p
s.
din),
which implies
and d
COROLLARY 2.1.
such that a
+
as
(ed-1 + d-2
GF(pn)
It follows
A.
d and u
m
Let X
mln,
p
+ 6
aX
m
(2.4)
,
then equation (2.4) has solutions
GF(p n) and all the solutions are included in
PROOF. GF(p
is a subfield of GF(pn). Equation (2.4) satisfies the conditions
in
i, 6
Z
P
a
1 and O(a)
GF(pm).
m)
of Theorem 2.1
up
m
a u
+ (a
over GF(p
-1
+
a-2
m)
and if u is a solution, then
+ 1)6
+
+
.
O
GF(pm)
Therefore
p -1
Note:
in
y r’
If 8 is a primitive element in
GF(pm).
+
P
r’
m
k’a
1
p-1
m
p -1
a
0
So, k a
8
r
p-i
k’.
a
+ i_-,
then y
0
is a primitive element
Therefore, the solutions of equation (2.4) are of the form:
where
pn-I
GF(pn),
p m -i
k
pn-I
a
,we have 8
p-I
(mod
pm
a
(8
and a
m
7
p
1
a
Since
p -1
n
p -1
k
_p.m_ 1 k’
1
m
p -i
p
m -I
p-l’
k
a
pn
-1
-I k’a
Op
Therefore, the solutions of (2.4) over GF(pn) are of the form
where
r
n
P -I
m
p -i
n
r
p -1
m
p -i
k
a
p-l- (mod
pn !)
m
p -i
_
378
Y. ABBAS AND J. J. LIANG
m
LEMMA 2.6. Equation (2.4) has p solutions in GF(pn).
PROOF. Corollary 2.1 insures that equation (2.4) has solution in GF(p
p -I
+
0
solution of (2.4).
O(a)[
and Theorem 2.2
is a solution and if a
a
a
p-i
m
From the previous note, Theorem 2.1
and a
m).
H
It is clear that H has
we have
0
8p -1
lGF(pm)}
{0
B +
pm
elements.
then a is a
THEOREM 2.4.
For every m divides n and every a E Z
1 such that
a
P
there exists a fundamental set in GF(p n) with index m and coefficient a.
n
m
PROOF.
By
Lemma 2.6,
x
p
ax has solutions in
n
p_ -1
primitive element in
is a solution.
an index m with coefficient a.
To prove this, we assume
coefficient b, therefore by Theorem 1.4 and Theorem 2.2
solution of x
p
bx and it is of the form
e
If
is a
ok
0 p -1
6F(p n) then
GF(pn).
es
We claim that
has an index
we conclude that
n
where s
has
m with
p -I
p -I
p-I nod
is a
pn_ 1)
p -i
So, there exists j where j > 0 such that:
n
k
p -1
_<
then 0
Since
j
p-1
p -1
pm -1
p-1
pro_ llp_ 1
pm_
i_ > P
ka
p -i
< p- 1 and i
_<
<_ J
<
0
we will have
p_--
P
and this is a contradiction.
Hence A--A
is a fundamental set with index m and coefficient a.
COROLLARY 2.2.
The minimum subfield which contains all the solutions of
GF(pm).
equation (2.4) is
The proof is a direct application of Theorem 1.3 and Theorem 2.4.
PROOF.
We will now review some known facts about the solutions of x
the field
j
p
m
x
+ b,
+
where b EGF(p n)
n
g.c.d. (m,
LEMMA 2.7.
If x
b in
d
in
GF(pn).
x
0
0
is a solution of
+ 8p
(2.5)
The following theorems were given in [i].
n) and r
i, 2,..., p -i,
element
x
GF(pn).
Let x
and d
p
-i
(2.5) in
is a solution of
GF(pn),
then for every
(2.5) where 8 is a primitive
FUNDAMENTAL SETS OVER A FINITE FIELD
PROOF.
m
n
Xo
+ 8p
379
d
m
x
-i
+ (O p
m
p
d
i
n_
(x
THEOREM 2.5.
THEOREM 2.6.
0+b)
+ e p -I
Xo
1
+ ep
+
b
(2.5) in GF(p n) is ether 0 or p
Equation (2.5) has solutions in GF(p n) if and only if
The number of solutions of
d
r-i
If we assume m divides n and b
Z
,
in the equation
P
x
p
m
x
+
(2.6)
b
then we can conclude the following:
(a)
(b)
d
g- c- d(m.n),
m
for every b
E Z
P
we have
n
E pm
b=--nb=O
m
.ffiO
if and only if p divides
n
m
Now, we can restate the following theorems:
(a) Theorem 2.7. Equation (2.6) has solution in GF(p n) if and only if p
divides
(b)
n
m
Theorem 2.8.
If equation (2.6)has a solution in
solutions, also if x
0
is a solution then x
0
GF(pn),
it ham
pm
+ 0 p -1
0,1,...,p -i is a solution where 0 is primitive in GF(pn).
2.9.
If (2.6) has a solution in GF(pn),then the minimum subfield that
THEOREM
j
,
contain all .the solutions is
PROOF.
p divides
Since equation
GF(ppm).
(2.6) has a solution and m divides n then by Theorem 2.7
therefore GF(p pro) is a subfield of
m
then we have
a
aP
a
+ b,
which implies
aP
GF(pn).
If a is a solution of (2.6),
pm
=
+ pb
=
so
=
GF(p pro) and
GF(pm)[b#0]
Let GF(p
Z)
be the minimum subfield which contains all the solutions of
(2.6)
therefore
GF(p)
and
#
m.
This implies
.
[pm.
GF(ppm)
But since equation (2.6) has p
and by Theorem 2.5 and Theorem 2.6,
Therefore, pm
C
g
c
d(m,)
m.
If a is a solution of (2.6) in GF(p n) where
i.
coefficient a then s divides m and a
LEMMA 2.8.
m
solutions in GF(p
)
Hence,
has an index s with
Y. ABBAS AND J. J. LIANG
380
Theorem 1 4
PROOF
P
e
Let m_
SIR.
implies
s
+
as
+
6 for
r
Hence a
b.
E Z
SORe
and
1
therefore
P r-i
r-2
+
a
a
P
P
+
+
Assumes a # 1 and
r.
s
m
rs
a
r
ar-i + ar-2+ ...+i)
+
,
0, but thls
0 which implies b
1
0 of equation (2.6)
contradicts the condition b
there
In GF(p n), for every m divides n and p divides
pm
m
GF(pm).
A
and
A
A
that
such
6 A
exist (p-l)p elements e where
COROLLARY 2.3
Zp*
m
PROOF. Equation (2.6) has p solutions over GF(pn) for fixed b
and there
are p-i different values for b.
In
COROLLARY 2.4.
GF(pn),
index m with coefficient a
PROOF.
This is a direct consequence of Lemma 2.8.
is a solution of (2.6) and
PROOF.
then (2.6) has p
Since p
has p
--_m (b),
s
hence s
solutions.
Let
1 where
lm and p
{Sl, s2,
solutions of (2.6) such that 1
_<
n
m
i
x
then a satisies
ps
x
<
sj
m
Sl + p s2 +
P < P
.
+
has
(2.7)
i
equation (2.7)
By Theorem 2.7 and Theorem 2.8
st}
s
Also by Lemma 2.8, if
solutions.
siR,
be the set of all the indices of the
< R for every
I
< j
+
p
Since
s _<
we have
is the greatest integer less than or equal to
where
where
then there exists an
has an index m with coefficient i.
an index s with coefficient a
s
__n
In equation (2.6) if p divides
THEOREM 2.10.
where 6
if m is a prime, then all solutions of (2.6) have
i.
s < p + p2 +
p
p-1
But since
P
k+l
p-I
p <
p2k
p
for k > i and
p
k+l
p-I
p
R
p-i
P
< p
2k + i
we have
p<p
and this is a contradiction.
COROLLARY 2.5.
In
GF(pn),
for every m divides n, and p divides
I.
exists a fundamental set with index m and coefficient a
3.
n
there
THE TOTAL NUMBER OF FUNDAMENTAL CLASSES IN SOME FINITE FIELDS.
In this section we will investigate the total number of fundamental classes
for some special finite fields From the previous study we conclude that If p
there exists a fundamental set with index m and coefficient 1 in GF(p n).
In
GF(pn),
for every m divides n, and every a
exists a fundamental set with index
n
c
g
p-l) # i.
R
Z*
P
and coefficient a.
a
# 1 where
Since a
m
n
m
a
1 there
1 then the
d(
If A is a fundamental set with index m then A has p(p- l)m elements.
FUNDAMENTAL SETS OVER A FINITE FIELD
381
In this section we will use the following notations:
Number of fundamental classes in GF(pn).
(a) 0(p n)
Number of fundamental classes in GF(pn) but not in any proper
n)
(b) A(p
GF(pn).
subfield of
Number of elements in GF(pn) with index m and coefficient a
GF(pn).
and none of these elements belonging to any proper subfield of
m
p
{x
ax +
a
Z
1}.
Z and
0 if a
E(m,a)
%(pn,m,a)
(c)
.
(d)
(e)
.
"
p
P
SE(m,a,n) is the set of all solutions of the equations of E(m,a) in
GF(pn) but not in
We shall investigate in the following the number of fundamental classes in
GF(p
q
), where q p-l, and q @ p.
LEMMA 3.1.
PROOF.
If
qp- l,then
q
t+l
pq
divides
t
[q-l]
0,i,2,
By Fermat Theorem the len,a
We will prove this 1emma by induction.
O.
is true for t
Assume it is true for t
pq
[q-l]_
(pq
s
[q-l]
s then
(pqS[q_l])
i
[
1)
i for every t
s
(pq [q-l])
1
s
q-i
q s-2
s
+
(pq [q-l])
+
+
eve J
0,I
I]
(pq [q-1])
[pq [n-ll, 11
]--1
pq
Since
s
s
[q-l]
i rood q, then
s
(pq [q-l])
q
E i
(mod q) for
q
s
S
(pq [q-l])
So,
q
s
-=
(mod q)
0 (mod q) which implies q
s+2
divides
j=l
[pq
s+l
[q_l]
1].
LEMMA 3.2.
t+l
where
qp-1
# p, we have
and q
t
t
A(p
In GF(p q
t+l) Pq [pq [q-l]_l]
t+l
p(p-l)q
PROOF.
Since g
d(q,p-l)
c
l,then for every a 6
,
Zp,
where a
s
@ I, a q
I
for every s
O, therefore equation (2.6) and equation (2.7) have no solutions in
t+l
t+l
q
q
h
but not in
for every m
GF(p
h > 0. Hence every element in GF(p
q
t+l
t
q
t+l
but not
which mplies that every fundamental set in GF(p
GF(p q has ndex q
t
t+l
q
So:
in GF(p
also has index q
A(pt+l
From Lemma 3.1
and g
COROLLARY 3.1
t
pq t+l p_q t pq.t [_pq_ [q-if_
t+l
p (p-l)q
c
d(p-l,q)
In GF(p q) where
O(p q)
i
p (p-l)q
11
t+l
1 we have A(p
gp-l,
+---P---=
p(p- l)q
t+l)
is an integer.
p we have:
q
i
p(pq-l_
+ p (p-l)q I)
(3.1)
382
Y. ABBAS AND J. J. LIANG
Since
O(p
q
t+l
O(p
q
t
t+l
+ A(p q
we conclude that:
-1
O(p q
1
A(p q
+
t+l
t--O
pqt [pq
-i
i
+
[q-l]
P(P-l)qt+l
t=0
where
t
qp-i, q # p.
s
We shall now study the number of fundamental classes in GF(p p ).
By Theorem 2.7, Theorem 2.8 and Theorem 2.9, the equation
x
pp
t
+
x
where b
t+l
p
GF(p
b
Z
,
and t < s has
P
).
LEMMA 3.3.
PROOF. Let
k < t and a
pp
pP
(3.2)
t
solutions and all solutions are included in
All the solutions of (3.2) have an index p
be a solution of (3.2) and has index m. We have m
k
+
c for some c
t
So
x(PP
t+l
p
For a fixed b E Z
PROOF.
k
for some
P
(3.2).
t+l
p
COROLLARY 3.2
In GF(p
pP
p
t
not a solution of
with index
t
,
t
I)
,
P
but not in GF(p
p
COROLLARY 3.3.
equation (3.2) has
If e E GF(p
t+l
p
t
pp t
(p-l)
elements in Z
p
pP
), there are (p-l)p p elements
t
solutions and we have (p-l)
GF(p t) then
but
has index p
t+l
or p
t
where t > I.
COROLLARY 3.4.
For t > i;
A’PP( t+l)
(p_l)p p
pp t+l_pp t _(p_l)pP
t
+
p(p-l)p
pp
p
t+l
pp
In
GF(pP),
where b 6 Z
,
P
the equation x
p
t
+
p(p-l)p
pp
t
-t-i
t+l
ppt (P-I)-I-I
(p-l)
t
-t-I
[pP (P-I)-I_I ]
(p-l)
x
+
+
i
b
LEMMA 3.4.
GF(pP)-z
P
wlth index p.
O(p p)
i
(3.3)
(3.4)
has p solutions and each solution has index p.
p(p-l) elements in
t
Therefore there are
This implies:
pP-p(p-l)-p_
+ p (p-l) + p(p-l)p
(3.5)
FUNDAMENTAL SETS OVER A FINITE FIELD
COROLLARY 3.5.
s-i
O(p
p
+
2
+
A(p
p
383
t+l
(3.6)
tl
s
In what follows we will study the number of fundamental classes in GF(p p "q
qp-I,
where
Since
s
p
p and
qp-i
s,
I.
>
eve
then for
m divides p
s
-q
and
eve
E Zp
a
with a
i, we
"q
m
have a
a
q
i.
# 1 and every
s
implies that SE(m, a, p q
So, Theorem 1.3
m lp
s
)
for every
q
By Theorem 2.7, Theorem 2.8 and Theorem 2.9, we conclude the following:
LEMMA 3.5.
For every t and h; 0 < t < s-i, 0 < h <
SE(ptq h, I, pSq) has
t
(p-l)p
GF(p
p
,
h
p "q
and
SE(ptq h,
LEMMA 3.6.
SE(ptqh
.
elements, all of them are contained in the minimum subfield
t+l h
q
PROOF.
is included in
Let u q SE
which implies u
t
GF(p
p "q
h
For every t, h such that 0 < t < s-i, 0 < h < -i,
pSq4)
1
pSq)N
i,
(ptqh,
pp t "q h + I
I,
SE(ptqh+l 1 pSq)
ptqh
pSq) Then u p
+ [q
u
], where [x]
a
+
[q- ]
Since
,
P’
is the least nonnegative residue
SE(p t q h+l i, p Sq).
It is clear that if u
SE(pt q 1, p Sq) ,then u has an index p t
some 0 < r < h. Since, SE(p
i, pSq A) C SE(p
i pSq) C
C SE
of x mod p.
q Z
6 for some
0 then
h,
t,
tq,
q
r
for
(ptqh+l, i p Sq)
we conclude that
X(P
p
t+l
h+l
"q
(p_l)pP
Therefore, in GF(.p p
t+l
I)
t
"q
hi
(p-l)p
p "q
h+l
t
h
t
pp qh [q-l]
(3.7)
h+l
q
there are
h
t
t
t h+l
p q
h
t
(p_l)pP "q [pP "q [q-l]_l ]=
P(P-I)pt’q h+l
ppt_
h
"q
[pp
p
t
t+l
h
.q [q-l]
"q
(3.8)
h+l
pt.q
h+l
fundamental classes with index
From Lemma 3.1
and the fact that
t
p
t + i, the number given in (3.8) is an integer.
t+l h+l
t h+l
"q
Since any proper subfield of GF(p p
is a subfield of GF(p p "q
or
GF(p
p
t+l
index p
"q
h
t+l
pp
So, in GF(p p
).
"q
t+l
h+l
.q
is equal
h+l
_pp
t
.q
t+l
"q
h+l
), the number of fundamental classes
t6
h+l
h
t
t h
h
_pp.t+l .q +pp .q... _(p_l)pP .q..
p(p-l)p
pP
t+l
.q
h
[pp
with
t+l
.q
h
t+l
.q
[q-l]_l]_pp
p (p-l) pt+l
Therefore, we have the following
t
[pp-q
h
[q-l]
h+l
t
-q
q
h
t
[pp-q
h+l
h
[q-l] -1]-
I:’
(3.9)
384
Y. ABBAS AND J. J. LIANG
s
In GF(p p "q ),
THEOREM 3.11.
(PP
t+l
"q
n+l
(3.8) + (3.9).
From it follows that
O(pP
t+l
"q
h+l
t+l
(pP
t
+ O(P p
"q
"q
h+l
+ O(p p
h+l
t+l
t
O(P
p "q
"q
h
h
(3.10)
We now study the general case, i.e. the number of fundamental classes in
First, let n
i
1,2
ql
r
qr
2
q2
i
such that
i
giP-I
# 0
and
qi # p
GF(pn).
for every
r.
Let N(s,p) be the number of elements in GF(p s) but not in any proper subfield
of GF (pS).
THEOREM 3.12..
(i)q j
The number N(s,p)
i-j=s
(See [2]).
Since for every m divides n and every a
E Z
where
is the Mblus functions.
.
,
the set SE(m,a,n)
It
EGF (pk) where GF (pk) is a subfield of GF(p n) and
doesn’t belong
to any proper subfield of GF(p
then = has an index
So, we have the following
implies that if
P
k),
THEOREM 3.13.
If GF(p
k)
pk.
C GF(p n) then A(p
N(k,p)
k)
p(p-l)p
Now let n
qi2P-I
k
p
ql
for every i
LEMMA 3 7
i
q2
2
qs
s
p
tl
t
ql
t2
t
i’
Since
i
LEMMA 3.8.
for i
1,2
then SE
(mqrl
GF(ppm),
p
,s and for some t
LEMMA 3.9.
If
COROLLARY 3.6.
,l,n) #
pk
r
t
t
"ql
t
r
SE(mqrl,l,n)
SE(mqr
# 0 and
.
.
t
1
<
qs s
where 0 < t < k-I, 0 < t
r -i, SE(m,l,n)
divides m, and m divides n then
If
i
(p-l)p elements, all of them are
SE(m,l,n) N GF(pm)
we have SE(m,l,n)
SE(m,l,pm).
For every m
SE(mqr,l,n).
"
where 0 < t < k-l,
pro)
SE(m,l,n)C
p’
t
qs s
m
q2
_< i _<
1,2,...,s then SE(m,l,n) has
contained in the minimum subfield GF(p
and
0
# O, qi #
such that k
1,2,...,s.
For every m
k
#
and
i
is a proper subset of
If
SE(mqr2,1;n)
1 n)
2
pt
SE(m,l,n) f SE(r,l,n)
divides m but not r, then
The proof is a direct application of Lemma 3.7.
COROLLARY 3.8.
p
If for some 0 < t < k-i
ptlm
and p
t+lr then
SE(m,l,n)
where (re,r)= g- c
SE(.r,l,n)
i
SE(m,l,n)
for some r # r
#
I
2
The proof is an immediate application of Lemma 3.7.
COROLLARY 3.7.
<
t
Ir but pt+l 2m and
SE((m,r),l,n)
d(m,r).
The proof is a direct application of Lemma 3.7, Corollary 3.6 and
.
FUNDAMENTAL SETS OVER A FINITE FIELD
3.8.
Corollary
We know from Theorem 2.10
SE(m,l,n) such that
a
SE(m,l,n)
0
_<
_<
t
385
Let m
k- i, t
# 0
Ir
p
tll t12
t
.ql
qtlh
"q12
I
l,...,h and h < s.
for r
factorization of n such that
qr
qlr
@ then there
is an
We want to find the number of elements in
a has an index m.
with index m.
SE(m,l,n)
if m divides n and
for r
and m divides n where
q’s
If we rearrange the
1,2,.
in the
,1,n)
,h, then SE(
h
for every i
i=iN
1,2,...,h, and
1,n)qi, SE(ql q2m’’’qh
SE(--"
l, n).
We shall use the following notations for the remaining of this section:
m
p
t
t
t
i
"ql q2
t
2
qh
h
1 < t.
h
h
and
m
m
I
qi
i=l
2
qiqj
i<j
h
r
qil qi2
.=
1 < i
qi
i
I
< i
2
r
r
||
||
qi
j=l
j =l
j
q.
for
i
If
R[SE(m,I,n)]
R[SE(m,I,n)]
O. Therefore m
r
h
we conclude the following.
Also, if B
m or pm.
is the number of elements in
m
m
and
+
If r > h then we define m
From previous Lemmas,
h
r
r
+
I
m
m
2
3
p
tl-i t2-1
t
SE(m,l,n)
qh
with index m then
+ (-i) hmh
+ (-l)rm +
+
th-1
"q2
"ql
r
GF(p pm) and not im any proper subfield of
Hence we have the following.
GF(ppm),
then
(3.11)
B
has index
THEOREM 3.]4.
&(ppm)
R[Sm(m,l,pm)]
(pm,p)-R[Sm(m,l,pm)]
p(p-l)m
p (p-l)pm
s
To determine the number of equivalence classes in GF(p q
to study
SE(m,a,q
SE(M,l,q s)
,
a
Z
p
if a
s
for all m dividing q
for every
q S/m
mlq s.
i then SE(m,a
a’s
in Z
Z*
P
and a
qS)
has
p(pm-l)
can one find?
order q
v
PROOF.
b
k
In Z
p’
if
qV
Z
P
mlq s
we need
and every a
# I,
One question we will try
a
# I such that
a
qs/m
I
E(m,a,q ).
divides (p-l) then there are q
(q-l) elements of
Let b be primitive element in Z
Assume for some k;
p
q
i. This implies k-q
0 mod(p-l). So, k
t-
is a solution x
qlp-l,
Another question is for fixed m and
P
a, how many m’ s, a’ s are there such that SE(m’,a’,q
LEMMA 3.10.
.
elements
m, is there an a
where
From Theorem 2.7 we have
Also we know that for every
to answer first is that for given
and then how many such
s
(3 12)
k
P-__!_
q
p
I,
for some
386
Y. ABBAS AND J. J. LIANG
p-1 for every
But also if r
1,2,...,q-I we have
q
r
and
I
b # 1 which implies that in Z we have (q-l) elements of
P
order q. Let p
i
q
h where the g
c
d(p-l,h)
I, then for every r such
that r
0 mod(q -v h) and 0 < r < qV we have b r is a solution of x
I, which
implies that in Z
there are qV- I elements such that x qv
i and x
For v
1
I,
P
v
2 we will have q 2
I- (q-l)
q(q-1) elements of order q 2
The same for
v
t
I, v t we will have
i
elements of order
3.11.
LEMMA
In Z
if for a fixed b where O(b)
and
i < p <
there exists
p
an a of order qV where
_> v > and a satisfies x qv- b
(3.13)
then there are q v- distinct elements in Z * of order qV satisfying (3.]3)
P
PROOF. Note that equation (3.13) has no repeated roots. If a is a solution, we
1,2,.
t
-(br) q
,q-l.
’p-I"
b
qv
qt
claim that a,a
qp+l
a
(qt-l-l)
2q+l
(qV--l)’q-i
a
prove our claim, first since O(a)
distinct c,l,.nrs in Z
g
d(rqU+ l,q)
c
r
Also (a
P
1 and O(a)
THEOREM 3.]5.
q
If b f Z
v
q
v
then
we
,ch that
are distinct solutions of (3.13). To
riq+l
a
q+l qV-
qt.
qt-l(q-l)
qP
r-q
a
aqV-
.ve
qP-
b
i
O(a rq+l)
O(b)
q v-U_l are
1,2
for r.
v
q
0 < v <
b.
Since
v
,
t|n for every v,
P
v
v-D e]rlnts
there
ord,r
are
q
_>
of
a and satisfying (3.]3).
v-i
v
PROOF. In Z we have q
(q-l) el,ments of order q
Let c e z and 0(c)
p
P
p
vq
q
and let c
d so d # 1 and d
i. Furthermore, the order of d is q
By
,
v
-
Lemma 3.11,
_q
v-i
q
we have q
q-l(q-l)
v- elements satisfying x qv-v
q
distinct d’s of order
PROOF.
If q divides p-i, then q
t
v
But there are
d for each d.
for which the above equation is
solvable and that is exactly the total number of elements in Z
LEMMA 3.12.
q
divides
pqt -i
P
having order
qV.
for every t > 0.
By induction.
In GF(p q) where q divides p-l, the set SE(l,a,q) #
if and only if O(a)
q.
q, the set
(q-l) elements of order q and for a fixed b with O(b)
SE(I b,q) has p(p-l) elements, therefore GF(p q) has p(p-l)(q-l)
(q-l) fundamental
p (p-l)
Since there are
classes with index i.
So we conclude:
O(p q)
Since P
-=
(q-l) +
pq-p-p(p-l)(q-l)+
(q-l) +
(--pq-l-l)
(q-l) +
pq-2 +pq-3 +
p (p-l) -q
I
+
(p-l)q
+1+1
(3.14)
q
I sod q.
q
i
+
>pq-r
1
+ (q-l)
sod q
0 sod q.
r=2
So, (3.14)
is an
iteger.
THEOREM 3.16.
In GF(p
q
where q divides (p-l), if O(a)
t+v
t, where 0 < t < s-v we have SE(
v+t=l
t
SE(qt ,a,q
Sm(q a,q
v+t)
qt ,a, qS)
SE( qt ,a,q
#
q
and
V
then for every
387
FUNDAMENTAL SETS OVER A FINITE FIELD
By Corollary 2.1 we have
PROOF.
SE
,a,
Theorem 2.4
and Corollary
q
solution a n with index m
set of this equation is
H (D GF(p q
H
will imply that X
t+v-i
2.2
t
{s
GF(Pq
n
and
+
O
pq
t
aX + b; b
GF(p
q
t
and it is clear that
It is possible that in some cases
This theorem is not true in general.
the minimum subfield that contains all the solutions of x
pm
subfield which contains some of thse solutions.
s
q
From ’Fheorem 3.
conclude that in CF(p ), w]ere
t
subset of SE(q ,a,q s) if and only if
v >
qV’
O(a)
q
v
t
t’ + v’
t
+ v and
+
has a proper
qlp-l, SE(q t’
t-t’
q
b
,b,q s) is a
a, wlere
v
Z * with order q where
p
Also for a fixed a
s-v.
ax
O, Theorem 3.15
insures the existence of exactly q elements "b" in Z such
P
v+l
q
v
q
and b
a. Hence, for a fixed a with order q and fixed t"
that 0(b)
< s-v there are q elements b in Z
such that:
p
SE(qt-i ,b,q s)
Zp
Therefore, if we start with a 0
a
Z where O(a
i)__
a
n
SE(qt- (i+l)
and
has a
P
i-a
NOTE:
O(b)
Z
By Theorem 2.1 the solution
).
/
#
,a,q
qV+n
ai+ I, q s)
>_
SE(q
t-i
(ai+l)q
and
,ai, q
s
qV
ao
So, we have
a..
for every i
then there are
0,1,2
,n-1 where t-n
(p-i).
Let p-1
every t
q
p
such that 0(
v+i
s
t
SE(q ,a,q ).
g
q
h where g
i such that O(a)
qV
,
d(h,q)
c
Then for every a
i.
i < v < E and t
+
SE(qt,a,qS),
In
elements with index q
t
(pq
there are exactly
and coefficient a.
From this lemma, we conclude that in GF(p q
t
(Pq -l)-(Pq
P
a
#
0
i and
v < s we have the following
Lemma.
LEMMA 3.13.
Z
>_
t
-1)p-(p
q
t-I
-1)p- g
t+v
there are
t-i
-i)
(p-l) -q
q
t
qV-l(q_l
(3.15)
fundamental classes and each class has an index q t and a coefficient with order qv
Lemma 3.12 insures that (3.15) has an integer value. We will use the notation:
F(m,t,v)
(3.15), where m
It is clear now that in GF(p q
(pq
s
t
+ v,
1 < v <
and t > i
s
where s > i, we have
-pq s-i )-p (pq s-i-i) (q-l)
p(p-l)q
s
(3.16)
Y. ABBAS AND J. J. LIANG
388
fundamental classes and each class has index q
Therefore, for s > 2,
O(q s)
O(q s-l) + (3.16)
+ 2 F(s
s-v
v) +
-Pq
6
(
s-6-1(q
(3.17)
p (p-l)q
v=l
If
with coefficient 1.
> 2 we conclude that:
s-6-1
where s-6-1
S
min{-l, s-l} and 6
max{O,s-}.
1 then we have:
s-i
0(q s)
0(q
s-I
+ (3 16) +
p(pq
-i) (q-l)
p(p-1)q
s-i
REFERENCES
i.
LIANG, Joseph J., On the solutions of trinomial equations over a Finite Field.
Bull. Calcutta Math. Soc. 70(1978), no. 6, pp. 379-382.
2.
LONG, Andrew F., Factorization of irreducible Polynomials over a Finite Field
with the substitution xq-x for X. Acta Arith. XXV 1973, pp. 65-80.
3.
BERT, A.A., Fundalnental
Concets of
Highher lgebra.
Phoenix Science Series,
The University of Chicago Press, 1956.
4.
ALAHEN, J.D. and KNUTH, Donald E., Tables of Finite Fields. Sankhya, The Indian
Journal of Statistics, Series A, Vol. 26, Dec. 1964, Part 4, pp. 305-328.