Steinitz classes of unimodular lattices
Robin Chapman School of Mathematical Sciences University of Exeter Exeter, EX4 4QE, UK [email protected] 9 December 2002
Abstract For each prime l with l 7 (mod 8), we define an action of the 1 ≡ + ring = Z[ 2 (1 + √ l)] on the unimodular lattice Dl+1 using a Paley O − + matrix. We determine the isomorphism class of Dl+1 as an -module. + O In particular we show that unless l = 7, Dl+1 is not a free -module. We note a consequence for the Leech lattice. O
1 Introduction
In [1] the author explained how given an n-by-n skew-symmetric conference matrix W one can give Rn a structure of a vector space over C. When n + is a multiple of 8, with respect to this structure the unimodular lattice Dn 1 becomes a module over the ring = Z[ 2 (1 + √1 n)]. Using this module structure we can derive other unimodularO lattices. One− natural question is to + determine the structure of Dn as a -module. Here we answer this question completely whenever W is a conferenceO matrix of Paley type. It turns out + that unless n = 8, Dn is not free over . As a corollary we show that the Leech lattice is not a free module withO respect to a natural action of Z[ 1 (1 + √ 23)]. 2 − 2 The Paley matrix
We fix a prime l congruent to 7 modulo 8 and define n to be l + 1.
1 We define the Paley matrix Wl as follows. Let
0 1 1 1 ··· − Wl = . . Wl0 1 − where the l-by-l matrix W 0 is the circulant matrix whose (i, j)-entry is the j i Legendre symbol −l . As the Paley matrix is intimately related to quadratic residue codes, we follow the usual practice with quadratic residue codes and label the entries of a typical vector of length n = l + 1 using the elements of the projective 1 line P (Fl) over Fl as follows: v = (v , v0, v1, v2, . . . , vl 1). We let e , e0, ∞ − ∞ e1,... el 1 denote the corresponding unit vectors, that is, eµ has a one in the position− labelled µ, and zeros elsewhere. A conference matrix of order n is a square matrix A with zeroes on the main diagonal and off-diagonal entries 1 satisfying AAt = (n 1)I. ± −
Lemma 1 (Paley) The matrix Wl is a skew-symmetric conference matrix of order l + 1.
Proof See for instance [3, Chapter 18]. 2 In particular Wl = lI. We define an action of C = R(√ l) on the Euclidean space Rn by (−a + b√ l)u = au + buW . Let = Z[ 1 (1− + √ l)]. − O 2 − Then is the maximal order of the quadratic number field K = Q(√ l) and so isO a Dedekind domain. In [1] the author proved that the unimodular− O + lattice Dn is a -module, under this action. We shall investigate its Steinitz class. O + n Let us recall the definition of Dn . It consists of all (a , a0, . . . , al 1) R ∞ − ∈ such that all 2aµ are integers of the same parity and µ aµ is even. It is an + P even unimodular lattice. As an -module, Dn is finitely generated and torsion-free. Hence it is a projectiveO -module. If is a finitely generated O r 1 M projective module over then ∼= − where is an ideal of . The Steinitz class of is theO classM [ ] of O in the⊕ I classgroupI of . It is uniquelyO M I I O determined by . We shall denote it by s( ). If 0 is a submodule of M M M M with / 0 torsion, then s( 0) = [ : 0 ]s( ). Here : 0 is the moduleM index:M if / Mk ( /|M) thenM | M: = |Mk M. | 0 ∼= j=1 j 0 j=1 j (For basic resultsM M on projectiveL O modulesI over|M DedekindM | Q domainsI see, for instance, [2, 9.6].) §
2 3 The main theorem
+ We state and prove a theorem giving the Steinitz class of Dn with respect to the Paley matrix Wl.
Theorem 1 Let n = l + 1 where l is prime and l 7 (mod 8). The Paley ≡ + 1 matrix Wl induces the structure of an -module on Dn where = Z[ 2 (1 + + O O √ l]. The Steinitz class of Dn with respect to this -module structure is [ n/− 8, 1 (1 √ l) ]. O 2 − − + Proof We identify a -submodule Λ of Dn for which we can calculate + O Dn :Λ . We shall prove that Λ is free over . | Let | O
+ Λ = (u , . . . , ul 1) Dn : u + + ul 1 0 (mod n/4) . { ∞ − ∈ ∞ ··· − ≡ } + + It is clear that Λ is a sublattice of Dn and that Dn /Λ is a cyclic group of order n/8 generated by the image of 2e . We need to show that Λ is a -module. 1 ∞ 1 O + For that it is sufficient that 2 (1 + √ l)Λ Λ. In fact, 2 (1 + √ l)Dn Λ. + − ⊆ 1 − ⊆ The lattice Dn is spanned by the vectors v = µ eµ and the e +eµ for µ 2 ∞ 1 1 P1 ∈ P (Fl). Now 2 (1+√ l)v = v (n/4)e Λ, 2 (1+√ l)2e = 2v Λ and 1 − − ∞ ∈ − ∞ ∈ each vector 2 (1+√ l)(e +ej) for 0 j < l has n/2 ones and n/2 zeroes and − ∞ ≤ + so lies in Λ. Thus Λ is an -module, and the -module Dn /Λ is annihilated 1 + O O by 2 (1 + √ l). But Dn /Λ is a cyclic group of order n/8 and so is a cyclic − 1 -module whose annihilator contains the ideal = n/8, 2 (1 + √ l) . As O1 2 J − 2 (1 + √ l) = n/4 is a multiple of n/8, this ideal has norm n/8. Hence | + − | Dn /Λ = / as -modules. ∼ O J O + The Steinitz class of Dn is
+ 1 s(D ) = [ ]− s(Λ) = [ ]s(Λ) = n/8, (1 √ l)/2 s(Λ). n J J hD − − Ei To prove the theorem it only remains to prove that Λ is a free -module. O We have already defined v = µ eµ. Define, for 0 j < l, uj = e ej. P ≤ ∞ − Note that v Λ and each uj Λ. We shall prove that the n/2 vectors v, ∈ ∈ u0, u1, . . . , un/2 2 generate Λ freely over . Let Λ0 be the -submodule of − O O Λ generated by v, u0, u1, . . . , un/2 2. It suffices to prove that Λ0 = Λ. The − lattice Λ0 is the set of integer linear combinations of the vectors
(1 √ l)v/2, u0, u1, . . . , un/2 2, − − − (1 + √ l)u0/2, (1 + √ l)u1/2,..., (1 + √ l)un/2 2/2, v. − − − −
3 Let M denote the n-by-n matrix with the above vectors as rows, in the order + given. Then Dn :Λ0 = det(M) and so the assertion that Λ0 = Λ is equivalent to the| assertion| that| det(M| ) = n/8. Now ± M1 0 M = M2 M3 where the matrices Mi are n/2-by-n/2. Hence det(M) = det(M1) det(M3). Also n/4 0 0 0 1 1 0 ··· 0 − ··· M = 1 0 1 0 1 − ··· ...... . . . . . 1 0 0 1 · · · − and so det(M1) = n/4. It suffices to prove that det(M3) = 1/2. ± ± The entries of M3 are given by
1 j i+(l 1)/2 2 1 − l− if i < n/2, (M3)i,j = ( h − i 1/2 if i = n/2.
It suffices to show that det(N) = 1 where N is the n/2-by-n/2 matrix with entries ± 1 j i+(l 1)/2 2 1 − l− if i < n/2, Ni,j = ( h − i 1 if i = n/2. To show that det(N) = 1, we shall show that the matrix N, which has integer entries, is nonsingular± considered as a matrix modulo every prime number p. We extend the matrix N to an n/2 by l matrix N 0 with entries
j i 0 if i < n/2, and −l = 1 j i 1 if i < n/2, and − = 1 N 0 = l i,j t if i < n/2, and i = j − 1 if i = n/2. where we shall specify t later. The matrix N is formed by taking the last n/2 columns of N 0. We claim that it is possible to choose t such that N 0 is the generator matrix of a cyclic code. For each field K and positive integer m, we can identify the vector space Km with the quotient ring K[X]/ Xm 1 by identifying the vec- h m−1 i j m tor (a0, a1, . . . , am 1) with the polynomial j=0− ajX . A subspace of K corresponding to− an ideal of K[X]/ Xm P1 is called a cyclic code. Each h − i 4 ideal of K[X]/ Xm 1 is principal, and equals g(X) where g is a monic factor of Xm 1.h If the− degreei of g is d then the dimensionh i of the correspond- ing cyclic code− is m d. The polynomial g is called the generator polynomial of the cyclic code. A− matrix whose rows form a basis of a cyclic code is called a generator matrix of the cyclic code. For p = l we take K to be the finite field Fp(ζ) where ζ is a primitive 6 l-th root of unity, and for p = l we take K = Fp and set ζ = 1. The first row of N 0 corresponds to the polynomial f(X) = t + Xj. X 0 All but the last row of N 0 is a cyclic shift of the first row, and they correspond to the Xif(X) K[X]/ Xl 1 for 0 i (l 3)/2. The last row corresponds to the∈ polynomial − ≤ ≤ − l 1 − Xl 1 g(X) = Xj = − . X X 1 j=0 − Set t = ζj X − 0 We aim to show that N 0 is a generator matrix for the cyclic code with gen- erator polynomial φ(X). We have φ(X) = (X ζj), g(X) = (X ζj), and Xl 1 = (X ζj). Y − Y − − Y − 0 (l 1)/2 and so f − (1) = 0. In this case too, the greatest common divisor of f(X) and g(X) is φ(X).6 The row space of the matrix N 0 corresponds to the set of polynomials of the form a(X)f(X) + rg(X) + c(X)(Xl 1) ( ) − ∗ where the degree of the polynomial a(X) is at most (l 3)/2, r K and c(X) − ∈ is a polynomial. To show that N 0 is the generator matrix of the cyclic code with generator matrix φ(X) we must show that each multiple b(X)φ(X) of φ(X) has the form ( ). Certainly ∗ b(X)φ(X) = u(X)f(X) + v(X)g(X) for some polynomials u(X) and v(X). Now f(X) = φ(X)f1(X) and g(X) = φ(X)g1(X) where g1(X) has degree (l 1)/2. We may write u(X) = a(X) + − q(X)g1(X) where a(X) has degree at most (l 3)/2. Then − b(X) = u(X)f1(X) + v(X)g1(X) = a(X)f1(X) + [q(X)f1(X) + v(X)]g1(X) and so b(X)ψ(X) = a(X)f(X) + w(X)g(X) 6 for some polynomial w(X). Now w(X) = w(1) + (X 1)c(X) for some polynomial c(X) and so − ψ(X) = a(X)f(X) + w(1)g(X) + c(X)(Xl 1) − as required. As the cyclic code generated by φ(X) has dimension (l + 1)/2, the rows of N 0 are linearly independent. Another generator matrix for this cyclic i code is the matrix B0 whose rows correspond to the polynomials X φ(X) for 0 i (l 1)/2. The matrices N 0 and B0 are row equivalent. Thus the matrices≤ ≤ N −and B are row equivalent where B is the matrix formed by the last (l + 1)/2 columns of B0. But B0 is lower triangular, having all diagonal entries 1. Thus B and also N are nonsingular over K and so N is nonsingular over Fp. As N is nonsingular over Fp for all primes p, then det(N) = 1. ± Thus det(M) = n/8, Λ = Λ0 is a free -module and ± O s(D+) = n/8, (1 √ l)/2 . n hD − − Ei + It is immediate that due to this theorem Dn is only free over when l = 7. O Corollary 1 The -module D+ is free if and only if the prime l = 7. O n 1 Proof The classgroup of Z[ 2 (1 + √ 7) is trivial, and so when l = 7 the + − -module D8 must be free. O + Suppose that l > 7. The -module Dn is free if and only if the ideal 1 O 2 = n/8, 2 (1 √ l) is principal. Suppose that = α . Then α = I − − I h i - 1 | | n/8 > 1. We cannot have α Z for then α > 1 and so α 2 (1 √ l) in . On the other hand if α∈ / Z then α 2 | |min(l, (l + 1)/4) = n/−4 which− O ∈ | | ≥ + is a contradiction. We conclude that is not principal and Dn is not free over . I O In [1] the Leech lattice Λ24 was constructed by considering the action of 1 + = Z[ 2 (1 + √ 23)] on D24 as given by the Paley matrix W23. With this O − 1/2 + 1 -module structure, Λ24 is 2− D24 where = 2, 2 (1 + √ l) . Conse- O 12 + I + J − quently, s(Λ24) = [ ] s(D24) = s(D24) as the classgroup of has order 3. + J O Thus Λ24 and D24 are isomorphic as -modules, and we obtain the amusing result that under a very natural actionO of the ring , the Leech lattice is not a free module. O 7 References [1] R. Chapman, ‘Conference matrices and unimodular lattices’ European J. Combin. 22 (2001), 1033–1045 [2] P. M. Cohn, Algebra vol. 2 (2nd ed.), John Wiley & Sons, 1989. [3] J .H. van Lint & R. M. Wilson, A Course in Combinatorics, Cambridge University Press, 1992. 8