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15J.-M. De Koninck et I. KátaiThe number of large prime factors of integers and normal numbers

R. Holowinsky, G. Ricotta et E. RoyerThe amplification method in the GL(3) Hecke algebra

S. LouboutinFundamental units for orders generated by a unit

A. SivieroRealisable classes, Stickelberger subgroup and its behaviour under change of the base field

S. WongThe maximal unramified extensions of certain complex Abelian number fields

Revue du Laboratoire de Mathématiques de Besançon (CNRS UMR 6623)

P r e s s e s u n i v e r s i t a i r e s d e F r a n c h e - C o m t é

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Jean-Robert Belliard, Université de Franche-Comté ________________________ [email protected]

Jean-Marc Couveignes, Université Bordeaux 1 ________________________ [email protected]

Vincent Fleckinger, Université de Franche-Comté _____________________________ [email protected]

Farshid Hajir, University of Massachusetts, Amherst (USA) ____________________________ [email protected]

Nicolas Jacon, Université de Reims Champagne-Ardenne ___________________________ [email protected]

Jean-François Jaulent, Université Bordeaux 1 _________________________ [email protected]

Henri Lombardi, Université de Franche-Comté __________________________________ [email protected]

Christian Maire, Université de Franche-Comté __________________________________ [email protected]

Ariane Mézard, Université Paris 6 _______________________________________________________ [email protected]

Thong Nguyen Quang Do, Université de Franche-Comté ___________________________ [email protected]

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Les Publications mathématiques de Besançon éditent des articles de recherche mais aussi des articles de synthèse, des actes, des cours avancés. Les travaux soumis pour publication sont à adresser à Christophe Delaunay [email protected] ou à l’un des membres du comité scientifique. Après acceptation, l’article devra être envoyé dans le format LaTeX 2e, de préférence avec la classe smfart. La version finale du manuscrit doit comprendre un résumé en français et un résumé en anglais.

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2015

Sommaire

J.-M. De Koninck et I. KátaiThe number of large prime factors of integers and normal numbers . . . . . . . . . . . 5-12

R. Holowinsky, G. Ricotta et E. RoyerThe amplification method in the GL(3) Hecke algebra . . . . . . . . . . . . . . . . . . . . . . . 13-40

S. LouboutinFundamental units for orders generated by a unit . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41-68

A. SivieroRealisable classes, Stickelberger subgroup and its behaviour under change of the

base field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69-92S. WongThe maximal unramified extensions of certain complex Abelian number fields 93-104

Publications mathématiques de Besançon – 2015, 5-12

THE NUMBER OF LARGE PRIME FACTORS OF INTEGERS ANDNORMAL NUMBERS

by

Jean-Marie De Koninck and Imre Kátai

Abstract. — In a series of papers, we constructed large families of normal numbers usingthe concatenation of the values of the largest prime factor P (n), as n runs through particularsequences of positive integers. A similar approach using the smallest prime factor function alsoallowed for the construction of normal numbers. Letting ω(n) stand for the number of distinctprime factors of the positive integer n, we then showed that the concatenation of the successivevalues of |ω(n)− blog log nc| in a fixed base q ≥ 2, as n runs through the integers n ≥ 3, yieldsa normal number. Here we prove the following. Let q ≥ 2 be a fixed integer. Given an integern ≥ n0 = max(q, 3), let N be the unique positive integer satisfying qN ≤ n < qN+1 and leth(n, q) stand for the residue modulo q of the number of distinct prime factors of n located inthe interval [log N, N ]. Setting xN := eN , we then create a normal number in base q using theconcatenation of the numbers h(n, q), as n runs through the integers ≥ xn0 .

Résumé. — Dans une série d’articles, nous avons construit de grandes familles de nombresnormaux en utilisant la concaténation des valeurs successives du plus grand facteur premierP (n), où n parcourt certaines suites d’entiers positifs. Une approche similaire en utilisant lafonction plus petit facteur premier nous a aussi permis de construire d’autres familles de nombresnormaux. En désignant par ω(n) le nombre de nombres premiers distincts de n, nous avonsmontré que la concaténation des valeurs successives de |ω(n) − blog log nc| dans une base fixeq ≥ 2, où n parcourt les entiers n ≥ 3, donne place à un nombre normal. Ici, nous démontronsle résultat suivant. Soit q ≥ 2 un entier fixe. Étant donné un entier n ≥ n0 = max(q, 3), soitN l’unique entier positif satisfaisant qN ≤ n < qN+1 et désignons par h(n, q) le résidu moduloq du nombre de facteurs premiers distincts de n situés dans l’intervalle [log N, N ]. En posantxN := eN , nous créons alors un nombre normal dans la base q en utilisant la concaténation desnombres h(n, q), où n parcourt les entiers ≥ xn0 .

Mathematical subject classification (2010). — 11K16, 11N37, 11N41.Key words and phrases. — Normal numbers, number of prime factors.Acknowledgements. — The research of the first author was supported in part by a grant from NSERC. Theresearch of the second author was supported by the Hungarian and Vietnamese TET 10-1-2011-0645.

6 The number of large prime factors of integers and normal numbers

1. IntroductionGiven an integer q ≥ 2, we say that an irrational number η is a q-normal number if theq-ary expansion of η is such that any preassigned sequence of length k ≥ 1, taken within thisexpansion, occurs with the expected limiting frequency, namely 1/qk.Even though constructing specific normal numbers is a very difficult problem, several authorspicked up this challenge. One of the first was Champernowne [2] who, in 1933, showed thatthe number made up of the concatenation of the natural numbers, namely the number

0.123456789101112131415161718192021 . . . ,

is normal in base 10. In 1946, Copeland and Erdős [4] proved that the same is true if onereplaces the sequence of natural numbers by the sequence of primes, namely for the number

0.23571113171923293137 . . .

In the same paper, they conjectured that if f(x) is any nonconstant polynomial whose valuesat x = 1, 2, 3, . . . are positive integers, then the decimal 0.f(1)f(2)f(3) . . ., where f(n) iswritten in base 10, is a normal number. Six years later, Davenport and Erdős [5] proved thisconjecture.Since then, many others have constructed various families of normal numbers. To name onlya few, let us mention Nakai and Shiokawa [15], Madritsch, Thuswaldner and Tichy [14] andfinally Vandehey [17]. More examples of normal numbers as well as numerous references canbe found in the recent book of Bugeaud [1].In a series of papers, we also constructed large families of normal numbers using the distribu-tion of the values of P (n), the largest prime factor function (see [6], [7], [8] and [9]). Recentlyin [10], we showed how the concatenation of the successive values of the smallest prime factorp(n), as n runs through the positive integers, can also yield a normal number. Letting ω(n)stand for the number of distinct prime factors of the positive integer n, we then showed thatthe concatenation of the successive values of |ω(n) − blog lognc| in a fixed base q ≥ 2, as nruns through the integers n ≥ 3, yields a normal number.Given an integer N ≥ 1, for each integer n ∈ JN := (eN , eN+1), let qN (n) be the smallestprime factor of n which is larger than N ; if no such prime factor exists, set qN (n) = 1. Fix aninteger Q ≥ 3 and consider the function f(n) = fQ(n) defined by f(n) = ` if n ≡ ` (mod Q)with (`,Q) = 1 and by f(n) = ε otherwise, where ε stands for the empty word. Then considerthe sequence (κ(n))n≥3 = (κQ(n))n≥3 defined by κ(n) = f(qN (n)) if n ∈ JN with qN (n) > 1and by κ(n) = ε if n ∈ JN with qN (n) = 1. Then, given an integer N ≥ 1 and writingJN = j1, j2, j3, . . ., consider the concatenation of the numbers κ(j1), κ(j2), κ(j3), . . ., thatis define

θN := Concat(κ(n) : n ∈ JN ) = 0.κ(j1)κ(j2)κ(j3) . . . .Then, set αQ := Concat(θN : N = 1, 2, 3, . . .) and let BQ = `1, `2, . . . , `ϕ(Q) be the set ofreduced residues modulo Q, where ϕ stands for the Euler function. In [11], we showed thatαQ is a normal sequence over BQ, that is, the real number 0.αQ is a normal number over BQ.Here we prove the following. Let q ≥ 2 be a fixed integer. Given an integer n ≥ n0 = max(q, 3),let N be the unique positive integer satisfying qN ≤ n < qN+1 and let h(n, q) stand for theresidue modulo q of the number of distinct prime factors of n located in the interval [logN,N ].Setting xN := eN , we then create a normal number in base q using the concatenation of thenumbers h(n, q), as n runs through the integers ≥ xn0 .

Publications mathématiques de Besançon – 2015

J.-M. De Koninck and I. Kátai 7

2. The main resultTheorem 2.1. — Let q ≥ 2 be a fixed integer. Given an integer n ≥ n0 = max(q, 3), let Nbe the unique positive integer satisfying qN ≤ n < qN+1 and let h(n, q) stand for the residuemodulo q of the number of distinct prime factors of n located in the interval [logN,N ]. Foreach integer N ≥ 1, set xN := eN . Then, Concat(h(n, q) : xn0 ≤ n ∈ N) is a q-ary normalsequence.

Proof. — For each integer N ≥ 1, let JN = (xN , xN+1). Further let SN stand for the set ofprimes located in the interval [logN,N ] and TN for the product of the primes in SN . Letn0 = max(q, 3). Given a large integer N , consider the function

(1) f(n) = fN (n) =∑

p|nlogN≤p≤N

1.

Let us further introduce the following sequences:UN = Concat (h(n, q) : n ∈ JN ) ,V∞ = Concat (UN : N ≥ n0) = Concat (h(n, q) : n ≥ xn0) ,Vx = Concat (h(n, q) : xn0 ≤ n ≤ x) .

Let us set Aq := 0, 1, . . . , q− 1. If we fix an arbitrary integer r, it is sufficient to prove thatgiven any particular word w ∈ Arq, the number of occurrences Fw(Vx) of w in Vx satisfies

(2) Fw(Vx) = (1 + o(1)) xqr

(x→∞).

For each integer r ≥ 1, considering the polynomialQr(u) = u(u+ 1) · · · (u+ r − 1).

and lettingρr(d) = #u mod d : Qr(u) ≡ 0 mod d,

it is clear that, since N is large,(3) ρr(p) = r if p ∈ SN .Observe that it follows from the Turán-Kubilius Inequality (see for instance Theorem 7.1 inthe book of De Koninck and Luca [12]), that for some positive constantC,

(4)∑

n∈JN(f(n)− log logN)2 ≤ CeN log logN.

Letting εN = 1/ log log logN , it follows from (4) that

(5) 1xN

#n ∈ JN : |f(n)− log logN | > 1εN

√log logN → 0 (N →∞).

This means that in the estimation of Fw(Vx), we may ignore those integers n appearing inthe concatenation h(2, q)h(3, q) . . . h(bxc, q) for which the corresponding f(n) is “far” fromlog logN in the sense described in (5).Let X be a large number. Then there exists a large integer N such that X

e< xN ≤ X. Letting

L =]X

e,X

], we write



L =]X

e, xN

]∪ ]xN , X] = L1 ∪L2,

say, and λ(Li) for the length of the interval Li for i = 1, 2.Given an arbitrary function δN which tends to 0 arbitrarily slowly, it is sufficient to considerthose L1 and L2 such that(6) λ(L1) ≥ δNX and λ(L2) ≥ δNX.The reason for this is that those n ∈ L1 (resp. n ∈ L2) for which λ(L1) < δNX (resp.λ(L2) < δNX) are o(x) in number and can therefore be ignored in the proof of (2).Let us first consider the set L2. We start by observing that any subword taken in the con-catenation h(n, q)h(n + 1, q) . . . h(n + r − 1, q) is made of co-prime divisors of TN (since notwo members of the sequence h(n, q), h(n+ 1, q), . . . , h(n+ r − 1, q) of r elements may havea common prime divisor p > logN). So, let d0, d1, . . . , dr−1 be co-prime divisors of TN andlet BN (L2; d0, d1, . . . , dr−1) stand for the number of those n ∈ L2 for which dj | n + j forj = 0, 1, . . . , r− 1 and such that

(Qr(n), TN

d0d1 · · · dr−1

)= 1. We can assume that each of the

dj ’s is squarefree, since the number of those n+j ≤ X for which p2 | n+j for some p > logNis X

∑

p>logN

1p2 = o(X).

In light of (4), we may assume that(7) ω(dj) ≤ 2 log logN for j = 0, 1, . . . , r − 1.By using the Eratosthenian sieve (see for instance the book of De Koninck and Luca [12])and recalling that condition (6) ensures that X − xN is large, we obtain that, as N →∞,

BN (L2; d0, d1, . . . , dr−1) = X − xNd0d1 · · · dr−1

∏

p|TN/(d0d1···dr−1)

(1− r

p

)

+o

xNd0d1 · · · dr−1

∏

p|TN/(d0d1···dr−1)

(1− r

p

) .(8)

Letting θN :=∏

p|TN

(1− r

p

), one can easily see that

(9) θN = (1 + o(1))(log logN)r(logN)r (N →∞).

Let us also introduce the strongly multiplicative function κ defined on primes p by κ(p) = p−r.Then, (8) can be written as

(10) BN (L2; d0, d1, . . . , dr−1) = X − xNκ(d0)κ(d1) · · ·κ(dr−1)θN + o

(xN

κ(d0)κ(d1) · · ·κ(dr−1)θN)

as N →∞. For each integer N > ee, let

RN :=[log logN −

√log logNεN

, log logN +√

log logNεN

].



Let `0, `1, . . . , `r−1 be an arbitrary collection of non negative integers < q. Note that thereare qr such collections. Our goal is to count how many times, amongst the integers n ∈ L2,we have f(n+ j) ≡ `j (mod q) for j = 0, 1, . . . , r−1. In light of (5), we only need to considerthose n ∈ L2 for which

f(n+ j) ∈ RN (j = 0, 1, . . . , r − 1).

Let

(11) S (`0, `1, . . . , `r−1) :=∑∗

f(dj)≡`j mod qdj |TN

j=0,1,...,r−1

1κ(d0)κ(d1) · · ·κ(dr−1) ,

where the star over the sum indicates that the summation runs only on those dj satisfyingf(dj) ∈ RN for j = 0, 1, . . . , r − 1.From (10), we therefore obtain that

#n ∈ L2 : f(n+ j) ≡ `j mod q, j = 0, 1, . . . , r − 1= (X − xN )θNS (`0, `1, . . . , `r−1) + o (xNθNS (`0, `1, . . . , `r−1))(12)

as N →∞. Let us now introduce the function

η = ηN =∑

p|TN

1κ(p) .

Observe that, as N →∞,

η =∑

logN≤p≤N

1p(1− r/p) =

∑

logN≤p≤N

1p

+O

∑

logN≤p≤N

1p2

= log logN − log log logN + o(1) +O

( 1logN

)

= log logN − log log logN + o(1).(13)

From the definition (11), one easily sees that

(14) S (`0, `1, . . . , `r−1) = (1 + o(1))∑

tj≡`j mod qtj∈RN

ηt0+t1+···+tr−1

t0!t1! · · · tr−1! (N →∞),

where we ignore in the denominator of the summands the factors κ(p)a (with a ≥ 2) sincetheir contribution is negligible.Moreover, for t ∈ RN , one can easily establish that

ηt+1

(t+ 1)! = (1 + o(1))ηt

t! (N →∞)

and consequently that, for each j ∈ 0, 1, . . . , r − 1,

(15)∑

tj≡`j mod qtj∈RN

ηtj

tj != (1 + o(1))1

q

∑

t∈RN

ηt

t! = (1 + o(1))eη

q(N →∞).



Using (15) in (14), we obtain that

(16) S (`0, `1, . . . , `r−1) = (1 + o(1))eηr

qr(N →∞).

Combining (12) and (16), we obtain that

#n ∈ L2 : f(n+ j) ≡ `j mod q, j = 0, 1, . . . , r − 1

= (X − xN )θNeηr

qr+ o

(xNθN

eηr

qr

)

= X − xNqr

+ o

(xN

1qr

)(N →∞),(17)

where we used (9) and (13).Since the first term on the right hand side of (17) does not depend on the particular collection`0, `1, . . . , `r−1, we may conclude that the frequency of those integers n ∈ L2 for whichf(n + j) ≡ `j (mod q) for j = 0, 1, . . . , r − 1 is the same independently of the choice of`0, `1, . . . , `r−1.The case of those n ∈ L1 can be handled in a similar way.We have thus shown that the number of occurrences of any word w ∈ Arq in h(n, q)h(n +

1, q) . . . h(n+ r − 1, q) as n runs over the bX −X/ec elements of L is (1 + o(1))(X −X/e)qr

.Repeating this for each of the intervals

]X

ej+1 ,X

ej

](j = 0, 1, . . . , blog xc),

we obtain that the number of occurrences of w for n ≤ x is (1 + o(1)) xqr

, as claimed.The proof of (2) is thus complete and the Theorem is proved.

3. Final remarksFirst of all, let us first mention that our main result can most likely be generalized in orderthat the following statement will be true:

Let a(n) and b(n) be two monotonically increasing sequences of n for n = 1, 2, . . . suchthat n/b(n), b(n)/a(n) and a(n) all tend to infinity monotonically as n → ∞. Letf(n) stand for the number of prime divisors of n located in the interval [a(n), b(n)]and let h(n, q) be the residue of f(n) modulo q; then, the sequence h(n, q), n = 1, 2, ...,is a q-ary normal sequence.

Secondly, let us first recall that it was proven by Pillai [16] (with a more general result byDelange [13]) that the values of ω(n) are equally distributed over the residue classes modulo qfor every integer q ≥ 2, and that the same holds for the function Ω(n), where Ω(n) := ∑

pα‖n α.We believe that each of the sequences Concat(ω(n) (mod q) : n ∈ N) and Concat(Ω(n)(mod q) : n ∈ N) represents a normal sequence for each base q = 2, 3, . . .. However, the proof



of these statements could be very difficult to obtain. Indeed, in the particular case q = 2,such a result would imply the famous Chowla conjecture

limx→∞

1x

∑

n≤xλ(n)λ(n+ a1) · · ·λ(n+ ak) = 0,

where λ(n) := (−1)Ω(n) is known as the Liouville function and where a1, a2, . . . , ak are kdistinct positive integers (see Chowla [3]).

Thirdly, we had previously conjectured that, given any integer q ≥ 2 and letting resq(n) standfor the residue of n modulo q, it may not be possible to create an infinite sequence of positiveintegers n1 < n2 < · · · such that

0.Concat(resq(nj) : j = 1, 2, . . .)is a q-normal number. However, we now have succeeded in creating such a monotonic sequence.It goes as follows. Let us define the sequence (mk)k≥1 by

mk = f(k) + k!,where f is the function defined in (1). In this case, we obtain that

mk+1 −mk = k! · k + f(k + 1)− f(k),a quantity which is positive for all integers k ≥ 1 provided(18) f(k + 1)− f(k) > −k! · k,that is if(19) f(k) < k! · k.But since we trivially have

f(k) ≤ ω(k) ≤ 2 log k ≤ k! · k,then (19) follows and therefore (18) as well.Hence, in light of Theorem 2.1, if we choose nk = mk, our conjecture is disproved.

4. AcknowledgmentThe authors are grateful to Professor Andrzej Schinzel, since the problem examined hereoriginated from an email exchange between the second author and Professor Schinzel. Theauthors are also grateful to the referee whose comments helped improve the quality of thispaper.

References[1] Y. Bugeaud, Distribution modulo one and Diophantine approximation. Cambridge Tracts in

Mathematics, 193. Cambridge University Press, Cambridge, 2012. xvi+300 pp.[2] D. G. Champernowne, The construction of decimals normal in the scale of ten, J. London Math.

Soc. 8 (1933), 254-260.[3] S. Chowla, The Riemann Hypothesis and Hilbert’s Tenth Problem, Gordon and Breach, New York,

1965.



[4] A. H. Copeland and P. Erdős, Note on normal numbers, Bull. Amer. Math. Soc. 52 (1946),857-860.

[5] H. Davenport and P. Erdős, Note on normal decimals, Can. J. Math. 4 (1952), 58-63.[6] J. M. De Koninck and I. Kátai, On a problem on normal numbers raised by Igor Shparlinski,

Bulletin of the Australian Mathematical Society 84 (2011), 337–349.[7] J. M. De Koninck and I. Kátai, Some new methods for constructing normal numbers, Annales

des Sciences Mathématiques du Québec 36 (2012), no.2, 349–359.[8] J. M. De Koninck and I. Kátai, Construction of normal numbers using the distribution of the

k-th largest prime factor, Bull. Australian Mathematical Society 88 (2013), 158-168.[9] J. M. De Koninck and I. Kátai, Using large prime divisors to construct normal numbers, Annales

Univ. Sci. Budapest, Sect. Comput. 39 (2013), 45-62.[10] J. M. De Koninck and I. Kátai, Normal numbers generated using the smallest prime factor func-

tion, Annales mathématiques du Québec (to appear).[11] J. M. De Koninck and I. Kátai, Constructing normal numbers using residues of selective prime

factors of integers, Ann. Univ. Sci. Budapest, Sect. Comp., Volume 42, 2014 (to appear).[12] J. M. De Koninck and F. Luca, Analytic Number Theory: Exploring the Anatomy of Integers,

Graduate Studies in Mathematics, Vol. 134, American Mathematical Society, Providence, RhodeIsland, 2012.

[13] H. Delange, Sur la distribution des valeurs de certaines fonctions arithmétiques, Colloque surla Théorie des Nombres, Bruxelles, 1955, pp. 147–161. Georges Thone, Liège; Masson and Cie,Paris, 1956.

[14] M. G. Madritsch, J. M. Thuswaldner and R. F. Tichy, Normality of numbers generated by thevalues of entire functions, J. of Number Theory 128 (2008), 1127-1145.

[15] Y. Nakai and I. Shiokawa, Normality of numbers generated by the values of polynomials at primes,Acta Arith. 81 (1997), no. 4, 345-356.

[16] S. S. Pillai, Generalisation of a theorem of Mangoldt, Proc. Indian Acad. Sci., Sect. A. 11 (1940),13–20.

[17] J. Vandehey, The normality of digits in almost constant additive functions, Monatshefte für Math-ematik, 2012, 1-17.

April 3, 2014

Jean-Marie De Koninck, Département de mathématiques et de statistique, Université Laval,1045 Avenue de la médecine, Québec, Québec G1V 0A6, Canada • E-mail : [email protected] : www.jeanmariedekoninck.mat.ulaval.ca

Imre Kátai, Computer Algebra Department, Eötvös Loránd University, 1117 Budapest, Pázmány PéterSétány I/C, Hungary • E-mail : [email protected]



THE AMPLIFICATION METHOD IN THE GL(3) HECKE ALGEBRA

by

Roman Holowinsky, Guillaume Ricotta and Emmanuel Royer

Abstract. — This article contains all the technical ingredients required to implement an ef-fective, explicit and unconditional amplifier in the context of GL(3) automorphic forms. Inparticular, several coset decomposition computations in the GL(3) Hecke algebra are explicitlydone.

Résumé. — Cet article contient tous les ingrédients techniques nécessaires à la mise en placeefficace, explicite et inconditionnelle de la méthode d’amplification dans le cadre des formesautomorphes de GL(3). En particulier, il y est donné plusieurs décompositions explicites desystèmes de représentants dans l’algèbre de Hecke de GL(3).

Contents1. Introduction and statement of the results 141.1. Motivation: sup-norms of GL(n) Hecke-Maass cusp forms 141.2. The GL(2) and GL(3) amplifier 141.3. Statement of the results 161.4. Organization of the paper 182. Background on the GL(3) Hecke algebra 193. Proof of the linearizations given in Theorem A 213.1. Linearization of Λdiag(1, 1, p)Λ ∗ Λdiag(1, p, p)Λ 213.2. Linearization of Λdiag(1, p, p2)Λ ∗ Λdiag(1, p, p2)Λ 24Appendix A. Decomposition of Λ-double cosets into Λ-cosets 28A.1. Decomposition and degree of Λdiag(1, 1, p)Λ 29A.2. Decomposition and degree of Λdiag(1, p, p)Λ 30A.3. Decomposition and degree of Λdiag(1, p, p2)Λ 30A.4. Degree of Λdiag(1, p2, p4)Λ 32A.5. Degree of Λdiag(1, p3, p3)Λ 36A.6. Degree of Λdiag(1, 1, p3)Λ 37References 39

Mathematical subject classification (2010). — 11F99, 11F60, 11F55.Key words and phrases. — Amplification method, Hecke operators, Hecke algebras.

14 The amplification method in the GL(3) Hecke algebra

1. Introduction and statement of the results1.1. Motivation: sup-norms of GL(n) Hecke-Maass cusp forms. — Let f be a GL(n)Maass cusp form and K be a fixed compact subset of SLn(R)/SOn(R) (see [Gol06]). Thegeneric or local bound for the sup-norm of f restricted to K is given by

||f |K ||∞ λn(n−1)/8f

where λf is the Laplace eigenvalue of f (see [Sar]). Note that F. Brumley and N. Templiernoticed in [BT] that the previous bound does not hold when n > 6 if f is not restricted to acompact.If f is an eigenform of the Hecke algebra, however, then the generic bound is not expected tobe the correct order of magnitude for the sup-norm of the restriction of f to a fixed compact.This is essentially due to the fact that the Hecke operators are additional symmetries on theambient space. In other words, we expect there to exist an absolute positive constant δn > 0such that(1.1) ||f |K ||∞ λ

n(n−1)/8−δnf .

H. Iwaniec and P. Sarnak proved the bound given in (1.1) in [IS95] when n = 2 for δ2 = 1/24.Note that improving this constant δ2 seems to be a very delicate open problem. The casen = 3 was completed by the authors in [HRR] with δ3 = 1/124. The general case was done,without an explicit value for δn, in a series of recent impressive works by V. Blomer andP. Maga in [BMb] and in [BMa].All of the above achievements (and much more) were made possible thanks to generalizationsof the amplification method developed by W. Duke, J. Friedlander and H. Iwaniec for GL(1)and GL(2) (see [FI92], [Iwa92] and [DFI94] for example). In particular, the proof of (1.1) forn = 3 with δ3 = 1/124 relies on Theorem B of this article which was stated without proof in[HRR] as Proposition 4.11. For the sake of completeness and future use, we provide the fulldetails of the proof of Theorem B, including computations, here in this article.

1.2. The GL(2) and GL(3) amplifier. —The general principle behind the constructionof an amplifier, is the existence of an identity which allows one to write a non-zero constantas a finite sum of Hecke eigenvalues. In the most basic context of GL(2) automorphic forms,this identity is(1.2) λf (p)2 − λf (p2) = 1where p is any prime and λf (n) is the n-th Hecke eigenvalue of a Hecke-Maass cusp form fof full level, i.e. Tnf = λf (n)f where

(Tnf)(z) = 1√n

∑

ad=n

d∑

b=1f

(az + b

d

).

One may interpret the above identity as the fact that the Rankin-Selberg convolution factorsas the product of the adjoint square and the Riemann zeta function and therefore has a poleat s = 1.1Theorem B and Proposition 4.1 in [HRR] are not entirely identical. Since releasing our first article [HRR],we have noticed a simplification in the construction of the amplifier. Therefore, Theorem B only contains theidentities in Proposition 4.1 of [HRR] which are necessary for the amplification method. The implied powersaving in the Laplace eigenvalue for the sup-norm bound remains the same.


R. Holowinsky and G. Ricotta and E. Royer 15

From the identity (1.2), one constructs an amplifier

Af :=∣∣∣∣∣∑

`

α`λf (`)∣∣∣∣∣

2

with

α` :=

λf0(`) if ` 6√L is a prime number,

−1 if ` 6 L is the square of a prime number,0 otherwise

for some fixed form f0. The advantage to constructing such an amplifier is that it is expectedto be small2 for general forms f while satisfying the lower bound Af ε L

1−ε when f = f0.Reinterpreting (1.2) in terms of Hecke operators, we may write

Tp Tp − Tp2 = Id.

In application to the sup-norm problem for GL(2) via a pre-trace formula argument, thistranslates into a need to geometrically understand the behavior of the following collection ofoperators on an automorphic kernel

Tp, Tp T ∗q , Tp T ∗q2 and Tp2 T ∗q2

both in the cases of primes p = q and p 6= q. Since the Hecke operators Tn in GL(2) are self-adjoint and computationally pleasant to work with due to their relatively simple compositionlaw, one quickly computes that the above collection of Hecke operators may be reduced tothe study of

Tp, Tpq, Tpq2 and Tp2q2 .

In truth, one has an opportunity to further reduce the collection of necessary Hecke operatorsthrough the simple inequality

(1.3) Af 6 2∣∣∣∣∣∑

p

αpλf (p)∣∣∣∣∣

2

+ 2∣∣∣∣∣∑

p

αp2λf (p2)∣∣∣∣∣

2

.

Indeed, an appropriate application of (1.3) (see for example [BHM]) allows one to furtherrestrict the set of necessary Hecke operators to

Tpq and Tp2q2

both in the cases of primes p = q and p 6= q.The case of GL(3) is much more computationally involved due to the lack of self-adjointnessof the Hecke operators and their multiplication law. Instead of looking at identities involvingHecke eigenvalues, we start immediately with the Hecke operators themselves (see §2 fordefinitions). Our fundamental identity now will be

(1.4) Tdiag(1,p,p) Tdiag(1,1,p) − Tdiag(1,p,p2) = (p2 + p+ 1)Id.

We set cf (p) = af (p, 1), cf (p)∗ = af (p, 1) and cf (p2) to be the eigenvalues of p−1Tdiag(1,1,p) =Tp, p−1Tdiag(1,p,p) = T ∗p and p−2Tdiag(1,p,p2) respectively when acting on a form f . See (2.3)

2This follows, at least conditionally, from a suitable version of the Generalized Riemann Hypothesis. However,we do not need to use this fact.



and (2.4) for the precise definitions. We construct the amplifier

Af :=∣∣∣∣∣∑

`

α`cf (`)∣∣∣∣∣

2

with3

α` :=

cf0(`)∗ if ` 6√L is a prime number,

−1 if ` 6 L is the square of a prime number,0 otherwise.

As in the case of GL(2), this amplifier will satisfy Af0 ε L1−ε and Af is otherwise expected

to be small for f 6= f0.Applying the inequality

(1.5) Af 6 2∣∣∣∣∣∑

p

αpcf (p)∣∣∣∣∣

2

+ 2∣∣∣∣∣∑

p

αp2cf (p2)∣∣∣∣∣

2

,

one is reduced to understanding the actions of

Tdiag(1,1,p) Tdiag(1,q,q) and Tdiag(1,p,p2) Tdiag(1,q,q2)

both in the cases of primes p = q and p 6= q on the relevant automorphic kernel. In thefollowing sections, we compute the above compositions as linear combinations of other Heckeoperators and state our main result as Theorem B. In the end, we shall see that the followingoperators are the relevant ones for our application

Tdiag(1,p,pq), Tdiag(1,pq,p2q2), Tdiag(1,p3,p3) and Tdiag(1,1,p3)

for primes p = q and p 6= q.

1.3. Statement of the results. —

Theorem A. — Let p be a prime number and Λ = GL3(Z).

– The set R1,1,p (respectively R1,p,p, R1,p,p2) defined in Proposition A.1 (respectively Propo-sition A.2, Proposition A.3) is a complete system of representatives for the distinct Λ-right cosets in the Λ-double coset of diag(1, 1, p) (respectively diag(1, 1, p), diag(1, p, p2))modulo Λ.

3One may also choose a variant, in the spirit of [Ven10], which involves the signs of cf0 (`).



– The following formulas for the degrees of Λ-double cosets hold.

deg (diag(1, 1, p)) = p2 + p+ 1,deg (diag(1, p, p)) = p2 + p+ 1,

deg(diag(1, p, p2)

)= p(p+ 1)(p2 + p+ 1),

deg (diag(p, p, p)) = 1,deg

(diag(1, p2, p4)

)= p5(p+ 1)(p2 + p+ 1),

deg(diag(1, p3, p3)

)= p4(p2 + p+ 1),

deg(diag(p, p, p4)

)= p4(p2 + p+ 1),

deg(diag(p, p2, p3)

)= p(p+ 1)(p2 + p+ 1),

deg(diag(p2, p2, p2)

)= 1.

– Finally,

(1.6) Λdiag(1, 1, p)Λ ∗ Λdiag(1, p, p)Λ = Λdiag(1, p, p2)Λ + (p2 + p+ 1)Λdiag(p, p, p)Λ.

and

(1.7) Λdiag(1, p, p2)Λ ∗ Λdiag(1, p, p2)Λ = Λdiag(1, p2, p4)Λ + (p+ 1)Λdiag(1, p3, p3)Λ+ (p+ 1)Λdiag(p, p, p4)Λ + (p+ 1)(2p− 1)Λdiag(p, p2, p3)Λ

+ p(p+ 1)(p2 + p+ 1)Λdiag(p2, p2, p2)Λ.

Remark 1.1. — In [Kod67], T. Kodama explicitely computed the product of other doublecosets in the slightly harder case of the Hecke ring for the symplectic group. The results statedin the previous theorem are similar in spirit.

Remark 1.2. — It is well-known that a Λ-double coset can be identified with its character-istic function χ. Under this identification, the multiplication law between Λ-double cosets isthe classical convolution between functions. If µ = (µ1, µ2, µ3) with µ1 > µ2 > µ3 > 0 andν = (ν1, ν2, ν3) with ν1 > ν2 > ν3 > 0 are two partitions of length less than n, then

χΛdiag(pµ1 ,pµ2 ,pµ3 )Λ ∗ χΛdiag(pν1 ,pν2 ,pν3 )Λ =∑

λ

gλµ,ν(p)χΛdiag(pλ1 ,pλ2 ,pλ3)Λ

for any prime number p where λ ranges over the partitions of length less than n and the gλµ,ν(p)are the Hall polynomials (see [Mac95, Equation (2.6) Page 295]). The sum on the right-handside of the above equality is finite since only a finite number of the Hall polynomials arenon-zero. However, determining which Hall polynomials vanish is not straightforward (see



[Mac95, Equation (4.3) Page 188]). Using Sage, one can check that

g(4,2,0)(2,1,0),(2,1,0)(p) = 1,

g(3,3,0)(2,1,0),(2,1,0)(p) = p+ 1,

g(4,1,1)(2,1,0),(2,1,0)(p) = p+ 1,

g(3,2,1)(2,1,0),(2,1,0)(p) = (p+ 1)(2p− 1),

g(2,2,2)(2,1,0),(2,1,0)(p) = p(p+ 1)(p2 + p+ 1)

and one can recover the coefficients occurring in (1.7). We prefer to give a different proof,which has the advantage of producing explicit systems of representatives for the Λ-right cosetsand formulas for the degrees.

Corollary B. — If p and q are two prime numbers then(1.8) Tdiag(1,p,p) Tdiag(1,1,q) = Tdiag(1,p,pq) + δp=q(p2 + p+ 1)Idand

(1.9) Tdiag(1,p,p2) Tdiag(1,q,q2) = Tdiag(1,pq,p2q2) + δp=q(p+ 1)(Tdiag(1,p3,p3) + Tdiag(1,1,p3)

)

+ δp=q(p+ 1)(2p− 1)Tdiag(1,p,p2) + δp=qp(p+ 1)(p2 + p+ 1)Id.

When p 6= q, the previous corollary immediately follows from (2.9) and (2.10). When p = q,it is a consequence of the previous theorem and of (2.9).

Remark 1.3. — As observed by L. Silberman and by A. Venkatesh in [SA] and used byV. Blomer and P. Maga in [BMb] and in [BMa], the precise formulas for the Hall polynomialsoccurring in (1.8) and in (1.9) are not really needed for the purpose of the amplificationmethod, since the Hall polynomials are easily well approximated for p and q large by themuch easier Schur polynomials. Nevertheless, the precise list of the Hecke operators relevantfor the amplification method, namely occurring in (1.8) and in (1.9), seems to be crucial inorder to obtain the best possible explicit result. For instance, G. Harcos and N. Templier usedsuch a list in order to prove the best known subconvexity exponent for the sup-norm of GL(2)automorphic forms in the level aspect in [HT13].

1.4. Organization of the paper. —The general background on GL(3) Maass cusp formsand on the GL(3) Hecke algebra is given in Section 2. The linearizations involved in Theo-rem A are detailed in Section 3. The proof requires decompositions of Λ-double cosets intoΛ-left and right cosets and computations of degrees as done in Appendix A.

Notations. — Λ stands for the group GL3(Z) of 3 × 3 invertible matrices with integercoefficients. If g is a 3× 3 matrix with real coefficients then tg stands for its transpose. Forg ∈ GL3(Q) we let Tg denote the Hecke operator associated to g (see §2). If a, b and c arethree rational numbers then

– diag(a, b, c) denotes the diagonal 3× 3 matrix with a, b and c as diagonal coefficients;

– La,b,c (respectively Ra,b,c) stands for a system of representatives for the decomposition ofthe Λ-double coset Λdiag(a, b, c)Λ into Λ-left (respectively right) cosets.



Acknowledgements. The authors would like to thank the referees for a very careful readingof the manuscript.They also want to thank J. Cogdell, É. Fouvry, E. Kowalski and P. Michel for fruitful discus-sions related to this work.This article was worked out at several places: Université Blaise Pascal (Clermont-Ferrand), theDepartment of Mathematics of The Ohio State University (Columbus), École PolytechniqueFédérale de Lausanne, Eidgenossische Technische Hochschule (Zurich), Forschungsinstitutfur Mathematik (Zurich). The authors would like to thank all of these institutions for theirhospitality and inspiring working conditions.The research of R. Holowinsky was partially supported by the Sloan fellowship BR2011-083and the NSF grant DMS-1068043. In-person collaboration with the other authors was madepossible through this funding.The research of G. Ricotta was partially supported by a Marie Curie Intra European Fel-lowship within the 7th European Community Framework Programme. The grant agreementnumber of this project, whose acronym is ANERAUTOHI, is PIEF-GA-2009-251271. Hewould like to thank ETH and its entire staff for the excellent working conditions. Last butnot least, the second author would like to express his gratitude to K. Belabas for his crucialbut isolated support for Analytic Number Theory among the Number Theory research teamA2X (Institut de Mathématiques de Bordeaux, Université de Bordeaux).

2. Background on the GL(3) Hecke algebraConvenient references for this section include [AZ95], [Gol06], [New72] and [Shi94].Let f be a GL(3) Maass cusp form of full level. Such f admits a Fourier expansion(2.1)

f(g) =∑

γ∈U2(Z)\SL2(Z)

∑

m1>1m2∈Z\0

af (m1,m2)m1|m2|

WJa

m1|m2|

m11

(γ

1

)g, νf , ψ1, m2

|m2|

for g ∈ GL3(R) (see [Gol06, Equation (6.2.1)]). Here U2(Z) stands for the Z-points of thegroup of upper-triangular unipotent 2× 2 matrices. νf ∈ C2 is the type of f , whose compo-nents are complex numbers characterized by the property that, for every invariant differentialoperator D in the center of the universal enveloping algebra of GL3(R), the cusp form f isan eigenfunction of D with the same eigenvalue as the power function Iνf , which is defined in[Gol06, Equation (5.1.1)]. ψ1,±1 is the character of the group of upper-triangular unipotentreal 3× 3 matrices defined by

ψ1,±1

1 u1,2 u1,31 u2,3

1

= e2iπ(u2,3±u1,2).

WJa(∗, νf , ψ1,±1) stands for the GL(3) Jacquet Whittaker function of type νf and characterψ1,±1 defined in [Gol06, Equation 6.1.2]. The complex number af (m1,m2) is the (m1,m2)-thFourier coefficient of f for m1 a positive integer and m2 a non-vanishing integer.For g ∈ GL3(Q), the Hecke operator Tg is defined by

Tg(f)(h) =∑

δ∈Λ\ΛgΛf(δh)



for h ∈ GL3(R) (see [AZ95, Chapter 3, Sections 1.1 and 1.5]). The degree of g or Tg definedby

deg(g) = deg(Tg) = card (Λ \ ΛgΛ)is scaling invariant, in the sense that(2.2) deg(rg) = deg(g)for r ∈ Q×. The adjoint of Tg for the Petersson inner product is Tg−1 . The algebra of Heckeoperators T is the ring of endomorphisms generated by all the Tg’s with g ∈ GL3(Q), acommutative algebra of normal endomorphisms (see [Gol06, Theorem 6.4.6]), which containsthe m-th normalized Hecke operator

(2.3) Tm = 1m

∑

g=diag(y1,y2,y3)y1|y2|y3y1y2y3=m

Tg

for all positive integers m. A Hecke-Maass cusp form f of full level is a Maass cusp form offull level, which is an eigenfunction of T. In particular, it satisfies(2.4) Tm(f) = af (m, 1)f and T ∗m(f) = af (1,m)f = af (m, 1)faccording to [Gol06, Theorem 6.4.11].The algebra T is isomorphic to the absolute Hecke algebra, the free Z-module generated bythe double cosets ΛgΛ where g ranges over Λ \ GL3(Q)/Λ and endowed with the followingmultiplication law. If g1 and g2 belong to GL3(Q) and

Λg1Λ =deg(g1)⋃

i=1Λαi and Λg2Λ =

deg(g2)⋃

j=1Λβj

then(2.5) Λg1Λ ∗ Λg2Λ =

∑

ΛhΛ⊂Λg1Λg2Λm(g1, g2;h)ΛhΛ

where h ∈ GL3(Q) ranges over a system of representatives of the Λ-double cosets containedin the set Λg1Λg2Λ and

m(g1, g2;h) = card ((i, j) ∈ 1, . . . ,deg(g1) × 1, . . . ,deg(g2), αiβj ∈ Λh) ,(2.6)

= 1deg(h)card ((i, j) ∈ 1, . . . ,deg(g1) × 1, . . . ,deg(g2), αiβj ∈ ΛhΛ) ,(2.7)

= deg(g2)deg(h) card (i ∈ 1, . . . ,deg(g1), αig2 ∈ ΛhΛ) .(2.8)

Confer [AZ95, Lemma 1.5 Page 96]. In particular,(2.9) Λdiag(r, r, r)Λ ∗ ΛgΛ = ΛrgΛfor g ∈ GL3(Q) and r ∈ Q× ([AZ95, Lemma 2.4 Page 107]). In addition, for p and q twodistinct prime numbers,(2.10) Λdiag(1, pα1 , pα2)Λ ∗ Λdiag(1, qβ1 , qβ2)Λ = Λdiag(1, pα1qβ1 , pα2qβ2)Λwhere α1, α2, β1, β2 are non-negative integers by [AZ95, Proposition 2.5 Page 107].



Every double coset ΛgΛ with g in GL3(Q) contains a unique representative of the form(2.11) rdiag(1, s1(g), s2(g))where r ∈ Q∗ and s1(g), s2(g) are some positive integers satisfying s1(g) | s2(g) (see [AZ95,Lemma 2.2]).Finally, let g = [gi,j ]16i,j63 be a 3 × 3 matrix with integer coefficients. Its determinantaldivisors are the non-negative integers given by

d1(g) = gcd(gi,j , 1 6 i, j 6 3),d2(g) = gcd(determinants of 2× 2 submatrices of g),d3(g) = |det(g)|.

and its determinantal vector is d(g) = (d1(g), d2(g), d3(g)). The determinantal divisors turnout to be useful since if h is another 3× 3 matrix with integer coefficients then h belongs toΛgΛ if and only if dk(h) = dk(g) for 1 6 k 6 3 (see [New72]).

3. Proof of the linearizations given in Theorem A3.1. Linearization of Λdiag(1, 1, p)Λ ∗ Λdiag(1, p, p)Λ. —This section contains the proofof (1.6).By (2.5), the product of these double cosets equals

∑

ΛhΛ⊂Λdiag(1,1,p)Λdiag(1,p,p)Λm

11

p

,

1p

p

;h

ΛhΛ

where h ∈ GL3(Q) ranges over a system of representatives of the Λ-double cosets containedin the set

Λ

11

p

Λ

1p

p

Λ.

Let us determine the matrices h occuring in this sum. Let h in GL3(Q) be such that ΛhΛ isincluded in the previous set. By (2.11), one has uniquely

ΛhΛ = Λελ1λ2

diag(1, s1, s2)Λ

with ε = ±1, λ1, λ2 > 0, (λ1, λ2) = 1, s1, s2 > 0, s1 | s2. The inclusion is equivalent toΛελ1diag(1, s1, s2)Λ = Λλ2δ1δ2Λ

for some matrices δ1 ∈ R1,1,p and δ2 ∈ L1,p,p by (A.3) and (A.5). So, both matrices have thesame determinantal divisors ie

ελ1 = λ2d1(δ1δ2),λ2

1s1 = λ22d2(δ1δ2),

ελ31s1s2 = λ3

2d3(δ1δ2) = λ32p

3.

One can check that the setδ1δ2, (δ1, δ2) ∈ R1,1,p × L1,p,p



is made exactly of the matricesp

pp

d(δ1δ2) = (p, p2, p3),

p d1 +D1

pp

d(δ1δ2) = ((p, d1 +D1), p(p, d1 +D1), p3),

p e1 + E1

p f1 + F1p

d(δ1δ2) = ((p, e1 + E1, f1 + F1), p(p, e1 + E1, f1 + F1), p3)

andp2 pE1

p F11

d(δ1δ2) = (1, p, p3),

p2 pD1

1p

d(δ1δ2) = (1, p, p3),

1 pd1p2

p

d(δ1δ2) = (1, p, p3),

p pd1 d1F1 + E1

p2 pF11

d(δ1δ2) = (1, p, p3),

1 pe1p pf1

p2

d(δ1δ2) = (1, p, p3),

p d2 pe1

1 pf1p2

d(δ1δ2) = (1, p, p3)

with 0 6 d1, e1, f1, D1, E1, F1 < p. As a consequence, only two cases can occur since

d(δ1δ2) ∈ (1, p, p3), (p, p2, p3).

First case: d(δ1δ2) = (1, p, p3).

ελ1 = λ2,

λ21s1 = λ2

2p,

ελ31s1s2 = λ3

2p3.

The first equation gives ε = λ1 = λ2 = 1 by the coprimality of λ1 and λ2. The secondequation gives s1 = p. The third equation gives s2 = p2. Thus,

ΛhΛ = Λdiag(1, p, p2)Λ.



Second case: d(δ1δ2) = (p, p2, p3).

ελ1 = λ2p,

λ21s1 = λ2

2p2,

ελ31s1s2 = λ3

2p3.

The first equation gives ε = λ2 = 1 and λ1 = p by the coprimality of λ1 and λ2. The secondequation gives s1 = 1. The third equation gives s2 = 1. Thus,

ΛhΛ = Λdiag(p, p, p)Λ.

As a consequence,

Λdiag(1, 1, p)Λ ∗ Λdiag(1, p, p)Λ = m1Λ

1p

p2

Λ +m2Λ

p

pp

Λ

where

m1 := m

11

p

,

1p

p

;

1p

p2

,

m2 := m

11

p

,

1p

p

;

p

pp

.

Let us compute the value of m2 first. By (2.6), (A.6) and (2.2),

m2 = (p2 + p+ 1)

∣∣∣∣∣∣

δ1 ∈ R1,1,p, δ1

1p

p

∈ Λ

p

pp

Λ

∣∣∣∣∣∣.

Let us compute the remaining cardinality. One can check that the setδ1

1p

p

, δ1 ∈ R1,1,p

is exactly made of the matricesp

pp

(d1, d2, d3) = (p, p2, p3),

1 pd1p2

p

(d1, d2, d3) = (1, p, p3),

1 0 pe1p pf1

p2

(d1, d2, d3) = (1, p, p3)



with 0 6 d1, e1, f1 < p. The fact that the determinantal vector of diag(p, p, p) is (p, p2, p3)implies that

δ1 ∈ R1,1,p, δ1

1p

p

∈ Λ

p

pp

Λ

=

p

pp

and is of cardinality 1 such that m2 = p2 + p+ 1.Now, let us compute the value of m1. By (2.6), (A.6) and (A.8),

m1 = 1p(p+ 1)

∣∣∣∣∣∣

δ1 ∈ R1,1,p, δ1

1p

p

∈ Λ

1p

p2

Λ

∣∣∣∣∣∣.

Let us compute the remaining cardinality. Both the analysis done for m2 and the fact thatthe determinantal vector of diag(1, p, p2) is (1, p, p2) imply thatδ1 ∈ R1,1,p, δ1

1p

p

∈ Λ

1p

p2

Λ

=

⋃

06d1<p

1 d1p

1

⋃

06e1,f1<p

1 e11 f1

p

which is of cardinality p(p+ 1) such that m1 = 1.

3.2. Linearization of Λdiag(1, p, p2)Λ∗Λdiag(1, p, p2)Λ. —This section contains the proofof (1.7).By (2.5), the product of these double cosets equals

∑

ΛhΛ⊂Λdiag(1,p,p2)Λdiag(1,p,p2)Λm

1p

p2

,

1p

p2

;h

ΛhΛ

where h ∈ GL3(Q) ranges over a system of representatives of the Λ-double cosets containedin the set

Λ

1p

p2

Λ

1p

p2

Λ.

Let us determine the relevant matrices h occuring in this sum. Let h in GL3(Q) be such thatΛhΛ is included in the previous set. By (2.11), one has uniquely

ΛhΛ = Λελ1λ2

diag(1, s1, s2)Λ

with ε = ±1, λ1, λ2 > 0, (λ1, λ2) = 1, s1, s2 > 0, s1 | s2. The inclusion is equivalent to

Λελ1diag(1, s1, s2)Λ = Λλ2δ1δ2Λ



for some matrices δ1 ∈ R1,p,p2 and δ2 ∈ L1,p,p2 by (A.7). So, both matrices have the samedeterminantal divisors ie

ελ1 = λ2d1(δ1δ2),λ2

1s1 = λ22d2(δ1δ2),

ελ31s1s2 = λ3

2d3(δ1δ2) = λ32p

6.

By (A.7), a straightforward but tedious computation ensures that the set

d(δ1δ2), (δ1, δ2) ∈ R1,p,p2 × L1,p,p2

is a subset of(1, p2, p6), (1, p3, p6), (p, p2, p6), (p, p3, p6), (p2, p4, p6).

Case 1: (d1, d2, d3) = (1, p2, p6).

ελ1 = λ2,

λ21s1 = λ2

2p2,

ελ31s1s2 = λ3

2p6.

The first equation gives ε = λ1 = λ2 = 1 by the coprimality of λ1 and λ2. The secondequation gives s1 = p2. The third equation gives s2 = p4. Thus,

ΛhΛ = Λdiag(1, p2, p4)Λ.

Case 2: (d1, d2, d3) = (1, p3, p6).

ελ1 = λ2,

λ21s1 = λ2

2p3,

ελ31s1s2 = λ3

2p6.

The first equation gives ε = λ1 = λ2 = 1 by the coprimality of λ1 and λ2. The secondequation gives s1 = p3. The third equation gives s2 = p3. Thus,

ΛhΛ = Λdiag(1, p3, p3)Λ.

Case 3: (d1, d2, d3) = (p, p2, p6).

ελ1 = λ2p,

λ21s1 = λ2

2p2,

ελ31s1s2 = λ3

2p6.

The first equation gives ε = λ2 = 1 and λ1 = p by the coprimality of λ1 and λ2. The secondequation gives s1 = 1. The third equation gives s2 = p3. Thus,

ΛhΛ = Λdiag(p, p, p4)Λ.

Case 4: (d1, d2, d3) = (p, p3, p6).

ελ1 = λ2p,

λ21s1 = λ2

2p3,

ελ31s1s2 = λ3

2p6.



The first equation gives ε = λ2 = 1 and λ1 = p by the coprimality of λ1 and λ2. The secondequation gives s1 = p. The third equation gives s2 = p2. Thus,

ΛhΛ = Λdiag(p, p2, p3)Λ.

Case 5: (d1, d2, d3) = (p2, p4, p6).

ελ1 = λ2p2,

λ21s1 = λ2

2p4,

ελ31s1s2 = λ3

2p6.

The first equation gives ε = λ2 = 1 and λ1 = p2 by the coprimality of λ1 and λ2. The secondequation gives s1 = 1. The third equation gives s2 = 1. Thus,

ΛhΛ = Λdiag(p2, p2, p2)Λ.

As a consequence,

Λdiag(1, p, p2)Λ ∗ Λdiag(1, p, p2)Λ = m1Λ

1p2

p4

Λ +m2Λ

1p3

p3

Λ

+m3Λ

p

pp4

Λ +m4Λ

p

p2

p3

Λ +m5Λ

p2

p2

p2

Λ

where

m1 := m

1p

p2

,

1p

p2

;

1p2

p4

,

m2 := m

1p

p2

,

1p

p2

;

1p3

p3

,

m3 := m

1p

p2

,

1p

p2

;

p

pp4

,

m4 := m

1p

p2

,

1p

p2

;

p

p2

p3

,

m5 := m

1p

p2

,

1p

p2

;

p2

p2

p2

.

Let us compute the value of m1. By (2.6), (A.8) and (A.9),

m1 = 1p4

∣∣∣∣∣∣

δ1 ∈ R1,p,p2 , δ1

1p

p2

∈ Λ

1p2

p4

Λ

∣∣∣∣∣∣.



Let us compute the remaining cardinality. One can check that the setδ1

1p

p2

, δ1 ∈ R1,p,p2

is exactly made of the matricesp2

p p2f1p3

(d1, d2, d3) = (p, p3, p6),

p pd2

p3

p2

(p | d2) (d1, d2, d3) = (p, p3, p6),

p pd1 p2e1

p2 p2f1p3

(d1f1 = 0, (d1, e1, f1) 6= (0, 0, 0)) (d1, d2, d3) = (p, p3, p6)

and

1 pd1 p2e2p2 p2f2

p4

(p | f2) (d1, d2, d3) = (1, p2, p6)

and

1 pd2 p2e1p3

p3

(d1, d2, d3) = (1, p3, p6)

andp p2e2

p p2f2p4

(p | e2) (d1, d2, d3) = (p, p2, p6)

andp2

p2

p2

(d1, d2, d3) = (p2, p4, p6)

where 0 6 d1, e1, f1 < p and 0 6 d2, e2, f2 < p2. The fact that the determinantal vector ofdiag(1, p2, p4) is (1, p2, p6) implies that m1 = 1.Let us compute the value of m2. By (2.6), (A.8) and (A.10),

m2 = p+ 1p3

∣∣∣∣∣∣

δ1 ∈ R1,p,p2 , δ1

1p

p2

∈ Λ

1p3

p3

Λ

∣∣∣∣∣∣.

Both the analysis done for m1 and the fact that the determinantal vector of diag(1, p3, p3) is(1, p3, p6) imply that m2 = p+ 1.



Let us compute the value of m3. By (2.6), (A.8), (2.2) and (A.11),

m3 = p+ 1p3

∣∣∣∣∣∣

δ1 ∈ R1,p,p2 , δ1

1p

p2

∈ Λ

p

pp4

Λ

∣∣∣∣∣∣.

Both the analysis done for m1 and the fact that the determinantal vector of diag(p, p, p4) is(p, p2, p6) imply that m3 = p+ 1.Let us compute the value of m4. By (2.6), (A.8), (2.2) and (A.8),

m4 = p+ 1p3

∣∣∣∣∣∣

δ1 ∈ R1,p,p2 , δ1

1p

p2

∈ Λ

p

p2

p3

Λ

∣∣∣∣∣∣.

Both the analysis done for m1 and the fact that the determinantal vector of diag(p, p2, p3) is(p, p3, p6) imply that m4 = (p+ 1)(2p− 1).Let us compute the value of m5. By (2.6), (A.8) and (2.2),

m5 = p(p+ 1)(p2 + p+ 1)

∣∣∣∣∣∣

δ1 ∈ R1,p,p2 , δ1

1p

p2

∈ Λ

p2

p2

p2

Λ

∣∣∣∣∣∣.

Both the analysis done for m1 and the fact that the determinantal vector of diag(p2, p2, p2)is (p2, p4, p6) imply that m5 = p(p+ 1)(p2 + p+ 1).

Appendix A. Decomposition of Λ-double cosets into Λ-cosetsBy [AZ95, Lemma 1.2 Page 94 and Lemma 2.1 Page 105], we know that every Λ-double cosetΛgΛ with g in GL3(Q) with integer coefficients is both a finite union of Λ-left cosets andΛ-right cosets. In addition, every Λ-right coset Λg contains a unique upper-triangular columnreduced representative, namely

Λg = Λ

a d e

b fc

where 0 6 d < b and 0 6 e, f < c by [AZ95, Lemma 2.7 Page 109].As a consequence, every Λ-left coset gΛ contains a unique upper-triangular row reducedrepresentative, namely

gΛ =

a d e

b fc

Λ

where 0 6 d, e < a and 0 6 f < b. More explicitely, if UW tgW = H is the upper-triangularcolumn reduced representative of the Λ-right coset ΛW tgW with W the anti-diagonal matrixwith 1’s on the anti-diagonal then gW tUW = W tHW is the upper-triangular row reducedrepresentative of the Λ-left coset gΛ.The previous fact also entails that

(A.1) ΛgΛ =⋃

δ∈RgΛδ ⇒ ΛgΛ =

⋃

δ∈W tRgW

δΛ



sinceΛgΛ = WΛgΛ = W t (ΛgΛ) = W

⋃

δ∈Rg

tδΛ =⋃

δ∈RgW tδWΛ.

Let us finish with a useful elementary practical remark for the computations done in thefollowing sections of the appendix. If H is an upper-triangular column reduced matrix ina Λ-double coset Λdiag(pα1 , pα2 , pα3)Λ where p is a prime number and α1, α2 and α3 arenon-negative integers then

(A.2) H =

pδ1 ∗ ∗

pδ2 ∗pδ3

,

3∑

j=1(αj − δj) = 0,∀j ∈ 1, 2, 3, 0 6 δj 6 max

16k63αk.

The fact that the diagonal cofficients of H are powers of p comes from the determinantequation. The condition on the exponents of these diagonal coefficients follows from the factthat pmax αk,16k63H−1 has integer coefficients.

A.1. Decomposition and degree of Λdiag(1, 1, p)Λ. —

Proposition A.1. — One has

(A.3) Λdiag(1, 1, p)Λ =⋃

δ∈R1,1,p

Λδ =⋃

δ∈L1,1,p

δΛ

where

R1,1,p = diag(p, 1, 1)⋃

06d1<p

1 d1p

1

⋃

06e1,f1<p

1 0 e11 f1

p

and

L1,1,p = diag(1, 1, p)⋃

06f1<p

1p f1

1

⋃

06d1,e1<p

p d1 e1

11

.

In particular,

(A.4) deg (diag(1, 1, p)) = p2 + p+ 1.

Proof of Proposition A.1. — The decomposition into Λ-right cosets implies the decomposi-tion into Λ-left cosets by (A.1). The possible upper-triangular column reduced matrices δthat can occur in the decomposition into Λ-right cosets are

1 0 e11 f1

p

d(δ) = (1, 1, p),

1 d1 0p 0

1

d(δ) = (1, 1, p),

p 0 0

1 01

d(δ) = (1, 1, p)



where 0 6 d1, e1, f1 < p. The fact that the determinantal vector of diag(1, 1, p) is (1, 1, p)implies the decomposition into Λ-left cosets given in (A.3) and the computation of the degreetoo.

A.2. Decomposition and degree of Λdiag(1, p, p)Λ. —


(A.5) Λdiag(1, p, p)Λ = ∪δ∈L1,p,pδΛ

where

L1,p,p = diag(1, p, p)⋃

06e1,f1<p

p e1

p f11

⋃

06d1<p

p d1

1p

.

In particular,

(A.6) deg (diag(1, p, p)) = p2 + p+ 1.

Proof of Proposition A.2. — By (A.2), the possible upper-triangular row reduced matrices δthat can occur in the decomposition into Λ-left cosets are

p d1 e1

p f11

d(δ) = (1, (p, d1, d1f1), p2),

p d1 e1

1p

d(δ) = (1, (p, e1), p2),

1p f1

p

d(δ) = (1, (p, f1), p2)

where 0 6 d1, e1, f1 < p. The fact that the determinantal vector of diag(1, p, p) is (1, p, p2)implies the decomposition into Λ-left cosets given in (A.5) and the computation of the degreetoo.

A.3. Decomposition and degree of Λdiag(1, p, p2)Λ. —


(A.7) Λdiag(1, p, p2)Λ = ∪δ∈R1,p,p2 Λδ = ∪δ∈L1,p,p2 δΛ



where

R1,p,p2 =⋃

06d1<p06e2,f2<p2

p|f2

1 d1 e2p f2

p2

⋃

06e1<p06d2<p2

1 d2 e1p2

p

⋃

06e2,f2<p2

p|e2

p e2

1 f2p2

⋃

06f1<p

p2

1 f1p

⋃

06d2<p2

p|d2

p d2

p2

1

⋃

06d1,e1,f1<pd1f1=0

(d1,e1,f1) 6=(0,0,0)

p d1 e1

p f1p

⋃

p2

p1

and

L1,p,p2 =⋃

06f1<p06d2,e2<p2

p|d2

p2 d2 e2

p f11

⋃

06e1<p06f2<p2

p e1

p2 f21

⋃

06d2,e2<p2

p|e2

p2 d2 e2

1p

⋃

06d1<p

p d1

1p2

⋃

06f2<p2

p|f2

1p2 f2

p

⋃

06d1,e1,f1<pd1f1=0

(d1,e1,f1)6=(0,0,0)

p d1 e1

p f1p

⋃

p2

p1

.

In particular,

(A.8) deg(diag(1, p, p2)

)= p(p+ 1)(1 + p+ p2).

Proof of Proposition A.3. — The decomposition into Λ-right cosets implies the decompo-sition into Λ-left cosets by (A.1). By (A.2), the possible upper-triangular column reducedmatrices δ that can occur in the decomposition into Λ-right cosets are

Type 1:

p d1 e1

p f1p

d(δ) = ((p, d1, e1, f1), (p2, pd1, pf1, d1f1 − pe1), p3)



and

Type 2:

1 d1 e2p f2

p2

d(δ) = (1, (p, f2), p3),

Type 3:

1 d2 e1p2 f1

p

d(δ) = (1, (p, f1), p3),

Type 4:

p e2

1 f2p2

d(δ) = (1, (p, e2), p3),

Type 5:

p2 e1

1 f1p

d(δ) = (1, (p, e1), p3),

Type 6:

p d2

p2

1

d(δ) = (1, (p, d2), p3),

Type 7:

p2 d1

p1

d(δ) = (1, (p, d1), p3)

where 0 6 d1, e1, f1 < p and 0 6 d2, e2, f2 < p2. Let us count the matrices among the previousones, whose determinantal vector is the same as the one of diag(1, p, p2), namely (1, p, p3).Let us consider the matrices of type 1. The condition on d2(δ) implies d1 6= 0. The conditionon d1(δ) implies that (d1, e1, f1) 6= (0, 0, 0). The condition on d2(δ) implies p | d1f1 such thatp | d1 or p | f1, namely d1 = 0 or f1 = 0. There are (p− 1)(2p+ 1) such matrices of type 1.Let us consider the matrices of type 2. The condition on d2(δ) implies p | f2. There are p4

such matrices of type 2.Let us consider the matrices of type 3. The condition on d2(δ) implies f1 = 0. There are p3

such matrices of type 3.Let us consider the matrices of type 4. The condition on d2(δ) implies p | e2. There are p3

such matrices of type 4.Let us consider the matrices of type 5. The condition on d2(δ) implies e1 = 0. There are psuch matrices of type 5.Let us consider the matrices of type 6. The condition on d2(δ) implies p | d2. There are psuch matrices of type 6.Let us consider the matrices of type 7. The condition on d2(δ) implies d1 = 0. There is 1 suchmatrix of type 7.One can recover the decomposition in Λ-right cosets given in (A.7) and the value of the degreegiven in (A.8) by summing all the contributions in the previous paragraphs.

A.4. Degree of Λdiag(1, p2, p4)Λ. —


(A.9) deg(diag(1, p2, p4)

)= p5(p+ 1)(p2 + p+ 1).



Proof of Proposition A.4. — By (A.2), the possible upper-triangular column reduced matri-ces δ that can occur in the decomposition into Λ-right cosets are

Type 1:

p4 d2

p2

1

d(δ) = (1, (p2, d2), p6),

Type 2:

p4 e2

1 f2p2

d(δ) = (1, (p2, e2), p6),

Type 3:

p2 d4

p4

1

d(δ) = (1, (p2, d4), p6),

Type 4:

p2 e4

1 f4p4

d(δ) = (1, (p2, e4), p6),

Type 5:

1 d4 e2p4 f2

p2

d(δ) = (1, (p2, f2, d4f2), p6),

Type 6:

1 d2 e4p2 f4

p4

d(δ) = (1, (p2, f4, d2f4), p6)

and

Type 7:

p4 d1 e1

p f1p

d(δ) = ((p, d1, e1, f1), (p2, pd1, d1f1 − pe1), p6),

Type 8:

p d4 e1

p4 f1p

d(δ) = ((p, d4, e1, f1), (p2, pf1, pd4, d4f1), p6),

Type 9:

p d1 e4

p f4p4

d(δ) = ((p, d1, e4, f4), (p2, pf4, d1f4 − pe4), p6)

and

Type 10:

p3 d3

p3

1

d(δ) = (1, (p3, d3), p6),

Type 11:

p3 e3

1 f3p3

d(δ) = (1, (p3, e3), p6),

Type 12:

1 d3 e3p3 f3

p3

d(δ) = (1, (p3, f3, d3f3), p6)



and

Type 13:

p3 d2 e1

p2 f1p

d(δ) = ((p, d2, e1, f1), (p3, pd2, d2f1 − p2e1), p6)

Type 14:

p3 d1 e2

p f2p2

d(δ) = ((p, d1, e2, f2), (p3, p2d1, d1f2 − pe2), p6)

Type 15:

p2 d3 e1

p3 f1p

d(δ) = ((p, d3, e1, f1), (p3, pd3, p

2f1, d3f1), p6)

Type 16:

p2 d1 e3

p f3p3

d(δ) = ((p, d1, e3, f3), (p3, d1f3 − pe3, p

2f3), p6)

and

Type 17:

p d3 e2

p3 f2p2

d(δ) = ((p, d3, e2, f2), (p3, pf2, p

2d3, d3f2), p6)

Type 18:

p d2 e3

p2 f3p3

d(δ) = ((p, d2, e3, f3), (p3, pf3, d2f3 − p2e3), p6)

and

Type 19:

p2 d2 e2

p2 f2p2

d(δ) = ((p2, d2, e2, f2), (p4, p2d2, p

2f2, d2f2 − p2e2), p6)

where 0 6 dj , ej , fj < pj for j = 1, 2, 3, 4. Let us count the matrices among the previous ones,whose determinantal vector is the same as the one of diag(1, p2, p4), namely (1, p2, p6).Let us consider the matrices of type 1. The condition on d2(δ) implies d2 = 0. There is 1relevant matrix of type 1.Let us consider the matrices of type 2. The condition on d2(δ) implies e2 = 0. There are p2

relevant matrices of type 2.Let us consider the matrices of type 3. The condition on d2(δ) implies p2 | d4. There are p2

relevant matrices of type 3.Let us consider the matrices of type 4. The condition on d2(δ) implies p2 | e4. There are p6

relevant matrices of type 4.Let us consider the matrices of type 5. The condition on d2(δ) implies e2 = 0 . There are p6

relevant matrices of type 5.Let us consider the matrices of type 6. The condition on d2(δ) implies p2 | f4 . There are p8

relevant matrices of type 6.Let us consider the matrices of type 7. The condition on d2(δ) implies d1 = e1 = 0 and thecondition on d1(δ) implies f1 6= 0. There are p− 1 relevant matrices of type 7.Let us consider the matrices of type 8. The condition on d2(δ) implies f1 = 0 and p | d4. Thecondition on d1(δ) implies e1 6= 0. There are p3(p− 1) relevant matrices of type 8.



Let us consider the matrices of type 9. The condition on d2(δ) implies p | f4 and p | d1f4/p−e4.One has d1 6= 0 since otherwise d1(δ) = 1 = (p, e4) and d2(δ) = p(p, e4) = p 6= p2. Thus, d1 isinvertible modulo p and f4/p ≡ e4d1 (mod p) such that f4/p can take p2 values. There are(p− 1)p6 relevant matrices of type 9.Let us consider the matrices of type 10. The condition on d2(δ) implies p2 || d3. There arep− 1 relevant matrices of type 10.Let us consider the matrices of type 11. The condition on d2(δ) implies p2 || e3. There are(p− 1)p3 relevant matrices of type 11.Let us consider the matrices of type 12. The condition on d2(δ) implies p2 || f3. There are(p− 1)p6 relevant matrices of type 12.Let us consider the matrices of type 13. Note that (e1, f1) 6= (0, 0) since otherwise d1(δ) =1 = (p, d2), which implies that d2(δ) = (pd2, p3) = p 6= p2. As a consequence, d1(δ) = 1 =(p, d2, e1, f1). The fact that d2(δ) = p2 implies that p | d2 and p | f1d2/p, namely f1 = 0 ord2 = 0. If d2 = 0 then d2(δ) = p2 = (p3, p2e1) such that e1 6= 0. There are p(p − 1) suchmatrices. If d2 6= 0 then f1 = 0, d2(δ) = p2(p, d2/p, e1) = p2 since d2/p is coprime with pand d1(δ) = 1 = (p, e1) such that e1 6= 0. There are (p− 1)2 such matrices. Finally, there are(p− 1)(2p− 1) relevant matrices of type 13.Let us consider the matrices of type 14. The fact that d2(δ) = p2 implies that p2 | d1f2− pe2.If d1 = 0 then p | e2 and d2(δ) = p2 = (p3, p2e2/p) if e2 6= 0. d1(δ) = 1 = (p, f2) impliesthat p - f2. There are (p − 1)(p2 − p) such matrices. If d1 6= 0 then the value of f2 is fixedby f2 ≡ pe2d1 (mod p2) and d1(δ) = (p, d1) = 1. There are p2(p− 1) such matrices. Finally,there are (p− 1)(2p2 − p) relevant matrices of type 14.Let us consider the matrices of type 15. The condition d2(δ) = p2 implies that p | d3 andp | f1d3/p. If f1 = 0 then d2(δ) = p2 = p2(p, d3/p) such that p || d3. The condition d1(δ) =1 = (p, e1) implies that e1 6= 0. There are (p2− p)(p− 1) such matrices. If f1 6= 0 then p2 | d3and d1(δ) = 1. There are p2(p− 1) such matrices. Finally, there are (p− 1)(2p2 − p) relevantmatrices of type 15.Let us consider the matrices of type 16. The condition d2(δ) = p2 implies that p2 | d1f3−pe3.If p | e3 then p2 | d1f3. If p | e3 and p | d1 then d1 = 0 and the condition d1(δ) = 1 = (p, f3)implies that p - f3 and d2(δ) = p2. There are p2(p3 − p2) such matrices. If p | e3 and p - d1then p2 | f3 then d2(δ) = p2(p, d1f3/p2 − e3/p) 6= p2 if and only if f3/p2 ≡ d1e3/p (mod p),which given d1 and e3/p can happen for only one value of f3/p2. There are (p− 1)p2(p− 1)such matrices. If p - e3 then d1(δ) = 1 = (p, d1, e3, f3). The condition p2 | d1f3 − pe3 impliesthat p2 - d1f3 and p - d1 but p | f3. The condition d2(δ) implies that p || d1f3/p − e3. Givend1 and e3, there are p choices for f3/p given by f3/p ≡ d1e3 (mod p) but one has to removethe value satisfying f3/p ≡ d1e3 (mod p2). There are (p − 1)(p3 − p2)(p − 1) such matrices.Finally, there are p3(p− 1)(2p− 1) relevant matrices of type 16.Let us consider the matrices of type 17. The condition d2(δ) = p2 implies that p | f2. If f2 = 0then d2(δ) = p2 = p2(p, d3) such that p - d3, which implies d1(δ) = 1. There are (p3 − p2)p2

such matrices. If f2 6= 0 then p | d3 since p | d3f2/p, in which case d2(δ) = p2. The conditiond1(δ) = 1 implies that p - e2. There are p2(p2 − p)(p − 1) such matrices. Finally, there arep3(p− 1)(2p− 1) relevant matrices of type 17.Let us consider the matrices of type 18. The condition d2(δ) = p2 implies that p | f3 andp | d2f3/p. If p2 | f3 then d2(δ) = p2 = p2(p, d2f3/p2 − e3). One has to remove the p2 valuesof e3 satisfying e3 ≡ d2f3/p2 (mod p). In this case, one has d1(δ) = (p, d2, e3) = 1 since ifp | (d2, e3) then (p, d2f3/p2 − e3) 6= 1. There are p2(p3 − p2)p such matrices. If p2 - f3 then



p | d2 and the conditions on d1(δ) and d2(δ) are satisfied. There are p(p3 − p2)(p2 − p) suchmatrices. Finally, there are p4(p− 1)(2p− 1) relevant matrices of type 18.Let us consider the matrices of type 19. The condition on d2(δ) implies that p2 | d2f2. Ifd2 = 0 then d2(δ) = p2 = p2(p4, e2, f2) and d1(δ) = 1 = (p2, e2, f2). One has to removethe couples (e2, f2) satisfying p | e2 and p | f2, namely p2 couples. There are p4 − p2 suchmatrices. If d2 6= 0 and f2 = 0 then d2(δ) = p2 = p2(p2, d2, e2) and d1(δ) = 1 = (p2, d2, e2).One has to remove the couples (d2, e2) satisfying p | d2 and p | e2, namely (p − 1)p couples.There are (p2 − 1)p2 − (p − 1)p such matrices. If d2 6= 0 and f2 6= 0 then d2(δ) = p2 =p2(p2, pd2/p, pf2/p, d2f2/p2 − e2) and d1(δ) = 1 = (p2, e2). Thus, p - e2 and p - d2f2/p2 − e2.Among the p2−p values of e2 satisfying p - e2, one has to remove these satisfying e2 ≡ d2f2/p2

(mod p) of cardinal p. There are (p − 1)(p2 − 2p)(p − 1) such matrices. Finally, there arep(p− 1)(3p2 − p+ 1) relevant matrices of type 19.One can recover the value of the degree given in (A.9) by summing all the contributions inthe previous paragraphs.

A.5. Degree of Λdiag(1, p3, p3)Λ. —


(A.10) deg(diag(1, p3, p3)

)= p4(p2 + p+ 1).

Proof of Proposition A.5. — By (A.2), the possible upper-triangular column reduced matri-ces δ that can occur in the decomposition into Λ-right cosets are

Type 1:

p3 d3

p3

1

d(δ) = (1, (p3, d3), p6),

Type 2:

p3 e3

1 f3p3

d(δ) = (1, (p3, e3), p6),

Type 3:

1 d3 e3p3 f3

p3

d(δ) = (1, (p3, f3, d3f3), p6)

and

Type 4:

p3 d2 e1

p2 f1p

d(δ) = ((p, d2, e1, f1), (p3, pd2, d2f1 − p2e1), p6),

Type 5:

p3 d1 e2

p f2p2

d(δ) = ((p, d1, e2, f2), (p3, p2d1, d1f2 − pe2), p6),

Type 6:

p2 d3 e1

p3 f1p

d(δ) = ((p, d3, e1, f1), (p3, pd3, p

2f1, d3f1), p6),

Type 7:

p2 d1 e3

p f3p3

d(δ) = ((p, d1, e3, f3), (p3, p2f1, d1f3 − pe3), p6),



Type 8:

p d3 e2

p3 f2p2

d(δ) = ((p, d3, e2, f2), (p3, p2d3, pf2, d3f2), p6),

Type 9:

p d1 e3

p2 f3p3

d(δ) = ((p, d1, e4, f4), (p3, pf3, d2f3 − p2e3), p6)

and

Type 10:

p2 d2 e2

p2 f2p2

d(δ) = ((p2, d2, e2, f2), (p4, p2d2, p

2f2, d2f2 − p2e2), p6)

where 0 6 dj , ej , fj < pj for j = 1, 2, 3. Let us count the matrices among the previous ones,whose determinantal vector is the same as the one of diag(1, p3, p3), namely (1, p3, p6).Let us consider the matrices of type 1. The condition on d2(δ) implies d3 = 0. There is 1relevant matrix of type 1.Let us consider the matrices of type 2. The condition on d2(δ) implies e3 = 0. There are p3

relevant matrices of type 2.Let us consider the matrices of type 3. The condition on d2(δ) implies f3 = 0. There are p6

relevant matrices of type 3.Let us consider the matrices of type 4. The condition on d2(δ) implies d2 = 0 and e1 = 0.Then, d1(δ) = 1 = (p, f1) such that f1 6= 0. There are p− 1 relevant matrices of type 4.Let us consider the matrices of type 5. The condition on d2(δ) implies d1 = 0 and e2 = 0.Then, d1(δ) = 1 = (p, f2) such that p - f2. There are p2 − p relevant matrices of type 5.Let us consider the matrices of type 6. The condition on d2(δ) implies f1 = 0 and p2 | d3.Then, d1(δ) = 1 = (p, e1) such that e1 6= 0. There are p(p− 1) relevant matrices of type 6.Let us consider the matrices of type 7. The condition on d2(δ) implies p | f3 and p | d1f3/p−e3.One has d1 6= 0 since otherwise p2 | e2 by the condition on d2(δ) such that d1(δ) = p 6= 1.Thus, d1 is invertible modulo p and f3/p ≡ e3d1 (mod p2) is fixed. There are (p−1)p3 relevantmatrices of type 7.Let us consider the matrices of type 8. The condition on d2(δ) implies p | d3 and f2 = 0.Then, d1(δ) = 1 = (p, e2) such that p - e2. There are p2(p2 − p) relevant matrices of type 8.Let us consider the matrices of type 9. The condition on d2(δ) implies p2 | f3 and p |d2f3/p − e3. If f3 = 0 then p | e3 and d1(δ) = 1 = (p, d2) such that p - d2. There are(p2 − p)p2 such matrices. If f3 6= 0 then d2 ≡ e3f3/p2 (mod p) can take p values. Then,d1(δ) = 1 = (p, e3) such that p - e3. There are p(p3 − p2)(p− 1) such matrices. Finally, thereare p4(p− 1) relevant matrices of type 9.Let us consider the matrices of type 10. The condition on d2(δ) implies p | d2, p | f2 andp | d2f2/p2−e2. One has d2 6= 0 since otherwise d2(δ) = p2(p2, e2) = p2d1(δ) = p2 6= p3. Thus,d2/p is invertible modulo p and f2 is fixed by f2/p ≡ e2d2/p (mod p). Then, d1(δ) = 1 =(p, e2) such that p - e2, p - f2/p and d2(δ) = p3. There are (p − 1)(p2 − p) relevant matricesof type 10.One can recover the value of the degree given in (A.10) by summing all the contributions inthe previous paragraphs.

A.6. Degree of Λdiag(1, 1, p3)Λ. —




(A.11) deg(diag(1, 1, p3)

)= p4(p2 + p+ 1).

Proof of Proposition A.6. — By (A.2), the possible upper-triangular column reduced matri-ces δ that can occur in the decomposition of the Λ-double coset Λdiag(1, 1, p3)Λ into Λ-rightcosets are

Type 1:

p3

11

d(δ) = (1, 1, p3),

Type 2:

1 d3p3

1

d(δ) = (1, 1, p3),

Type 3:

1 e31 f3

p3

d(δ) = (1, 1, p3)

and

Type 4:

1 d1 e2p f2

p2

d(δ) = (1, (p, f2), p3),

Type 5:

1 d2 e1p2 f1

p

d(δ) = (1, (p, f1), p3),

Type 6:

p e2

1 f2p2

d(δ) = (1, (p, e2), p3),

Type 7:

p2 e1

1 f1p

d(δ) = (1, (p, e1), p3),

Type 8:

p d2

p2

1

d(δ) = (1, (p, d2), p3),

Type 9:

p2 d1

p1

d(δ) = (1, (p, d1), p3)

and

Type 10:

p d1 e1

p f1p

d(δ) = ((p, d1, e1, f1), (p2, pd1, pf1, d1f1 − pe1), p3)

where 0 6 dj , ej , fj < pj for j = 1, 2, 3. Let us count the matrices among the previous ones,whose determinantal vector is the same as the one of diag(1, 1, p3), namely (1, 1, p3).Let us consider the matrices of type 1. There is 1 relevant matrix of type 1.



Let us consider the matrices of type 2. There are p3 relevant matrices of type 2.Let us consider the matrices of type 3. There are p6 relevant matrices of type 3.Let us consider the matrices of type 4. The condition on d2(δ) implies p - f2. There arep3(p2 − p) relevant matrices of type 4.Let us consider the matrices of type 5. The condition on d2(δ) implies f1 6= 0. There arep3(p− 1) relevant matrices of type 5.Let us consider the matrices of type 6. The condition on d2(δ) implies p - e2. There arep2(p2 − p) relevant matrices of type 6.Let us consider the matrices of type 7. The condition on d2(δ) implies e1 6= 0. There arep(p− 1) relevant matrices of type 6.Let us consider the matrices of type 8. The condition on d2(δ) implies p - d2. There are p2−prelevant matrices of type 7.Let us consider the matrices of type 9. The condition on d2(δ) implies d1 6= 0. There are p−1relevant matrices of type 9.Let us consider the matrices of type 10. One has d1 6= 0 since otherwise p | d2(δ). Thus,d1(δ) = 1. In addition, f1 6= 0 since otherwise p | d2(δ). There are p(p− 1)2 relevant matricesof type 10.One can recover the value of the degree given in (A.11) by summing all the contributions inthe previous paragraphs.

References[AZ95] A. N. Andrianov and V. G. Zhuravlëv, Modular forms and Hecke operators, volume 145 of

Translations of Mathematical Monographs. American Mathematical Society, Providence, RI,1995. Translated from the 1990 Russian original by Neal Koblitz.

[BHM] V. Blomer, G. Harcos, and D. Milićević, Bounds for eigenforms on arithmetic hyperbolic3-manifolds. Available at http://arxiv.org/abs/1401.5154.

[BMa] Valentin Blomer and Péter Maga, Subconvexity for sup-norms of automorphic forms onPGL(n). Available at http://arxiv.org/pdf/1405.6691.pdf.

[BMb] Valentin Blomer and Péter Maga, The sup-norm problem for PGL(4). Available at http://arxiv.org/pdf/1404.4331.pdf.

[BT] Farrell Brumley and Nicolas Templier, Large values of cusp forms on GL(n). available athttp://arxiv.org/abs/1411.4317.

[DFI94] W. Duke, J. B. Friedlander, and H. Iwaniec, Bounds for automorphic L-functions. II, Invent.Math., 115 (2):219–239, 1994.

[FI92] J. Friedlander and H. Iwaniec, A mean-value theorem for character sums, Michigan Math.J., 39 (1):153–159, 1992.

[Gol06] Dorian Goldfeld, Automorphic forms and L-functions for the group GL(n,R), volume 99of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge,2006. With an appendix by Kevin A. Broughan.

[HRR] Roman Holowinsky, Guillaume Ricotta, and Emmanuel Royer, On the sup-norm of SL(3)Hecke-Maass cusp form. Available at http://arxiv.org/abs/1411.4317.

[HT13] Gergely Harcos and Nicolas Templier, On the sup-norm of Maass cusp forms of large level.III, Math. Ann., 356 (1):209–216, 2013.



[IS95] H. Iwaniec and P. Sarnak, L∞ norms of eigenfunctions of arithmetic surfaces, Ann. of Math.(2), 141 (2):301–320, 1995.

[Iwa92] Henryk Iwaniec, The spectral growth of automorphic L-functions, J. Reine Angew. Math.,428:139–159, 1992.

[Kod67] Tetsuo Kodama, On the law of product in the Hecke ring for the symplectic group, Mem. Fac.Sci. Kyushu Univ. Ser. A, 21:108–121, 1967.

[Mac95] I. G. Macdonald, Symmetric functions and Hall polynomials, Oxford Mathematical Mono-graphs. The Clarendon Press, Oxford University Press, New York, second edition, 1995. Withcontributions by A. Zelevinsky, Oxford Science Publications.

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Publications of the Mathematical Society of Japan. Princeton University Press, Princeton,NJ, 1994. Reprint of the 1971 original, Kanô Memorial Lectures, 1.

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February 10, 2015

Roman Holowinsky, Department of Mathematics, The Ohio State University, 100 Math Tower, 231 West18th Avenue, Columbus, OH 43210-1174 • E-mail : [email protected]

Guillaume Ricotta, Université de Bordeaux, IMB, 351 cours de la libération, 33405 Talence Cedex, FranceE-mail : [email protected]

Emmanuel Royer, Université Blaise Pascal, Laboratoire de mathématiques, Les Cézeaux, BP 80026, 63171Aubière Cedex, France • E-mail : [email protected]



FUNDAMENTAL UNITS FOR ORDERS GENERATED BY A UNIT

by

Stéphane R. Louboutin

Abstract. — Let ε be an algebraic unit for which the rank of the group of units of the orderZ[ε] is equal to 1. Assume that ε is not a complex root of unity. It is natural to wonder whetherε is a fundamental unit of this order. It turns out that the answer is in general positive, and thata fundamental unit of this order can be explicitly given (as an explicit polynomial in ε) in therare cases when the answer is negative. This paper is a self-contained exposition of the solutionto this problem, solution which was up to now scattered in many papers in the literature. Wealso include the state of the art in the case that the rank of the group of units of the order Z[ε]is greater than 1 when now one wonders whether the set ε can be completed in a system offundamental units of the order Z[ε].

Résumé. — Soit ε une unité algébrique pour laquelle le rang du groupe des unités de l’ordreZ[ε] est égal à 1. Supposons que ε ne soit pas une racine complexe de l’unité. Il est alors naturelde se demander si ε est une unité fondamentale de cet ordre. Nous montrons que la réponse esten général positive et que, dans les rares cas où elle ne l’est pas, une unité fondamentale de cetordre peut être explicitement donnée (comme polynôme en ε). Nous présentons ici une expositioncomplète de la solution à ce problème, solution jusqu’à présent dispersée dans plusieurs articles.Nous incluons l’état de l’art de ce problème dans le cas où la rang du groupe des unités del’ordre Z[ε] est strictement plus grand que 1, où la question naturelle est maintenant de savoirsi on peut adjoindre à ε d’autres unités de l’ordre Z[ε] pour obtenir un système fondamentald’unités de cet ordre.

Contents1. Introduction and Notation 422. The real quadratic case 443. The non-totally real cubic case 453.1. Statement of the result for the cubic case 453.2. Sketch of proof 463.3. Bounds on discriminants 473.4. Being a square 49

Mathematical subject classification (2010). — 11R16, 11R27.Key words and phrases. — Cubic unit, cubic orders, quartic unit, quartic order, fundamental units.

42 Fundamental units for orders generated by a unit

3.5. Computation of explicit nth roots 494. The totally imaginary quartic case 494.1. Statement of the result for the quartic case 514.2. Sketch of proof 524.3. Being a square 534.4. The case that µ(ε) is of order 8, 10 or 12 544.5. The case that ε = ζ4η2 or ε = ζ3η3 554.6. Bounds on discriminants 565. The totally real cubic case 595.1. Cubic units of type (T+) 605.2. Bounds on discriminants 605.3. Being a square 615.4. Statement and proof of the result for the totally real cubic case 626. A conjecture in a quartic case of unit rank 2 637. The general situation 65References 67

1. Introduction and NotationLet ε be an algebraic unit for which the rank of the group of units of the order Z[ε] is equalto 1. Then either (i) ε is a totally real quadratic unit, or (ii) ε is a non-totally real cubic unitor (iii) ε is a totally imaginary quartic unit. In these three situations, if we assume that εis not a complex root of unity, it is natural to ask whether ε is a fundamental unit of theorder Z[ε]. And in case it is not, it is natural to construct one from it. In Section 2, in thevery simple case of the real quadratic units, we introduce the method that we will also usein Sections 3 and 4 to answer to this question in the more difficult remaining cases of cubicunits of negative discriminants and totally imaginary quartic units. Now, assume that thethe rank of the group of units of the order Z[ε] is greater than 1. The natural question isnow to wonder whether the set ε can be completed in a system of fundamental units of theorder Z[ε]. In Section 5, we will answer to this question in the case of totally real cubic units,the only situation where to date this question has been answered. We conclude this paper bygiving in Section 6 a conjecture for the case of quartic units of negative discriminants and byshowing in Section 7 that the solution to this problem for units of degree greater than 4 isbound to be more complicated.In order to put the previously published elements of solution to our natural question ina general framework that might lead to solutions in presently unsolved cases, we introduceheights for polynomials and algebraic units. They will enable us to formulate our mains resultsin (8), (14), (18), (20) and (21) in a clear and uniform way. Let Πα(X) = Xn − an−1Xn−1 +· · ·+ (−1)n−1a1X + (−1)na0 ∈ Z[X] be the minimal polynomial of an algebraic integer α. Itis monic and Q-irreducible. Let dα > 0 be the absolute value of its discriminant Dα 6= 0. Letβ ∈ Z[α] and assume that Q(β) = Q(α), i.e. that deg Πβ(X) = deg Πα(X). Then

(1) Dβ = (Z[α] : Z[β])2Dα and hence dβ = (Z[α] : Z[β])2dα.

In particular, Z[β] = Z[α] if and only if dβ = dα.


S. Louboutin 43

Now, assume that α is an algebraic unit, i.e. assume that a0 = ±1. For p > 0 a real numberand p =∞ we define the heights of α and of its minimal polynomial Πα(X) by

Hp(α) = Hp(Πα(X)) := max1≤k≤n

(|αk|nkp + |αk|−nkp)1/p > 1

andH∞(α) = H∞(Πα(X)) := max

1≤k≤nmax(|αk|nk , |αk|−nk) ≥ 1,

where α1, · · · , αn are the complex roots of Πα(X) and nk = 1 if αk is real and nk = 2is αk is not real. Notice that p 7→ Hp(P (X)) is a decreasing function of p > 0 such thatlimp→∞Hp(P (X)) = H∞(P (X)). Notice also that H∞(α) > 1 as soon as α is not a com-plex root of unity (e.g. see [Was, Lemma 1.6]). The most useful property of our height,as compared to the usual height H(P (X)) = max0≤i≤n−1 |ai| and the Malher measureM(P (X)) = ∏

1≤i≤n max(1, |αi|) of P (X) = ∏ni=1(X − αi) = Xn − an−1Xn−1 + · · · +

(−1)n−1a1X + (−1)na0 ∈ C[X], is that for any algebraic unit α and any 0 6= m ∈ Z wehave(2) H∞(αm) = H∞(α)|m|,provided that Q(αm) = Q(α), a property akin to the one satisfied by the canonical height onan elliptic curve.Assume that we have proved that for all algebraic units α of a given degree n > 1 we have(3) CaH∞(α)a ≤ dα ≤ CbH∞(α)b

where 0 < a < b and Ca, Cb > 0 depend only on the numbers of real and complex conjugatesof α (see Lemma 3, Theorems 9, 24, 33 and Conjecture 39). Since there are only finitely manyalgebraic units of a given degree of a bounded height, there exists a unit η0 of degree n suchthat H∞(α) ≥ H∞(η0) > 1 for all algebraic units α of degree n that are not a complex root ofunity. Let N be the least rational integer greater than b/a. If an algebraic unit ε of degree nthat is not a complex root of unity is such that ε = ±ηm for some η ∈ Z[ε] and some m ≥ N ,then Z[η] = Z[ε] and Dη = Dε (Lemma 2). In particular, ε and η are of the same degree andhave the same numbers of real and complex conjugates. Using (2) and (3), we obtain

CaH∞(η)Na ≤ CaH∞(η)ma = CaH∞(ε)a ≤ dε = dη ≤ CbH∞(η)b.Hence, H∞(η) ≤ (Cb/Ca)1/(Na−b) is bounded and there are only finitely many such η’s.Moreover, the inequalities

H∞(η0)ma ≤ H∞(η)ma ≤ (Cb/Ca)H∞(η)b ≤ (Cb/Ca)Na/(Na−b)

show that m is bounded and there are only finitely many such ε’s. In conclusion, assumingthat (3) holds true, we obtain that if ε = ±ηm for some η ∈ Z[ε] and some 0 6= m ∈ Z then|m| ≤ b/a, appart from finitely many sporadic algebraic units ε’s. This line of reasoning isclearly a key step toward solving our general question. Indeed, as it is rather easy to settlethe case m = 2 (see Lemmas 13, 20 and 35), our problem is almost solved in the situationswhere b/a < 3. However, solving the problem ε = ±η3 is not that easy. In fact, we do noteven know how to solve it for totally imaginary quartic units of negative discriminant (seeConjecture 38). Hence our problem is probably hard to solve in the situations where the bestexponents in (3) turn out to satisfy b/a ≥ 3.Anyway, clearly our main tool to tackle our problem will thus to obtain lower and upperbounds of the form (3) for the absolute discriminants dα of algebraic units α. We could have



restrained ourselves to the use of the single height H∞, but the proof of Corollary 12 makesit clear that using the height H1 yields better bounds.Notice that if ε is an algebraic unit, then Z[ε] = Z[−ε] = Z[1/ε] = Z[−1/ε], Dε = D−ε =D1/ε = D−1/ε and Hp(ε) = Hp(−ε) = Hp(1/ε) = Hp(−1/ε) for p > 0 or p = ∞. Thefour monic polynomials Πε(X) = Xn − an−1Xn−1 + · · · + (−1)n−1a1X + (−1)na0 ∈ Z[X],with a0 ∈ ±1, Π−ε(X) = (−1)nΠα(−X), Π1/ε(X) = (−1)na0XnΠε(1/X) and Π−1/ε(X) =(−1)na0XnΠε(−1/X) are called equivalent. They have the same discriminant and the sameheight. At least one of them is such that |a1| ≤ an−1. We call it reduced. By an appropriatechoice of the root of Πε(X), we may also assume that |ε| ≥ 1. If ε is real, we may insteadassume that ε > 1.For example, take Πε(X) = X4 +3X3 +6X2 +4X+1. Since Πε(X) is not reduced, we set ε′ =−1/ε = ε3+3ε2+6ε+4 = P (ε), for which Πε′(X) = X4Πε(−1/X) = X4−4X3+6X2−3X+1is reduced. By Point 2(c)iii of Theorem 18, η = ε′3 − 3ε′2 + 3ε′ = ε3 + 2ε2 + 4ε + 1 isa fundamental unit of Z[ε′] = Z[ε] and ε′ = −1/η4 yields ε = −1/ε′ = η4. Moreover,Πη(X) = X4 −X3 + 1 is reduced, dε = dη = 229 and Z[ε] = Z[η].

2. The real quadratic caseWe introduce in this very simple situation the three tools and method we will use to solvethe more difficult cubic and quartic cases.Let ε be an algebraic quadratic unit which is not a complex root of unity and for which therank of the group of units of the quadratic order Z[ε] is equal to 1. Hence, ε is totally real.We may and we will assume that ε > 1.

Lemma 1. — The smallest quadratic unit greater than 1 is η0 := (1 +√

5)/2.

Lemma 2. — Let ε be an algebraic integer. If ε = ±ηn with n ∈ Z and η ∈ Z[ε], thenZ[η] = Z[ε]. Hence, Dη = Dε and dη = dε.

Proof. — Notice that Z[η] ⊆ Z[ε] = Z[±ηn] ⊆ Z[η] and use (1).

Lemma 3. — Let α be a real quadratic unit. Then

(4)(|α| − |α|−1

)2≤ dα ≤

(|α|+ |α|−1

)2.

Proof. — We have dα = (α− α′)2, where α′ = ±1/α be the conjugate of α.

Theorem 4. — A real quadratic unit ε > 1 is always the fundamental unit of the quadraticorder Z[ε], except if ε = (3 +

√5)/2, in which case ε = η2, where 1 < η = (1 +

√5)/2 =

ε− 1 ∈ Z[ε] is the fundamental unit of Z[ε] = Z[η], and dε = dη = 5.

Proof. — Assume that ε > 1 is not the fundamental unit of Z[ε]. Then there exist a quadraticunit 1 < η ∈ Z[ε] and n ≥ 2 such that ε = ηn, which yields dε = dη (Lemma 2). Using (4) weobtain

0 < η − η−1 = η2 − η−2

η + η−1 ≤ηn − η−nη + η−1 = ε− ε−1

η + η−1 ≤√dε/dη = 1.

But 0 < η − η−1 ≤ 1 implies 1 < η ≤ (1 +√

5)/2 and η = η0, by Lemma 1. We now obtain1 = η0 − η−1

0 = (η20 − η−2

0 )/(η0 + η−10 ) ≤ (ηn0 − η−n0 )/(η0 + η−1

0 ) ≤ 1, which yields n = 2.


S. Louboutin 45

3. The non-totally real cubic caseThe aim of this Section 3 is to prove Theorem 8. It was first proved in [Nag]. However,while working on class numbers of some cubic number fields, we come up in [Lou06] with acompletely different proof of Nagell’s result. Our proof was based on lower bounds on absolutediscriminants of non-totally real algebraic cubic units (see Theorem 9), proof then simplifiedin [Lou10].

Definition 5. — A cubic polynomial of type (T) is a monic cubic polynomial P (X) = X3−aX2+bX−1 ∈ Z[X] which is Q-irreducible (⇔ b 6= a and b 6= −a−2), of negative discriminantDP (X) < 0 and whose only real root εP satisfies εP > 1 (⇔ P (1) < 0 ⇔ b ≤ a − 1). In thatsituation, Hp(P (X)) = (εpP + ε−pP )1/p and H∞(P (X)) = εP .

Let ε be an algebraic cubic unit for which the rank of the group of units of the cubic orderZ[ε] is equal to 1. Hence, ε is not totally real. We may and we will assume that ε is real andthat ε > 1, i.e. that Πε(X) is a cubic polynomial of type (T) (notice that if ε is of type (T)and ε = ηn for some odd n ≥ 3 and some η ∈ Z[ε], then η is clearly also of type (T), whereasif ε is reduced then it is not clear whether η is also necessarily reduced):

Lemma 6. — Let εP > 1 be the only real root of a cubic polynomial P (X) = X3 − aX2 +bX − 1 ∈ Z[X] of type (T). Then(5) 0 ≤ a < εP + 2 and |b| ≤

√4a+ 4.

Proof. — Let ε−1/2P eiφ and ε

−1/2P e−iφ be the non-real complex roots of P (X). Then a =

εP + 2ε−1/2P cosφ ≥ εP − 2 > −1 and b = 2ε1/2

P cosφ+ ε−1P . Hence,

(6) 4a− b2 = 4εP sin2 φ+ 4ε−1/2P cosφ− ε−2

P > −5and (5) holds true. Notice that (5) makes it easy to list all the cubic polynomials of type (T) whose real rootsare less than or equal to a given upper bound B. Taking B = 2, we obtain:

Lemma 7. — The real root η0 = 1.32471 · · · of Π(X) = X3−X − 1 is the smallest real butnon-totally real cubic unit greater than 1.

3.1. Statement of the result for the cubic case. —

Theorem 8. — Let ε > 1 be a real cubic algebraic unit of negative discriminant Dε =−dε < 0. Let η > 1 be the fundamental unit of the cubic order Z[ε]. Then ε = η, except inthe following cases:

1. The infinite family of exceptions for which Πε(X) = X3 −M2X2 − 2MX − 1, M ≥ 1,in which case ε = η2 where η = ε2 −M2ε−M ∈ Z[ε] is the real root of X3 −MX2 − 1,Z[η] = Z[ε] and dε = dη = 4M3 + 27.

2. The 8 following sporadic exceptions:(a) (i) Πε(X) = X3 − 2X2 +X − 1, in which case ε = η2 where η = ε2 − ε ∈ Z[ε].

(ii) Πε(X) = X3 − 3X2 + 2X − 1, in which case ε = η3 where η = ε− 1 ∈ Z[ε].(iii) Πε(X) = X3−2X2−3X−1, in which case ε = η4 where η = ε2−2ε−2 ∈ Z[ε],



(iv) Πε(X) = X3−5X2 +4X−1, in which case ε = η5 where η = ε2−4ε+1 ∈ Z[ε].(v) Πε(X) = X3−12X2−7X−1, in which case ε = η9 where η = −3ε2+37ε+10 ∈

Z[ε].In these five cases, η > 1 is the real root of Πη(X) = X3 −X − 1, Z[η] = Z[ε] anddη = dε = 23.

(b) (i) Πε(X) = X3−4X2 +3X−1, in which case ε = η3 where η = ε2−3ε+1 ∈ Z[ε].(ii) Πε(X) = X3−6X2−5X−1, in which case ε = η5 where η = −2ε2 +13ε+5 ∈

Z[ε].In these two cases, η > 1 is the real root of Πη(X) = X3 −X2 − 1, Z[η] = Z[ε] anddη = dε = 31.

(c) Πε(X) = X3−7X2 +5X−1, in which case ε = η3, where 1 < η = −ε2 +7ε−3 ∈ Z[ε]is the real root of Πη(X) = X3 −X2 −X − 1, Z[η] = Z[ε] and dη = dε = 44.

3.2. Sketch of proof. — Let ε > 1 be a real but non-totally real cubic unit. Then ε is nota fundamental unit of the order Z[ε] if and only if there exists p ≥ 2 a prime and η ∈ Z[ε]such that ε = ηp (the main feature that makes it easier to deal with this cubic case than withthe totally imaginary quartic case dealt with below is that −1 and +1 are the only complexroots of unity in Z[ε]). Now, if ε = ηn for some η ∈ Z[ε], then Z[η] = Z[ε] and dε = dη, byLemma 2.1. Assume that ε = ηn for some non-totally real cubic unit 1 < η ∈ Z[ε] and some n ≥ 3.Using η ≥ η0 = 1.32471 · · · (Lemma 7) and a double bound (7) for dε similar to (4), we willobtain in Corollary 12 that 1 < η ≤ 4.5 and n ≤ 10.2. In contrast with the quadratic case, this double bound (7) does not prevent ε from beinginfinitely many often a square in Z[ε]. Hence, we characterize in Lemma 13 when this isindeed the case.3. Finally, to determine all the 1 < ε’s that admit a p-th root in Z[ε] for some p ∈ 3, 5, 7,and to determine this p-th root η > 1, we make a list of all the cubic polynomials Πη(X) =X3 − AX2 + BX − 1 ∈ Z[X] of type (T) with 0 ≤ A ≤ 6 < 4.5 + 2, by (5), for whichthere exist p ∈ 3, 5, 7 such that Z[η] = Z[ηp], i.e. such that Dη = Dηp , where Πηp(X) =X3 − aX2 + bX − 1 is computed as the resultant of Πη(Y ) and X − Y n, considered as poly-nomials of the variable Y . A Maple Program 1 settling this step is given below. We found 6such occurrences, of discriminants −23, −31 or −44. Taking also into account the points 2and 3 of Lemma 13, both of discriminant −23, and singling out the only case of point 1 ofLemma 13 of discriminant in −23,−31,−44, namely the case M = 1 of discriminant −31,we obtain Table 1, which completes the proof of Theorem 8:


S. Louboutin 47

p Πη(X) Πηp(X) Dη = Dηp

2 X3 −X − 1 X3 − 2X2 + X− 1 −232 X3 − 2X2 + X− 1 X3 − 2X2 − 3X − 1 −233 X3 −X − 1 X3 − 3X2 + 2X− 1 −233 X3 − 3X2 + 2X− 1 X3 − 12X2 − 7X − 1 −235 X3 −X − 1 X3 − 5X2 + 4X − 1 −232 X3 −X2 − 1 X3 −X2 − 2X − 1 −313 X3 −X2 − 1 X3 − 4X2 + 3X − 1 −315 X3 −X2 − 1 X3 − 6X2 − 5X − 1 −313 X3 −X2 −X − 1 X3 − 7X2 + 5X − 1 −44

Table 1.

Program 1:for A from 0 to 6 by 1 doborneB := isqrt(4A+ 4):for B from −borneB to min(borneB,A− 1) by 1 dop := x3 −A · x2 +B · x− 1;if irreduc(p) thenDp := discrim(p, x);if Dp < 0 thenfor n in [3,5,7] doq := resultant(subs(x = y, p), x− yn, y);Dq := discrim(q, x);if Dq = Dp then print(n, sort(p, x), sort(q, x), Dp) end ifend doend ifend ifend doend do:

3.3. Bounds on discriminants. —

Theorem 9. — Let α be a real cubic algebraic unit of negative discriminant. Then(7) max(|α|3/2, |α|−3/2)/2 ≤ dα ≤ 4(|α|+ |α|−1)3 ≤ 32 max(|α|3, |α|−3).Hence, if P (X) = X3 − aX2 + bX − c ∈ Z[X], c ∈ ±1, is Q-irreducible and of negativediscriminant DP (X) < 0, then

(8) H∞(P (X))3/2/2 ≤ |DP (X)| ≤ 4H1(P (X))3 ≤ 32H∞(P (X))3.

Proof. — Clearly, (8) follows from (7): if α (real), β and β (not real) are the roots of P (X),then |α||β|2 = 1, hence Hp(P (X)) = (|α|p + |α|−p)1/p = (|β|2p + |β|−2p)1/p and H∞(P (X)) =max(|α|, |α|−1) = max(|β|2, |β|−2).Since (7) remains unchanged if we change α into −α, 1/α and −1/α, we may assume thatα > 1, i.e. that Πα(X) is of type (T). Let β = α−1/2eiφ and β = α−1/2e−iφ be the non-realcomplex roots of Πα(X) = X3 − aX2 + bX − 1 ∈ Z[X]. Then(9) dα = −(α− β)2(α− β)2(β − β)2 = 4(α3/2 − 2 cosφ+ α−3/2)2 sin2 φ.



Hence, setting Xα = α3/2 + α−3/2 ≥ 2, we havedα = X2

α + 4− (Xα cosφ+ 2 sin2 φ)2 − 4 cos2 φ ≤ X2α + 4 = α3 + α−3 + 6 ≤ (α+ α−1)3.

Let us now prove the lower bound on dα. By (9) we have

dα ≥ 4(α3/4 − α−3/4)4 sin2 φ.

Assume that α > 16.2.First, assume that sin2 φ ≥ 2α−3/2. Then we obtain dα ≥ 7α3/2.Secondly, assume that sin2 φ < 2α−3/2. By (6), we have

−1 < −4α−1/2 − α−2 ≤ 4a− b2 = 4α sin2 φ+ 4α−1/2 cosφ− α−2 < 12α−1/2 < 3.Since 4a − b2 ≡ 0 or 3 (mod 4), we obtain 4a = b2 and cosφ < 0 (otherwise 4a − b2 ≥4α−1/2 sin2 φ+4α−1/2 cos2 φ−α−2 > 0). Hence, 0 = 4a−b2 = 4α sin2 φ−4α−1/2√1− sin2 φ−α−2. Therefore, sin2 φ = α−3/2−α−3/4, cosφ = −1+α−3/2/2 (hence b = 2α1/2 cosφ+α−1 =−2(α1/2 − α−1) < 0 and Πα(X) = X3 −B2X2 − 2BX − 1 for some B ≥ 1) and (9) yields

dα = 4(α3/2 + 2)2(α−3/2 − α−3/4) > 4α3/2.

Therefore, dα ≥ 4α3/2 for α > 16.2.Finally, if 1 < α ≤ 16.2, then Πα(X) = X3 − aX2 + bX − 1 ∈ Z[X] is of type (T) with0 ≤ a ≤ 18, by (5). Using (5) we obtain that there are 211 such cubic polynomials. Bycomputing approximations to the real root α > 1 of each of these 211 cubic polynomials,we check that the lower bound on dα given in (7) holds true for each of these 211 cubicpolynomials.

Remark 10. — The exponents 3/2 and 3 in (7) are optimal. Indeed,if Πα(X) = X3 −M2X2 − 2MX − 1, M ≥ 1, then M2 < α < M2 + 1, and dα = 4M3 + 27is asymptotic to 4α3/2.If Πα(X) = X3−MX2− 1, M ≥ 1, then M < α < M + 1, and dα = 4M3 + 27 is asymptoticto 4α3.

Remark 11. — We can reformulate the lower bound on dα in (7) as follows: let γ be a non-real cubic algebraic unit of negative discriminant satisfying |γ| > 1. Then |=(γ)| |γ|−1/2

(explicitly). We wish we understood beforehand why such a lower bound must hold true.

Corollary 12. — Let ε > 1 be a real cubic algebraic unit of negative discriminant. If ε = ηn

for some 1 < η ∈ Z[ε] and some n ≥ 3, then η ≤ 4.5 and n ≤ 10. In particular, by (5), ifΠη(X) = X3 − aX2 + bX − 1 ∈ Z[X] is of type (T), then 0 ≤ a ≤ 6 and |b| ≤

√4a+ 4.

Proof. — By (7) we have

η9/2/2 ≤ η3n/2/2 = ε3/2/2 ≤ dε = dη ≤ 4(η + η−1

)3,

that implies η ≤ 4.5. Moreover, we have η ≥ η0 = 1.32471 · · · (Lemma 7). Hence, by (7), wehave

1 = dη/dε ≤4(η + η−1)3

ε3/2/2= 8

(η + η−1

ηn/2

)3

≤ 8(η0 + η−1

0

ηn/20

)3

,

that implies n < 11.


S. Louboutin 49

3.4. Being a square. —

Lemma 13. — Let ε > 1 be a real cubic algebraic unit of negative discriminant. Thenε = η2 for some 1 < η ∈ Z[ε] if and only if we are in one of the following three cases:

1. Πε(X) = X3 − M2X2 − 2MX − 1 with M ≥ 1, in which case ε = η2 where η =ε2 −M2ε−M ∈ Z[ε], Πη(X) = X3 −MX2 − 1 and dη = dε = 4M3 + 27.

2. Πε(X) = X3 − 2X2 − 3X − 1, in which case ε = η2 where η = −ε2 + 3ε + 2 ∈ Z[ε],Πη(X) = X3 − 2X2 +X − 1 and dη = dε = 23.

3. Πε(X) = X3 − 2X2 + X − 1, in which case ε = η2 where η = ε2 − ε ∈ Z[ε], Πη(X) =X3 −X − 1 and dη = dε = 23.

Proof. — Assume that ε = η2 for some 1 < η ∈ Z[ε] with Πη(X) = X3−aX2+bX−1 ∈ Z[X]of type (T). Then Z[ε] = Z[η] and dε = dη (Lemma 2). Clearly, the index (Z[η] : Z[η2]) is equalto |ab−1|, where Πη(X) = X3−aX2 + bX−1. Hence, we must have |ab−1| = 1, and we willhave η = (ε2−(a2−b)ε−a)/(1−ab) and Πε(X) = Πη2(X) = X3−(a2−2b)X2+(b2−2a)X−1.First, assume that ab = 2. Then a = 2 and b = 1 (for a ≥ 0 and b ≤ a − 1), Πη(X) =X3 − 2X2 +X − 1 and Πε(X) = X3 − 2X2 − 3X − 1.Secondly, assume that ab = 0. If a = 0, then b ≤ a − 1 = −1 and dη = 4b3 + 27 > 0 yieldsb = −1, Πη(X) = X3 − X − 1, and Πε(X) = X3 − 2X2 + X − 1. If a 6= 0, then b = 0,Πη(X) = X3 − aX2 − 1, dη = 4a3 + 27 and Πε(X) = X3 − a2X2 − 2aX − 1.

3.5. Computation of explicit nth roots. — Let θ be an algebraic integer of degree n.Suppose that θ = αm for some α ∈ Q(θ) and some m ≥ 2. Let us explain how to computethe coordinates of α in the canonical Q-basis of Q(θ), provided that Πα(X) is known. First,we compute the matrix P := [pi,j ]1≤i,j≤m ∈Mn(Z) such that

θj−1 = αm(j−1) =n∑

i=1pi,jα

i−1 (1 ≤ j ≤ n).

Since Q(α) ⊆ Q(θ) = Q(αn) ⊆ Q(α), we have Q(α) = Q(θ). Therefore, detP 6= 0 and we cancompute P−1 = [qi,j ]1≤i,j≤n ∈Mn(Q). Clearly, we have α = ∑n

i=1 qi,2αm(i−1) = ∑n

i=1 qi,2θi−1.

For example, if Πα(X) = X3 − uX2 + vX − w ∈ Z[X] is the minimal polynomial of a cubicalgebraic number α, then θ = α2 is also a cubic algebraic number, Πα2(X) = X3 − (u2 −2v)X2 + (v2 − 2uw)X − w2 ∈ Z[X] and α = (θ2 + (v − u2)θ − uw)/(w − uv).

4. The totally imaginary quartic caseThe aim of this Section 4 is to prove Theorem 18. Indeed, after having found a completelydifferent proof of Nagell’s result we thought it should now be possible to settle this third casewhere the rank of the group of units of the order Z[ε] is equal to 1. In [Lou08a] we partiallysolved this problem and conjectured Theorem 18. We could not prove it because we could notcome up with lower bounds on discriminants of totally imaginary quartic algebraic units (seeTheorem 24). Such a lower bound was then obtained in [PL] and their proof was simplifiedin [Lou10].



Let ε be an algebraic quartic unit which is not a complex root of unity, and for which the rankof the group of units of the quadratic order Z[ε] is equal to 1. Hence, ε is totally imaginaryand |ε| 6= 1 (use [Was, Lemma 1.6]). Notice that if ε1, ε2 = ε1, ε3 and ε4 = ε3 are the fourcomplex conjugates of ε, then 1 = ε1ε2ε3ε4 = |ε1|2|ε3|2). By changing ε into −ε, 1/ε, or −1/εif necessary, we may and we will assume that its minimal polynomial Πε(X) is of type (T):

Definition 14. — A quartic polynomial of type (T) is a Q-irreducible monic quartic poly-nomial P (X) = X4 − aX3 + bX2 − cX + 1 ∈ Z[X] which satisfies |c| ≤ a and which has noreal root (see Lemma 15 for a characterization). It is of positive discriminant DP (X).

Lemma 15. — Let εP be any complex root of a quartic polynomial P (X) = X4 − aX3 +bX2 − cX + 1 ∈ Z[X] of type (T). Then

(10) − 1 ≤ b ≤ |εP |2 + 1/|εP |2 + 4 and |c| ≤ a ≤√

4b+ 5.

Proof. — Let ρeiφ, ρe−iφ, ρ−1eiψ and ρ−1e−iψ be the four complex roots of P (X). Thena = 2ρ cosφ+ 2ρ−1 cosψ and(11) b = ρ2 + ρ−2 + 4(cosφ)(cosψ).Hence,(12) 4b− a2 = 4(sinφ)2ρ2 + 4(sinψ)2ρ−2 + 8(cosφ)(cosψ) > −8.Since 4b− a2 ≡ 0 or 3 (mod 4), we have 4b− a2 ≥ −5.

Lemma 16. — Let P (X) = X4 − aX3 + bX2 − cX + d ∈ Q[X] be Q-irreducible. ThenP (X) has no real root if and only if DP (X) > 0 and either A := 3a2 − 8b < 0 or B :=3a4 − 16a2b+ 16ac+ 16b2 − 64d < 0.Assume moreover that d = 1 and let η = ρeiα, η, η′ = ρ−1eiβ and η′ be these four non-realroots. Then ρ2 + 1/ρ2 ≥ 2, 2 cos(α+ β) and 2 cos(α− β) are the roots of

R(X) = X3 − bX2 + (ac− 4)X − (a2 − 4b+ c2) ∈ Q[X],of positive discriminant DR(X) = DP (X) = dη.

Proof. — Write P (X) = (X − α1)(X − α2)(X − α3)(X − α4) in C[X].1. Set β1 = (α1 + α2 − α3 − α4)2, β2 = (α1 − α2 + α3 − α4)2 and β3 = (α1 − α2 − α3 + α4)2.Then Q(X) := (X − β1)(X − β2)(X − β3) = X3−AX2 +BX −C, where A and B are givenin the statement of Lemma 16 and C = (a3 − 4ab+ 8c)2. Moreover, DQ(X) = 212DP (X).(i) If P (X) has two real roots, say α1 and α2, and two non-real roots, say α3 and α4 = α3,then β1 > 0, whereas β2 and β3 = β2 are non-real. Hence DQ(X) < 0. (ii). If P (X) has fournon-real roots, say α1, α2 = α1, α3 and α4 = α3, then Q(X) has three real roots β1 > 0,β2 < 0 and β3 < 0. Hence DQ(X) > 0 and Q′(X) = 3X2 − 2AX +B has a negative real rootγ ∈ (β2, β3), which implies A < 0 or B < 0. (iii). If P (X) has four real roots, say α1, α2, α3and α4, then Q(X) has three real roots β1 > 0, β2 > 0 and β3 > 0. Hence DQ(X) > 0 andQ′(X) = 3X2 − 2AX + B has two positive real roots γ1 and γ2, which implies A ≥ 0 andB ≥ 0.The proof of the first part is complete.2. Set γ1 = α1α2 + α3α4, γ2 = α1α3 + α2α4 and γ3 = α1α4 + α2α3. Then R(X) := (X −γ1)(X − γ2)(X − γ3) = X3 − bX2 + (ac− 4d)X − (a2d− 4bd+ c2), and DR(X) = DP (X). In


S. Louboutin 51

our situation, d = 1 and we may assume that α1 = η, α2 = η, α3 = η′ and α4 = η′. Hence,γ1 = |η|2 + 1/|η|2 ≥ 2, γ2 = 2<(ηη′) = 2 cos(α+ β) and γ3 = 2<(ηη′) = 2 cos(α− β). Notice that (10) and the first part of Lemma 16 make it easy to list of all the quarticpolynomials of type (T) whose roots are of absolute values less than or equal to a givenupper bound B. Using the second part of Lemma 16 we can compute these absolute values.Taking B = 2, we obtain:

Lemma 17. — Let η be a totally imaginary quartic unit. If |η| > 1 then |η| ≥ |η0| =1.18375 · · · , where Πη0(X) = X4 −X3 + 1.

4.1. Statement of the result for the quartic case. —Theorem 18. — Let ε be a totally imaginary quartic unit with Πε(X) of type (T). Assumethat ε is not a complex root of unity. Let η be a fundamental unit of Z[ε]. We can chooseη = ε, except in the following cases:

1. The infinite family of exceptions for which Πε(X) = X4− 2bX3 + (b2 + 2)X2− (2b−1)X + 1, b ≥ 3, in which cases ε = −1/η2 where η = ε3− 2bε2 + (b2 + 1)ε− (b− 1) ∈ Z[ε]is a root of Πη(X) = X4 − X3 + bX2 + 1 of type (T), Z[η] = Z[ε] and dη = dε =16b4 − 4b3 − 128b2 + 144b+ 229.

2. The 14 following sporadic exceptions(a) (i) Πε(X) = X4−3X3 +2X2 +1, in which case ε = −η−2 where η = −ε2 +ε+1 ∈

Z[ε] is a root of Πη(X) = X4 − 2X3 + 2X2 −X + 1 of type (T), Z[η] = Z[ε]and dε = dη = 117.

(ii) Πε(X) = X4 − 3X3 + 5X2 − 3X + 1, in which case ε = η2 where η = −ε3 +2ε2 − 2ε ∈ Z[ε] is a root of Πη(X) = X4 − X3 − X2 + X + 1 of type (T),Z[η] = Z[ε] and dε = dη = 117.

(iii) Πε(X) = X4−5X3 +8X2−4X+1, in which case ζ3 = ε3−4ε2 +5ε−2 ∈ Z[ε]and ε = ζ3η3 where η = −ε2 + 3ε− 1 ∈ Z[ε] is a root of Πη(X) = X4− 2X3 +2X2 −X + 1 is of type (T), Z[η] = Z[ε] and dη = dε = 117.

(b) Πε(X) = X4 − 5X3 + 9X2 − 5X + 1, in which case ε = −η2 where η = −ε3 + 4ε2 −6ε+ 2 ∈ Z[ε] is a root of Πη(X) = X4 −X3 + 3X2 −X + 1 of type (T), Z[η] = Z[ε]and dη = dε = 189.

(c) (i) Πε(X) = X4−X3+2X2+1, in which case ε = η2 where η = −ε3+ε2−ε ∈ Z[ε].(ii) Πε(X) = X4−3X3 +3X2−X+1, in which case ε = 1/η3 where η = −ε+1 ∈

Z[ε].(iii) Πε(X) = X4 − 4X3 + 6X2 − 3X + 1, in which case ε = −1/η4 where η =

ε3 − 3ε2 + 3ε ∈ Z[ε].(iv) Πε(X) = X4 − 5X3 + 5X2 + 3X + 1, in which case ε = −η6 where η =

ε2 − 2ε− 1 ∈ Z[ε].(v) Πε(X) = X4 − 7X3 + 14X2 − 6X + 1, in which case ε = −1/η7 where η =

ε2 − 4ε+ 2 ∈ Z[ε].In these five cases, Πη(X) = X4 −X3 + 1 is of type (T), Z[η] = Z[ε] and dη = dε =229.

(d) (i) Πε(X) = X4 − 2X3 + 3X2 − X + 1, in which case ε = −1/η2 where η =ε3 − 2ε2 + 2ε ∈ Z[ε].



(ii) Πε(X) = X4−3X3 +X2 +2X+1, in which case ε = 1/η3 where η = ε2−2ε ∈Z[ε].

(iii) Πε(X) = X4−5X3 +7X2−2X+1, in which case ε = −η4 where η = ε2−2ε ∈Z[ε].

In these three cases, Πη(X) = X4 − X3 + X2 + 1 is of type (T), Z[η] = Z[ε] anddη = dε = 257.

(e) Πε(X) = X4−4X3 +7X2−4X+1, in which case ζ4 = −ε3 +4ε2−6ε+2 ∈ Z[ε] andε = ζ4η2, where η = −ε3 +3ε2−4ε+1 ∈ Z[ε] is a root of Πη(X) = X4−2X3 +X2 +1of type (T), Z[η] = Z[ε] and dη = dε = 272.

(f) Πε(X) = X4−13X3 +43X2−5X+1, in which case ε = −η3 where η = −ε3 +6ε2 +3ε ∈ Z[ε] is a root of Πη(X) = X4 − 2X3 + 4X2 −X + 1 of type (T), Z[η] = Z[ε]and dη = dε = 1229.

4.2. Sketch of proof. — Let ε be a totally imaginary quartic unit which is not a complexroot of unity. Compared with the non-totally real cubic case, this quartic case is more tricky.The first problem is that the totally imaginary quartic order Z[ε] may contain complex roots ofunity. Let µ(ε) denote the cyclic group of order 2N ≥ 2 of complex roots of unity containedin Z[ε]. Since the cyclotomic field of conductor 2N and degree φ(2N) is contained in thequartic field Q(ε), we obtain that φ(2N) divides 4, hence that 2N ∈ 2, 4, 6, 8, 10, 12.We will devote Section 4.4 to the case that 2N ∈ 8, 10, 12 and will settle our problem inthis situation.Hence we may and we now assume that 2N ∈ 2, 4, 6.We want to determine when ε = ζηp for some ζ ∈ µ(ε), some η ∈ Z[ε] and some primep ≥ 2. We may and we will assume that η is also a totally imaginary quartic unit which isnot a complex root of unity (if η is not totally imaginary then it is a real quadratic unit andζ 6= ±1, and we have ε = ζ ′η′p with η′ = ζη ∈ Z[ε] a totally imaginary quartic unit andζ ′ = ζ1−p ∈ µ(ε)).Clearly there are three subcases.

1. For p = 2 we determine when ε = ±η2 (Lemma 20), and when ε = ζ4η2 where ζ4 is acomplex root of unity of order 4 in Z[ε] (Lemma 22).

2. For p = 3 we determine when ε = η3 (next subcase), and when ε = ζ3η3 where ζ3 is acomplex root of unity of order 3 in Z[ε] (Lemma 22).

3. For p ≥ 3 we determine when ε = ηp for some η ∈ Z[ε]. Using |η| ≥ |η0| = 1.18375 · · ·(Lemma 17) and a double bound (13) for dε similar to (4) and (7), we will obtain inCorollary 29 that p ∈ 3, 5, 7.

In the cubic case, if ε = ηp > 1 and Πε(X) is of type (T), then so is Πη(X). This is no longertrue in the present quartic case. For example, if Πε(X) = X4 − 3X3 +X2 + 2X + 1, of type(T), and η = −ε2 + ε+ 1 ∈ Z[ε], then ε = η3 and Πη(X) = X4 +X2 −X + 1 is not of type(T). Since we want η to be of type (T) in all our statements, we may have to present ourresults in using −η, 1/η or −1/η instead, as in Lemma 20.Putting together the results of Lemma 20, Lemma 22 and Corollary 29, we obtain Table 2(similar to Table 1 of section 3.2, where we single out the cases b = 1 and b = 2 of point1 of Lemma 20, of discriminants D = 257 and D = 229), which completes the proof ofTheorem 18:


S. Louboutin 53

p Πη(X) Q(X) D2 X4 −X3 −X2 +X + 1 Πη2(X) = X4 − 3X3 + 5X2 − 3X + 1 1172 X4 − 2X3 + 2X2 −X + 1 Π−1/η2(X) = X4 − 3X3 + 2X2 + 1 1173 X4 − 2X3 + 2X2 −X + 1 Πζ3η3(X) = X4 − 5X3 + 8X2 − 4X + 1 1172 X4 −X3 + 3X2 −X + 1 Π−η2(X) = X4 − 5X3 + 9X2 − 5X + 1 1893 X4 −X3 + 1 Π1/η3(X) = X4 − 3X3 + 3X2 −X + 1 2292 X4 − 3X3 + 3X2 −X + 1 Π−1/η2(X) = X4 − 5X3 + 5X2 + 3X + 1 2292 X4 −X3 + 1 Πη2(X) = X4 −X3 + 2X2 + 1 2292 X4 −X3 + 2X2 + 1 Π−1/η2(X) = X4 − 4X3 + 6X2 − 3X + 1 2293 X4 −X3 + 2X2 + 1 Π−η3(X) = X4 − 5X3 + 5X2 + 3X + 1 2297 X4 −X3 + 1 Π−1/η7(X) = X4 − 7X3 + 14X2 − 6X + 1 2292 X4 −X3 +X2 + 1 Π−1/η2(X) = X4 − 2X3 + 3X2 −X + 1 2572 X4 − 2X3 + 3X2 −X + 1 Π−1/η2(X) = X4 − 5X3 + 7X2 − 2X + 1 2573 X4 −X3 +X2 + 1 Π1/η3(X) = X4 − 3X3 +X2 + 2X + 1 2572 X4 − 2X3 +X2 + 1 Πζ4η2(X) = X4 − 4X3 + 7X2 − 4X + 1 2723 X4 − 2X3 + 4X2 −X + 1 Π−η3(X) = X4 − 13X3 + 43X2 − 5X + 1 1229

Table 2.

Remark 19. — Contrary to the non-totally real cubic case, here ε = ζηn, with η ∈ Z[ε]and ζ ∈ µ(ε), does not always imply Z[ε] = Z[η]. For example for ε = ζ5η = 1 + ζ2

5 withη = ζ5 + ζ4

5 ∈ Q(√

5) we have Z[ε] = Z[ζ5] 6= Z[η] = Z[(1+√

5)/2] (notice that ζ5 = (ε−1)3 ∈Z[ε] and η = ζ5 + ζ4

5 ∈ Z[ζ5] ⊆ Z[ε]). Indeed, for this conclusion to hold true we need to haveQ(ε) = Q(η), i.e. η must also be a totally imaginary quartic unit. But it is not quite clearto us whether this necessary condition for the conclusion to hold true is also sufficient. Butin the case that η is also a totally imaginary quartic unit, then Z[η] ⊆ Z[ε] implies that dεdivides dη.

4.3. Being a square. —Lemma 20. — Let ε be a totally imaginary quartic algebraic unit which is not a complexroot of unity, with Πε(X) of type (T). Then ±ε is a square in Z[ε] if and only if we are inone of the seven following cases:

1. Πε(X) = X4−2bX3 + (b2 + 2)X2− (2b−1)X+ 1, b ≥ 1, in which cases ε = −1/η2 whereη = ε3− 2bε2 + (b2 + 1)ε− (b− 1) ∈ Z[ε] is a root of Πη(X) = X4−X3 + bX2 + 1 of type(T), and dε = dη = 16b4 − 4b3 − 128b2 + 114b+ 229.

2. Πε(X) = X4 −X3 + 2X2 + 1, in which case ε = η2 where η = −ε3 + ε2 − ε ∈ Z[ε] is aroot of Πη(X) = X4 −X3 + 1 of type (T), and dε = dη = 229.

3. Πε(X) = X4 − 3X3 + 2X2 + 1, in which case ε = −η−2 where η = −ε2 + ε+ 1 ∈ Z[ε] isa root of Πη(X) = X4 − 2X3 + 2X2 −X + 1 of type (T), and dε = dη = 117.

4. Πε(X) = X4−3X3 +5X2−3X+1, in which case ε = η2 where η = −ε3 +2ε2−2ε ∈ Z[ε]is a root of Πη(X) = X4 −X3 −X2 +X + 1 of type (T), and dε = dη = 117.

5. Πε(X) = X4−5X3 +9X2−5X+1, in which case ε = −η2 where η = −ε3 +4ε2−6ε+2 ∈Z[ε] is a root of Πη(X) = X4 −X3 + 3X2 −X + 1 of type (T), and dε = dη = 189.



6. Πε(X) = X4−5X3 +5X2 +3X+1, in which case ε = −η−2 where η = −ε2 +2ε+2 ∈ Z[ε]is a root of Πη(X) = X4 − 3X3 + 3X2 −X + 1 of type (T), dη = dε = 229 .

7. Πε(X) = X4−5X3+7X2−2X+1, in which case ε = −η−2 where η = ε3−4ε2+4ε ∈ Z[ε]is a root of Πη(X) = X4 − 2X3 + 3X2 −X + 1 of type (T), and dη = dε = 257.

Proof. — Assume that ε = ±η2 or ±η−2 for some η ∈ Z[ε], with Πη(X) = X4−aX3 +bX2−cX + 1 ∈ Z[X] of type (T). Hence,

−1 ≤ b and |c| ≤ a ≤√

4b+ 5.The index (Z[η] : Z[η2]) is equal to |a2 + c2 − abc|. Hence, we must have

|a2 + c2 − abc| = 1,and we will have Πη2(X) = X4−AX3 +BX2−CX+ 1, where A = a2−2b, B = b2−2ac+ 2and C = c2 − 2b.Assume that c = 0. Then 1 = |a2 + c2 − abc| = a2, hence a = 1 and we are in the first or thesecond case.Assume that c 6= 0. Then a ≥ 1 and a2 + c2 − abc = ±1 yield |b| ≤ f(|c|), where f(x) =ax + x

a + 1ax is convex. Hence,

|b| ≤ g(a) := max(f(1), f(a)) = max(a+ 2/a, 2 + 1/a2) = a+ 2/a.Since g is convex we obtain |b| ≤ max(g(1), g(

√4b+ 5)) = max(3, 4b+7√

4b+5). Hence, b ≤ 5.There are 9 triplets (a, b, c) satisfying −1 ≤ b ≤ 5 and 1 ≤ |c| ≤ a ≤

√4b+ 5 for which

|a2+c2−abc| = 1. Getting rid of the three of them for which Πη(X) = X4−aX3+bX2−cX+1is of negative discriminant and of the one of them for which η is a 5th complex root of unity,namely (a, b, c) = (1, 1, 1), we fall in one of the five remaining last cases.Finally, by choosing between the four units ε = ±η2 or ε = ±η−2 the ones for which Πε(X) =X4 −AX3 +BX2 − CX + 1 satisfies |C| ≤ A, we complete the proof of this Lemma.

4.4. The case that µ(ε) is of order 8, 10 or 12. —

Lemma 21. — Let ε be a totally imaginary quartic algebraic unit which is not a complexroot of unity. If ζ2N ∈ Z[ε], with 2N ∈ 8, 10, 12, then ε is a fundamental unit of the orderZ[ε].

Proof. — We have Z[ε] ⊆ Z[ζ2N ] (Z[ζ2N ] is the ring of algebraic integers of Q(ζ2N )). Hence,Z[ε] = Z[ζ2N ] and dε = dζ2N . Let η2N be a fundamental unit of Z[ζ2N ]. Then η2N ∈ Z[ε] andε = ζm2Nη

n2N , with m ∈ Z and 0 6= n ∈ Z. We want to prove that |n| = 1. We prove that

|n| ≥ 2 implies dε > dζ2N . Set ρ = |η2N |. Let σt be the Q-automorphism of Q(ζ2N ) such thatσt(ζ2N ) = ζt2N , where gcd(t, 2N) = 1 and t 6≡ ±1 (mod 2N), i.e. σt is neither the identity northe complex conjugation. Then ε1 = ε, ε2 = σt(ε), ε3 = ε1 and ε4 = ε2 are the four complexconjugates of ε. Since |ε1|2|ε2|2 = ε1ε2ε3ε4 = NQ(ζ2N )/Q(ε) = 1, we have |ε2| = 1/|ε1| = 1/ρnand

dε = ((ε1 − ε2)(ε1 − ε3)(ε1 − ε4)(ε2 − ε3)(ε2 − ε4)(ε3 − ε4))2

= 16=2(ε1)=2(ε2)|ε1 − ε2|4|ε1 − ε2|4 ≥ 16=2(ε)=2(σt(ε))|ρn − 1/ρn|8.Notice that if η2N is real, then 16=2(ε)=2(σt(ε)) = 16 sin2(πmN ) sin2( tπmN ).


S. Louboutin 55

1. If 2N = 8, we may take η8 = 1 +√

2 = ρ and t = 3 and we obtain

dε ≥ 16 sin2(πm4 ) sin2(3πm4 )(ρn − ρ−n)8 ≥ 4(ρ2 − ρ−2)8 = 222 > 256 = dζ8

(since ε is totally imaginary, we have m 6≡ 0 (mod 4)).2. If 2N = 10, we may take η10 = (1 +

√5)/2 = ρ and t = 3 and we obtain

dε ≥ 16 sin2(πm5 ) sin2(3πm5 )(ρn − ρ−n)8 ≥ 5(ρ2 − ρ−2)8 = 55 > 125 = dζ10

(since ε is totally imaginary, we have m 6≡ 0 (mod 5)).3. If 2N = 12, then ε0 = 2 +

√3 = |1 + ζ12|2 = NQ(ζ12)/Q(

√3)(1 + ζ12) is the fundamental unit

of Z[√

3]. Hence, we may take η12 = 1 + ζ12, ρ = ε1/20 and t = 5. Noticing that η12 = ζ24(ζ24 +

ζ−124 ) = ζ24ρ and σ(η12) = 1 + ζ5

12 = ζ312

1+ζ12= ζ3

12/η12 = ζ524/ρ, we obtain 16=2(ε)=2(σt(ε)) =

16 sin2( (2m+n)π12 ) sin2(5(2m+n)π

12 ) ∈ 1, 4, 9, 16 and

dε ≥ (εn/20 − ε−n/20 )8 ≥ (ε0 − 1/ε0)8 = 1442 > 144 = dζ12

(since ε is totally imaginary, we have 2m+ n 6≡ 0 (mod 12)).

4.5. The case that ε = ζ4η2 or ε = ζ3η3. —

Lemma 22. — Let ε be a totally imaginary quartic unit which is not a complex root ofunity such that Πε(X) = X4 − aX3 + bX2 − cX + 1 satisfies |c| ≤ a.

1. Assume that ε = ζ3η3 for some totally imaginary quartic unit η ∈ Z[ε] and some complexroot of unity ζ3 ∈ Z[ε] of order 3. Then Πε(X) = X4 − 5X3 + 8X2 − 4X + 1, in whichcase ε = ζ3η3, where η = −ε2 + 3ε−1 ∈ Z[ε] and ζ3 = ε3−4ε2 + 5ε−2 ∈ Z[ε]. Moreover,Πη(X) = X4 − 2X3 + 2X2 −X + 1 is of type (T) and dε = dη = 117 and Z[ε] = Z[η].

2. Assume that ε = ζ4η2 for some totally imaginary quartic unit η ∈ Z[ε] and some complexroot of unity ζ4 ∈ Z[ε] of order 4. Then Πε(X) = X4 − 4X3 + 7X2 − 4X + 1, in whichcase ε = ζ4η2, where η = −ε3 + 3ε2 − 4ε+ 1 ∈ Z[ε] and ζ4 = −ε3 + 4ε2 − 6ε+ 2 ∈ Z[ε].Moreover, Πη(X) = X4 − 2X3 +X2 + 1 is of type (T), dε = dη = 272 and Z[ε] = Z[η].

Proof. — Set K := Q(ζ3) and A := Z[ζ3]. Since η is quadratic over K, there exist α and β inA such that η2 − αη + β = 0. Clearly, β ∈ A∗ and α 6= 0. Moreover, A[η] ⊆ A[ε] = A[ζ3η3] =A[η3] ⊆ A[η] yields A[η] = A[η3]. Since, η3 = (α2 − β)η − αβ and since η3 6∈ A (otherwise ηwould be a complex root of unity), we obtain α2−β ∈ A∗. Now, 1 ≤ |α|2 ≤ |α2−β|+ |β| = 2yields |α|2 = 1 (there is no element of norm 2 in A) and α ∈ A∗. Hence, α, β and α2−β are inA∗. Setting β = −α2γ with γ ∈ A∗, we have 1+γ ∈ A∗ = ±1,±ζ3,±ζ2

3. Hence, γ ∈ ζ3, ζ23

and η2−αη−α2γ = 0 yields that ε = ζ3η3 is a root of P (X) = X2−δζ3(3γ+1)X−ζ23 ∈ K[X]

(use α6 = γ3 = 1), where δ = α3 ∈ ±1. Hence,

Πε(X) = P (X)P (X) =X4 + 4δX3 + 8X2 + 5δX + 1 if γ = ζ3,X4 − 5δX3 + 8X2 − 4δX + 1 if γ = ζ2

3 .Set K := Q(ζ4) and A := Z[ζ4]. Since η is quadratic over K, there exist α and β in A such thatη2 − αη + β = 0. Clearly, β ∈ A∗ and α 6= 0. Moreover, A[η] ⊆ A[ε] = A[ζ4η2] = A[η2] ⊆ A[η]yields A[η] = A[η2]. Since, η2 = αη − β and since η2 6∈ A we obtain α ∈ A∗. Hence, α and β



are in A∗. Setting β = −α2γ with γ ∈ A∗, we obtain η2 − αη − α2γ = 0, with α, γ ∈ A∗ =±1,±ζ4. It follows that ε = ζ4η2 is a root of P (X) = X2 − δζ4(1 + 2γ)X − γ2 ∈ K[X] (useα4 = 1), where δ = α2 ∈ ±1. Hence,

Πε(X) = P (X)P (X) =

X4 −X2 + 1 if γ = −1,X4 + 4δX3 + 7X2 + 4δX + 1 if γ = ζ4,

X4 − 4δX3 + 7X2 − 4δX + 1 if γ = −ζ4,

X4 + 7X2 + 1 if γ = 1.If γ = −1 then ε is a complex root of unity of order 12, a contradiction. If γ = 1 thenζ4 ∈ ±(ε3 + 8ε)/3 is not in Z[ε].

Remark 23. — Whereas there are only finitely many cases for which the quartic order Z[ε]contains a complex root of unity of order 8, 10 or 12, by Lemma 21, it happens infinitely oftenthat it contains a complex root of unity of order 3 or 4. For example, if Πε(X) = X4−2AX3+(A2−1)X2 +AX+1, A ≥ 0, then ζ3 = −ε2 +Aε ∈ Z[ε]. If Πε(X) = X4−2AX3 +A2X2 +1,A ≥ 0, then ζ4 = ε2 −Aε ∈ Z[ε].


Theorem 24. — Let α be a totally imaginary quartic algebraic unit. Then(13) 7 max(|α|4, |α|−4)/10 ≤ dα ≤ 16(|α|2 + |α|−2)4 ≤ 256 max(|α|8, |α|−8).Hence, if P (X) = X4− aX3 + bX2− cX + 1 ∈ Z[X] is Q-irreducible of positive discriminantDP (X) > 0 and with no real root, then

(14) 7H∞(P (X))2/10 ≤ DP (X) ≤ 16H1(P (X))4 ≤ 256H∞(P (X))4.

Proof. — Clearly, (14) follows from (13): if α, α, β and β are the (non-real) roots of P (X),then |α|2|β|2 = 1, hence Hp(P ) = (|α|2p + |α|−2p)1/p = (|β|2p + |β|−2p)1/p and H∞(P (X)) =max(|α|2, |α|−2) = max(|β|2, |β|−2).Since both terms of (13) remain unchanged if we change α into −α, 1/α and −1/α, we mayassume that Πα(X) = X4 − aX3 + bX2 − cX + 1 ∈ Z[X] satisfies |c| ≤ a. By taking anappropriate choice of the root of Πα(X), we may also assume that ρ := |α| ≥ 1.Let α = ρeiφ, α, α′ = ρ−1eiψ and α′ be the four complex roots of Πα(X) (use |α|2|α′|2 = 1).Then

dα =((α− α)(α− α′)(α− α′)(α− α′)(α− α′)(α′ − α′))2

= 16(sinφ)2(sinψ)2|ρ− ρ−1ei(ψ−φ)|4|ρ− ρ−1ei(ψ+φ)|4.Setting X = ρ2 + ρ−2, we have dα = 4 (F (cos(ψ − φ), cos(ψ + φ)))2 ≤ 16X4, by Lemma 25below.(Notice that dα = 256 = 16(|α|2 + |α|−2)4 = 256 max(|α|8, |α|−8) if α = ζ8).Asssume that ρ ≥

√3 and a ≥ 37.

Then |2ρ−1 cosφ+ 2ρ cosψ| = |c| ≤ a = 2ρ cosφ+ 2ρ−1 cosψ implies cosφ ≥ | cosψ| anddα ≥

(4(ρ− ρ−1)4 sin2 φ

)2.

First, if sin2 φ ≥ 34ρ−2, then dα ≥

(3(1− ρ−2)

)2ρ4 ≥ 4ρ4 (use ρ ≥

√3).


S. Louboutin 57

Secondly, assume that sin2 φ < 34ρ−2. Since ρ ≥ 1, we have

−8 < 4b− a2 < 3 + 4 sin2 ψ + 8√

1− sin2 ψ ≤ 11,

by (12) (α is totally imaginary, hence sinφ 6= 0 and sinψ 6= 0, and√

1− t ≤ 1−t/2 for 0 ≤ t =sin2 ψ ≤ 1). Since 4b−a2 ≡ 0 or 3 (mod 4), we obtain J := 4b−a2 ∈ −5,−4,−1, 0, 3, 4, 7, 8and we are in the situation of Lemma 26. By Lemma 27 we have dα ≥ 4 max(|α|4, |α|−4) = 4ρ4.Finally, since |c| ≤ a ≤

√4b+ 5 and −1 ≤ b ≤ ρ2 + ρ−2 + 4, it is easy to list all the possible

polynomials Πα(X) for which 1 ≤ ρ ≤√

3 or for which a ≤ 36 and b = (a2 + J)/4 withJ ∈ −5,−4,−1, 0, 3, 4, 7, 8 and also to check that the lower bound for dα in (13) holds truefor these polynomials, by using Lemma 16.

Lemma 25. — For X ≥ 2 we have sup|x|,|y|≤1 F (x, y) ≤ 2X2, where F (x, y) := |x− y|(X −2x)(X − 2y).

Proof. — First, for X ≥ 2 we haveS1 := sup

|x|,|y|≤1xy≥0

F (x, y) = sup0≤x≤y≤1

(y − x)(X + 2x)(X + 2y)

= sup0≤x≤1

(1− x)(X + 2x)(X + 2) = X(X + 2) ≤ 2X2.

Secondly, for X ≥ 2 we haveS2 := sup

|x|,|y|≤1xy≤0

F (x, y)

= sup0≤x,y≤1

(x+ y)(X − 2x)(X + 2y) = sup0≤x≤1

(x+ 1)(X − 2x)(X + 2).

IfX ≥ 6 then S2 = 2(X−2)(X+2)≤ 2X2. If 2 ≤ X ≤ 6 then f(x) := (x+1)(X−2x)(X+2) ≤f((X − 2)/4) = (X + 2)3/8 ≤ 2X2.

Lemma 26. — Fix J ∈ −5,−4,−1, 0, 3, 4, 7, 8. For a ∈ Z with a2 ≡ −J (mod 4), i.e.with a even if J ≡ 0 (mod 4) and a odd if J ≡ 3 (mod 4), set

ΠJ(X) = X4 − aX3 + a2 + J

4 X2 − cX + 1 ∈ Z[X].

Assume |c| ≤ a, that a ≥ 15 and that ΠJ(X) is Q-irreducible with no real roots, hence is ofpositive discriminant D(a, J, c) (a quartic polynomial in c). Set

B :=

a− 1 if J = 8Ja/8 if J ∈ −4, 0, 4(Ja− 1)/8 if J ∈ −5,−1, 3, 7.

Then −a ≤ c ≤ B and D(a, J, c) ≥ F (a, J) := min(D(a, J,−a), D(a, J,B)).

Proof. — Assume that a = 2A is even. Then J = 4j and ΠJ(A) = jA2−cA+1 > 0 (for ΠJ(X)has no real root) yields c ≤ jA = Ja/8. Moreover, since X4−2AX3 +(A2 +2)X2−2AX+1 =(X2 −AX + 1)2 is not irreducible, we obtain that c 6= a = Ja/8 for J = 8. Hence, c ≤ B.Now, assume that a is odd. Then 0 < 16ΠJ(a/2) = Ja2 − 8ac + 16 ≡ 3 (mod 4) yieldsJa2 − 8ac+ 16 ≥ 3. Hence, 8c ≤ Ja+ 13/a < Ja+ 1 for a ≥ 15. Hence, c ≤ (Ja− 1)/8 = B.



Numerical investigations suggest that D(a, J, c) as a function of c has four real roots close to−a, Ja/8, a and a3/54 + Ja/24. Set e = 0 if J is even and e = 1 if J is odd, and

∆(a, J, c) = −27(c+ a)(c− Ja− e

8

)(c− (a− 1 + e

8))(c− a3

54 −Ja

24 − 1).

Using any software of symbolic computation, we check that D(a, J, c) = ∆(a, J, c)+P (a, J, c),where P (a, J, c) = α(a, J)c2 +β(a, J)c+ γ(a, J) is a quadratic polynomial in c whose leadingcoefficient

α(a, J) =

−a3

2 +(3 + J2

64

)a2 +

(27− 9J

8

)a− J3

16 + 36J − 27 J even−a3

2 +(3 + J2

64

)a2 +

(1898 − 45J

64

)a− J3

16 + 36J − 153964 J odd

is less than or equal to 0 for J ∈ −5,−4,−1, 0, 3, 4, 7, 8 and a ≥ 14.Clearly, we have ∆(a, J, c) ≥ 0 for −a ≤ c ≤ B (notice that a3/54 − Ja/24 − 1 ≥ a3/54 −a/3− 1 ≥ a− 1 ≥ B for a ≥ 9). Hence, D(a, J, c) ≥ P (a, J, c). Since α(a, J) ≤ 0, we have

P (a, J, c) ≥ min(P (a, J,−a), P (a, J,B)) for − a ≤ c ≤ B.Finally, we have ∆(a, J,−a) = ∆(a, J,B) = 0. So, D(a, J,−a) = P (a, J,−a) and D(a, J,B) =P (a, J,B). The desired result follows.

Lemma 27. — Assume that Πα(X) = X4 − aX3 + a2+J4 X2 − cX + 1 ∈ Z[X] is a quartic

polynomial of type (T) with J ∈ −5,−4,−1, 0, 3, 4, 7, 8, a ∈ Z and a ≡ J (mod 2). Assumethat a ≥ 37. Then dα ≥ 4 max(|α|4, |α|−4).

Proof. — Set M = max(|α|, |α|−1). Using (11), we obtain(15) (a2 + J − 16)/8 ≤M2 ≤ (a2 + J + 16)/4.By Lemma 26, for J ∈ −4, 0, 4, 8 we have

D(a, J, c) ≥ min(D(a, J,−a), D(a, J,B))≥ D(a, 4, a/2) = 9((a2 − 8)2 + 192)/16

((i) check the quartic polynomials with positive leading coefficientD(a, J,−a)−D(a, 4, a/2) =(J+5)(J+11)

16 a4 + · · · are non-negative for J ∈ −4, 0, 4, 8 and a ≥ 1, (ii) check that D(a, 8, a−1)−D(a, 4, a/2) ≥ 0 for a ≥ 0 and (iii) check that the quartic polynomials with non-negativeleading coefficient D(a, J, Ja/8)−D(a, 4, a/2) = (16−J2)(112−J2)

4096 a4 + · · · are non-negative forJ ∈ −4, 0, 4 and a ∈ Z).Using (15), we have M2 ≥ (a2 − 20)/8 > 8 and a2 − 8 ≥ 4(M2 − 8) > 0. Hence, dα ≥9((M2 − 8)2 + 12) ≥ 4M4.In the same way, for J ∈ −5,−1, 3, 7 we have

D(a, J, c) ≥ min(D(a, J,−a), D(a, J, (Ja− 1)/8))≥ D(a,−5,−a) = 9((a2 + 19)2 − 192)/16

(check (i) that the quartic polynomials with non-negative leading coefficient D(a, J,−a) −D(a,−5,−a) = (J+5)(J+11)

16 a4 + · · · are non-negative for J ∈ −5,−1, 3, 7 and a ≥ 1and (ii) that the quintic polynomials with positive leading coefficient D(a, J, (Ja − 1)/8) −D(a,−5,−a) =

(1− J2

64

)a5

16 + · · · are positive for J ∈ −5,−1, 3, 7 and a ≥ 36).


S. Louboutin 59

Using (15), we have M2 ≥ (a2 − 21)/8 > 1 and 4(M2 − 1) ≤ a2 + 19. Hence, dα ≥9((M2 − 1)2 − 12) ≥ 4M4.

Remark 28. — The exponents 4 and 8 in (13) are optimal. Indeed, if Πα(X) = X4−2bX3+(b2 + 2)X2 − (2b − 1)X + 1, b ≥ 3, then |α|2 + 1/|α|2 is the root greater than 2 of R(X) =X3−(b2 +2)X2 +(4b2−2b−4)X−4b2 +4b+7 (Lemma 16), hence is asymptotic to b2. Hence,max(|α|, |α|−1) is asymptotic to b and dα is asymptotic to 16b4, hence to 16 max(|α|4, |α|−4).If Πβ(X) = X4 −X3 + bX2 + 1, b ≥ 3, then β2 = α and dβ = dα. Hence, dβ is asymptoticto 16 max(|β|8, |β|−8).

Corollary 29. — Let ε be a totally imaginary quartic algebraic unit which is not a complexroot of unity. If ε = ηn for some η ∈ Z[ε] and some n ∈ Z, then max(|η|, |η|−1) < 2.27 and|n| ≤ 9. In particular, by (10), if Πη(X) = X4 − aX3 + bX2 − cX + 1 ∈ Z[X] is of type (T),then −1 ≤ b ≤ 9 and |c| ≤ a ≤

√4b+ 5.

Proof. — We may assume that |ε| ≥ 1, that |η| ≥ 1 and that n ≥ 3. Notice that η isnecessarily a totally imaginary quartic algebraic unit which is not a complex root of unity.By (13), we have

7|η|12/10 ≤ 7|η|4n/10 = 7|ε|4/10 ≤ dε = dη ≤ 16(|η|2 + |η|−2

)4,

that implies |η| ≤ 2.27. Moreover, we have |η| ≥ |η0| = 1.18375 · · · (Lemma 17). Hence, by(13), we have

1 = dη/dε ≤16(|η|2 + |η|−2)4

7|ε|4/10 = 1607

(|η|2 + |η|−2

|η|n

)4

≤ 1607

(|η0|2 + |η0|−2

|η0|n

)4

,

that implies n < 10.

5. The totally real cubic caseThe aim of this Section 5 is to prove Theorem 36. After having settled in Sections 2, 3 and4 the problem of the determination of a fundamental unit of any order Z[ε] generated by aunit ε whenever the rank of the group of units is equal to 1 we attacked the general situationin [Lou12] . We solved this problem in the case that ε is a totally real cubic algebraic unit,in which case the rank of the group of units of the order Z[ε] is equal to 2. We managed toprove the best result one could expect, namely that there should exist a second unit η ∈ Z[ε]such that ε, η is a system of fundamental units of the cubic order Z[ε] (and unbeknownto us this problem was simultaneously solved in [BHMMS] and [MS]). We present here astreamlined proof of this result, once again based on lower bounds for discriminants of totallyreal algebraic cubic units (Theorem 33).Let ε be a totally real cubic unit of Q-irreducible minimal polynomial Πε(X) = X3 − aX2 +bX − c ∈ Z[X], c ∈ ±1, of positive discriminant Dε = dε = DΠε(X). Since the rank of thegroup of units of the cubic order Z[ε] is equal to 2, the natural question becomes: does thereexist a unit η ∈ Z[ε] such that ε, η is a system of fundamental units of Z[ε]? Clearly, thereexists η ∈ Z[ε] such that ε, η is a system of fundamental units of Z[ε] if and only if ±ε arenot an nth power in Z[ε] for any n ≥ 2 (e.g., see [Lou12, end of proof of Lemma 7]).



5.1. Cubic units of type (T+). —We may assume that |ε| ≥ |ε′| ≥ |ε′′| > 0, where ε, ε′and ε′′ are the three conjugates of ε, i.e. the three real roots of Πε(X). By changing ε into1/ε′′, we may assume that |ε| > 1 > |ε′| ≥ |ε′′| > 0 (use εε′ε′′ = c = ±1). Finally, by changingε into −ε, we obtain that at least one of the four polynomials equivalent to Πε(X) is of type(T+) (notice that if ε is of type (T+) and ε = ηn for some odd n ≥ 3 and some η ∈ Z[ε],then η is clearly also of type (T+), whereas if ε is reduced then it is not clear whether η isalso necessarily reduced):

Definition 30. — A cubic polynomial of type (T+) is a monic cubic polynomial P (X) =X3−aX2+bX−c ∈ Z[X], c ∈ ±1, which is Q-irreducible (⇔ b 6= a+c−1 and b 6= −a−c−1),of positive discriminant DP (X) > 0 and whose three real roots εP , ε′P and ε′′P can be sortedso as to satisfy

(16) εP > 1 > |ε′P | ≥ |ε′′P | > 0.

In that situation, Hp(P (X)) = (εpP + ε−pP )1/p and H∞(P (X)) = εP .

Lemma 31. — (See [Lou12, Lemma 3]). P (X) = X3−aX2 + bX− c ∈ Z[X] with c ∈ ±1is of type (T+) if and only if (i) DP (X) > 0 and (ii) −a − c ≤ b ≤ a + c − 2. It implies0 ≤ a ≤ H∞(P (X)) + 2.

Lemma 31 makes it easy to list of all the cubic polynomials of type (T+) whose real rootsare less than or equal to a given upper bound B. Taking B = 3, we obtain:

Lemma 32. — The real root η0 = 2.24697 · · · greater than one of Π(X) = X3−2X2−X+1of type (T+) is the smallest totally real cubic unit greater than 1 that satisfies (16).


Theorem 33. — Let α be a totally real cubic algebraic unit satisfying (16). Then

(17) α3/2/2 ≤ Dα ≤ 4(α+ α−1)4 ≤ 64α4.

Hence, if P (X) = X3 − aX2 + bX − c ∈ Z[X], c ∈ ±1, is Q-irreducible and of positivediscriminant DP (X) > 0, then

(18) H∞(P (X))3/2/2 ≤ DP (X) ≤ 4H1(P (X))4 ≤ 64H∞(P (X))4.

Proof. — Write α′ = t/√α and α′′ = c/t

√α, with 1 ≤ |t| ≤ √α. Then a = α+ (t+ c/t)/

√α,

b = (t+ c/t)√α+ c/α and

(19) b2 − 4ac = (t− c/t)2α− 2c(t+ c/t)/√α+ 1/α2.

Let us prove the upper bound. We have

Dα = (α− α′)2(α− α′′)2(α′ − α′′)2.


S. Louboutin 61

Hence, writing x =√α > 1, and 1 ≤ y = |t| ≤ x, we have

√Dα ≤ (

x2 + y

x

)(x2 + 1

xy

)(yx

+ 1xy

)

= (x3 + y + 1/y + 1/x3)(y + 1/y)≤ (x3 + x+ 1/x+ 1/x3)(x+ 1/x)= (x2 + 1/x2)(x+ 1/x)2 ≤ 2(x2 + x−2)2 = 2H1(P (X))2.

Let us now prove the lower bound. We haveDα ≥ (α− 1)4(α′ − α′′)2 = (α− 1)4(t− c/t)2/α.

Assume that α ≥ 44.First, assume that | t− c/t|2 ≥ 1/(3α). Then Dα ≥ (α− 1)4/(3α2) ≥ 2α3/2.Secondly, assume that |t − c/t|2 < 1/(3α). Then |t − c/t| < 2. Hence c = 1, |t| ≥ 1 and|t− 1/t| ≤ 5/6, hence 1 ≤ |t| ≤ 3/2 and 2|t+ c/t| = 2|t+ 1/t| ≤ 13/3 and (19) yields

−1 < −13/3√α< b2 − 4a < 1

3 + 13/3√α

+ 1α2 < 1.

Hence, b2 − 4a = 0, T = t+ 1/t 6∈ (−2, 2), αT 2 − 2T/√α + 1/α2 − 4α = 0, T = 2 + 1/α3/2,

t− 1/t =√

4/α3/2 + 1/α3 and

Dα = 4α3/2(

1− 2α5/2 + 1

α3 −1α4

)2 (1 + 1

4α3/2

)> 2α3/2

(and b = 2B = (t + 1/t)√α + 1/α = 2

√α + 2/α > 2 is even, a = B2 and P (X) =

X3 −B2X2 + 2BX − 1, B ≥ 3).Therefore, Dα ≥ 2α3/2 for α ≥ 44.Finally, using Lemma 31 to find the 3236 cubic polynomials P (X) = X3 − aX2 + bX − c ∈Z[X] of type (T+) with 0 ≤ a ≤ 45 (for 1 < εP < 44 implies a < 46), and checking thatDP (X) ≥ H∞(P (X))3/2/2 for each of these polynomials (which is equivalent to checking thatP ((2DP (X))2/3) > 0), we end up with the desired result.

Remark 34. — The exponents 3/2 and 4 in Theorem 33 are optimal. Indeed, if P (X) =X3−B2X2+2BX−1, B ≥ 3, (of type (T+) by Lemma 31), then DP (X) = 4B3−27 is asymp-totic to 4B3 and H∞(P (X)) is asymptotic to B2, i.e. DP (X) is asymptotic to 4H∞(P (X))3/2.If P (X) = X3 − aX2 − (a+ 3)X − 1, a ≥ −1, (of type (T+) by Lemma 31), then DP (X) =(a2 +3a+9)2 is asymptotic to a4 and H∞(P (X)) is asymptotic to a, i.e. DP (X) is asymptoticto H∞(P (X))4.

5.3. Being a square. —

Lemma 35. — Let ε be a reduced totally real algebraic cubic unit, i.e. such that Πε(X) =X3 − uX2 + vX − w ∈ Z[X] with Dε > 0, w ∈ ±1 and |v| ≤ u. Then ε = ±η2 for someη ∈ Z[ε] if and only if we are in one of the two following cases:

1. Πε(X) = X3 − 6X2 + 5X − 1, in which case ε = η2, where η = ε2 − 5ε + 2, Πη(X) =X3 − 2X2 −X + 1 and Dε = Dη = 49.



2. Πε(X) = X3−B2X2+2BX−1 with B ≥ 3, in which case ε = η2, where η = −ε2+B2ε−B,Πη(X) = X3 −BX2 + 1 and Dε = Dη = 4B3 − 27.

In both cases, η is reduced and η is not a square in Z[η] = Z[ε].

Proof. — Assume that ε = ±η2 for some η ∈ Z[ε]. By changing η into −η if necessary,we may assume that Πη(X) = X3 − aX2 + bX − 1 ∈ Z[X], which gives Πη2(X) = X3 −(a2 − 2b)X2 + (b2 − 2a)X − 1 and Π−η2(X) = X3 + (a2 − 2b)X2 + (b2 − 2a)X + 1. ThenZ[η2] = Z[ε] = Z[η] (Lemma 2), hence (Z[η] : Z[η2]) = |ab − 1| = 1, and we will haveη = (ε2 ∓ (a2 − b)ε− a)/(1− ab).First, assume that ab = 2. We are in one of the following eight cases, with X3−6X2 +5X−1being the only reduced polynomials of positive discriminant among these eight polynomials:

a b Πη2 (X) Π−η2 (X)2 1 X3 − 2X2 − 3X − 1 X3 + 2X2 − 3X + 11 2 X3 + 3X2 + 2X − 1 X3 − 3X2 + 2X + 1−2 −1 X3 − 6X2 + 5X − 1 X3 + 6X2 + 5X + 1−1 −2 X3 − 5X2 + 6X − 1 X3 + 5X2 + 6X + 1

Secondly, assume that ab = 0.If a = 0, then Πε(X) = Πη2(X) = X3+2bX2+b2X−1 with |b2| ≤ −2b implies b ∈ −2,−1, 0,and Πε(X) = Π−η2(X) = X3− 2bX2 + b2X + 1 with |b2| ≤ 2b implies b ∈ −0, 1, 2. In thesesix cases, either Πε(X) is reducible, a contradiction, or of negative discriminant, anothercontradiction.If b = 0 and a 6= 0, then Πε(X) = Π−η2(X) = X3 + a2X2 + 2aX + 1 with 0 6= |2a| ≤ −a2

is impossible, and Πε(X) = Πη2(X) = X3 − a2X2 − 2aX − 1 with 0 6= |2a| ≤ a2 andDε = −4a3 − 27 > 0 implies a = −B with B ≥ 3.

5.4. Statement and proof of the result for the totally real cubic case. —

Theorem 36. — Let ε be a reduced totally real algebraic cubic unit, i.e. such that Πε(X) =X3 − uX2 + vX − w ∈ Z[X] with Dε > 0, w ∈ ±1 and |v| ≤ u.If Πε(X) = X3 − 6X2 + 5X − 1, set ξ1 = ε2 − 5ε+ 2 for which ξ2

1 = ε.If Πε(X) = X3 −B2X2 + 2BX − 1 with B ≥ 3, set ξ1 = −ε2 +B2ε−B for which ξ2

1 = ε.Otherwise, set ξ1 = ε.Hence, ξ1 is always reduced, by Lemma 35.Then there exists another unit ξ2 ∈ Z[ε] such that ξ1, ξ2 is a system of fundamental unitsof the cubic order Z[ε].

Proof. — By Lemma 35, it suffices to prove that if ξ satisfies (16), then there does not existany prime p ≥ 3 and any η ∈ Z[ξ] such that ξ = ηp. Notice that such an η must also satisfy(16). We would have Z[ξ] = Z[η] and Dξ = Dη. Theorem 33 would yield

η9/2/2 ≤ η3p/2/2 ≤ ξ3/2/2 ≤ Dξ = Dη ≤ 4(η + η−1)4,

that implies 1 < η ≤ 64.2. If η0 is as in Lemma 32, we then obtain

1 = Dη/Dξ ≤4(η + η−1)4

ξ3/2/2= 8

(η + η−1

η3p/8

)4

≤ 8(η0 + η−1

0

η3p/80

)4

,

that implies p < 5, hence p = 3, ξ = η3 and Dη = Dη3 .


S. Louboutin 63

It remains to computationally check that Dη3 6= Dη for the 7271 cubic polynomial Πη(X) =X3 − aX2 + bX − c ∈ Z[X] of type (T+) with 0 ≤ a ≤ 66 (for 1 < εP ≤ 64.2 implies0 ≤ a ≤ 66). Noticing that

Dη3/Dη = |a3c+ b3 − a2b2|2and using Lemma 31, this is a straightforward verification Remark 37. — For several parametrized families of totally real cubic orders, an explicitsystem of 2 fundamental units is known (e.g., see [Tho]). By [Lou12, Theorem 2], if ε is atotally real cubic unit for which (i) the cubic number field Q(ε) is Galois and (ii) the orderZ[ε] is invariant under the action of the Galois group Gal(Q(ε)/Q), then it was reasonableto conjecture that ε and any one of its two other conjugates ε′ should form a system offundamental units of the cubic order Z[ε] (allowing for safety a finite number of exceptions).Surprisingly, in [LL15] we found that such a cubic order Z[ε] is almost never invariant underthe action of the Galois group Gal(Q(ε)/Q). And proving that ε, ε′ is in general a systemof fundamental units of the larger Galois-invariant totally real cubic order Z[ε, ε′] is hopeless(see however [LL14] for related results). In contrast, we still think that if ε > 1 is a real cubicunit of negative discriminant with complex conjugates η and η, then the group < −1, ε, η >is of bounded index in the group of units of the totally imaginary sextic order Z[ε, η, η] (see[LL14] and use Theorem 8 for a partial solution to this problem).

6. A conjecture in a quartic case of unit rank 2Let ε be a quartic algebraic unit which is neither totally real nor totally imaginary, i.e. aquartic unit of negative discriminant Dε < 0. Here again, the rank of the group of units ofthe quartic order Z[ε] is equal to 2 and the only complex root of unity in Z[ε] are ±1. As inSection 5 we would like to prove that, in general, there exists a second unit η ∈ Z[ε] such thatε, η is a system of fundamental units of Z[ε]. Since we can assume that Πε(X) is reduced,we would like to prove the very precise statement Conjecture 38:

Conjecture 38. — Let ε be a reduced quartic unit of negative discriminnant, i.e. such thatΠε(X) = X4 −AX3 +BX2 − CX +D ∈ Z[X] with Dε < 0, D ∈ ±1 and |C| ≤ A. Let usdefine a unit ξ1 ∈ Z[ε] as follows:

1. An infinite family: if Πε(X) is one of the four polynomials equivalent to Πξ2

1(X) = X4 −

(a2−2b)X3 +(b2−2ac+2d)X2−(c2−2bd)X+1, where Πξ1(X) = X4−aX3 +bX2−cX+

d ∈ Z[X] with d ∈ ±1 and |c| ≤ a is a Q-irreducible reduced polynomial of negativediscriminant different from the ones that appear in the two sporadic cases 3f and 3g belowand such that |a2d + c2 − abc| = 1, then we may assume that ε ∈ ±ξ2

1 ,±1/ξ21 and we

have

ξ1 = −aξ61 + (a3 − 2ab+ c)ξ4

1 + (a2c− ab2 − ad+ bc)ξ21 − (ab− c)d

a2d+ c2 − abc ∈ Z[ε],

2. An infinite family: if Πε(X) = X4 − a3X3 + 3a2dX2 − 3aX + d, d ∈ ±1 and a ≥ 2,set ξ1 = ε3 − a3ε2 + 3a2dε − 2a ∈ Z[ε], for which ε = ξ3

1, Πξ1(X) = X4 − aX3 + d and

Dε = Dξ1= −27a4 + 256d.

3. The following 8 sporadic cases:



(a) If Πε(X) = X4 − 5X3 − 9X2 − 5X − 1, set ξ1 = ε3 − 6ε2 − 3ε ∈ Z[ε], for whichε = ξ3

1, Πξ1(X) = X4 − 2X3 +X − 1 and Dε = Dχ1 = −275.

(b) If Πε(X) = X4 − 3X3 + 3X2 −X − 1, set ξ1 = −ε+ 1 ∈ Z[ε] for which ε = −1/ξ31,

Πξ1(X) = X4 −X3 − 1 and Dε = Dχ1 = −283.

(c) If Πε(X) = X4 − 5X3 + 7X2 −X − 1, set ξ1 = ε2 − 2ε ∈ Z[ε], for which ε = −ξ31,

Πξ1(X) = X4 −X3 +X2 +X − 1 and Dε = Dχ1 = −331.

(d) If Πε(X) = X4−11X3+15X2−7X+1, for which ε = ξ31 with ξ1 = ε3−10ε2+5ε ∈ Z[ε]

and Πξ1(X) = X4 − 2X3 −X + 1, in which case Dε = Dχ1 = −643.

(e) If Πε(X) = X4− 19X3 + 91X2− 7X − 1, set ξ1 = −ε3 + 10ε2− 5ε ∈ Z[ε], for whichε = −ξ3

1, Πξ1(X) = X4 − 2X3 + 4X2 +X − 1 and Dε = Dχ1 = −5987.

(f) If Πε(X) = X4 − 5X3 + 6X2 − 4X + 1, set ξ1 = ε3 − 4ε2 + 2ε− 1 ∈ Z[ε], for whichε = ξ4

1, Πξ1(X) = X4 −X3 − 1 and Dε = Dχ1 = −283.

(g) If Πε(X) = X4 − 15X3 + 17X2 − 7X + 1, set ξ1 = ε3 − 15ε2 + 17ε − 5 ∈ Z[ε], forwhich ε = ξ4

1, Πξ1(X) = X4 −X3 −X2 −X − 1 and Dε = Dχ1 = −563.

(h) If Πε(X) = X4 − 8X3 − 14X2 − 7X − 1, set ξ1 = 5ε3 − 43ε2 − 44ε − 9 ∈ Z[ε], forwhich ε = ξ7

1, Πξ1(X) = X4 −X3 − 1 and Dε = Dχ1 = −283.

4. Otherwise, set ξ1 = ε.

Hence, ξ1 is always reduced.Then there exists another unit ξ2 ∈ Z[ε] such that ξ1, ξ2 is a system of fundamental unitsof the quartic order Z[ε].

The main problem is that we have not yet been able to prove useful lower bounds for dis-criminants of such quartic units, even though we think it reasonable to conjecture that:

Conjecture 39. — Let P (X) = X4 − aX3 + bX2 − cX + d ∈ Z[X], d ∈ ±1, be a Q-irreducible quartic polynomial of negative discriminant DP (X) < 0. Then

(20) H∞(P (X))4/3 |DP (X)| H1(P (X))6,

where the implicit contants do not depend on P (X).

At least, we know that the exponent on the left hand side of (20) must be less than or equalto 4/3 (take n = 3 in Theorem 44) and we can prove the easiest part of Conjecture 39:

Lemma 40. — Let P (X) = X4−aX3 + bX2− cX+d ∈ Z[X], d ∈ ±1, be a Q-irreduciblequartic polynomial of negative discriminant DP (X) < 0. Then

|DP (X)| ≤ 212H1(P (X))6.

This exponent 6 is optimal: if P (X) = X4 − aX3 + X2 − aX + 1, a ≥ 2, then H1(P (X)) =(a+

√a2 + 4)/2 and −DP (X) = (4a2 − 9)(a2 + 4)2 > 0 is asymptotic to 4H1(P (X))6.

Proof. — Let ρ1 and ρ2 be the absolute values of the two real roots ε1 and ε2 of P (X). Let ηand η be its two non-real roots. Set ρ = |η| = 1/√ρ1ρ2. By changing P (X) into its associatedX4P (1/X), we may assume that ρ1 ≥ 1 and 1/ρ1 ≤ ρ2 ≤ ρ1. Then

H1(P (X)) ≥ H∞(P (X)) = max (ρ1, ρ1ρ2)


S. Louboutin 65

andD(X) =

((ε1 − ε2)|ε1 − η|2|ε2 − η|2(η − η)

)2

yields|D(X)| ≤

((2ρ1)((2ρ1)2(ρ2 + ρ)2(2ρ))

)2= 28ρ5

1(ρ2 + 1/√ρ1ρ2)4/ρ2.

First, if ρ2 ≤ 1/√ρ1ρ2, then ρ2 ≤ 1, H∞(P (X)) = ρ1, ρ2 +1/√ρ1ρ2 ≤ 2/√ρ1ρ2 and |D(X)| ≤212ρ3

1/ρ32 ≤ 212ρ6

1 = 212H∞(P (X))6.Secondly, if 1/√ρ1ρ2 ≤ ρ2 ≤ 1, then H∞(P (X)) = ρ1, ρ2 + 1/√ρ1ρ2 ≤ 2ρ2 and |D(X)| ≤212ρ5

1ρ32 ≤ 212ρ5

1 = 212H∞(P (X))5.Thirdly, if 1 ≤ ρ2 ≤ ρ1, then then H∞(P (X)) = ρ1ρ2, ρ2 + 1/√ρ1ρ2 ≤ 2ρ2 and |D(X)| ≤212ρ5

1ρ32 ≤ 212(ρ1ρ2)5 = 212H∞(P (X))5.

If Conjecture 39 is true, then ε can be infinitely often an nth-power only for |n| ∈ 2, 3, 4.Hence a first step towards proving Conjecture 38 would be (i) to prove that ε > 1 cannot beinfinitely many often a 4th power, (ii) to prove that ε is infinitely often a 3rd power if andonly if we are in case 2 of Conjecture 38 and (iii) to write case 1 of Conjecture 38 much moreexplicitly, i.e. to give a necessary and sufficient condition easy to check on the coeffcients ofΠε(X) = X4 −AX3 +BX2 − CX +D ∈ Z[X].

Remark 41. — Regarding Conjecture 39, we also did some numerical computation. For the38 413 452 quartic Q-irreducible polynomials P (X) = X4 − aX3 + bX2 − cX + d ∈ Z[X],d ∈ ±1, |c| ≤ a ≤ 400, |b| ≤ 400, of negative discriminant DP (X) < 0 we have

|DP (X)| ≥ 15H∞(P (X))4/3,

except for the following 6 polynomials:P (X) |DP (X)|/H∞(P (X))4/3

X4 − 8X3 − 14X2 − 7X − 1 13.97761 · · ·X4 − 8X3 + 10X2 + 7X + 1 8.06342 · · ·X4 − 9X3 + 22X2 − 8X + 1 5.81131 · · ·X4 − 24X3 − 26X2 − 9X − 1 3.86062 · · ·X4 − 95X3 + 64X2 − 14X + 1 1.49757 · · ·

X4 − 252X3 + 120X2 − 19X + 1 0.30925 · · ·

7. The general situationThe situation for number fields of degree ≥ 5 can be even worse. For example, nothingprevents a quintic algebraic unit ε from being infinitely many often a square, a cube or afourth power in Z[ε]:

Proposition 42. — Let m > n ≥ 2 be coprime. Assume thatXm − (aX + 1)n

is Q-irreducible. Then any of its complex roots ε is an n-th power in Z[ε].

Proof. — Ifmu−nv = 1 withm,n ≥ 1 rational integers, then ε = ηn with η = (aε+1)u/εv ∈Z[ε]. Notice that η is a root of Xm−aXn− 1. Hence, Xm− (aX + 1)n is Q-irreducible if andonly if Xm − aXn − 1 is Q-irreducible



To be able to deal with the general situation of algebraic units of any degree, it is not thatunreasonable to put forward a conjecture generalizing Lemma 3 and Theorems 9, 24, and 33:

Conjecture 43. — Let P (X) = Xn − an−1Xn−1 + · · · + (−1)n−1a1X + (−1)na0 ∈ Z[X]range over the monic Q-irreducible polynomials of a given degree n ≥ 3 with a0 ∈ ±1. Thenfor some positive constants β(n) ≥ α(n) > 0 it holds that

(21) H∞(P (X))α(n) |DP (X)| H1(P (X))β(n),

where the implicit contants depend on n only.

Clearly, the upper bound in (21) holds true with β(n) = n(n − 1) (any complex root α ofP (X) satisfies |α| ≤ H1(P (X))). More interestingly, if Conjecture 43 were true, then wewould have α(n) ≤ (n+ 1)/n:

Theorem 44. — Fix n ≥ 2. Let a range over the rational integers greater than 2. Thenthe polynomials Rn(X) = Xn+1 − (aX + 1)n ∈ Z[X] are Q-irreducible, of discriminantsDRn(X) = (−1)n(n−1)/2(nnan+1 + (n + 1)n+1) of absolute values asymptotic to nnan+1 andH∞(Rn(X)) is asymptotic to an. Hence, DRn(X) is asymptotic to nnH∞(Rn(X))(n+1)/n.Moreover, any root ε of Rn(X) is an algebraic unit of degree n such that ε = ηn with η =a+ ε−1 ∈ Z[ε].

Proof. — By Perron’s criterium Pn(X) = Xn+1−aXn− 1 ∈ Z[X] is Q-irreducible for a ≥ 3.If η is a root of Pn(X) then ε := ηn that satisfies εn+1 = (ηn+1)n = (aηn + 1)n = (aε+ 1)n isa root of Rn(X). Hence, Rn(X) is Q-irreducible, DRn(X) = DPn(X) = (−1)n(n−1)/2(nnan+1 +(n+ 1)n+1) and H∞(Rn(X)) = H∞(Pn(X))n is asymptotic to an, by Lemma 45.

Lemma 45. — Fix n ≥ 2, an integer, and a primitive 2n-th complex root of unity ζ2n. Leta ≥ 3 range over the integers. Then the n + 1 complex roots ρ(k)

a , 0 ≤ k ≤ n, of Pn(X) =Xn+1 − aXn − 1 ∈ Z[X] can be sorted so as to satisfy ρ

(0)a = a + O(a−n) and ρ

(k)a :=

ζ2k−12n a−1/n +O(a−1), 1 ≤ k ≤ n. Hence, H∞(Pn(X)) is asymptotic to a.

Proof. — If An(X) = ∑n+1k=0 αk(a)Xk ∈ C[X], a ≥ 3, we write An(X) = O(ac) if αk(a) =

O(ac) for 0 ≤ k ≤ n+ 1. Set

Qn(X) := (X − a)n∏

k=0(X − θ(k)

a ), where θ(k)a = ζ2k−1

2na1/n + ζ

2(2k−1)2n

na(n+2)/n .

We want to prove that the coefficients of Qn(X) are very close to those of Pn(X) as a goes toinfinity, and that it implies that the roots of Pn(X) are close to that of Qn(X), hence impliesthe desired result. Set

Rn(X) =n∏

k=1

(X − ζ2k−1

2na1/n

)= Xn + a−1 and Sn(X) :=

n∏

k=1(X − θ(k)

a ).

Clearly, we have

Sn(X) = Rn(X)−n∑

k=1

Rn(X)

X − ζ2k−12na1/n

ζ2(2k−1)2n

na(n+2)/n +O(a−2(n+2)/n).


S. Louboutin 67

Using the partial fraction decomposition over C

X

Rn(X) = X

Xn + a−1 = −an∑

k=1

ζ2(2k−1)2nna2/n

X − ζ2k−12na1/n

,

we obtain Sn(X) = Xn + a−2X + a−1 + O(a−2(n+2)/n) and Qn(X) − Pn(X) = O(a−2).Applying [Lou08b, (4) of Section 3] with α = 2, β = −1/n, γ = −1 and δ = −1/n we obtainthe approximations of the ρ(k)

a for 1 ≤ k ≤ n. Using Pn(a) = −1 < 0 and Pn(a + a−n) =(1 + a−n−1)n − 1 > 0, we obtain the approximations of the ρ(0)

a .

References[BHMMS] J. Beers, D. Henshaw, C. Mccall, S. Mulay and M. Spindler, Fundamentality of a cubic

unit u for Z[u], Math. Comp., 80 (2011), 563–578, Corrigenda and addenda, Math. Comp. 81(2012), 2383–2387.

[Coh] H. Cohen, A course in computational algebraic number theory, Graduate Texts in Mathematics,138, Springer-Verlag, Berlin, 1993.

[LL14] J. H. Lee and S. Louboutin, On the fundamental units of some cubic orders generated by units,Acta Arith., 165 (2014), 283–299.

[LL15] J. H. Lee and S. Louboutin, Determination of the orders generated by a cyclic cubic unit thatare Galois invariant, J. Number Theory, 148 (2015), 33–39.

[Lou06] S. Louboutin, The class-number one problem for some real cubic number fields with negativediscriminants, J. Number Theory, 121, (2006), 30–39.

[Lou08a] S. Louboutin, The fundamental unit of some quadratic, cubic or quartic orders, J. Ramanu-jan Math. Soc. 23, No. 2 (2008), 191–210.

[Lou08b] S. Louboutin, Localization of the complex zeros of parametrized families of polynomials, J.Symbolic Comput., 43, (2008), 304–309.

[Lou10] S. Louboutin, On some cubic or quartic algebraic units, J. Number Theory, 130, (2010),956–960.

[Lou12] S. Louboutin, On the fundamental units of a totally real cubic order generated by a unit, Proc.Amer. Math. Soc., 140, (2012), 429–436.

[MS] S. Mulay and M. Spindler, The positive discriminant case of Nagell’s theorem for certain cubicorders, J. Number Theory, 131, (2011), 470–486.

[Nag] T. Nagell, Zur Theorie der kubischen Irrationalitäten, Acta Math., 55, (1930), 33–65.

[PL] S.-M. Park and G.-N. Lee, The class number one problem for some totally complex quartic numberfields, J. Number Theory, 129, (2009), 1338–1349.

[Tho] E. Thomas, Fundamental units for orders in certain cubic number fields , J. Reine Angew.Math., 310, (1979), 33-55.



[Was] L. C. Washington, Introduction to Cyclotomic Fields, Graduate Texts in Mathematics, 83,Springer-Verlag, Second Edition, 1997.

June 17, 2015

Stéphane R. Louboutin, Aix Marseille Université, CNRS, Centrale Marseille, I2M, UMR 7373, Institut deMathématiques de Marseille, Aix Marseille Université, 163 Avenue de Luminy, Case 907, 13288 MarseilleCedex 9, France • E-mail : [email protected]



REALISABLE CLASSES, STICKELBERGER SUBGROUP AND ITSBEHAVIOUR UNDER CHANGE OF THE BASE FIELD

by

Andrea Siviero

Abstract. — Let K be an algebraic number field with ring of integers OK and let G be afinite group. We denote by R(OK [G]) the set of classes in the locally free class group Cl(OK [G])realisable by rings of integers in tamely ramified G-Galois K-algebras. McCulloh showed that,for every G, the set R(OK [G]) is contained in the so-called Stickelberger subgroup St(OK [G])of Cl(OK [G]).In this paper first we describe the relation between St(OK [G]) and Cl(OK [G]), where Cl(OK [G])is the kernel of the morphism Cl(OK [G]) −→ Cl(OK), induced by the augmentation mapOK [G] −→ OK . Then, as an example of computation of St(OK [G]), we show, just using itsdefinition, that St(Z[G]) is trivial, when G is a cyclic group of order p or a dihedral group oforder 2p, where p is an odd prime number.Finally we prove that St(OK [G]) has good functorial behaviour under change of the base field.This has the interesting consequence that, given an algebraic number field L, if N is a tameGalois L-algebra with Galois group G and St(OK [G]) is known to be trivial for some subfieldK of L, then ON is stably free as an OK [G]-module.

Résumé. — Soient K un corps de nombres d’anneau des entiers OK et G un groupe fini.On note R(OK [G]) l’ensemble des classes dans le groupe des classes des modules localementlibres Cl(OK [G]) qui peuvent être obtenues par l’anneau des entiers des K-algèbres galoisiennesmodérément ramifiées de groupe de Galois G. McCulloh a prouvé que, pour tout G, l’en-semble R(OK [G]) est contenu dans le soi-disant sous-groupe de Stickelberger St(OK [G]) dansCl(OK [G]).Dans ce papier d’abord nous nous focalisons sur la relation entre St(OK [G]) et Cl(OK [G]),où Cl(OK [G]) est le noyau du morphisme Cl(OK [G]) −→ Cl(OK), induit par l’augmentationOK [G] −→ OK . Puis, comme exemple de calcul du groupe St(OK [G]), nous montrons, en uti-lisant sa définition, que St(Z[G]) est trivial si G est soit un groupe cyclique d’ordre p soit ungroupe diédral d’ordre 2p, avec p premier impair.Enfin, nous montrons la fonctorialité de St(OK [G]) par rapport au changement du corps debase. Ceci implique que, soit L est un corps de nombres, si N est une L-algèbre galoisienne mo-dérément ramifiée, de groupe de Galois G, et St(OK [G]) est connu être trivial pour un certainsous-corps K de L, alors ON est un OK [G]-module stablement libre.

Mathematical subject classification (2010). — 11R33, 11R04, 11R18, 11R29, 11R32, 11R65.Key words and phrases. — Galois module structure, Realisable classes, Locally free class groups, Fröhlich’sHom-description of locally free class groups, Stickelberger’s theorem.

70 Real. classes, Stickel. subgr. and its beh. under change of the base field

1. IntroductionLet K be an algebraic number field with ring of integers OK and let G be a finite group.Given a Galois field extension N/K with Galois group isomorphic to G, we can consider thering of integers ON as an OK [G]-module. The classical Noether criterion implies that, whenN/K is tamely ramified, then ON is a locally free OK [G]-module of rank 1 and determines aclass in the locally free class group Cl(OK [G]) (for a precise definition see (1) in Section 2).Noether’s criterion holds in general for G-Galois K-algebras (a K-algebra N is G-Galois if itis étale and G acts on the left on N as a group of automorphisms, such that [N : K] = |G| andNG = K). Hence, as for field extensions, to every tame G-Galois K-algebra we can associatea class in Cl(OK [G]).LetKc be a chosen algebraic closure ofK, then it is well-known that, if Ωt

K denotes the Galoisgroup of the maximal tame extension Kt/K in Kc, then the set of isomorphism classes oftame G-Galois K-algebras is in bijection with H1(Ωt

K , G) (see [Ser94, Chapter I, §5] for aprecise definition), the first cohomology set of Ωt

K with coefficients in G (where ΩtK acts

trivially on G). Hence, thanks to Noether’s criterion, we can consider the following morphismof pointed sets:

R : H1(ΩtK , G) −→ Cl(OK [G])

[N ] 7−→ [ON ],where [N ] (resp. [ON ]) denotes the associated class of the G-Galois K-algebra N (resp. ON )in H1(Ωt

K , G) (resp. Cl(OK [G])).The set of realisable classes, denoted by R(OK [G]), is defined as the image of R, i.e. it isthe set of all classes in Cl(OK [G]) which can be obtained from the rings of integers of tameG-Galois K-algebras. The problem of realisable classes consists in the study of the structureof this set. One of the main question which is still open nowadays is the following:

Question. — Is R(OK [G]) a subgroup of Cl(OK [G])?

Note that R, when not trivial, is not a priori a group homomorphism. Indeed, if G is notabelian, the domain of R is just a pointed set, but even if G is abelian (and so H1(Ωt

K , G)is a group), it is not difficult to find an example which explains why R is not a group homo-morphism (see [Siv13, Appendix]).

When the base field equals Q and G is abelian it follows from a result by Taylor ([Tay81]),proving a conjecture of Fröhlich, that R(Z[G]) is trivial. By [Tay81], the same holds if K = Qand G is a non-abelian group with no irreducible symplectic characters. More generally Tay-lor proved that any element in R(Z[G]) has order at most 2 in Cl(Z[G]).

Over a general number field K, when G is abelian, a positive answer to the previous questionis given by Leon McCulloh in [McC87]. Given a finite group G, he introduced a subgroupSt(OK [G]) of Cl(OK [G]) (the notation used here differs from the original one by McCulloh),defined in terms of some Stickelberger maps (see Section 2), and he proved the followingresult.

Theorem 1.1. — Let G be a finite abelian group. Then, for every algebraic number field K,we have

R(OK [G]) = St(OK [G]).


A. Siviero 71

When G is non-abelian, he managed to prove the following inclusion (this is an unpublishedresult announced in a talk given in Oberwolfach in February 2002, [McC] - for a detailedproof of it, see [Siv13, Chapter 2]).

Theorem 1.2. — For every algebraic number field K and finite group G, the inclusionR(OK [G]) ⊆ St(OK [G]) holds.

The proof of the reverse inclusion is still unknown. Indeed, when G is non-abelian andK 6= Q,determining if R(OK [G]) forms a subgroup is in general an open problem.Nevertheless, starting from the inclusion R(OK [G]) ⊆ Cl(OK [G]) (where the subgroupCl(OK [G]) is the kernel of the augmentation map from Cl(OK [G]) to Cl(OK), see Sec-tion 3), some non-abelian results have been achieved. In particular, it has been proved thatR(OK [G]) = Cl(OK [G]) in the following cases: G = D8, the dihedral group of order 8, withthe assumption that the ray class group modulo 4OK of OK has odd order (see [BS05a]); andG = A4, without any restriction on the base field K (see [BS05b]).Recently in [BS13], Nigel P. Byott and Bouchaïb Sodaïgui, under the assumption that Kcontains a root of unity of prime order p, showed that R(OK [G]) is a subgroup of Cl(OK [G]),when G is the semidirect product V o C of an elementary abelian group V of order pr byany non-trivial cyclic group C which acts faithfully on V and makes V into an irreducibleFp[C]-module (where Fp is the finite field with p elements). This last result contains as acorollary the main result of [BS05b].More recently, Adebisi Agboola and Leon McCulloh showed that R(OK [G]) is a subgroupof Cl(OK [G]) when G is a nilpotent group subject to certain conditions (see [AC15] for theoriginal preprint and further details).In the non-abelian context, more has been done in describing a weaker form of R(OK [G]). IfM denotes a maximal order in K[G] containing OK [G], then, as done for R(OK [G]), we candefine R(M) to be the subset in Cl(M) of the classes [M ⊗OK[G] OL], where L runs throughthe tame G-Galois algebras over K. The two sets R(M) and R(OK [G]) are linked by theextension of scalars Ex : Cl(OK [G]) −→ Cl(M) and in fact one has R(M) = Ex(R(OK [G])).For a complete list of the works on R(M) we refer to [Siv13, Introduction].

After a general review of the main definitions and tools, the first aim of this paper is to studyexplicitly the relation between St(OK [G]) and Cl(OK [G]). In Section 3, we shall prove thefollowing proposition.

Proposition 1. — For every algebraic number field K and finite group G, the inclusionSt(OK [G]) ⊆ Cl(OK [G]) holds.

We shall exhibit an abelian counterexample showing that the reverse inclusion is not in gen-eral true.

In Section 4 we shall give the first main result of this paper. As an example of computationof the group St(OK [G]), just using its definition, we shall prove the following theorem.

Theorem 1. — Given a prime number p 6= 2. If G = C2, a cyclic group of order 2, orG = Cp, a cyclic group of order p or G = Dp, a dihedral group of order 2p, then St(Z[G]) istrivial.

This result will immediately imply, just using Theorem 1.2, the following corollary.



Corollary 1. — In the cases of the theorem above, R(Z[G]) is trivial.

Remark 1.3. — For p = 2 we have that D2 is the Klein four group. Also in this caseTheorem 1 holds, thanks to the fact that Cl(Z[D2]) is trivial (cf. [CR87, Corollary 50.17]).

Note that the result of Corollary 1 was already included in the more general result by Taylorcontained in [Tay81]. The proof of Theorem 1 given here does not assume [Tay81], however,but instead uses the definition of St(Z[G]) and its connection to the classical Stickelbergertheorem.In the dihedral case, the result of Theorem 1 is genuinely new, since the inclusion R(Z[Dp]) ⊇St(Z[Dp]) is not known a priori because the dihedral group Dp is not abelian. The resultobtained for the dihedral case is the more interesting one, since in particular it goes in the di-rection of extending the equality of Theorem 1.1 to non-abelian groups, giving a non-abelianexample where the equality R(Z[G]) = St(Z[G]) holds.

In Section 5 we shall obtain the second main result of this paper concerning the behaviourof St(OK [G]) under change of the base field K. Namely, if L is an algebraic number fieldcontaining K, considering every OL[G]-module as an OK [G]-module, there is a restrictionmap NL/K : Cl(OL[G]) −→ Cl(OK [G]) (see [CR87, §52]). We shall prove the following result.

Theorem 2. — For every finite group G, we haveNL/K(St(OL[G])) ⊆ St(OK [G]).

This will have some nice consequences, such as a new proof of a result by Taylor, containedin [Tay81], which says that the ring of integers of an abelian tame G-Galois K-algebra is free(of rank [K : Q]) over Z[G].A final remark will explain that, if we consider the homomorphism given by extension of thebase field extL/K : Cl(OK [G]) −→ Cl(OL[G]), an analogous result to the previous theoremholds.

Notation and conventions. — Let K be an algebraic number field and OK its ring of inte-gers. Given a place p of K, we denote by Kp its completion with respect to the metric definedby p: if p is a finite place, then Kp is a non-archimedean field which is a finite extension ofQp (where p is the characteristic of OK/p); if p is an infinite place, then Kp is isomorphic toR or C.We choose an algebraic closure Kc (resp. Kc

p) of K (resp. Kp) and let ΩK (resp. ΩKp) denotethe Galois group of Kc/K (resp. Kc

p/Kp).The symbol Ωnr

K (resp. ΩnrKp

) will denote the Galois group of the maximal unramified (at finiteplaces) extension Knr/K (resp. Knr

p /Kp) in Kc (resp. Kcp) and Ωt

K (resp. ΩtKp

) will be theGalois group of the maximal tame extension Kt/K (resp. Kt

p/Kp) in Kc (resp. Kcp). At the

infinite places we take Knrp = Kt

p = Kcp .

If p is a finite place, let OK, p be the completion of OK with respect to p (which also coincideswith the ring of integers of the completion Kp) and Oc

K, p the integral closure of OK in Kcp . If

p is an infinite place, we define OK, p to be Kp.

Let J(K) denote the group of idèles of K, i.e. the restricted direct product of K×p p with re-spect to O×K, pp, where p runs through the places of K (both finite and infinite). We consider


A. Siviero 73

K× diagonally embedded in J(K) and U(OK) will denote the product ∏pO×K, p. Similarly,

given a finite group G, the idèle group J(K[G]) is the restricted direct product of Kp[G]×pwith respect to OK, p[G]×p, where p runs through the places of K, and U(OK [G]) stands forthe product ∏pOK, p[G]×. The symbol J(Kc) (resp. U(OcK)) will denote the direct limit ofthe idèle groups J(L) (resp. U(OL)), as L runs over all finite Galois extensions ofK insideKc.

When we consider a representative for a class in a class group, we use the brackets [− ] todenote its class (e.g. [OL] denotes the class in Cl(OK [G]) corresponding to the ring of integersOL of a tame G-Galois algebra L/K).

Throughout this paper, G will denote a finite group, Gab its abelianization and G its set ofconjugacy classes. Given an element s in G, we denote by s its conjugacy class.The set of irreducible complex characters of G will be denoted by Irr(G) and RG will stand forthe ring of virtual characters of G, i.e. the ring of Z-linear combinations of elements in Irr(G).

If Y is a group acting on the left on a set X, we denote the action of y ∈ Y on x ∈ X bythe symbol xy (occasionally by y · x); note in particular that (xy)z = xzy. If Y is a groupacting on two groups H and H ′, we denote by HomY (H,H ′) the set of all group homomor-phisms from H to H ′ fixed by the action of Y ; in other words, let f ∈ Hom(H,H ′), thenf ∈ HomY (H,H ′) if and only if f(hy) = f(h)y, for all y ∈ Y and h ∈ H. If we choose anembedding of Kc in C, then the absolute Galois group ΩK naturally acts on the left on Irr(G)by χω(s) = χ(s)ω, where ω ∈ ΩK , χ ∈ Irr(G) and s ∈ G. We extend this action by linearityto RG. When we have an ΩK-action on a set, we can always consider a ΩKp-action on thesame set, considering ΩKp embedded into ΩK .

Given a rational number a ∈ Q, the symbol bac (resp. a) denotes its integer (resp. frac-tional) part.

2. Review of the main definitions and toolsIn this section we briefly recall the main tools needed to describe McCulloh’s results onrealisable classes. We will not focus on any proof in particular, referring to the original worksfor more details. We recall from the Introduction that K is an algebraic number field and Ga finite group.

2.1. Locally free class group. — Let M be an OK [G]-lattice (i.e. an OK [G]-module,finitely generated and projective as OK-module) and let Mp denote the tensor productOK, p⊗OK M . We say that M is in the genus of OK [G] (or M is a locally free OK [G]-moduleof rank 1) if, for every maximal ideal p of OK , there is an isomorphism Mp

∼= OK, p[G]. Wedenote by g(OK [G]) the set of all OK [G]-lattices in the genus of OK [G]. Moreover given twoOK [G]-lattices M and N , we say that M is stably isomorphic to N and we denote it byM ∼=s N , if M ⊕OK [G]n ∼= N ⊕OK [G]n, for some n ≥ 1.

Remark 2.2. — Note that in general M ∼=s N ; M ∼= N (see [Swa62] for an explicitexample). Nevertheless there are many groups G for which two OK [G]-lattices which are stably



isomorphic are isomorphic too (e.g. abelian groups, dihedral groups, groups of odd order). Werefer to [CR87, §51C] for more details.

If now [M ]s denotes the stable isomorphism class of M , we define the set(1) Cl(OK [G]) := [M ]s : M ∈ g(OK [G])and we call it the locally free class group of OK [G]. Now, on this set, we can define a well-defined operation. Given [M1]s and [M2]s in Cl(OK [G]), let us set(2) [M1]s + [M2]s := [M3]s,where, M3 is an OK [G]-lattice in g(OK [G]) such that M1⊕M2 ∼= OK [G]⊕M3 (the existenceof such an OK [G]-lattice M3 follows from a lemma of Roiter, see [CR81, 31.6]).As one can see, under this operation, Cl(OK [G]) is an abelian group (see [CR87, §49A] formore details). Moreover, Jordan–Zassenhaus theorem (see [Rei03, Theorem 26.4]) tells usthat the number of isomorphism classes of OK [G]-lattices in g(OK [G]) is finite, so a fortioriCl(OK [G]) is finite.

Remark 2.3. — Given an OK [G]-lattice M , its associated class in Cl(OK [G]) is trivial ifand only if M ∼=s OK [G]. Note that by the previous remark, if G is abelian, dihedral or ofodd order, then this is equivalent to say that M ∼= OK [G].

2.4. Hom-description. — Fröhlich gave a useful description of the locally free class groupCl(OK [G]) in terms of some ΩK-equivariant homomorphisms. This description is well-knownas the Hom-description of Cl(OK [G]).Let us consider the determinant map Det : U(OK [G]) −→ Hom ΩK

(RG, J(Kc)), obtainedcomponentwise by the linear extension of the map on Irr(G) given by Det(a)(χ) = det (Tχ(a)),where Tχ is a complex representation affording the character χ and det is the usual determi-nant of a matrix.The Hom-description of Cl(OK [G]) is given by the following isomorphism

(3) Cl(OK [G]) ∼=Hom ΩK

(RG, J(Kc))Hom ΩK

(RG, (Kc)×) ·Det(U(OK [G])) .

Remark 2.5. — For a given class in Cl(OK [G]) to be trivial means that the representativehomomorphism f ∈ Hom ΩK

(RG, J(Kc)), under the previous isomorphism, can be written asf = gd, where g ∈ Hom ΩK

(RG, (Kc)×

), a global homomorphism fixed by ΩK , and d belongs

to Det (U (OK [G])).

2.6. The map RagK . —Given a character χ ∈ Irr(G), we define a map detχ on the groupG as

detχ(s) = det(Tχ(s)),where Tχ is, as above, a complex representation associated to χ. Note that detχ can beconsidered as a character of G of degree 1 (or equivalently as a character of Gab).This definition is independent of the choice of the representation Tχ and we can in turnconsider the homomorphism det : RG −→ Irr(Gab) defined by

det

∑

χ∈ Irr(G)aχχ

=

∏

χ∈ Irr(G)(detχ)aχ .


A. Siviero 75

Let AG be the kernel of this map, we shall call it the augmentation kernel. Then we canconsider the following short exact sequence of groups:

(4) 0 // AG// RG

det // Irr(Gab) // 1.

Remark 2.7. — When the group G is abelian we have an explicit Z-basis of AG. Fromthe proof of [McC87, Theorem 2.14], if Irr(G) has a basis χ1, . . . , χk, with χi of order ei,for i = 1, . . . , k and every χ ∈ Irr(G) is written uniquely as χ = ∏k

i=1 χri(χ)i , where 0 ≤

ri(χ) < ei; a Z-basis of AG is given by the non-zero elements in the collection eiχi| i =1, . . . , k ∪ χ−∑k

i=1 ri(χ)χi|χ ∈ Irr(G).For every finite place p of K, applying the functor Hom(−, (Oc

K, p)×) to the short exactsequence (4), we get the following short exact sequence

1 // Hom(Irr(Gab), (Oc

K, p)×) // Hom(RG, (Oc

K, p)×) rag// Hom

(AG, (Oc

K, p)×) // 1

where the map rag is just the restriction map to the augmentation kernel (this also explainsits name). Now using the local analog of the functor Det, previously defined, we have thefollowing proposition.

Proposition 2.8. — For every finite place p of K, there is a commutative ΩKp-diagram(every map preserves the ΩKp-action) of pointed sets with exact rows:

1 // G //

Det

OcK, p[G]× //

Det

OcK, p[G]×/G //

Det

1

1 // Gab // Hom(RG, (Oc

K, p)×) rag

// Hom(AG, (Oc

K, p)×)

// 1

Proof. — If s ∈ G, then Det(s)(χ) = detχ(s). So, from the definition of the map Det andfrom the identification of Hom

(Irr(Gab), (Oc

K, p)×)

with Gab, the map Det : G −→ Gab

coincides with the natural quotient map G −→ Gab which sends s ∈ G to its associated cosetin G/[G,G]. Thus Det induces a map Det : Oc

K, p[G]×/G −→ Hom(AG, (Oc

K, p)×), making

the diagram commute.

Let us define the pointed set(5) H(OK, p[G]) := (Oc

K, p[G]×/G)ΩKp

Taking now the ΩKp-invariants we deduce the following proposition.

Proposition 2.9. — For every finite place p of K, we have the following commutativediagram with exact rows:

1 // G //

Det

OK, p[G]× //

Det

H(OK, p[G]) //

Det

H1(ΩnrKp, G) //

1

1 // Gab // Hom ΩKp

(RG, (Oc

K, p)×) rag// Hom ΩKp

(AG, (Oc

K, p)×) // Hom(

ΩnrKp, Gab

)// 1 .

Moreover the set Det(H(OK, p[G])) is a subgroup of Hom ΩKp

(AG, (Oc

K, p)×).



Proof. — The first part of the proposition follows from Proposition 2.8, applying ΩKp-cohomology. In particular exactness at the top row comes from the fact that any local un-ramified extension has a normal integral basis generator, while exactness at the bottom rowis a consequence of the fact that the map H1(Ωnr

Kp, G) −→ Hom(Ωnr

Kp, Gab), induced by the

natural map from Hom(ΩnrKp, G) to Hom(Ωnr

Kp, Gab), is surjective as one can check using the

description of ΩnrKp

as a procyclic group.For the second part of the proposition, we observe that, for every finite place p, there is aone-to-one correspondence between OK, p[G]×\H(OK, p[G]) and H1(Ωnr

Kp, G). Thus, applying

the map Det and using the fact that the map H1(ΩnrKp, G) −→ Hom(Ωnr

Kp, Gab) is surjective,

we getrag(Det(OK, p[G]×))\Det(H(OK, p[G])) ∼= Hom(Ωnr

Kp, Gab).

From this and using the fact that Hom(ΩnrKp, Gab) is an abelian group, the group structure

of Det(H(OK, p[G])) follows. For more details on this proof we refer to [Siv13, Propositions2.2.6 and 2.2.7].

Now, if we write

(6) U(OK [G]) :=∏

p

Det(H(OK, p[G])) ⊆ Hom ΩK(AG, J(Kc))

and we define the group

(7) MCl(OK [G]) :=Hom ΩK

(AG, J(Kc))Hom ΩK

(AG, (Kc)×) · U(OK [G]) ,

we see that the restriction map rag : Hom ΩK(RG, J(Kc)) −→ Hom ΩK

(AG, J(Kc)) and Proposi-tion 2.9, yield a group homomorphism

(8) RagK :Hom ΩK

(RG, J(Kc))Hom ΩK

(RG, (Kc)×) ·Det(U(OK [G])) −→Hom ΩK

(AG, J(Kc))Hom ΩK

(AG, (Kc)×) · U(OK [G]) .

Using the Hom-description of Cl(OK [G]) (see (3)), this can be written asRagK : Cl(OK [G]) −→ MCl(OK [G]).

2.10. The Stickelberger map. —We introduce now one of the main ingredient of Mc-Culloh’s results, the so called Stickelberger map. The original definition of the Stickelbergermap, when G is abelian, is contained in [McC87], while its extension to the non-abelian casewas presented for the first time by McCulloh in a talk given in Oberwolfach in 2002 ([McC]).Let us start defining the Stickelberger pairing.

We define a Q-pairing 〈−,−〉 : Q⊗Z RG ×Q[G] −→ Q as follows:

? Characters of degree 1: If χ is a character of degree 1 and s ∈ G, 〈χ, s〉 is the rationalnumber defined by

χ(s) = e2πi〈χ,s〉,

such that 0 ≤ 〈χ, s〉 < 1. This was the original definition contained in [McC87] (in theabelian case every irreducible character is of dimension 1). If G is abelian, this alreadydefines, extending it by Q-bilinearity, a Q-pairing 〈−,−〉 : Q⊗Z RG ×Q[G] −→ Q.


A. Siviero 77

? Characters of higher degree: If χ is a character of degree bigger than 1, then we define〈χ, s〉 := 〈resG〈s〉 χ, s〉,

where resG〈s〉 χ is the restriction of the character χ to the cyclic group generated by s.Extending it by Q-bilinearity, we have the required pairing for a generic finite group G.

Thanks to this pairing, the Stickelberger map ΘG : Q⊗Z RG −→ Q[G] is defined as

ΘG(α) :=∑

s∈G〈α, s〉s, for α ∈ Q⊗Z RG.

An important property of AG is given by the following proposition.

Proposition 2.11. — Let α ∈ Q ⊗Z RG, then ΘG(α) ∈ Z[G] ⇐⇒ α ∈ AG. In particularΘG induces a homomorphism ΘG : AG −→ Z[G].

Proof. — See [McC, Proposition 1].

Up to now we have not considered the ΩK-action. If we let ΩK act trivially on G, it is easyto see that the Stickelberger map does not preserve the ΩK-action. In order to have such aninvariant property we have to introduce a non-trivial ΩK-action on G.

Definition. — Let m be the exponent of G and let µm be the group of m-th roots of unity.Restricting ΩK to Gal (K (µm) /K), we consider the map κ : ΩK −→ (Z/mZ)× defined viathe formula ζω = ζκ(ω), for ζ ∈ µm. We denote by G(−1) the group G with an ΩK-actiondefined via the inverse of κ:

sω := sκ−1(ω).

If we take a character χ of degree 1, we have χ(s) ∈ µm and, since χω(s) equals χ (s)ω, weget

χω(s) = χ (s)ω = χ (s)κ(ω) = χ(sκ(ω)).By bilinearity, we deduce that, for all α ∈ Q⊗Z RG and for all s ∈ G(−1),

(9) 〈αω, s〉 = 〈α, sκ(ω)〉 = 〈α, sω−1〉.Applying this to the Stickelberger map, we get

ΘG(αω) =∑

s∈G(−1)〈αω, s〉 s =

∑

s∈G(−1)〈α, sω−1〉 s =

∑

s∈G(−1)〈α, s〉 sω;

from which we deduce the following proposition.

Proposition 2.12. — The map ΘG : Q⊗ZRG −→ Q[G(−1)] is an ΩK-homomorphism, i.e.ΘG(αω) = ΘG (α)ω , for all α ∈ Q⊗Z RG and ω ∈ ΩK .

The pairing 〈χ, s〉 just depends on the conjugacy class of s ∈ G and hence ΘG(Q⊗Z RG) ⊆Z(Q[G]), where Z(Q[G]) is the centre of the group algebra Q[G], with basis the conjugacyclass sum of G. If we denote by G the set of conjugacy classes of G, then the action of ΩK

via κ−1 preserves conjugacy classes and it induces an ΩK-action on Z[G]; we denote thisΩK-module by Z[G(−1)]. Thus, defining the Stickelberger pairing on the set of conjugacyclasses as

〈χ, s〉 := 〈χ, s〉,



we denote by ΘG the map, defined by Q-bilinearity as:

ΘG : Q⊗Z RG −→ Q[G](10)α 7−→

∑

s∈G〈α, s〉s.

Again we have ΘG(α) ∈ Z[G]⇐⇒ α ∈ AG. Thus, transposing the map ΘG : AG → Z[G(−1)],we get the ΩK-equivariant homomorphism

ΘtG,K

: Hom(Z[G(−1)], (Kc)×

)−→ Hom

(AG, (Kc)×

).

Hence, if we write(KΛ

)×:= Hom ΩK

(Z[G(−1)], (Kc)×

),

Λ× := Hom ΩK

(Z[G(−1)], (Oc

K)×)

;

the map ΘtG,K

induces a homomorphism

ΘtG,K

:(KΛ

)×−→ Hom ΩK

(AG, (Kc)×

).

For every place p of K, we get a local analog just replacing K with Kp:

ΘtG,Kp

:(KpΛp

)×−→ Hom ΩKp

(AG, (Kc

p)×).

Moreover ΘtG,Kp

(Λ×p ) ⊆ Hom ΩKp

(AG, (Oc

K, p)×). At the infinite places, since we set OK, p =

Kp, we have Λp = KpΛp.Thus, defining the idèle group J(KΛ) as the restricted product of (KpΛp)×p with respectto Λ×p p, the homomorphisms Θt

G,Kpcombine to give an idelic transpose Stickelberger ho-

momorphism:(11) Θt

G,K: J(KΛ) −→ Hom ΩK

(AG, J (Kc)) .

Remark 2.13. — We can also define J(KΛ) as Hom ΩK

(Z[G(−1)], J(Kc)

); for details see

[McC87, Remark 6.22].

Remark 2.14. — If G is abelian, we can remove the “bar” from all our notation, sinceG = G. In the sequel, if G is abelian, we will adopt this simplification in the notation.

2.15. The Stickelberger subgroup and McCulloh’s results. —We can finally definethe Stickelberger subgroup St(OK [G]) and state McCulloh’s main results on realisable classes.Thanks to the definitions given in the previous parts, we have the following group homomor-phisms

Cl(OK [G])RagK// MCl(OK [G]) J(KΛ),

ΘtG,K

oo

where the map on the right is the natural map given by the composition of the map ΘtG,K

:J(KΛ) −→ Hom ΩK

(AG, J (Kc)) with the quotient map Hom ΩK(AG, J(Kc)) −→ MCl(OK [G])

(and we will denote it again by ΘtG,K

).


A. Siviero 79

Definition. — The Stickelberger subgroup St(OK [G]) is defined as

St(OK [G]) := Rag−1K

(Im(ΘtG,K

)).

Remark 2.16. — The description and the notation used for St(OK [G]) do not reflect Mc-Culloh’s original choice, but they are rather inspired from some later informal notes by A.Agboola.

Denoting by Rnr(OK [G]), the set of realisable classes obtained from unramified G-GaloisK-algebras, McCulloh’s results then follow (the next theorem is a more detailed version ofTheorem 1.1 and Theorem 1.2 in the Introduction).

Theorem 2.17. — For every finite group G and algebraic number field K, we getRnr(OK [G]) ⊆ Ker(RagK),(12)

R(OK [G]) ⊆ St(OK [G]).(13)Furthermore, when G is abelian, we have

Rnr(OK [G]) = Ker(RagK),(14)R(OK [G]) = St(OK [G]).(15)

Proof. — The equalities concerning the abelian case are proved in [McC87], while the firstinclusions are claimed in some unpublished notes on two talks given by McCulloh in Oberwol-fach in 2002 ([McC]) and in Luminy in 2011. For a precise proof of them we refer to [Siv13,Chapter 2].

3. Comparison between St(OK [G]) and Cl(OK [G])

The group Cl(OK [G]) is defined as the kernel of ε? : Cl(OK [G]) −→ Cl(OK), the grouphomomorphism induced by the augmentation map ε : OK [G] −→ OK , which sends an element∑s∈G ass to ∑s∈G as. McCulloh proved the following result.

Proposition 3.1. — For every algebraic number field K and finite group G,R(OK [G]) ⊆ Cl(OK [G]).

Proof. — The original proof is contained in [McC77] (see also [McC83]).

Remark 3.2. — In terms of the Hom-description, we have the following isomorphism

Cl(OK [G]) ∼=HomΩK (RG, J(Kc))

HomΩK (RG, (Kc)×) ·Det (U (OK [G])) .

The superscript “ ” means that we are considering the homomorphisms f such that f(χ0) =1, where χ0 is the trivial character of G (see [BS05a]).

Considering the inclusion R(OK [G]) ⊆ St(OK [G]) (cf. Theorem 2.17), a natural questionarises: what is the link between the two groups St(OK [G]) and Cl(OK [G])? Are they equal?

A first answer to these questions is given by the following result.



Proposition 1. — For every algebraic number field K and finite group G, we getSt(OK [G]) ⊆ Cl(OK [G]).

Proof. — Let us consider a class c ∈ St(OK [G]) represented in terms of the Hom-descriptionby f ∈ Hom ΩK

(RG, J(Kc)). In order to prove that c belongs to Cl(OK [G]), we need toshow that f(χ0) ∈ K× · U(OK) (see [BS05a, Proposition 2.1]).Since for every finite group G, the trivial character χ0 belongs to AG, in order to get f(χ0)we can compute the value of rag(f)(χ0). By the definition of St(OK [G]), we have

rag(f) ∈ Hom ΩK

(AG, (Kc)×

) · U(OK [G]) ·ΘtG,K

(J(KΛ)),so we can split the computation on χ0 into the three components on the right.Let us compute these values:

? Hom ΩK(AG, (Kc)×): If we take g ∈ Hom ΩK

(AG, (Kc)×), the fact that it is ΩK-equi-variant means that, for every ω ∈ ΩK , we have g (χ)ω = g(χω). Thus, when we considerthe value g(χ0), we get that, for every ω ∈ ΩK ,

g (χ0)ω = g(χω0 ) = g(χ0).Then g(χ0) ∈ K×, since it is fixed by ΩK . So we have shown that every element in thegroup Hom ΩK

(AG, (Kc)×) evaluated at χ0 gives a an element in K×.

? U(OK [G]): We look at each place p separately and we compute the values at χ0. Bydefinition of the map Det and considering an element a := ∑

s∈G ass ∈ Kc[G], we obtainDet(a)(χ0) = Tχ0(a) = ∑

s∈G as (where Tχ0 is the trivial representation).If we take xp ∈ (Oc

K, p[G]×/G)ΩKp represented by ap ∈ OcK, p[G]×, we have

Det(xp)(χ0) = rag(Det(ap))(χ0) = Det(ap)(χ0).Now, for every ω ∈ ΩKp , we have aωp = ap · s′, where s′ ∈ G. Thus, applying Tχ0 , we getTχ0(ap)ω = Tχ0(ap). Hence, for each p, the element Det(ap)(χ0) belongs to O×K, p. Thus,every element in U(OK [G]), when evaluated at χ0, gives an element in U(OK).

? ΘtG,K

(J(KΛ)): Given h ∈ J(KΛ), by definition, we have that ΘtG,K

(h)(χ0) equal toh(ΘG (χ0)

). Moreover ΘG(χ0) = 0, since χ0(s) = 1 for every s ∈ G. Thus, every element

in ΘtG,K

(J(KΛ)) evaluated at χ0 is trivial.

Combining all together, we can now see that, if we take c ∈ St(OK [G]) and we consider arepresentative of it f ∈ Hom ΩK

(RG, J(Kc)), we obtain f(χ0) ∈ K× · U(OK), as we wantedto prove. After this proposition, one may wonder if the reverse inclusion also holds. This is the case forsome groups (e.g. G = A4, see [BS05b]), but is not in general true as the next counterexampleshows.

Counterexample. Given a prime number p, take G = Cp, a cyclic group of order p; then, asshown in [Rim59], we have Cl(Z[Cp]) ∼= Cl(Z[ζp]), where ζp is a primitive p-th root of unity.Since Cl(Z) is trivial, we get Cl(Z[Cp]) = Cl(Z[Cp]).


A. Siviero 81

We know that R(Z[Cp]) is trivial and so, by McCulloh’s results, the same is true for St(Z[Cp]).Thus, just taking a cyclic group Cp, with a prime number p such that the class number ofCl(Z[ζp]) is not one (e.g. p = 23, see [Was97, Chapter 11]), we have a simple example of agroup G and a number field K for which St(OK [G]) ( Cl(OK [G]).

4. Computing the Stickelberger subgroupIn this section we explicitly compute St(OK [G]) in some special cases, just using its algebraicdefinition and the classical Stickelberger theorem. In particular, we shall prove the next result,already announced in the Introduction.

Theorem 1. — Given a prime number p 6= 2. If G = C2, a cyclic group of order 2, orG = Cp, a cyclic group of order p or G = Dp, a dihedral group of order 2p, then St(Z[G]) istrivial.

This result, as explained in the Introduction, implies the following corollary.

Corollary 1. — In the cases of the theorem above, R(Z[G]) is trivial.

4.1. The classical Stickelberger theorem. — First, we briefly recall here some annihi-lation results for class groups.Let N/Q be a finite abelian extension, by the Kronecker–Weber theorem, N ⊆ Q(ζn) (wheren is assumed to be the minimal integer with this property and ζn is a primitive n-th root ofunity). If H = Gal(N/Q), then it can be viewed as a quotient of (Z/nZ)× and we denote byσµ, where µ ∈ (Z/nZ)×, both the element of Gal(Q(ζn)/Q) which sends ζn to its µ-th powerand its restriction to N . Then the Stickelberger element of N is defined as

Ψ :=∑

µ∈ (Z/nZ)×

µ

n

σ−1µ ∈ Q[H].

We have the following classical theorem.

Theorem 4.2. — (Stickelberger’s theorem). Let I be a fractional ideal of N , let β ∈ Z[H],and suppose βΨ ∈ Z[H]. Then (βΨ) · I is principal.

Proof. — [Was97, Theorem 6.10].

Another useful relation for ideal classes of cyclotomic extensions is given by the next theorem.

Theorem 4.3. — Let L be the cyclotomic extension Q(ζn), where ζn is a primitive n-throot of unity, and denote by σµ, for µ ∈ (Z/nZ)×, the automorphism defined above. Let p bea prime number, such that p - n and let us consider a prime ideal P|p in OL. For positiveintegers a, b such that ab(a+ b) 6≡ 0 mod n, let us write

Ψa, b :=∑

µ∈ (Z/nZ)×

(⌊(a+ b)µn

⌋−⌊aµ

n

⌋−⌊bµ

n

⌋)σ−1µ .

Then (Ψa, b) ·P is principal. Since any ideal class contains infinitely many primes, this givesa relation on the ideal class group of Q(ζn).

Proof. — [Lan94, Chapter IV, §4, Theorem 11].



4.4. Background on Cp and Dp. — Let Cp be a cyclic group of order a prime number pwith generator denoted by t and let the group Irr(Cp) be generated by χp, where χp(t) := e

2πip .

We denote by χ0 the trivial character (χpp = χ0).The following result, that we have already used in the counterexample of the previous section,describes Cl(Z[Cp]) in terms of the class group of the cyclotomic extension Q(ζp) and is dueto D. S. Rim. Using the Hom-description for the class group Cl(Z[Cp]), we can state it in thefollowing way.

Lemma 4.5. — Let p be a prime number and let ζp be a primitive p-th root of unity. Thefollowing group isomorphism holds:

(16) L : Cl(Z[Cp])∼=−→ Cl(Z[ζp])

c = [f ] 7−→ [f(χp)] .

Proof. — This is a result contained in [Rim59] and here rewritten in terms of the Hom-de-scription, after having recalled the idelic representation of the ideal class group

Cl(Z[ζp]) ∼= J (Q(ζp)) /(Q(ζp)× · U(Z[ζp])

).

Note that the element f , representative of c, belongs to Hom ΩQ

(RCp , J(Qc)

).

The dihedral group Dp is the group of symmetries of a regular polygon with p sides, includingboth rotations and reflections. It has order 2p and it can be represented as

Dp := 〈r, s | rp = s2 = 1, s−1rs = r−1〉.

We will just consider p ≥ 3, note that D2 is the Klein four-group.If p ≥ 3, the set Irr(Dp) consists of two characters ψ0 and ψ′0 of dimension 1 and (p − 1)/2characters ψj (with j = 1, . . . , (p − 1)/2) of dimension 2. The character ψ0 is the trivialcharacter, while ψ′0 sends rk to 1 and srk to −1, for k = 0, . . . , p− 1. The characters ψj , forj = 1, . . . , (p− 1)/2, are defined as

ψj :

rk 7−→ 2 cos

(2πjkp

), k = 0, . . . , p− 1 ;

srk 7−→ 0, k = 0, . . . , p− 1.

For Dp an analogous result to Lemma 4.5 follows.

Lemma 4.6. — Let p be an odd prime number and let ζp be as above. The following groupisomorphism holds:

J : Cl(Z[Dp])∼=−→ Cl(Z(ζp + ζ−1

p ))[f ] 7−→ [f(ψ1)] .

Proof. — This follows from the Wedderburn decomposition Q[Dp] ∼= Q×Q×M2(Q(ζp+ζ−1p ))

and the isomorphism Cl(Z[Dp]) ∼= Cl(M) ∼= Cl(Z(ζp + ζ−1p )), where M denotes a maximal

order in Q[Dp] containing Z[Dp]. See [CR87, Theorem 50.25] for more details.


A. Siviero 83

4.7. The Stickelberger map for Cp and Dp. — For every prime number p, in the cycliccase Cp, it is easy to see that 〈χ0, tj〉 is equal to 0 and 〈χip, tj〉 =

ij

p

, for j = 0, . . . , p − 1

and i = 1, . . . , p− 1. Hence

ΘCp :

χ0 7−→ 0 ,

χip 7−→i

pt+

2ip

t2 + · · ·+

(p− 1)ip

tp−1 , for 1 ≤ i ≤ p− 1.

While for the dihedral group Dp (with p ≥ 3), first we have to think about the restriction ofthe irreducible characters over the cyclic subgroups 〈r〉 (of order p) and 〈srk〉 (of order 2),for k = 0, . . . , p − 1. We adopt the same notation of the characters of Cp used above for thecharacters of 〈r〉 and we denote by φ0, φ′0 the trivial and the non-trivial character of 〈srk〉,respectively. Then, for the characters of dimension 1 we clearly have

resDp〈r〉ψ0 = χ0 resDp〈srk〉ψ0 = φ0

resDp〈r〉ψ′0 = χ0 resDp〈srk〉ψ

′0 = φ′0

while, for the characters of dimension 2, using the inner products and some computations,we get

resDp〈r〉ψj = χjp + χp−jp , for j = 1, . . . , (p− 1)/2,

and for the subgroups 〈srk〉, where k = 0, . . . , p− 1, we have

resDp〈srk〉ψj = φ0 + φ′0, for j = 1, . . . , (p− 1)/2.

We easily deduce the values of the Stickelberger pairings on the elements of Irr(Dp):

〈ψ0, rk〉 = 〈ψ0, srk〉 = 0, for k = 0, . . . , p− 1,〈ψ′0, rk〉 = 0, 〈ψ′0, srk〉 = 1/2, for k = 0, . . . , p− 1,

〈ψj , 1〉 = 0, for j = 1, . . . , (p− 1)/2,〈ψj , rk〉 = 1, for k = 1, . . . , p− 1 and j = 1, . . . , (p− 1)/2,〈ψj , srk〉 = 1/2, for k = 0, . . . , p− 1 and j = 1, . . . , (p− 1)/2.

We can now consider the Stickelberger map on the conjugacy classes ΘDp: Q⊗ZRDp → Q[Dp]

(cf. (10)). There are (p+ 3)/2 conjugacy classes of Dp:

1, rk, r−k, for k = 1, . . . , (p− 1)/2, and s, sr, sr2, . . . , srp−1;

then, since 〈χ, s〉 was defined as 〈χ, s〉, it is easy to see that we obtain:

ΘDp:

ψ0 7−→ 0 ,ψ′0 7−→ 1

2s ,

ψj 7−→∑(p−1)/2k=1 rk + 1

2s, for j = 1, . . . , (p− 1)/2.



4.8. The augmentation kernels ACp and ADp. —As we have already seen in Remark2.7, we have the following lemma.

Lemma 4.9. — Let p be a prime number, thenAC2 = 〈χ0, 2χ2〉 ,ACp = 〈χ0, jχp − χjp, pχp〉, for 2 ≤ j ≤ p− 1, with p 6= 2.

An analogous result for the dihedral group Dp follows.

Lemma 4.10. — Let p be an odd prime number, then

ADp = 〈ψ0, 2ψ′0, ψ′0 − ψj〉, for j = 1, . . . , (p− 1)/2.

Proof. — Consider an element α ∈ RDp and write it as

α = α0ψ0 + α′0ψ′0 +

(p−1)/2∑

j=1αjψj .

As det(ψj) = ψ′0, we have

det(α) =(ψ′0)α′0+

∑(p−1)/2j=1 αj ,

hence, by the definition of ADp ,

α ∈ ADp ⇐⇒ α′0 +(p−1)/2∑

j=1αj ≡ 0 mod 2.

Thus, writing

α = α0ψ0 + 2bψ′0 −(p−1)/2∑

j=1αj(ψ′0 − ψj),

where b ∈ Z such that α′0 +∑(p−1)/2j=1 αj = 2b, we get our claim.

4.11. The triviality of ΘtCp,Q and Θt

Dp,Q. —Once we know the structure of the augmen-

tation kernel ACp , we can apply the classical Stickelberger theorem for the computation ofΘtCp,Q

(Hom ΩQ

(Z[Cp(−1)], J(Qc))), as the following proposition explains.

Proposition 4.12. — For every prime number p,

ΘtCp,Q

(Hom ΩQ

(Z[Cp(−1)], J(Qc)))⊆ Hom ΩQ

(ACp ,Q(ζp)× · U (Z[ζp])

).

Proof. — The group Hom ΩQ(Z[Cp(−1)], J(Qc)) is equal to Hom ΩQ

(Z[Cp(−1)], J (Q(ζp)))(think about the ΩQ-action) and, given an element h ∈ Hom ΩQ

(Z[Cp(−1)], J (Q(ζp))), weimmediately understand that, thanks to the ΩQ-action, it is uniquely determined by h(1)and h(t) (where t is the chosen generator of Cp). Indeed σ−1

i · t = ti (remember the twistin the definition of the ΩQ-action on Cp(−1)), where, as before, the automorphism σi ∈Gal(Q(ζp)/Q) is such that σi(ζp) = ζip, for i = 1, . . . , p − 1. Thus if h(t) = x ∈ J(Q(ζp)),considering the ΩQ-invariance, we have h(ti) = σ−1

i · x.


A. Siviero 85

Now, using the description of the Stickelberger map given above, on the generators of ACp

we get

ΘCp :

χ0 7−→ 0 ,pχp 7−→ t+ 2t2 + · · ·+ (p− 1)tp−1 ,

jχp − χjp 7−→⌊2jp

⌋t2 + · · ·+

⌊(p− 1)jp

⌋tp−1 , for 2 ≤ j ≤ p− 1;

where the last line is not considered if p = 2.We can now compute the transpose of the Stickelberger map (again when p = 2 we justconsider the first two lines) on h ∈ Hom ΩQ

(Z[Cp(−1)], J (Q(ζp))), obtaining

ΘtCp,Q(h) :

χ0 7−→ 1 ,pχp 7−→

(∑p−1i=1 iσ

−1i

)· x ,

jχp − χjp 7−→(∑p−1

i=1

⌊ijp

⌋σ−1i

)· x , for 2 ≤ j ≤ p− 1.

Now, using the idelic representation of Cl(Z[ζp]) recalled in the proof of Lemma 4.5, we imme-diately deduce from Theorem 4.2 that Θt

Cp,Q(h)(pχp) is trivial once considered in Cl(Z[ζp]).This means that Θt

Cp,Q(h)(pχp) ∈ Q(ζp)× ·U(Z[ζp]), which proves the proposition for p = 2.When p 6= 2, for the other generators jχp − χjp, we use Theorem 4.3 on the cyclotomicextension Q(ζp) and we proceed by induction. Starting with j = 2, we get

ΘtCp,Q(h)(2χp − χ2

p) =

p−1∑

i=1

⌊2ip

⌋σ−1i

· x

and using Theorem 4.3, with a = b = 1, we get ΘtCp,Q(h)(2χp − χ2

p) ∈ Q(ζp)× · U(Z[ζp])(proving the result for p = 3).For p > 3, let j be a natural number in 2, . . . , p− 1, denote Θt

Cp,Q(h)(jχp − χjp) by xj andassume that xj ∈ Q(ζp)× · U(Z[ζp]), then we have

xj+1xj

=

p−1∑

i=1

(⌊(j + 1)ip

⌋−⌊ji

p

⌋)σ−1i

· x ,

which belongs to Q(ζp)× · U(Z[ζp]), applying Theorem 4.3 with a = j and b = 1. Thus wededuce that, if xj ∈ Q(ζp)× · U(Z[ζp]), then xj+1 ∈ Q(ζp)× · U(Z[ζp]), which by inductionconcludes the proof.

We do exactly the same for Dp and an analogous result follows.

Proposition 4.13. — Let p be an odd prime number. Then

ΘtDp,Q

(Hom ΩQ

(Z[Dp(−1)], J(Qc)

))⊆ Hom ΩQ

(ADp ,Q(ζp + ζ−1

p )× · U(Z[ζp + ζ−1

p ])).

Proof. — Going back to the definition of the ΩQ-action onDp(−1), we see that Stab (s) equalsΩQ, since s is of order 2, while Stab

(rk)

= Gal(Qc/Q(ζp+ζ−1p )), for all k = 1, . . . , (p−1)/2.

Thus, Hom ΩQ

(Z[Dp(−1)], J(Qc)

)is equal to the set Hom ΩQ

(Z[Dp(−1)], J

(Q(ζp + ζ−1

p )))

.Given h ∈ Hom ΩQ

(Z[Dp(−1)], J(Qc)

), then Θt

Dp,Q(h) in Hom ΩQ

(ADp , J(Qc)

)is defined by

the values it assumes on the set of basis elements of ADp which we studied above. In particular,



if we denote by x ∈ J(Q) the element h(s) and by y ∈ J(Q(ζp + ζ−1p )) the element h(r), we

have

ΘtDp,Q

(h) :

ψ0 7−→ 1 ,2ψ′0 7−→ x ,

ψ′0 − ψj 7−→ −(∑σ∈Gal(Q(ζp+ζ−1p )/Q) σ) · y, for j = 1, . . . , (p− 1)/2.

where in the last computation we used the fact that Gal(Q(ζp + ζ−1p )/Q) acts transitively on

the set of conjugacy classesrkk=1, ..., (p−1)/2

.We see that Θt

Dp,Q(h)(ψ0) and Θt

Dp,Q(h)(2ψ′0) are in J(Q) so they can be written as a

product of a global and a unit element (Cl(Z)=1). The same holds for ΘtDp,Q

(h)(ψ′0 − ψj),with j = 1, . . . , (p−1)/2, since (∑σ∈Gal(Q(ζp+ζ−1

p )/Q) σ) ·y = NQ(ζp+ζ−1p )/Q(y). This concludes

the proof.

4.14. Proof of Theorem 1. — Let us consider the isomorphism L, given in Lemma 4.5.For every ω ∈ ΩQ and c ∈ Cl(Z[Cp]), represented in terms of the Hom-description byf ∈ Hom ΩQ

(RCp , J(Qc)

), we have:

L(c)ω = [f(χp)]ω = [f(χp)ω] = [f(χωp

)],

where, in the last equality, we use the ΩQ-equivariance of f .Using again the classical Stickelberger theorem and the isomorphism L between the locallyfree class group Cl(Z[Cp]) and the ideal class group Cl(Z[ζp]), we can now prove the followingproposition.

Proposition 4.15. — Let p be a prime number and let f be in Hom ΩQ

(RCp , J(Qc)

), such

thatrag(f) ∈ Hom ΩQ

(ACp ,Q(ζp)× · U (Z[ζp])

).

If c := [f ] ∈ Cl(Z[Cp]), then c is trivial.

Proof. — If p = 2, then Cl(Z[C2]) ∼= Cl(Z) = 1, so there is nothing to prove and in our proofwe can assume p 6= 2. Using the isomorphism L, we have

(17) L(c)p = [f(χp)]p = [f(χp)p](a)= [f(pχp)]

(b)= [rag(f)(pχp)]

(c)= 1,

where (a) is given by the fact that f is a homomorphism, (b) follows since pχp ∈ ACp and(c) is given by hypothesis and thanks to the idelic representation of the ideal class groupCl(Z[ζp]).If σj ∈ Gal(Q(ζp)/Q) is such that σj(ζp) = ζjp , for j = 1, . . . , p− 1, we also get

σj · (L(c)) = [f(jχp − (jχp − χjp))] = [f(jχp)][f(jχp − χjp)]−1 (d)= [f(jχp)] = L(c)j ,

where in (d) we use the fact that jχp − χjp ∈ ACp and the idelic representation of Cl(Z(ζp)).Once we know the action of σj on L(c), we can apply Stickelberger’s theorem to the elementof Cl (Z[ζp]) given by L(c):

(18) 1 =p−1∑

j=1jσ−1

j · (L(c))(e)=

p−1∏

j=1L(c)jj−1 = L(c)p−1,


A. Siviero 87

where the first equality is exactly Stickelberger’s theorem and (e) is assured by σ−1j = σj−1 ,

where j−1 is the inverse of j in (Z/pZ)× belonging to 1, . . . , p−1. Note that the last equalityfollows from the fact that jj−1 ≡ 1 mod p and from (17).Finally, putting together (17) and (18), we have L(c) = 1, which, thanks to the isomorphism(16), implies the triviality of c in Cl(Z[Cp]).

Using the isomorphism J given in Lemma 4.6, an analogous result for the dihedral case follows.

Proposition 4.16. — Let p be an odd prime and let f be in Hom ΩQ

(RDp , J(Qc)

), such

thatrag(f) ∈ Hom ΩQ


p )× · U(Z[ζp + ζ−1

p ])).

If c := [f ] ∈ Cl(Z[Dp]), then c is trivial.

Proof. — Given c ∈ Cl(Z[Dp]) as in the hypothesis, then

J(c) = [f(ψ1)] = [f(ψ′0 − (ψ′0 − ψ1))] = [f(ψ′0)][f(ψ′0 − ψ1)]−1.

Now since ψ′0 − ψ1 is contained in ADp , by hypothesis we have [f(ψ′0 − ψ1)] = 1 and so weget J(c) = [f(ψ′0)]. Since f(ψ′0) ∈ J(Q), this concludes the proof.

We can finally prove Theorem 1.Proof of Theorem 1. We consider the case G = Dp, the Cp case is analogous.A class c = [f ] ∈ Cl(Z[Dp]) belongs to St(Z[Dp]) if and only if

rag(f) = g · w ·ΘtDp,Q

(h),

where g is in Hom ΩQ

(ADp , (Qc)×

), w ∈ U(Z[Dp]) ⊆ Hom ΩQ

(ADp , U(Zc)

)(see (6) for the

original definition of U(Z[Dp])) and h is in Hom ΩQ

(Z[Dp(−1)], J

(Q(ζp + ζ−1

p

))).

SinceHom ΩQ

(ADp , (Qc)×

)= Hom ΩQ


p )×)

Hom ΩQ

(ADp , U(Zc)

)= Hom ΩQ

(ADp , U(Z[ζp + ζ−1

p ])),

clearly g · w ∈ Hom ΩQ


p )× · U(Z[ζp + ζ−1

p ]))

. Thus, using Proposition 4.13and Proposition 4.16, we finally get that c is trivial, as we wanted to show.

5. Some functorial properties of St(OK [G])

In this section, we shall study the behaviour of the subgroup St(OK [G]) under change of thebase field.

Given K a subfield of a number field L, as already explained in the Introduction, we havea restriction map NL/K : Cl(OL[G]) −→ Cl(OK [G]). In terms of the Hom-description it isexpressed by the norm map

NL/K :Hom ΩL

(RG, J(Qc))Hom ΩL

(RG, (Qc)×) ·Det(U(OL[G])) −→Hom ΩK

(RG, J(Qc))Hom ΩK

(RG, (Qc)×) ·Det(U(OK [G]))[f ] 7−→ [NL/K(f)]



where NL/K(f)(α) := ∏ω∈ΩK/ΩL f

ω(α) = ∏ω∈ΩK/ΩL f

(αω−1)ω

(by the definition of the leftΩQ-action on Hom (RG, J(Qc))), for every α ∈ RG. Instead of taking Kc (resp. Lc) in theHom-description of Cl(OK [G]) (resp. Cl(OL[G])), we consider Qc in order to homogenize thenotation (we can do it as L and K are both algebraic extensions of Q - cf. (3)).From Theorem 2.17, we know that R(OL[G]) (resp. R(OK [G])) is contained in the Stickel-berger subgroup St(OL[G]) (resp. St(OK [G])), with equality when the group G is abelian.An interesting question naturally arises: is the Stickelberger subgroup functorial under thismap? Or, more precisely, does the inclusion NL/K(St(OL[G])) ⊆ St(OK [G]) hold?In this section we are going to give an affirmative answer to this question, which will havesome nice consequences, as explained in the last part.

5.1. Changing the base field for the Stickelberger subgroup. —Using the grouphomomorphisms

Cl(OL[G])RagL // MCl(OL[G]) J(LΛ),

ΘtG, L

oo

we defined St(OL[G]) as Rag−1L (Im(Θt

G,L)). Analogously St(OK [G]) is Rag−1

K (Im(ΘtG,K

)).The norm map NL/K induces the following well-defined group homomorphisms (for which wewill use the same name):

NL/K : MCl(OL[G]) −→ MCl(OK [G]),

NL/K : Hom ΩL

(Z[G](−1), J(Qc)

)−→ Hom ΩK

(Z[G](−1), J(Qc)

).

Thus we can prove the next result.

Proposition 5.2. — The following diagram commutes:

Cl(OL[G])RagL //

NL/K

MCl(OL[G])

NL/K

Hom ΩL

(Z[G(−1)], J(Qc)

)ΘtG,L

oo

NL/K

Cl(OK [G])RagK //MCl(OK [G]) Hom ΩK

(Z[G(−1)], J(Qc)

).

ΘtG,K

oo

Proof. — First of all we claim that the following diagram commutes

Cl(OL[G])RagL //

NL/K

MCl(OL[G])

NL/K

Cl(OK [G])RagK //MCl(OK [G]).

Given a homomorphism f ∈ Hom ΩK(RG, J(Qc)) and α := ∑

χ∈ Irr(G) aχχ in AG, using thedefinition of NL/K , we have

NL/K(RagL(f))(α) =∏

ω∈ΩK/ΩL

RagL(f)(αω−1)ω

=∏

ω∈ΩK/ΩL

f(αω−1)ω


A. Siviero 89

= NL/K(f)(α)= RagK(NL/K(f))(α)

which proves our claim. From this, we have NL/K(Ker(RagL)) ⊆ Ker(RagK).We pass now to the proof of the commutativity of the following diagram

(19) MCl(OL[G])

NL/K

Hom ΩL

(Z[G(−1)], J(Qc)

)ΘtG,L

oo

NL/K

MCl(OK [G]) Hom ΩK

(Z[G(−1)], J(Qc)

).

ΘtG,K

oo

Given g ∈ Hom ΩL

(Z[G(−1)], J(Qc)

)and an element α ∈ AG, we have

ΘtG,K

(NL/K(g))(α) = NL/K(g)

∑

s∈G(−1)

〈α, s〉s

=∏

ω∈ΩK/ΩL

g

∑

s∈G(−1)

〈α, s〉s

ω−1

ω

=∏

ω∈ΩK/ΩL

∏

s∈G(−1)

g(sω−1)〈α,s〉

ω

=∏

ω∈ΩK/ΩL

∏

t∈G(−1)

g (t)〈α,tω〉

ω

using the fact that every ω acts as an automorphism. On the other side

NL/K(ΘtG,L

(g))(α) =∏

ω∈ΩK/ΩL

ΘtG,L

(g)(αω−1)ω

=∏

ω∈ΩK/ΩL

g

∑

s∈G(−1)

〈αω−1, s〉s

ω

=∏

ω∈ΩK/ΩL

∏

s∈G(−1)

g (s)〈αω−1

,s〉

ω

=∏

ω∈ΩK/ΩL

∏

s∈G(−1)

g (s)〈α,sω〉

ω

where in the last equality we used the relation 〈αω−1, s〉 = 〈αω−1

, s〉 = 〈α, sω〉 = 〈α, sω〉 =〈α, sω〉, which one can get using the definition of the action of ω on G(−1), the definition ofthe Stickelberger pairing for the set of conjugacy classes and property (9). This proves the



commutativity of (19).The previous two diagrams combine to prove that the following diagram commutes

Cl(OL[G])RagL //

NL/K

MCl(OL[G])

NL/K

Hom ΩL

(Z[G(−1)], J(Qc)

)ΘtG,L

oo

NL/K

Cl(OK [G])RagK //MCl(OK [G]) Hom ΩK

(Z[G(−1)], J(Qc)

).

ΘtG,K

oo

Thus the following result now easily follows (this is a refined version of Theorem 2 in theIntroduction).

Theorem 5.3. — Given a finite group G and a subfield K of an algebraic number field L,then

NL/K(Ker(RagL)) ⊆ Ker(RagK),NL/K(St(OL[G])) ⊆ St(OK [G]).

Proof. — The first inclusion is already included in the proof of Proposition 5.2. For thesecond one it is sufficient to have in mind the definition of the Stickelberger subgroup anduse Proposition 5.2. A first consequence of Theorem 5.3 in the abelian case follows.

Corollary 5.4. — Let G be a finite abelian group and let K be a subfield of an algebraicnumber field L. Then NL/K(Rnr(OL[G])) ⊆ Rnr(OK [G]) and NL/K(R(OL[G])) ⊆ R(OK [G]).Proof. — This follows from Theorem 5.3 and from the equalities in the abelian case of Theo-rem 2.17: Rnr(OL[G]) = Ker(RagL) (respectively Rnr(OK [G]) = Ker(RagK)) and R(OL[G]) =St(OL[G]) (respectively R(OK [G]) = St(OK [G])). The following result is valid for every finite group G.

Corollary 5.5. — Let G be a finite group and K be a subfield of an algebraic number fieldL, such that St(OK [G]) = 1. Then for every tame G-Galois L-algebra N , its ring of integersON is a stably free OK [G]-module.Proof. — Clear from Theorem 5.3 and from the fact that the class [ON ] in the class groupCl(OK [G]) is trivial if and only if ON is stably free when seen as an OK [G]-module (cf.Remark 2.3). From this we deduce the two following corollaries which are contained in a more general resultby Taylor ([Tay81]).

Corollary 5.6. — Given an algebraic number field L and an abelian tame G-Galois L-algebra N , its ring of integers ON is a free Z[G]-module.Proof. — It follows from Corollary 5.5 with K = Q and from the fact that, since G isabelian, St(Z[G]) = R(Z[G]) = 1, by [Tay81] and McCulloh’s results. Moreover, note that inthe abelian case to be a stably free Z[G]-module is equivalent to be a free Z[G]-module (seeRemark 2.3).


A. Siviero 91

Corollary 5.7. — Let Dp be a dihedral group of order 2p, where p is an odd prime number.Given a number field L and a tame Dp-Galois L-algebra N , its ring of integers ON is a freeZ[Dp]-module.

Proof. — The proof is a direct consequence of Theorem 1, Corollary 5.5 and the statementconcerning the dihedral groups in Remark 2.3. As the following remark explains, an analogous result to Theorem 5.3 holds when we considerthe behaviour of St(OK [G]) under extension of the base field.

Remark 5.8 (Extension of the base field). — Let L be a finite extension of K asabove. There is a homomorphism extL/K : Cl(OK [G]) −→ Cl(OL[G]) obtained consideringthe extension of scalars via the tensor product OL⊗OK−. In terms of the Hom-description thisfunctor is induced by the canonical injection Hom ΩK

(RG, J(Qc)) −→ Hom ΩL(RG, J(Qc)).

Analogously as before, the following diagram commutes

Cl(OL[G])RagL //MCl(OL[G]) Hom ΩL

(Z[G(−1)], J(Qc)

)ΘtG,L

oo

Cl(OK [G])RagK //

extL/K

OO

MCl(OK [G])

extL/K

OO

Hom ΩK

(Z[G(−1)], J(Qc)

),

ΘtG,K

oo

extL/K

OO

indeed, if f ∈ Hom ΩK(RG, J(Qc)), g ∈ Hom ΩK

(Z[G(−1)], J(Qc)

), and α ∈ AG, we have

extL/K(RagK)(f)(α) = RagK(f)(α) = RagL(f)(α) = RagL(extL/K)(f)(α),extL/K(Θt

G,K)(g)(α) = Θt

G,K(g)(α) = g(ΘG(α)) = Θt

G,L(extL/K(g))(α).

From this we deduce the following inclusionsextL/K(Ker(RagK)) ⊆ Ker(RagL),extL/K(St(OK [G])) ⊆ St(OL[G]).

Acknowledgments. — I am very grateful to Philippe Cassou-Noguès, Bart de Smit andBoas Erez for their suggestions and precious help. Further I would like to thank AlessandroCobbe for his interest on this paper. Finally, I also wish to thank Adebisi Agboola, NigelP. Byott and Jean Gillibert for all the remarks and advices they gave me as part of my thesiscommittee.

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76:55–61, 1962.[Tay81] M. J. Taylor, On Fröhlich’s conjecture for rings of integers of tame extensions, Invent. Math.,

63(1):41–79, 1981.[Was97] L. C. Washington, Introduction to cyclotomic fields, volume 83 of Graduate Texts in Math-

ematics. Springer-Verlag, New York, second edition, 1997.

July 15, 2014

Andrea Siviero, Université Bordeaux I – 351, cours de la Libération – F 33405 Talence cedex –Universiteit Leiden – Niels Bohrweg, 1 – 2333 CA Leiden • E-mail : [email protected]



THE MAXIMAL UNRAMIFIED EXTENSIONS OF CERTAINCOMPLEX ABELIAN NUMBER FIELDS

by

Siman Wong

Abstract. — We combine root discriminant bounds with a ramification argument to showunconditionally that Q(

√−7,√

61) has no nontrivial unramified extension, a result first provedby Yamamura under the generalized Riemann hypothesis (GRH). This renders unconditionalhis determination of the maximal unramified extensions of the complex quadratic fields withclass number 2. Assuming the GRH, we prove an analogous result for the degree 14 subfieldof the cyclotomic field Q(ζ49), a case previously not handled by conditional root discriminantbounds alone.

Résumé. — Nous combinons les minorations des discriminants avec des considérations por-tant sur la ramification pour montrer, inconditionnellement, que le corps Q(

√−7,√

61) n’a pasd’extension non-ramifiée non-triviale (ce résultat a été montré par Yamamura avec l’aide deGRH). Cela rend inconditionnelle la détermination des extensions non-ramifiées maximales descoprs quadratiques complexes de nombre de classes 2. Sous GRH, nous montrons un résultatanalogue pour le sous-corps de degré 14 de Q(ζ49) (corps non étudié même sous GRH).

1. IntroductionThere are many examples of real quadratic fields with class number one that admit non-trivialextensions unramified at all finite places (for examples and discussion see [8, p.121], [16], [19],[22]), but we have no analogous example of complex Abelian number fields with class numberone [17, p. 914ff]. Yamamura [17] shows that there are 172 complex Abelian number fieldswith class number one. Using the unconditional Odlyzko bound of root discriminants [10], wefind that 132 of these fields K are unramified-closed, i.e. K = Kur, the maximal unramifiedextension of K. The stronger form of the Odlyzko bound under the generalized Riemannhypothesis for zeta functions of number fields (GRH) shows that an additional 23 of these172 fields are also unramified-closed. Combining root discriminant bounds with the theoryof group extensions, Yamamura [21] has since verified unconditionally that five additional

Mathematical subject classification (2010). — 11R20, 11R21, 11R29, 20E22.Key words and phrases. — Abelian fields, group extensions, root discriminants, unramified extensions.Acknowledgements. — Siman Wong’s work is supported in part by NSF grant DMS-0901506.

94 Maximal unramified extensions of complex Abelian fields

complex Abelian fields are unramified-closed (and two more under GRH). In this paper weaugment this approach with a ramification argument to study Kur for two additional cases.

Theorem 1. — (a) Unconditionally the complex Abelian number field Q(√−7,√

61) isunramified-closed.(b) Assume the generalized Riemann hypothesis for the zeta functions of number fields. Thenthe degree 14 subfield of the cyclotomic field Q(ζ49) is unramified closed.

Note that Q(√−7,√

61) is the Hilbert class field of Q(√−427). By [9], there are 18 complex

quadratic fields Q(√−d) with class number 2; specifically

−d = 15, 20, 24, 35, 40, 51, 52, 88, 91, 115, 123, 148, 187, 232, 235, 267, 403, 427.

In a later work, Yamamura [18] shows that except for −d = 115, 235, 403 and 427, the Hilbertclass field H−d of these quadratic fields (which turn out to be the genus fields) are unramifiedclosed, and that for each of −d = 115, 235 and 403, the second Hilbert class field of Q(

√−d)

(i.e. the Hilbert class fields of H−d) is unramified closed. Finally, using the GRH form of theOdlyzko bound, he shows that H−427 is unramified-closed. Thanks to theorem 1 we can nowremove this GRH condition.

Theorem 2. — The Hilbert class field of Q(√−427) has no nontrivial unramified extension,

whence the maximal unramified extensions of imaginary quadratic fields of class number 2 asdetermined by Yamamura are valid unconditionally.

We now give an outline of the proof. Thanks to the unconditional Odlyzko bound, [Kur : K]is finite for Q(

√−7,√

61); the same is true for the field in theorem 1(b) under GRH, thanksto the conditional Odlyzko bound. Suppose Kur 6= K. Since K has class number one and iscomplex, Gal(Kur/K) must admit a simple quotient. From the root discriminant of K wefind that Gal(Kur/K) is either A5 or PSL2(F7). Using the theory of group extensions andexplicit knowledge of the groups Gal(Kur/K) and Gal(K/Q), we deduce from the hypothesisKur 6= K the existence of a subfield k/Q of degree ≤ 8 and with known Galois closure. Forthe two fields in Theorem 1, careful analysis of their ramification data leads to a sharp boundof |Disc(k/Q)|. For the degree 14 field, the bound is sharp enough that we can rule out theexistence of k/Q by looking up tables of number fields [7]. For Q(

√−7,√

61), we are ledto a hypothetical field k of degree 8 and of discriminant ±74614 ∼ ±3.3 × 1010, which liesoutside the range of [7]. To eliminate this remaining case we combine our construction withan argument of Roberts [13].There are four more fields in [17] which are known to be unramified closed only under GRH,and 17 more which are not known to be unramified closed even conditionally. To handle thesefields requires new ideas; see Section 6 for details.

2. Preliminaries on group theoryIn this section we recall the basic theory of group extensions and perform a calculation forlater use. For more details, see [14, Chap. 11].For any integer n > 0, denote by Cn the cyclic group of order n. For any positive integersn1, . . . , nk, set Cn1,...,nk

:= Cn1 × · · · × Cnk. All groups will be written multiplicatively. In


S. Wong 95

particular, the identity element of Cn is denoted 1. An extension of a group N by anothergroup H is a short exact sequence(1) 1 −→ N −→ E −→ H −→ .

A computation using the theory of group extensions and outer automorphisms of finite simplegroups yields the following result [18, Prop. 2, Prop. 3].

Proposition 1 (Yamamura). — (a) For each of the following pairs of groups N,H, theisomorphism classes of groups G which are extensions of N by H are as follows:

N = A5H C2,2 C14G A5 × C2,2 A5 × C7

S5 × C2 S5 × C7

N = PSL2(F7)H C2,2 C7G PSL2(F7)× C2,2 PSL2(F7)× C7

PGL2(F7)× C2 PGL2(F7)× C7

(b) With the notation as above,

– if N ' A5 then G always contains an index 5 subgroup;

– if G ' PSL2(F7)× C2,2 or PSL2(F7)× C7 then G contains an index 7 subgroup;

– if G ' PGL2(F7)× C2 or PGL2(F7)× C7 then G contains an index 8 subgroup.

The following elementary facts about PGL2(F7) will be needed later on.

Lemma 1. — (a) PGL2(F7) has two conjugacy classes of order 2 elements. One of the twoclasses is contained in PSL2(F7). The other class is disjoint from PSL2(F7), has size 28,and the normalizer of any one of them is conjugate to(2) H := 〈( ∗ 0

0 1),( 0 1

1 0)〉 ⊂ PGL2(F7).

(b) PGL2(F7) has 28 C6 subgroups. They are pairwise conjugate.(c) Let B ⊂ PGL2(F7) be the projective image of a Borel subgroup. For any C6 subgroupS ⊂ PGL2(F7), either S ⊂ B or S ∩B is trivial.(d) PGL2(F7) has 42 C2,2 subgroups not contained in PSL2(F7); they are pairwise conjugate.Every order 2 element in PGL2(F7)− PSL2(F7) is contained in three C2,2 subgroups.

Proof. — (a) Every order 2 element in PGL2(F7) is represented by a matrix m ∈ GL2(F7)such that m2 = ±I. So the possible choices for eigenvalues of m are ±1,±

√−1 ∈ F72 . Since

m is not a scalar matrix and√−1 6∈ F7, the characteristic polynomial of m (necessarily F7-

rational) must be one of x2 ± 1. Both cases occur: Consider for example the matrices( 0 −1

1 0)

and γ :=(−1 0

0 1). Thus we get two PGL2(F7)-conjugacy classes of order 2 elements, one of

which is in PSL2(F7) and the other one is disjoint from PSL2(F7). The latter contains theprojective image of γ which as a matrix in GL2(F7) is contained in a split Cartan subgroup.So the PGL2(F7)-normalizer of γ is (2) above (cf. [15, prop. 17]). Since #H = 12, thePGL2(F7)-class of γ has size 336/12 = 28.(b) The C6 subgroups of PGL2(F7) are projective image of split Cartan subgroups. Thusthey are pairwise conjugate, and the number of such subgroups is equal to the number ofunordered pairs of distinct lines through the origin of a 2-dimensional F7-vector space. Thereare 8 such lines, so there are 8 · (8− 1)/2 = 28 such unordered pairs.



(c) Let S′ be a non-trivial subgroup of a C6 subgroup S ⊂ PGL2(F7). Then S′ is the projectiveimage of a non-trivial, non-cyclic subgroup S′ of a split Cartan subgroup S ⊂ GL2(F7) (whoseprojective image is S). Now, S corresponds to a unique, unordered pair of distinct lines `1, `2of F2

7 (and two S correspond to the same unordered pair if and only if both subgroups arecontained in the same (maximal) split Cartan subgroup). Since S′ is not cyclic, with respectto the ordered basis `1, `2 the only Borel subgroups B ⊂ GL2(F7) that contain S′ are〈( 1 ∗

0 1)〉 and 〈( 1 0

∗ 1)〉, in which case S ⊂ B as well, whence S ⊂ B.

(d) Let T be a C2,2 subgroup of PGL2(F7) not contained in PSL2(F7). By part (a), Tcontains a conjugate of γ =

(−1 00 1

), and hence T is conjugate to a C2,2 subgroup of the

centralizer of γ. Since γ has order 2, this centralizer is in fact the normalizer of γ. We readilycheck that (2) contains the following three C2,2 subgroups:

(3)(±1 0

0 1),( 0 ±1

1 0)

;(±1 0

0 1),( 0 ±2

1 0)

;(±1 0

0 1),( 0 ±4

1 0).

So every order 2 element in PGL2(F7)− PSL2(F7) is contained in three C2,2 subgroups.Since PSL2(F7) has index 2 in PGL2(F7), a C2,2 subgroup not in PSL2(F7) contains exactlytwo elements not in PSL2(F7), and these two unordered pair of elements uniquely determinethis C2,2 subgroup. Combine part (a) with the previous paragraph and we see that there are(28× 3)/2 = 42 C2,2 subgroups not in PSL2(F7).Finally, from the description (3) we see that the C2,2 subgroups not in PSL2(F7) are theprojective image of a split Cartan subgroup, and hence they are pairwise conjugate.

3. Proof of the theorem: Basic setupLet K be one of the fields in Theorem 1. Denote by Kur the maximal unramified extensionof K. In the table below we list the root discriminant of K, as well as an upper bound ofthe degree of Kur/Q furnished by the unconditional (resp. conditional) Odlyzko bound ([4,§2.2]; [10]). For odd m and for n|ϕ(m), denote by Lnm the degree n subfield of Q(ζm), so L14

49is the degree 14 subfield of Q(ζ49).

K Gal(K/Q) root discriminant [Kur : Q] [Kur : K] Odlyzko boundQ(√−7,

√61) C2,2

√7 · 61 = 20.664 < 786 < 196 unconditional

L1449 C14 7 25/14 = 32.293 < 4800 < 343 conditional

By Galois theory, Gal(Kur/K) is an extension of Gal(Kur/K) by Gal(K/Q). The next resultputs restrictions on Gal(Kur/K) (cf. also [18, Prop. 2]).

Lemma 2. — Let K be one of the fields in the table. Assume GRH if K = L1449.

(a) Let K ′/K be a non-Abelian, simple, unramified finite Galois extension. Then K ′/Q isGalois.(b) Suppose the extensions K ′/K in part (a) do not exist. Then Kur = K.

Proof. — (a) Suppose otherwise; denote by M/Q the Galois closure of K ′/Q. The simplicityof Gal(K ′/K) then implies that the intersection of any two conjugates of K ′/Q is exactlyK, so [M : Q] ≥ [K ′ : K]2[K : Q]. But M is the compositum of all conjugates of K ′/Q andK ′/K is unramified, so M/K is also unramified. Thus M has the same root discriminant as


S. Wong 97

K, which is too small, by the unconditional Odlyzko bound [4] for K = Q(√−7,√

61), andby the conditional Odlyzko bound [11] for K = L14

49.(b) If Gal(Kur/K) is not trivial then it has a simple quotient, by Jordan-Hölder. But Khas class number 1 and is totally complex, so Gal(Kur/K) has trivial Abelianization. Thehypothesis then implies that K = Kur. Let K be one of the Abelian number fields in the table above, so [Kur : K] < 343. Soif Kur 6= K then Gal(Kur/K) is either A5 or PSL2(F7). Thanks to Proposition 1 andLemma 2, we are reduced to study number fields k/Q unramified outside the bad primes ofK/Q, such that

(4) [k : Q] = 5 if [Kur : K] = 60,

7 if Gal(Kur/Q) contains a direct factor of PSL2(F7),8 otherwise.

Moreover, when [k : Q] = 8 the Galois group of k/Q is PGL2(F7). In each case we exploit thearithmetic of K and the group theoretical properties of the Galois group of these hypotheticalfields k to show that k cannot exist, and hence K must be unramified-closed.

4. The case of L1449

Lemma 3. — Let F/Q be a degree 7 number field with Galois group PSL2(F7). ThenDisc(F/Q) is a perfect square.

Proof. — By [3], S7 contains a single conjugacy class of transitive subgroups isomorphicto PSL2(F7). Furthermore, such subgroups are all contained in A7. So if f ∈ Q[x] is aseptic polynomial with F as its splitting field, then disc(f) is a perfect square. Polynomialdiscriminant differs from the field discriminant by a square, so we are done. For the rest of this section we will focus on the case K = L14

49. This C14 extension is totallyramified at 7 and is unramified at all other finite primes. Suppose K 6= Kur, and consider theassociated extension k/Q in (4) (furnished under GRH). If [k : Q] = 5 then Disc(k/Q) mustdivide 74. By the database of Jones and Roberts [7] (where the result is proven completein this case), there is no such quintic field. Next, suppose [k : Q] > 5 and that 7 is tamelyramified in k/Q. Then the ramification index of any prime of k lying above 7 is ≤ 2, whenceDisc(k/Q) divides 74. Again this is not possible, thanks to [7].Finally, suppose [k : Q] = 7 or 8 and that 7 is wildly ramified. Then exactly one prime p ink/Q lying above 7 ramifies, with ramification index 7. The completion kp of k at p is containedin the completion of L14

49 at its unique prime above 7, so kp/Q7 is a degree 7 Abelian extensionwith conductor 72. The conductor-discriminant formula then says that this Abelian degree7 extension kp/Q7 has discriminant (72)6, and hence |Disc(k/Q)| = 712. By the database ofJones and Roberts [7] (where the result is proven complete in this case), there is no such fieldof degree of 7. To handle the case where [k : Q] = 8 we now give an argument applicable tothe case [k : Q] = 7 as well.Denote by L/Q the Galois closure of k/Q. Since Gal(L/Q) ⊂ PGL2(F7), that meansGal(L/Q) has no order 14 element. Since k/Q already has a prime with ramification in-dex 7 and the inertia group of any ramified prime in Kur/Q is C14, the inertia group of anyramified prime in L/Q is C7. Since Q has no non-trivial extension unramified at all finiteplaces, the Galois group of any finite Galois extension of over Q is generated by the inertia



groups of the extension. Since the order 7 subgroups of PGL2(F7) are transvections and theygenerate PSL2(F7), it follows that Gal(L/Q) ' PSL2(F7). Thus [k : Q] = 7 and L14

49L/k isunramified, whence the root discriminant of L14

49L/Q is 712/7 = 28.102. That is too small, bythe GRH Odlyzko bound [11]. Thus k/Q does not exist, whence Kur = K.

Remark 1. — Recall that in the outset we need to use the GRH Odlyzko bound to deducethat [Kur : K] is finite for K = L14

49, so even if we replace the last line above with the databasesearch as at the end of the previous paragraph, our argument for K = L14

49 would still beconditional.

5. The case of Q(√−7,√

61)

For the rest of this section we take K = Q(√−7,√

61). Suppose K 6= Kur. Let k/Q beas in (4). Then Kur/K is unramified, and the ramification index of 7 or 61 in K/Q is 2.Consequently,(5) the ramification index of each ramified prime in k lying above 7 or 61 is 2.When [k : Q] = 5, by (5) we see that k has either at most two ramified primes of residualdegree 1 lying above p, or it has exactly one ramified prime of residual degree 2. ThusDisc(k/Q) divides 72612. When [k : Q] = 7, Lemma 3 plus (5) together imply that Disc(k/Q)divides 72612 as well. There is no such quintic or septic field, by [7].Finally, suppose [k : Q] = 8. The argument above shows that Disc(k/Q) divides 74614, whichis too large for us to handle. By Proposition 1, Kur/Q has Galois group PGL2(F7)× C2. Ithas a unique PGL2(F7) subfield L/Q which is the Galois closure of k/Q. To get a sharperestimate of Disc(k/Q) we now analyze closely the ramification of L/Q.

Lemma 4. — Let p ⊂ OL be a prime lying above p ∈ 7, 61.(a) The inertia group I(p) has order 2 and is not contained in PSL2(F7).(b) The decomposition group D(p) is isomorphic to one of C2, C2,2 or C6.

Proof. — (a) Let p ⊂ OL be a prime lying above p ∈ 7, 61. Since L ⊂ Kur and Kur/Kis unramified, so #I(p) = 2. Inertia groups of conjugate primes are PGL2(F7)-conjugate,so I(p) ⊂ PSL2(F7) if and only if I(p′) ⊂ PSL2(F7) for every p′ lying above p. ButGal(Kur/K) ' PSL2(F7) and Gal(Kur/Q) ' PGL2(F7), so if I(p) ⊂ PSL2(F7) then pis unramified in K/Q, a contradiction. Thus I(p) 6⊂ PSL2(F7).(b) D(p) normalizes I(p) ' C2 and I(p) 6⊂ PSL2(F7), so by Lemma 1(a), it is conjugate to asubgroup of H := 〈( a 0

0 d),( 0 1

1 0)〉 ⊂ PGL2(F7). In particular, #D(p) = 2m where m = 1, 2, 3,

or 6. Since D(p)/I(p) is cyclic and I(p) ' C2, we see that D(p) is one of the followingsubgroups of the order 12 dihedral group H:

m D(p)1 C2

2 C2,2, C4

3 C6

Since I(p) 6⊂ PSL2(F7) and PGL2(F7)−PSL2(F7) contains no order 4 element, we can ruleout C4.


S. Wong 99

Lemma 5. — We have Disc(k/Q) = ±73613.

Proof. — Let p ∈ 7, 61, and let p ⊂ OL be a prime lying above p. Note that we can takeGal(L/k) to be the normalizer of

( 1 10 1)in PGL2(F7), i.e. a projective Borel subgroup. Such a

subgroup has at least one but not all C6 subgroups, so by Lemma 1(c), if D(p) ' C6 then atleast one prime in k lying above p has inertia degree 1 and ramification index 1, and at leastone other prime in k has inertia degree 3 and ramification index 2. As [k : Q] = 8, that meanspOk = P2

0P1P2 with Norm(P0) = p3 and Norm(P1) = Norm(P2) = p, whence p3||Disc(k/Q).To handle the two remaining cases of D(p), we use the classical fact (cf. [6, Lemma 5]) thatif

(6) PGL2(F7) =∐

τ∈TGal(L/k) τ I(p),

is the double-coset decomposition of PGL2(F7) by Gal(L/k) and I(p), then there are exactly#T primes in k lying above p, and that the ramification index of the prime corresponding toτ ∈ T is the index

(7) [τI(p)τ−1 : Gal(L/k) ∩ τI(p)τ−1].

To compute this double-coset decomposition, we will use the fact [2, p. 213] that PGL2(F7)is realizable as a subgroup of S8 generated by the permutations γ := (2687453) and δ :=(13867542).First, take D(p) = I(p) ' C2. Using the computer algebra package GAP [5] we find that N ,the normalizer of the 7-cycle γ in 〈γ, δ〉, contains the order two element µ := (16)(24)(58).Since PSL2(F7) is the commutator subgroup of PGL2(F7), and hence necessarily even,while µ is odd, it follows that µ 6∈ PSL2(F7). By Lemma 1(d), the order 2 elements inPGL2(F7) − PSL2(F7) are pairwise conjugate, so we can µ take to be the generator ofI(p) = D(p). Using this explicit description

PGL2(F7) = 〈γ, δ〉, Gal(L/k) = N, and D(p) = I(p) = 〈µ〉,we use GAP to find that the double coset decomposition (6) has size 5. Thus there are exactlyfive primes in k lying above p, each with inertia degree 1 and ramification index ≤ 2. Since[k : Q] = 8, the only possibility is that pOk = P2

1P22P2

3P4P5 with every Norm(Pi) = p,whence p3||Disc(k/Q). Note that we do not need to invoke (7).Finally, suppose D(p) ' C2,2. Using GAP we find that the centralizer of µ contains the per-mutation ω := (14)(26)(37). Taking 〈µ, ω〉 as a model of D(p) we now find that the doublecoset decomposition has size 3, with representatives

τ1 = (), τ2 = (13)(27)(58), τ3 = (15628473).

Using (7), we check that the ramification indices of the corresponding primes of k are 1, 2and 2, respectively. Since [k : Q] = 8 and each such prime has inertia degree ≤ 2, the onlypossibility is that pOk = P1P2

2P23 with Norm(P1) = Norm(P2) = p2 and Norm(P3) = p,

whence p3||Disc(k/Q). This completes the proof of the lemma.

Lemma 5 improves the trivial estimate Disc(k/Q)|(74614), but it still lies outside the rangeof the database [7]. We now augment this with an argument of Roberts [13] to eliminate k.Fix an order 2 element σ ∈ PGL2(F7)−PSL2(F7), and denote by Lσ the fixed field of L byσ.



Lemma 6. — (a) The field discriminant of Lσ/Q is ±7786181,±7816178, or ±7816181.(b) Suppose Lσ is not totally real and that the unique quadratic subfield of L/Q is complexquadratic. Then Lσ has at least six distinct real places.

Proof. — (a) Let p be either 7 or 61, and let p ⊂ OL be a prime lying above p.First, supposeD(p) = C2. By Lemma 1(a), PGL2(F7) has 28 order 2 subgroups not containedin PSL2(F7). For any one of them, call it J2, there are (336/2)/28 = 6 primes in L withJ2 as its inertia group. Exactly one of the 28 choices of J2 is Gal(L/Lσ). Thus there are(28− 1) · 6 · 1

2 = 81 ramified primes in Lσ lying above p, each with ramification index 2 andinertia degree 1. Thus the p-part of the field discriminant of Lσ is p81.Next, suppose D(p) = C6. By Lemma 1(b), there are 28 C6 subgroups J6, each one beingthe decomposition group of (336/6)/28 = 2 primes in L. Thus there are (28− 1) · 2 · 1

2 = 27ramified primes in Lσ lying above p, each with ramification index 2 and inertia degree degree3. Thus the p-part of the field discriminant of Lσ is (p3)27 = p81.Finally, suppose D(p) = C2,2. By Lemma 1(d), there are (336/4)/42 = 2 primes in L witha given C2,2 as its decomposition group, and three of the 42 C2,2 contain σ. Thus there are(42− 3) · 2 · 1

2 = 39 ramified primes in Lσ lying above p, each with ramification index 2 andinertia degree degree 2. Thus the p-part of the field discriminant of Lσ is (p2)39 = p78.(b) Since the unique quadratic subfield of L/Q is complex, complex conjugation gives rise toan order 2 element in Gal(L/Q) ' PGL2(F7) not contained in PSL2(F7). Conjugate numberfields have the same number of real places, and the conjugates of Lσ are precisely Lσ′ whereσ′ are PGL2(F7)-conjugate to σ. Recall Lemma 1(a) and we see that to prove part (b) wecan take σ ∈ Gal(L/Q) to be complex conjugation. Then Lσ ⊂ R, and we are reduced tofind six distinct field automorphisms of Lσ.In the course of proving the D(p) ' C2 case of Lemma 5, we saw that PGL2(F7) is realizableas a subgroup of S8 generated by the permutations (2687453) and (13867542), and thatµ := (16)(24)(58) is an order 2 element of this permutation representation of PGL2(F7) notcontained in PSL2(F7). Using the computer algebra system GAP, we find that the centralizer ofµ has order 12. That means there are twelve elements α1, . . . , α12 in Gal(L/Q) that commutewith σ.Fix a a normal basis of L/Q, i.e. fix an element ω ∈ L so that gω : g ∈ Gal(L/Q) is a Q-basis of L. Let g1, . . . , g168 be a complete set of right coset representatives of 〈σ〉 ⊂ Gal(L/Q).Then the elements

(8) giω + σgiω (1 ≤ i ≤ 168)

are Q-linearly independent, and hence they form a Q-basis of Lσ/Q. Since each αi abovecommutes with σ, left-multiplication by αi takes the set of elements in (8) to itself. We claimthat

(9) αm and αn induce the sameaction on the elements (8) ⇐⇒ αm = σiαn for some i ∈ 0, 1.

The set of αi, being the centralizer of σ, is closed under multiplication by σ. It then followsthat the restriction of these twelve αi ∈ Gal(L/Q) to Lσ define six pairwise distinct fieldautomorphisms of Lσ ⊂ R, and hence Lσ has at least six distinct real embeddings.


S. Wong 101

It remains to verify the claim (9). To say that αm and αn induce the same action on theelements (8) is to say that

(αm + αmσ − αn − αnσ)(giω) = 0 for all i.Since σ has order 2, it follows that

(αm + αmσ − αn − αnσ)(σgiω) = 0 for all i.Recall that the gi is a complete set of right coset representatives of 〈σ〉 ⊂ Gal(L/Q) and thatgω : g ∈ Gal(L/Q) is a Q-basis of L/Q, that means

(αm + αmσ − αn − αnσ)(x) = 0 for all x ∈ L.By the linear indepdendency of field automorphisms [1, Cor. on p. 84], that means αm = σiαnfor some i ∈ 0, 1, as desired. Finally, we apply Roberts’ argument [13] to deduce a contradiction from Lemma 6, andtherefore Q(

√−7,√

61) must be unramified-closed.First, suppose |Disc(Lσ/Q)| 6= ±7816181. Then the root discriminant of Lσ is ≤ 17.912. Thisis not possible, since Diaz y Diaz [4] shows unconditionally that a degree 168 extension hasroot discriminant ≥ 17.98.Next, suppose Lσ/Q has field discriminant ±7816181. Note that Lσ has a unique quadraticsubfield which is one of Q(

√−7× 61),Q(

√61) or Q(

√−7). In the first two cases, Lσ/Q is

linearly disjoint from Q(√−7), and hence Lσ(

√−7)/Q(

√−7) is unramified at the prime of

Q(√−7) above 7. Then in these two cases the root discriminant of Lσ(

√−7)/Q is

(716861162)1/336 = 19.20 . . .This is not possible, since Diaz y Diaz [4] shows unconditionally that a degree 336 extensionhas root discriminant ≥ 19.47, so we are done.Finally, suppose |Disc(Lσ/Q)| = 7816181 and that Q(

√−7) is the unique quadratic subfield

of the PGL2(F7)-extension L/Q. Then lemma 6(b) says that Lσ/Q has at least six distinctreal places. For any degree 168 field with r1 real places and r2 pairs of complex places, usingb = 9.000 in [11, Table 4] yields the unconditional root discriminant lower bound (see [11,description of tables] for details)

> 53.047 r1/16820.710 2r2/168e−24.001/168.

In particular, if r1 ≥ 6 then the root discriminant is ≥ 18.566, contradicting our hypothesis|Disc(Lσ/Q)|1/168 = (7816181)1/168 = 18.5456. This completes the proof of the theorem.

6. Remaining casesAs we pointed out in the introduction, among the 172 complex Abelian number fields withclass number one, 132 of them are known to be unramified closed under the unconditionalOdlyzko, 23 of them are known to be unramified closed under the conditional Odlyzko bound(we will call these GRH fields), and the status of the 17 remaining fields are open (we willcall these unknown fields). In this paper we determine the unramified closure of three of the23 GRH fields and three of the 17 unknown field. We now discuss the remaining fields.The unconditional Odlyzko bound for the root discriminant of a degree n field is |D|1/n >22.38161+o(1) [12, (2.5)]. The root discriminant of 16 of the GRH fields exceed this value, so



for these 16 fields we do not even know unconditionally if [Kur : Q] is finite. Of the sevenremaining GRH fields, Yamamura [21] has since verified unconditionally that Kur = K forQ(√−53− 2

√53), Q(

√−3,√−11, cos(2π/7)), and Q(

√−61− 6

√61). Theorem 1(a) handles

the case Q(√−7,√

163). The three remaining GRH fields are as follows (in this table thedegree [Kur : Q] is computed using the unconditional Odlyzko bound):

K Gal(K/Q) root discriminant [Kur : Q] [Kur : K]L3

43(√−3) C6 (33434)1/4 = 21.26 < 1524 < 254

Q(√−3,√−163) C2,2

√3 · 163 = 22.11 < 13538 < 3384

Q(√−11, i sin(2π/8)) C2,4 (222114)1/8 = 22.31 < 102183 < 12772

The smallest case above, L343(√−3), is almost reachable by the techniques here. The remaining

cases seem to be out of reach by current technology.We now turn to the 17 unknown fields. Four of them,

Q(√−43,

√−67), Q(

√−19,

√−163), Q(

√−43,

√−163), Q(

√−67,

√−163)

have root discriminants that exceed the GRH Odlyzko bound [12, (2.6)]. For these fields wedo not even know if Kur is a finite extension. Yamamura [21] has since verified uncondi-tionally that Kur = K for Q(

√−1,√−163) and Q(

√−11,

√−67), and under GRH, the field

L437(√−1). Theorem 1(b) resolves the case L14

49 under GRH, and we list the nine remainingunknown fields in the table below. In this table the degree [Kur : Q] is computing using theconditional Odlyzko bound; specifically, for root discriminant up to 41.122 we use [11, Table1], and for root discriminant > 41.122 we use the bound

|Disc(M/Q)|1/[M :Q] > Be−E/[M :Q],

where B = 43.425 and E = 3.5263 × 108 are given by [11, Table 3] using b = 25 (see [11,description of tables] for details.) These fields seem to be beyond current technology.

K Gal(K/Q) root discriminant [Kur : Q] [Kur : K]Q(√−7,√−163) C2,2

√7 · 163 = 33.78 < 10000 < 2500

Q(√−19,

√−67) C2,2

√19 · 67 = 35.68 < 31970 < 7993

Q(√−11,

√−163) C2,2

√11 · 163 = 43.34 < 1012 < 1012/4

L667 C6 675/6 = 33.25 < 4800 < 800

L429(√−2) C2,4 293/423/2 = 35.35 < 31970 < 3997

L1261 C12 6111/12 = 43.31 < 1012 < 1012/5

L643(√−3) C2,6 435/631/2 = 39.79 < 106 < 83334

L37L

313(√−2) C2,2,3 (7686138)1/12 = 41.37 < 1012 < 1012/9

L1443 C14 4313/14 = 32.97 < 4800 < 343


S. Wong 103

Acknowledgment. — I would like to thank Professor Ken Yamamura for sending me areprint of [18] and for detailed comments and references; Professor Farshid Hajir for usefuldiscussions; and Professor David Roberts for showing us the beautiful argument in [13].

References[1] E. Artin, Galois theory, 2nd ed. University of Notre Dame Press, 1944.[2] J. E. Burns, The abstract definitions of groups of degree 8, Amer. J. Math. 37 (1915) 195-214.[3] G. Butler and J. McKay, The transitive subgroups of degree up to eleven, Comm. Algebra 11

(1983) 863-911.[4] F. Diaz y Diaz, Tables minorant la racine m-ème du discriminant d’un corps de degré n, Publ.

Math. d’Orsay, 1980.[5] The GAP Group, GAP – Groups, Algorithms, and Programming, Version 4.7.4; 2014.

http://www.gap-system.org

[6] F. Hajir and S. Wong, Specializations of one-parameter families of polynomials, Annales del’Institut Fourier 56 (2006) 1127-1163.

[7] J. W. Jones and D. P. Roberts, A database of number fields. Preprint, May 2014. Accompanyingsearchable online database: http://hobbes.la.asu.edu/NFDB/

[8] S. Lang, Algebraic number theory, Springer-Verlag, 1986.[9] H. L. Montgomery and P. J. Weinberger, Notes on small class numbers, Acta. Arith. 24 (1974)

529-542.[10] A. M. Odlyzko, On conductors and discriminants, in Algebraic number fields, A. Fröhlich, ed.

1977.[11] A. M. Odlyzko, Unpublished tables titled Discriminant bounds, November 29, 1976. Available

from http://www.dtc.umn.edu/∼odlyzko/unpublished/index.html

[12] A. M. Odlyzko, Bounds for discriminants and related estimates for class numbers, regulators andzeros of zeta functions: a survey of recent results, Sém. Théorie des Nombres, Bordeaux 2 (1990)119-141.

[13] D. Roberts, Email communication. May 8, 2014.[14] D. Robinson, A course in the theory of groups. 2nd ed. Springer-Verlag, 1996.[15] J. P. Serre, Propriétés galoisiennes des points d’ordre fini des courbes elliptiques, Invent. Math.

15 (1972) 259-331.[16] K. Yamamura, On unramified Galois extensions of real quadratic number fields, Osaka J. Math.

23 (1986) 471-478.[17] K. Yamamura, The determination of the imaginary Abelian number fields with class number one.

Math. Comp. 62 (1994) 899-921.[18] K. Yamamura, The maximal unramified extension of the imaginary quadratic fields with class

number two. J. Number Theory 60 (1996), no. 1, 42-50.[19] K. Yamamura, On quadratic number fields each having an unramified extension which properly

contains the Hilbert class field of its genus field, Galois groups and modular forms (Saga, 2000).Kluwer Acad. Publ., 2003, p. 271-286.



[20] K. Yamamura, Maximal unramified extensions of imaginary quadratic number fields of smallconductors, II, J. Théor. Nombres Bordeaux 13 (2001) 633-649.

[21] K. Yamamura, Table of the imaginary Abelian number fields with class number one,http://tnt.math.se.tmu.ac.jp/pub/CDROM/CM-fields/imab.dvi

[22] K. Yamamura, Unpublished table titled Examples of real quadratic number fields with class num-ber one having an unramified nonsolvable Galois extension.

June 11, 2014

Siman Wong, Department of Mathematics & Statistics, University of Massachusetts. Amherst, MA 01003E-mail : [email protected]


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