Numerical Verification of the Stark-Chinburg
Conjecture for Some Icosahedral Representations
Arnaud Jehanne, Xavier-François Roblot, Jonathan Sands
October 6, 2003
Abstract
In this paper, we give fourteen examples of icosahedral representations
for which we have numerically verified the Stark-Chinburg conjecture.

1

Introduction

Let K/k be a Galois extension of number fields, with Galois group G = Gal(K/k),and suppose ρ : G → GLn (C) is a non-trivial irreducible representation of G.
Stark’s conjectures [Tate 1984] aim to unravel the arithmetic information encoded in the leading coefficient of the Taylor series for the Artin L-function
L(s, ρ) of ρ at s = 0. When G is abelian and one modifies the Artin L-function
by removing the factors in the Euler product at primes in a finite set S whichcontains all of the infinite primes, Stark formulated an especially precise conjecture for the case of a first-order zero at 0 [Stark 1980]. It states that the exact
value of this coefficient may be obtained from an “L-function evaluator” element in K which is an S-unit in the typical case. Rubin [Rubin 1996], Popescu[Popescu 2003], Burns [Burns 2001], Sands [Sands 1987] and others have formulated similarly precise conjectures for abelian L-functions with any order of zeroat s = 0.
In the general non-abelian case with L(s, ρ) possessing a zero at s = 0 oforder r = r(ρ), the conjecture states that the L-function coefficient equals an
algebraic factor multiplied by the determinant of a regulator matrix defined interms of a set of r special units and the representation ρ. But this algebraic
factor is not fully specified and in particular may be multiplied by any nonzero
rational factor without affecting the truth of the conjecture. Hence the conjecture in this generality is considered to be a conjecture “over Q”, as opposed tothe more precise conjectures “over Z” mentioned above in the abelian case.
Chinburg [Chinburg 1983] has formulated a conjecture “over Z,” in the nonabelian case when the order of the zero at 0 is r(ρ) = 1, the base field is
k = Q, and the dimension of the irreducible representation ρ is n = 2. We willshow that this conjecture is closely related to a Question of Stark appearing in[Stark 1981], and hence we will use the term “Stark-Chinburg conjecture.” Here1the regulator matrix is 1 by 1, and involves a single special unit, so we have thepossibility of actually constructing this special unit from the first derivatives
at 0 of certain Artin L-functions. This method of constructing S-units while
simultaneously gaining numerical confirmation of the conjecture at hand appearsin [Dummit et al. 1997] and [Roblot 2000] for the abelian case. A difference inthis paper is that the extension field K is no longer a class field which can be
explicitly constructed from abelian L-functions by means of the conjecture. We
will choose our non-abelian extension field K beforehand in order to define the
L-functions.
Irreducible two-dimensional representations are classified according to theisomorphism type of their images in PGL2 (C), the four possible types being dihedral, tetrahedral (A4 ), octahedral (S4 ), and icosahedral (A5 ). Stark[Stark 1981] has provided illuminating examples in the dihedral cases; Chinburg [Chinburg 1983] has confirmed the conjecture numerically for five tetrahedral representations; and Fogel [Fogel 1998] has confirmed it numerically foreight octahedral representations with K of degree 48. Our aim in this paper isto provide the first numerical confirmation of the Stark-Chinburg conjecture forsome icosahedral representations. As we will see, the minimal type of field K
providing such an example is a complex field of degree 240 over Q, while the
Stark unit ε lies in a subfield K+ of degree 120 admitting a real embedding. Thissubfield K+ is Galois over a field M of degree 30. We identify ε by obtaining
its minimal polynomial over M .
The outline of the article is the following: in section 2 we state the StarkChinburg conjecture, but also a question of Stark related to the same situation.
In section 3 we look at Â5 -extensions which provide the simplest cases for testing
the conjecture on icosahedral representations, where Â5 is a central extension of
A5 by a cyclic group of order 4 (see Section 3 for details). We briefly explain how
to construct those extensions and how to compute the value of the derivative of
the corresponding L-functions at s = 0. Finally, in the last section, we describe
the computations performed, give some remarks on the results obtained andconclude with an example.

2

The Stark-Chinburg conjecture

2.1

Odd Representations

A standard formula [Tate 1984, p. 24] for the order r(ρ) of the zero of L(s, ρ)at s = 0 calls for the choice of a single prime w of K above each infinite prime
v of k. One then defines τv to be the generator of the decomposition group of
the prime w over v, which is thus either the identity or a complex conjugation.
Assuming that ρ is a non-trivial irreducible representation, r(ρ) may then be
obtained by taking the dimension of the eigenspace of ρ(τv ) corresponding to
the eigenvalue 1, and summing over v. Now suppose that our representation ρ
is as in the Stark-Chinburg conjecture. Since k = Q, there is a single infinite
prime v = ∞. Since τ = τv has order 1 or 2, all eigenvalues of ρ(τ ) must be ±1.

2

Since ρ is 2-dimensional and r(ρ) = 1, τ must be a complex conjugation of order
2 and ρ(τ ) must have eigenvalues 1 and -1. Thus det ρ(τ ) = −1, a condition
which is described by saying that ρ is “odd.” The associated character ψ definedas the trace of ρ then clearly takes the value ψ(τ ) = 0. Conversely, it is easy
to see that if ψ(τ ) = 0, for the character ψ of degree 2, then the corresponding
representation ρ is odd.

2.2

Regulators

For simplicity, in this section, we continue to assume (with Stark and Chinburg)that k = Q. We also fix a subfield K of C which is Galois over Q with groupG, and assume ρ is an irreducible non-trivial n-dimensional representation of
G = Gal(K/Q) with character ψ. Let τ denote the restriction of complex
conjugation to K, which may be trivial, and k k denote the normalized absolutevalue on K corresponding to the embedding of K in C (this is the square of theusual one if K is complex). Also let w be the infinite prime of K corresponding
to this absolute value, so the decomposition group of this prime is Gw = hτ i.
Following Stark, we may assume by conjugating the representation that ρ(τ ) isdiagonal and the diagonal elements consist of a certain number a of 1’s followed
by a certain number b = n − a of -1’s. In [Stark 1975, p. 62], Stark introduces aregulator which we will denote (with Tate) as R(ψ, ε); this also calls for a choiceof element ε ∈ K. Then!
X
σR(ψ, ε) = det
ρa (σ) log kε k ,σ∈Gwhere ρa (σ) denotes the a × a matrix in the top left corner of ρ(σ).
Like Tate and unlike Stark, our convention will be that G acts on K on theleft, so that εστ = σ(τ (ε)).
In this regulator, Stark uses a choice
Q of a unit ε ∈ K ∩ R for which the only
relation among the conjugates εσ is σ∈G/Gw εσ = ±1. Such a unit is called a“Minkowski unit,” since its existence is guaranteed by [Minkowski 1900].
On the other hand, Tate’s regulator R(ψ, F ) ([Tate 1984]) is attached toa choice of a Q[G]-isomorphism F . Let U = UK denote the unit group of
K, QU = Q ⊗Z U , QY be the Q-vector space with basis consisting of theinfinite primes of K, and QX be the subspace of elements whose coordinatesin this basis sum up to 0. There exists a Q[G]-isomorphism F : QX → QU
(by a theorem of Herbrand [Herbrand 1930, Herbrand 1931], in general), andthis is used to define R(ψ, F ). We will not repeat the construction of thisregulator found in [Tate 1984] but wish to note the important connection with
Stark’s regulator, described on page 41 there. The unique Q[G]-homomorphismfrom QY to QU which sends the fixed infinite prime w of K to ε induces anisomorphism Fε : QX → QU , andR(ψ, Fε ) = |Gw |a R(ψ, ε).

3

Now we make the connection between a regulator of the form R(ψ, π), for
π in K and the conjecture considered in this paper. So assume that ρ is a2-dimensional irreducible
representation which is odd. As seen above, we may0take ρ(τ ) = 10 −1. So we now have a = 1, and ρ1 (σ) is the single entry in the
top left corner of ρ(σ). Using the fact that the absolute value k k is fixed by τ ,it follows thatX1X
Tr 20 00 ρ(σ) log kπ σ kR(ψ, π) =ρ1 (σ) log kπ σ k =
2
σ∈G
σ∈G
X



1
0
=
Tr { 10 01 + 10 −1}ρ(σ) log kπ σ k2σ∈G1XTr ({ρ(1) + ρ(τ )}ρ(σ)) log kπ σ k
=
2
σ∈G
1X
Tr (ρ(σ) + ρ(τ σ)) log kπ σ k
=
2
σ∈G
1X
=
(Tr(ρ(σ)) + Tr(ρ(τ σ))) log kπ σ k
2
σ∈G
1X
=
(ψ(σ) + ψ(τ σ)) log kπ σ k2
σ∈G1X(ψ(σ) log kπ σ k + ψ(τ σ) log kπ τ σ k)
=
2
σ∈G
X
=
ψ(σ) log kπ σ kσ∈GThis computation reconciles the difference in appearance between equations
in [Stark 1975] and [Stark 1981] (see the comment in the middle of page 263 of[Stark 1981]).

2.3

Stark’s Non-Abelian Question

The preceding computation relates specifically to the following question of Stark
from [Stark 1981].
Question 2.1 (Stark). Suppose that K is a complex Galois extension of Q withgroup G, and W is the number of roots of unity in K. Fix a rational integer
f divisible by the conductor of every character ψ of G which corresponds to an
odd irreducible representation ρ of dimension 2, and let L(s, ψ, f ) be the Artin
L-function of ρ with the Euler factors at primes dividing f removed. Is there
an algebraic integer π in K such that
1. π σ /π is a unit for each σ ∈ G, and some power of π is real,42. π σ /π p is a W th power in K whenever p is a prime not dividing W f times
the discriminant of K and whose associated Frobenius automorphisms areconjugate to σ in G, and3. For every ψ corresponding to an odd irreducible representation of dimension 2, we haveL0 (0, ψ, f ) =−1 X
ψ(σ) log ||π σ ||?
2W

(2.1)

σ∈G

Remark 1. From our preceding discussion, one can see that condition (2.1) of
Question 2.1 does indeed refine the conjecture of [Stark 1975], specifying thatL0 (0, ψ, f ) =−1−1
R(ψ, π) =R(ψ, Fπ ).
2W
4W

It is a question “over Z.”
Remark 2. Since ψ(στ ) = ψ(τ σ) and τ fixes the absolute value k k, it is clearthat an affirmative answer to the question would imply thatL0 (0, ψ, f ) =−1 X(ψ(σ) + ψ(στ )) log kπ σ k.
4Wσ∈GIn several cases where K is a class field of a real quadratic field, Stark has
confirmed numerically that his Question has an affirmative answer [Stark 1976,
Stark 1980]. But the Question itself does not suggest an effective means ofconstructing the distinguished element π because it does not provide enoughinformation about the conjugates of π. However, the Stark-Chinburg conjecture
does provide such information about a special element of K.
Stark’s contribution to this conjecture is as follows. Fix a character ψ asin Stark’s Question. The field E = Q(ψ), obtained by adjoining to Q all thevalues of the character ψ, is contained in a cyclotomic extension of Q. We let Γ
denote the abelian Galois group of E over Q. Following the formulation of his
Question in [Stark 1981], Stark gave an argument which implies the following.
Proposition 2.2. Suppose the Question
has an affirmative answer with π real.
P
If d ∈ E has the property that γ∈Γ dγ ψ(σ)γ ∈ Z for all σ ∈ G, then there
exists a positive real unit εf (d) ∈ K such thatXγ∈Γ1dγ L0 (0, ψ γ , f ) = − log(εf (d)) = − log kεf (d)k2

Indeed εf (d) can be defined byεf (d)W =

Y

π

σ∈G

5

P

γ
γ
γ∈Γ d ψ(σ) σ.

Remark 3. When π is real, and therefore fixed by τ , it is clear that we canalso writeY Pγγεf (d)2W =π γ∈Γ d (ψ(σ)+ψ(στ )) σ .
σ∈GWe now derive a strengthening of Proposition 2.2 which incorporates the
conjugates of εf (d) and thus leads to the conjecture formulated by Chinburg.
First we record a preliminary step.
Lemma 2.3. In the setting of Stark’s non-abelian question, let σ0 and σ beelements of G, and let ψ be an odd character of degree 2 corresponding to arepresentation ρ of G. Then(ψ(σ0 ) + ψ(σ0 τ )) (ψ(σ) + ψ(στ )) = ψ(σ0 σ) + ψ(σ0 στ ) + ψ(σ0 τ σ) + ψ(σ0 τ στ ).
0Proof. We may again assume that ρ(τ ) = 10 −1. Then(ψ(σ0 ) + ψ(σ0 τ )) (ψ(σ) + ψ(στ ))= Tr (ρ(σ0 )(ρ(1) + ρ(τ ))) Tr (ρ(σ)(ρ(1) + ρ(τ )))= Tr ρ(σ0 ) 20 00 Tr ρ(σ) 20 00= (2ρ1 (σ0 ))(2ρ1 (σ)) = Tr 2ρ1 (σ0∗)2ρ1 (σ) 00




= Tr 2ρ1∗(σ0 ) 00 2ρ1∗(σ) 00 = Tr ρ(σ0 ) 20 00 ρ(σ) 20 00= Tr (ρ(σ0 ) (ρ(1) + ρ(τ )) ρ(σ) (ρ(1) + ρ(τ )))= ψ(σ0 σ) + ψ(σ0 στ ) + ψ(σ0 τ σ) + ψ(σ0 τ στ )Proposition 2.4. Assume that
P Stark’s question has an affirmative answer with
π real. Fix d ∈ E such that γ∈Γ dγ ψ(σ)γ ∈ Z for all σ ∈ G. Define εf (d) as
in Proposition 2.2. Then for each σ0 ∈ G, we haveX−1γ(d(ψ(σ0 ) + ψ(σ0 τ ))) L0 (0, ψ γ , f ) = − log kεf (d)σ0 kγ∈Γ6Proof.
Pγ(d(ψ(σ0 ) + ψ(σ0 τ ))) L0 (0, ψ γ , f )γ∈Γ=−1 X X γγγ
d (ψ(σ0 ) + ψ(σ0 τ )) (ψ(σ) + ψ(στ )) log ||π σ || (by Remark 2)
4Wγ∈Γ σ∈G====−1 X X γγd (ψ(σ0 σ) + ψ(σ0 στ )) log kπ σ k
4W γ σ1 XX γγ−d (ψ(σ0 τ σ) + ψ(σ0 τ στ )) log kπ σ k (by the lemma)
4W γ σ−1−1 X X γγd (ψ(σ) + ψ(στ )) log kπ σ0 σ k (on replacing σ by σ0−1 σ)
4W γ σ−1
1 XX γγd (ψ(σ) + ψ(στ )) log kπ (σ0 τ ) σ k (on replacing σ by (σ0 τ )−1 σ)−
4W γ σ−1−1−2W
2Wlog kεf (d)σ0 k −log kεf (d)τ σ0 k by Remark 3
4W
4W−1− log kεf (d)σ0 k since the chosen absolute value is fixed by τ

In the Stark-Chinburg conjecture, there is a further restriction on the choiceof d which is formulated in terms of the Dirichlet series for L(s, ψ). In returnfor this restriction, one gains in that the conjecture concerns the primitive Artin
L-series without Euler factors removed.

2.4

Dirichlet Series

For this section, we need only assume that K/k is a finite Galois extension withgroup G, and that ρ is a representation of G with character ψ. Artin’s expression
for his L-series involves the choice of a representative Frobenius element σp ∈ G
for each prime ideal p of k. That is, one picks a prime ideal P of K above p andselects σp ∈ G which acts as the Frobenius in the corresponding residue field
extension. Since ψ is a class function, ψ(σpn ) is well-defined for p unramified in
K/k. For ramified p, define ψ(σpn ) to be the average over the coset of σpn by theappropriate inertia group Ip = IP/p ⊆ G. Then for <(s) > 1,!
∞XX
ψ(σpn )L(s, ρ) = L(s, ψ) = exp
n N(p)ns
p n=1
!m
∞
∞
∞
X
X
an
1 X X ψ(σpn )
=
,
=
ns
m!
n
N(p)
ns
p n=1
n=1
m=0

7

which clearly shows that each an lies in Q(ψ), and thatap = ψ(σp )when p is a rational prime which is the norm of a first degree prime p of k not
ramifying in K.
On the other hand, one can express L(s, ρ) using the eigenvalues λp,i (listedwith multiplicity) for ρ(σp ) acting on the subspace Vp fixed by the inertia group
Ip . These eigenvalues are necessarily roots of unity and hence algebraic integers.
Again for <(s) > 1,Y−1
L(s, ρ) =det I − N(p)−s ρ(σp ) |Vp
p

=

YY
p

=

−1

i



∞
X
λjp,i


N(p)js
i
j=0

YY
p

=

1 − λp,i N(p)−s

∞
X

an
ns
n=1

This expression clearly shows that each an is an algebraic integer. Thus we havesupplied a proof of an important fact.
Proposition 2.5. For any finite Galois extension K/k with group G and any
representation ρ of G with character ψ, the Artin L-function L(s, ρ) = L(s, ψ)P anwhose coefficients an are algebraic integers lying in
has a Dirichlet seriesns
Q(ψ), and ap = ψ(σp ) when p is a rational prime which is the norm of a first
degree prime p of k not ramifying in K.

2.5

The Stark-Chinburg Conjecture

Assume from now on that ρ is an irreducible 2-dimensional odd representationof the group G = Gal(K/Q), with associated character ψ. The statement of theconjecture involves the Dirichlet series expansion
X anL(s, ρ) = L(s, ψ) =nsnfor <(s) > 1.
For d ∈ E = Q(ψ) and Γ = Gal(E/Q), define the functionX
fd (s) :=dγ L(s, ψ γ ).
γ∈ΓThus fd0 (0) = γ∈Γ dγ L0 (0, ψ γ ), which acts as an analog for the primitive Lfunction of the quantity in Proposition 2.2 in which the Euler factors for theprimes dividing f have been removed.
P

8For <(s) > 1, we also have the expressionX
fd (s) =An n−sn≥1

with An =Pγ(dan ) = TrE/Q (dan ) ∈ Q.

γ∈ΓConjecture 2.6 (Stark-Chinburg). Assume that d ∈ E is such that all thecoefficients An are in fact rational integers. Then there exists a unit ε(d) in
K+ = K ∩ R, the so-called Stark unit, such that, for all σ ∈ G
log kε(d)σ−10
k = fd(ψ(σ)+ψ(στ)) (0).

Furthermore, the real conjugates of ε(d) are positive.
Remark 4. Note that the unit ε(d), if it exists, is unique and it is given by theformulaon0(0).
ε(d) = exp fd(ψ(1)+ψ(τ))
Remark 5. The condition on d in the conjecture can be restated as the requirement that d lie in the product of the co-different and the inverse of the ideal
generated by the coefficients an .
This conjecture is to be compared with Proposition 2.4. It is stronger inas much as the L-functions are primitive. However, in general the conjecture
places a slightly greater restriction on the choice of d ∈ E. For the Čebatorev
density theorem shows that, given any σ ∈ G, there are infinitely many primes
p such that σp = σ, and Proposition 2.5 then gives ap = ψ(σ) for such a p. But
in many cases, including all examples arising as ours do from G ∼= Â5 , the setof possible d is the same in the Stark-Chinburg conjecture as it is in Stark’s
Question. Indeed this happens whenever the character values generate the ringof integers OE of E as a Z-module. For Proposition 2.5 also shows that all an
are in the set OE , and we can easily deduce the following result.
Proposition 2.7. Suppose d ∈ E, and let D(E)−1 be the co-different of E,
defined by D(E)−1 = {d ∈ E | TrE/Q (da) ∈ Z ∀a ∈ OE }. Thend ∈ D(E)−1 ⇒ An = TrE/Q (dan ) ∈ Z ∀n ⇒ TrE/Q (dψ(σ)) ∈ Z ∀σ ∈ G.
Also, if the values of ψ(σ) as σ ranges over G generate OE as a Z-module, then
the third condition implies the first, so all three are equivalent.
To summarize, the Stark-Chinburg conjecture is a precise conjecture “overZ” designed to be a close analog for primitive L-functions of a consequence of anaffirmative answer to Stark’s non-abelian question for imprimitive L-functions.
It should be noted that we have not stated the most general form of the conjecture formulated by Chinburg in [Chinburg 1983]. The conjecture there applies
to a finite linear combination of L-functions for odd irreducible 2-dimensional
characters such that the resulting Dirichlet series has integral coefficients. For
the specific group G used in the computations in this paper, all four charactersof the appropriate type are conjugate, and the two conjectures are equivalent inthis case.
9

3

Icosahedral representations

3.1

Minimal Icosahedral Representations

In this section, we briefly determine the minimal degree of an extension K/Q
supporting an icosahedral representation of the type appearing in Chinburg’s
conjecture, namely one for which the associated Artin L-function L(s, ρ) has afirst order zero at s = 1.
So suppose that ρ is an odd icosahedral representation of G = Gal(K/Q).
This means that the image of ρ(G) in PGL2 (C) is isomorphic to the alternatinggroup A5 ∼= PSL2 (F5 ), which has order 60 and can also be identified with thegroup of symmetries of the icosahedron. By minimality, we may assume that ρ
is faithful, i.e. has trivial kernel, so G ∼= ρ(G). We have an exact sequence
1 → A → ρ(G) → A5 → 1,
where the kernel A may be described as ρ(G) ∩ C∗ , and is therefore a finite
cyclic group lying in the center of ρ(G). We seek the minimal possible order forA.
The kernel A cannot have order 1, because A5 has no irreducible representation of degree 2.
Let Cn denote the cyclic group of order |Cn | = n and V4 denote a 2-Sylow
subgroup of A5 , isomorphic to the Klein 4-group. If A has order 2, then ρ(G)represents an element of H 2 (A5 , C2 ), which we will show is isomorphic to C2 .
The group H 2 (A5 , C2 ) has exponent 2 since C2 does. Then by the restriction
map, H 2 (A5 , C2 ) = H 2 (A5 , C2 )2 ⊂ H 2 (V4 , C2 ). This last group classifies central
extensions of V4 by C2 , of which there are 8: one trivial class represented by C23 ,three classes represented by C2 × C4 , three classes represented by D8 , and oneclass represented by Q8 . Of these, the classes fixed by the action of A5 actuallyconstitute H 2 (A5 , C2 ), by [Brown 1994, III.10.3]. There are two fixed classes:
those of C23 and Q8 . This gives two possibilities for G ∼= ρ(G) when |A| = 2,
namely G ∼= A5 × C2 and G ∼= SL2 (F5 ). Each irreducible character of A5 × C2
is obtained as the product of an irreducible character of A5 and an irreduciblecharacter of C2 , and thus cannot be of degree 2.
On the other hand, SL2 (F5 ) admits two conjugate characters of degree 2,
but these have value -2 on each element of order 2 (see [Buhler 1978, p. 135] forthe character table). Hence r(ρ) = 0, in violation of the assumption r(ρ) = 1.
So |A| = 2 is impossible.
If |A| = 3 then ρ(G) represents an element of H 2 (A5 , C3 ) = 0. This time, thegroup H 2 (A5 , C3 ) has exponent 3 and is isomorphic to a subgroup of H 2 (C3 , C3 ).
The latter group classifies central extensions of C3 by C3 of which there are: onetrivial class represented by C3 × C3 and two classes represented by C9 . Thus
H 2 (C3 , C3 ) has order 3, while the only class fixed by the action of A5 is thetrivial one. So we must have G ∼= A5 × C3 . Again the irreducible characters of
this group are products and so none are of degree 2.
Thus the minimal order for A is at least 4, and A must be cyclic. The case of|A| = 4 is indeed realized by G ∼= ESL2 (F5 ) = {M ∈ GL2 (F5 ) : det(M ) = ±1},10for which an odd irreducible representation ρ exists. One can show that there
is only one equivalence class of extensions of A5 by C4 ; a representative of thisequivalence class is usually denoted by Â5 . (Just check that Q8 × C2 represents
the only non-trivial class in H 2 (V4 , C4 ) which is fixed by the action of A5 .) We
have chosen a convenient realization ESL2 (F5 ) ∼= Â5 .

3.2

Â5 extensions

From now on, we assume that K/Q is a Galois extension of group G isomorphic
to Â5 . As mentioned in the last section, the group Â5 , hence also G, can be
identified with the group ESL2 (F5 ) of 2 × 2 matrix with coefficients in F5 and
0 1 and 2 1 .
determinant ±1. This group is generated by the two matrices:
1
0
3
2

We choose such an identification of G so that τ = 01 10 (recall that τ is thecomplex conjugationin G). The center of G is cyclic of order 4 and generatedby z = 20 02 .
The group G has 18 conjugacy classes, listed in the following table. For eachconjugacy class C, we list its cardinality |C|, an representative element κ ∈ C,
the order of κ and the values ψ(κ) and χ(κ) where ψ is the character of ρ andχ is the abelian character obtained by composing ρ with the determinant map.
1
1

#
|C|
κ

10
01

order

1



2
30
τ

3
12
01
11

2

20

4
20


01
12

ψ(κ)

2

0

χ(κ)

1

−1

−1

−1

#
|C|
κ

10
20

11
12

12
12

13
12

01
44

order

3

01
43

10



ψ(κ)

−1

√
1− 5
2

χ(κ)

1

1

02
22

5





12

√
−i 1+2 5



5
12
01
14

6
20


20

−i

√
i 1+2 5

02
23



√
1+ 5
2

1

1

01
40

12



4

01
41

6

9
12


01
42
√



5

i

0

1

−1

−1

1

1

1

14
12

15
12

16
1
z

17
1
z2

18
1
z3

4

2

4

−2i

−2

2i

−1

1

−1





02
34

5−1
2

√
i 1−2 5

20

√

i



8
20

5−1
2

02
31

10

√
− 1+2 5

01
13

7
30

−1

20

−1
√

The field E generated√over Q√by the values of ψ is Q(i, 5), an integral basis
of E is given by 1, i, 1+2 5 , i 1+2 5 , and its Galois group Γ is generated by γ1
and γ2 with√√
5
γ1 (i) = −i, γ1 (√5) = √γ2 (i) = i,γ2 ( 5) = − 5√√
A Z-basis of D(E)−1 is 12 , 2i , 5+20 5 , i 5+20 5 . If d1 and d2 are two elementsof D(E)−1 for which Conjecture 2.6 is true, so the units ε(d1 ) and ε(d2 ) exist,
then the Conjecture is also true of md1 (m ∈ Z) and d1 + d2 simply by taking
ε(md1 ) = ε(d1 )m and ε(d1 + d2 ) = ε(d1 )ε(d2 )11Hence, Conjecture
2.6 is√true for all d ∈ D(E)−1 if and only if it is true for√d = 12 , 2i , 5+20 5 , and i 5+20 5 .
Moreover, one can readily prove that iψ(σ) = ψ(z 3 σ) for all σ ∈ G. Thus
fid(ψ(σ)+ψ(στ )) (s) = fd(ψ(z3 σ)+ψ(z3 στ )) (s),and the truth of Conjecture 2.6 for some d ∈ E implies the truth of the conjecture for di simply by taking
ε(di) = ε(d)z .
We have proved the following
Lemma 3.1. Conjecture 2.6 is true if and only if it is true for
√
5+ 5
1
.
d = and d =
2
20

3.3

Computations of L0 (0, ρ)In order to test Conjecture 2.6, we have to compute the value of L0 (0, ργ ) toa high accuracy. For this, we follow the method used by Stark [Stark 1977].
Recall thatX
L(s, ρ) =an n−sn≥1

is the expansion as a Dirichlet series of the L-function for <(s) > 1. Let C be
the conductor of ρ, then the functionξ(s, ρ) =

C
4π 2

s/2
Γ(s)L(s, ρ)satisfies the functional equation
ξ(s, ρ) = wξ(1 − s, ρτ ),(3.2)

where w, the so-called Artin Root Number, is a complex number of modulus 1.
Let


X
2πn
f (t, ρ) =an exp − √ t .
C
n≥1
Then

Z ∞
ξ(s, ρ) =ts−1 f (t, ρ)dt

(3.3)

0

for <(s) > 1, that is ξ is the Mellin transform of f . By the inverse Mellintransform formula, we get
Z1
f (t, ρ) =ξ(s, ρ)t−s ds,
2iπ <(s)=σ12

where σ is a real number > 1. Now, we haveZ1f (t−1 , ρ) =ξ(s, ρ)ts ds
2iπ <(s)=σ
=

1
2iπ

Z

w−1 t
=
2iπ

ξ(1 − s, ρ)t1−s ds<(s)=1−σZξ(s, ρτ )t−s ds(by the functional equation).

<(s)=1−σ

We now assume that ξ(s, ρτ ) is holomorphic over C (actually, we only need that
there exists  > 0 such that ξ(s, ρτ ) is holomorphic on the half-plane <(s) > −,but, by the functional equation (3.2), this is equivalent to the holomorphy of
ξ(s, ρτ ) over C). Then, we have for two numbers σ1 < σ2 , see for example[Cohen 2000], Lemma 10.3.5:
Z σ2 +iTξ(s, ρτ )t−s ds = 0,lim
|T |→∞

σ1 +iTand thus we obtain
Z
Z
1
1
τ −s
ξ(s, ρ )t ds =ξ(s, ρτ )t−s ds = f (t, ρτ )2iπ <(s)=1−σ2iπ <(s)=σand sof (t−1 , ρ) = w−1 tf (t, ρτ ).

(3.4)

For any fixed u > 0, we break the integral in (3.3) into two pieces: the first one
from 0 to u, the second one from u to ∞. In the first integral, we replace t by
t−1 and use (3.4) to get:Z ∞Z ∞ξ(s, ρ) = w−1
t−s f (t, ρτ )dt +ts−1 f (t, ρ)dt.
u−1

u

So finallyL0 (0, ρ) = ξ(0, ρ)
Z ∞
Z ∞
dt
−1
τ
=w
f (t, ρ )dt +f (t, ρ)
t
−1
u
√u




−1
X Z ∞
w
C X an2πn2πndt=
exp − √+
anexp − √ t
.
2π
n
t
u C
C
u
n≥1
n≥1
We have proved the following result:
Proposition 3.2. Assume the L-function L(s, ρ) is holomorphic over C. Then
there exists a complex number w of modulus 1, such that, for any u > 0
√

 X


2πn
2πun
w−1 C X an
exp − √+
an Ei √
L0 (0, ρ) =
2π
n
u C
C
n≥1
n≥1
13

where Ei(x) =R +∞ −te dt/t is the exponential integral function.
x

Remark 6. The proposition requires the hypothesis that L(s, ρ) is holomorphic
over C, that is, that ρ satisfies the Artin conjecture. This conjecture is not provedin general, but it has been proved in all examples we work with in this paper using either results of Kiming and Wang [Kiming and Wang 1994], or the work of
Buzzard, Dickinson, Shepherd-Baron and Taylor which proves the Artin conjecture for infinitely many icosahedral odd representations [Buzzard et al. 2001], or
results of Jehanne and Müller [Jehanne and Müller 2000, Jehanne and Müller 2001].
The formula given by Proposition 3.2 can be used to compute approximationsof L0 (0, ρ), or of L0 (0, ργ ), if one knows how to compute the coefficients an andthe value of Artin Root Number w. The former can be computed using themethod of [Jehanne 2001] (see also the next section), the latter can be found(following Stark) by computing the two sums in the formula for two different
values of u and solving the system. This also provides a neat check of thecomputations since the complex number found must be of modulus 1.

3.4

Construction of Â5 extensions

In this section, we explain how one can construct Â5 extensions with an odd
irreducible degree 2 representation with a quadratic determinant, that is representation whose determinant is a quadratic character. Any such extension K
contains a quintic field K of A5 -type. Furthermore, since the representation
is odd, K is a complex field. A table of quintic complex fields of A5 -type of
discriminant less than 40272 has been computed by J. Basmaji [Basmaji 2002]
using the methods of [Basmaji and Kiming 1994].
Such a A5 -type complex quintic field K yields two projective representations
corresponding to the two embeddings of A5 in PGL2 (C). Let ρproj be oneof these two representations; methods to decide whether or not ρproj has alifting ρ of given conductor and determinant are described in [Crespo 1992] or[Jehanne 2001]. Assume from now on such a lifting exists and call it ρ. Thenthe set of all liftings of ρproj isE(ρ) = {ρ ⊗ ν with ν a Dirichlet character of GQ }.
Some of these ρ ⊗ ν may have the same conductor as ρ. Also, det(ρ ⊗ ν) = det ρ
if and only if ν 2 = 1. In particular, if det ρ is quadratic, then ρ ⊗ det ρ has thesame conductor and the same determinant as ρ.
By looking at a table of irreducible characters of Â5 (which can be easily
constructed from [Buhler 1978] p. 135), we see that there are four √charactersof degree 2, and that they are conjugate under the action of Gal(Q( 5, i)/Q).
The characters of ρ and ρ ⊗ det ρ are conjugate by the complex conjugation,and the two other representations correspond to the other embedding of A5 in
PGL2 (C).
To find all the representations with odd quadratic determinant and conductor up to 3676, we use the table of [Basmaji 2002] and the following result,
14

K
4

2 hτ iK+ = K ∩PPR
PPP
PPP
5
P

4 hzi
N UUUU
S
UUUU2
UUUU
2
M = KF
~
~
~~
~~ 5
00
12
~
~
nn F
n
6 ~~
n
~
nnn
~~
nnn 2
~
F
~~
~~
~
~
K JJ
JJ
JJ
6
JJ
JJ
J
hhhh kJJ
52hhhhh
JJ
h
h
JJ
hh
hhhh
Qh

120

Figure 1: Some subfields of K
obtained by local computations (for the computations of conductors, we refer
to [Kiming 1994]).
Proposition 3.3. Let N be an A5 -extension, let ∆ be the discriminant of aquintic field contained in K and let C be the conductor of √a representation ρwith quadratic determinant corresponding to N . Then C ≥ ∆.
Table 1 lists all icosahedral representations with odd quadratic determinantand conductor up to 36762 . For each representation ρ, we read on this table: theconductor C, a polynomial for the corresponding quintic
field, and the square√
free integer δ such that det(ρ) fixes the field k = Q( δ).
As mentioned above, we used the method described in [Jehanne 2001] tocompute the coefficients of the L-series of the representations . We briefly
explain this method (see also Figure 1). Let f (X) be one of the polynomials in
Table 1, and let K be the field defined by an arbitrary (fixed) complex root x1of f . Thus K is a complex quintic field of A5 -type. Let N be the Galois closureof K/Q, so Gal(N/Q) ' A5 , and let x2 , . . . , x5 ∈ N be the other roots of f .
Define θ by
θ = (x1 − x2 )2 (x2 − x3 )2 (x3 − x4 )2 (x4 − x5 )2 (x5 − x1 )2
and let F = Q(θ). The field F is a degree 6 field with Galois closure N . The
degree 30 field M = KF will be play an important part in the computations (see
15

C
1948
2083
2336
2336
2707
2863
2863
3004
3203
3547
3548
3587
3587
3676

polynomial defining K
x − 7x3 − 17x2 + 18x + 73
x5 + 8x3 + 7x2 + 172x + 53
x5 + 2x3 − 4x2 − 2x + 4
x5 + 2x3 − 4x2 − 2x + 4
x5 − x4 + 9x3 − 6x2 − 32x + 93
x5 + 12x3 + 21x2 + 22x + 7
x5 + 12x3 + 21x2 + 22x + 7
x5 − 8x3 + 10x2 + 160x + 128
x5 + 8x3 + 5x2 − 4x + 1
x5 − 8x3 − 2x2 + 31x + 74
x5 + 10x3 + 10x2 + 44x + 56
x5 + 3x3 + 24x2 − 20x + 131
x5 + 3x3 + 24x2 − 20x + 131
x5 − 8x3 + 28x2 − 40x + 48
5

δ
−487
−2083
−73
−146
−2707
−409
−2863
−751
−3203
−3547
−887
−311
−3587
−919

Table 1: The first icosahedral representations with odd quadratic determinant√
next section). Then we define F 00 = Q( θδ). In [Jehanne 2001], it is provedthat there exists a quadratic extension S of F 00 such that its Galois closure is
the field K we are looking for. Since we know the conductor, the determinant
00
and the image in PGL2 (C) of ρ, we know the ramification of S/F√ and thus
can construct a finite set B of elements of F 00 such that S = F 00 ( β) for someβ ∈ B. We then use an explicit criterion to decide which element is the rightone. Once the field S has been found, we can use the explicit decomposition ofprime ideals in S (and possibly also some other subfields of K) to compute thecoefficients an of the L-function of ρ.

4Computations

4.1

Numerical determination of the Stark unit

Let d be a fixed element of D(E)−1 . In this section we explain how to findthe conjectural Stark unit ε(d), assuming from now on that it exists, using
numerical approximations. (Actually, according to √lemma 3.1, we have only
performed these computations for d = 12 and d = 5+20 5 .)As mentioned in the introduction, we find ε(d) by constructing its minimal
polynomial over the field M . This field has degree 30 and signature (2, 14).
Also, it is a subfield of K+ . More precisely it is the subfield of K+ fixed byz, that is to say by the center of G. Since ε(d) is real, its conjugates over M ,i.e. z l (ε(d)) with 0 ≤ l ≤ 3, are real too and positive by the second part of the16Conjecture. Thus they are given by the formulano0z l (ε(d)) = exp fd(ψ(zl )+ψ(z l τ )) (0)for 0 ≤ l ≤ 3.
Hence, we can compute approximations of L0 (0, ρ) and get from them approximations of the conjugates of ε(d) over M , and then form the monic polynomial P̃ whose roots are these approximations. This polynomial is thus anapproximation of the minimal polynomial P (X) of ε(d) over M . Now, sinceψ(z 3 σ) = iψ(σ), it follows that ψ(z 2 σ) = −ψ(σ), and thus z 2 (ε(d)) = ε(d)−1 ,z 3 (ε(d)) = z(ε(d))−1 , and one can write
P (X) = X 4 + aX 3 + bX 2 + aX + 1
with a, b ∈ OM . We now need to be able to recover the coefficients a and b fromthe corresponding coefficients ã and b̃ of P̃ .
The problem can be stated in a more general setting as follows: given areal number x̃, and two positive real numbers C1 and C2 , find, if it exists, analgebraic integer x in OM such that|x̃ − x| < C1

and |x0 | < C2 where x0 is any conjugate of x (6= x).

(4.5)

Note that here and it what follows, when we talk about the conjugates of x,
we mean the conjugates of x different from x. It is not difficult to show that if
we choose C1 small enough, then we can make sure that there is at most one
algebraic integer in M satisfying these conditions.
The method we used is a generalization of a method due to H. Cohen (see[Cohen 2000, section 6.2.4]). Let r1 , r2 be the two real embeddings of M , with
r1 being the identity, and let c1 , . . . , c14 be a complete fixed set of non-conjugate
complex embeddings of M . For y ∈ M and 1 ≤ l ≤ 30, we define
if l = 1 or 2
rl (y)(l)y = <(cl−2 (y)) + =(cl−2 (y))
if 3 ≤ l ≤ 16<(cl−16 (y)) − =(cl−16 (y)) if 17 ≤ l ≤ 30
Let v(y) be the 30-dimensional vector whose l-component is y (l) . Then themap from M to R30 which sends y to v(y) sends the ring of integers OM to afull lattice in R30 of determinant the absolute value of the discriminant of M .
We will need the following lemma whose proof is direct.
Lemma 4.1. Let x ∈ M and assume that all the conjugates of x have absolutevalue less than C2 . Then√x(2) < C2 andx(l) < 2C2 for 3 ≤ l ≤ 30.
In the reverse direction, ifx(l) < C2 for 2 ≤ l ≤ 30
then all the conjugates of x have absolute value less than C2 .
17

Let {ω1 , . . . , ω30 } be an integral basis of OM . For a fixed real number x̃,consider the following quadratic form on Z31 :

2

2
30
30
30
X
X
X
(1)
(l)

Q(v0 , v1 , . . . , v30 ) = C22 v02 + (C2 /C1 )2 
vj ωj − v0 x̃ +vj ωj 
j=1

j=1

l=2

If x = x1 ω1 + · · · + x30 ω30 ∈ OM is a solution of (4.5) thenQ(1, x1 , . . . , x30 ) <C22 + (C2 /C1 )2 (x − x̃)2 + x(2)

2

+

30
P

x(l)

2

l=3

<

C22 + C22 + C22 + 2

30
PC22 = 59C22

l=331

Conversely, let (x0 , . . . , x30 ) ∈ Z

be such that

Q(x0 , . . . , x30 ) < 59C22 .
Then C22 x20 < 59C22 so |x0 | ≤ 7. If x0 is actually equal to ±1, then we can set
30
Xx∗ = x0 
xj ωj  ∈ OM .
j=1√

and x∗ satisfies|x∗ − x̃| < 59C1 and all its conjugates are of absolute value√less than 59C2 . Therefore, solutions toQ(x0 , . . . , x30 ) < 59C22(4.6)

with x0 = ±1 are not too far from being solutions to our original problem, andthere are only a small number of solutions to (4.6) when C1 is small enough.
In order to find solutions to (4.5), one can use the Fincke-Pohst algorithm[Cohen 1993, 2.7.3] to find solutions to (4.6), then discard those for which x0 6=±1 (or even better, modify the algorithm in such a way that it only considers
vectors with x0 = ±1). Then for each solution found (with x0 = ±1), computethe corresponding algebraic integer and check whether or not it satisfies thestronger conditions of (4.5).
In practice, this method works very well for small enough values of C1 andgives only a few vectors satisfying (4.6), only one of those satisfying (4.5). For
information, the size of C1 , that is the precision used, was between 10−100 and
10−200 for most examples, with a precision up to 400 decimal places in one
case (N = 3004). However, the computations used to compute the value of the
L-functions was higher, around 600 decimal places.
Once the (conjectural) unit ε(d) has been found, we need to check that is
satisfies the conjecture. Of course, one part of the check involves testing whether
or not two real numbers, given by approximations, are equal, which is an impossible computational task. So we will not be able to prove the conjecture inthese cases but only to give evidence pointing toward the truth of the conjecture. These checks are described in section 4.4 together with an example. We
summarize these computations in the following result.
18

Theorem 4.2. For the fourteen icosahedral representations with odd quadratic
determinant listed in Table 1, Conjecture 2.6 has been numerically verified up
to the precision of the computation.

4.2

Square-root of the Stark unit

In all the examples, we have found that the unit ε(d) was a square in K. In fact,
in almost all examples, it is actually a fourth power (see below). We have used
this fact to simplify the computation. Indeed, in all examples, we have started
by assuming that it was a fourth power, and instead of trying to recognize
the coefficients of the minimal polynomial of ε(d) over M , we searched for thecoefficients of the minimal polynomial of ε(d)1/4 . In doing so, we always took
the positive fourth root as conjugates of ε(d)1/4 . If we were not able to findthose, then we searched for that of the minimal polynomial of ε(d)1/2 , assuming
again that all the conjugates were positive. As stated above, in all cases, wewere able to find these coefficients. Not only does this method prove directlythat the unit ε(d) is a fourth power (resp. a square), but it also greatly simplifies
the computations since we had to deal with numbers having one fourth (resp.
one half) as many digits! Of course, if we failed to recognize the coefficients ofthe minimal polynomial of ε(d)1/4 that did not prove that it was not a fourth
power, since we arbitrarily decided to consider only the positive fourth root.
So, in those cases, we did check once the unit had been found that it was not a
fourth power.
The following table sums up the information mentioned above. For eachconductor N , an entry 2 (resp. 4) means that the unit ε(d) was a square (resp.
√a fourth power) in K. We do not specify in the table the value of d ( 21 or 5+20 5 ),or the representation (if there are more than one to test of the same conductor)
since in all the examples we have found that this property does not depend onthese.
N
power
1948
4
2863
4
3548
2

4.3

N
power
2083
4
3004
2
3587
4

N
power
2336
2
3203
4
3676
2

N
power
2707
4
3547
4

The abelian condition

The so-called abelian rank one Stark conjecture for an abelian extension K1 /K2of number fields (see [Tate 1984, Chap. IV]) predicts the existence of a unit ∈ K1 satisfying conditions similar to that of Conjecture 2.6 and such that
1/e defines an abelian extension of K2 , where e is the number of roots of unity
in K1 . A similar condition for the Stark-Chinburg conjecture has not yet beenstated. However, following a suggestion made by Stark, we have checked in all
fourteen of our examples that the fourth root of the Stark unit always generates
an abelian extension of M (of course, this is trivially satisfied when the Stark
unit is already of fourth power). We ask the following question:
19

Question 4.3. In this setting, is it true that the extension K(ε(d)1/4 )/M isalways an abelian extension?

4.4

X

An example

120

− 14X

119

+ 10X

− 5891390X

118

111

+ 528X− 2148339186X+ 4235000319554X+ 318871476120728X

106

102

− 67266744140X

98

94

116

− 12186X+ 105936396X+ 17880418336X

105+ 1311505408046X+ 45100336757892X+ 744008040343916X

109

115

− 1022X+ 15711872585X

101

97

104

− 89291738681X+ 272500X

108− 1682328456079649X− 28166525613664065X

82− 538617388539670046X

81− 1405725781756458029X

78− 3855005474440939128X+ 2827558653955188558X+ 9089153383531781324X

74
71+ 22886584508143805334X− 38951665554067587963X

68− 37074726381332624176X

65+ 89065895937691036822X

62

77

99

80

76

− 31520286306610904124X− 74983502912940337331X

64− 10418769031801600904X

61+ 56660611016658159176X+ 27894404791194859718X− 94322409774714428665X

66
63
60

− 74983502912940337331X

56− 37074726381332624176X

55

53− 38951665554067587963X

52− 31520286306610904124X

51

49− 15279740471791884367X

48+ 24920013640589593880X− 14516564995099084468X− 3855005474440939128X

47

43− 538617388539670046X+ 9089153383531781324X+ 637495087229463472X

35+ 4496703792844188X

31+ 744008040343916X+ 7510163330929880X

27− 2148339186X

14

19+ 511431X+ 272500X

7

13

18

+ 1586294443053165584X+ 144746005476226278X− 3766815912764636X

30− 2135167290906516X

26− 198789238356076X

22+ 206807275X− 1022X+ 56660611016658159176X+ 8024368738725683150X− 8494623508910X− 168995037912X

6+ 27894404791194859718X

34+ 4235000319554X− 67266744140X− 1534364982X
8

38+ 318871476120728X
23

46

42

+ 90442638583264834X+ 45100336757892X+ 1311505408046X+ 22886584508143805334X+ 2827558653955188558X

39− 22239156146839804X

50

12− 12186X

17

37− 721X

4

11

57
54

− 5224727512198411575X− 1405725781756458029X− 255223687887713211X

33− 28166525613664065X

29− 1682328456079649X

25− 44699345526371X

21− 89291738681X+ 15711872585X+ 105936396X
5

45

41

16

+ 528X

+ 10X

2

10

44

40

36

32

28

24

20

+ 17880418336X− 13723194X
3

75

69

59

+ 37417793964722184658X

58

79

72

− 10418769031801600904X+ 89065895937691036822X

83+ 8024368738725683150X

70+ 37417793964722184658X

87+ 1586294443053165584X− 15279740471791884367X

67

91+ 144746005476226278X

73+ 24920013640589593880X

95− 3766815912764636X

84− 5224727512198411575X− 14516564995099084468X

103− 2135167290906516X

88

112

107

− 198789238356076X

92− 255223687887713211X+ 4496703792844188X+ 511431X− 8494623508910X

96

85

86

113

− 1534364982X− 168995037912X

100− 44699345526371X

93

114

+ 206807275X− 22239156146839804X+ 7510163330929880X

90− 721X110

89+ 90442638583264834X+ 637495087229463472X

117− 13723194X

15− 5891390X

9− 14X + 1√

Figure 2: The irreducible polynomial over Q of ε( 5+20 5 )1/4
We conclude with an example of a computation. We will look at the representation of conductor N = 2863 and with determinant
the quadratic character√√of the field Q( −409), and the value d = 5+20 5 . (This example is the one forwhich the irreducible polynomial over Q of the Stark unit
has the smallest coef√
5+ 5
ficients.) In what follows we will write ε instead of ε( 20 ) to denote the Stark
unit.

20

First, we compute the field F using the explicit formula for θ and then findthat the field M is generated over Q by a (fixed) real root of the polynomial:
X 30 − 11X 29 + 60X 28 − 184X 27 + 282X 26 − 93X 25 + 1155X 24 − 15102X 23+ 81876X 22 − 295153X 21 + 824690X 20 − 1918902X 19 + 3838834X 18− 6617268X 17 + 9651756X 16 − 11548871X 15 + 10886632X 14− 7709825X 13 + 3980211X 12 − 1749801X 11 + 1033046X 10− 526435X 9 − 55897X 8 + 112042X 7 + 213353X 6 − 221284X 5− 31311X 4 + 78204X 3 + 4802X 2 − 12005X − 2401
0
We compute the values of fd(ψ(g)+ψ(gτ)) (0) for all g ∈ G with a precision of
620 decimal places for g ∈ hzi and of 100 decimal places for g 6∈ hzi. Using the(positive) fourth root of these values, we find that, if these choices are correct,
then ε1/4 must be a root of the following polynomial which must have coefficientsin OM if the fourth root belongs to K+ : †X 4 − 11.0733582927400638184932897075796398...X 3+ 26.4538517976073658614124922380428030...X 2− 11.0733582927400638184932897075796398...X + 1.
Also, we find that the other conjugates of the coefficient a of X 3 (which is alsothat of X) are bounded in absolute value by 14, and those of the coefficient bof X 2 are bounded by 44. These bounds are also found by using Conjecture
2.6. Using the method explained above, we recognize the two coefficients aand b using C1 = 10−120 . We will not list those since each one would require
several pages just to write it down! However, once these coefficients have beenfound, we construct the corresponding polynomial and compute its roots to a
precision of 600 decimal digits. We then check than these values agree with theone computed via the conjecture. This is already a first good check since onlya precision of 120 decimal digits was used to recover the coefficients. (Actually,
the first good check is that there are indeed elements a and b in OM satisfying
the conditions we imposed.)
We then compute the irreducible polynomial of ε1/4 over Q; it is a degree 120
polynomial with quite big coefficients even though we are only looking at the
fourth root of the Stark unit (see Figure 2). We compute its roots (ei )1≤i≤120 toa precision of 100 decimal digits. We then look for a one-to-one correspondence0
between the absolute values of the ei ’s and the values of exp{fd(ψ(g)+ψ(gτ)) (0)},
for g in a set of representatives of G/hτ i, such that corresponding values agree
up to the precision. Such a correspondence must exist if the conjecture is true
and, indeed, we find that it does. This is also quite a good check since we usedthe values of the other conjugates of ε which are not conjugates over M only
† In this example, we will, of course, give all the numerical results with a much smaller
precision than the one used in the computations.

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through the upper bound that they provide on the conjugates of the coefficientsa and b. Actually, if the conjecture is true, this correspondence should give us
explicitly the Galois action of G on the conjugates of the Stark unit. However, it
is not practical to recover this (conjectural) correspondence by trying to matchthe values of the absolute values of the conjugates with the values predictedby the conjecture since there are too many conjugates of the Stark unit withthe same absolute value and thus too many possible correspondences (in thisexample, the number of possible correspondences is around 1026 ).
Finally, we check that the Stark unit generates the field K+ over Q in thefollowing way. Recall that the method of [Jehanne 2001] gives us an explicit
construction of S. Now, looking at Figure 1, it is clear that K+ = SK = SM .
We find a primitive element α of S over F , so K+ = M (α). Next, we compute thecompositum field over M of M (ε) and M (α) using the method of [Cohen 2000,
2.1.3] and find that it has degree 4. Thus M (ε) = M (α) = K+ .

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Arnaud Jehanne, A2X, Université Bordeaux I, 351, cours de la Libération, 33405
Talence cedex, FRANCE (jehanne@math.u-bordeaux.fr)
Xavier-François Roblot, IGD, Université Claude Bernard, 43, boulevard du
11 Novembre 1918, 69622 Villeurbanne cedex, FRANCE (roblot@euler.univlyon1.fr)
Jonathan Sands, Department of Mathematics, University of Vermont, 16 Colchester Ave, Burlington, VT 05405, USA (sands@math.uvm.edu)

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