2-designs over GF(2^m)

Graphs and Combinatorics 6, 293-296 (1990)

Graphsand Combinatorics © Springer-Verlag 1990

2-Designs over

GF(2m)

H i r o s h i Suzuki Mathematical Sciences Division, Department of Sciences and Arts Osaka Kyoiku University, Tennoji Osaka 543, Japan

Dedicated to Professor Tuyosi Oyama on his 60th Birthday (n, k, 2; q) design is a collection ~ of k-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 called a design over a finite field or a q-analogue t-design. The first nontrivial example for t > 2 was given by S. Thomas. Namely, he constructed a series of 2 - (n, 3, 7; 2) design for all n > 7 satisfying (n, 6) = 1. Under the same restriction on n, we show that the base field of Thomas' design is extensible to GF(2m), i.e., we construct a 2 - (n, 3,2 TM + 2" + 1;2") design for all m > 1. A b s t r a c t . A t --

1. I n t r o d u c t i o n

In [2], S. T h o m a s defined a n o t i o n of a special triangle a n d showed that the collection of 3 - d i m e n s i o n a l spaces s p a n n e d b y special triangles forms a 2 - (n, 3, 7; 2) design, if (n, 6) = 1 a n d n > 7. A t the e n d of the second section in [2], he gave an alternative definition of ~ . Let V = GF(2") be an n - d i m e n s i o n a l vector space over GF(2), which also has a structure of a field. Let ~ be the set of all 2 - d i m e n s i o n a l subspaces of V. F o r each U = {0, a, b, c} ~ Y, a -1, b -1 a n d c -1 s p a n a 3 - d i m e n s i o n a l subspace, say U*. Then

{u'Iv N o w it's n o t h a r d to check t h a t U* = (y, yx, yx2), where y = a-lbc -x a n d x = ab -1. So

= {(y, yx, yx2)]O ~ y ~ Gr(Z"),x E G F ( 2 " ) - GF(2)}. W e reached a p o i n t to define o u r ~ . Let K = GF(q") with q = 2" a n d k = GF(q). T h e n K is an n - d i m e n s i o n a l vector space over k. F o r each y ~ K × = K - {0} a n d x ~ K - k, L(y, x) denotes the s u b s p a c e over k s p a n n e d b y y, yx a n d yx 2. Let

= {L(y,x)[yeK×,xeK-

k}.

In the following, we show t h a t ~ is a 2 - (n, 3, q2 4. q + 1; q) design u n d e r the c o n d i t i o n t h a t (n, 6) = 1. A t - (n, k, 2; q) design ~ is said to be nontrivial, if 2 ¢ 0 a n d at least one s u b s p a c e of d i m e n s i o n k is n o t a m e m b e r of ~ .

H. Suzuki

294

Theorem. If (n,6) = 1, there is a 2 - (n, 3,q2 + q + t;q) design. The design is

nontriviaI if n > 7.

2. Proof of the Theorem Let k = GF(q") with q = 2 m, and let k, L(y, x) and ~ be as in Introduction.

L e m m , 1. Let f(t) be a nonzero potynomiat over t~ of degree at most 4. If f(t) has a root c~in K, ~ belongs to k.

Proof. Since (n, 4!) = (n, 6) -- 1, there is no nontrivial field extension of k in K of degree less than 5. The assertion follows. L e m m a 2. ~ is a collection of subspaces of dimension 3.

Proof. Suppose c%y + cqyx + azyx z = O, with % , ~1, ~2 ff k. Since y ~ K ×, it yields

% + ~ax + ~2x 2 = O. As x is not an element of k, we must have c% = el = ~2 = 0 by L e m m a 1. L e m m a 3. Let L(y, x) and L(b, a) be members of N. Then L(y, x) = L(b, a) if and

only if

y = (c~a + fl)Zb,

x = (Ta + 6)/(~a + fl),

where c¢ fl, 7, 6 ~ k with ~6 - ? f l ¢ O. Proof. Suppose y = (eta + fl)2b, and x = (Ta + 5)/(aa + fl). Then L(y,x) = ((aa + fl)2b,(Ta + 6)(aa + fl)b,(Ta + 6)2b}. Since the characteristic of the field K is 2,

(~a + ~)2 = ~2a2 + ~2,

(~a + ~)2 = ~ a 2 + ~:

and

aO q- yfl =/=0 • ~202 -- 72fl 2, we have L(y,x) = L(b,a). Conversely, assume L ( y , x ) = L(b,a). N o w we m a y assume that b = 1, for L(b-ly, x) = t ( 1 , a ) . Since

(y, yx) fl ( 1 , a 2) ¢ O, there is a nonzero element (aa + fl)2 in (y, yx). As L(1, x) = L ( y - l , a ) = L(y-l(aa + fl)2,(7a + c~)/(aa + fl)) with suitable Y, 6 e k, there exist polynomials f, 9 and h in k[t] such that

2-Designs over GF(2")

295

y(x) = y - l ( . a +

g(x) = y-l(c~a + fl)(ya + 6),

and

h(x) = y-i(Ta + 6) 2. Since (ca + fl)2 belongs to (y, yx), we may assume that deg(f) < 1,

deg(g) _< 2

and

deg(h) < 2.

On the other hand

J(x)h(x) = g(x)2. Since x does not belong to k and the degrees of the polynomials f . h and g2 are at most 4, the coefficients of the equal power of x in the equation above must coincide by Lemma 1. So d e g ( f , h) < 3 implies deg(g) _< 1. As the subspace L(1, x) = (f(x), O(x), h(x)) is of dimension 3, we have deg(f) = 0 and deg(g) = 1. Now the assertion easily follows.

Corollary 4. If L(y, x) = L(b, a), (y, yx 2) = (b, ba2). Corollary 5.

(qn _ 1)(qn _ q) 1~[ -- (q2 1)(q2 _ q)

Let U be a 2-dimensional subspace of K and = I{B

V c B}I.

Our goal is to show that 2(U) does not depend on the choice of U and actually the number is equal to q2 + q + 1. Then the design is nontrivial by Corollary 5, ifn > 7. Since ~ is invariant under the multiplications by any element of K ×, we may assume that U contains 1, namely U = ( 1 , a ) with a e K - k. Let U c B = L(y,x) e ~. Thanks to Corollary 4, we can define a 2-dimensional subspace W = (y, yx 2 ) of B, which is independent of the choices of y and x. Lemma 6. U = W, if and only if B = L(1,b), where b 2 = a.

Proof. 'If' part is clear. Suppose (1, a) = (y, yx2). Since 1 belongs to (y, yx2), we may assume y = 1 by Lemma 3. Then x 2 E W, and the assertion easily follows from Lemma 3. Assume U ¢ W. Then U n W is a one dimensional subspace of U. By Lemma 3, we may choose y so that ( y ) = U A W. Let c be an element of K such that (y, yc) = U. Then (1, c) = y - l U c y-lL(y,x) = L(1, x), and

(l,c) = y-lU # y-lW = (l, x2). Hence there exist uniquely determined elements "o, cq and e2 in k such that ~2x 2 + c q x + c ~ o + c = O with cq ¢ O. In particular we have the following Lemma.

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H. Suzuki

Lemma 7. I f

O~2 :

O, there are q(q -- 1) possibilities of x's.

Finally we treat the case ~2 ~ 0. Lemma 8. For each fixed o~2 ¢

O,

there are q(q - 1) possibilities of x's.

Proof. Substituting z = ~-1~2x we have the following equation. Z 2 - - Z = ~0(Z12(X2 -'~ ~ 1 - 2 ~ 2 C .

By Hilbert's Theorem 90 (see for example [1]), this equation has solutions ( and ( + 1 in K, if and only if

Trr/v(~o~2~2 + e~-2~2c) = 0, where F = GF(2) and TrK/F denotes the trace function of the field extension K/F. Since (n, 2) • 1,

Trr/p(%~;2~2) = Trklv(TrK/k(%~;2~2))= n" Trk/e(Cto~-~2~2) =

Hence for fixed ~1 and ~2, q/2 choices of ~o give the value 0 and q/2 choices of ~0 give the value 1 for Trk/F(%~T2~2). Thus TrKIV(%~72~2 + ~]-2~2c) = 0 holds for q/2 choices of %, in which case we have two solutions in K. Therefore for each fixed ~2 there are q(q - 1) possibilities of x's. Now we can evaluate 2(U). By Lemma 3, for a fixed y we have q(q - 1)choices ofx expressing the same member B = L(y, x) o f ~ , we may simply divide the number of the possibilities of x's by q(q - 1). Suppose W ¢ U then there are q + 1 choices of W rGU = ( y ) . So we have 2(U)= l+(q+

1)(l+(q--1))=q2+q+l,

by Lemma 6, 7, and 8 as required.

References 1. Lang,S.: Algebra. Addison-Wesley(1965) 2. Thomas, S.: Designs over finite fields. Geom. Dedicata 24, 237-242 (1987)

Received: June 18, 1989 Revised: April 20, 1990