A table of quintic number fields

mathematics of computationvolume 63, number 207july 1994, pages 361-376

A TABLE OF QUINTIC NUMBER FIELDS

A. SCHWARZ, M. POHST, AND F. DIAZ Y DIAZ

Abstract. All algebraic number fields F of degree 5 and absolute discriminant

less than 2 x 107 (totally real fields), respectively 5 x 106 (other signatures)

are determined. We describe the methods which we applied and list significant

data.

1. 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 discrim-

inant 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 poly-nomial. 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 esti-

mates 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 polynomialsof incorrect signature are removed. Then reducible polynomials are eliminated.

All remaining ones are generating polynomials for number fields F . We com-

pute 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 whichare 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.

Received by the editor June 25, 1992 and, in revised form, December 30, 1992.

1991 Mathematics Subject Classification. Primary 11Y40. @I994 American Mathematical ^¡^0025-5718/94 $1.00+$.25 per page

361

License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use

362 a. schwarz, m. pohst, and f. diaz y diaz

2. 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 char-

acteristic 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 •We introduce the function

T2:F^R^°:x~J2\xii)\2-i=l

With x e F represented by a fixed basis, T2(x) becomes a positive definitequadratic form in the coefficients. Also, we need the /eth power sum of x ,

n

Sk:F* -+l:rH 5^x(/)* (fc e Z).¡=i

We note that for xé^f and /c > 0 the /eth power sum of x is a rational

integer. Newton's relations

fc-i

Sk + Y^aisk-i + kak = 0 (l<k<n);'=1

allow us to calculate the power sums from the coefficients of the characteristicpolynomial fx(t) = t" + axtn~x H-h a„ e Z[t] of the algebraic integer x e F,

and vice versa. Hence, we can easily estimate the coefficients of fx(t) if we

have bounds for the power sums of x e F . Additionally, we note that

S_x = (-lyft!.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 fifth-

degree 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


A TABLE OF QUINTIC NUMBER FIELDS 363

F = Q(p) and characteristic polynomial fp(t) satisfying

aie{0, 1,2}

and5 (4 ^1/4

T2(p) <U2:=-+ (jB

As a consequence, we also get bounds for the coefficients a2, as of the char-

acteristic polynomial of that element p . Newton's relation a2 = (a2 - S2)/2immediately 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

5/2

1 < N = \\pU)7=1

<-mRemark 2.2. For totally real fields F, i.e., rx = 5 , we have T2(p) = S2(p) > §«[15] and therefore much better upper bounds

^ ! 2 15 . . . ̂ fS2\a2< 2«i --4- and lfl5l < I y )5/2

Estimates for a3, a4 are derived from bounds for S3, S4, S_x, and the latterare obtained by calculating global maxima and minima of the functions

n

Sk: C5^E: x^]T;tf1=1

for k e {3, 4, -1} and fixed values of Sx, S2 and as and under the additional

constraint

T2(x):=|*i|2 + --- + l*5|2<t/2.

These extrema are determined by the Lagrange multiplier method.

In the sequel we assume that Sx, S2, as e Z and f/2 € R>0 are fixed.

2.1. Polynomials with signature rx = 5 and r2 = 0. We determine extrema of

the functions

n

Sk:R5 ̂ R:x^Y,xï> ¿te{3,4,-l},1=1

under the restrictions

gx(x) = xx + x2 + x3 + x4 + xs - Sx =0,

g2(x) = x¡ + x¡ + x2 + x¡ + x¡-S2 = 0,

gi(x) = xxx2X}X4xs + as = 0.


364 A. SCHWARZ, M. POHST, AND F. DIAZ Y DIAZ

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 com-pact. 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]).

Proposition 2.3. (a) Each of the functions S3 and S-X has a global maximum

and a global minimum on the set G at a vector x - (xx, ... , xs) whose coor-

dinates satisfyxx= x2 = X3 or xx = x2 A x?, = x4.

(b) The function S4 has a global maximum and a global minimum on the set

G at a vector x = (xx,... , xs) whose coordinates satisfy

xx = x2 = Xt, or xx = x2 A x-} = x4

or, in case Sx ̂ 0,

xx = x2 = Sx and xx, X3, x4, xs are pairwise distinct.

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 xsfrom the system of equations

3xi + x4 + Xs — Sx

3xx + x4 4- x5 — S2

X] X4X5 + ¿Z5

we get

x? - is,x, + ^(S2- -S2)x,3 + X-as = 0.

For every real solution xx of that equation we compute X5 from

xj + (3x, - Sx)x5 + 6x2 - 3Sxxx + ^(S¡ -S2) = 0

and, finally, X4 fromX4 = 01 — 3xi — X5.

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, thenwe 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),

r. / \ 1 1 1 ~ Xd5_,(x) := — + — + — + 2

Xi x2 x3 xj + x2

= 0,

= 0,

= 0,



The constraints are

gx (x) = xx + x2 + x3 + 2x4 - Sx = 0,

g2(x) = x\ + x22 + x32 + 2(x| - x52) -S2 = 0,

g3(x) = xix2x3(x| + X2) + «5 = 0,

g4(x) = x2 + x2 + x2 + 2(x2 + x2) - U2 < 0.

We define sets

G := {x g M5|£4(x) < 0 A gj(x) = 0, j = 1,2,3},

G3:={xeR5\gj(x) = 0, j= 1,2,3},

C74:={xeK5|s;(x) = 0, j= 1,2,3,4}.

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]).

Proposition 2.4. (a) Each of the functions S3 and S-X has its local extrema in

the set G3 only at points x = (xx,..., x$) which satisfy one of the conditions

(1) X) = X2 = X3 ,(2) X) = x4 A x5 = 0,

(3) xi = x2 A x5 = 0.

(b) 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,

(5) x4 = Sx A X5 = 0 A Xi, x2, X3, x4 pairwise distinct.

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. Thus, it is necessary to calculate Lagrange multipliers explicitly, ifthe Jacobi matrix

dgj(x)\

°Xi /1<K5,1<J<4

has rank four. Because of S2 < U2, we have

0¿x2=l-(U2-S2)

in G4.

Remark 2.5. If X5 ̂ 0 and Xi, x2, X3 are pairwise distinct, then the rank ofJ(x) is four.

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. Comput-

ing 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 as3xf + Xx + 2X2xx -A3-h 2A4xi = 0,

Xl

3x2 + Xx + 2X2x2 — A3-h 2A4x2 = 0,x2

3X3 + Xx + 2X2X3 — A3-h 2A4x3 = 0,X3

3(x4 - x2) +XX + 2X2x4 - A3x4 2 5 2 + 2A4x4 = 0,x4 +X5

-6x4 — 2A2 — A3—^-, + 2A4 = 0.X^ ~r X5

Eliminating Xi, x2, X3, we obtain

x^ + X5

A4 = 3x4 - - ( —-j-—j-T + Sx ) ,4V(x2 + x2)2 /

3X2= - X4 — -(Sx - 2x4),

Xx = 1((SX - 2x4)2 - (S2 - 2(x¡ - x2))).

These results are inserted into the fourth equation above to obtain an equationof 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 func-

tions

53(x) := xf + 2(x2 - 3x2x|) + 2(x4 - 3x4x2),

54(x) := xf + 2(x2 - 6x2x2 + x4) + 2(x4 - 6x|x| + x4),

5_,(x):= 1 + 2^^ + 2:A] Jij 1 A^-5 JCa "T" -V«

subject to the restrictions

gi (x) := xi + 2x2 + 2x4 - Sx =0,

g2(x) := x2 + 2(x¡ - xj) + 2(x¡ - x¡) -S2 = 0,

gi(x) := xi(x22 + x32)(x2 + x52) + a5 = 0,

g4(x) := x2 + 2(x2 + x32) + 2(x2 + x52) - U2 < 0.

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,

(3) x2 = x4 A X3 = X5.

(b) 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 ofX-, ~T Xn X4 "T X5Remark 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.

3. 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 ofthis section should be really fast. In a first step we compute the discriminant

and the signature of each polynomial simultaneously. For a monic «th-degree

polynomial f(t) e Z[t] we define the quadratic form

qf(xx ,...,x„):= ^2 Si+j-iXiXj = xuQfxl<i,j<n

with coefficient matrix

/So Sx S2 ■■■ Sn-X ^

sx s2

Q/= s2

\Sn-X . S2n-2J

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 lemmafrom [13].

Lemma 3.1. Let M be a real symmetric « x n-matrix, and det(M;) ^ 0 for

M¡ := (mjk)x<j k<¡ (1 </'<«). 77te« the number of negative eigenvalues of M

is equal to the number of sign changes in the sequence 1, det(Af! ),..., det(Af„).



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 ratio-

nal integers. The advantage of operating exclusively with integers was already

discussed in [1].

Proposition 3.2. Let M = (wj? )i</j<n be a real symmetric « x n-matrix with

det(Af) ^ 0, and let q(x) := ]£?,,/= i m\j x¡xj be the corresponding quadratic

form. Then for each k e {0, ... , « - 1} the form q(x) is equivalent to the

quadratic formk m{k) , n

m(k) ■*' T m(k) Z- mU "l~* ',=1 '"/_i),_i mkk ¡,j=k+l

defining mW := 1 and

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.

Algorithm 3.3. (Computation of polynomial discriminant and signature)

Input. Degree « of the polynomial / and the polynomial coefficients.

Output. Polynomial discriminant D(f) and the number of real roots of / in

case D(f) ¿ 0.(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 thenIf 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 thenset 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 < « .



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]):

Proposition 3.4. Let f(t) = t" + axtn~x +-Y an e Z[t] be a polynomial of

degree « > 2 and define a := max({|a,| 11 < i < «} U {1}). Then(a) Algorithm 3.3 requires at most

4(« - l)2 + 2«(« - 1) + f «(« - 1)(« + 1) + 2 = 0(n3)

arithmetical operations.

(b) The absolute values of the intervening integers are bounded by

23(«-l)(a+l)4(«-l)2.

This method of computing the polynomial discriminant and the signaturesimultaneously 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 considerationwe must check them for irreducibility. Factoring them modulo small prime

numbers and comparing the degrees of potential factors often proves irreducibil-

ity 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 theremaining 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 frequencyof 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:

Ss

As

Hol(C5)

D5

IC5

To decide which subgroup of Ss is the Galois group of a given monic irreducible

polynomial f(t) e Z[t], we use the following criteria which are contained in

(or easily deducible from) [17, pp. 202-204].

Lemma 3.5. Let f(t) e Z[t] be a monic irreducible polynomial of degree n, and

let Yf ç Sn be the Galois group of f. Let p be a prime number not dividing

the discriminant of f, and let f = /i • • • f mod p be a congruence factorization

into monic irreducible polynomials. Then Yf contains an element n = nx---nr,

where the iij are disjoint cycles of length deg(/}) (1 < j < r).

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 hasexactly 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 con-

tained 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.

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.

Let

gx(xx, ... ,x5)

:= (X[X2 + X2X3 + X3X4 + X4X5 + X5X1 — X1X3 — X3X5 — X5X2 — X2X4 — X4X))

and Vx be a full set of representatives of Ss/Ho^Cs), for example,

Vx = {(1), (12435), (15243), (12453),(12543),(12)(34)}.

Assume that the values gx(xxW, ... , xT(5)) are distinct for x e Vx. Then the

Galois group Yf is contained in Hol(C5) if and only if y = gx(xK(X), ... , xn(s))

is a rational integer for some n e Vx.

Similarly, we define

g2(xi, ... , x5) := xix2 + x2x| + x3x4 + x4x| + x5x,2



and

K2:={(1),(12)(35)}.

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.

4. Numerical results

All computations were carried out on Apollo workstations (CPU Motorola

68020, 68030, 68040). The use of the software package KANT [14] was abso-lutely essential.

4.1. Totally real fields with dF < 20000000. According to Proposition 2.1,we have to compute all characteristic polynomials of algebraic integers p withT2(p) < 64.50. Our method generates 110476 592 polynomials, 35 758 669of which are totally real. There remain 22 740 nonisomorphic fields. In Table

1 we list the first 50 field discriminants and the coefficients of a correspondinggenerating 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 symmetricGalois group Ss (99.72%). We list the remaining 64 fields.

Table 1

No dF f No dF

1234

567

89

1011121314

1516171819202122232425

146412421736497385696565770601815098158989417

101833106069117688122821124817126032135076138136138917144209147109149169153424157457160801161121

-3

110

-1-2

1

4-4

-5-2-4-4

60

-1

-3

21

-5

3-2-5-4

-3

26272829

30313233343536373839404142

4344

454647

484950

170701173513176281176684179024180769181057186037195829202817205225207184210557216637218524220036220669223824223952224773230224233489236549240133240881

-6 0-5 -3-5 -3-6 1-8-7-7-6-5-5

-4

-11-7

-6-6-4 -6-6 -1-6 -7-6 -4

04

-2-2-6-4

-6 -3-6 -2-6 -4 8-7 2 3

-5

5-5

22

-2

2-1-2-2

21

11

-1



Table 2

Galois group Hol(C5)

No No /23820322450000369800067258976889792690312571290888804429

0 -90 -101 -81 -8

4 170 20

1 -11 -9

13131

0 -10 0 201 -12 -4 182 -18 22 -2

-12

109

-3

1

92

-3

9101112131415

11122069112500001505112516200000191209761950312519827925

2

01 -130 -151 -14

0 -10

12 -24 820 0 80

-1 23

0 45-6 3

0 20

23169

301

7

1 -24 -56 29 104

Galois group As

No dF f No

31046445184729616032461801967017201740928483059241079122511744329

-11 -1-7 -11

-15 15

15

0 -172112

-12

30-19-9 -8

-14 1-15 -31

1211

62

-4

17 22

496

411

-4

2-7

-25

4-41

5

101112

131415161718

118129691227100913060996145847611578472916386304166627241808800919096900

0 -2510

11-9

-21 -48-13-9

-22-17 -13-10 -23-21 -35

-2-13 13-40 57

7 -19 -1

-2 38

34 -197

52-6

6

-2

5249

Galois group Ds

No / No

1234

5

67

89

1011

1213

160801667489

11946491940449204204126928813083536359860939840164330561463540984564649740641

-5-6

1 -72 -121 -102 -10

-13-9

-11-8

-15-9

-4-5

3-6-3-1

33

10

312

21-14 21-8 27

4 10-28 -6-7 3-4 36-8 3

11

-4

14

-9

16-1-4

91

-16

1

14

1516

17181920212223242526

9790641101187611058200912852225129672011542918415976009167854091867104118948609189834491972248119749136

1 -112 -1 -

-28-17-28-15

11 -1515 13

177

329

-17 -41

101 -230 -191 -231 -25

-9-15

321929

21221321132721

-18

3120

91424

-9

17-11-9

1-1-7

9

1-9

-28

1

2

Galois group C5

No

14641390625923521

282576113845841

/-4

-10-12-16-24

-3 3

5 10-21 1

5 21-17 41

11

5-9-13

There are 61 discriminants below 20 000 000 belonging to two nonisomor-

phic fields, namely,



181096918913773060944335067345698084602229464123264705936499197

73332327834576876028992621179344997

10782801107965171102265611160412

117905441205514912096169121745921232916812500560129651841307283713075749

135201961389879714009328141184731418644814282192143131361432123214433201

146004161473114515873344159101011620938116509232169591931736251217515184

179460251794606418038480183717211853612118623952186405921865232918871504

18908613193036451944401619595088196302371972001319984208

All other fields are uniquely determined by their discriminant.

4.2. Fields with signature rx = 3 and r2 = 1 and \dF\ < 5 000000. ApplyingProposition 2.1, we compute all characteristic polynomials of algebraic integers

p with T2(p) < 45.98. We get 502250482 polynomials, 248231495 ofwhich are of correct signature. There remain 79 394 nonisomorphic fields with

a discriminant of absolute value below 5 000 000. All of them have Galois

group S5 according to Lemma 3.7.

In Table 3 we list the first 50 fields, each of which has a power integral basis.

Table 3

No

1

234

5

67

89

10111213

14151617

18192021

22232425

dp-4511-4903-5519-5783-7031-7367-7463-8519-8647-9439-9759-10407-11119-11243-11551-12447-13219-13523-13799-13883-14103-14631-14891-14911-15536

1 -11 -1

2-1

2-2-1

-2-2

23

1 -1 -30 0-2

-3 2

120

1-3 1-1 -1

3 1-2 -3 -1

0 1 -2-2 -2

033

-3

0 -2

No

26272829303132333435363738394041

424344454647

4849

50

dp-15919-16816-17151-17348-18063-18463-18583-18839-19015-19951-21191-21227-22331-22424-22448-22583-22687-22935-23103-23119-23339-23679-23831-23891-24299

/-3 2

4 0-1 0-2 1-2 -1 -1-3 2 2-2 -1 -1-3 0 1

0 -21 1

1 -23 -4

00

-4

-2-2

01 1

0 -11

0 -31 02 00 -31 -3

1 -12 11 -21 -25 3

0 -1 -3021

21

-3

-3

-2

-3

1-2

2

There are various discriminants for which nonisomorphic fields exist. Ta-ble 4 gives a short account on this phenomenon. If there are more than 10

different discriminants for which p nonisomorphic fields of that discriminantoccur (which happens to be the case for p = 2, 3), we only list the 10 largestdiscriminants.



Table 4

number of

nonisomorphic

fields

discriminants -28976-100656-107467-112919-125391-159196-171184-174608-195203-210352

-428976-447232-490288-643127-646704-818096-853456-956623

-1051056-1068336

-2025648-2170800-2711232-3223600-3539376-3830976-3912624

1673136

total 2139 106

Table 5

No dp No

1

234567

89

1011

1213141516171819202122232425

1609164917772209229726172665286930173089323333693857388941694261440944174429443244774549459747574817

0 -2 -

2627

282930313233343536373839404142

4344454647484950

48975025516454375501558456535753586459136241

644965816757679370967177726573337373737676727684

77177909

1 -1 -

1 -21 -21 10 -1

-1 -2

1 -2

-1

2-1

31

2-1

-2-1

3

10

-2

1 -11 1

1 -3-1-3

23

-2

02

-2

4.3. Fields with signature rx = 1 and r2 = 2 and dF < 5 000000. We com-pute all characteristic polynomials of algebraic integers p with T2(p) < 45.98.

This yields 670725 968 polynomials, 534 326207 of which have exactly onereal zero. There remain 186906 nonisomorphic fields of a discriminant below5 000 000. We list in Table 5 the smallest 50 fields, all of them with a powerbasis as integral basis.

There are 81 fields with Galois group Hol(C5), 258 fields with A5, and129 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 casean integral basis is a power basis.



Group /

Hol(C5)

As

Ds

160935152184962209

0-1 1 1-112 4 111 0-2-2 -20-1 2-2 1

Table 6 gives some information on discriminant values for which nonisomor-

phic 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

16757204323412937584378924091245009477974862949744

9333

1774448592

206928213840214272216432223737251984254148255312

756

225872258768514512587088640629752976880848939600

10488961057104

103

721872115387213335841350864166497617102881862352234849624153122592000

29

16722722016576205934422459682546640308966437078564482000

24325924081104

4050000

Bibliography

1. E. H. Bareiss, Sylvester's identity and multistep integer preserving Gaussian elimination,

Math. Comp. 22 (1968), 565-578.

2. J. Buchmann and D. J. Ford, On the computation of totally real quartic fields of small

discriminant, Math. Comp. 52 (1989), 161-174.

3. J. Buchmann, D. Ford, and M. Pohst, Enumeration of quartic fields of small discriminant,

Math. Comp. 61 (1993), 873-879.

4. F. Diaz y Diaz, A table of totally real quintic number fields, Math. Comp. 56 (1991),801-808.

5. D. Ford, Enumeration of totally complex quartic fields of small discriminant, Computational

Number Theory, Proc. Colloq. on Comp. Number Theory (Debrecen, Hungary, 1989) (A.

Pethö, M. E. Pohst, H. C. Williams, and H. G. Zimmer, eds.), de Gruyter, Berlin and NewYork, 1991, pp. 129-138.

6. J. Hunter, The minimum discriminant of quintic fields, Proc. Glasgow Math. Assoc. 3(1957),

57-67.

7. N. ObreschkofF, Verteilung und Berechnung der Nullstellen reeller Polynome, VEB Deutscher

Verlag der Wissenschaften, Berlin, 1963.

8. M. Pohst, Berechnung kleiner Diskriminanten total reeller algebraischer Zahlkörper, J. Reine

Angew. Math. 278/279 (1975), 278-300.

9. _, On the computation of number fields of small discriminants including the minimum

discriminants of sixth degree fields, J. Number Theory 14 (1982), 99-117.

10. _, On computing isomorphisms of equation orders, Math. Comp. 48 (1987), 309-314.

U.M. Pohst, J. Martinet, and F. Diaz y Diaz, The minimum discriminant of totally real octic

fields, J. Number Theory 36 (1991), 145-159.



12. M. Pohst and H. Zassenhaus, Algorithmic algebraic number theory, Cambridge Univ. Press,

Cambridge, 1989.

13. A. Schwarz, Berechnung von Zahlkörpen fünften Grades mit kleiner Diskriminante, Diplom-

arbeit, Heinrich-Heine-Universität Düsseldorf, 1991.

14. J. Graf von Schmettow, KANT-a tool for computations in algebraic number fields, Compu-

tational Number Theory (A. Pethö, M. E. Pohst, H. C. Williams, and H. G. Zimmer, eds.),de Gruyter, Berlin and New York, 1991, pp. 321-330.

15. C. L. Siegel, The trace of totally positive and real algebraic integers, Ann. of Math. (2) 46

(1945), 302-312.

16. R. P. Stauduhar, The determination of Galois groups, Math. Comp. 27 (1973), 981-996.

17. B. L. van der Waerden, Algebra. I, 8th ed., Heidelberger Taschenbücher 12, Springer-Verlag,

Berlin, Heidelberg, New York, 1971.

18. H. Zassenhaus, On the second round of the maximal order program, Applications of Number

Theory to Numerical Analysis, Academic Press, New York, 1972, pp. 398-431.

Itterstr. 107, 40589 Düsseldorf, Germany

Fachbereich 3 Mathematik, TU Berlin, Strasse des 17. Juni 136, 10623 Berlin, Ger-

many

Mathématiques et Informatique, Université de Bordeaux I, 351, cours de la Libéra-

tion, F-33405 Talence Cedex, France


A table of quintic number fields

Documents

Transcript of A table of quintic number fields