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

 
Some Results on The Complete Arcs in Three Dimensional 

Projective Space Over Galois Field 
 

A.SH. Al- Muk htar  
Departme nt of Mathematics, College of Education I bn-Al-Haitham,Unive rsity 
of Baghdad  
Received in : 11   May  2011 
Accepte d in : 16  June  2011 
 
Abstract 
        The aim of this p aper is t o introduce the definition of p rojective 3-sp ace over Galois field 
GF(q), q = p

m
, for some prime number p and some integer m. 

        Also t he definitions of (k,n)-arcs, comp lete arcs, n-secants, t he index of the point and the 
p rojectively  equivalent arcs are given. 
        M oreover some theorems about these notations are p roved. 
 
Keywords: arcs, index, p lane. 

 
 

Introduction: [1] 
        A p rojective 3 – sp ace PG(3,K) over a field K is a 3 – dimensional p rojective sp ace 
which consist s of p oints, lines and planes with the incidence relation between them. 

        The projective 3 – sp ace satisfies the following axioms: 

A.  Any  two dist inct p oints are contained in a unique line. 

B. Any  three distinct non-collinear p oints, also any line and p oint not on the line are 
contained in a unique p lane. 

C. Any  two dist inct coplanar lines intersect in a unique p oint. 

D. Any  line not on a given p lane intersects t he plane in a unique point. 

E.  Any  two dist inct p lanes intersect in a unique line. 

        A p rojective sp ace PG(3,q) over Galois field GF(q), q = p
m
, for some p rime number p 

and some integer m, is a 3 – dimensional p rojective sp ace. 

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

The orem 1:  
        Every  line in PG(3,q) contains exactly q + 1 p oints. 
 
The orem 2: 
        Every  p oint in PG(3,q) is on exactly q + 1 lines. 
 
The orem 3:  
        Every  p lane in PG(3,q) contains exactly q

2
 + q + 1 p oints. 

 
The orem 4:  
        Every  p lane in PG(3,q) contains exactly q

2
 + q + 1 lines. 

 



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

 
The orem 5:  
        Every  p oint in PG(3,q) is on exactly q

2
 + q + 1 p lanes. 

 
The orem 6:  
        There exist q

3
 + q

2
 + q + 1 p oints in PG(3,q). 

 
The orem 7: 
        There exist q

3
 + q

2
 + q + 1 p lanes in PG(3,q). 

 
The orem 8: 
        Any  line in PG(3,q) is on exactly q +1 p lanes. 
 
De fini tion 1: [1] 
        A (k,n) – arc A in PG(3,q) is a set of k p oints such that at most n p oints of which lie in 
any p lane, n  3. n is called the degree of the (k,n) – arc. 
 
De fini tion 2:  
        In PG(3,q), if A is any (k,n) – arc, then an (m-secant) of A is a p lane ℓ such that                  
ℓ  A= m. 
 
De fini tion 3: [1,2] 
        A p oint 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. 
 
De fini tion 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 comp lete iff C0 = 0. Thus 
the (k,n)-arc is comp lete iff every  p oint of PG(3,q) lies on some n-secant of the (k,n)-arc. 
 
De fini tion 5: [1,3] 
        Let Ti be the total number of the i – secants of a (k,n) – arc A, t hen the ty p e of A denoted 
by  (Tn, Tn – 1, , T0).  
 
De fini tion 6: [1] 
        Let (k1,n) – arc A is of t y p e (Tn, Tn – 1,, T0) and (k2,n) – arc B is of ty p e (Sn,Sn – 1,,S0), 
then A and B have the same ty p e iff Ti = Si , for all i, in this case they are p rojectively  
equivalent. 
 
The orem 9:  
        Let t(P) rep resents the number of 1-secants (p lanes) through a point P of a (k,n) – arc A 
and let T i represent t he numbers of i – secants (p lanes) for the arc A in PG(3,q), t hen: 

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.  S CI.            VOL. 24 (3) 2011 
  

 

5. Tn = 
( 1) ( 1)

!

  k k k n

n
 

6. T0 = q
3
 + q

2
 + 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                q
2
 + q + 1 p lanes through P, t hen the number of the 1-secants through P: 

t(P) = q
2
 + q + 1 – (k – 1) – 

1

2

 
 
 

k
 –  – 

1

1

 
 

 

k

n
 

      = q
2
 + 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 t o 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 p assing through any two 

p oints of A. Hence T2 = 
2

 
 
 

k
 = 

( 1)

2

k k
. 

 
4. T3 = the number of 3-secants of A, which is the number of p lanes p assing through any 

three p oints of A. Hence T3 = 
3

 
 
 

k
 = 

( 1)( 2)

3!

 k k k
. 

 

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

k

n
 = 

( 1) ( 1)

!

  k k k n

n
. 

 
6. q

3
 + q

2
 + q + 1  represents the number of all p lanes, then in a (k,n) – arc of PG(3,q),              

q
3
 + q

2
 + q + 1 = T0 + T 1 + T 2 + T 3 +  + T n 

T0 = q
3
 + q

2
 + q + 1 – T 1 – T2 – T3 –  – Tn  

So 

T0 = q
3
 + q

2
+q+1–k t – 

( 1)

2

k k
 – 

( 1) ( 2)

3!

 k k k
 –  – 

( 1) ( 2) ( 1)

!

   k k k k n

n
. 

 
The orem 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

Ti = q
3
 + q

2
 + q + 1 

2. 
1


n

i

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



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

 

3. 
2


n

i

i (i – 1) Ti = 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

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

p lanes in the sp ace. Hence 
0


n

i

Ti = q
3
 + q

2
 + q + 1. 

 

2. T1 = k t, t = q
2
 + 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 ! Ti = T 1 + 2 ! T 2 + 3 ! T 3 +  + 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) Ti = 2 T 2 + 6 T 3 + 12 T 4 +  + 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) 

 
The orem 11: 
        Let Ri = Ri(P) represents the number of the i – secants (p lanes) through a p oint P of a 
(k,n) – arc A, in PG(3,q) then the following equations are satisfied: 

1. 
1


n

i

Ri = q
2
 + 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 t he number of the p lanes through P. T hus, 

1


n

i

Ri = q
2
 + q + 1. 

 
 
 



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

 

2. 
2


n

i

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

From p roof (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) 

 
The orem 12: 
        Let Si = Si(Q) rep resent 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 = q
2
 + 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 t hrough a p oint Q 

not in A, which is equal to the number of all p lanes through Q. T hus 
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 t he numbers of t he i – secants of the arc A t hrough the p oint Q not 
in A.  
S1 is the number of the 1-secants t o A, each one passes t hrough one point of A.  
S2 is the number of the 2-secants t o A, each one passes t hrough two p oints of A.  
S3 is the number of the 3-secants t o A, each one passes t hrough three p oints of A.  
Also, Sn is the number of the n – secants to A, each one p asses through n p oints 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.  S CI.            VOL. 24 (3) 2011 
  

 
The orem 13: 
        Let Ci be the number of p oints of index i in S = PG(3,q) which are not on a comp lete 
(k,n) – arc A, t hen the constants Ci of A satisfy  the following equations: 

(i)   




Ci = q
3
 + q

2
 + q + 1 – k 

(ii)   




i Ci = 
( 1) ( 1)

!

  k k k n

n
(q

2
 + q + 1 – n) 

where  is the smallest i for which Ci  0,  be the largest i for which Ci  0. 
Proof : 
        The equations exp ress in different ways t he 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 p oints in the sp ace which are not in A, then                          




 Ci = q3 + q2 + q + 1 – k, and in (ii) 





i Ci represents all points in the sp ace not in A, 

which are on n – secants of A, t hat is, each n – secant contains q
2
 + q + 1 – n p oints, 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). 

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

2
 + q + 1 p lanes through P contain at most              

(n – 1)(q
2
 + q) + m p oints of A. 

 
 

Re ferences 
1. Al-M ukhtar, A.Sh., (2008) Complete Arcs and Surfaces in three Dimensional Projective 

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

Projective 3-Sp ace, 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
 
 

ثالثي االبعاد          فضاء اسقاطي االقواس الكاملة في  حولبعض النتائج 
  حول حقل كالوا

  
  

  آمال شھاب المختار
جامعة بغداد،ابن الھیثم  - كلیة التربیة،قسم الریاضیات   

 
 

2011آیار  11:استلم البحث في   
  2011حزیران  16  :قبل البحث في 

  
  
  

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

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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