IBN AL- HAITHAM J. FOR PURE & APPL.  SCI.            VOL.24 (3) 2011 
  

 
Some Results on The Complete Arcs in Three Dimensional 

Projective Space Over Galois Field 
 

A.SH. Al- Mukhtar  
Department of Mathematics, College of Education Ibn-Al-Haitham,University 
of Baghdad  
Received in : 11   May  2011 
Accepted in : 16  June  2011 
 
Abstract 
        The aim of this paper is to introduce the definition of projective 3-space over Galois field 
GF(q), q = pm

, for some prime number p and some integer m. 
        Also the definitions of (k,n)-arcs, complete arcs, n-secants, the index of the point and the 
projectively equivalent arcs are given. 
        Moreover some theorems about these notations are proved. 
 
Keywords: arcs, index, plane. 

 
 

Introduction: [1] 
        A projective 3 – space PG(3,K) over a field K is a 3 – dimensional projective space 
which consists of points, lines and planes with the incidence relation between them. 

        The projective 3 – space satisfies the following axioms: 

A. Any two distinct points are contained in a unique line. 

B. Any three distinct non-collinear points, also any line and point not on the line are 
contained in a unique plane. 

C. Any two distinct coplanar lines intersect in a unique point. 

D. Any line not on a given plane intersects the plane in a unique point. 

E.  Any two distinct planes intersect in a unique line. 

        A projective space PG(3,q) over Galois field GF(q), q = p
m, for some prime number p 

and some integer m, is a 3 – dimensional projective space. 

        Now, some theorems on PG(3,q) p roved in [1] and [2] are given in the following. 

Theorem 1:  
        Every line in PG(3,q) contains exactly q + 1 points. 
 
Theorem 2: 
        Every point in PG(3,q) is on exactly q + 1 lines. 
 
Theorem 3:  
        Every plane in PG(3,q) contains exactly q

2 + q + 1 points. 
 
Theorem 4:  
        Every plane in PG(3,q) contains exactly q

2
 + q + 1 lines. 

 



IBN AL- HAITHAM J. FOR PURE & APPL.  SCI.            VOL.24 (3) 2011 
  

 
Theorem 5:  
        Every point in PG(3,q) is on exactly q2 + q + 1 p lanes. 
 
Theorem 6:  
        There exist q

3 + q2 + q + 1 points in PG(3,q). 
 
Theorem 7: 
        There exist q3 + q2 + q + 1 p lanes in PG(3,q). 
 
Theorem 8: 
        Any line in PG(3,q) is on exactly q +1 planes. 
 
Definition 1: [1] 
        A (k,n) – arc A in PG(3,q) is a set of k points such that at most n points of which lie in 
any plane, n  3. n is called the degree of the (k,n) – arc. 
 
Definition 2:  
        In PG(3,q), if A is any (k,n) – arc, then an (m-secant) of A is a plane ℓ such that                  
ℓ  A= m. 
 
Definition 3: [1,2] 
        A point N not on a (k,n)-arc A has index i if there exists exactly i (n –secants) of A 
through N, one can denote the number of points N of index i by Ci. 
 
Definition 4:  
        (k,n)-arc A is complete if it is not contained in any (k + 1,n)-arc. 
 
        From definitions 3 and 4, it is concluded that the (k,n)-arc is complete iff C0 = 0. Thus 
the (k,n)-arc is complete iff every point of PG(3,q) lies on some n-secant of the (k,n)-arc. 
 
Definition 5: [1,3] 
        Let T i be the total number of the i – secants of a (k,n) – arc A, then the type of A denoted 
by (Tn, Tn – 1, , T0).  
 
Definition 6: [1] 
        Let (k1,n) – arc A is of type (Tn, Tn – 1,, T0) and (k2,n) – arc B is of type (Sn,Sn – 1,,S0), 
then A and B have the same type iff T i = Si , for all i, in this case they are projectively 
equivalent. 
 
Theorem 9:  
        Let t(P) represents the number of 1-secants (planes) through a point P of a (k,n) – arc A 
and let T i represent the numbers of i – secants (p lanes) for the arc A in PG(3,q), then: 

1. t = t(P) = q
2 + q + 2 – k – 

( 1) ( 2)

2

 k k
 –  – 

( 1) ( 2) ( ( 1))

( 1)!

   



k k k n

n
 

2. T1 = k t 

3. T2 = 
( 1)

2

k k
 

4. T3 = 
( 1)( 2)

3!

 k k k
 

 



IBN AL- HAITHAM J. FOR PURE & APPL.  SCI.            VOL.24 (3) 2011 
  

 

5. Tn = 
( 1) ( 1)

!

  k k k n

n
 

6. T0 = q3 + q2 + q + 1 – k t – 
( 1)

2

k k
 – 

( 1)( 2)

3!

 k k k
 –  – 

( 1)( 2) ( 1)

!

   k k k k n

n
 

Proof : 

1. there exist (k – 1) 2-secants to A through P and there exist 
1

2

 
 
 

k
 (3-secants) to A 

through P, and so there exist 
1

1

 
  

k

n
 n - secants to A through P, and since there exist 

exactly                q2 + q + 1 p lanes through P, then the number of the 1-secants through P: 

t(P) = q2
 + q + 1 – (k – 1) – 

1

2

 
 
 

k
 –  – 

1

1

 
  

k

n
 

      = q2 + q + 2 – k – 
( 1) ( 2)

2

 k k
 –  – 

( 1) ( 2) ( 1)

( 1)!

   



k k k n

n
 = t. 

 
2. T1 = the number of 1-secants to A, since each point of A has t (1-secants) and the number 

of the points is k, then T1 = k t. 
 

3. T2 = the number of 2-secants to A, which is the number of planes passing through any two 

points of A. Hence T2 = 
2

 
 
 

k
 = 

( 1)

2

k k
. 

 
4. T3 = the number of 3-secants of A, which is the number of planes passing through any 

three points of A. Hence T3 = 
3

 
 
 

k
 = 

( 1)( 2)

3!

 k k k
. 

 

5. Tn = the number of n – secants p lanes to A, Tn = 
 
 
 

k

n
 = 

( 1) ( 1)

!

  k k k n

n
. 

 
6. q3 + q2 + q + 1  represents the number of all planes, then in a (k,n) – arc of PG(3,q),              

q3 + q2 + q + 1 = T0 + T1 + T2 + T3 +  + Tn 
T0 = q3 + q2 + q + 1 – T1 – T2 – T3 –  – Tn  
So 

T0 = q3 + q2+q+1–k t – 
( 1)

2

k k
 – 

( 1)( 2)

3!

 k k k
 –  – 

( 1)( 2) ( 1)

!

   k k k k n

n
. 

 
Theorem 10: 
        Let Ti represents the total number of the i – secants for a (k,n) – arc A in PG(3,q), then 
the following equations are satisfied: 

1. 
0


n

i

T i = q
3
 + q

2
 + q + 1 

2. 
1


n

i

i ! T i = k t + k (k – 1 ) + k (k – 1)(k – 2) +  + k (k – 1)  (k – n) 



IBN AL- HAITHAM J. FOR PURE & APPL.  SCI.            VOL.24 (3) 2011 
  

 

3. 
2


n

i

i (i – 1) T i = k (k – 1 ) + k (k – 1)(k – 2) + 
1

2
 k (k – 1)(k – 2)(k – 3) +  +     

1

( 2)!n
 

k (k – 1)  (k – n). 
 
Proof : 

1. 
0


n

i

T i represents the sum of numbers of all i – secants to A, which is the number of all 

planes in the space. Hence 
0


n

i

Ti = q3 + q2 + q + 1. 

 

2. T1 = k t, t = q2 + q + 2 – k – 
( 1)( 2)

2

 k k
 –  – 

( 1) ( 1)

( 1)!

  



k k n

n
,  

T2 = 
( 1)

2

k k
, T3 = 

( 1)( 2)

3!

 k k k
, T4 = 

( 1)( 2)( 3)

4!

  k k k k
, ,                                

Tn = 
( 1) ( 1)

!

  k k k n

n
 

1


n

i

i ! T i = T1 + 2 ! T2 + 3 ! T3 +  + n ! Tn 

                = k t + k (k – 1 ) + k (k – 1)(k – 2) +  + k (k – 1)  (k – n + 1) 
 

3. 
2


n

i

i (i – 1) T i = 2 T2 + 6 T3 + 12 T4 +  + n (n – 1) Tn 

                              = k (k – 1 ) + k (k – 1)(k – 2) + 
1

2
 k (k – 1)(k – 2)(k – 3) +  +             

1

( 2)!n
 k (k – 1)  (k – n + 1) 

 
Theorem 11: 
        Let Ri = Ri(P) represents the number of the i – secants (planes) through a point P of a 
(k,n) – arc A, in PG(3,q) then the following equations are satisfied: 

1. 
1


n

i

Ri = q2 + q + 1 

2. 
2


n

i

(i – 1)! Ri = (k – 1) + (k – 1)(k – 2) +   + (k – 1)(k – 2)  (k – n – 1)  

                             = 
1

1





n

i

(k – 1)  (k – i) 

Proof : 

1. 
1


n

i

Ri = R1 + R2 +   + Rn, 
1


n

i

Ri represents the sum of numbers of all the i – secants 

through a point P of the arc A, which is the number of the planes through P. Thus, 

1


n

i

Ri = q
2
 + q + 1. 

 
 
 



IBN AL- HAITHAM J. FOR PURE & APPL.  SCI.            VOL.24 (3) 2011 
  

 

2. 
2


n

i

(i – 1)! Ri = R2 + 2! R3 + 3! R4 +  + (n – 1)! Rn 

From proof (1) of theorem 9, there exist (k – 1) 2-secants to A through P, and there exist 
1

2

 
 
 

k
 3-secants to A through P, and so there exist 

1

1

 
  

k

n
 n-secants to A through P. 

Thus R2 = k – 1, R3 = 
1

2

 
 
 

k
, R4 = 

1

3

 
 
 

k
, , Rn = 

1

1

 
  

k

n
 

R3 = 
( 1)!

2!( 3)!





k

k
, R4 = 

( 1)!

3!( 4)!





k

k
, , Rn = 

( 1)!

( 1)!( )!



 

k

n k n
 

R3 = 
( 1)( 2)

2

 k k
, R4 = 

( 1)( 2)( 3)

3!

  k k k
, , Rn = 

( 1) ( ( 1))

( 1)!

  



k k n

n
 

2

n

i 

 (i – 1)! Ri = k – 1 + 
2!( 1)( 2)

2!

 k k
 + 

3!( 1)( 2)( 3)

3!

  k k k
 +  +  

                           
( 1)!( 1)( 2) ( ( 1))

( 1)!

    



n k k k n

n
 

                       = (k – 1) + (k – 1)(k – 2) +(k –1)(k –2)(k –3)+  + (k –1)(k – 2)  (k – (n–1))  

                             = 
1

1





n

i

(k – 1)  (k – i) 

 
Theorem 12: 
        Let Si = Si(Q) represent the numbers of the i – secants (p lanes) of a (k,n) – arc A through 
a point Q not in A, then the following equations are satisfied: 

1. 
0


n

i

Si = q2 + q + 1 

2. 
1


n

i

i Si = k 

Proof : 

1. 
0


n

i

Si  represents the sum of the total numbers of all i – secants to A through a point Q 

not in A, which is equal to the number of all planes through Q. Thus 
0


n

i

Si = q
2
 + q + 1. 

2. 
1


n

i

i Si = S1 + 2 S2 + 3 S3 +  + n Sn 

S1, S2, , Sn represent the numbers of the i – secants of the arc A through the point Q not 
in A.  
S1 is the number of the 1-secants to A, each one passes through one point of A.  
S2 is the number of the 2-secants to A, each one passes through two points of A.  
S3 is the number of the 3-secants to A, each one passes through three points of A.  
Also, Sn is the number of the n – secants to A, each one passes through n points of A. 

Since the number of points of the (k,n) – arc A is k, then 
1


n

i

i Si = k. 

 
 
 
 



IBN AL- HAITHAM J. FOR PURE & APPL.  SCI.            VOL.24 (3) 2011 
  

 
Theorem 13: 
        Let Ci be the number of points of index i in S = PG(3,q) which are not on a complete 
(k,n) – arc A, then the constants Ci of A satisfy  the following equations: 

(i)   




Ci = q3 + q2 + q + 1 – k 

(ii)   




i Ci = 
( 1) ( 1)

!

  k k k n

n
(q2 + q + 1 – n) 

where  is the smallest i for which Ci  0,  be the largest i for which Ci  0. 
Proof : 
        The equations express in different ways the cardinality  of the following sets 
(i) {Q  Q  S \ A} 
(ii) {(Q,)  Q   \ A,  an n – secant of A} 

for in (i), 




Ci represents all points in the space which are not in A, then                          




 Ci = q

3
 + q

2
 + q + 1 – k, and in (ii) 





i Ci represents all points in the space not in A, 

which are on n – secants of A, that is, each n – secant contains q
2
 + q + 1 – n points, and the 

number of the n – secants is 
 
 
 

k

n
, then  






i Ci =
 
 
 

k

n
(q

2
 + q + 1–n) = 

( 1) ( 1)

!

  k k k n

n
(q

2
 + q + 1 – n). 

 
Theorem 14: 
        If P is a point of a (k,n)-arc A in PG(3,q), which lies on an m-secant (plane) of A, then 
the planes through P contain at most (n – 1 ) q (q + 1) + m points of A. 
Proof : 
        If P in A lies on an m – secant (plane), then every other plane through P contains at most 
n – 1 points of A distinct from P. Hence the q

2
 + q + 1 planes through P contain at most              

(n – 1)(q
2 + q) + m points of A. 

 
 

References 
1. Al-Mukhtar, A.Sh., (2008) Complete Arcs and Surfaces in three Dimensional Projective 

Space Over Galois Field, Ph.D. Thesis, University  of Technology, Iraq. 
2. Kirdar,M.S. and Al-Mukhtar, A. Sh., (2009), Engineering and Technology Journal, On 

Projective 3-Space, Vol.27(8): 
3. Hirschfeld, J. W. P., (1998), Projective Geometries Over Finite Fields, Second Edition, 

Oxford University  Press. 
 
 
 
 
 
 
 
 
 
 



  2011) 3( 24المجلد           قیة     مجلة ابن الھیثم للعلوم الصرفة والتطبی
 
 

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  حول حقل كالوا

  
  

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بحث  من ھدفال         ، لبعض قیم GF(q)  ،q = pmحول حقل كالوا  تقدیم تعریف الفضاء الثالثي االسقاطيھو ھذا ال

p    وm  ان  اذp   عدد أولي وm عدد صحیح.  

طة، واالقواس المتكافئة ، دلیل النقn –، االقواس الكاملة، القاطع  (k,n) –كذلك أعطیت تعاریف االقواس         

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