ISSN 2148-838Xhttp://dx.doi.org/10.13069/jacodesmath.729440 J. Algebra Comb. Discrete Appl. 7(2) • 141–160 Received: 7 October 2019 Accepted: 4 December 2019 Journal of Algebra Combinatorics Discrete Structures and Applications Trace forms of certain subfields of cyclotomic fields and applications∗ Research Article Agnaldo José Ferrari, Antonio Aparecido de Andrade, Robson Ricardo de Araujo, José Carmelo Interlando Abstract: In this work, we present a explicit trace forms for maximal real subfields of cyclotomic fields as tools for constructing algebraic lattices in Euclidean space with optimal center density. We also obtain a closed formula for the Gram matrix of algebraic lattices obtained from these subfields. The obtained lattices are rotated versions of the lattices Λ9, Λ10 and Λ11 and they are images of Z-submodules of rings of integers under the twisted homomorphism, and these constructions, as algebraic lattices, are new in the literature. We also obtain algebraic lattices in odd dimensions up to 7 over real subfields, calculate their minimum product distance and compare with those known in literatura, since lattices constructed over real subfields have full diversity. 2010 MSC: 11H06, 11H31, 11R80, 97N70 Keywords: Cyclotomic fields, Algebraic lattices, Twisted homomorphism, Signal design 1. Introduction Lattices have been considered in different areas, especially in coding theory and more recently in cryptography. Algebraic lattices are lattices obtained via the ring of integers of a number field and they have been studied in several papers and from different points of view [1–7, 10–13, 15, 16, 18]. ∗ This work was supported by Fapesp 2013/25977-7 and CNPq 429346/2018-2. Agnaldo José Ferrari; Department of Mathematics, School of Sciences, São Paulo State University (Unesp), Bauru-SP, Brazil (email: agnaldo.ferrari@unesp.br). Antonio Aparecido de Andrade (Corresponding Author); Department of Mathematics, Institute of Biosciences, Humanities and Exact Sciences (Ibilce), São Paulo State University (Unesp), São José do Rio Preto-SP, Brazil (email: antonio.andrade@unesp.br). Robson Ricardo de Araujo; São Paulo Federal Institute at Cubatão, São Paulo, Brazil (email: dearaujorobison- ricardo@gmail.com). José Carmelo Interlando; Department of Mathematics & Statistics, San Diego University, San Diego, California, USA (email: interlan@sdsu.edu). 141 https://orcid.org/0000-0002-1422-1416 http://orcid.org/0000-0001-6452-2236 https://orcid.org/0000-0002-1357-9926 https://orcid.org/0000-0002-1357-9926 https://orcid.org/0000-0003-4928-043X A. J. Ferrari et al. / J. Algebra Comb. Discrete Appl. 7(2) (2020) 141–160 The classical sphere packing problem is to determine how densely a large number of identical spheres can be packed together in the Euclidean space. The packing density of a lattice Λ is the proportion of the space covered by the non-overlapping spheres of maximum radius centered at the points of Λ. The densest known lattice packings in dimensions 1 through 8 and 24 are also the optimal ones, see [12, p. 12] for n = 1, 2, . . . , 8, and [11] for n = 24. Those lattice packings are unique. For all other dimensions, it is not known whether the current records are optimal. A lattice Λ has diversity equal to k if k is the maximum number such that any non-zero vector y ∈ Λ has at least k non-zero coordinates. Given an n-dimensional lattice Λ ⊆ Rn of full diversity, the minimum product distance of Λ is defined as dp,min(Λ) = min{ ∏n i=1 |yi|; y = (y1,y2, . . . ,yn) ∈ Λ, y 6= 0}. Usually the problem of finding good signal constellations for a Gaussian channel is associated with the search for lattices with high packing density, see [12, Chapter 3]. On the other hand, for a Rayleigh fading channel, the efficiency of the signal constellation, measured by the error probability in the transmission, is strongly related to the lattice diversity and its minimum product distance, see [10, Section III]. For this purpose the lattice parameters we consider here are packing density, diversity, and minimum product distance. The approach in this work, following [2, 3] is the use of algebraic number theory to construct lattices which have good performance on both channels. For general lattices the packing density and the minimum product distance are usually hard to estimate [17]. Those parameters can be calculated in certain cases of lattices associated to number fields through algebraic properties. In [4–6] some families of rotated Zn- lattices of full diversity and high minimum product distance are studied for transmission over Rayleigh fading channels. In [13] some families of rotated Zn-lattices of full diversity in dimensions power of 3 are studied and lower bounds for the minimum product distances of such construction are also presented. In [7] the lattices Ap−1, p prime, E6, E8, K12 and Λ24 were realized as full diversity ideal lattices via some subfields of cyclotomic fields. In this work we construct the lattices D3,E5,E7, Λ9, Λ10 and Λ11, calculate (or estimate) their minimum product distance and compare the obtained values with those known in literature, mainly Zn-lattices given in [5, 18]. In [14, Theorem 1] a trace form for cyclotomic fields Q(ζn) via the Minkowski homomorphism is derived. In this work, we generalize the result for the maximal real subfields Q(ζn + ζ−1n ) via the twisted homomorphism. In [6, Proposition 4.1(a)] and [8] the Gram matrix of the algebraic lattice constructed via the Minkowski homomorphism over Q(ζpr +ζ−1pr ) is determined, but in this work we use a different aproach. In this work, we generalize the result to the case Q(ζn + ζ−1n ) considering the twisted homomorphism. Trace forms are used to calculate the packing radius of algebraic lattices. As an application, we present constructions of algebraic lattices with optimal center density in dimensions 3, 5, 7, 9, 10 and 11. The paper is organized as follows. In Sections 2, we collect some results on number fields and algebraic lattices. In Section 3, we present a explicit trace form for the maximal real subfields via the twisted homomorphism. We also present a closed formula of Gram matrix for the lattice σα(OK), where K = Q(ζn + ζ−1n ). In Section 4, we construct algebraic lattices in Euclidean space with optimal center density in dimensions 3, 5, 7, 9, 10 and 11 and calculate (estimate) their minimum product distance. Finally, in Section 5, we draw our conclusions. 2. Background on number fields and algebraic lattices If L is a number field of degree n, that is, a field that is a finite degree extension of Q, then L = Q(α), where α ∈ C is a root of a monic irreducible polynomial p(x) ∈ Z[x]. The n distinct roots of p(x), namely, α1,α2, . . . ,αn, are the conjugates of α. If σ : L → C is a Q-homomorphism, then σ(α) = αi for some i = 1, 2, . . . ,n. Furthermore, there are exactly n Q-homomorphisms σi, for i = 1, 2, . . . ,n, of L in C, where r1 are real monomorphisms and 2r2 are complex monomorphisms with n = r1 + 2r2. An element α ∈ L is called an algebraic integer if there is a monic polynomial f(x) with integer coefficients such that f(α) = 0. The set OL = {α ∈ L : α is an algebraic integer} 142 A. J. Ferrari et al. / J. Algebra Comb. Discrete Appl. 7(2) (2020) 141–160 is a ring, called ring of algebraic integers of L [20, 21]. It can be shown that OL, as a Z-module, has a basis {α1,α2, . . . ,αn} over Z, called integral basis, where n is the degree of L. In other words, every element α ∈OL can be uniquely written as α = n∑ i=1 aiαi, where ai ∈ Z for all i = 1, 2, . . . ,n. The trace and the norm of an element α ∈ L over Q are defined as the rational numbers TrL/Q(α) = n∑ i=1 σi(α) and NL/Q(α) = n∏ i=1 σi(α). If α ∈ OL, then TrL/Q(α) and NL/Q(α) are algebraic integers. The discriminant of L over Q is defined by dL = d(α1,α2, . . . ,αn) = det 1≤i,j≤n (σi(αj)) 2, where {α1,α2, . . . ,αn} is an integral basis of L. A lattice Λ is a discrete additive subgroup of Rn considered as the standard real vector space, that is, Λ ⊆ Rn is a lattice if there are linearly independent vectors α1,α2, . . . ,αm ∈ Rn such that Λ = { m∑ i=1 aiαi; ai ∈ Z, i = 1, 2, . . . ,m } . The classical sphere packing problem is to find out how densely a large number of identical spheres can be packed together in the Euclidean space. The packing density, ∆(Λ), of a lattice Λ is the proportion of the space Rn covered by the non-overlapping spheres of maximum radius centered at the points of Λ. The densest possible lattice packings have only be determined in dimensions 1 to 8 and 24 [12, p. 12]. It is also known that these densest lattice packings are unique. Let {α1,α2 . . . ,αm} be a set of linearly independent vectors in Rn and Λ = { ∑m i=1 aiαi; ai ∈ Z} a lattice. The set {α1,α2, . . . ,αm} is called a basis for Λ. A matrix M whose rows are these vectors is said to be a generator matrix for Λ whereas G = MMt = (〈αi,αj〉) m i,j=1 is called the Gram matrix of Λ. The determinant of Λ, denoted by det Λ, is equal to det G and it is an invariant under change of basis. The volume of Λ is equal to √ det(Λ). The packing density of Λ is the proportion occupied by the spheres centered in the points of the lattice and having radius min{||x−y||; x,y ∈ Λ, x 6= y}/2 relative to the entire space Rn. If ∆(Λ) is the packing density of Λ, then δ(Λ) = ∆(Λ)/Vn is the center density of the lattice, where Vn is the volume of an n-dimensional sphere of radius 1 [12, p. 9]. Let α ∈ L such that αi = σi(α) > 0 for all i = 1, . . . ,n. If R(x) and I(x) denote, respectively, the real part and the imaginary part of x, the homomorphism σα : L −→ Rn defined by σα(x) = (√ α1σ1(x), . . . , √ αr1σr1 (x), . . . , √ αr1+r2R(σr1+r2 (x)), √ αr1+r2I(σr1+r2 (x)) ) , for every x ∈ L, is called twisted homomorphism [2, 3]. When α = 1 the twisted homomorphism is the Minkowski homomorphism. If M is a Z-module in L of rank n with Z-basis {w1,w2, . . . ,wn}, then the set Λ = σα(M) is a complete lattice in Rn with basis {σα(w1),σα(w2), . . . ,σα(wn)}. If L is a totally real number field then G = ( TrL/Q(αwiwj) )n i,j=1 is a Gram matrix for σα(M). From [2], det(Λ) = [OK : M]2NL/Q(α)|dL|, so the center density of Λ is given by δ(Λ) = ρn√ det(Λ) = tn/2 2n[OL : M] √ NL/Q(α)|dL| , 143 A. J. Ferrari et al. / J. Algebra Comb. Discrete Appl. 7(2) (2020) 141–160 where dL denotes the discriminant of the number field L, [OL : M] denotes the index of M and t = min { TrL/Q(αx 2) : x ∈M, x 6= 0 } . If M is a Z-module in L of rank m, m < n, with Z-basis {w1,w2, . . . ,wm}, then the set Λ = σα(M) is a lattice of rank m in Rn with basis {σα(w1),σα(w2), . . . ,σα(wm)} and Q = ( TrL/Q(αwiwj) )n i,j=1 is a Gram matrix for σα(M). The center density of Λ is given by δ(Λ) =   tm/2 2m √ det(Q) , if L is totally real. tm/2 23m/2 √ det(Q) , if L is totally complex. (1) If K is a totally real field number with [K : Q] = n and M⊆ K a free Z-module of rank n, then the minimum product distance of Λ = σα(M) is defined as dp,min(Λ) = √ NK/Q(α) min 06=y∈M |NK/Q(y)|. (2) In particular, by [5], if M⊆ K is a principal ideal then dp,min(Λ) = √ det(Λ) |dK| . (3) The relative minimum product distance of Λ, denoted by dp,rel(Λ), is the minimum product distance of a scaled version of Λ with unitary minimum norm vector. Thus, if Λ1 is a scaled version of Λ of dimension n with scale factor √ k, i. e., Λ1 = √ kΛ and the minimum norm of Λ is µ, then the relative minimum product distance of Λ1 is given by dp,rel(Λ1) = ( 1 √ kµ )n dp,min(Λ1). (4) 3. Trace forms for cyclotomic fields In this section, we present explicit trace forms for maximal real subfiel K = Q(ζn + ζ−1n ) via twisted homomorphism. We also present a closed formula of Gram matrix for the lattice σα(OK). A related result, through a different approach, can be found in [6] and [8], where the authors use abelian fields of odd prime power conductor. A cyclotomic field is a number field L such that L = Q(ζn), where ζn is a primitive n-th root of unity. It can be shown that [L : Q] = ϕ(n), where ϕ is the Euler function, OL = Z[ζn] is the ring of algebraic integers of Z[ζn], {1,ζn,ζ2n, . . . ,ζ ϕ(n)−1 n } is an integral basis of L. Let K = Q(ζn + ζ−1n ) be the maximal real subfield of a cyclotomic field Q(ζn). In this case, [K : Q] = ϕ(n)/2, OK = Z[ζn + ζ−1n ] and {1,ζn + ζ−1n ,ζ2n + ζ−2n , . . . ,ζ ϕ(n) 2 −1 n + ζ −ϕ(n) 2 +1 n } is an integral basis of OK. [21]. Lemma 3.1. [14] Let j,n be integers. If gcd(j,n) = d, then TrQ(ζn)/Q(ζ j n) = ϕ(n) ϕ(n/d) TrQ(ζn/d)/Q(ζ j/d n/d ). Lemma 3.2. [14] If j,ai are integers, ai ≥ 1 and pi is a prime number such that gcd(j,piai) = 1, then TrQ(ζpiai )/Q (ζ j pi ai ) = { −1, if ai = 1. 0, if ai > 1. 144 A. J. Ferrari et al. / J. Algebra Comb. Discrete Appl. 7(2) (2020) 141–160 Lemma 3.3. [14] Let n = pa11 · · ·p as s , where ak ≥ 1, for k = 1, . . . ,s. If j is a prime number and gcd(j,n) = d, then TrQ(ζn)/Q(ζ j n) = ϕ(n) ϕ(n/d) µ(n/d), where µ is the Moebius function. Lemma 3.4. Let n = pa11 · · ·p as s , where aj ≥ 1, for j = 1, . . .s. If i is an integer such that i < ϕ(n) and d = gcd(i,n), then TrQ(ζn)/Q(ζ i n) 6= 0 ⇔ d = (n/P)tk and i = (n/P)k, where P = p1 · · ·ps, tk = gcd(k,P) and k = 0, 1, 2, . . . ,ϕ(P) − 1. Proof. If d = (n/P)tk, where tk = gcd(k,P), then the values that tk can assume are 1 and pα1 · · ·pαt, where 1 ≤ αr ≤ s, for r = 1, 2, . . . , t and αr 6= αl if r 6= l and 1 ≤ t < s. So, d = pa1−11 · · ·p as−1 s or d = pa1−11 · · ·p aα1 α1 · · ·p aαt αt · · ·pas−1s . Thus, n/d = p1 · · ·ps or n/d = p1 · · ·pα1−1pα1+1 · · ·pαt−1pαt+1 · · ·ps, and therefore, µ(n/d) = ±1 6= 0. But, from Lemma 3.3, it follows that TrQ(ζn)/Q(ζ i n) = ϕ(n) ϕ(n/d) µ(n/d) 6= 0. Now, if i = (n/P)k, then gcd(i,n) = d, for k = 0, 1, 2, . . . ,ϕ(P) − 1. In fact, if gcd(i,n) = d′, then gcd((n/P)k,n) = d′. Thus, gcd(((n/P)k)/(n/P)n/(n/P)) = d′/(n/P), that is, tk = gcd(k,P) = (P/n)d′. So, d′ = (n/P)tk = d and k = 0, 1, 2, . . . ,ϕ(P) − 1. From the Euler function, it follows that ϕ(n) ϕ(P) = ϕ(p a1 1 ···p as s ) ϕ(p1...ps) = ϕ(p a1 1 )···ϕ(p as s ) ϕ(p1)···ϕ(ps) = [p1 a1−1(p1−1)]···[psas−1(ps−1)] (p1−1)···(ps−1) = pa1−11 · · ·p as−1 s = n P . Thus, if k ≥ ϕ(P), then i = (n/P)k = ((ϕ(n))/(ϕ(P))k ≥ ((ϕ(n))/(ϕ(P))ϕ(P) = ϕ(n), which is a contradiction. Therefore, k = 0, 1, 2, . . . ,ϕ(P) − 1. Furthermore, d = (n/P)tk if and only if i = (n/P)k. Reciprocically, suppose that TrQ(ζn)/Q(ζ i n) 6= 0 com d 6= (n/P)tk. Thus, n/d is not square free. So, from definition of Euler function, it follows that µ(n/d) = 0, and therefore, TrQ(ζn)/Q(ζ i n) = 0, which is a contradiction. Lemma 3.5. Let n = pa11 · · ·p as s , where ar ≥ 1, for r = 1, . . .s. If i and j are integers such that i,j < ϕ(n) and d = gcd(i− j,n), then TrQ(ζn)/Q(ζ i−j n ) 6= 0 ⇔ d = (n/P)tk and |i− j| = (n/P)k, where P = p1 · · ·ps, tk = gcd(k,P) and k = 0, 1, 2, . . . ,ϕ(P) − 1. Proof. It is enough to observe that TrQ(ζn)/Q(ζ i−j n ) = TrQ(ζn)/Q(ζ j−i n ), gcd(i − j,n) = gcd(j − i,n) and as i,j < ϕ(n) then |i− j| < ϕ(n). Therefore, by Lemma 3.4, it follow the result. Lemma 3.6. Let n = pa11 · · ·p as s , where ar ≥ 1, for r = 1, . . .s. If i and j are integers such that i,j < ϕ(n) and d = gcd(i + j,n), then TrQ(ζn)/Q(ζ i+j n ) 6= 0 ⇔ d = (n/P)tk and i + j = (n/P)k, where P = p1 · · ·ps, tk = gcd(k,P) and k = 0, 1, . . . , 2ϕ(P) − 2 if n = P and k = 0, 1, . . . , 2ϕ(P) − 1, otherwise. 145 A. J. Ferrari et al. / J. Algebra Comb. Discrete Appl. 7(2) (2020) 141–160 Proof. For n = P , if k ≥ 2ϕ(P) − 1, then i + j = (n/P)k = k ≥ 2ϕ(P) − 1 = 2ϕ(n) − 1, which is a contradiction, since i,j ≤ ϕ(n) − 1 implies i + j ≤ 2ϕ(n) − 2. Thus, k = 0, 1, . . . , 2ϕ(P) − 2. For n > P, if k ≤ 2ϕ(P) − 1, then i + j = (n/P)k ≤ (ϕ(n)/ϕ(P))(2ϕ(P) − 1) = 2ϕ(n) − (n/P) < 2ϕ(n)−1. So, i + j ≤ 2ϕ(n) − 2. If k ≥ 2ϕ(P), then i + j = (n/P)k ≥ (ϕ(n)/ϕ(P))2ϕ(P) = 2ϕ(n), which is a contradiction, since i + j ≤ 2ϕ(n) − 2. Therefore, k = 0, 1, . . . , 2ϕ(P) − 1. Lemma 3.7. Let n = pa11 · · ·p as s , where ar ≥ 1, for r = 1, 2, . . .s. If i and j are integers such i,j ≤ ϕ(n)/2 − 1 and d = gcd(i + 2j,n), then TrQ(ζn)/Q(ζ i+2j n ) 6= 0 ⇔ d = (n/P)tk and i + 2j = (n/P)k, where P = p1 · · ·ps, tk = gcd(k,P) and k = 0, 1, . . . ,b3ϕ(P)/2 − 3P/nc, where byc is the greater integer less than or equal to y. Proof. If k > 3ϕ(P)/2 − 3P/n, then i + 2j = (n/P)k > (n/P) (3ϕ(P)/2 − 3P/n) = 3ϕ(n)/2 − 3, which is a contradiction, since i,j ≤ ϕ(n)/2 − 1 implies i + 2j ≤ 3ϕ(n)/2 − 3. Thus, k = 0, 1, . . . ,b3ϕ(P)/2 − 3P/nc. Lemma 3.8. Let n = pa11 · · ·p as s , where ar ≥ 1, for r = 1, 2, . . .s. If i and j are integers such i,j ≤ ϕ(n)/2 − 1 and d = gcd(−i + 2j,n), then TrQ(ζn)/Q(ζ −i+2j n ) 6= 0 ⇔ d = (n/P)tk and |− i + 2j| = (n/P)k, where P = p1 · · ·ps, tk = gcd(k,P) and k = 0, 1, 2, . . . ,bϕ(P) − 3P/nc. Proof. It is enough to observe that TrQ(ζn)/Q(ζ −i+2j n ) = TrQ(ζn)/Q(ζ i−2j n ), gcd(−i + 2j,n) = gcd(i−2j,n). If k > ϕ(P)−3P/n, then |−i+2j| = (n/P)k > (n/P) (ϕ(P) − 3P/n) = ϕ(n)−3, which is a contradiction, since i,j ≤ ϕ(n)/2−1 implies |−i+2j| ≤ ϕ(n)−3. Thus, k = 0, 1, . . . ,bϕ(P) − 3P/nc. Proposition 3.9. Let L = Q(ζn) and K = Q(ζn +ζ−1n ) be its maximal real subfield, where n = p a1 1 . . .p as s , with aj ≥ 1, for j = 1, 2, . . .s, m = ϕ(n). Let α = α0+α1(ζn+ζ−1n )+α2(ζ2n+ζ−2n )+· · ·+αm/2−1(ζ m/2−1 n + ζ −m/2+1 n ) be a totally positive element of Z[ζn + ζ−1n ], i. e., σi(α) > 0, for all i = 1, 2, . . . ,m/2, where σi are the m/2 distinct Q-homomorphisms from K to C. If x = a0 + a1(ζn + ζ−1n ) + a2(ζ 2 n + ζ −2 n ) + · · · + 146 A. J. Ferrari et al. / J. Algebra Comb. Discrete Appl. 7(2) (2020) 141–160 am 2 −1(ζ m/2−1 n + ζ −m/2+1 n ) is an element of Z[ζn + ζ−1n ], then TrK/Q(αx 2) = n P  ϕ(P)2 α0a20 + α0 u1∑ k=l1 n P k:even a2nk 2P ρ(tk) + 2α0a0 u2∑ k=l2 ank P ρ(tk) + 2α0 u3∑ k=l3 Bnk P ρ(tk) + ϕ(P)α0 m/2−1∑ j=1 a2j + 2α0 u4∑ k=l2 Ank P ρ(tk) + a 2 0 u2∑ k=l2 αnk P ρ(tk) + ∑ n6=i+2j= nk P l3≤k≤u5 1≤i≤m 2 −1 1≤j≤m 2 −1 2∑ i αia 2 j ρ(tk) + ϕ(P) 2∑ i+2j=n 1≤i≤m 2 −1 1≤j≤m 2 −1 2∑ i αia 2 j + ∑ 0≤|−i+2j|= nk P 0≤k≤u3 1≤i≤m/2−1 1≤j≤m/2−1 2∑ i αia 2 j ρ(tk) + 2a0 ∑ i+j= nk P l1≤k≤u1 1≤i≤m/2−1 1≤j≤m/2−1 2∑ i αiaj ρ(tk) + +2a0 ∑ 0≤|i−j|= nk P 0≤k≤u4 1≤i≤m/2−1 1≤j≤m/2−1 2∑ i αiaj ρ(tk) + 2 ∑ n6=i+j= nk P l4≤k≤u7 1≤i≤m/2−1 3≤j≤m−3 2∑ i αiBj ρ(tk) + 2ϕ(P) ∑ i+j=n 1≤i≤m/2−1 3≤j≤m−3 2∑ i αiBj + 2 ∑ 0≤|i−j|= nk P 0≤k≤u8 1≤i≤m/2−1 3≤j≤m−3 αiBj ρ(tk) + 2 u2∑ k=l2 m/2−1∑ j=1 αnk P a2jρ(tk) + 2 ∑ i+j= nk P l1≤k≤u3 1≤i≤m/2−1 1≤j≤m 2 −2 2∑ i αiAj ρ(tk) + 2 ∑ 0≤|i−j|= nk P 0≤k≤u4 1≤i≤m/2−1 1≤j≤m 2 −2 2∑ i αiAj ρ(tk) u1∑ k=l1 n P k:even   . where P = p1 · · ·ps, tk = gcd(k,P), dye is the smaller integer greater than or equal to y, byc is the greater integer less than or equal to y, l1 = d2P/ne, l2 = dP/ne, l3 = d3P/ne, l4 = d4P/ne, u1 = bϕ(P) − 2P/nc, u2 = bϕ(P)/2 −P/nc, u3 = bϕ(P) − 3P/nc, u4 = bϕ(P)/2 − 2P/nc, u5 = b3ϕ(P)/2 − 3P/nc, u6 = b(m− 2)/4c, u7 = b3ϕ(P)/2 − 4P/nc, u8 = bϕ(P) − 4P/nc, ρ(tk) = µ( Ptk )ϕ(tk) with µ the Mobius function and ϕ the Euler function, Aj = a1aj+1+a2aj+2+· · ·+am 2 −1−jam/2−1, Bj = ∑ k≥1 k 0, for all i = 1, 2, . . . ,m/2, where σi are the m/2 distinct Q-homomorphisms from K to C. A Gram matrix for the lattice Λ = σα(OK) is given by G = ( TrK/Q(αeiej) )m/2−1 i,j=0 , where (a) TrK/Q(αe0e0) = mα0 2 + n P u1∑ k=l1 αnk p ρ(tk), (b) For j ≥ 1, TrK/Q(αe0ej) = ϕ(n) ϕ(n/d1) µ(n/d1)α0 + n P ∑ r+j= nk P l2≤k≤u2 1≤r≤m/2−1 αrρ(tk) + n P ∑ |r−j|= nk P 0≤k≤u3 1≤r≤m/2−1 αrρ(tk), (c) For i,j ≥ 1, TrK/Q(αeiej) = ϕ(n) ϕ(n/d2) µ(n/d2)α0 + ϕ(n) ϕ(n/d3) µ(n/d3)α0 + n P ∑ r+i+j= nk P l3≤k≤u4 1≤r≤m/2−1 αrρ(tk) + n P ∑ |r−(i+j)|= nk P 0≤k≤u5 1≤r≤m/2−1 αrρ(tk) + n P ∑ |r+(i−j)|= nk P 0≤k≤u5 1≤r≤m/2−1 αrρ(tk) + n P ∑ |r+(j−i)|= nk P 0≤k≤u5 1≤r≤m/2−1 αrρ(tk), where P = p1 · · ·ps, tk = gcd(k,P), dye is the smaller integer greater than or equal to y, byc is the greater integer less than or equal to y, l1 = dP/ne, l2 = d2P/ne, l3 = d3P/ne, u1 = bϕ(P)/2 −P/nc, u2 = bϕ(P) − 2P/nc, u3 = bϕ(P)/2c, u4 = b3ϕ(P)/2 − 3P/nc, u5 = bϕ(P) − 3P/nc, d1 = gcd(j,n), d2 = gcd(i + j,n), d3 = gcd(i − j,n), ρ(tk) = µ( Ptk )ϕ(tk) with µ the Mobius function and ϕ the Euler function, and any sum must be disregarded if the lower bound of k is greater than the upper bound. Proof. From Lemmas 3 to 6 and following the same steps of the proof of the Proposition 3.9, the result follows. The next proposition, stated and proved in [19, Corollary 2.3], will be used in the next section. Before recalling it, we need a few assumptions. Let L/Q be a Galois extension of prime degree p such that p is unramifed in OL, the ring of integers of L. Denote the conductor of L by n, that is, n is the smallest positive integer such that L ⊆ Q(ζn). Then {σ(i)(θ)} p−1 i=0 is an integral basis for L where θ = TrQ(ζn)/L(ζn) and σ is any generator of Gal(L/Q). Proposition 3.11. Let x = p−1∑ i=0 aiσ (i)(θ) be any element in OL. Then TrL/Q(x 2) = n p−1∑ i=0 a2i − n− 1 p ( p−1∑ i=0 ai )2 . 151 A. J. Ferrari et al. / J. Algebra Comb. Discrete Appl. 7(2) (2020) 141–160 4. Constructions of algebraic lattices In this section, we construct algebraic lattices in Euclidean space with op- timal center density in dimensions 9, 10 and 11 which are rotated ver- sions of the lattices Λ9, Λ10 and Λ11 via twisted embeddings applied to Z-modules of the ring of integers of a number field K. We believe these constructions, as alge- braic lattices, are new in the literature. Constructions of rotated D3, D5 and E7-lattices via ideals and free Z-modules that are not ideals are also presented. The same lattices are also constructed in [15, 16], through a different approach, where the authors construct these lattices by shifting ideal lattices constructed over cyclotomic fields via ideal or module in the maximal totally real subfields of cyclotomic fields. 4.1. Construction of the D3-lattice If K = Q(ζ9 + ζ−19 ), then [K : Q] = 3 and dK = 3 4. If α = 1, then α is a totally positive element of Z[ζ9 + ζ−19 ] and N(α) = 1. If M is a submodule of OK generated by {2 + e1 + e2,−e1 + e2, 2 − e2}, where ej = ζ j 9 + ζ −j 9 , for j = 1, 2, then M is a submodule of OK of index 6 and δ(σα(M)) = tn/2 2n[OL : M] √ N(α)|dK| = 183/2 236 √ |34| = 1 4 √ 2 , i.e., with the same center density of the lattice D3. The norm equation |NK/Q(y)| = 2 has no solution in OK [9], however |NK/Q(y)| = 3 when y = 2 − e2 ∈ M. Thus, min 06=y∈M |NK/Q(y)| = 3. and the minimum norm in D3 is µ = 2. As NK/Q(α) = 1 and σα(M) is a scaled version of D3 with scale factor √ 9, from Equation (4), it follows that dp,rel(σα(M)) = ( 1 √ 18 )3 √ 1 × 3 = 0.03928, and therefore, 3 √ dp,rel(σα(M)) = 0.33994. 4.2. Construction of the D5-lattice If K = Q(ζ11 + ζ−111 ), then [K : Q] = 5 and dK = 11 4. If α = 2 − e1, where e1 = ζ11 + ζ−111 , then α is a totally positive element of Z[ζ11 + ζ−111 ] and N(α) = 11. If M is a submodule of OK generated by {2 +e1,−e1,−e2,−e3,−e4}, where ej = ζ j 11 +ζ −j 11 , for j = 1, 2, 3, 4, then σα(M) is a lattice of rank 5 and δ(σα(M)) = tn/2 2n[OL : M] √ N(α)|dK| = 225/2 26 √ 115 = 1 8 √ 2 , i.e., with the same center density of the lattice D5. In this case, min 06=y∈M |NK/Q(y)| = 1, because NK/Q(y) = 1, where y = 2 + e1 ∈M. The minimum norm in D5 is µ = 2. As NK/Q(α) = 11 and σα(M) is a scaled version of D5 with scale factor √ 11, by Equation (4), it follows that dp,rel(σα(M)) = ( 1 √ 22 )5 √ 11.1 = 0.00146, 152 A. J. Ferrari et al. / J. Algebra Comb. Discrete Appl. 7(2) (2020) 141–160 and therefore, 5 √ dp,rel(σα(M)) = 0.27097. 4.3. Construction of the E7-lattice If K is a subfield of L = Q(ζ29) such that [K : Q] = 7 and 〈σ〉 = Gal(K/Q), where σ : ζ29 7→ ζ729, then K = Q(θ), where θ = TrL/K(ζ29) = ζ29 + ζ −1 29 + ζ 12 29 + ζ −12 29 . If α = NL/K(1 − ζ29) = ∏ i∈{1,−1,12,−12} (1 − ζi29) = = −(6θ + 4σ(θ) + 4σ2(θ) + 4σ3(θ) + 4σ4(θ) + 3σ5(θ) + 4σ6(θ)), then TrK/Q(α) = NK/Q(α) = 29. If M = {a0θ + a1σ(θ) + · · · + a6σ6(θ) ∈OK; aj ≡ 0 (mod 2), for j = 0, 1, 2, 4}, then M is a Z-submodule of OK and M is not an ideal of OK, because for x = σ(θ) ∈ OK and y = σ3(θ) ∈ M it follows that xy = −3θ − 2σ(θ) − 4σ2(θ) − 2σ3(θ) − 3σ4(θ) − 3σ5(θ) − 4σ6(θ) /∈ M. Since dK = 29 6, it follows that δ(σα(M)) = tn/2 2n[OK : M] √ N(α)|dK| = (22.29)7/2 27.16 √ 29|296| = 1 16 , i.e., with the same center density of the lattice E7. As E7 is the only lattice with such center density in R7, it follows that σα(M) is a rotated version of E7. In this case, min 06=y∈M |NK/Q(y)| = 1, since for y = −2θ − 2σ2(θ) + σ3(θ) − σ6(θ) ∈ M, NK/Q(y) = 1. Now, the minimum norm in E7 is µ = 2. As NK/Q(α) = 29 and σα(M) is a scaled version of E7 with scale factor √ 58, by Equation (4), it follows that dp,rel(σα(M)) = ( 1 √ 116 )7 √ 29.1 = 3.203 × 10−7, and consequently, 7 √ dp,rel(σα(M)) = 0.11809. 4.4. Construction of the Λ9 and Λ10-lattice Let K be a number field such that K = Q(ζ180 + ζ−1180). In this case, [K : Q] = 24, taking e0 = 1 and ej = ζ j 180 + ζ −j 180, for j = 1, 2, . . . , 23, we have that {1,e1,e2, . . . ,e22,e23} is a basis of K. Let α = 165 + 129e2 + 153e4 + 120e6 + 119e8 + 105e10 + 67e12 + 82e14 + 25e16 + 49e18 + 3e20 + 17e22 be a 153 A. J. Ferrari et al. / J. Algebra Comb. Discrete Appl. 7(2) (2020) 141–160 totally positive element of Z[ζ180 + ζ−1180] and let M1 be a submodule of OK generated by the linearly independent vectors {w1,w2, · · · ,w10}, where w = Mv,w = (w1,w2, · · · ,w10)T ,v = (1,e1,e2, · · · ,e23)T and the matrix M is given by   8 −24 8 8 0 8 16 −16 24 0 −16 0 8 −24 16 8 −32 32 −8 8 −8 0 −24 8 8 −16 −8 24 −16 0 8 −72 8 24 −16 64 8 −56 16 8 −16 40 −8 −48 16 8 8 32 −8 −16 8 24 16 16 −8 −8 −8 24 16 16 −8 −40 −16 −8 16 24 8 −16 −16 −24 −8 0 0 −16 −8 24 0 8 8 −40 0 24 8 40 8 −48 −8 0 0 32 −24 −32 0 −8 16 16 −8 48 0 −32 16 −8 0 96 −8 −16 24 −56 0 88 −16 −24 8 −88 −16 32 −16 −16 8 −48 −16 −8 −8 24 16 16 −16 0 −32 32 40 24 −8 −32 −32 −16 40 8 −8 −24 −8 −32 32 −8 −56 −16 −16 16 56 8 −56 −24 −96 16 128 24 −32 −32 −104 0 136 24 −8 −16 −40 −8 80 8 0 8 −16 0 −16 8 −8 40 −16 8 0 −16 0 16 0 −16 16 −24 0 8 24 −24 32 −24 24 48 16 −52 −8 8 24 84 44 −36 −32 −48 12 88 24 −20 −56 −72 0 40 −8 0 −48 −36 −16 52 0 −60 8 −4 −24 104 −16 −36 20 −56 −12 96 −12 −28 28 −80 48 0 4 −8 20 −44   . In this case, σα(M1) is a lattice of rank 10 in R24 and forall x ∈M1 we have that σα(x) = γTA, where γ = (a1,a2,a3,a4,a5,a6,a7,a8,a9,a10), T = 4   2 −6 8 2 0 2 4 −4 6 0 −4 0 2 −6 4 2 −8 8 −2 2 −2 0 −6 2 2 −4 −2 6 −4 0 2 −18 2 6 −4 16 2 −14 4 2 −4 10 −2 −12 4 2 2 8 −2 −4 2 6 4 4 −2 −2 −2 6 4 4 −2 −10 −4 −2 4 6 2 −4 −4 −6 −2 0 0 −4 −2 6 0 2 2 −10 0 6 2 10 2 −12 0 0 −2 8 −6 −8 0 −2 4 4 −2 12 0 −8 4 −2 0 24 −2 −4 6 −14 0 22 −4 −6 2 −22 −4 8 −4 −4 2 −12 −4 −2 −2 6 4 4 −4 0 −8 8 10 6 −2 −8 −8 −4 10 2 −2 −6 −2 −8 8 −2 −14 −4 −4 4 14 2 −14 −6 −24 4 32 6 −8 −8 −26 0 34 6 −2 −4 −10 −2 20 2 0 2 −4 0 −4 2 −2 10 −4 2 0 −4 0 4 0 −4 4 −6 0 2 6 −6 8 −6 6 12 4 −13 −2 −2 6 21 11 −9 −8 −12 3 22 6 −5 −14 −18 0 10 −2 0 −12 −9 −4 13 0 −15 2 −1 −6 26 −4 −9 5 −14 −3 24 −3 −7 7 −20 0 12 1 −2 5 −11   and A =   √ σ1(α)σ1(1) · · · √ σ24(α)σ24(1)√ σ1(α)σ1(e1) · · · √ σ24(α)σ24(e1) ... ... ...√ σ1(α)σ1(e23) · · · √ σ24(α)σ24(e23)   , where A is a generator matrix of lattice σα(OK), since {1,e1,e2, . . . ,e22,e23} is a Z-basis for OK. We have that B1 = TA is a generator matrix for σα(M1), and as B1 has rank 10, it follows that the lattice σα(M1) has rank 10. A Gram matrix for the lattice σα(M1) is Q = TGTt, where G = ( TrK|Q(αeiej) )23 i,j=0 is a 154 A. J. Ferrari et al. / J. Algebra Comb. Discrete Appl. 7(2) (2020) 141–160 Gram matrix for the lattice σα(OK). From Proposition 3.10, it follows that the matrix G is given by 90   47 0 68 0 89 0 72 0 80 0 76 0 69 0 76 0 62 0 71 0 55 0 64 0 0 162 0 157 0 161 0 152 0 156 0 145 0 145 0 138 0 133 0 126 0 119 0 , 112 68 0 183 0 140 0 169 0 148 0 149 0 152 0 131 0 147 0 117 0 135 0 103 0 0 157 0 166 0 148 0 165 0 141 0 156 0 138 0 140 0 131 0 126 0 119 0 111 89 0 140 0 174 0 144 0 158 0 148 0 142 0 147 0 124 0 140 0 110 0 127 0 0 161 0 148 0 170 0 137 0 165 0 134 0 151 0 131 0 133 0 124 0 , 118 0 109 72 0 169 0 144 0 163 0 144 0 151 0 143 0 135 0 140 0 117 0 132 0 100 0 0 152 0 165 0 137 0 170 0 130 0 160 0 127 0 144 0 124 0 125 0 114 0 109 80 0 148 0 158 0 144 0 156 0 139 0 144 0 136 0 128 0 132 0 107 0 123 0 0 156 0 141 0 165 0 130 0 165 0 123 0 153 0 120 0 136 0 114 0 116 0 106 76 0 149 0 148 0 151 0 139 0 149 0 132 0 137 0 128 0 118 0 123 0 99 0 0 145 0 156 0 134 0 160 0 123 0 158 0 116 0 145 0 110 0 127 0 106 0 102 69 0 152 0 142 0 143 0 144 0 132 0 142 0 124 0 127 0 119 0 110 0 109 0 0 145 0 138 0 151 0 127 0 153 0 116 0 150 0 106 0 136 0 102 0 113 0 100 76 0 131 0 147 0 135 0 136 0 137 0 124 0 132 0 115 0 119 0 105 0 104 0 0 138 0 140 0 131 0 144 0 120 0 145 0 106 0 141 0 98 0 122 0 96 0 96 62 0 147 0 124 0 140 0 128 0 128 0 127 0 115 0 124 0 101 0 113 0 88 0 0 133 0 131 0 133 0 124 0 136 0 110 0 136 0 98 0 127 0 92 0 105 0 93 71 0 117 0 140 0 117 0 132 0 118 0 119 0 119 0 101 0 118 0 84 0 110 0 0 126 0 126 0 124 0 125 0 114 0 127 0 102 0 122 0 92 0 110 0 89 0 87 55 0 135 0 110 0 132 0 107 0 123 0 110 0 105 0 113 0 84 0 115 0 66 0 0 119 0 119 0 118 0 114 0 116 0 106 0 113 0 96 0 105 0 89 0 92 0 82 64 0 103 0 127 0 100 0 123 0 99 0 109 0 104 0 88 0 110 0 66 0 108 0 0 112 0 111 0 109 0 109 0 106 0 102 0 100 0 96 0 93 0 87 0 82 0 80   , and consequently, Q = TGTt = 5760   4 −2 0 0 0 0 0 0 0 0 −2 4 −2 2 0 0 0 0 0 0 0 −2 4 0 0 2 0 0 0 0 0 2 0 4 2 2 0 0 0 0 0 0 0 2 4 2 0 0 2 1 0 0 2 2 2 4 2 2 1 2 0 0 0 0 0 2 4 2 0 2 0 0 0 0 0 2 2 4 0 2 0 0 0 0 2 1 0 0 4 2 0 0 0 0 1 2 2 2 2 4   , such that det(Q) = 278321510. By Proposition 3.9, the trace form of x ∈M1 is given by TrK/Q(αx 2) = 23040a21 − 23040a1a2 + 23040a22 − 23040a2a3 + 23040a23 + 23040a2a4 + 23040a24 +11520a10a5 +23040a4a5 +23040a 2 5 +23040a10a6 +23040a3a6 +23040a4a6 +23040a5a6 + 23040a26 + 23040a10a7 + 23040a6a7 + 23040a 2 7 + 23040a10a8 + 23040a6a8 + 23040a7a8 + 23040a28 + 23040a10a9 + 23040a5a9 + 11520a6a9 + 23040a 2 9 + 23040a 2 10. Thus, t = min{TrK/Q(αx2); x ∈ M1, x 6= 0} = 23040 with a1 = 1 and aj = 0, for j 6= 1. By Equation (1) it follows that the center density of lattice σα(M1) is given by δ(σα(M1)) = tm/2 2m √ det(Q) = (23040)10/2 210 √ 278321510 = 1 16 √ 3 . Therefore, σα(M1) is a lattice of rank 10 with the same center density of Λ10. As 1 5760 Q is a standard Gram matrix of Λ10 [12], it follows that 1 √ 5760 σα(M1) is a rotated version of Λ10. Using a computer, 155 A. J. Ferrari et al. / J. Algebra Comb. Discrete Appl. 7(2) (2020) 141–160 we observed that min 0 6=y∈M |NK/Q(y)| ≤ 4.13 × 1020. Now, the minimum norm in Λ10 is µ = 4. As NK/Q(α) = 8.3 × 109 and σα(M) is a scaled version of Λ10 with scale factor √ 5760, by Equation (4) dp,rel(σα(M)) ≤ ( 1 √ 23040 )24 √ 8.3 × 109 × 4.13 × 1020 = 1.68 × 10−27, and consequently, 10 √ dp,rel(σα(M)) ≤ 0.00210. Now, let M2 be a submodule of OK generated by the linearly independent vectors {w1,w2, . . . ,w9}, following the same steps as the construction of Λ10, it follows that T = 4   2 −6 8 2 0 2 4 −4 6 0 −4 0 2 −6 4 2 −8 8 −2 2 −2 0 −6 2 2 −4 −2 6 −4 0 2 −18 2 6 −4 16 2 −14 4 2 −4 10 −2 −12 4 2 2 8 −2 −4 2 6 4 4 −2 −2 −2 6 4 4 −2 −10 −4 −2 4 6 2 −4 −4 −6 −2 0 0 −4 −2 6 0 2 2 −10 0 6 2 10 2 −12 0 0 −2 8 −6 −8 0 −2 4 4 −2 12 0 −8 4 −2 0 24 −2 −4 6 −14 0 22 −4 −6 2 −22 −4 8 −4 −4 2 −12 −4 −2 −2 6 4 4 −4 0 −8 8 10 6 −2 −8 −8 −4 10 2 −2 −6 −2 −8 8 −2 −14 −4 −4 4 14 2 −14 −6 −24 4 32 6 −8 −8 −26 0 34 6 −2 −4 −10 −2 20 2 0 2 −4 0 −4 2 −2 10 −4 2 0 −4 0 4 0 −4 4 −6 0 2 6 −6 8 −6 6 12 4 −13 −2 −2 6 21 11 −9 −8 −12 3 22 6 −5 −14 −18 0 10 −2 0 −12 −9   , and consequently a Gram matrix for the lattice σα(M2) is given by Q = TGTt = 5760   4 −2 0 0 0 0 0 0 0 −2 4 −2 2 0 0 0 0 0 0 −2 4 0 0 2 0 0 0 0 2 0 4 2 2 0 0 0 0 0 0 2 4 2 0 0 2 0 0 2 2 2 4 2 2 1 0 0 0 0 0 2 4 2 0 0 0 0 0 0 2 2 4 0 0 0 0 0 2 1 0 0 4   , such that det(Q) = 27231859. By Proposition 3.9 the trace form of x ∈M2 is given by TrK/Q(αx 2) = 23040a21 − 23040a1a2 + 23040a22 − 23040a2a3 + 23040a23 + 23040a2a4 + 23040a24 + 23040a4a5 + 23040a 2 5 + 23040a3a6 + 23040a4a6 + 23040a5a6 + 23040a 2 6 + 23040a6a7 + 23040a27 + 23040a6a8 + 23040a7a8 + 23040a 2 8 + 23040a5a9 + 11520a6a9 + 23040a 2 9. Thus, t = min{TrK/Q(αx2); x ∈ M2, x 6= 0} = 23040 with a1 = 1 and aj = 0, for j 6= 1. By Equation (1) it follows that the center density of lattice σα(M2) is given by δ(σα(M2)) = tm/2 2m √ det(Q) = (23040)9/2 29 √ 27231859 = 1 16 √ 2 . Therefore, σα(M2) is a lattice of rank 9 with the same center density of Λ9. As 1 5760 Q is a standard Gram matrix of Λ9 [12], it follows that 1 √ 5760 σα(M2) is a rotated version of Λ9. Using a computer, we observed 156 A. J. Ferrari et al. / J. Algebra Comb. Discrete Appl. 7(2) (2020) 141–160 that min 06=y∈M |NK/Q(y)| ≤ 2.28 × 1021. Now, the minimum norm in Λ9 is µ = 4. As NK/Q(α) = 8.3 × 109 and σα(M) is a scaled version of Λ9 with scale factor √ 5760, by Equation (4) dp,rel(σα(M)) ≤ ( 1 √ 23040 )24 √ 8.3 × 109 × 2.28 × 1021 = 9.29 × 10−27, and consequently, 9 √ dp,rel(σα(M)) ≤ 0.00128. 4.5. Construction of the Λ11-lattice Let L = Q(ζ23), K the subfield of L such that [K : Q] = 11, and 〈σ〉 = Gal(K/Q), where σ : ζ23 7→ ζ223. Then K = Q(θ) where θ = TrL/K(ζ23) = ζ23 + ζ −1 23 . A Z-basis for OK, the ring of integers of K, is {σi(θ)}10i=0. By Proposition 3.11, the trace form of K is given by TrK/Q(x 2) = 23(a20 + a 2 1 + · · · + a 2 10) − 2(a0 + a1 + · · · + a10) 2, where x = a0θ + a1σ(θ) + · · · + a10σ10(θ). Let γ = NL/K(1 − ζ23) = (1 − ζ23)(1 − ζ−123 ) = −θ − 2 10∑ i=0 σi(θ). It follows that TrK/Q(γ) = NK/Q(γ) = 23. Furthermore, let � = 6 10∑ i=0 σi(θ) + 4σ6(θ) + σ7(θ) be a unit in OK. The element α = �γ is totally positive and NK/Q(α) = 23. Let Λ0 denote the lattice σα(OK), where σα : K → R11 is the twisted homomorphism given by σα(x) = ( √ σ0(α)σ0(x), √ σ(α)σ(x), . . . , √ σ10(α)σ10(x)) and M be the Z-submodule of OK defined by M = {a0θ + a1σ(θ) + · · · + a10σ10(θ) ∈OK; aj ≡ 0 (mod 2), for j = 2, 4, 6, 8, 10}, then the Gram matrix of σα(M) is given by G = 23   4 −2 0 2 −2 −2 0 2 0 0 −2 −2 4 0 −1 4 0 0 −1 −4 −1 2 0 0 12 0 0 −2 0 4 −8 −2 −4 2 −1 0 4 −6 −2 −2 0 0 2 0 −2 4 0 −6 24 0 0 0 −4 −6 0 −2 0 −2 −2 0 4 4 −2 0 −1 0 0 0 0 −2 0 4 12 −2 −4 −4 0 2 −1 4 0 0 −2 −2 4 −2 0 −4 0 −4 −8 0 −4 0 −4 −2 16 4 4 0 −1 −2 2 −6 −1 −4 0 4 4 0 −2 2 −4 0 0 0 0 −4 4 0 8   . 157 A. J. Ferrari et al. / J. Algebra Comb. Discrete Appl. 7(2) (2020) 141–160 We have that t = min{TrK/Q(αx2) : x ∈M,x 6= 0} = 22 × 23, with a0 = 1, a1 = a2 = · · · = a10 = 0 and [OK : M] = 32. Since dK = 2310, the center density of the lattice σα(M) is given by δ(σα(M)) = tn/2 2n[OK : M] √ N(α)|dK| = (22.23)11/2 211 × 32 √ 23|2310| = 1 32 , i.e., the same as that of the Λmax11 lattice [12, Chapter 6, Section 4]. The above lattice is a rotated version of Λmax11 . Indeed, G ′ = 1 23 UGUt, where G′ =   4 −2 0 0 0 0 0 0 0 0 0 −2 4 −2 2 0 0 0 0 0 0 0 0 −2 4 0 0 2 0 0 0 0 0 0 2 0 4 2 2 0 0 0 0 0 0 0 0 2 4 2 0 0 2 1 0 0 0 2 2 2 4 2 2 1 2 0 0 0 0 0 0 2 4 2 0 2 0 0 0 0 0 0 2 2 4 0 2 0 0 0 0 0 2 1 0 0 4 2 0 0 0 0 0 1 2 2 2 2 4 2 0 0 0 0 0 0 0 0 0 2 4   is a Gram matrix of Λmax11 and U =   −1 0 0 3 1 2 1 3 0 2 1 0 0 0 −2 −1 −2 −1 −2 0 −2 −1 −1 −2 −1 −1 0 −2 0 −1 −1 0 0 1 0 1 1 1 2 0 1 0 2 1 3 2 2 1 1 3 0 −1 1 2 0 0 0 0 −1 0 −2 0 −2 0 0 −1 −2 0 −1 −2 −1 −4 0 −2 0 −2 −2 −1 0 −1 −2 −1 −4 0 −2 0 −2 −2 −1 1 0 2 0 0 0 0 1 −2 −1 −2 0 −1 −1 −1 −4 0 −2 0 −3 −2 −2 −2 −1 −2 −1 −4 0 −2 −1 −2 −1   is an element of GL(11,Z). Thus, min 06=y∈M |NK/Q(y)| = 1, since for y = −θ ∈M, NK/Q(y) = 1. Now, the minimum norm in Λ11 is µ = 4. As NK/Q(α) = 23 and σα(M) is a scaled version of Λ11 with scale factor√ 23, by Equation (4) dp,rel(σα(M)) = ( 1 √ 92 )11 √ 23.1 = 7.58 × 10−11, and consequently, 11 √ dp,rel(σα(M)) = 0.12022. 5. Conclusions In this section, we assess the performance of the lattices presented in this paper in terms of center density and relative minimum product distance. These parameters are associated with the efficiency in signal transmission over Gaussian and Rayleigh fading channels, respectively. Table 1 shows a comparison between the best known relative product distance of rotated Zn-lattices, obtained via cyclotomic, cyclic and mixed constructions [5] and via Krüskemper method [18] (First column), the densest lattices Λ0 158 A. J. Ferrari et al. / J. Algebra Comb. Discrete Appl. 7(2) (2020) 141–160 (D3, D5 and E7) obtained in [15, 16] (Second column), and the densest lattices Λn (Λ3 = D3, Λ5 = D5, Λ7 = E7 and Λ12 = K12) from our construction (Third column). The center density δ of these lattices are also displayed. For the lattice D3 the minimum product distance 0.36964 presented in [16, Table 3] is higher than that from our construction, while for dimension 5 our construction yields the same value. In dimension 7 (see [15, Remark 4.13]) the authors obtained a lower bound on the minimum product distance whereas for our construction the bound holds. For dimensions 9 and 10, we constructed the lattices Λ9 and Λ10 with an upper bound for their minimum product distance. Although the bounds are not very high, the lattices in those two dimensions and in dimension 11 are new and are not known in the literature (as algebraic lattices). In particular, the value that we obtained in dimension 11 is half of the one obtained for Zn (cyclotomic and cyclic constructions in [5]). A broader question to be investigated is whether algebraic constructions of lattices, mainly in dimen- sions 9, 10 and 11, as the ones approached here, can provide greater relative minimum product distance for rotated densest lattices. It is noticed that the relative minimum product distances dp,rel(Λn) of the rotated lattices obtained in the present paper are smaller than the relative minimum product distances dp,rel(Zn) of rotated Zn- lattices constructed for the Rayleigh channels in [5, 18]. Nevertheless, if the goal is to construct lattices which have good performance on both Gaussian and Rayleigh channels, were may assert that taking into account the trade-off center density versus product distance, there are some advantages in considering the rotated Λn-lattices instead of rotated Zn-lattices. Table 1. 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Washington, Introduction to Cyclotomic Fields, Springer-Verlag, 1995. 160 https://doi.org/10.5802/jtnb.500 https://doi.org/10.5802/jtnb.500 https://doi.org/10.1007/s00013-005-1469-9 https://doi.org/10.1007/s00013-005-1469-9 https://doi.org/10.1007/s00208-013-0932-3 https://doi.org/10.1007/s00208-013-0932-3 https://doi.org/10.1006/jsco.1996.0125 https://doi.org/10.1006/jsco.1996.0125 https://doi.org/10.1109/18.485720 https://doi.org/10.1109/18.485720 https://www.jstor.org/stable/25662173 https://www.jstor.org/stable/25662173 https://doi.org/10.1142/S021949881850175X https://doi.org/10.1142/S021949881850175X https://doi.org/10.1142/S0219498815500450 https://doi.org/10.1142/S0219498815500450 https://doi.org/10.1016/j.jalgebra.2014.12.044 https://doi.org/10.1016/j.jalgebra.2014.12.044 https://doi.org/10.1016/j.jnt.2012.05.002 https://doi.org/10.1016/j.jnt.2012.05.002 https://doi.org/10.1109/ISIT.2003.1228051 https://doi.org/10.1109/ISIT.2003.1228051 https://doi.org/10.1216/RMJ-2017-47-4-1075 https://doi.org/10.1216/RMJ-2017-47-4-1075 Introduction Background on number fields and algebraic lattices Trace forms for cyclotomic fields Constructions of algebraic lattices Conclusions References