2-Designs over GF(q)

Graphs and Combinatorics (1992) 8:381 389

Graphs and Combinatorics 9 Springer-Verlag 1992

2-Designs over

GF(q)

Hiroshi Suzuki Mathematical Sciences Division, Department of Arts and Sciences Osaka Kyoiku University, Kashiwara, Osaka, 582 Japan

Abstract. We give a construction of a series of 2-(n, 3, q2 ~_ q + 1; q) designs of vector spaces over a finite field GF(q) of odd characteristic. These designs correspond to those constructed by Thomas and the author for even characteristic. As a natural generalization we give a collection of m-dimensional subspaces which possibly become a 2-(n, m, 2; q) design.

1. Introduction

A t-(n, m, 2; q) design, or a t-design over GF(q), is a nonempty collection ~ of m-dimensional subspaces of an n-dimensional vector space over GF(q) with the property that any t-dimensional subspace is contained in exactly 2 members of ~. It is also known as the q-analogue of the classical t-design. It is easy to see that if ~ is the collection of all m-dimensional subspaces, automatically becomes a t-design over GF(q) for any t _< m. This design is called trivial. The first nontrivial example for t > 2 was given by Thomas [-2] when q = 2 and extended to the case q = 2r by the author [1]. Let K be a finite field GF(q") with q" elements and k be a subfield GF(q) with q elements, where q is a power of a prime p. Then K is an n-dimensional vector space

et[m]dnottoco,letiooofallm diensonalubsacesofanl [:

be its cardinality. It is well known that

qnqi i=O

IK]

Let U be a member of 2 number r let

' i.e., a 2-dimensional subspace of K. For each natural

Lr(U) = (al""a, lai~ U,i = 1,2 .... ,r), the subspace of K generated by all products of r elements of U, and

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H. Suzuki

Question. Suppose (n,(2r)[) = 1. W h e n does ~ r b e c o m e a 2- n,r + 1,

2

;q

design? It is not hard to see that the designs of T h o m a s and the author are in this setting with r = 2 and q even. In this p a p e r I show that ~3~ is actually a collection of r + 1 dimensional subspaces of K and that the average n u m b e r of the m e m b e r s of ~3r containing a 2-dimensional subspace is equal to J r + '2 l . So it is easy to see that if Nr becomes

(

a 2-design over GF(q), it is a 2- n,r + 1,

[r+ l ) 2

;q

design. M o r e o v e r if

2 _< r _< n - 4, the design is nontrivial. F o r r = 2 and q odd, we show that ~2 is actually a 2-design over GF(q). So by [1] N2 is always a 2-design over GF(q) for any prime p o w e r q under the condition (n, 4!) = (n, 6) = 1 in the question. In addition we construct a classical 2- ( n, r + 1, ( r +2 1 ) ) design for any natural n u m b e r r satisfying (n, (2r)!) = 1 using a finite abelian g r o u p of order n. This design is based on an analogy of our setting corresponding to the case q = 1. Schram at the O h i o State University informed the author in a recent letter that he obtained similar results.

2. Preliminary L e m m a s

I collect several l e m m a s in this section; in particular I show that ~ , is a collection of r + 1 dimensional subspaces if (n, r!) = 1. L e m m a 2.1. Suppose (n, sI) = 1.

(1) Let F(x, y) e k [x, y] be a homogeneous polynomial of degree at most s, and a and b be linearly independent elements of K over k. I f F(a, b) = O, F(x, y) is the zero polynomial. (2) Let F(x) ~ k[x] be a polynomial of degree at most s and a be an element of K not 9in k. I f F(a) = O, F(x) is the zero polynomial.

Proof. Let c = ab -1. T h e n c is an element of K not in k and it is a root of a polynomial F(x, 1) e k[x] of degree at m o s t s. So (2) implies (1). S u p p o s e F(x) ~ O. Since a is a root-of a polynomial F(x) ~ k[x] of degree at most s, the degree of the field extension [k(a) : k] divides s!. O n the other hand k(a) ~ K implies that [k(a) : kl divides n. Hence Ek(a):k] = 1 or a e k as (n,s!) = 1. A contradiction. Corollary 2.2. Suppose (n,r!) = 1. Then dimLy(U) = r + 1 and ~ c 9

Proof. Let U = (a, b). Then L , ( U ) - - (ar, ar-lb .... ,abr-l, br).

[Kl r+l

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383

Since a a n d b are linearly i n d e p e n d e n t over k, {ar, ar-lb .... , a b ' - l , b r} is a linearly i n d e p e n d e n t set b y L e m m a 2.1. L e m m a 2.3. Suppose (n,s!) = i. Then k K s = K, where K s = {aSia ~ K}.

Proof. Let s = tu. S u p p o s e k K t = kK" = K. T h e n k K s = kk'(Kt)" = k(kKt)" = kK" = K. So we m a y assume t h a t s is a p r i m e n u m b e r , a n d a divisor of q" - 1. Let a be an element of K such t h a t a s = 1. T h e n b y L e m m a 2.1 a is in k. So s divides q - 1. Let m = (q" - 1)/(q - 1). Since m = q , - 1 + ... + 1 -- n ( m o d s), s does n o t divide m. T h u s we have k K ~ = K as a Sylow s - s u b g r o u p of K x is c o n t a i n e d in k x. T h e following two L e m m a s on a q u a d r a t i c form over GF(q) of m i n u s t y p e are well k n o w n . But for c o m p l e t e n e s s a n d for the convenience of the readers who are not familiar with the facts we give proofs of them. L e m m a 2.4. Let e be a nonsquared element of k, i.e., e ~ k - k 2 . Let f(t, u) = t 2 - e u 2 be a quadratic form on a 2-dimensional vector space W = {(~, fl)[~, fl ~ k}. (1) A ( f ) = {f(~,fl)l~,fl E k} = k. (2) For any nonzero element c in k, f is equivalent to cf, i.e., there is a nonsingular

matrix(:

~)overksuchthat c "f(t, u) = f ( , t + 7u, fit + 6u).

Proof. (1) Sincef(ct, cu) = cZf(t, u), b y the definition o f f x i ( f ) = k 2 or A ( f ) = k. N o t e that k = k2U ek 2. S u p p o s e A ( f ) -- k 2. Since A ( f ) ~ - e k 2, - 1 does n o t b e l o n g to k 2. So we m a y a s s u m e t h a t e = - 1 a n d k 2 + k 2 c k 2. As k 2 - {0} is a finite multiplicative g r o u p , k 2 is a subfield of k with (q + 1)/2 elements. A c o n t r a d i c tion. (2) By (1) we find ~ a n d fl in k such t h a t c = f(~, fl). Let (~, 6) be a n o n z e r o vector o r t h o g o n a l to (a, fl) with respect to the bilinear form a t t a c h e d to f. Let d = f(7, 6). T h e n

So cd e - e k 2. Let cd = - ~ e 2. R e p l a c i n g (7, 6) by (Tc/e, 6c/e), we have

(;

fl

1

.)G 0

ct

0

1

0)

as desired. L e m m a 2.5. Let f be a quadratic form in Lemma 2.4. Let

Then I O ( f ) l = 2(q + 1).

o)}

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H. Suzuki

Proof.

Since f(a, f l ) = a 2 - - e f l 2 = O So U S 1 a n d So N S 1 = ~ , where

Si

=

implies a = f l = 0 ,

{(v>[veW--{0}}=

{ < v ) [ f ( ~ , f l ) e e'k 2 for a n y (~,fl) e
i = 0, 1. N o w O ( f ) acts transitively on the sets $1 a n d $2 (Witt's Theorem). F o r if

f(u) = f(v) # 0 a n d u # v, a linear m a p p i n g a u_~(w) = w

2B(u - v, w) ~ f ( u ~ v)

tu - v) cor-

r e s p o n d s to an e l e m e n t of O ( f ) sending u to v. H e r e B d e n o t e s the bilinear form a t t a c h e d to f. If v # 0, W is the o r t h o g o n a l direct sum of
3. [~rl The aim of this section is to p r o v e the following.

Proposition 3.1. Suppose (n, (2r)!) = 1. Let U, w e [ K ] .

Then Lr(U) ~ L r ( W ) implies

U=W. Proof. Let a be a n o n z e r o element in K. T h e n L r ( a U ) -- arLr(U). So we m a y assume that l e U a n d L r ( U ) - - < l , a , a 2 .... ,a~). Let W = < x , y ) a n d a s s u m e that L r ( W ) = L~(U). Since <1, a , . . . , a ~-1 ) f-) ( x ~, x r - l y ) @ 0, there is a n o n z e r o element ~x + fly in W such t h a t x r - l ( ~ x + fly) e ( 1 , a . . . . . a ~ - i ). Let F~(t) be a p o l y n o m i a l in k[t] of degree less t h a n or e q u a l to r such t h a t Fi(a) = x ~ - i y i. T h e n by our o b s e r v a t i o n a b o v e we m a y a s s u m e t h a t either deg(Fo) o r d e g ( F l ) is at m o s t r - 1 by a suitable c h a n g e of the basis of W. M o r e o v e r we claim t h a t Fi(t)" F i + 2 ( t ) = f i +1 (t) 2 in k [t], where i = 0, 1. . . . . r - 2. For

Fi(a)Fi+z(a) = x r - i y i x r - i - 2 y i + 2

~_ x Z r - Z i - Z y 2i+2

= ( x ' - i - l y i + l ) 2 = Fi+l(a) 2, F~(t)F~+z(t)

- F i + l ( t ) 2 is a p o l y n o m i a l of degree at m o s t 2r h a v i n g a as a root. So the claim follows from L e m m a 2.1 as a is n o t in k. Let r i = deg(Fi(t)). S u p p o s e r = r o. Since r i + ri+ z = 2ri+1 or r e - - r i + 1 ---r i ~ 1 - - r i + 2 , w e have

r=ro>r-l>_rl>r2>...>r

~.

Hence we have r~ = r - i as d i m L , ( W ) = r + 1. Similarly if r > ro, we have rj = j. Therefore b y replacing x a n d y if necessary, we m a y a s s u m e t h a t

x ~ - i y i e < l , a , a 2. . . . . ai). In particular, x ~ e k. So x belongs to k by L e m m a 2.1. N o w x ~ - l y e ( 1 , a ) = U implies y belongs to U. Therefore we have W = ( x , y ) c < l , a ) = U, or U = W, as desired.

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Corollary 3.2. Suppose (n, (2r)!) = 1. Then

Corollary 3.3. Suppose (n, (2r)!) = 1. If arLr(U) = L~(U) for some a in K x, Then a is in k. Proof. Since a~L,(U) = L,(aU), we have aU = U. Let b be a nonzero element of U. Then {ab, a2b, a3b} c U is linearly dependent. So a is a root of a quadratic equation ove k. It follows from Lemma 2.1 that a is in k. Remark. Let 5~ = t(W, B) Then by Corollary 3.2,

Hence

In other words, the average number of the members of ~r containing a 2-dimensional subspace is equal to

2 E~ ' ]

, as we mentioned in the introduction.

4. Proof of the Case r = 2, q Odd

The goal of this section is to prove the following. Theorem 4.1. Suppose (n,6)= 1 and q odd. Then ~2 is a 2-(n, 3,[~];q) design.

Moreover if n >_ 7, the design is nontrivial. Throughout this section assume (n, 6) = 1 and q odd. Let M be ~2 and L(U) be L2(U) for short. Evaluating the number of elements in the set

we need only to show the following inequality: 2(W) = I{B ~ ~ I W ~ B}[ < [32] = q2 + q + 1, for all w e [ K ] .

See also the remark in the last section.

386

H. Suzuki

L e m m a 4.2. Let W ~ L(U). Then there are three possibilities: (I) W "= (X2,y2) and U = ( x , y ) (II) W = ( x z , x y ) and U = ( x , y ) (III) W = ( x y , x 2 + ey 2) and U = ( x , y ) , where e is a f i x e d nonsquared element in

k,i.e.,eEk-k

z.

Proof. Since P r o p o s i t i o n 3.1 g u a r a n t e e s the uniqueness of the expression L(U), L(U) 9 ~ d e t e r m i n e s U a n d the set

s(u)

= {x21x 9 u}.

Suppose d i m ( ( S ( U ) N W ) ) = 2. T h e n 8(U) c o n t a i n s a basis of IV, say {x2,y2}. Since x a n d y are linearly i n d e p e n d e n t , U = ( x , y ) , a n d we have (I). Next assume t h a t d i m ( ( S ( u ) n W ) ) = 1. Let x 2 be a n o n z e r o element of $(U)N W, a n d c h o o s e y so t h a t ( x , y ) = W. Since

U1 = ( x y , y 2) = {(~x + flY)Yl~,fl 9 k} is a 2 - d i m e n s i o n a l s u b s p a c e of L(U), there is a n o n z e r o e l e m e n t (c~x + fly)y in U1NWas 4 = d i m U1 + d i m W > d i m L(U) = 3. See C o r o l l a r y 2.2. Since d i m ( ( S ( U ) N W ) ) = 1, yZ is n o t in W. So we m a y assume that (x + fly)y 9 U 1 N W. As

c~x2 + (x + fly)y = cox2 + x y + fly2 is not an element of S(U), 1 - 4~fl c a n n o t be z e r o for all c~in k. Hence fl = 0. N o t e that q is o d d by o u r a s s u m p t i o n . Therefore W = ( x 2, x y ) a n d U = ( x , y ) . W e have (II). F i n a l l y assume t h a t d i m ( ( S ( U ) N W ) ) = 0. Let U = ( x , y ) a n d UI = ( x y , y2). Then there is a n o n z e r o element (~x + fly)y in U 1 N W. Since d i m ( ( S ( U ) N W ) ) = 0, # 0. So we m a y replace ~x + fly by x a n d a s s u m e t h a t x y is in W. Let

U z = ( ( x + y ) x , ( x + y)y). Then by a similar a r g u m e n t we c a n find a n o n z e r o e l e m e n t (x + y)(c~x + fly) in U2 N W. By the s y m m e t r i c choice of x a n d y, we m a y a s s u m e t h a t c~ = 1 a n d

W = ( x y , ( x + y)(x + fly)) = ( x y , x 2 + fly2). Since S(U) N W = {0}, a n y e l e m e n t of the form x 2 + ~xy + flyZ does n o t b e l o n g to S(U). S ~ ~2 _ 4fl c a n n o t be zero for a n y element a in k. H e n c e fl 9 k - k z a n d fl can be written as fl = ey z, where 7 9 kX. R e p l a c i n g y b y 7-1y, if necessary, we have

W = ( x y , x 2 + ~yZ) a n d U = ( x , y). Hence we o b t a i n three cases d e s c r i b e d in L e m m a 4.2. L e m m a 4.3. The following holds:

N(I) = I{L(U) 9 N I U = ( x , y ) , W = ( x 2 , y 2 ) } I

-<(q+21)_q(q+21)

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387

Proof. Since N ( I ) is less than or equal to the number of unordered pairs of the projective points of W,, we have the bound. Remark. By the application of L e m m a 2.3 with a little further argument we can show the equality in L e m m a 4.3.

L e m m a 4.4. The following holds. N ( I I ) = I{L(U) e ~ I U

= (x,y),W

= ( x 2 , x y ) } l <_ q + 1.

Proof. Let W = ( x 2, a). Then by replacing y by a suitable element in U, we m a y assume that x y = a. Hence N ( I I ) is less than or equal to the number of projective points in W. Thus we have the bound. Remark. By L e m m a 2.3, we have the equality in L e m m a 4.4.

L e m m a 4.5. The followin9 holds. N ( I I I ) = I { L ( U ) 9 ~ ] U = ( x , y ) , W = ( x y , x 2 + ey2>}l

Proof. Let U = (x, y ) and W = ( x y , x 2 + eyE). Then (x 2 -- ey2) z = (x z + eyZ) 2 -- 4e(xy) 2

is an element of L ( W ) . Let N = I{(a 2 - e b 2 ) l ( a , b ) = W}l. Since L ( U ) = ( x y , x 2 + ey2,x 2 -- ey2), N(II1) is at most N. Since Nx is less than or equal f(t, u) = t z - eu 2,

=

[{a 2 -- e b Z l ( a , b ) = W}I

to the number

of quadratic

1GL(2,q)[ (q2_ 1)(q2_q) NI_<__ [O(f)l 2(q + 1)

forms equivalent

to

q ( q _ 1)z

2

by L e m m a 2.5. Since the quadratic form f(t, u) is equivalent to a. f(t, u) for any nonzero a in k by L e m m a 2.4, N-

1 q-1

q ( q - 1)

N1< - 2

'

as required. P r o o f o f Theorem 4.1. N o w we can evaluate the number 2(W). It follows from L e m m a 4.2 that each L ( U ) containing W satisfies one of the conditions (I), (II) or (III). The numbers of L ( U ) of each type have been bounded in L e m m a 4.3, 4.4, and 4.5. Therefore

2(U) _< -q(q- ++ 1)( q + l ) + q(q - - < 1) 2 2 -

-

_ q2

+ q - ~ 1.

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H. Suzuki

Hence ~ is a 2-(n, 3,q 2 + q + 1;q) design. I f n _> 7, the design is nontrivial because

We have Theorem 4.1.

5. q = 1 c a s e

( ( r + l ) ) In this section we construct a 2- n, r + 1, 2

design by a similar idea.

Let G be a finite abelian group of order n. Let ( rG ) denote the collection of all

r-elementofG. F o r U e ( G 2 ) , l e t L j U ) = { a l . . . a ~ l a i e U ,

i = l .... , r}. Finally let

Note that L~(U) and ~ are different from ones we defined in the introduction, but they are natural q = 1 analogies of them. 1

T h e o r e m 5.1.

Lemma {xr, x r - l y

5.2. .....

Suppose

ondr

-

(n,r!)=l.

r ,sa 2,nr+l k

Let

1,r+l,,l'' es, n k 2

U={x,y}e[~i.

tz/

//

Then

L~(U) =

y'} e ( r +G l ) "

Proof. We may assume y = I. Then the assertion is trivial. Note that G is abelian. L e m m a 5.3.

Suppose (n,(2r)I)= 1: Let U, W e l ; \

). Then Lr(U ) = Lr(W)implies f

U=W. Proof. We may assume U = {1,a} and W = {x,y}. Let a i = x'-~(i)y e~i), where 0 <_d(i) <_r and 0 _< i <_ r. Then X2r-d(i)-d(i+2)yd(i)+d(i+2) : aia I+2 = (ai+l) 2 - - x2r-2d(i+l)y2d(i+2). yd(i)+d(i+E)-Ed(i+l), or d(i) + d(i + 2) - 2d(i + l) = 0 by our assumption. So d(i) - d(i + 1) equals d(i + 1) - d(i + 2). Thus we have d(i) = i or d(i) = r - i. Either case yields { 1, a} = {x, y}. So

x d(i)+d(i+E)-Ed(i+l) =

Lemma

5.4. Let 2(U) = I{B e ~IU c B} I. If 2({1, a}) >

r +2 11 for any element a

i n G - {1},~risa2- ( n,r + 1, (r +2 l))designundertheassumptionof Theorem5.1.

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P r o o f . Let x be an element in G. Then by our a s s u m p t i o n there is an element y in G

such that y" = x. So xL,(U)

= L,(yU).

Therefore Mr is invariant under the translations by the elements of G. Hence 2({1,a}) = 2 ( { x , x a } ) . N o w the assertion is an easy consequence of the fact that

P r o o f o f T h e o r e m 5.1. Let Uij = {ai/'J,a"-')/'J}, each U~ is well defined. As

1 _<j _< i _< r. Since (IGl,(2r)[) -- 1,

(ai/O)r-i(a(i-')/'J)i= (al/~

1,

i - j = a,

contains the set {1, a}. By L e m m a 5.4, it suffices to show that Uffs are all different. Suppose a ~/'j = a t/'u and a"-r)/rJ = a,-r)/r,. Then i/j = t/u, (i - r)/j = (t - r)/u, or r/j = flu. So j = u. Hence we have i = t and j -- u in this case. Suppose a i/rJ a (t-')/'" and a (i-')/'j = a t/". Then i/j = (t - r)/u, (i - r)/j = t/u. We obtain - r / j = r/u, a contradiction. Thus L(Uo)

=

,{U,i,l < j < i < r } , _

=(r+2

1).

This proves T h e o r e m 5.1.

References 1. Suzuki, H. (1990): 2-Designs over GF(2"), Graphs and Combinatorics 6, 293-296 2. Thomas, S. (1987): Designs over finite fields, Geometricae Dedicata 24, 237-242

Received: September 13, 1989