Diskrete Mathematik und Optimierung

Transcription

1 Diskrete Mathematik und Optimierung Steffen Hitzemann and Winfried Hochstättler: On the Combinatorics of Galois Numbers Technical Report feu-dmo Contact: FernUniversität in Hagen Lehrgebiet Mathematik Lehrstuhl für Diskrete Mathematik und Optimierung D Hagen

2 2000 Mathematics Subject Classification: 05A10, 51D25 Keywords: interval decomposition, lattice of subspaces, Galois numbers

3 On the Combinatorics of Galois Numbers Steffen Hitzemann Winfried Hochstättler January 29, 2009 Abstract We define interval decompositions of the lattice of subspaces of vector spaces of finite dimension. We show that such a decomposition exists if and only if there exists a family of linear forms with certain properties. As applications we prove that all finite-dimensional real vector spaces admit an interval decomposition while GF (2) n has an interval decomposition if and only if n 4. On the other hand we present an interval decomposition of GF (3) 5. This partially answers a question of Faigle [4, 1]. 1 Introduction Goldmann and Rota [2] defined the Galois numbers G q n as the total number of linear subspaces of GF (q) n and showed that they satisfy the recursion G q 0 = 1, G q 1 = 2. G q n = 2G q n 1 + (q n 1 1)G n 2 for n 2. Ulrich Faigle [4, 1] asked whether this has an immediate combinatorial interpretation in the following sense: Is it always possible to partition the lattice of subspaces of GF (q) n into two intervals of length n 1 and (q n 1 1) intervals of length n 2, for n 2? We consider such interval decompositions for vector spaces F n of finite dimension over arbitrary fields F and show that the existence of such a decomposition is equivalent to the existence of what we call pointwise irreflexive and antisymmetric linear forms (Theorem 1). This immediately implies that for n 3 an interval decomposition of F n exists only if F n 1 admits an 1

4 interval decomposition. Also we can show that R n always has an interval decomposition. Considering finite fields, we present an algorithm that, given all (canonical) interval decompositions of GF (q) n 1 constructs all (canonical) interval decompositions of GF (q) n if they exist. This is used to show that GF (2) n has a unique (canonical) interval decomposition for n 4 and no such decomposition if n 5. On the other hand we present an interval decomposition of GF (3) 5 and report on an implementation of a special version of our algorithm that shows the existence of 52 such (canonical) decompositions with a certain structure. The paper is organized as follows. In the next section we introduce the notation and prove the main results. In Section 4 we derive the algorithms, which are applied to GF (2) and GF (3) in the following section. We conclude with some remarks and open problems. Our notation should be fairly standard. If not explicitly defined otherwise, F will denote an arbitrary field, F := F\{0}, V a vector space of finite dimension n 2 over F, and for a set or X V we denote by X the linear closure of X, if X = {q} we simply write q. 2 The Main Theorem 2.1 Interval Decompositions Definition 1. An interval decomposition of the lattice of subspaces of V is a triple (p 0, H 0, m) where 1. p 0 V \ {0} is a point, i.e. it generates a one-dimensional subspace U 0 = p 0 of V. 2. H 0 is a co-dimension one subspace, i.e. a hyperplane of V such that p 0 H 0 and 3. m : Q H is a bijection between the set Q of one-dimensional subspaces, disjoint from U 0 and H 0, represented by suitable points q, and the set H of hyperplanes different from H 0 that do not contain p 0, such that q m( q ) for all i I and the intervals [ q, m( q )] in the lattice of subspaces of V are pairwise disjoint. Example 1. Let n = 2, p 0, h 0 V \ {0} and m : V \ {p 0, h 0 } V \ {p 0, h 0 } the identity map. Then clearly, (p 0, h 0, m) is an interval decomposition. 2

5 Assume we are given an interval decomposition (p 0, H 0, m). Let (p i ) i I denote the generators of the one-dimensional subspaces of H 0, i.e. I is a suitable index set and each one-dimensional subspace of H 0 is generated by a p i for a unique i I. Proposition 1. For all i j I p 0, p i, p j are linearly independent. Proof. Assume λ 0 p 0 +λ i p i +λ j p j = 0. Then λ 0 p 0 = (λ i p i +λ j p j ) U 0 H 0 = {0}. Hence λ 0 = 0 and the claim follows since p i, p j span different subspaces of H 0. Proposition 2. For all i, j I and all β, β F where either i j or β β we have that p i + βp 0 and p j + β p 0 are linearly independent. Proof. 0 = λ(p i + βp 0 ) + µ(p j + β p 0 ) = λp i + µp j + (λβ + µβ )p 0. And the claim follows from Proposition 1. Proposition 3. For all i I and β F the subspace p i +βp 0 does neither lie in the filter generated by U 0 in the lattice of subspaces of V nor in the ideal of H 0. Proof. Obviously, p i + βp 0 is not a multiple of p 0. Assuming p i + βp 0 H 0 implies β 1 (p i + βp 0 p i ) = p 0 H 0, a contradiction. Hence, the points q i,β := p i + βp 0 for i I generate pairwise different one-dimensional subspaces that are disjoint from p 0 and H 0. On the other hand assume that q generates such a one-dimensional subspace of V. By the rank formula of linear algebra there exists a point 0 p := α 1 p 0 + α 2 q H 0. Clearly, we must have α 2 0 α 1. Hence q = α2 1 p + α2 1 ( α 1 )p 0. We conclude that there exists some 0 γ F and some i I such that q = γq i,β. Thus, we have Q = { q i,β i I}. 2.2 Pointwise irreflexive and antisymmetric linear forms Denote by H i,β = m( q i,β ) the hyperplane which matches q i,β = p i + βp 0 in the interval decomposition. Then H i,β is the kernel of the linear form σ i,β : V F defined by 0 if p H i,β H 0 σ i,β (p) := 0 if p = p i + βp 0 (1) β if p = p i. 3

6 Lemma 1. For p j H 0 we have p j + β p 0 H i,β σ i,β (p j ) = β. Proof. By definition σ i,β (p 0 ) = 1 and hence σ i,β (p j + β p 0 ) = σ i,β (p j ) β. Definition 2. Let H 0 be a hyperplane of V and p i for i I as above denote a set of generators of the points of H 0. Let S = {σ i,β : V F i I, β F } be a set of linear forms. We say that S is pointwise irreflexive and antisymmetric on H 0, if for all i I we have σ i,β (p i ) 0 and for all i j I σ j,β (p i ) = σ i,β (p i ) σ i,β (p j ) σ j,β (p j ). Example 2. Let V = R n be a finite-dimensional vector space over the reals with Euclidean norm 2, e 0 the first unit vector and H 0 = e 0 its orthogonal complement. As generators for the one-dimensional subspaces of H 0 we choose those p i H 0 such that p i = 1 and the first non-zero coordinate is positive. For such a p = p i and β R we define σ p,β : V F as follows: σ p,β (s) := ( e 0 + βp )s. (2) Then the linear forms are irreflexive, since for p as above we have σ p,β (p) = β p 2 = β. Now let p H 0 be another vector of unit length where the first non-zero coordinate is positive and assume that β (p p) = σ p,β (p) = σ p,β(p) = β. Then σ p,β (p ) = βp p = β (p p ) 2. Hence, σ p,β (p ) = σ p,β (p ) = β (p p ) 2 = 1. By Cauchy-Schwarz inequality this implies p = ±p. Since both are vectors of unit length where the first non-zero coordinate is positive we necessarily have p = p and β = β. Hence, the linear forms are also pointwise antisymmetric. Proposition 4. Let H 0, p i, i I and S be as in Definition 2. Let W H 0 be a subspace of V of dimension at least two. Define H W 0 := H 0 W, I W := {i I p i H W 0 } and S W = {(σ i,β ) W : W F i I W, β F }. Then S W is pointwise irreflexive and antisymmetric on H W 0. Proof. Clearly, the p i, i I W are a set of generators of the points of H W 0 and the validity of the other two axioms is inherited. 4

7 2.3 The equivalence Theorem 1. Let p 0 V \{0} be a point, H 0 a hyperplane of V not containing U 0 = p 0. Then there exists a bijection m : Q H such that (p 0, H 0, m) is an interval decomposition of the lattice of subspaces of V if and only if there exists a set S := {σ i,β : V F i I, β F } of linear forms that is pointwise irreflexive and antisymmetric on H 0. Proof. First assume there is an interval decomposition and define S = {σ i,β } i,β as in (1). Then by definition σ i,β (p i ) = β 0. To verify the second condition we assume for a contradiction that for some i j I σ j,β (p i ) = σ i,β (p i ) = β and σ i,β (p j ) = σ j,β (p j ) = β. By Lemma 1 we have p i + βp 0 H j,β as well as p j + β p 0 H i,β. Hence {p i + βp 0, p j + β p 0 } [p i + βp 0, H i,β ] [p j + β p 0, H j,β ], contradicting the properties of an interval decomposition. Now assume that a pointwise irreflexive and antisymmetric set of linear forms is given and define H i,β = m(p i + βp 0 ) := {p i + βp 0 } (H 0 ker(σ i,β )). We claim that (p 0, H 0, m) is an interval decomposition. Clearly, none of the p i + βp 0 is contained in H 0. The assumption p 0 H i,β yields p 0 = λ(p i + βp 0 ) + z for some 0 z H 0 ker(σ i,β ) and thus the contradiction p 0 H 0. (Note, that λp i ker(σ i,β ) z). Hence it suffices to verify [p i + βp 0, H i,β ] [p j + β p 0, H j,β ] for all (i, β) (j, β ). Assume for a contradiction there exists W [p i + βp 0, H i,β ] [p j + β p 0, H j,β ]. We conclude that p i + p j + (β + β )p 0 W. Since p 0 W there exists some k I, λ F such that λp k = p i + p j and Lemma 1 implies that σ i,β (p k ) = λ 1 (β + β ) = σ j,β (p k ). Since σ i,β (p k ) = λ 1 (σ i,β (p i ) + σ i,β (p j )) and σ i,β (p i ) = β we conclude that σ i,β (p j ) = β. By symmetry we also have σ j,β (p i ) = β. Hence σ j,β (p i ) = σ i,β (p i ) = β and σ i,β (p j ) = σ j,β (p j ) = β contradicting that S is pointwise irreflexive and antisymmetric. 5

8 Theorem 1 and Proposition 4 imply: Corollary 1. If F n has an interval decomposition, then so has F k 2 k n. for all And Example 2 yields Corollary 2. If n 2, then R n admits an interval decomposition. Remark 1. We might as well consider the set of linear forms as in Definition 2 as linear forms defined only on H 0, since their value outside of H 0 does not matter. We conclude this section by showing that an interval decomposition yields a partition of the lattice of subspaces. Theorem 2. Let (p 0, H 0, m) be an interval decomposition and W V a subspace of V. Then either p 0 W, W H 0 or there exists q Q such that q W m( q ). Proof. We proceed by induction on the dimension n 2 of V. If n = 2 the assertion is immediate. Thus assume n > 2. We may assume that p 0 W H 0. If dim(w ) = n 1, then W m(q) and we are done. Otherwise, let H be a hyperplane of V containing H and I H, S H as in Proposition 4. By induction there exist some i I H and β F such that W [q i,β, {q i,β } (ker((σ i,β ) H ) H 0 H )) ]. Hence q i,β W m( q i,β ). 3 Algorithms Theorem 1 and Proposition 4 enable us to derive an algorithm to compute interval decompositions, if they exist, by computing a set of irreflexive and antisymmetric linear forms from the corresponding forms for the projections. It will provide helpful to choose a basis {b 1,..., b n 1 } of H 0 such that the matrix (σ bi,1(b j )) i,j is lower triangular. Definition 3. Let S = {σ i,β : V F i I, β F } be a set of linear forms that is pointwise irreflexive and antisymmetric on H 0 and (b 1,..., b n 1 ) an ordered basis of H 0. We say that S is in canonical form with respect to (b 1,..., b n 1 ) if 1 i < k n 1 : σ bi,1(b k ) = 0. 6

9 Proposition 5. If S = {σ i,β : V F i I, β F } is a set of linear forms that is pointwise irreflexive and antisymmetric on H 0 then there exists an ordered basis (b 1,..., b n 1 ) of H 0 such that S is in canonical form with respect to (b 1,..., b n 1 ). Proof. Choose b 1 = p i H 0 \ {0} arbitrarily and then for 2 i n 1 b i H 0 i 1 j=1 Such a choice is always possible since ( dim H 0 i 1 j=1 ker(σ bj,1) \ {0}. ker(σ bj,1) ) n i. Using σ bj,1(b j ) 0 and the above choice it is easy to show that the b i are linearly independent and hence form a basis of H 0. Note that (p 0, b 1,..., b n 1 ) is an ordered basis of V. The following is also immediate: Proposition 6. Let S be in canonical form with respect to (b 1,..., b n 1 ) and H i = {p 0, b 1,..., b i 1, b i+1,..., b n 1 }. Then S H i is in canonical form with respect to (b 1,..., b i 1, b i+1,..., b n 1 ). From now on we assume that F is a finite field. We may assume that (p 0, b 1,..., b n 1 ) is an ordered basis of F n. If we know the set S of all possible sets of irreflexive and antisymmetric linear forms S on a hyperplane H 0 of F n 1 which are in canonical form with respect to a basis, then considering the projections of such linear forms for F n on the H i for i = 1,..., n as in the above proposition will yield an element of S n 1. Hence, we examine all this (n 1)-tuples whether they generate a suitable set of linear forms for F n. Given such a tuple (S 1,..., S n 1 ) first for all i j we check whether S i Hi H j = S j Hi H j then we construct a suitable S and check whether it is pointwise irreflexive and antisymmetric. We will demonstrate this algorithm in the next section and apply it in the case F = GF (2). Before doing so, we will introduce more structure into our linear forms, to reduce the computational effort in our search if F > 2. Definition 4. Let S = {σ i,β : V F i I, β F } be a set of linear forms that is pointwise irreflexive and antisymmetric on H 0. We call S structured if for all i I and all β, β F we have ker(σ i,β ) H 0 = ker(σ i,β ) H 0. 7

10 Such a structured set of linear forms may be considered as a bijection between the points and hyperplanes of H 0. Also note, that in the case of F = GF (2) any set of irreflexive and antisymmetric linear forms is structured and the same holds for the linear forms in Example 2. The following is again immediate: Proposition 7. A projection of a structured set of irreflexive and antisymmetric linear forms is structured. Thus, in our above algorithm we may restrict our search to structured sets of linear forms. We will report on an implementation of this method for F = GF (3) in the following section. 4 GF (2) and GF (3) 4.1 GF(2) Since GF (2) = {1} has only one element we omit the subscript β in this subsection. For n = 2 there is only one non-zero linear form σ b1 : H 0 GF (2). If n = 3 let b 1, b 2 denote a basis of H 0 and (d 0, d 1, d 2 ) the ordered basis of linear forms dual to (p 0, b 1, b 2 ). Considering only linear forms, which are in canonical form with respect to (p 0, b 1, b 2 ) irreflexivity implies σ b1 = d 1. Now, irreflexivity and antisymmetry yields σ b1 +b 2 = d 2 and, finally, σ b2 = d 1 + d 2. Now, let n = 4 and (d 0, d 1, d 2, d 3 ) be the basis dual to (p 0, b 1, b 2, b 3 ). Considering the projections on H 3, H 2, H 1 and using the property of the canonical form we find that σ b1 = d 1, σ b2 = d 1 + d 2, σ b1 +b 2 = d 2 + α 1 d 3, σ b3 = d 1 + β 1 d 2 + d 3, σ b1 +b 3 = β 2 d 2 + d 3, σ b3 = γ 1 d 1 + d 2 + d 3, σ b2 +b 3 = γ 2 d 1 + d 3. To make this compatible we have to set β 1 = γ 1 = 1. Hence, σ b3 = d 1 +d 2 +d 3. Irreflexivity allows for σ b1 +b 2 +b 3 only the choices d i for i = 1, 2, 3 or d 1 +d 2 +d 3. By antisymmetry we are left with σ b1 +b 2 +b 3 = d 2 or σ b1 +b 2 +b 3 = d 3. Assuming the latter we find that σ b1 +b 2 +b 3 (b 3 ) = σ b3 (b 3 ) = 1 = σ b1 +b 2 +b 3 (b 1 + b 2 + b 3 ) = σ b3 (b 1 + b 2 + b 3 ) contradicting antisymmetry. Hence we are left with σ b1 +b 2 +b 3 = d 2 and conclude that α 1 = 1, β 2 = 0 and γ 2 = 1. Altogether, we have b 1 b 2 b 1 + b 2 b 3 b 1 + b 3 b 2 + b 3 b 1 + b 2 + b 3 d 1 d 1 + d 2 d 2 + d 3 d 1 + d 2 + d 3 d 3 d 1 + d 3 d 2 and verify that this set of linear forms is indeed irreflexive and antisymmetric on H 0. Summarizing we find: 8

11 Proposition 8. For n = 2, 3, 4 there exists one unique set of linear forms which is irreflexive and antisymmetric on H 0 and in canonical form with respect to (p 0, b 1 ), (p 0, b 1, b 2 ), resp. (p 0, b 1, b 2, b 3 ). The existence of an interval decomposition for GF (2) n for n = 2, 3, 4 is already proven in [4]. The same approach using projections on H 4, H 3, H 2 resp. H 1 and canonical form for n = 5 yields the following table: b 1 b 2 b 1 + b 2 b 3 d 1 d 1 + d 2 d 2 + d 3 + α 1 d 4 d 1 + d 2 + d 3 b 1 + b 3 b 2 + b 3 b 1 + b 2 + b 3 b 1 + b 2 d 3 + α 2 d 4 d 1 + d 3 + α 3 d 4 d 2 + α 4 d 4 d 2 + β 1 d 3 + d 4 b 4 b 1 + b 4 b 2 + b 4 b 1 + b 2 + b 4 d 1 + d 2 + β 2 d 3 + d 4 β 3 d 3 + d 4 d 1 + β 4 d 3 + d 4 d 2 + β 5 d 3 b 1 + b 3 b 4 b 1 + b 4 b 3 + b 4 γ 1 d 2 + d 3 + d 4 d 1 + γ 2 d 2 + d 3 + d 4 γ 3 d 2 + d 4 d 1 + γ 4 d 2 + d 4 b 1 + b 3 + d 4 b 2 + b 3 b 4 b 2 + b 4 γ 5 d 2 + d 3 δ 1 d 1 + d 3 + d 4 δ 2 d 1 + d 2 + d 3 + d 4 δ 3 d 1 + d 4 b 3 + b 4 b 2 + b 3 + b 4 δ 4 d 1 + d 2 + d 4 δ 5 d 1 + d 3 To make this compatible we conclude that σ b1 +b 2 = d 2 + d 3 + d 4, σ b1 +b 3 = d 3 + d 4, σ b1 +b 4 = d 4, σ b2 +b 3 = d 1 + d 3 + d 4 which leaves σ b2 +b 4 = d 1 + d 4, σ b3 +b 4 = d 1 + d 2 + d 4, σ b4 = d 1 + d 2 + d 3 + d 4 b 1 + b 2 + b 3 b 1 + b 2 + b 4 b 1 + b 3 + b 4 b 2 + b 3 + b 4 b 1 + b 2 + b 3 + b 4 d 2 + α 4 d 4 d 2 + β 5 d 3 γ 5 d 2 + d 3 δ 5 d 1 + d 3? and d 2, d 3, d 1 + d 3, d 2 + d 3, d 2 + d 4 as linear forms. By irreflexivity, we must have σ b1 +b 2 +b 3 +b 4 {d 2, d 3 }. If σ b1 +b 2 +b 3 +b 4 = d 2 then σ b1 +b 2 (b 1 +b 2 +b 3 +b 4 ) = σ b1 +b 2 (b 1 + b 2 ) = 1 = σ b1 +b 2 +b 3 +b 4 (b 1 + b 2 + b 3 + b 4 ) = σ b1 +b 2 +b 3 +b 4 (b 1 + b 2 ) contradicting antisymmetry. Hence, we must have σ b1 +b 2 +b 3 +b 4 = d 3 which yields the contradiction σ b3 (b 1 +b 2 +b 3 +b 4 ) = σ b3 (b 3 ) = 1 = σ b1 +b 2 +b 3 +b 4 (b 1 +b 2 +b 3 +b 4 ) = σ b1 +b 2 +b 3 +b 4 (b 3 ). Hence, we have proven 9

12 Proposition 9. The lattice of subspaces of GF (2) 5 does not admit an interval decomposition. We summarize the results of this subsection as Theorem 3. The lattice of subspaces of GF (2) n admits an interval decomposition if and only if 2 n GF(3) In this subsection we will show, that the situation becomes a lot richer considering other fields. In particular we will present a structured interval decomposition of the lattice of subspaces of GF (3) 5. Considering only structured forms allows us to continue omitting the subscript β for the σ p. There is only one structured set of linear forms for GF (3) 2. If H 0 is a hyperplane of GF (3) 3 and (b 1, b 2 ) an ordered basis of H 0, (d 0, d 1, d 2 ) a corresponding dual basis, any such set of linear forms which is in canonical form with respect to (b 1, b 2 ) has to satisfy σ b1 = d 1. For σ b2 we have three choices d 2, d 1 + d 2 and 2d 1 + d 2. In turns out that we can complete all these choices to structured, irreflexive and antisymmetric sets S 1, S 2, S 3 of linear forms. p H 0 σ p S 1 σ p S 2 σ p S 3 b 1 d 1 d 1 d 1 b 2 d 2 d 1 + d 2 d 1 + 2d 2 b 1 + b 2 d 1 + d 2 d 2 d 1 + d 2 b 1 + 2b 2 d 1 + 2d 2 d 1 + 2d 2 d 2 We implemented the algorithm described in the last section and found 26 structured interval decompositions for the lattice of subspaces of GF (3) 4 and 52 for GF (3) 5. We list three of the former in Table 1, which are used to compute one of the latter. The full lists can be found in the Appendix of [3]. Using S 1, S 6 once and S 3 two times as projection we discovered the set of linear forms in Table 2. It is possible to check by hand that these indeed are irreflexive and antisymmetric. Theorem 4. There exists an interval decomposition of the lattice of subspaces of GF (3) 5. 10

13 p H 0 σ p S 1 σ p S 3 σ p S 6 b 1 d 1 d 1 d 1 b 2 d 2 d 2 d 2 b 3 d 3 d 3 d 3 b 1 + b 2 d 1 + d 2 + 2d 3 d 1 + d 2 + d 3 d 1 + d 2 b 1 + 2b 2 d 1 + 2d 2 + d 3 d 1 + 2d 2 + d 3 d 1 + 2d 2 b 1 + b 3 d 1 + d 2 + d 3 d 1 + d 3 d 1 + 2d 2 + d 3 b 1 + 2b 3 d 1 + 2d 2 + 2d 3 d 1 + 2d 3 d 1 + 2d 2 + 2d 3 b 2 + b 3 d 2 + d 3 d 1 + 2d 2 + 2d 3 d 1 + d 2 + d 3 b 2 + 2b 3 d 2 + 2d 3 d 1 + d 2 + 2d 3 d 1 + d 2 + 2d 3 b 1 + b 2 + b 3 d 1 + d 2 d 2 + d 3 d 1 + d 3 b 1 + b 2 + 2b 3 d 1 + 2d 3 d 1 + d 2 d 1 + 2d 3 b 1 + 2b 2 + b 3 d 1 + d 3 d 2 + 2d 3 d 2 + 2d 3 b 1 + 2b 2 + 2b 3 d 1 + 2d 2 d 1 + 2d 2 d 2 + d 3 Table 1: Three structured interval decompositions for GF (3) 4 5 Conclusion and Open Problems While we could completely settle the problem of existence of interval decompositions for vector spaces of finite dimension over GF (2) and over the reals, the situation seems to become more difficult for other finite fields. On one hand the additional choices for linear forms provide a lot more flexibility and enable us to construct several interval decompositions for GF (3) 5 while an interval decomposition is impossible for GF (2) 5. On the other hand our argument used for real vector spaces applies the Cauchy-Schwarz inequality which is not applicable for finite fields. Using matching theory (see e.g. [5] Corollary 16.2b) it is immediate that there is an interval decomposition of GF (q) 3 for all prime powers q. Thus all in all we get the following table on the existence of an interval decomposition. Dimension GF (2) GF (3) GF (4) GF (q), q 5 R 2,3 yes yes yes yes yes 4 yes yes yes? yes 5 no yes?? yes 6 no??? yes We tried to fill some of the question marks doing more extensive computations using our algorithm. We found 11 structured decompositions of GF (4) 3 and 53 for GF (5) 3. Alas, already the search for structured decompositions of GF (4) 4 turned out to be too costly. Imposing even more structure 11

14 b 1 b 2 b 3 b 4 d 1 d 2 d 3 d 4 b 1 + b 2 b 1 + 2b 3 b 1 + b 3 b 1 + 2b 3 d 1 + d 2 + 2d 3 + d 4 d 1 + 2d 2 + d 3 + d 4 d 1 + d 2 + d 3 + d 4 d 1 + 2d 2 + 2d 3 + d 4 b 1 + b 4 b 1 + 2b 4 b 2 + b 3 b 2 + 2b 3 d 1 + d 4 d 1 + 2d 4 d 2 + d 3 d 2 + 2d 3 b 2 + b 4 b 1 + 2b 4 b 3 + b 4 b 3 + 2b 4 d 1 + 2d 2 + d 3 + 2d 4 d 1 + d 2 + 2d 3 + 2d 4 d 1 + 2d 2 + 2d 3 + 2d 4 d 1 + d 2 + d 3 + 2d 4 b 1 + b 2 + b 3 b 1 + b 2 + 2b 3 b 1 + 2b 2 + b 3 b 1 + 2b 2 + 2b 3 d 1 + d 2 + 2d 4 d 1 + 2d 3 + 2d 4 d 1 + d 3 + 2d 4 d 1 + 2d 2 + 2d 4 b 1 + b 2 + b 4 b 1 + b 2 + 2b 4 b 1 + 2b 2 + b 4 b 1 + 2b 2 + 2b 4 d 2 + d 3 + d 4 d 1 + d 2 + d 3 d 2 + d 3 + 2d 4 d 1 + 2d 2 + 2d 3 b 1 + b 3 + b 4 b 1 + b 3 + 2b 4 b 1 + 2b 3 + b 4 b 1 + 2b 3 + 2b 4 d 2 + 2d 3 + 2d 4 d 1 + 2d 2 + d 3 d 2 + 2d 3 + d 4 d 1 + d 2 + 2d 3 b 2 + b 3 + b 4 b 2 + b 3 + 2b 4 b 2 + 2b 3 + b 4 b 2 + 2b 3 + 2b 4 d 1 + d 2 + d 4 d 1 + 2d 2 + d 4 d 1 + 2d 3 + d 4 d 1 + d 3 + d 4 b 1 + b 2 + b 3 + b 4 b 1 + b 2 + b 3 + 2b 4 b 1 + b 2 + 2b 3 + b 4 b 1 + b 2 + 2b 3 + 2b 4 d 1 + d 3 d 3 + 2d 4 d 1 + d 2 d 2 + 2d 4 b 1 + 2b 2 + b 3 + b 4 b 1 + 2b 2 + b 3 + 2b 4 b 1 + 2b 2 + 2b 3 + b 4 b 1 + 2b 2 + 2b 3 + 2b 4 d 1 + 2d 2 d 2 + d 4 d 1 + 2d 3 d 3 + d 4 Table 2: An interval decomposition of GF (3) 5 we managed to find six decompositions of GF (4) 4 with simpler structure, indicated by a yes in the table. It is impossible, though, to combine these into a decomposition of GF (4) 5 with that simpler structure. The structured decompositions we found for GF (3) 5 do not seem to indicate a way to construct interval decompositions in the general case. Moreover, to our surprise, they cannot be combined into a structured decomposition of GF (3) 6. References [1] Ulrich Faigle, personal communication, [2] Jay Goldman and Gian-Carlo Rota, The number of subspaces of a vector space, Recent Progress in Combinatorics (Proc. Third Waterloo Conf. on Combinatorics, 1968), Academic Press, New York, 1969, pp [3] Steffen Hitzemann, Über die Kombinatorik der Galoiszahlen, Master s thesis, FernUniversität in Hagen, October 2008, p

15 [4] Eva Kruse, Symmetrische Kettenzerlegungen von Verbänden und Intervallzerlegung des linearen Verbandes, Master s thesis, Universität zu Köln, August 2004, p [5] Alexander Schrijver, Combinatorial optimization: Polyhedra and efficiency, 1 ed., Springer, July