A table of quintic number fields

Introduction

In the last few years several extensive lists of number fields were calculated. In particular, we mention the calculation of fourth-degree fields up to a discriminant bound of one million by D. Ford, J. Buchmann, and the second author [2,3,5]. The huge amount of computation time showed that similar tables for primitive fields in higher dimensions would require refined techniques. A first attack on the totally real quintic case was done by the third author about 2 years ago [4]. At the same time the determination of the minimum discriminant for totally real octic fields by Pohst, Martinet, and Diaz y Diaz [11] was successful because of much better estimates for several coefficients of a generating polynomial. In this paper we take up those ideas and apply them to fifth-degree polynomials of arbitrary signature.

In §2 we describe the generation of the polynomials and develop new estimates for their coefficients for each of the three possible signatures. In §3 we discuss the processing of those polynomials. Their discriminants are calculated integrally; bounds on the required number of arithmetical operations and on the size of the occurring integers are also given. At the same time polynomials of incorrect signature are removed. Then reducible polynomials are eliminated.

All remaining ones are generating polynomials for number fields F . We compute an integral basis for F (hence, also the discriminant dF of F) with the ROUND-2 algorithm of Zassenhaus [18]. Redundant fields (i.e., those which are isomorphic to a field which was already obtained) are removed with the help of the isomorphy test of [10]. Finally, the Galois group of F is computed with the resolvent method. In the last section we present various numerical data. We found a total of • 22 740 totally real fields with discriminant less that 2 x 107, • 79 394 fields with 2 complex conjugates and discriminant larger than -5x 106, • 186 906 fields with 4 complex conjugates and discriminant less than 5x 106. All data can be obtained from the second author.

Generation of polynomials

Let F be an algebraic number field of degree « and discriminant dF , and let tfp be the ring of integers of F. Let F = Q(p) for a root p of a monic irreducible polynomial fp(t) e Z[t] with deg(//>) = « . Then fp(t) is the characteristic polynomial of p, and we denote its zeros in C by p = /?(1), ... , //"'. As usual, we choose pW, ... , p(r^ e R and p^+V, ... , p^+r^ e C\R with p(r¡+r2+j) = p{rt+j) for 1 < / < r2, so that « equals rx + 2r2 . For any element x e F, say x = £"=i 9//''-1 (í¡ e Q, 1 < i < n), the conjugates of jc are given by x& := E"=i Ï«PU)'_I • an Improving methods in earlier publications, for example [2,4,6,8,9], we use all our information about the values of Sx, S2 and an to determine good bounds for Sk (k e {3, ... , n-1, -1}). Thus, the number of generated polynomials is reduced drastically. Without these improved bounds it would not be possible to compute such large sets of algebraic number fields in reasonable CPU-time.

In the sequel, F = Q(p) denotes an algebraic number field of degree five with prescribed signature (rx, r2). For an upper bound B on the absolute value of the discriminant of F we choose B = 2 x 107 for rx = 5, and B = 5 x 106 otherwise. For each of the three signatures we construct a set M of monic fifthdegree polynomials such that any quintic number field F of correct signature and of absolute discriminant \dF\ < B contains an integer p (e F\Q) whose characteristic polynomial fp(t) = t5 + axt4 + a-¡t} + a^t2 + a4t + as e Z[t] is contained in M. We proceed in analogy to [6,9]. Proposition 2.1. Let F be an algebraic number field of degree « = 5 with discriminant dF , \dF\ < B. Then there exists an algebraic integer p in F with License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use F = Q(p) and characteristic polynomial fp(t) satisfying aie{0, 1,2} and 5 (4 ^1/4 T2(p) <U2:=-+ (jB As a consequence, we also get bounds for the coefficients a2, as of the characteristic polynomial of that element p . Newton's relation a2 = (a2 -S2)/2 immediately yields a2 > (a] -U2)/2.

Applying the Cauchy-Schwarz inequality to the vectors (/?(1), ... , pW), (1,..., 1), we obtain a2 < 5(U2 + S2)/2 ; hence, a2 < (u2 + L2) /2.

5

The inequality between arithmetic and geometric means implies

For totally real fields F, i.e., rx = 5 , we have T2(p) = S2(p) > §« [15] and therefore much better upper boundŝ ! 2 15 . . . ^ fS2\ a2< 2«i --4-and lfl5l < I y ) 5 Because of T2(x) = S2(x) for all x e R5, we do not need the side condition T2(x) < U2 in this case. The set G := {x e R5\gj(x) = 0, j -1,2,3} is compact. Hence, Sk attains a maximum and a minimum in G for k e {3, 4, -1} . The following proposition is a consequence of Lagrange's multiplier method (see [13]). We thus must consider three cases for which we calculate all x e G subject to one of the conditions of the last proposition. We give a short example how to determine all vectors x e G with xi = x2 = x-$. Eliminating x4 and xs from the system of equations

we get

x? -is,x, + ^(S2--S2)x,3 + X-as = 0. For all solutions x e G we compute Sk(x) for k e {3, 4, -1}. In the other two cases we proceed analogously.

The required extrema are among the Sk -values for the finitely many x e G obtained. All polynomials for xi are of degree at most five. Using the bounds for S3 we can estimate 03 by Newton's relations. The bounds for S4 and S-X yield estimates for a4 . Additional bounds from [6,8] are used to possibly further reduce the ranges for 03 and a4 .

2.2. Polynomials with signature rx = 3 and r2 = 1. If we denote the real roots by xx, x2, X3 and the complex roots by X4 ± ix$ with X4, X5 e R, then we have to search for extrema of the real functions S-}(x) := x\ + x2 + X3 + 2(x4 -3x4X5), S4(x) := xf + x24 + x34 + 2(x4 -6x42x| + x4),

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The constraints are gx (x) = xx + x2 + x3 + 2x4 -Sx = 0, g2(x) = x\ + x22 + x32 + 2(x| -x52) -S2 = 0,

We define sets

Since G is compact, we have global extrema in G. If x e E5 is an extremum, then it is either an element of G$ which satisfies g4(x) < 0, or it is an element of G4 . So we must determine all local extrema in G3 and G4. Again, we apply the method of Lagrange multipliers and obtain the following proposition ( [13]).

The function S4 has its local extrema in the set G$ only at points x = (xi, ... , X5) which satisfy either one of the conditions (l)-(3) of (a) or, in case Sx t¿ 0, one of the conditions (4) xx = x2 = Sx A Xi t¿ x3 A x5 ^ 0,

There are only finitely many local extrema. The degree of the polynomials whose roots we must compute is at most five. We only consider solutions x e K5 with g4(x) < 0.

To determine the extrema in G4 is somewhat more difficult, because we do not have similar conditions for the coordinates of a solution x as in the totally real case. Hence, if J(x) is not of maximal rank, then two of the coordinates xi, x2, X3 must be equal, and without loss of generality we can assume Xi = x2. Computing resultants of the polynomials gj(xx, xx, X3, x4, \(U2 -S2)) (1 < j < 3), we obtain an equation of degree ten for x4.

In the sequel we assume that J(x) has rank four.

/( H Determining the extrema of S3 by the Lagrange multiplier method yields Xx,... , X4 e R subject to 1 as

Eliminating Xi, x2, X3, we obtain

These results are inserted into the fourth equation above to obtain an equation of degree six in x4. For the functions S4 and S-X we proceed similarly and get equations of degree seven in x4 in each case.

Having calculated x4 , we form resultants of the polynomials £,(Xi,X2,X3,X4,X5), 1<7<3, in the variables xi and x2 . This yields an equation of degree three for X3.

2.3. Polynomials with signature rx = 1 and r2 = 2. We apply the same method as before. We denote the zeros of a generating polynomial by Xi, x2 ± ¿X3, x4 ± ¿X5. Hence, we need to determine maxima and minima of the functions Let the sets G, Gj, and G4 be defined as in the preceding case. For the extrema in (73 we get the following result (see [13]). Proposition 2.6. (a) Each of the functions S3 and S-X has its local extrema in the set G3 only at points x = (xi, ... , x5) which satisfy one of the conditions (1) x3 = x5 = 0, (2) Xi = x2 A x3 = 0,

The function S4 has its local extrema in the set G3 only at points x = (xx, ... , Xs) which satisfy either one of the conditions (l)-(3) of (a) or, in case Sx ± 0, the condition (4) xi ¿ x2 = Sx A x3 = 0 A x5 ± 0.

Again we obtain only finitely many local extrema. Solving the corresponding system of equations we must calculate the zeros of polynomials of degree at most five.

For finding the global extrema in G4 we have to check if the Jacobi matrix has rank four. For this, the following remark is helpful. and X3 t¿ 0 t¿ X5, the rank of X-, ~T Xn X4 "T X5 Remark 2.7. For x2 j= x4 . the Jacobi matrix is four.

The cases x2 = x4 or x\ + x2 = x2 + x2 or x3 = 0 are not difficult to handle. We notice that X3 and X5 cannot both be zero. If the rank of the Jacobi matrix is four, then we compute the Lagrange multipliers as functions of Xi and obtain a polynomial equation for xi. We note that in this case the degree can become as large as 17.

Processing of generated polynomials

Since the set M of polynomials generated by the ideas of the preceding section turns out to be quite large for each signature (see §4), the methods of this section should be really fast. In a first step we compute the discriminant and the signature of each polynomial simultaneously. i.e., the coefficients are the power sums of the zeros of f(t). We note that det(Q/) is the discriminant of f(t) and that the number of real roots of /(/) equals the difference of the numbers of positive and negative eigenvalues of Qf [7]. The latter are easily computable by an application of the following lemma from [13]. We remark that in our case det(ß/) ^ 0, because otherwise we can discard f(t) immediately, and therefore we can always achieve that the principal minors of Qf are not zero. Hence, using a Cholesky-type method [9] for evaluating the determinant of Qf, we obtain the signature of f(t) at the same time for free. The following proposition from [ 13] allows us to do all calculations with rational integers. The advantage of operating exclusively with integers was already discussed in [1]. m mg-'<-"-»g-""g-"

for k > 0 andk < i, j < n . If we have M e Znxn , then also m{k) eZ for 0 < k < « -1, k <i, j < n.

We remark that the principal minors of M satisfy det(Mjfc) = m(kkX) = m(k)k (i<k<n).

The following algorithm is immediate. (1) (Initialization) Compute the power sums Sk for 0 < k < 2« -2 via Newton's relations and initialize the Hankel matrix Qf = (m¡j)x<ij<n <-(SVh/_2)i<( j<" . Set s <-0 and d <-1 .

(2) For k = 1 to « -1 do If mkk = 0 then If there is an index / with k < I < « and m¡¡ ¿ 0, then interchange mki and «?/, for k < i < n and then interchange mik and w,/ for k < i < n. Else if «z" = 0 for all i (k < i < n) and there is an index / with k < I < n and mk¡ ¿ 0 then set mki <-mki + m/, for k < i < n and then set mik <-mik + m,/ for k < i < n .

Else set D(f) <-0 and terminate.

For i = k + 1 to « do Set m¡j <-(mkkm¡j -mikmkj)/d for i < j < « .

License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use Set mki <-0 for k + 1 < i < « . Set mji <-m¡j for k < i < n -1 and i + 1 < j < n .

Set 5 <-s + sign(mkk) • si%n(d) and d <-mkk . (3) Set D(f) <-m"" , s «-s + sign(m"") • si%n(d) and terminate.

Regarding the complexity of the algorithm we get (see [13]): This method of computing the polynomial discriminant and the signature simultaneously is very fast in practice. Comparisons with the PARI system showed that for our polynomials, Algorithm 3.3 is about ten times faster.

After the calculation of all polynomials of the signature under consideration we must check them for irreducibility. Factoring them modulo small prime numbers and comparing the degrees of potential factors often proves irreducibility in a fast way. If proper factors can still exist, we easily obtain estimates for their coefficients and test all remaining candidates.

The irreducible polynomials then generate fields of correct signature. Next we must check whether the field discriminant lies within the given bounds. We use the Dedekind criterion (see [12], for example) to detect index divisors. If there are none, then we know that the equation order is maximal and the discriminant of the polynomial coincides with the field discriminant. In the remaining cases we compute an integral basis of the corresponding maximal order by a specialized version of the ROUND-2 algorithm of Zassenhaus [18].

Our methods sometimes yield several generating polynomials for one field. Hence, a final task is to reject all redundant ones. An easy test to check whether several polynomials generate isomorphic fields is given in [10].

3.1. Computation of Galois groups. As a prerequisite we list all transitive subgroups of the symmetric group S$, and for each of them the frequency of cycle distributions. For example, " 15 x (2, 2, 1)" means that the group contains 15 elements which decompose into two cycles of length two and one cycle of length one.

Ss lx(l,l,l,l,l),10x(2,l,l,l),15x(2,2,l),20x(3,l,l), 20x (3,2), 30 x (4, l),24x(5) Hol(Cs) 1 x(l, 1, 1, 1, l),5x(2,2, 1), 10 x (4, l),4x(5) As 1 x(l, 1, 1, 1, 1), 15x(2,2, l),20x(3, 1, l),24x(5) D5 lx(l, 1, 1,1, l),5x(2,2, l),4x(5) C5 lx(l, 1, 1,1, l),4x (5) The groups include each other in the following way: Lemma 3.6. Let f(t) be as in the preceding lemma. If Tf contains cycles of length 2 and deg(/), respectively, then Yf is the symmetric group Sn . Lemma 3.7. Let f(t) be as in Lemma 3.5 but of prime degree p > 3. If f has exactly two nonreal roots, then its Galois group Yf is the symmetric group Sp . Lemma 3.8. Let f(t) be as in Lemma 3.5 . Then its Galois group Yf is contained in the alternating group A" if and only if the discriminant of f is a square in Z. Lemma 3.9. Let f(t) be as in Lemma 3.5 but of degree « = 5. If its Galois group Yf is the cyclic group Cs, then f has five real roots.

Ss

Hence, we decompose a given monic irreducible polynomial f(t) modulo pZ[t] into its prime factors for a few (usually not more than 10) small prime numbers p . From these results we can already guess the corresponding Galois group in most cases. For the remaining ones we use indicator functions (see [12,16]) which tell us whether Yf is contained in Hol(Cs), C5, respectively. Assume that the values g2(x^X), ... , xT (5)) are distinct for x e V2. Then the Galois group Yf is contained in (hence equal to) C5 if and only if y = c?2(*jt(i), ■■■ , *7t (5)) is a rational integer for some n e V2.

Numerical results

All computations were carried out on Apollo workstations (CPU Motorola 68020, 68030, 68040). The use of the software package KANT [14] was absolutely essential. Table 1 we list the first 50 field discriminants and the coefficients of a corresponding generating polynomial in each case. An integral basis is always a power basis for these generators. Next, in Table 2 (next page), we consider the distribution of the fields with respect to their Galois group. Among all 22 740 fields, 22 676 have symmetric Galois group Ss (99.72%). We list the remaining 64 fields. There are 61 discriminants below 20 000 000 belonging to two nonisomorphic fields, namely, License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use In Table 3 we list the first 50 fields, each of which has a power integral basis.

There are various discriminants for which nonisomorphic fields exist. Table 4 gives a short account on this phenomenon. If there are more than 10 different discriminants for which p nonisomorphic fields of that discriminant occur (which happens to be the case for p = 2, 3), we only list the 10 largest discriminants. 4.3. Fields with signature rx = 1 and r2 = 2 and dF < 5 000000. We compute all characteristic polynomials of algebraic integers p with T2(p) < 45.98. This yields 670725 968 polynomials, 534 326207 of which have exactly one real zero. There remain 186906 nonisomorphic fields of a discriminant below 5 000 000. We list in Table 5 the smallest 50 fields, all of them with a power basis as integral basis.

There are 81 fields with Galois group Hol(C5), 258 fields with A5, and 129 fields with D$. The group C5 cannot occur as Galois group. For every possible Galois group we present the field of smallest discriminant. In each case an integral basis is a power basis. Table 6 gives some information on discriminant values for which nonisomorphic fields occur. If there are more than 10 different discriminants for a fixed number of nonisomorphic fields, only the 10 smallest discriminants are listed. Table 6 number of nonisomorphic fields discriminants total

Table 3