Graphs and Combinatorics (1992) 8:381 389
Graphs and Combinatorics 9 Springer-Verlag 1992
2-Designs over
GF(q)
Hiros...
8 downloads
480 Views
336KB Size
Report
This content was uploaded by our users and we assume good faith they have the permission to share this book. If you own the copyright to this book and it is wrongfully on our website, we offer a simple DMCA procedure to remove your content from our site. Start by pressing the button below!
Report copyright / DMCA form
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
382
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
2-Designs over GF(q)
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)}
384
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 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.
2-Designs over GF(q)
385
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
-