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 + p 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 + p1 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 + p 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 0 1 1 ··· − Wl = B . C B . Wl0 C B C @ 1 A − 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, 1 − 1 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(p l) on the Euclidean space Rn by (−a + bp l)u = au + buW . Let = Z[ 1 (1− + p l)]. − O 2 − Then is the maximal order of the quadratic number field K = Q(p 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 1 − 2 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 ( =jM) thenM j M: = jMk M. j 0 ∼= j=1 j 0 j=1 j (For basic resultsM M on projectiveL O modulesI overjM DedekindM j Q domainsI see, for instance, [2, 9.6].) x 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 p l]. The Steinitz class of Dn with respect to this -module structure is [ n=− 8; 1 (1 p 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 . j Let j O + Λ = (u ; : : : ; ul 1) Dn : u + + ul 1 0 (mod n=4) : f 1 − 2 1 ··· − ≡ g + + 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 1 O + For that it is sufficient that 2 (1 + p l)Λ Λ. In fact, 2 (1 + p l)Dn Λ. + − ⊆ 1 − ⊆ The lattice Dn is spanned by the vectors v = µ eµ and the e +eµ for µ 2 1 1 1 P1 2 P (Fl). Now 2 (1+p l)v = v (n=4)e Λ, 2 (1+p l)2e = 2v Λ and 1 − − 1 2 − 1 2 each vector 2 (1+p l)(e +ej) for 0 j < l has n=2 ones and n=2 zeroes and − 1 ≤ + so lies in Λ. Thus Λ is an -module, and the -module Dn =Λ is annihilated 1 + O O by 2 (1 + p 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 + p l) . As O1 2 J − 2 (1 + p l) = n=4 is a multiple of n=8, this ideal has norm n=8. Hence j + − j Dn =Λ = = as -modules. ∼ O J O + The Steinitz class of Dn is + 1 s(D ) = [ ]− s(Λ) = [ ]s(Λ) = n=8; (1 p 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 ≤ 1 − Note that v Λ and each uj Λ. We shall prove that the n=2 vectors v, 2 2 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 p l)v=2; u0; u1; : : : ; un=2 2; − − − (1 + p l)u0=2; (1 + p l)u1=2;:::; (1 + p 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 thej assertionj thatj det(Mj ) = 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 0 1 1 1 0 ··· 0 − ··· M = B 1 0 1 0 C 1 B − ··· C B . .. C B . C B C @ 1 0 0 1 A · · · − 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 8 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.