PRODUCT ALGEBRAS

GR ´EGORY BERHUY, RICHARD SLESSOR

Abstract. In this paper, we explain how to construct reliable codes for wireless communication channels using crossed product division algebras, and we prove the optimality of the codes already constructed on cyclic algebras and biquadratic crossed products.

Contents

Introduction 1

1. Space-time coding and crossed product algebras 3

1.1. Algebraic space-time codes. 3

1.2. Codes based on crossed product K-algebras. 6 1.3. Minimum determinant of a crossed product based code 10 2. Optimality of codes based on cyclicK-algebras 12

2.1. Preliminaries 12

2.2. The case n= 4 14

2.3. The case n= 6 18

3. Optimality of codes based on biquadratic crossed products 22

Appendix A. Complex ideal lattices 31

A.1. Generalities on hermitian lattices 31

A.2. Complex ideal lattices 33

References 38

Introduction

Within the last few years we have seen a notable increase in the use of wireless communication, which has led to the need for higher data rates. In view of this multiple antenna communication systems have been investigated, which can provide very high data rates particularly

1

when there is perfect channel state information at the receiver. The problem addressed in the design of space-time codes in the coherent case (that is, in the case where the receiver knows the properties of the channel) can be summarized as follows: find a set C of complex n×n matrices such that the minimum determinant

δ_{min}(C) = inf

X6=X^{0}∈C|det(X−X^{0})|^{2}

is maximal. Of course, the first step is to ensure that δ_{min}(C) is not
zero. When this is the case, we will say that C is fully diverse. One
natural way to achieve this is to use division algebras. Indeed, any
division algebra D whose centerK is a subfield of Cmay be identified
to a subring of a matrix algebra M_{n}(C). In particular, D^{×} may be
identified to a subgroup of GL_{n}(C), and taking C to be a subset of D
yields a fully diverse algebraic space-time code.

The use of division algebras for space-time coding started with the seminal work by B. A. Sethuraman and B. Sundar Rajan [16]. Number fields and cyclic algebras were discussed, which have been a favourite tool for space-time design. Some surveys are by now available [9, 15], and we refer the interested reader to them for further details. Other algebras have also been explored, such as crossed product algebras [2]

or non-associative algebras [13].

Finding codes that are fully diverse led to an interest in constructing codes from division algebras [16], in particular cyclic division algebras.

This work generated a lot of interest and in [17] constructions of codes based on crossed product algebras were given that included the codes given in [16] as a subset. An approach based on cyclic division algebras, which differs from [16] was given in [11]. This paper introducedperfect codes (PSTBCs). These codes satisfy a large number of properties including a shaping constraint that is related to the cubic lattice. In the paper the authors give examples of perfect codes in dimensions 2,3,4 and 6. Codes from non-cyclic division algebras have also been investigated. In [2] the authors consider biquadratic crossed product algebras and construct a code with good performance in dimension 4.

In [3] it is shown that PSTBCs only exist in these dimensions, although by relaxing the definition slightly PSTBCs can exist for any number of antennas [5]. We will concentrate on the former case. The optimality of perfect codes has been studied and in [8] it is shown that the golden code, a PSTBC of dimension 2 presented in [1], is optimal with respect to the coding gain.

In this paper we will prove the optimality of the perfect codes of dimen- sion 4 and 6, as well as the optimality of the biquadratic code presented in [2]. We will also generalise bounds on the minimum determinant of

codes based on cyclic division algebras to the case of codes based on crossed product algebras.

This paper is organised as follows. In Section 1 some basic aspects of coding theory and the wireless channel are introduced as well as the bound on the pairwise probability of error. It then explains how division algebras can be used to give fully diverse codes and gives a description of codes based on crossed product algebras. The section also introduces the energy constraint and its link with the cubic lat- tice. This leads to Section A which introduces complex ideal lattices and gives results that will be necessary in deciding when it is possible to construct the cubic lattice. Bounds on the minimum determinant of codes based on crossed product algebras are also derived. Section 2 is then concerned with proving the optimality of the PSTBCs of dimen- sion 4 and 6 given in [11]. Finally in Section 3 we prove that the code constructed in [2] is optimal.

1. Space-time coding and crossed product algebras 1.1. Algebraic space-time codes. As explained in the introduction, a code C will have a better performance if we design the codebook in such a way that the minimum determinant is as large as possible.

Of course, the first step is to ensure that δ_{min}(C) > 0. The main
difficulty comes from the non-linearity of the determinant. The idea is
then to choose C to be a (large) finite subset of a subringD of M_{n}(C),
which is also a division ring. In this case, the difference of two distinct
codewords will lie in D^{×} ⊂GLn(C). But how to find explicitly such a
division ring?

This is where (central) simple K-algebras come into play. Assume that
A is a simple finite dimensional K-algebra, and let L/K be a field
extension such that L is a K-subalgebra of A (in particular L/K has
finite degree). The product law induces on A the structure of a finite
dimensional right L-vector space of dimension r. In particular, if we
denote by M_{a} ∈ M_{r}(L) the matrix of left multiplication by a ∈ A
in a fixed basis of the right L-vector space A, we get a K-algebra
homomorphism

ϕ: A−→M_{r}(L)⊂M_{r}(C)

a7−→M_{a} ,

which is injective since A is simple.

In particular, if A is a divisionK-algebra, ϕ(A) is a subring of M_{r}(C)
which has the required properties, and we may take our codebook C to
be a finite subset of C_{A,L} ={M_{a} |a∈A} ⊂M_{r}(C).

The previous way of encoding will introduce enough redundancy to prevent the loss of too much information during transmission. However

we cannot introduce too much redundancy either, since sending an
information symbol has an energy cost. One goal is of course to encode
as much information as possible in a single matrix without losing too
much information after transmission. The rate of a code C ⊂ M_{r}(C)
is the ratio of information symbols per coefficients sent. Of course, it
is in our best interest to design codes with high rates.

Let us compute the rate r(C) of a code C ⊂ CA,L. The information
symbols that we would like to transmit are elements of K, which may
be used to define elements of A. Each element of a may then carry
dim_{K}(A) information symbols. However, an r×r matrix may contain
r^{2} information symbols, so we have

r(C) = dimK(A)
r^{2} ,

wherer = dim_{L}(A). Now since dim_{K}(A) = dim_{L}(A)[L:K] =r[L:K],
this may be rewritten as

r(C) = [L:K]^{2}
dim_{K}(A),

so we should choose L/K such that [L:K] is as large as possible. IfA
is a central simple K-algebra, it is known that [L:K] ≤deg(A) (see
[4, Lemma IV.1.4] for example). In particular, if Ais a central division
K-algebra andL is a maximal subfield of A (that is a subfield L such
that dim_{K}(A) = [L: K]^{2}), the corresponding code has a rate equal to
1, which is the maximal possible value in this case. Such a maximal
subfield always exists by [4, Corollary IV.1.9].

Once full-diversity is achieved, encoding need to satisfy two extra con- ditions.

First of all, encoding needs to satisfy theshaping constraint, in order to preserve the energy cost:

An element X ∈ C represents some information which is already en- coded. In fact, if we write

X=M_{a}, a=e_{1}a_{1}+. . .+e_{n}a_{n}, a_{i} ∈L

the real information which is transmitted is represented by the elements
a_{ij} ∈ O_{K} defined by

a_{i} =

n

X

j=1

a_{ij}ω_{j}, i= 1, . . . , n.

Sending the n^{2} information symbols a_{ij} has an energy cost, which is
represented by the real number

X

i,j

|a_{ij}|^{2}.

Now sending X= (x_{ij})∈M_{n}(L) will have an energy cost equal to
X

i,j

|xij|^{2}.

Shaping constraint: Encoding the information symbols a_{ij} into the
matrix M_{a} needs to preserve the energy cost.

We refer the reader willing to know more about shaping constraints to
[10]. Notice that, once an L-basis of A and anO_{K}-basis are fixed, if x
and a are the two vectors whose coordinates are thex_{ij}’s and the a_{ij}’s
respectively, one may write

x=Ma,

for some M ∈ M_{n}^{2}(L) which does not depend on a. The energy con-
straint then states

M^{t}M = I_{n}^{2}.

Now let us introduce the second condition. One major problem is thatC
may have in practical applications a large number of elements. Thus, it
could turn out thatδ_{min}(C) is very close to 0 ifC is not chosen carefully,
simply because C_{A,L} contains matrices of arbitrary small determinant.

To avoid this problem, we will ask forδmin(C) to be bounded away from
zero, independently from the size of C. When K is equal to Q or to
a totally imaginary quadratic extension of Q, one way to achieve this
is to ask the coordinates a_{1}, . . . , a_{n} to lie in an ideal I of the ring of
integers O_{L} of L (see [4, Corollary IV.3.8]). In other words, C will be
a subset of

C_{A,I} ={M_{a}|a ∈Λ_{A,I} =e_{1}I⊕ · · · ⊕e_{n}I},
where e_{1}, . . . , e_{n} is an L-basis of the right L-vector spaceA.

Notice that in practice, a signal is represented by an element of Q(i)
or Q(j), wherej =e^{2iπ/3} (see [6] for more details), so the condition on
K is automatically fulfilled.

We will assume therefore in the sequel that K = Q(i) or Q(j), which
both satisfy the assumptions of the previous corollary. Notice that in
this case, O_{K} is a principal ideal domain, and any ideal I ofO_{L}has an
O_{K}-basis ω_{1}, . . . , ω_{n}.

Of course, in order to deal with this constraint, we first need examples of central simpleK-algebras for which we can find an explicit maximal commutative subfield, and an explicit L-basis. In [11], the authors use cyclic algebras to construct codes with good performances, as well as computations or estimations of the minimum determinant. Observe now that cyclic algebras are particular examples of crossed product algebras.

The goal of the next section is to give a general construction of space- time codes on a crossed product K-algebras, to derive the shaping constraint in more tractable terms, and to generalize the estimations of the minimum determinant obtain in the case of cyclic algebras in [11].

1.2. Codes based on crossed product K-algebras. LetL/K be a Galois field extension of degree n, with Galois group G. If σ ∈G and λ ∈L, we set

λ^{σ} =σ^{−1}(λ).

Let ξ : G×G → L^{×} be a map satisfying the cocyclicity conditions,
namely

ξ_{Id,τ} =ξ_{σ,Id} = 1, ξ_{σ,τ ρ}ξ_{τ,ρ} =ξ_{στ,ρ}ξ_{σ,τ}^{ρ} for all σ, τ, ρ∈G.

The crossed productK-algebra (ξ, L/K, G) is theK-algebra generated
as a right L-vector space by elements (e_{σ})_{σ∈G} subject to the relations

e_{σ}e_{τ} =e_{στ}ξ_{σ,τ}, λe_{σ} =e_{σ}λ^{σ} for all σ, τ ∈G, λ∈L.

One can show that (ξ, L/K, G) is a central simpleK-algebra of degree n, with maximal commutative subfield L (see [4, Lemma VI.1.7], for example).

In particular, we get an injective K-algebra homomorphism
ϕ: A−→M_{n}(L)

a7−→Ma

associated to this particular K-algebra. In order to derive explicitly the shaping constraint associated to this K-algebra, we first need to compute the multiplication matrix of an element a∈A.

Lemma 1.1. Let A = (ξ, L/K, G), and let a = X

σ∈G

e_{σ}a_{σ}. Then the
matrix M_{a} of left multiplication by a, relative to the L-basis (e_{σ})_{σ∈G} is

Ma= (ξ_{στ}^{−1}_{,τ}a^{τ}_{στ}−1)σ,τ.
Proof. For all τ ∈G, we have

aeτ = X

σ∈G

eσaσeτ

= X

σ∈G

e_{σ}e_{τ}a^{τ}_{σ}

= X

σ∈G

e_{στ}ξ_{σ,τ}a^{τ}_{σ}

= X

σ∈G

e_{σ}ξ_{στ}^{−1}_{,τ}a^{τ}_{στ}−1

This concludes the proof.

We now derive the shaping constraint we are looking for. As explained
in the previous section, from now on, we assume that K = Q(i) or
Q(j). Notice that in this case,O_{K} is a principal ideal domain, and any
ideal I of O_{L} has an O_{K}-basis ω_{1}, . . . , ω_{n}.

Since any ideal of O_{L} is a free O_{K}-module of rank n = |G|, we will
index any O_{K}-basis ofI with the elements of G.

Let A = (ξ, L/K, G) and let I be an ideal of OL. If we choose an
OK-basis (ωσ)σ∈G of I, we will (temporarily) encode n^{2} information
symbols (a_{σ,τ})_{σ,τ∈G} into the matrix M_{a} ∈ C_{A,I}, where

a =X

σ∈G

e_{σ}(X

τ∈G

a_{σ,τ}ω_{τ}).

Proposition 1.2. With this way of encoding, the shaping constraint is satisfied if and only if the two following conditions are fulfilled:

(1) |ξ_{σ,τ}|^{2} = 1 for all σ, τ ∈G

(2) The matrix W = (ω^{σ}_{τ})_{σ,τ} is unitary.

Proof. For σ∈G, set a_{σ} =X

τ∈G

a_{σ,τ}ω_{τ}. We would like to have

X

σ,τ∈G

|a_{σ,τ}|^{2} = X

σ,τ∈G

|ξ_{στ}^{−1}_{,τ}a^{τ}_{στ}−1|^{2} = X

σ,τ∈G

|ξ_{σ,τ}a^{τ}_{σ}|^{2}

for all a_{σ,τ} ∈ O_{K}. For σ ∈ G, we consider the two column vectors of
L^{n}

X_{σ} = (ξ_{σ,ρ}a^{ρ}_{σ})ρ∈G, A_{σ} = (a_{σ,ρ})ρ∈G.

Let Dσ be the diagonal matrix of Mn(L) whose non-zero entry at col- umn ρ is ξσ,ρ. Since aσ,τ ∈K, we have

ξ_{σ,τ}a^{τ}_{σ} =X

ρ∈G

ξ_{σ,τ}a^{τ}_{σ,ρ}ω_{ρ}^{τ} =X

ρ∈G

ξ_{σ,τ}a_{σ,ρ}ω_{ρ}^{τ}.

Now if W = (ω_{ρ}^{τ})_{τ,ρ}, then D_{σ}W = (ξ_{σ,τ}ω_{ρ}^{τ})_{τ,ρ}, and therefore we get

X_{σ} =D_{σ}W A_{σ} for all σ ∈G.

Let xand abe the block column vectors defined by

x=

...
X_{σ}

...

,a=

...
A_{σ}

...

,

and let M ∈M_{n}^{2}(L) be the block diagonal matrix

M =

. ..

D_{σ}W
. ..

.

Then we have x = Ma. Since x contains all the entries of M_{a} and a
contains all the information symbols, fulfilling the energy constraint is
equivalent to ask forM to be unitary. It is equivalent to say thatD_{σ}W
is unitary for all σ ∈ G. Since D_{Id} =I_{n}, it is equivalent to ask for W
to be unitary and for Dσ to be unitary for all σ ∈ G. In view of the
definition of Dσ, this is equivalent to conditions (1) and (2).

Finding anO_{K}-basis ofI satisfying condition (2) is not easy. In order to
simplify the problem, we will make the extra assumption that complex
conjugation induces a Q-automorphism on L, which commutes with
every element of Gal(L/K).

In this case, it is easy to check that W^{t}W = (Tr_{L/K}(ω_{σ}ω_{τ}))_{σ,τ}. Hence,
we may find an OK-basis of I for which encoding is energy-preserving
if and only if the hermitian OK-lattice

I×I −→ O_{K}

(x, y)7−→Tr_{L/K}(xy)

is isomorphic to the cubic latticeO_{K}^{n} (see next section for the definition
of a hermitian O_{K}-lattice).

This has very few chances to happen, so we now modify the encoding
process as follows. Let λ∈L^{×} satisfying the following conditions:

(a)λ=λ.

(b)λ^{σ} is a positive real number for allσ ∈G.

(c) Tr_{L/K}(λxy)∈ OK for all x, y ∈I.

Notice that in this case, NL/K(λ) and TrL/K(λ) are positive real num- bers.

Let D_{λ} ∈ M_{n}(R) be the diagonal matrix whose diagonal entries are
the real numbers √

λ^{σ}, σ ∈ G. If (ω_{σ})σ∈G is an O_{K}-basis of I, we will
encode n^{2} information symbols (a_{σ,τ})_{σ,τ}∈G into the matrix

M_{a}D_{λ} = (√

λ^{τ}ξ_{στ}^{−1}_{,τ}a^{τ}_{στ}−1)_{σ,τ},
where a_{σ} =X

τ∈G

a_{σ,τ}ω_{τ} for all σ∈G.

The reader will check that this new way of encoding simply replaces
W by W_{λ} = D_{λ}W in the proof of the previous proposition. Now
W^{t}_{λ}Wλ = (Tr_{L/K}(λωσωτ))σ,τ.

Hence we may find an O_{K}-basis of I for which encoding is energy-
preserving if and only if the hermitian O_{K}-lattice

I×I −→ O_{K}

(x, y)7−→Tr_{L/K}(λxy)
is isomorphic to the cubic lattice O_{K}^{n}.

We then set

C_{A,λ,I} ={M_{a}D_{λ} |a ∈Λ_{A,I}}.

Clearly, we haveδ_{min}(C_{A,λ,I}) =N_{L/K}(λ)δ_{min}(C_{A,I}), and therefore is also
bounded by a positive constant if A is a division K-algebra.

At this point, we would like to summarise what we have done so far.

Let K = Q(i) or Q(j). Let L/K be a Galois extension with Galois group G satisfying the following conditions:

(1) Complex conjugation induces a Q-automorphism of L which com- mutes with every element of G.

(2) There exists λ ∈ L^{×} and I an ideal of OL satisfying the following
conditions:

(a)λ=λ, that is λ∈R

(b)λ^{σ} is a positive real number for allσ ∈G
(c) Tr_{L/K}(λxy)∈ O_{K} for all x, y ∈I

(d) The hermitian O_{K}-lattice

h_{λ}: I×I −→ O_{K}

(x, y)7−→Tr_{L/K}(λxy)
is isomorphic to the cubic lattice O_{K}^{n}.

Then for any orthonormal O_{K}-basis (ω_{σ})_{σ∈G} of (I, h_{λ}), and for any
crossed product division K-algebra A = (ξ, L/K, G) such that

|ξ_{σ,τ}|^{2} = 1 for all σ, τ ∈G,
the encoding map

O^{n}_{K} −→ C_{A,λ,I}
(aσ,τ)σ,τ∈G 7−→(

√

λ^{τ}ξ_{στ}^{−1}_{,τ}a^{τ}_{στ}−1)σ,τ,
where a_{σ} = X

τ∈G

a_{σ,τ}ω_{τ} for all σ ∈ G, is energy-preserving. Moreover,
δ_{min}(C_{A,λ,I}) is bounded by a positive constant.

Notice that the conditions above also imply that other coding con- straints are fulfilled. We will not go into details and let the reader refer to [10].

The hermitian lattice h_{λ} is acomplex ideal lattice. Basic results on
such lattices and necessary conditions for a complex ideal lattice to be
isomorphic to the cubic lattice are given in Appendix A.

The next step now is δ_{min}(C_{A,λ,I}) in terms of the data, in order to
choose A and I such that δ_{min}(C_{A,λ,I}) is as large as possible.

1.3. Minimum determinant of a crossed product based code.

We would like to evaluate the performance of our code. In order to
do this, we have to estimate δ_{min}(C_{A,λ,I}) = infa6=0|det(X_{a})|^{2}. Let us
introduce some notation first. The set

E_{ξ}^{(τ)} ={c∈ O_{K} |cξ_{στ}^{−1}_{,τ} ∈ O_{L} for all σ ∈G}

is an ideal of OK. We will denote by ∆^{(τ)}_{ξ} the norm of this ideal.

Equivalently, we have

∆^{(τ)}_{ξ} =N_{K/}_{Q}(c^{(τ)}_{ξ} ) = |c^{(τ)}_{ξ} |^{2},

for any generatorc^{(τ)}_{ξ} ofE_{ξ}^{(τ)}. Notice that by the definition of a cocycle,
we have ∆^{(Id)}_{ξ} = 1.

By definition, we have

∆^{(τ)}_{ξ} = 1 ⇐⇒ ξ_{στ}^{−1}_{,τ} ∈ O_{L} for all σ ∈G.

We then set

∆_{ξ} = Y

σ∈G

∆^{(σ)}_{ξ} .
We then have

∆_{ξ}= 1 ⇐⇒ ξ_{σ,τ} ∈ O_{L} for all σ, τ ∈G.

We set

Nmin(I) = min

x∈I\{0}|NL/Q(x)|.

Notice that N_{min}(I) = N_{L/}_{Q}(I) if I is a principal ideal. In general,
equality does not hold.

Finally, recall that the relative discriminant d_{L/K} of L/K is
d_{L/K} = det(Tr_{L/K}(w_{i}w_{j})).

This is a positive integer by Remark A.16.

Proposition 1.3. Assume that A is a division K-algebra. With the previous notation, we have

1

∆_{ξ}d_{L/K} ≤δmin(CA,λ,I)≤N_{L/K}(λ)Nmin(I).

If moreover ∆_{ξ} = 1 and I is principal, we have
δmin(CA,λ,I) = 1

d_{L/K}.

Proof. Let x∈I, x6= 0 with minimal absolute norm. Then we have
X_{e}_{Id}_{x}=M_{e}_{Id}_{x}D_{λ} =xM_{e}_{Id}D_{λ} =xD_{λ}.

It follows that

det(X_{e}_{Id}_{x}) = Y

τ∈G

x^{τ}√

λ^{τ} =N_{L/K}(x)
q

N_{L/K}(λ),

and therefore |det(X_{e}_{Id}_{x})|^{2} = N_{L/}_{Q}(x)N_{L/K}(λ) = N_{min}(I)N_{L/K}(λ),
since we have

|N_{L/K}(x)|^{2} =N_{K/Q}(N_{L/K}(x)) =N_{L/Q}(x).

The upper bound then follows from the definition of δ_{min}(C_{A,λ,I}).

Now let X_{a} = M_{a}D_{λ} ∈ C_{A,λ,I}. To establish the lower bound, notice
that

|det(X_{a})|^{2} =|det(M_{a})|^{2}N_{L/K}(λ).

Recall that we have det(M_{a}) = Nrd_{A}(a) ∈ K. Now let c^{(τ)}_{ξ} be a
generator ofE_{ξ}^{(τ)}, letCbe the invertible diagonal matrix whose diagonal
entry at column τ isc^{(τ)}_{ξ} , and let M_{a}^{0} =M_{a}C. By definition of c^{(τ)}_{ξ} , we
have M_{a}^{0} = (c^{(τ)}_{ξ} ξ_{στ}^{−1}_{,τ}a^{τ}_{στ}−1)_{σ,τ} ∈M_{n}(O_{L}). Thus det(M_{a}^{0})∈ O_{L}∩K =
O_{K} and |det(M_{a}^{0})|^{2} ∈Z.

Since each coefficient in the τ^{th}-column of M_{a}^{0} lies in I^{τ}, the definition
of the determinant and the previous observation show that we have
det(M_{a}^{0})∈(Y

τ∈G

I^{τ})∩ O_{K} =N_{L/K}(I). It follows that

|det(M_{a}^{0})|^{2} ∈ N_{L/K}(I)N_{L/K}(I)∩Z=N_{L/Q}(I) =N_{L/Q}(I)Z.
Now we have

|det(X_{a})|^{2} =|det(M_{a}^{0}C^{−1}D_{λ})|^{2} = 1

∆ξ

N_{L/K}(λ)|det(M_{a}^{0})|^{2},
and thus

|det(X_{a})|^{2} ∈ 1

∆_{ξ}N_{L/K}(λ)N_{L/}_{Q}(I)Z.
Since det(X_{a})6= 0 ifa 6= 0, we get

|det(X_{a})|^{2} ≥ 1

∆_{ξ}N_{L/K}(λ)N_{L/}_{Q}(I) for alla ∈Λ_{A,I}, a6= 0.

Using Proposition A.15 and the definition of the minimum determinant, we get the desired lower bound.

Finally, ifI is a principal ideal, thenN_{min}(I) =N_{L/Q}(I). Using Propo-
sition A.15 again, we see that the two bounds are equal whenever

∆ξ = 1 andI is principal.

Remark 1.4. Notice that if x ∈I, x 6= 0 is an element with minimal
norm, then the first isomorphism theorem applied to the surjective
morphism O_{L}/xO_{L} → O_{L}/I shows that N_{min}(I) = N_{L/}_{Q}(I)[I : xO_{L}].

Hence the equation in the previous proposition may be rewritten as 1

∆_{ξ}d_{L/K} ≤δ_{min}(C_{A,λ,I})≤ [I :xO_{L}]
d_{L/K} .

This shows that maximising δ_{min}(C_{A,λ,I}) is essentially equivalent to
minimising d_{L/K}. The lower bound also shows that it is in our interest
to choose the cocycle values to be algebraic integers whenever it is
possible.

2. Optimality of codes based on cyclic K-algebras In [11] examples of codes with good performance were given for n = 4 and n = 6 that are based on a specific type of crossed product K- algebra, namely cyclic K-algebras. In this section, we will establish the optimality of these codes.

2.1. Preliminaries. First, let us recall the definition of a cyclic K-
algebra. Let L/K be a cyclic Galois extension of degreen with Galois
group G generated byσ, and let γ ∈K^{×}.

The map

ξ^{γ}:

G×G−→L^{×}

(σ^{i}, σ^{j})7−→

1 if i+j < n γ if i+j ≥n is a 2-cocycle.

We denote by (γ, L/K, σ) the corresponding crossed product. Such a K-algebra is called a cyclic K-algebra. This K-algebra is generated by a single element e(= eσ) subject to the relations

e^{n} =γ, λe=eλ^{σ} for all λ∈L.

Codes based on cyclic division K-algebras that satisfy the conditions
discussed at the beginning of this paper are called Perfect Space
Time Block Codes(PSTBC). It has been shown that ifγ ∈ O_{K}, then
these codes only exist in dimension 2,3,4 and 6 [3]. If the condition
that γ ∈ O_{K} is dropped then perfect codes exist for any dimension [5].

The case of the optimal PSTBC in dimension 2 has been addressed in [8]. In this section we deal with the dimension 4 and 6 cases.

In order to study the optimality of these codes, we will need to study the ramification of some Kummer extensions of K.

Let L=K(√^{n}

d) be a Kummer extension of K ⊃µ_{n} of degree n. After
multiplying by a suitablen^{th}-power of an element ofK, we may assume
that d∈ O_{K} and that 0≤v_{π}(d)≤n−1 for every prime element π of
O_{K}.

The following result is well-known.

Lemma 2.1. The prime elements π of K which eventually ramify are
those dividing d or n. Every prime π | d ramifies and if v_{π}(d) and n
are relatively prime, then π totally ramifies.

If furthermore π- n, then π totally ramifies if and only if v_{π}(d) and n
are relatively prime.

We continue these preliminaries by giving a necessary and sufficient condition ondto have complex conjugation commuting with the Galois group of L/K.

Lemma 2.2. Let L=K(√^{n}

d) be a Kummer extension of K of degree
n. Then complex conjugation induces a Q-automorphism of L which
commutes with Gal(L/K) if and only if dd∈K^{×n}.

Proof. Let ζ_{n} ∈K be a primitive n^{th}-root of 1. Then a generator σ of
Gal(L/K) is given by

σ: L−→L
α 7−→ζ_{n}α,
where α = √^{n}

d. Assume first that complex conjugation induces a Q- automorphism ofLwhich commutes with Gal(L/K). Thenα∈L, and we have

σ(αα) = σ(α)σ(α) =αα,
since ζ_{n}ζ_{n}= 1. Thus αα∈K^{×}. Now we have

(αα)^{n} =α^{n}α^{n}=dd,

so dd∈K^{×n}. Conversely, assume thatdd ∈K^{×n}. Then we have
(αα)^{n}=dd=c^{n} for some c∈K^{×},

and therefore α = cζ^{0}

α for some ζ^{0} ∈ µ_{n}. In particular, α ∈ L and
complex conjugation is therefore a Q-automorphism of L. Moreover,
we have

σ(α) = cζ^{0}

ζ_{n}α =ζ_{n}cζ^{0}

α =ζ_{n}α,

that isσ(α) =σ(α). Hence complex conjugation commutes withσand hence with Gal(L/K). This completes the proof.

We finish this section by computing ∆_{ξ}^{γ}.
Assume that K =Q(i) or Q(j). Write γ = γ_{1}

γ_{2}, where γ_{1}, γ_{2} ∈ O_{K} are
relatively prime. For all 1≤j ≤n−1, we have

ξ_{σ}^{γ}j−1(σ^{j})^{−1},σ^{j} =ξ_{σ}^{γ}n−1,σ^{j} =γ.

This implies easily that E_{ξ}^{(σ}γ^{j}^{)} = (γ_{2}). We then have ∆^{σ}_{ξ}γ^{j} = |γ_{2}|^{2}, and
therefore

∆_{ξ}^{γ} =|γ_{2}|^{2(n−1)}.

Thus, if C_{A,λ,I} is a codebook built on a cyclic division K-algebra A =
(γ, L/K, σ), then by Proposition 1.3 and the considerations of the pre-
vious paragraph, we have

1

δ_{min}(C_{A,λ,I}) ≤ |γ_{2}|^{2(n−1)}d_{L/K}.

Moreover, if |γ2|= 1, that isγ ∈ OK, and I is principal, we have δmin(CA,λ,I) = 1

dL/K

,

which is in fact the results obtained in [11], rephrased differently.

2.2. The case n = 4. In [11], Oggier et al. constructed a suitable code on the cyclic division Q(i)-algebra

(i,Q(i)(ζ_{15}+ζ_{15}^{−1})/Q(i), σ).

The cyclic extensionQ(i)(ζ15+ζ_{15}^{−1})/Q(i) has relative discriminant 1125
as we will see later on. Hence for any C ⊂ CA,λ,I, we get

1

δ_{min}(C) ≤1125.

We will show here that this bound is optimal, in the following sense:

Theorem 2.3. If C ⊂ C_{A,λ,I} is a PSTBC built on a cyclic division
K-algebra A = (γ, L/K, σ) of degree4, then we have

|γ_{2}|^{6}d_{L/K} ≥1125.

We will need several intermediate results. Assume that we may build
a PSTBC on a cyclic division K-algebra A= (γ, L/K, σ), where K =
Q(i) or Q(j), such that|γ_{2}|^{6}d_{L/K} <1125.We would like to notice first
that the case K =Q(i) andγ =i is the only one worth considering.

Assume that |γ_{2}|^{6}d_{L/K} <1125. Since we need the existence of an ideal
trace lattice on L isomorphic to the cubic lattice structure, we have
d_{L/K} ≥4^{4}by Corollary A.24. Taking into account that|γ|^{2}is a positive
integer, we get easily from the previous inequality that|γ2|^{2} = 1. Since

K/Q is quadratic imaginary, this implies that γ_{2} is a unit, soγ ∈ O_{K}.
Now |γ|^{2} = 1, and thus γ is also a unit of O_{K}.

Now if K = Q(j), then by the previous point γ = ±1,±j,±j^{2}, so
γ^{6} = 1. In particular, we have

6[A] = [(γ^{6}, L/K, σ)] = [(1, L/K, σ)] = 0∈Br(K).

Since A has degree 4, we also have 4[A] = 0 and so 2[A] = 0. Hence A has exponent 1 or 2, and since K is a number field, it implies that A has index 1 or 2. Thus, A is not a divisionK-algebra.

Assume now that K =Q(i). In this case,γ =±1,±i. Since the index
of A equals its exponent, A will be a division K-algebra if and only
if 2[A] 6= 0. Now if L = K(√^{4}

d), we have 2[A] = [(γ, d)]. Since −1
is a square in K, we have (γ, d) ' (−γ, d), and (γ, L/K, σ) is then a
division K-algebra if and only if (−γ, L/K, σ) is. Hence, it is enough
to consider the two cases γ = 1, i. The first case has to be discarded
since it yields the split K-algebra M_{4}(K).

Therefore, we may assume without loss of generality that K = Q(i)
and A= (i, L/K, σ). In this case, we are reduced to show that there is
no PSTBC on a cyclic division K-algebra (i, L/K, σ) satisfyingd_{L/K} <

1125.

We now assume once and for all that K =Q(i) and that L=K(√^{4}
d),
where d∈ O_{K} is not divisible by any 4^{th}-power of an element of O_{K}.
We denote by S_{3}, S_{1}, S_{1} the following subsets of O_{K}:

S_{3} ={p≡3[4], p prime number}

S1 ={π =a+bi |0< a < b, pπ =a^{2}+b^{2}, pπ prime number}

S1 ={π |π∈S1}.

Then any prime element of O_{K} is associate either to 1−ior to exactly
one element of S_{3}, S_{1} or S_{1}.

Decomposing d into a product of a unit and of prime elements and using Lemma 2.2 then immediately gives the following result:

Lemma 2.4. Let L = K(√^{4}

d). Then complex conjugation induces a Q-automorphism of L which commutes with Gal(L/K) if and only if all the following conditions are satisfied:

(1) v1−i(d) = 0.

(2) v_{p}(d) = 0 or 2, for allp∈S_{3}.

(3) (v_{π}(d), v_{π}(d)) = (1,3),(2,2) or (3,1) for all π ∈S_{1} dividing d.

We will assume from now on that the conditions above are satisfied. In this case, we have the following estimation of the relative discriminant.

Lemma 2.5. The odd part of d_{L/K} is

( Y

p∈S_{3}
p|d

p)^{2}( Y
π ∈S_{1}
v_{π}(d) = 1,3

p_{π})^{3}( Y
π∈S_{1}
v_{π}(d) = 2

p_{π})^{2}.

Proof. Let p ∈S_{3} dividing d. By Lemma 2.4, v_{p}(d) = 2, so p ramifies
but does not totally ramify by Lemma 2.1. Hence

(p) = P^{2}_{0} or P^{2}_{1}P^{2}_{2} in OL,

where P_{1},P_{2} form an orbit under the action of Gal(L/K). Now since
p is odd, p tamely ramifies, and thus v_{P}_{i}(D_{L/K}) = 2−1 = 1.

If (p) =P^{2}_{0}, we haveN_{L/Q}(P_{0}) = p^{4} and by Lemma A.18,v_{p}(d_{L/K}) = 2.

If (p) = P^{2}_{1}P^{2}_{2}, we have N_{L/}_{Q}(P_{i}) = p^{2} and by Lemma A.18, we also
get v_{p}(d_{L/K}) = 2 in this case as well.

Assume now that π ∈ S_{1} divides d with an odd valuation. Then π
also divides d with an odd valuation by Lemma 2.4. In this case, π
and π totally ramify by Lemma 2.1. We then have (π) = P^{4} and
(π) = P^{4}. Thus N_{L/Q}(P) = N_{L/Q}(P) = p_{π}. Once again π and π
are tamely ramified, and reasoning as before shows that v_{p}_{π}(d) = 3.

Finally, assume thatvπ(d_{L/K}) = 2 and so vπ(d) = 2 by Lemma 2.4. By
Lemma 2.1, π and π ramify but do not totally ramify. We then have

(π) = P^{2}_{0} and (π) =P^{2}_{0}
or

(π) = P^{2}_{1}P^{2}_{2} and (π) =P^{2}_{1}P^{2}_{2}

In the first case we have N_{L/}_{Q}(P_{0}) = N_{L/}_{Q}(P_{0}) = p^{2}_{π}. In the second
case we have N_{L/}_{Q}(P_{i}) = N_{L/}_{Q}(P_{i}) = p_{π}. We now finish the proof as

before.

Example 2.6. The cyclic extension Q(i)(ζ_{15}+ζ_{15}^{−1})/Q(i) has relative
discriminant 1125. Indeed, the only prime elements which ramify here
are the prime elements lying above 3 and 5, and we may apply the
previous lemma to conclude.

We now give an explicit criterion to decide whether or notAis a division K-algebra.

Lemma 2.7. The cyclic K-algebra A = (i, L/K, σ) is a division K-
algebra if and only if there exists a prime element π ∈ S_{1} dividing d
with an odd valuation such that p_{π} ≡5[8].

Proof. As already pointed out, A is a division K-algebra if and only if
the quaternion K-algebra (i, d) does not split. In view of the previous
lemma, d is congruent to uπ_{1}π_{1}· · ·π_{r}π_{r} modulo squares, where u is
a unit and π_{1}, . . . , π_{r} are the elements of S_{1} dividing d with an odd
valuation. Thus, we have

(i, d)'(i, uπ_{1}π_{1}· · ·π_{r}π_{r}), u=±1,±i.

This implies that ifp∈S_{3}, or ifπ ∈S_{1}dividesdwith an even valuation,
the Hasse symbols (i, d)_{p},(i, d)_{π} and (i, d)_{π} are trivial. Ifπ ∈S_{1} divides
d with an odd valuation, the Hasse symbol is the image of i^{(p}^{π}^{−1)/2} in
F^{×}pπ. Since i^{4} = 1, it is equal to 1 if p_{π} ≡1[8] and to −1 ifp_{π} ≡5[8].

We are now ready to conclude. If d is divisible by a prime element
π ∈ S_{1} with an odd valuation such that p_{π} ≡ 5[8], then the Hasse
symbol (i, d)_{π} is not trivial, so (i, d) does not split and A is a division
K-algebra in this case. Ifdis not divisible by a prime element satisfying
the previous conditions, then all Hasse symbols (i, d)_{π}^{0}, π^{0} 6= 1−i are
trivial. The remaining Hasse symbol is then trivial by the product
formula, hence (i, d) splits and A is not a division K-algebra.

We may now finish the proof of Theorem 2.3. Assume that we may
build a suitable code on a cyclic division K-algebra A = (i, L/K, σ)
with d_{L/K} < 1125. By Lemma 2.7, there exists π ∈ S_{1} dividing d
with an odd valuation such that p_{π} ≡ 5[8]. In particular, p_{π} ≥ 5.

Then by Lemma 2.5, we have p^{3}_{π} | d_{L/K}. Thus, we necessarily have
π = 1 + 2i and 125 | d_{L/K} (Otherwise, we would have p_{π} ≥ 13 and
d_{L/K} ≥ 13^{3} > 1125). Now if π^{0} ∈ S_{1}, π^{0} 6= 1 + 2i divides d, we would
have 125·p^{2}_{π}0 | d_{L/K}, and thus p^{2}_{π}0 ≤ 9, which is a contradiction since
p_{π}^{0} ≥5. Similarly, ifp∈S_{3} divides d, we have 125·p^{2} |d_{L/K} and thus
necessarily p= 3.

Hence the only possible prime divisors for d are 1−i,3 and 1±2i.

Noticing that conjugate values of d generate the same field extension, we see that the remaining possibilities for d are

d=u·3^{m}(1 + 2i)(1−2i)^{3}, m= 0,2, u=±1,±i.

Using PARI GP [18] to compute the relative discriminants of the cor-
responding extensions, we see that the only possibility to have d_{L/K} <

1125 is

d= (1 + 2i)(1−2i)^{3}.

In this case, d_{L/K} = 125 < 4^{4}. Hence, using Corollary A.24, we see
that no complex ideal lattice on L/K will be isomorphic to the cubic
lattice, and this completes the proof.

2.3. The case n = 6. If C ⊂ C_{A,λ,I} is a codebook built on a cyclic
division K-algebra A= (γ, L/K, σ), then this time we have

1

δmin(C) ≤ |γ_{2}|^{10}d_{L/K},
where we have written γ = γ_{1}

γ_{2} with γ_{1}, γ_{2} ∈ O_{K} relatively prime.

In [11], Oggier et al. constructed a PSTBC on the cyclic division Q(j)- algebra

(−j,Q(j)(ζ_{28}+ζ_{28}^{−1})/Q(j), σ).

The cyclic extension Q(j)(ζ28 + ζ_{28}^{−1})/Q(j) has relative discriminant
2^{6}·7^{5}. Hence for any C ⊂ CA,λ,I, we get

1

δ_{min}(C) ≤2^{6}·7^{5}.
Once again, this bound is optimal:

Theorem 2.8. If C ⊂ C_{A,λ,I} is a PSTBC built on a cyclic division
K-algebra A = (γ, L/K, σ) of degree6, then we have

|γ_{2}|^{10}d_{L/K} ≥2^{6} ·7^{5}.

Arguing as in the previous section, we see that we may assume without
loss of generality that K =Q(j) and A = (−j, L/K, σ). In this case,
we are reduced to show that there is no PSTBC on Asuch thatd_{L/K} <

2^{6}·7^{5}.

We now assume that K =Q(j) and that L =K(√^{6}

d), where d ∈ O_{K}
is not divisible by any 6^{th}-power of an element of O_{K}.

We denote by T_{2}, T_{1}, T_{1} the following subsets of O_{K}:
T_{2} ={p≡2[3], p >2 prime number }

T_{1} ={π =a+bj |0< a < b, p_{π} =a^{2}+b^{2}−ab, p_{π} ≡1[3], p_{π} prime }

T_{1} ={π |π∈T_{1}}.

Then any prime element of O_{K} is associate either to 1− j, 2 or to
exactly one element of T2, T1 orT1.

We then have:

Lemma 2.9. Let L = K(√^{6}

d). Then complex conjugation induces a Q-automorphism of L which commutes with Gal(L/K) if and only if all the following conditions are satisfied:

(1) v1−j(d) = 0.

(2) v_{p}(d) = 0 or 3, for allp∈T_{2} or p= 2.

(3) (v_{π}(d), v_{π}(d)) = (1,5),(2,4),(3,3),(4,2) or (5,1) for all π∈T_{1}
dividing d.

We will assume from now on that the conditions above are satisfied. In this case, we have the following estimation of the relative discriminant.

Lemma 2.10. The prime-to-6 part of d_{L/K} is

( Y

p∈T_{2}
p|d

p)^{3}( Y
π ∈T_{1}
v_{π}(d) = 1,5

pπ)^{5}( Y
π ∈T_{1}
v_{π}(d) = 3

pπ)^{3}( Y
π ∈T_{1}
v_{π}(d) = 2,4

pπ)^{4}.

Moreover, the following holds:

(1) If 2 ramifies in L/K, then v_{2}(d_{L/K})≥6
(2) If 1−j ramifies in L/K, then v_{3}(d_{L/K})≥4.

Proof. Let p ∈T_{2} dividing d. By Lemma 2.9, v_{p}(d) = 3, so p ramifies
but does not totally ramify by Lemma 2.1. Moreover, ptamely ramifies
sincepis prime to 6. Writed=d^{0}p^{3} withd^{0} ∈ O_{K}, p-d^{0}. Thenptotally
ramifies in K_{2} = K(√

d) = K(√

d^{0}p) by the same lemma, so we may
write

(p) =p^{2}_{0},

for some ideal p_{0} of O_{K}_{2}. Since L/K_{2} is a Galois extension of prime
degree 3 , either p_{0} is totally ramified, inert or totally split. Since p_{0}
cannot be totally ramified (otherwise p would be totally ramified), we
finally have

(p)O_{L}=p^{2}_{0} orp^{2}_{1}p^{2}_{2}p^{2}_{3} inO_{L},

where p_{1},p_{2},p_{3} form an orbit under the action of Gal(L/K_{2}).

If (p) = p^{2}_{0}, we have N_{L/}_{Q}(p_{0}) = p^{6}, and since v_{p}_{0}(D_{L/K}) = 1, we get
v_{p}(d_{L/K}) = 3.

If (p) = p^{2}_{1}p^{2}_{2}p^{2}_{3}, we have N_{L/Q}(pi) = p^{2}, and since vpi(D_{L/K}) = 1, we
also get vp(dL/K) = 3 in this case.

Now, let π ∈ T_{1} such that v_{π}(d) = 1 or 5. Then v_{π}(d) = 5 or 1
respectively. By Lemma 2.1, π and π totally ramify, so we have

(π) = p^{6} and (π) = p^{6} inOL,

with N_{L/}_{Q}(p) =p_{π}. We then have v_{p}(D_{L/K}) = 5 =v_{p}(D_{L/K}) and thus
v_{p}_{π}(d_{L/K}) = 5.

Letπ ∈T_{1} such thatv_{π}(d) = 3. Then we also havev_{π}(d) = 3. Reason-
ing as above, we see that we have

(π) =p^{2}_{0} or p^{2}_{1}p^{2}_{2}p^{2}_{3} inO_{L},
and similarly for (π), so that v_{p}_{π}(d_{L/K}) = 3.

Finally, let π ∈T_{1} such that v_{π}(d) = 2 or 4, and set K_{3} =K(√^{3}
d). In
this case, we have

(π) =p^{3}_{0} or p^{3}_{1}p^{3}_{2} inO_{L},

where p_{1},p_{2} form an orbit under the action of Gal(L/K_{3}). One may
check as before that in both cases, we have v_{p}_{π}(d_{L/K}) = 4.

We now examine the case of the wildly ramified primes. Let us start with 1−j. Since 1−j does not divide d or 2, it does not ramify in K2. Then it necessarily totally ramifies inK3, so we have

(1−j) =p^{3}_{0} inO_{K}_{3},

where N_{K}_{3}_{/}_{Q}(p_{0}) = 3. Now since 1−j wildly ramifies, v_{p}_{0}(D_{K}_{3}_{/K}) ≥
3. Therefore, 3^{3} divides N_{K}_{3}_{/Q}(D_{K}_{3}_{/K}) = d^{2}_{K}

3/K. Since d_{K}_{3}_{/K} is an
integer, we get that 3^{2} |d_{K}_{3}_{/K}. By Corollary A.19, we get 3^{4} |d_{L/K}.
Assume now that 2 ramifies in L/K. Since v2(d) = 0 or 3, it does not
ramify in K_{3}, hence it totally ramifies in K_{2}. We then have

(2) =p^{2}_{0} inO_{K}_{2},

so that N_{K}_{2}_{/}_{Q}(p_{0}) = 2^{2}. Since 2 wildly ramifies, v_{p}_{0}(D_{K}_{2}_{/K}) ≥ 2, and
we get as before that 2^{4} | d^{2}_{K}

2/K. Hence 2^{2} | d_{K}_{2}_{/K}, and by Corollary

A.19, we get 2^{6} |dL/K.

As before we now give an explicit criterion to decide whether or not A is a division K-algebra.

Lemma 2.11. The cyclic K-algebra A = (−j, L/K, σ) is a division
K-algebra if and only if there exist (not necessarily distinct) prime
elements π, π^{0} ∈ T_{1} such that v_{π}(d) is odd, v_{π}^{0}(d) is prime to 3 and
satisfying the following conditions:

(1) p_{π} ≡7[12].

(2) pπ^{0} ≡4 or 7[9].

Proof. Since K is a number field, the index of A equals the exponent of A. Thus A will be a division K-algebra if and only if 2[A] 6= 0 and 3[A] 6= 0. Notice that A is nothing but the symbol K-algebra {−j, d}6. Hence 3[A] = [(−j, d)] and 2[A] = [{−j, d}3]. Since (−j, d)

and {−j, d}_{3} have prime degrees, we conclude thatA will be a division
K-algebra if and only if A_{2} = (−j, d) and A_{3} ={−j, d}_{3} are not split.

Using Lemma 2.9 and properties of quaternionK-algebras, we see that we have

A2 = (−j, u2^{m}π1π1. . . πrπrp1· · ·ps),

where u is a unit, m = 0 or 1, π_{1}, . . . , π_{r} ∈ T_{1} and p_{1}, . . . , p_{s} ∈T_{2} are
prime elements dividing d with an odd valuation.

It follows in particular that if π 6=π_{i}, p_{j}, then the corresponding Hasse
symbol is trivial. If π = p ∈ T_{2}, then the number of elements of the
residue field κ(π) is qπ =p^{2}. The corresponding Hasse symbol is then
equal to

−j

p2−1

2 ∈κ(π)^{×}.

Since p≡2[3] andp is odd, we havep≡5[6], so p^{2}−1 is a multiple of
12. Since (−j)^{6} = 1, we deduce that the Hasse symbol is trivial in this
case as well.

Assume now that π ∈T_{1}. In this case q_{π} =p_{π}, and the corresponding
Hasse symbol is then equal to

−j^{pπ−1}^{2} ∈κ(π)^{×}.

This Hasse symbol will then be trivial if and only if 6 | ^{p}^{π}_{2}^{−1}, that is
p_{π} ≡ 1[12]. In other words, since p_{π} ≡ 1[3] and p_{π} is odd, the Hasse
symbol will be non-trivial if and only if p_{π} ≡ 7[12]. Notice also that
the Hasse symbols corresponding to π and π are equal.

It follows from our computations above that if no prime element π∈T_{1}
such that pπ ≡ 7[12] divides d with an odd valuation, then all the
Hasse symbols of A2 are trivial, except maybe for the Hasse symbol
corresponding to 2. By the product formula, this last symbol is also
trivial, and A_{2} is split in this case. If however such aπ exists, then A_{2}
has at least one non-trivial Hasse symbol, and is therefore not split.

Thus A_{2} is not split if and only if there exists π ∈ T_{1} such that p_{π} ≡
7[12] dividing d with an odd valuation.

Using Lemma 2.9 and properties of symbol K-algebras, we see that we have

A_{3} ={j, uπ_{1}π^{2}_{1}. . . π_{r}π^{2}_{r}π_{1}^{0}^{2}π^{0}_{1}· · ·π^{0}_{s}^{2}π^{0}_{s}}_{3},

where u is a unit and π_{1}, . . . , π_{r}, π_{1}^{0}, . . . , π_{s}^{0} ∈ T_{1} are prime elements
dividing d with a valuation prime to 3.

The Hasse symbol corresponding to each of these prime elements has
the form j^{m}

pπ−1

3 , where m= 1 or 2. This symbol is then trivial if and
only if p_{π} ≡ 1[9]. In other words, the corresponding Hasse symbol is
not trivial if and only if pπ ≡4,7[9].

Reasoning as above, we see that A_{3} is not split if and only if there
exists π ∈T_{1} such that p_{π} ≡4,7[9] dividing d with a valuation prime

to 3. This concludes the proof.

Assume now that we may build an energy-preserving code on a cyclic
division K-algebra A= (−j, L/K, σ), where d_{L/K} <2^{6}·7^{5}.

Let π ∈T_{1} divide d with an odd valuation, such thatp_{π} ≡7[12] (such
a π exists by the previous result). If p_{π} > 7, then p_{π} ≥ 19. In this
case, Lemma 2.10 implies that 19^{3} | d_{L/K}. Now let π^{0} ∈ T_{1} divide
d with a valuation which is prime to 3, such that p_{π}^{0} ≡ 4,7[9]. We
then have p_{π}^{0} ≥ 7 and thus Lemma 2.10 implies that 7^{4} |d_{L/K}. Hence
7^{4}·19^{3} |d_{L/K}, which is a contradiction since 7^{4}·19^{3} >2^{6}·7^{5}.

Therefore, p_{π} = 7. We then get 7^{3} |d_{L/K}. If p_{π}^{0} >7, we have p_{π}^{0} ≥13
and thus 7^{3}·13^{4} |d_{L/K}, which is again a contradiction. Hence pπ^{0} = 7.

Hence we have proved that π = π^{0} = 2 + 3j. Moreover, since 2 + 3j
divides d_{L/K} with an odd valuation, which is also prime to 3, then
v_{2+3j}(d_{L/K}) = 1 or 5, and thus 7^{5} |d_{L/K}.

Assume that p ∈ T_{2} ramifies in L/K. Since p ≡ 2[3], we have p ≥ 5,
and by Lemma 2.10, we get 5^{3} | d_{L/K}. We then get a contradiction,
since 5^{3} ·7^{5} >2^{6}·7^{5}. If π^{00} ∈ T_{1}∪T_{1}, π^{00}- 7 ramifies in L/K, we have
p_{π}^{00} ≥ 13 since p_{π}^{00} ≡ 1[3]. In this case, we obtain that 13^{3} | d_{L/K},
which again yields a contradiction. Notice that 1 −j and 2 do not
ramify in L/K either, since otherwise we would have 3^{4}·7^{5} | dL/K or
2^{6}·7^{5} |d_{L/K}, which is a contradiction.

The computations above shows that d_{L/K} = 7^{5} < 6^{6}, so we may not
construct the cubic lattice as a complex ideal lattice on L/K by Corol-
lary A.24.

This proves Theorem 2.8.

3. Optimality of codes based on biquadratic crossed products

Let L/K be a biquadratic extension, and fix two generators σ, τ of
G = Gal(L/K). Then any G-crossed product is isomorphic to the K-
algebra (a, b, u, L/K, σ, τ), for somea, b, u ∈L^{×}, which is generated by
two elements e and f subject to the relations

a^{σ} =a, b^{τ} =b, uu^{σ} = a

a^{τ}, uu^{τ} = b^{σ}
b .
Notice that these relations imply that (abu^{τ})^{στ} =abu^{τ}.