TABLES OF OCTIC FIELDS WITH A QUARTIC SUBFIELD

H. Cohen, F. Diaz y Diaz and M. Olivier

Abstract. We describe the computation of extended tables of degree 8 fields with a quartic subfield, using class field theory. In particular, we find the minimum discriminants for all signatures and for all the possible Galois groups. We also discuss some phenomena and statistics discovered while making the tables, such as the occurence of 11 non-isomorphic number fields having the same discriminant, or several pairs of non-isomorphic number fields having the same Dedekind zeta function.

1. Introduction

The construction of tables of number fields has a long history, and is useful in two ways. First, to test the algorithms available for such constructions, and second and probably most importantly, to give to researchers a vast amount of data that they can examine and on which they can make conjectures. Large tables are now available on the internet by anonymous ftp, either from the Kant group

(ftp.math.tu-berlin.de) or from our group, the Pari group (megrez.math.u-
bordeaux.fr).

The tables are built in the following way. In degree 2, we simply need to find squarefree numbers up to a certain bound, and considering the bounds that we must take to be realistic, this is not a problem. In degree 3, it was previously thought that only the general methods were available, but it was observed recently by K. Belabas (see [Be]) that one can adapt methods of Davenport and Heilbronn to the algorithmic construction of cubic fields in time comparable to that of quadratic fields. Thus, computations of tables of cubic fields is now also easy (tables have been built up to discriminant 10^{11}). In higher degrees, the only methods known to construct general tables of number fields are based on the geometry of numbers. These methods are highly inefficient, since they require looking at orders of magnitudes more polynomials than necessary, but they are the only ones available in general. Large tables exist for degrees 4, 5 and 6, small tables in degree 7, and in degree 8, only the first few minima are known in signature (8,0) and (0,4), and the minima for the other signatures are not even known without assuming the GRH.

If one considers imprimitive fields, the situation improves considerably, since, thanks to a relative version of the theorem used in the absolute case due to J. Martinet, we can entirely work using relative data. For example, although requiring a huge amount of work, we now have large tables of degree 9 fields containing a cubic

Typeset by AMS-TEX

Key words and phrases. class field theory, discriminant, number field.

1991 Mathematics Subject Classification. Primary 11R37, 11Y40.

subfield (see [Di-Ol1]). One could reasonably do this for degree 8 fields containing a quadratic or quartic subfield, but it has not been done.

Another approach to the computation of imprimitive fields is to use class field theory. Recall that class field theory gives an explicit description of all finite Abelian extensions of a number field. Thus, for example, we can use class field theory to compute all quadratic extensions of a given base field, since such extensions are Abelian, and this is what will be described in this paper when the base field has degree 4, thus giving number fields of degree 8. One could of course also apply this to base fields of degree 2 or 3, but extensive tables have already been done in this case, or to base fields of degree larger than 4.

In the following sections, we describe the methods used for constructing the tables, and then give a number of observations, statistical or otherwise, obtained when constructing the tables (some of these observations are not visible on the tables themselves, but have been made during the construction of the tables).

2. Construction of the tables

(A) Results from Class Field Theory

Let K be a fixed base number field. For any modulus 𝔪 (i.e. a pair formed by an integral ideal 𝔪_{0} and a set of real places 𝔪_{∞} of K), let I_{𝔪}(K) be the group of fractional ideals of K coprime to 𝔪, let P_{𝔪}(K) be the group of principal ideals generated by an element α ≡ 1 (mod ^{∗}𝔪), and let Cl_{𝔪}(K) = I_{𝔪}(K)/P_{𝔪}(K) be the ray class group.

A congruence subgroup is any group of ideals C such that P_{𝔪}(K) ⊂ C ⊂ I_{𝔪}(K). One of the fundamental theorems of class field theory tells us that equivalence classes of pairs (𝔪, C) (for a suitable equivalence relation) are in canonical one to one correspondence with K-isomorphism classes of finite Abelian extensions L/K, satisfying in particular

Gal(L/K) ≃ I_{𝔪}(K)/C ≃ Cl_{𝔪}(K)/\overline{C} where \overline{C} = C/P_{𝔪}(K) .

In [Co-Di-Ol], we described an efficient method to compute the ray class group Cl_{𝔪}(K), and to compute the discriminant (relative or absolute), signature and conductor of the corresponding Abelian extension as above. Briefly described, we proceed as follows. We first note the following (easy) exact sequence:

U(K) ⟶ (ℤ_{K}/𝔪)^{∗} ⟶ Cl_{𝔪}(K) ⟶ Cl(K) ⟶ 1 .

We first compute the ordinary class group Cl(K) and unit group U(K) using the methods that can be found in [Co] or in [Po-Za]. We then compute (ℤ_{K}/𝔪)^{∗} using the methods explained in [Co-Di-Ol]. Finally, using the methods explained in that paper for dealing with exact sequences which are not necessarily split, we obtain a complete description of the group Cl_{𝔪}(K) by minimal generators and relations.

Let (𝔪, C) be a congruence subgroup and let L/K be the Abelian extension which corresponds to (𝔪, C) by class field theory. Assume that 𝔪 is the conductor, i.e. that (𝔪, C) is minimal in its equivalence class. We know in particular that Gal(L/K) ≃ Cl_{𝔪}(K)/\overline{C}, and that the prime ideals of K which ramify in L are exactly those dividing 𝔪. We also know precisely the signature and discriminant of L, thanks to the following theorem proved in [Co-Di-Ol].

Theorem 1. Let L/K be the (equivalence class of) Abelian extension corresponding to the congruence subgroup (𝔪, C) by class field theory, where 𝔪 is not necessarily assumed to be the conductor.

Denote by 𝔡(L/K) the relative discriminant ideal of L/K, and write

𝔪 = ∏_{𝔭|𝔪} 𝔭^{v_{𝔭}} .

For any divisor 𝔫 of 𝔪, let h_{𝔫,C} be the cardinality of Cl_{𝔫}(K)/s(\overline{C}), where s is the natural surjection from Cl_{𝔪}(K) to Cl_{𝔫}(K) (in particular, [L : K] = h_{𝔪,C}). Then 𝔡(L/K) = ∏_{𝔭|𝔪} 𝔭^{a_{𝔭}}, with

a_{𝔭} = v_{𝔭}h_{𝔪,C} − ∑_{1≤i≤v_{𝔭}} h_{𝔪/𝔭^{i},C} .

If, in addition, 𝔪 is the conductor, we have the following:

(1) If h_{𝔪,C} = [L : K] = ℓ is prime, then 𝔡(L/K) = 𝔪^{ℓ−1}.

(2) Denote by m_{∞} be the number of real places in the modulus 𝔪, and by (r_{1}, r_{2}) (resp. (R_{1}, R_{2})) the signature of K (resp. L). Then we have R_{2} = h_{𝔪,C}(m_{∞}/2 + r_{2}).

(3) The absolute discriminant d(L) is given by

d(L) = (−1)^{m_{∞}h_{𝔪,C}/2}d(K)^{h_{𝔪,C}} N (𝔡(L/K)) .

(B) Finding all suitable pairs (𝔪, C)

Let K be a fixed quartic field and let L be an octic field obtained by class field theory as a quadratic extension of K corresponding to the pair (𝔪, C). Since we will restrict to the case where 𝔪 is the conductor, Theorem 1 tells us that 𝔪 = 𝔡(L/K), hence that |d(L)| = d(K)^{2} N_{K/ℚ}(𝔪). Thus, if we want to compute octic fields such that |d(L)| ≤ B for a certain bound B, it is enough to consider all quartic fields K such that |d(K)| ≤ B^{1/2}, and for each such quartic field all moduli such that N_{K/ℚ}(𝔪) ≤ B/d(K)^{2}. Furthermore, to obtain the correct signature, we have to include in 𝔪 exactly the correct number of ramified real places (more precisely, if we want R_{1} real places in L, we must take K with r_{1} ≥ R_{1}/2 real places and exactly r_{1} − R_{1}/2 of those real places dividing 𝔪).

For a given quartic field K, we construct all integral ideals of norm up to a given bound by a recursive procedure similar to the sieve of Erathosthenes. We initialize the procedure by saying that there is a single ideal of norm 1, i.e. ℤ_{K}. Then, we split each prime number up to the bound into a product of prime ideals in K, and multiply each of the already obtained ideals by successive powers of the prime ideals, as long as the norm of the results are less than the given bound. This gives a very efficient procedure, although the amount of storage can become large (for some of our computations, we have had to use bound close to 10^{6}, and this can give storage of several hundred megabytes).

For each of the moduli, we then compute the class group Cl_{𝔪}(K) using the method explained in [Co-Di-Ol]. We can also do this in a similar recursive procedure to speed up the computation. Furthermore, since we do this recursively, it is easy to eliminate from the list of moduli those for which (𝔪, P_{𝔪}) is not a conductor. Indeed, a very easy lemma tells us that if 𝔪 is the conductor for some congruence group C, then it is the conductor for P_{𝔪}.

For each remaining modulus 𝔪, we compute all the subgroups \overline{C} of Cl_{𝔪}(K) of index 2, using the method explained in [Co-Di-Ol]. If E is the diagonal matrix in Smith normal form giving the structure of Cl_{𝔪}(K), this amounts to finding all integer matrices H in Hermite normal form such that H^{−1}E is an integer matrix, and det(H) = 2. If the diagonal entries of E are (e_{i}) with e_{i+1} | e_{i}, let r be the largest index such that e_{i} is even (0 if none exist), so that r is equal to the 2-rank of Cl_{𝔪}(K). It is easy to show that the possible matrices H = (h_{i,j}) are exactly those matrices such that there exists an integer i_{0} satisfying the following conditions.

(1) i_{0} ≤ r.

(2) h_{i,i} = 1 if i ≠ i_{0} and h_{i_{0},i_{0}} = 2.

(3) h_{i_{0},j} = 0 or 1 if j ≤ r.

(4) h_{i,j} = 0 for all other pairs (i, j).

There are 2^{r} − 1 such matrices, corresponding to the 2^{r} − 1 subgroups of index 2 of a group of 2-rank equal to r. However, we want only those subgroups \overline{C} for which 𝔪 is equal to the conductor of (𝔪, C), hence we keep only those subgroups satisfying this condition.

By class field theory, the pairs (𝔪, C) thus found are in one-to-one correspondence with K-isomorphism classes of quadratic extensions of K. Note that, even though two different pairs will correspond to two fields L which are not K-isomorphic, they may of course be ℚ-isomorphic. Furthermore, different quartic fields may give isomorphic octic fields. However, at this stage we do not worry about this problem, since, contrary to the methods using the geometry of numbers, the amount of polynomials that we will have to remove because of isomorphisms will not be very large.

(C) Finding the polynomial equations

For a given base quartic field K, we now have a large number of pairs (𝔪, C), and we know that 𝔪 is the conductor of the equivalence class of (𝔪, C). We must now find the relative and absolute equation of the quadratic extension L/K corresponding to (𝔪, C). We do this by using Kummer theory, which is the standard method for this kind of computations (although one can also use Stark’s conjecture, see [Ro]). We are here in a very favorable case, since the 2-nd roots of unity are already in our field K, so we do not need to adjoin them.

We need some notations.

If 𝔭 is a prime ideal of K, let z(𝔭) = 2e(𝔭/2) + 1, where e(𝔭/2) = 0 when 𝔭 is not a prime above 2.

We define 7 sets of prime ideals associated to 𝔪 as follows.

(1) S_{𝔪,1} (resp. S_{𝔪,2}, resp. S_{𝔪,3}) is the set of prime ideals 𝔭 of K which divide both 2 and 𝔪, and such that v_{𝔭}(𝔪) = z(𝔭) (resp. v_{𝔭}(𝔪) < z(𝔭), resp. v_{𝔭}(𝔪) > z(𝔭)).

(2) S_{2} (resp. S_{𝔪}) is the set of all prime ideals dividing 2 and not 𝔪 (resp. dividing 𝔪 and not 2).

(3) S_{∅} is the set of all prime ideals dividing neither 𝔪 nor 2.

(4) S = S_{𝔪} ∪ S_{𝔪,1}.

Let

Cl(K) = ⨁_{i=1}^{g}(ℤ/d_{i}ℤ)\overline{𝔞_{i}}

be the ordinary class group of K, where the d_{i} are ordered so that d_{i+1} | d_{i}, and assume that the representatives of the ideal classes \overline{𝔞_{i}} are chosen to be integral ideals coprime to the ideal 2𝔪 (this can always be done). Let α_{i} be elements such that 𝔞_{i}^{d_{i}} = α_{i}ℤ_{K}. These α_{i} can easily be found using the solution to the principal ideal problem (see [Co]), and by our assumption will be coprime to 𝔪. Let h be the largest index (0 if none exist) such that d_{i} is even.

For each 𝔭 ∈ S, we can write

𝔭 = β_{𝔭} ∏_{1≤i≤g} 𝔞_{i}^{p_{i,𝔭}}

again by using the solution to the principal ideal problem.

By Theorem 1, since 𝔪 is the conductor and the relative degree is equal to 2, the relative discriminant 𝔡(L/K) is equal to 𝔪. Using Hecke’s theorem on ramification in Kummer extensions of prime degree, it is not difficult to prove the following theorem (see for example [Co-Di-Ol]).

Theorem 2. With the above notations, let L/K be a quadratic extension of conductor equal to 𝔪. We have L = K(\sqrt{α}), where α is of the following form:

α = ϵ ∏_{1≤i≤h} α_{i}^{n_{i}} ∏_{𝔭∈S} β_{𝔭} ,

and ϵ is a unit.

In addition, the following conditions must be satisfied.

(1) S_{𝔪,3} = ∅.

(2) If 𝔭 ∈ S_{𝔪,2}, then v_{𝔭}(𝔪) ≥ 2.

(3) If 𝔭 ∈ S_{𝔪}, then v_{𝔭}(𝔪) = 1.

(4) For all i ≤ h, we must have ∑_{𝔭∈S} p_{i,𝔭} ≡ 0 (mod 2).

(5) For all i ≤ h, we have n_{i} = 0 or 1.

(6) For 𝔭 ∈ S_{𝔪,2}, the largest k such that the congruence

x^{2} ≡ α (mod 𝔭^{k})

has a solution must be equal to z(𝔭) − v_{𝔭}(𝔪).

(7) For each 𝔭 ∈ S_{2}, the congruence

x^{2} ≡ α (mod 𝔭^{z(𝔭)−1})

has a solution.

(8) For each σ ∈ 𝔪_{∞}, we have σ(α) < 0, while for each real embedding σ such that σ ∉ 𝔪_{∞}, we have σ(α) > 0.

(9) α ∉ (K^{∗})^{2} (this condition is necessary only if 𝔪 = ℤ_{K}).

Conversely, if α is chosen as above and the conditions are satisfied, then L/K is a quadratic extension of conductor equal to 𝔪.

The solubility of the congruences needed in the above theorem can easily be established by computing the structure of the groups (ℤ_{K}/𝔭^{k})^{∗} as explained in [Co-Di-Ol].

If all the conditions are satisfied, then K(\sqrt{α}) will be a quadratic extension of K which has the correct signature and relative discriminant 𝔡(L/K). However, it may not be the field that we are looking for, since we must also ask that its norm group is equal to C.

To terminate, for each α that we find we compute the norm group of K(\sqrt{α}) and we check that we obtain once and only once all the subgroups of I_{𝔪} of index 2 (otherwise there is an error), and this gives us precisely to which group corresponds which α. If we do not care about the norm group, we do not need to do this, since we know by class field theory that up to squares there will be exactly the same number of α as subgroups of index 2 and conductor 𝔪 of I_{𝔪}(K). Thus, to check the correctness of our computations, it is sufficient to check that the number is the same, not the exact correspondence. Note also that it is quite easy to compute the norm group explicitly by using its characterization in terms of the decomposition of unramified prime ideals.

Once we know that L = K(\sqrt{α}), it is a trivial matter to find the absolute equation of L/ℚ. This equation has usually quite large coefficients, and we then use a strong polynomial reduction algorithm (analogous to the Polred algorithm described in [Co-Di], but searching for minima using the Fincke-Pohst algorithm, instead of simply small vectors using LLL), and we finally obtain an absolute equation with quite small coefficients

(D) Removing isomorphisms

We now have a large list of degree 8 polynomials (in our tables, we chose the bounds so as to have between 13000 and 18000 polynomials at this stage). We must remove the polynomials defining isomorphic number fields over ℚ. The use of the strong polynomial reduction algorithm mentioned above is already a very big step in this direction, since it is not frequent that two different strongly reduced polynomials define isomorphic number fields, although it may of course happen.

We sort the table acccording to increasing absolute value of discriminant, and for two entries having the same discriminant, we do the following. If the reduced polynomials are the same, evidently the fields are isomorphic, so we remove one of the two. Otherwise, we factor the two polynomials modulo small primes not dividing the indices of one or the other (so that the factorization modulo the prime reflects the factorization into prime ideals). If the factorizations differ, the fields are not isomorphic. If they are equal, then we use an absolutely certain but slow method, i.e. we factor one polynomial in the number field defined by the other. Once this is done, we also compute the Galois group of the Galois closure of the corresponding number field, since we will be interested in this later.

It must be noticed that our method using class field theory is much less wasteful than the methods based on the geometry of numbers also with respect to isomorphisms: in our methods, we found rarely more than two distinct polynomials defining isomorphic number fields, while in methods based on the geometry of numbers, there are usually many more isomorphic number fields.

In fact, using our methods, we can find distinct polynomials defining isomorphic number fields only when the octic field contains at least two non-isomorphic quartic fields, and this rarely happens.

In performing this routine and apparently uninteresting task, we have in fact stumbled onto two quite interesting phenomena.

First, the fields which cannot be distinguished by prime factorizations only are rare, but do occur. In fact, we have looked closely at these examples, and they are all arithmetically equivalent fields, i.e. number fields having the same Dedekind ζ function. We refer to the literature on the subject (see [Ga], [Pe]), but note two facts.

Even though equality of the ζ function implies the equality of many arithmetic quantities, it does not imply the identity of the ramification exponents, only the residual indices. We have several examples of arithmetically equivalent fields which can be distinguished by factoring a ramified prime.

The second fact is that it is well known that arithmetic equivalence is a purely group-theoretical property. In the case of degree 8 fields, one can show that there are exactly two Galois groups among all the possible 50 transitive groups of degree 8 having this property (see below). These are the groups Hol(C_{8}) (denoted T_{15} in [Bu-Mc]) of order 32, equal to the semidirect product of the cyclic group C_{8} by its group of automorphisms acting in the natural way, and the group GL_{2}(𝔽_{3}) (denoted T_{23} in [Bu-Mc]) of order 48.

A list of these fields will be given in the next section.

The second phenomenon which we have observed (and which we call the “mirror effect”) will be described in detail below.

3. Description of the tables

(A) Length of the tables

We have chosen the discriminant bounds so that the tables are roughly of the same length, and contain at least 10000 non-isomorphic octic fields. As indicated above, there are very few isomorphic number fields corresponding to distinct polynomials, hence we could easily judge at an early stage whether our bounds were sufficient.

The following table gives the chosen bounds, as well as the number of nonisomorphic octic fields up to that bound, corresponding to the five possible signatures.

Signature Bound for discriminants Number of non

isomorphic fields

(0, 4)  5000 00000  11639
(2, 3)  −16000 00000  12301
(4, 2)  35000 00000  13077
(6, 1)  −5 00000 00000  11680
(8, 0)  25 00000 00000  13726

(B) Galois groups

We have computed the Galois group of every octic field that we have found, using the methods described in [Ei] and [Ei-Ol]. The Galois groups which we obtain are necessarily among those corresponding to octic fields containing a quartic subfield, and this corresponds to 36 of the 50 possible Galois groups in degree 8 (see the table in signature (0, 4) or (8, 0) for the complete list).

Even though we found many octic fields, there was no reason that we would find all the possible combinations of signatures and Galois groups, and indeed we found only 97 of the 114 possible combinations.

For the 17 missing ones, we used several methods.

The field denoted by (a) in the tables was kindly computed for us by J. Martinet (it is the totally complex cyclic octic field of smallest discriminant).

We found 1 field (denoted by (b)) in the literature, the field T_{5}^{+} in signature (0, 4), due to S. Y. Kwon (see [Kw]).

A second missing field (T_{23} in signature (0, 4)) can be found in [Je], unfortunately, contrary to what is claimed in that paper, this field does not have the minimum discriminant. However, we did use it as an upper bound for the search for the minimum, given in the table below.

For each of the 15 remaining cases (denoted by (c)), we proceeded in the following way. Using various methods described below, we first computed an example of a field having the desired signature and Galois group. The discriminant of this field gave us a (possibly quite large) upper bound for the minimal discriminant. We then used a systematic search up to this bound as was done for the main tables, using the specific properties of the group that we were looking for. For example, the search was largely simplified when the group was even since we needed to use only quartic fields with square discriminant.

The methods that we used to find at least one example were the following.

(1) Specializations of the parametrized solutions given in the literature, for example in [Sm]. Of course, these solutions do not necessarily have the correct signature.

(2) The “mirror effect” already mentioned above: an octic field L having a quartic subfield K, can be defined by an even polynomial P (X^{2}) ∈ ℤ[X]. Let D be a rational integer such that \sqrt{D} is not in the Galois closure N of L over ℚ and denote by G the Galois group of the extension N/ℚ. The field obtained by adjoining to the rationals the roots {±θ_{1}, ±θ_{2}, ±θ_{3}, ±θ_{4}} of P (X^{2}) as well as \sqrt{D} is Galois over ℚ and its Galois group is isomorphic to G × C_{2}. In this group, the intersection of the stabilizers of the elements \sqrt{D}θ_{j} is trivial if there does not exist σ ∈ G such that σ(θ_{j}) = −θ_{j} for j = 1, 2, 3, 4, and it is equal to H = {(1, 1), (σ, 1)} ≃ C_{2} otherwise. In this case, since σ is a central element in G we have (G × C_{2})/H ≃ G as an abstract group.

Thus, replacing P (X^{2}) with P ((X\sqrt{D})^{2}) = P (DX^{2}) gives a polynomial whose Galois group is (as an abstract group) isomorphic to G × C_{2} in the first case above, and to G in the second case.

Using this, one can prove (Y. Eichenlaub, personal communication) that the Galois group of the Galois closure of L/ℚ is not changed except when G is the group T_{4}^{+} (resp. T_{14}^{+}) in which case the new group becomes T_{4}^{+} or T_{9}^{+} (resp. T_{24}^{+}).

The most useful case of the above transformation corresponds to D = −1. In that case, starting with an octic field of signature (r_{1}, r_{2}), we obtain an octic field of signature (r′_{1}, r′_{2}) with the following possibilities. If r_{1} = 8, we obtain r′_{1} = 0. If r_{1} = 6, we obtain r′_{1} = 2. If r_{1} = 4, we can have r′_{1} = 4 or r′_{1} = 0. If r_{1} = 2, we can have r′_{1} = 2 or r′_{1} = 6. Finallty, if r_{1} = 0, we can have r′_{1} = 0, r′_{1} = 4 or r′_{1} = 8.

(3) The direct study of the group structure. For example, the group T_{14}^{+} is the group S_{4} considered as a transitive group of degree 8. It is not difficult to prove that an octic field having such a Galois group is obtained by taking a quartic field of Galois group S_{4}, and adjoining the square root of the discriminant of the quartic field (which of course belongs to the Galois closure).

(4) Pushing this idea further, we adjoined to quartic fields square roots of divisors of the discriminant, and we obtained in this way practically all the missing groups and signatures.

The following tables give, for all signatures and for all possible Galois groups of the Galois closure corresponding to this signature:

(1) The name of the Galois group in the notation of [Bu-Mc]. We chose not to use the more recent (but more complex) notation used in [Co-Hu-Mc].

(2) The number of non-isomorphic number fields having this Galois group and signature in the limits of our main tables. In particular, when the minimal discriminant was obtained by specific methods ((a), (b) or (c)), the indicated number is 0.

(3) The minimal discriminant (in absolute value) corresponding to this Galois group.

TABLES OF OCTIC FIELDS WITH A QUARTIC SUBFIELD 11

Signature (0,4)

G #  d_{min}  G #  d_{min} G #  d_{min}
T_{1}  0 3 32374 32513^{(a)} T_{13}^{+} 9  178 50625 T_{26} 26 187 53525
T_{2}^{+} 14  12 65625 T_{14}^{+} 13  608 86809 T_{27} 592  15 78125
T_{3}^{+} 10  53 08416 T_{15} 8  314 43200 T_{28} 20 378 79808
T_{4}^{+} 31  17 50329 T_{16} 5  94 53125 T_{29}^{+} 201  35 04384
T_{5}^{+} 0 1 22305 90464^{(b)} T_{17} 55  12 57728 T_{30}  1 2153 78125
T_{6} 14  41 02893 T_{18}^{+} 194  60 36849 T_{31} 1017  15 13728
T_{7}  0 51200 00000^{(c)} T_{19}^{+} 3  671 08864 T_{32}^{+}  0 11424 40000^{(c)}
T_{8}  0 10 49633 09568^{(c)} T_{20}^{+} 7  262 65625 T_{35} 4194  13 27833
T_{9}^{+} 209  32 11264 T_{21} 7  335 54432 T_{38}  4 1671 86432
T_{10}^{+} 27  18 90625 T_{22}^{+} 31  254 01600 T_{39}^{+} 262  42 27136
T_{11}^{+} 38  32 40000 T_{23} 0 3 39710 01237^{(c)} T_{40} 10 120 08989
T_{12}^{+} 0 7059 11761^{(c)} T_{24}^{+} 510 17 63584 T_{44} 4127 13 61513

Signature (2,3)

G # d_{min} G # d_{min} G # d_{min}
T_{6} 18 −42 86875 T_{26} 51 −74 86875 T_{35} 2798  −44 61875
T_{8} 5 −1071 71875 T_{27} 40 −746 71875 T_{38}  0 −49413 82327^{(c)}
T_{15} 20 −409 60000 T_{30} 18 −214 34375 T_{40} 25 −226 65187
T_{23} 46 −226 65187 T_{31} 21 −793 60000 T_{44} 9259  −47 11123

12 H. COHEN, F. DIAZ Y DIAZ AND M. OLIVIER

Signature (4,2)

G #  d_{min} G #  d_{min} G #  d_{min}
T_{7}  2  569 53125 T_{20}^{+} 12 2684 35456 T_{31} 50 1049 60000
T_{9}^{+} 60  409 60000 T_{21} 0 88305 03125^{(c)} T_{32}^{+}  0 4 51783 52704^{(c)}
T_{10}^{+} 25  640 00000 T_{22}^{+} 15 3686 40000 T_{35} 2899  173 18125
T_{11}^{+} 16  230 40000 T_{24}^{+} 344 393 75625 T_{38}  0 56472 94088^{(c)}
T_{15} 0 1 16625 89952^{(c)} T_{26} 7 3219 78368 T_{39}^{+} 576  205 02784
T_{16} 3 3200 00000 T_{27} 105 713 03168 T_{40}  0 74950 14493^{(c)}
T_{17} 6 152 43125 T_{28} 67 318 78125 T_{44} 8515 152 97613
T_{18}^{+} 183  193 60000 T_{29}^{+} 188 257 55625
T_{19}^{+} 0 2 77102 63296^{(c)} T_{30} 4 10768 90625

Signature (6,1)

G # d_{min} G # d_{min} G # d_{min}
T_{27} 1205 −746 71875 T_{35} 5657 −688 56875 T_{44} 4045 −1034 05923
T_{31} 764 −2494 95552 T_{38}  9 −49413 82327

TABLES OF OCTIC FIELDS WITH A QUARTIC SUBFIELD 13

Signature (8,0)

G #  d_{min} G #  d_{min} G #  d_{min}
T_{1}  3 4103 38673 T_{13}^{+} 3 6 05238 72256 T_{26} 12  25760 88125
T_{2}^{+} 28 3240 00000 T_{14}^{+} 0 82 13869 40416^{(c)} T_{27} 1407  12922 03125
T_{3}^{+} 9 33177 60000 T_{15} 6 11 94356 44125 T_{28} 57  81608 00000
T_{4}^{+} 20 4420 50625 T_{16} 3 4 78975 78125 T_{29}^{+} 100  50694 40000
T_{5}^{+} 1 1 22305 90464 T_{17} 49  2823 00416 T_{30}  4 12 39119 40625
T_{6}  8 59101 06112 T_{18}^{+} 163 41634 75625 T_{31} 795  19481 60000
T_{7}  8 51200 00000 T_{19}^{+} 1 8 75781 16096 T_{32}^{+}  0 315 03303 56889^{(c)}
T_{8}  3 1 60984 53125 T_{20}^{+} 13 1 48840 00000 T_{35} 6139  3095 93125
T_{9}^{+} 142 15341 32224 T_{21} 1 17 51562 32192 T_{38}  6 6 28261 46729
T_{10}^{+} 79 10643 90625 T_{22}^{+} 8 1 80633 60000 T_{39}^{+} 125  52400 22544
T_{11}^{+} 43 4326 40000 T_{23} 8 2 14154 71433 T_{40}  0 46 31434 05393^{(c)}
T_{12}^{+} 6 58873 39441 T_{24}^{+} 295 23936 55625 T_{44} 4251  11527 84549

(C) Non-isomorphic fields having the same discriminant

We have observed that non-isomorphic fields having the same discriminant occur very frequently. In the limits of our tables, the maximum number observed is a set of 11 such fields in signature (0, 4). This corresponds to discriminant 4840 00000 = 2^{8} · 5^{6} · 11^{2}. We give below the 11 polynomials defining these fields, together with their Galois group.

X^{8} − X^{6} + X^{4} + 24X^{2} + 16 : T_{9}^{+}
X^{8} − 7X^{6} + 34X^{4} − 88X^{2} + 121 : T_{10}^{+}
X^{8} + X^{6} + 16X^{4} + 66X^{2} + 121 : T_{10}^{+}
X^{8} − X^{6} + 16X^{4} − 66X^{2} + 121 : T_{10}^{+}
X^{8} + 5X^{6} + 20X^{4} + 50X^{2} + 25 : T_{18}^{+}
X^{8} + 4X^{6} + X^{4} + 4X^{2} + 1 : T_{18}^{+}
X^{8} + 7X^{6} + 19X^{4} + 28X^{2} + 16 : T_{18}^{+}
X^{8} − 12X^{6} + 64X^{4} − 143X^{2} + 121 : T_{20}^{+}
X^{8} + 12X^{6} + 64X^{4} + 143X^{2} + 121 : T_{20}^{+}
X^{8} − 11X^{6} + 51X^{4} − 121X^{2} + 121 : T_{20}^{+}
X^{8} + 11X^{6} + 51X^{4} + 121X^{2} + 121 : T_{20}^{+}

Observe once again in this table the “mirror effect”. For polynomials number 3, 4, 8, 9, 10, 11, changing X into X\sqrt{−1} changes the number field into a nonisomorphic one having the same discriminant, Galois group and signature. For polynomials number 5, 6, 7 (having Galois group T_{18}^{+}), it changes the number field into a number field having same discriminant and Galois group, but in signature (4, 2). For polynomial number 1, we obtain a reducible polynomial, but this is clearly irrelevant to the problem. Finally, for polynomial number 2, it changes the number field into a number field having the same Galois group and the same signature, but a discriminant which is 256 times smaller. Different choices for D would lead to similar results.

Since we have obtained 3 fields having the same discriminant and Galois group in signature (4, 2) by using the mirror effect, we observed in our tables that in signature (4, 2) there are two more number fields having the same discriminant:

X^{8} − 5X^{6} + 5X^{4} − 25X^{2} + 25 : T_{10}^{+}
X^{8} − 3X^{6} + 4X^{4} − 7X^{2} + 1 : T_{10}^{+}

Note that in certain cases (but not in the examples that we have just given), we could obtain isomorphic number fields by this method.

The following table gives, for each signature, the number of doublets, triplets, etc... of non-isomorphic number fields having the same discriminant.

Signature 2 3 4 5 6 7 8 9 10 11
(0, 4) 2009 134 210 22 15 6 12 1 4 1
(2, 3) 2775 92 307 9 18 2 12 0 2 0
(4, 2) 1799 141 108 7 14 3 3 1 0 0
(6, 1) 1868 169 288 6 12 7 10 0 0 0
(8, 0) 1160 28 35 1 0 0 0 0 0 0

(D) Arithmetically equivalent fields

In this section we give all the examples of non isomorphic octic fields having the same Dedekind ζ-function found in the tables (such fields are called arithmetically equivalent).

The following theorem gives a necessary and sufficient condition for the existence of arithmetically equivalent fields (see [Ga]).

Theorem 3. Let K_{1} and K_{2} be two number fields (assumed to be in a fixed algebraic closure of ℚ). The fields K_{1} and K_{2} are arithmetically equivalent if and only if the following two conditions are satisfied.

(1) The fields have a common Galois closure N.

(2) Let G = Gal(N/ℚ), G_{1} = Gal(N/K_{1}) and G_{2} = Gal(N/K_{2}). Then for each conjugacy class C in G, we must have |G_{1} ∩ C| = |G_{2} ∩ C| (where | | denotes cardinality).

From this theorem, it is not difficult to deduce the following (Y. Eichenlaub, personal communication).

Corollary. If K_{1} and K_{2} are non-isomorphic arithmetically equivalent octic fields, their Galois group is isomorphic to T_{15} or to T_{23}. Conversely, if an octic field K_{1} has Galois group isomorphic to T_{15} or to T_{23} there exists a non-isomorphic field K_{2} arithmetically equivalent to K_{1}.

Since T_{15} cannot occur in signature (6, 1), we cannot have arithmetically equivalent fields with such a Galois group in that signature. In the limits of our tables, we found no example in signature (4, 2), but the desired (minimal) example was found during the search for the minimal discriminant with Galois group T_{15}.

Similarly, T_{23} cannot occur in signature (4, 2) and (6, 1), hence we cannot have arithmetically equivalent fields with such a Galois group in these signatures. In the limits of our tables, we found no example in signature (0, 4), but the desired (minimal) example was found during the search for the minimal discriminant with Galois group T_{23} using the upper bound from [Je].

For each signature we give all the examples found in the limits of our tables (plus the example of T_{15} in signature (4, 2) and T_{23} in signature (0, 4)). For each pair of fields having the same ζ-function, we give octic polynomials generating the corresponding fields. In some cases, ramified ideals do not have the same decomposition as a product of prime ideals in both fields. When this occurs, whe denote by 𝔭, 𝔭′, … prime ideals of residual degree equal to 1 and by 𝔮, 𝔮′, … prime ideals of residual degree equal to 2.

In the limits of our tables, we have found two examples of quadruplets of number fields having the same discriminant, signature and Galois group, forming two pairs of arithmetically equivalent fields (all having T_{23} as Galois group). These examples occur for discriminants −150730227 and −1327373299.

Two arithmetically equivalent fields have the same product h(K)R(K) of the class number by the regulator. Since the class number is very often equal to 1, it is usually the case that the class numbers and the regulators are equal. It has however been noticed by several authors (see for example [Sm-Pe]) that the class numbers (hence the regulators) of arithmetically equivalent fields may be different.

Two of the 18 pairs of arithmetically equivalent fields with T_{15} as Galois group that we have found give such examples. Both are in signature (2, 3). The pairs are for discriminant −518711875 and −1097440000, for which the second field given below has class number 2 while the first has class number equal to 1. In both of these cases, the narrow class number of the fields coincide and is equal to 2.

Signature (0,4)

d_{K}  Polynomial  Different

factorisation

31443200  X^{8} − X^{6} − 4X^{5} − 2X^{4} + 4X^{3} + 12X^{2} + 6X + 1 5ℤ_{K} = 𝔭^{2}𝔭′^{2}𝔮𝔮′

17ℤ_{K} = 𝔭^{2}𝔭′𝔭′′𝔮^{2}

31443200 X^{8} − 2X^{7} + 3X^{6} − 6X^{5} + 13X^{4} − 20X^{3} + 21X^{2} − 14X + 5 5ℤ_{K} = 𝔭^{2}𝔭′𝔭′′𝔮^{2}

17ℤ_{K} = 𝔭^{2}𝔭′^{2}𝔭′′^{2}𝔮

70304000  X^{8} − 4X^{7} + 7X^{6} − 2X^{5} − 8X^{4} + 8X^{3} + 2X^{2} − 4X + 1 5ℤ_{K} = 𝔭^{2}𝔭′^{2}𝔭′′^{2}𝔮

13ℤ_{K} = 𝔭^{2}𝔭′𝔭′′𝔮^{2}

70304000  X^{8} + X^{6} − 2X^{5} + 5X^{4} − 2X^{3} + X^{2} + 1 5ℤ_{K} = 𝔭^{2}𝔭′𝔭′′𝔮^{2}

13ℤ_{K} = 𝔭^{2}𝔭′^{2}𝔭′′^{2}𝔮

143327232  X^{8} + 6X^{6} + 15X^{4} + 12X^{2} + 3
143327232  X^{8} + 3X^{4} + 3
212556032  X^{8} − 4X^{7} + X^{6} + 8X^{5} + 5X^{4} − 14X^{3} − 13X^{2} + 4X + 17
212556032 X^{8} − 2X^{7} + 5X^{6} − 10X^{5} + 23X^{4} − 36X^{3} + 51X^{2} − 48X + 17

TABLES OF OCTIC FIELDS WITH A QUARTIC SUBFIELD 17

Signature (2,3)

d_{K} Polynomial Different

factorisation

−40960000  X^{8} + 4X^{6} + 5X^{4} + 2X^{2} − 1
−40960000  X^{8} − X^{4} − 1
−131274675  X^{8} − X^{7} + X^{6} − 4X^{5} + X^{4} − 4X^{3} + X^{2} − X + 1 5ℤ_{K} = 𝔭^{2}𝔭′^{2}𝔮𝔮′
−131274675  X^{8} − 2X^{7} + 4X^{6} − 5X^{5} + X^{4} − 5X^{3} + 4X^{2} − 2X + 1 5ℤ_{K} = 𝔭^{2}𝔭′𝔭′′𝔮^{2}
−342102016  X^{8} + 8X^{4} − 1
−342102016  X^{8} + 4X^{6} + 5X^{4} + 2X^{2} − 4
−359661568  X^{8} − 4X^{6} − 4X^{5} + 2X^{4} − 2 7ℤ_{K} = 𝔭^{2}𝔭′^{2}𝔭′′^{2}𝔮
−359661568  X^{8} − 4X^{5} + 8X^{3} + 4X^{2} + 4X + 1 7ℤ_{K} = 𝔭^{2}𝔭′𝔭′′𝔮^{2}
−518711875  X^{8} − 6X^{6} − X^{5} + 4X^{4} + 13X^{3} + 9X^{2} + 10X − 5 19ℤ_{K} = 𝔭^{2}𝔭′^{2}𝔭′′^{2}𝔮
−518711875 X^{8} − X^{7} − 6X^{6} + 8X^{5} + 11X^{4} − 15X^{3} − 29X^{2} + 55X − 25 19ℤ_{K} = 𝔭^{2}𝔭′𝔭′′𝔮^{2}
−1024000000  X^{8} − 15X^{4} − 50X^{2} − 25
−1024000000  X^{8} + 5X^{4} − 25
−1097440000  X^{8} − 5X^{6} + 3X^{4} + 15X^{2} − 19 19ℤ_{K} = 𝔭^{2}𝔭′𝔭′′𝔮^{2}
−1097440000  X^{8} + 5X^{6} + 3X^{4} − 15X^{2} − 19 19ℤ_{K} = 𝔭^{2}𝔭′^{2}𝔭′′^{2}𝔮
−1119744000  X^{8} − 2X^{7} + X^{6} − 8X^{5} + X^{4} − 8X^{3} + X^{2} − 2X + 1 5ℤ_{K} = 𝔭^{2}𝔭′𝔭′′𝔮^{2}
−1119744000  X^{8} − 2X^{7} + X^{6} + 4X^{5} − 5X^{4} + 4X^{3} + X^{2} − 2X + 1 5ℤ_{K} = 𝔭^{2}𝔭′^{2}𝔭′′^{2}𝔮
−1344252672  X^{8} − 3X^{6} + 6X^{2} − 3
−1344252672  X^{8} + 3X^{6} − 6X^{2} − 3
−1517535243  X^{8} − X^{7} − 5X^{6} + 8X^{5} + 4X^{4} − 25X^{3} − 5X^{2} + 11X − 5 17ℤ_{K} = 𝔭^{2}𝔭′𝔭′′𝔮^{2}
−1517535243  X^{8} − X^{7} + 4X^{6} − X^{5} + 7X^{4} − 10X^{3} − 8X^{2} + 14X − 5 17ℤ_{K} = 𝔭^{2}𝔭′^{2}𝔭′′^{2}𝔮

18 H. COHEN, F. DIAZ Y DIAZ AND M. OLIVIER

Signature (4,2)

d_{K} Polynomial Different

factorisation

11662589952  X^{8} − 4X^{7} + 2X^{6} − 4X^{5} + 12X^{4} + 12X^{3} − 4X − 2
11662589952 X^{8} − 4X^{7} + 6X^{6} + 8X^{5} − 36X^{4} + 32X^{3} + 14X^{2} − 24X + 1

Signature (8,0)

d_{K} Polynomial Different

factorisation

119435644125 X^{8} − 4X^{7} − 3X^{6} + 23X^{5} − 3X^{4} − 37X^{3} + 8X^{2} + 15X − 5 5ℤ_{K} = 𝔭^{2}𝔭′^{2}𝔭′′^{2}𝔮

17ℤ_{K} = 𝔭^{2}𝔭′^{2}𝔭′′^{2}𝔮

119435644125 X^{8} − X^{7} − 11X^{6} + 4X^{5} + 21X^{4} − 4X^{3} − 11X^{2} + X + 1 5ℤ_{K} = 𝔭^{2}𝔭′𝔭′′𝔮^{2}

17ℤ_{K} = 𝔭^{2}𝔭′𝔭′′𝔮^{2}

131153375232  X^{8} − 12X^{6} + 45X^{4} − 54X^{2} + 12
131153375232  X^{8} − 9X^{6} + 24X^{4} − 21X^{2} + 3
186601439232  X^{8} − 14X^{6} + 44X^{4} − 46X^{2} + 13 13ℤ_{K} = 𝔭^{2}𝔭′^{2}𝔭′′^{2}𝔮
186601439232  X^{8} − 10X^{6} + 32X^{4} − 38X^{2} + 13 13ℤ_{K} = 𝔭^{2}𝔭′𝔭′′𝔮^{2}

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Laboratoire A2X, Université Bordeaux I, 351 cours de la Libération, 33405 Talence Cedex, FRANCE

E-mail address: cohen@math.u-bordeaux.fr, diaz@math.u-bordeaux.fr, olivier@math.u-bordeaux.fr