Stark’s Conjectures and Hilbert’s Twelfth
ProblemXavier-François Roblot
Laboratoire A2X, Université Bordeaux I,
351 cours de la Libération,
33405 Talence Cedex, FRANCE
September 19, 2013
Abstract
We give a constructive proof of a theorem given in [Tate 84] whichstates that (under Stark’s Conjecture) the field generated over a totallyreal field K by the Stark units contains the maximal real Abelian extensionof K. As a direct application of this proof, we show how one can compute
explicitly real Abelian extensions of K. We give two examples.

In a series of important papers [Stark 71, Stark 75, Stark 76, Stark 80] H. M.
Stark developed a body of conjectures relating the values of Artin L-functions
at s = 1 (and hence, by the functional equation, their leading terms at s = 0)
with certain algebraic quantities attached to extensions of number fields. For
example, in the case of Abelian L-functions with a first-order zero at s = 0, theconjectural relation is between the first derivative of the L-functions and thelogarithmic embedding of certain units in ray class fields known as Stark units,
which are predicted to exist.
The use of these conjectures to provide explicit generators of ray class fields,
and thus to answer Hilbert’s famous Twelfth Problem was one of the original
motivations for their formulation. It has been noticed by several people (including Stark himself [Stark 76]) that they could provide a new way to construct
ray class fields of totally real fields.
In particular, if K is a totally real field, the field extension generated over
K by the Stark units (see below for details) contains the maximal real Abelian
extension of K. This result is a direct consequence of Proposition 3.8 (Chap.
IV) of [Tate 84].
Using the ideas given in [Stark 76], in Section 2 we give a constructive proofof this result, i.e. for each finite real Abelian extension L/K we constructexplicit generators of L over K using Stark units. This proof has a direct
application since we can use it to explicitly compute real class fields of a totallyreal field. This is discussed in Section 3. Since this construction is based on a1

conjecture, we also explain in that section how to check the correctness of theresult. Finally, we end the paper by giving two examples of such a construction.
I would like to thank David Solomon for his careful reading of the original
manuscript and for his comments.

1

The Abelian rank one Stark conjecture

The main reference for Stark’s conjectures is the book of Tate [Tate 84].
Let N/K be an Abelian extension of number fields, and let G and f denoterespectively its Galois group and its conductor.
For m an admissible modulus for N/K (that is, m is a multiple of f), we
let IK (m) be the group of fractional (non-zero) ideals of K coprime to m, andPK (m) the group of fractional (non-zero) principal ideals which have a generator
multiplicatively congruent to 1 modulo m. The ray class group modulo m is thengiven by ClK (m) = IK (m)/PK (m). The norm group of N/K modulo m is definedas the subgroup of IK (m) generated by the norms from N to K of the fractional
ideals of N coprime to mON and the group PK (m). Let NN/K (m) denote thisnorm group, it is known from Class Field Theory that IK (m)/NN/K (m) ∼=Gby the Artin isomorphism. Thus, for a given admissible modulus m, the norm
group NN/K (m) defines N uniquely.
Let S be a fixed finite set of places of K containing the infinite places of K
and the finite places ramified in N/K.
To an element σ ∈ G, one associates the partial zeta function defined for acomplex number s with R(s) > 1 by the Dirichlet seriesXζS (s, σ) =N a−s(a,S)=1,σa =σwhere a runs through the integral ideals of K not divisible by any (finite) prime
ideal contained in S and such that the Artin symbol σa is equal to σ.
To a character χ over G, one associates the Artin L-function defined for acomplex number s with R(s) > 1 by the Euler productY
LS (s, χ) =(1 − χ(p)N p−s )−1p∈S/

where p runs through the (finite) prime ideals of K not contained in S.∗
These functions can be analytically continued to meromorphic functions on
the whole complex plane (L-functions can even be continued to holomorphic
functions if the character χ is non-trivial). As is well-known, they are also
∗ Here and in the sequel, by abuse of notation we consider that the character χ is not only
defined on the Galois group G but also on the ray class group ClK (f) and the group IK (f) of
the non-zero fractional ideals of K coprime to the finite part of f2

related to partial zeta functions by the two equivalent identitiesζS (s, σ)=LS (s, χ)=X
1
LS (s, χ)χ(σ),[N : K]bχ∈GX
ζS (s, σ)χ(σ).
σ∈GLet χ be a character of G. If χ is the trivial character 1 we set r(1) =Card(S) − 1, otherwise r(χ) is the number of places v ∈ S such that the decomposition group Dv of v in N/K is contained in the kernel of χ, in otherwords such that χ|Dv = 1. The following result can be found in [Martinet 77]
or [Tate 84].
Proposition 1.1 The order of vanishing at s = 0 of the Artin L-function
LS (s, χ) is equal to r(χ).
We now assume that there exists an infinite place v which is totally split in
N/K and we fix w, a place of N dividing v. We also assume that Card(S) ≥ 2.
It follows from Proposition 1.1 that LS (0, χ) = 0 for every character, thus thepartial zeta functions ζS are all zero at s = 0.
Conjecture 1.2 (Stark) Let m be the number of roots of unity contained in
N.
Then, there exists an S-unit ε ∈ N such that for every σ ∈ G
log |σ(ε)|w = −mζS0 (0, σ),or equivalently
1 X
χ(σ) log |σ(ε)|wm
σ∈G√
for any character χ over G. Furthermore N ( m ε)/K is an Abelian extension
and if Card(S) ≥ 3 then ε is a unit.
L0S (0, χ) = −Remark 1 We denote by ε(N/K, S, w) the unit ε appearing in the Conjecture
if it exists (note that this is an abuse of language since this unit may not be
unique, however in what follows the place w will be a real place and we makeε(N/K, S, w) unique by assuming that w(ε(N/K, S, w)) > 0). When the set S
is chosen to be minimal, i.e. S is the set of infinite places of K together withthe finite places ramified in N/K, we simply write ε(N/K, w).

2

Application to Hilbert’s twelfth problem

Let K be a totally real field distinct from Q and let v be a fixed infinite placeof K. We identify K with its image v(K) in R. From now on, we assume
that Conjecture 1.2 is true for any finite Abelian extensions N/K in which v is
totally split and any choice of the place w of N dividing v.
3

Let K Stark denote the subfield of C generated over K by all the units
ε(N/K, w) where N/K runs through the finite Abelian extensions of K in which
v is totally split, and w runs through the infinite places of N dividing v. Then,
we have the following
Theorem 2.1 The maximal real Abelian extension of K is contained in K Stark .
Equivalently, for any finite real Abelian extension L/K, there exist Stark
units ε1 , . . . , εr such that L ⊂ K(ε1 , . . . , εr ).
Remark 2 As we have already said in the introduction, this theorem is a consequence of Proposition 3.8 (Chap. IV) of [Tate 84].
We will prove the theorem by proving the second assertion. For that purpose,
we will construct the units ε1 , . . . , εr . In fact, we will prove a little more sincethese units will verify−1
L = Q(ε1 + ε−1
1 , . . . , εr + εr ).

(†)

For a prime ideal p of K we define an integer rp as follows. If p does notdivide 2 then rp = 2, otherwise rp = np + 2 where np is the degree of the local
extension Kp /Q2 .
Proposition 2.2 Let L/K be a finite Abelian extension of totally real fields.
Let v be a infinite place of K and let T be a finite set of prime ideals of K suchthat for each prime p in T , the 2-rank of the decomposition group Dp of p in
L/K is strictly less than rp .
Then there exists a quadratic extension N/L verifying the following three
conditions:A. The extension N/K is Abelian.
B. All the infinite places of K except v become complex in N ,C. The prime ideals of L above T do not split in N/L.
Remark 3 The maximal value for the 2-rank of the decomposition group Dp ofa prime ideal p in any Abelian extension of K is rp .
Remark 4 The conditions (A-C) are very important for the construction. Conditions (A) and (B) allow us to apply Conjecture 1.2 to the extension N/K.
Condition (C) is necessary to ensure that L0S (0, χ) is not going to vanish for
too many characters χ, and so make sure that the Stark unit that we obtain isa generator of N (see below).
Proof. Let p be a prime ideal in T and fix a prime ideal P in L dividing p.
Let sp denote the 2-rank of the decomposition group of p in L/K. Then Galois
Theory tells us that the number of quadratic extensions of Kp contained in LP
is 2sp − 1. On the other hand, Kummer Theory tells us that the number of4

quadratic extensions of Kp is 2rp − 1. Since sp < rp , there exists (at least) one
quadratic extension, say Ep /Kp , such that Ep is not contained in LP and thus√a p-adic integer in Kp , say κp , such that Ep = Kp ( κp ). In particular, κp is
not a square in LP .
For each prime ideal p ∈ T , choose such an element κp and let
mp = vP (κp ) + vP (2) + 1
where vP denotes the valuation associated to P (note that mp does not dependon the choice of P since κp is an element of Kp ). By the Approximation
Theorem, one can find an algebraic integer κ in K such that
1. v(κ) > 0,
2. v 0 (κ) < 0 for any infinite place v 0 of K distinct from v,
3. κ ≡ κp (mod pmp ) for any prime ideal p ∈ T .
√Then I claim that N = L( κ) satisfies the properties (A-C). First, it is clearthat N/K is Abelian
since it is the compositum of the twop Abelian extensions√
L/K and K( κ)/K, so (A)isverified.
Second,sincev 0 (κ) is a complex
pnumber for v 0 6= v whereas v(κ) is real, it follows that v is the only infinite
place of K which remains real in N/K and this gives (B). Third, let p be ae aprime ideal in T and suppose that p splits in N/L. Denote by P (resp. P)
eprime ideal in L (resp. N ) dividing p and such that P | P. Then the local
fields NP
e and LP are the same and thus κ is a square in LP . Now, consider
the quadratic polynomial X 2 − κp with coefficients in LP . This polynomial has
a simple root modulo Pmp , namely κ, and thus it follows by Hensel’s Lemma
that it has a root in LP . But this is impossible since we know that κp is not asquare in LP and thus p cannot split in N/L and (C) is also verified.
We now prove Theorem 2.1. Assume first that L/K is a cyclic extension. We
want to construct a quadratic extension N/L satisfying conditions (A-C), andsuch that not too many derivatives of L-functions associated to this extension
vanish at s = 0 since, otherwise, Conjecture 1.2 would be useless. Looking at
the formulae for r(χ) before Proposition 1.1, one way to do this is to ensurethat no prime ideals in S split in N/L. Now, with our choice of S as minimal,
the prime ideals in S are exactly the prime ideals that ramify in N/K. Let p
be such a prime ideal. If p is not ramified in L/K, then p must be ramified in
N/L and thus is not split. If p is ramified in L/K, then we want to make surethat it is not going to split in N/L, so we let p be an element of T . Hence, wechoose T be the set of the prime ideals of K which are ramified in L/K. For
each prime ideal p in T , the 2-rank of its decomposition group in L/K is equal
to 1, so we can apply Proposition 2.2 and obtain a quadratic extension N/Lverifying conditions (A-C). We fix an infinite place w in N dividing v and let
ε = ε(N/K, w).
Let τ denote the unique non-trivial automorphism of the quadratic extension
N/L, and let p be a prime ideal in S. Since p does not split in N/L, Dp contains
5τ . In particular, if χ is a character of G such that χ(τ ) 6= 1 then r(χ) = 1 and
we deduce from Proposition 1.1 that L0S (0, χ) 6= 0. Finally, we apply Theorem
1 of [Stark 76] that we restate in our situation with our notations for the sakeof completeness.
Theorem 2.3 (Stark) Assume Conjecture 1.2 is true. Let Γ be the quotientgroup G/{1, τ }, thus Γ is the Galois group of L/K, and assume that for every
character ψ of G which is not induced by a character of Γ, one has L0 (0, ψ) 6= 0.
Then, N = Q(ε) and L = Q(ε−1 + ε).
When L/K is not cyclic, we can split L/K as the compositum of cyclic
extensions L1 /K, . . . , Lr /K and apply to each of these cyclic extensions theabove construction to obtain quadratic extensions Ni /Li and infinite places wisuch that εi = ε(Ni /K, wi ) verifies Li = K(εi + ε−1i ). Assertion (†) follows since
L is generated over K by the elements εi + ε−1,
proving the theorem.
iRemark 5 Note that conditions (A-C) on the extension N/K of Proposition
2.2 (with T containing the set of prime ideals ramified in L/K) are enough toprove the theorem, and that we may obtain an extension N/L verifying thoseconditions by other means than those given in the proof of the proposition whichconstructs√ N as the compositum of the extension L/K with the quadratic extension K( κ)/K (see last section for such an example).
We end this section with a very useful lemma.
Lemma 2.4 The Stark unit ε appearing in the above construction is a unit
(and not merely an S-unit) and verifies|ε|w0 = 1
for any infinite place w0 of N which does not divide v (that is, for any infinite
complex place w0 of N ).
Proof. To prove the first assertion, it suffices to prove that Card(S) ≥ 3
and the result will follow from Conjecture 1.2. Since S must contain the infinite
places of K, the case Card(S) < 3 can only happen when K is a real quadratic
field and N/K is unramified at all the finite places. But this is possible only if
N is the Hilbert Class Field or the Narrow Hilbert Class Field of K; in the firstcase, no infinite places of K are ramified in N/K, in the second the two infinite
places of K are ramified in N/K, and neither of these cases apply here sinceexactly one infinite place must be ramified in N/K.
The proof of the second assertion can be found on page 74 of [Stark 76] in aslightly different form.

3

An overview of the computational aspect

In this section we will briefly describe how one can use the proof of Theorem
2.1 to compute real class fields of a totally real ground field (see [Roblot 97] for6

a complete exposition). Similar computations for checking Stark’s Conjecture
over a real cubic field can be found in [Dummit et al. 97]. In a more specialized
case, namely when K is a real quadratic field and L its Hilbert Class Field, it ispossible to obtain a much more powerful algorithm, see [Cohen and Roblot 98]
for details.
Let K be a totally real field of degree N ≥ 2 and discriminant dK , let L bea finite real Abelian extension of K of conductor m (hence m is an integral idealof K). Assume for the sake of simplicity that there exist quadratic extensions
N/L verifying conditions (A-C) of Proposition 2.2 (e.g. if L/K is cyclic, or moregenerally if for every prime ideal p of K which divides m, one has sp < rp with
the notations of Proposition 2.2).
First, we need to compute a quadratic extension N/L verifying conditions(A-C). Of course, the proof of Proposition 2.2 gives us a direct way to do
this, but, as quoted in Remark 5, we cannot obtain by this method all thequadratic extensions N/L verifying (A-C). Furthermore, heuristics and numerical evidences seem to show that the Stark unit tends to grows more or less likethe exponential of the square root of the discriminant times the norm of the
conductor of N/K and thus we want to lower this norm as much possible. The
best way to find N is then to construct explicitly the class group of conductor
f0 f∞ where f0 = am and a runs through the integral ideals of K by increasing
norm and f∞ contains all the infinite places of K but one, and look for the firstone of these class groups which contains a subgroup defining a suitable extension
N/L. These computations involve only class groups and can be done using thetools of [Cohen et al. 98].
Assume we have found such a suitable extension N/L, let f be the conductorof N/K, G its Galois group, v the only unramified infinite place, and let ε =ε(N/L, w) denote the corresponding Stark unit. We want to compute ε, or more
precisely the element α = ε + ε−1 which verifies L = K(α). Note that thanks
to Lemma 2.4, we know that α is in fact an algebraic integer.
Now, we need to compute the values of ζS0 (0, σ) for σ ∈ G to high precision.
In fact, it is simpler to compute the values of L0S (0, χ) and deduce from themthe values of ζS0 (0, σ) using the formulae given in the first section. Let χ bea character of G, using Proposition 1.1, it is easy to prove that L0S (0, χ) = 0
if χ(τ ) = 1 (recall that τ is a generator of Gal(N/L)) and this term does notcontribute to the computation of ζS0 (0, χ), thus we assume that χ(τ ) = −1. Let
L(s, χ) denote the primitive L-function associated to χ which is defined by
Y
L(s, χ) =(1 − χ(p)N p−s )−1 ,pwhere p runs through all the prime ideals of K and where χ(p) is set to be equal
to zero whenever p divides the conductor f(χ) of χ. We have
L0S (0, χ) = A(χ)L0 (0, χ)

7whereA(χ) =

Y(1 − χ(p)).
p|f

The value of the first derivative of L(s, χ) at s = 0 can be easily determined
thanks to the functional equation. Let Λ(s, χ) be the “enlarged” L-function
given byN −1
L(s, χ)Λ(s, χ) = C(χ)s Γ(s/2)Γ( s+12 )pwhere C(χ) = π −N dK N f(χ). Then for any s ∈ C
Λ(1 − s, χ) = W (χ)Λ(s, χ̄).
The constant W (χ) is a complex number of modulus one, the so-called Artin
Root Number (see [Martinet 77] for a complete exposition of these results inthe more general case of Artin L-functions). Letting s tends to zero in thisfunctional equation yields the relation
Λ(1, χ̄)
L0S (0, χ) = A(χ) √2 π N −1 W (χ̄)
where A(χ) is the constant defined above.
Thus the computation of L0S (0, χ) (hence of ζS0 (0, σ)) boils down to the computation of three quantities: A(χ), W (χ̄) and Λ(1, χ̄). The computation ofA(χ) is direct using the methods of [Cohen et al. 98] and one can use explicit
formulae to compute W (χ̄) (see [Dummit et al. 98] for example). However, thecomputation of Λ(1, χ̄) requires much more work. We use the following resultof Friedman [Friedman 87].
P−s
Theorem 3.1 (Friedman) Let L(s, χ) =be the expressionn≥1 an (χ)nfor L(s, χ) as a Dirichlet series for R(s) > 1. Then one hasXΛ(1, χ) =[an (χ)f (C(χ)/n, 1) + W (χ)an (χ̄)f (C(χ)/n, 0)]
n≥1

where
f (x, t) =12iπ

Z δ+i∞

xz

δ−i∞

N −1
Γ(z/2)Γ( z+1
2 )
dz
z−t

for any real number δ > R(t).
There exist various methods to compute these integrals. One was developedby E. Tollis and can be found in [Tollis 97]. A quite similar method is used in[Dummit et al. 97].
One of the most time consuming part of the algorithm is the computation of
the coefficients an (χ). We explain briefly how to do this. Assume that the prime
ideals of K have been ordered in a sequence (pi )i≥1 such that N pi+1 ≥ N pi ,8and set p0 = OK . Let In,h (χ) denote the set of all integral ideals in K of normn, prime with f(χ) and divisible only by prime ideals pi with i ≤ h. We setXχ(a).
an,h (χ) =a∈In,h (χ)

Then it is clear that an (χ) = limh→∞ an,h (χ), and even more an (χ) = an,h (χ)
if N ph > n. Now, the coefficients an,h (χ) are computed using a sieve and thefollowing lemma.
Lemma 3.2 We have a1,0 (χ) = 1 and an,0 (χ) = 0 for n ≥ 2, and for h ≥ 1mn,han,h (χ) =X

an/qh k ,h−1 (χ)χ(ph )k ,k=0

where qh = N ph and mn,h is the largest integer m such that qhm | n.
Proof. This is a direct application of the Euler product formulaXY
−1an (χ)n−s =1 − χ(ph )/N p−s
.
h
n≥1

h>1

Once we have computed the value Λ(1, χ) for all characters χ such that
χ(τ ) = −1, we obtain the values of L0S (0, χ) and the values of ζS0 (0, σ) for allσ ∈ G. In order to deduce from these values the conjugates of ε, we need toremove the absolute value appearing in Conjecture 1.2. Since we have alreadyassumed that w(ε) √> 0, it is a direct consequence of the second assertion of theconjecture (i.e. N ( ε)/K is Galois) that all the others conjugates of ε over K
are also positive at w. Hence, we can approximate the irreducible polynomialof α over K by writingYPe(X) =
X − e−2Zσ + e2Zσσwhere σ runs through a system of representatives of G/ < τ > (which is isomorphic to Gal(L/K)) and Zσ denotes the approximation of ζS0 (0, σ) which has
been computed (note that Zσ = −Zτ σ ).
We need to recognize the coefficients of Pe(X) as the v-embedding of algebraic
integers of K. For the other embeddings, Lemma 2.4 gives us |α|w0 ≤ 2 for any
infinite place w0 of N which does not divide v, and thus provides bounds onthe embeddings of these coefficients at the others infinite places of K. Now
there exist finitely many algebraic integers of a given degree such that all theirconjugates are bounded, and thus finitely many algebraic integers in K which
are very close to a given real number at the infinite place v and bounded at all
the others, and we can list them (in fact, if the precision is sharp enough, weobtain only one candidate for each coefficient). Once we have recognized all thecoefficients of Pe, we obtain a polynomial P (X) ∈ OK [X].
9Finally, we need to prove that P is indeed the irreducible polynomial of agenerating element of L. It is quite easy to prove that P is irreducible. If so, lete denote the field it defines. It is also easy to check that the field L
e is totallyLereal and to compute the relative discriminant of L/K(see [Cohen et al. 96] foralgorithms to perform these tasks). Once all these checks have been done, weestill need to prove that the extension L/K
is Abelian and that its norm groupis the same as the one of L/K. Although this can be done quite easily when the
degree of the extension is small, it is a difficult task in general and we will not
go into details here (see [Roblot 97]). However, under the Generalized RiemannHypothesis (GRH), it is possible to find an algorithm which at the same time
proves that the extension is Abelian and computes its norm group. We justquote the key result here (which rely on the theorem of Bach and Sorenson,[Bach and Sorenson 96]), and we refer the interested reader to [Roblot 97].
e
Theorem 3.3 Assume GRH is true. Let L/K
be a finite extension of totallyreal number fields, and let d, d denote respectively the absolute discriminant ofe and the relative discriminant of L/K.
e
L,
Let
C=

2

5 e
: Q] + 5 ,4 log d + [L2eand let S denote the set of prime ideals of K of degree 1, unramified in L/K,eand of absolute norm smaller than C. Then L/K is an Abelian extension if andonly ife dividing p ∈ S have the same residual degree fp ,(i) all prime ideals P in L(ii) the group N generated by pfp , as p ranges through S, and PK (d) has index
e : K] in IK (d).
[L
Furthermore, if conditions (i) and (ii) are verified, then N is the norm group of
e
L/K.

4

Two Examples

This method has been used to compute the Hilbert Class Field of totally real
fields of degrees 2, 3, and 4 and various ray class fields (see [Roblot 97]) using
the PARI package [GP 99]. One can download a table of the Hilbert Class Fieldof all real quadratic fields of discriminant less than 10 000, real cubic fields of
discriminant less then 150 000 and real quartic fields of discriminant less than
600 000 (a total of 3303 non-principal fields) at the following URL
http://www.math.u-bordeaux.fr/~roblot/resources/hilb.gp
We now give two examples of the construction of real Abelian extensions ofa totally real field using Theorem 2.1. Similar use of Stark’s Conjectures to construct class fields can be found in [Cohen and Roblot 98], [Dummit et al. 97],10

and [Stark 76, Stark 80]. One can also use Kummer Theory to compute classfields, see for example [Pohst and Daberkow 95], [Fieker 97].
These two examples where computed with the latest version of GP/PARI(v.2.0.13 - alpha) on a DEC Alpha 2100 300MHz with 512Mb of memory. Computation times are provided at the end of each example.
• Let K be the real quadratic field generated over Q by a square root ω of 82.
Its discriminant is 328, and {1, ω} forms a Z-basis of its ring of integers OK .
The field we wish to construct is L = HK , the Hilbert Class Field of K, i.e.
the maximal Abelian extension of K unramified everywhere, which is a cyclic
extension of degree 4 of K.
A quadratic extension N/L verifying conditions (A-C) and with minimal(norm of) conductor is given by the ray class field modulo f = p3 v1 where
p3 = 3OK + (2ω − 1)OK
is a prime ideal dividing 3, and v1 is the real place sending ω to the negativesquare root of 82 in R. The extension N/K is a cyclic extension of degree 8, so
this extension cannot be constructed using the proof of Proposition 2.2. In fact,
the first extension that one can construct using Proposition 2.2 has a conductorof norm 16, and it is certainly more efficient not to use this one but the former(see remark below).
Let G denote the Galois group of N/K, let σ be a generator of G and letτ = σ 4 be the unique element of order 2 (τ generates the Galois group of N/L).
It is easy to prove that all characters χ of G such that χ(τ ) = −1 have conductorf, hence LS (s, χ) = L(s, χ) for any such character. We compute the values ofL0 (0, χ) and obtain the values of the derivatives of partial zeta functions†ζS0 (0, 1) =
ζS0 (0, σ) =ζS0 (0, σ 2 ) =ζS0 (0, σ 3 ) =−1.855345769803922 . . . ,ζS0 (0, τ ) =0−0.811399495928109 . . . , ζS (0, τ σ) =−1.056128108731457 . . . , ζS0 (0, τ σ 2 ) =0.597654704583391 . . . , ζS0 (0, τ σ 3 ) =−ζS0 (0, σ),−ζS0 (0, σ),−ζS0 (0, σ 2 ),−ζS0 (0, σ 3 ).

Remark 6 Note that the logarithmic height of ε (see [Lang 83] for a definition)is
2 |ζS0 (0, 1)| + |ζS0 (0, σ)| + |ζS0 (0, σ 2 )| + |ζS0 (0, σ 3 )| = 8.64105615809373 . . .
(since we know that the other conjugates of ε have an absolute value of 1 byLemma 2.4). If instead, we had used the class field of conductor 16 for thisconstruction, the logarithmic height of ε would have been 16.985931238837 . . . ,that is to say pnearly 2 times larger (the heuristics quoted at the beginning ofsection 3 give 16/3 ≈ 2.3). This illustrates why we need to find a suitable N
with minimal conductor to speed up the computations.
† Note that the computations have been made with much more precision than given in this
paper

11

The polynomial Pe (with the above notation) is
X 4 − 58.16615541441224X 3 + 799.4369460780463X 2− 3980.184730390231X + 6515.938649729469
and we “recognize” this polynomial as the embedding of the following polynomial of OK [X]
P (X) = X 4 −(3ω+31)X 3 +(44ω+401)X 2 −(220ω+1988)X +(360ω+3256).
Although the discriminant of this polynomial is far from trivial (its norm is
28 312 732 ), it nevertheless defines an unramified extension of K of degree 4.
Thus, to prove that this extension is indeed the Hilbert Class Field of K, it
remains to prove that it is Abelian. For that purpose, it suffices to prove that
it is Galois since the only Galois groups of order 4 of quartic extensions are
Abelian groups (namely the cyclic group of order 4 and the Klein group). There
are various way to prove this. One way is to compute an absolute (and in thiscase reduced) polynomial over Q defining the same field and to compute its
Galois group. It our case, we obtain the polynomial
X 8 − 4X 7 − 14X 6 + 56X 5 + 49X 4 − 196X 3 + 28X 2 + 80X − 25
whose Galois group is the dihedral group of order 8 which proves that√the field
it defines is Abelian over K and thus is the Hilbert Class Field of Q( 82).
This example was computed in 4 seconds. The computation of the polynomial P (X) took 3 seconds, and the verification (computation of the relative
discriminant, and of the Galois group) took less then one second.
• Let K be the field generated over Q by a root α of the polynomial
X 3 − 4X − 1.
This is a real cubic field of discriminant 229 and with ring of integers OK = Z[α].
Let m denote the prime ideal above 37 defined by
m = 37OK + (8 + α)OK ,we want to construct the ray class field L of K modulo m. This is an Abelian
extension of K of degree 3.
Using the proof of Proposition 2.2, one can choose for N the field generated
over L by a square root of κ = α. This yields an Abelian extension of K ofdegree 6 and conductor mv1 v2 where v1 (resp. v2 ) is the infinite place sending
α to −1.860805 . . . (resp. −0.254101 . . . ). Here, it is clear that N is of minimal
norm. As usual, we denote by G the Galois group of N/K, by τ the non-trivial
element of Gal(N/L) and by σ a generator of G. Let χ denote the character ofG such that χ(σ) = exp(2iπ/6). This is a generator of the group of characters
of G.

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The characters of G which are non-trivial on τ are χ, χ3 and χ5 , since
χ(τ ) = −1. For these, we compute the Artin Root Numbers
W (χ) = −0.367664745 · · · − 0.92995840 . . . i,

W (χ3 ) = 1,

W (χ5 ) = W (χ),

and the corrective factors
A(χ) = 1,

A(χ3 ) = 2,

A(χ5 ) = 1.

This gives us the following values for the partial zeta functions
ζS0 (0, 1) = 1.96011188229224 . . .
ζS0 (0, σ) = 1.57294437150264 . . .
ζS0 (0, σ 2 ) = 0.72454531436117 . . .

ζS0 (0, τ ) =ζS0 (0, τ σ) =ζS0 (0, τ σ 2 ) =−ζS0 (0, 1)−ζS0 (0, σ)−ζS0 (0, σ 2 )and the polynomial
X 3 − 78.20893338606214X 2 + 1505.492174458384X − 5276.952425687298
which can be seen to be very close to the following polynomial
P (X) = X 3 −(9α2 +17α+2)X 2 +(160α2 +338α+75)X −(560α2 +1185α+266).
One can check that the field extension L generated by a root of this polynomial isa totally real field of degree 9 and that its relative discriminant is m2 . Since thisis a square and since L/K is a cubic extension, it follows from Galois Theory that
it is an Abelian extension (Galois group of cubic extensions are S3 or A3 ' C3 ,and the latter occurs if and only if the discriminant is a square). Since L/K is an
Abelian extension of prime degree l = 3, the Führerdiskriminantenproduktformeltells us that its discriminant is equal to its conductor raised to the power l − 1 =2. Thus, m is the conductor of L/K, and that finishes the proof that this isindeed the field L that we wanted. For the sake of completeness, we give thefollowing reduced polynomial which defines L over Q
X 9 + 2X 8 − 9X 7 − 11X 6 + 28X 5 + 18X 4 − 34X 3 − 8X 2 + 13X − 1.
This example was computed in 24 seconds. The computation of the polynomial P (X) took 23 seconds, and the verification (computation of the relative
discriminant) took less than one second.

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