Journal de Théorie des Nombres de Bordeaux

Xavier Caruso

David Savitt

Poids de l’inertie modérée de certaines représentations cristallines

Michael Coons

(Non)Automaticity of number theoretic functions

Alina Firicel

Sur le développement en fraction continue d’une généralisation de la cubique de Baum et Sweet

Bianca Viray

Bianca VIRAY
A family of varieties with exactly one pointless rational fiber
Tome 22, no 3 (2010), p. 741-745.
<http://jtnb.cedram.org/item?id=JTNB_2010__22_3_741_0>
© Université Bordeaux 1, 2010, tous droits réservés.
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bres de Bordeaux » (http://jtnb.cedram.org/), implique l’accord
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org/legal/). Toute reproduction en tout ou partie cet article sous
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cedram
Article mis en ligne dans le cadre du
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Journal de Théorie des Nombres
de Bordeaux 22 (2010), 741-745
A family of varieties with exactly one pointless
rational fiber
par Bianca VIRAY
Résumé. On construit un exemple concret d’une famille à un pa-
ramètre de variétés lisses, projectives, et géométriquement intègres
sur un sous-schéma ouvert de P1Q, de sorte qu’il y ait précisément
une fibre rationnelle sans point rationnel. Ceci rend explicite une
construction de Poonen.
Abstract. We construct a concrete example of a 1-parameter
family of smooth projective geometrically integral varieties over
an open subscheme of P1Q such that there is exactly one rational
fiber with no rational points. This makes explicit a construction
of Poonen.
1. Introduction
We construct a family of smooth projective geometrically integral sur-
faces over an open subscheme of P1Q with the following curious arithmetic
property: there is exactly one Q-fiber with no rational points. Our proof
makes explicit a non-effective construction of Poonen [6, Prop. 7.2], thus
giving “an extreme example of geometry not controlling arithmetic” [6,
p.2]. We believe that this is the first example of its kind.
Theorem 1.1. Define P0(x) := (x2−2)(3−x2) and P∞(x) := 2x4+3x2−1.
Let π : X → P1Q be the Châtelet surface bundle over P1Q given by
y2 + z2 =
(
6u2 − v2
)2
P0(x) +
(
12v2
)2
P∞(x),
where π is projection onto (u : v). Then π(X(Q)) = A1Q(Q).
Note that the degenerate fibers of π do not lie over P1(Q) so the family of
smooth projective geometrically integral surfaces mentioned above contains
all Q-fibers.
The non-effectivity in [6, Prop. 7.2] stems from the use of higher genus
curves and Faltings’ theorem. (This is described in more detail in [6, §9]).
Manuscrit reçu le 4 septembre 2009.
This research was supported by the Mentored Research Award from University of California,
Berkeley.
742 Bianca Viray
We circumvent the use of higher genus curves by an appropriate choice of
P∞(x).
2. Background
This information can be found in [6, §3,5, and 6]. We review it here for
the reader’s convenience.
Let E be a rank 3 vector sheaf on a k-variety B. A conic bundle C over B is
the zero locus in PE of a nowhere vanishing zero section s ∈ Γ(PE ,Sym2(E)).
A diagonal conic bundle is a conic bundle where E = L1 ⊕ L2 ⊕ L3 and
s = s1 + s2 + s3, si ∈ Γ(PE ,L⊗2i ).
Now let α ∈ k×, and let P (x) ∈ k[x] be a separable polynomial of degree
3 or 4. Consider the diagonal conic bundle X given by B = P1,L1 =
O,L2 = O,L3 = O(2), s1 = 1, s2 = −α, s3 = −w4P (x/w). This smooth
conic bundle contains the affine hypersurface y2 − αz2 = P (x) ⊂ A3 as an
open subscheme. We say that X is the Châtelet surface given by
y2 − αz2 = P (x).
Note that since P (x) is not identically zero, X is an integral surface.
A Châtelet surface bundle over P1 is a flat proper morphism V → P1
such that the generic fiber is a Châtelet surface. We can construct them in
the following way. Let P,Q ∈ k[x,w] be linearly independent homogeneous
polynomials of degree 4 and let α ∈ k×. Let V be the diagonal conic bundle
over P1(a:b) × P
1
(w:x) given by L1 = O,L2 = O,L3 = O(1, 2), s1 = 1, s2 =
−α, s3 = −(a2P + b2Q). By composing V → P1 × P1 with the projection
onto the first factor, we realize V as a Châtelet surface bundle. We say that
V is the Châtelet surface bundle given by
y2 − αz2 = a2P (x) + b2Q(x),
where P (x) = P (x, 1) and Q(x) = Q(x, 1). We can also view a, b as rela-
tively prime, homogeneous, degree d polynomials in u, v by pulling back by
a suitable degree d map φ : P1(u:v) → P
1
(a:b).
3. Proof of Theorem 1.1
By [5], we know that the Châtelet surface
y2 + z2 = (x2 − 2)(3− x2)
violates the Hasse principle, i.e. it has Qv-rational points for all completions
v, but no Q-rational points. Thus, π(X(Q)) ⊆ A1Q(Q). Therefore, it remains
to show that X(u:1), the Châtelet surface defined by
y2 + z2 = (6u2 − 1)2P0(x) + 122P∞(x),
has a rational point for all u ∈ Q.
A family of varieties with exactly one pointless rational fiber 743
If P(u:1) := (6u2− 1)2P0(x) + 122P∞(x) is irreducible, then by [3], [4] we
know that X(u:1) satisfies the Hasse principle. Thus it suffices to show that
P(u:1) is irreducible and X(u:1)(Qv) 6= ∅ for all u ∈ Q and all places v of Q.
3.1. Irreducibility. We prove that for any u ∈ Q, the polynomial
P(u:1) (x) is irreducible in Q[x] by proving the slightly more general state-
ment, that for all t ∈ Q
Pt(x) := (2x4 + 3x2 − 1) + t2(x2 − 2)(3− x2)
= x4(2− t2) + x2(3 + 5t2) + (−6t2 − 1)
is irreducible in Q[x]. We will use the fact that if a, b, c ∈ Q are such that
b2−4ac and ac are not squares in Q then p(x) := ax4 +bx2 +c is irreducible
in Q[x].
Let us first check that for all t ∈ Q,
(
3 + 5t2
)2 − 4 (2− t2) (−6t2 − 1)
is not a square in Q. This is equivalent to proving that the affine curve
C : w2 = t4 + 74t2 + 17 has no rational points. The smooth projective
model, C : w2 = t4 + 74t2s2 + 17s4 in weighted projective space P(1, 1, 2),
has 2 rational points at infinity. Therefore C is isomorphic to its Jacobian.
A computation in Magma shows that Jac(C)(Q) ∼= Z/2Z [1]. Therefore, the
points at infinity are the only 2 rational points of C and thus C has no
rational points.
Now we will show that
(
−6t2 − 1
) (
2− t2
)
is not a square in Q for any
t ∈ Q. As above, this is equivalent to determining whether C ′ : w2 = (−6t2−
1)(2−t2) has a rational point. Since 6 is not a square in Q, this is equivalent
to determining whether the smooth projective model, C ′, has a rational
point. The curve C ′ is a genus 1 curve so it is either isomorphic to its
Jacobian or has no rational points. A computation in Magma shows that
Jac (C ′) (Q) ∼= Z/2Z [1]. Thus #C ′ (Q) = 0 or 2. If (t, w) is a rational point
of C ′, then (±t,±w) is also a rational point. Therefore, #C (Q) = 2 if and
only if there is a point with t = 0 or w = 0 and one can easily check that
this is not the case.
3.2. Local solvability.
Lemma 3.1. For any point (u : v) ∈ P1Q, the Châtelet surface X(u:v) has
R-points and Qp-points for every prime p.
Proof. Let a = 6u2−v2 and let b = 12v2. We will refer to a2P0(x)+b2P∞(x)
both as P(a:b) and P(u:v).
R-points: It suffices to show that given (u : v) there exists an x such that
P(a:b) = x4(2b2 − a2) + x2(3b2 + 5a2) + (−6a2 − b2)
is positive. If 2b2 − a2 is positive, then any x sufficiently large will work.
So assume 2b2 − a2 is negative. Then α = −(3b
2+5a2)
2(2b2−a2) is positive. We claim
744 Bianca Viray
P(a:b)(
√
α) is positive.
P(a:b)(
√
α) = α2(2b2 − a2) + α(3b2 + 5a2) + (−6a2 − b2)
= (3b
2 + 5a2)2
4(2b2 − a2)
+ −(3b
2 + 5a2)2
2(2b2 − a2)
+ (−6a2 − b2)
= 1
4(2b2 − a2)
(
4(2b2 − a2)(−6a2 − b2)− (3b2 + 5a2)2
)
= 1
4(2b2 − a2)
(
−17b4 − 74a2b2 − a4
)
Since 2b2−a2 is negative by assumption and −17b4− 74a2b2−a4 is always
negative, we have our result.
Qp-points:
p ≥ 5: Without loss of generality, let a and b be relatively prime inte-
gers. Let X(a:b) denote the reduction of X(a:b) modulo p. We claim
that there exists a smooth Fp-point of X(a:b) that, by Hensel’s
lemma, we can lift to a Qp-point of X(a:b).
Since P(a:b) has degree at most 4 and is not identically zero mod-
ulo p, there is some x ∈ Fp such that P(a:b) (x) is nonzero. Now let
y, z run over all values in Fp. Then the polynomials y2, P(a:b) (x)−z2
each take (p+ 1)/2 distinct values. By the pigeonhole principle, y2
and P(a:b) (x) − z2 must agree for at least one pair (y, z) ∈ F2p and
one can check that this pair is not (0, 0). Thus, this tuple (x, y, z)
gives a smooth Fp-point of X(a:b). (The proof above that the qua-
dratic form y2 + z2 represents any element in Fp is not new. For
example, it can be found in [2, Prop 5.2.1].)
p = 3: From the equations for a and b, one can check that for any (u :
v) ∈ P1Q, v3(b/a) is positive. Since Q3(
√
−1)/Q3 is an unramified
extension, it suffices to show that given a, b as above, there exists an
x such that P(a:b)(x) has even valuation. Since v3(b/a) is positive,
v3(2b2−a2) = 2v3(a). Therefore, if x = 3−n, for n sufficiently large,
the valuation of P(a:b)(x) is −4n+ 2v3(a) which is even.
p = 2: From the equations for a and b, one can check that for any
(u : v) ∈ P1Q, v2(b/a) is at least 2. Let x = 0 and y = a. Then we
need to find a solution to z2 = a2(−7 + (b/a)2). Since v2(b/a) > 1,
−7 + (b/a)2 ≡ 12 mod 8. By Hensel’s lemma, we can lift this to a
solution in Q2.

A family of varieties with exactly one pointless rational fiber 745
Acknowledgements
I thank my advisor, Bjorn Poonen, for suggesting the problem and many
helpful conversations. I thank the referee for suggesting the pigeonhole prin-
ciple argument in Lemma 3.1. I also thank Daniel Erman and the referee
for comments improving the exposition.
References
[1] Wieb Bosma, John Cannon, Catherine Playoust, The Magma algebra system. I. The
user language. J. Symbolic Comput. 24 (1997), 235–265.
[2] Henri Cohen Number theory. Vol. I. Tools and Diophantine equations Springer, 2007.
[3] Jean-Louis Colliot-Thélène, Jean-Jacques Sansuc, Peter Swinnerton-Dyer Inter-
sections of two quadrics and Châtelet surfaces. I. J. Reine Angew. Math. 373 (1987), 37–
107.
[4] Jean-Louis Colliot-Thélène, Jean-Jacques Sansuc, Peter Swinnerton-Dyer Inter-
sections of two quadrics and Châtelet surfaces. II. J. Reine Angew. Math. 374 (1987),
72–168.
[5] V.A. Iskovskih A counterexample to the Hasse principle for systems of two quadratic forms
in five variables Mat. Zametki 10 (1971), 253–257
[6] Bjorn Poonen Existence of rational points on smooth projective varieties J. Eur. Math.
Soc. (JEMS) 3 529–543
Bianca Viray
Mathematics Department
Box 1917
Brown University
Providence, RI 02912 USA
E-mail: bviray@math.brown.edu
URL: http://math.brown.edu/~bviray


A family of varieties with exactly one pointless rational fiber

David Adam

Fonctions à valeurs entières et module de Carlitz

Hung Manh Bui

Gaps between zeros of the derivative of the Riemann $\xi $-function

Georges Gras

Georges GRAS
Stickelberger’s congruences for absolute norms of relative discriminants
Tome 22, no 2 (2010), p. 397-402.
<http://jtnb.cedram.org/item?id=JTNB_2010__22_2_397_0>
© Université Bordeaux 1, 2010, tous droits réservés.
L’accès aux articles de la revue « Journal de Théorie des Nom-
bres de Bordeaux » (http://jtnb.cedram.org/), implique l’accord
avec les conditions générales d’utilisation (http://jtnb.cedram.
org/legal/). Toute reproduction en tout ou partie cet article sous
quelque forme que ce soit pour tout usage autre que l’utilisation à
fin strictement personnelle du copiste est constitutive d’une infrac-
tion pénale. Toute copie ou impression de ce fichier doit contenir la
présente mention de copyright.
cedram
Article mis en ligne dans le cadre du
Centre de diffusion des revues académiques de mathématiques
http://www.cedram.org/
Journal de Théorie des Nombres
de Bordeaux 22 (2010), 397-402
Stickelberger’s congruences for absolute norms of
relative discriminants
par Georges GRAS
Résumé. Nous généralisons un résultat de J. Martinet sur les
congruences de Stickelberger pour les normes absolues des discri-
minants relatifs des corps de nombres, en utilisant des arguments
classiques du corps de classes.
Abstract. We give an improvement of a result of J. Martinet
on Stickelberger′s congruences for the absolute norms of relative
discriminants of number fields, by using classical arguments of
class field theory.
1. Introduction
Let L/K be a finite extension of number fields. Denote by dL/K the
relative discriminant of L/K and by c the number of complex infinite places
of L which lie above a real place of K.
The absolute norm of an ideal a of K is a positive rational denoted
NK/Q (a); it is the positive generator of NK/Q (a), where NK/Q is the arith-
metic norm. If α ∈ K×, we define the absolute norm of α (or (α)) by
NK/Q (α) := |NK/Q (α) | (this has some importance in class field theory).
In [Ma], J. Martinet proved the following result about NK/Q (dL/K):
Proposition 1. If K contains a primitive 2m+1th root of unity (m ≥ 0)
and if L/K is not ramified at 2, then (−1)cNK/Q (dL/K) ≡ 1 mod (4 · 2m).
In [Pi], S. Pisolkar proved, in connection with the previous result:
Proposition 2. Let p ≥ 2 be any prime number. Let Kv be the completion
of K at a place v | p (or any finite extension of Qp); we suppose that Kv
contains a primitive ph+1th root of unity, h ≥ 0. Let Kv( p
√
α ), α ∈ K×v , be
an unramified Kummer extension of Kv. Then NKv/Qp (α) ≡ 1 mod (ph+2).
In this paper we give a synthetic proof of these results with some gener-
alization of the hypothesis (especially for the case p = 2); see Theorem 2.
Manuscrit reçu le 27 mars 2010, révisé le 8 avril 2010.
Mots clefs. Number fields, Discriminants, Stickelberger congruences, Class field theory, Kum-
mer theory.
Classification math.. 11R29, 11R37.
398 Georges Gras
2. Prerequisites on discriminants
Classical proofs of Stickelberger’s congruences make use of the fact that
any odd discriminant ideal dL/K is canonically associated with the discrimi-
nant of a quadratic extension ofK, unramified at 2. This essential reduction
is summarized in the following proposition (see [Ma, § 3]).
Proposition 3. Let L/K be a finite extension of number fields and let
αK×2, in K×/K×2, be the image of the discriminant α of a K-base of L.
Then:
(i) The class αK×2 does not depend on the choice of the K-base.
(ii) Let K ′ := K(
√
α ); then there exists an integral ideal a of K such
that dL/K = dK′/K a2.
(iii) If 2 is unramified in L/K it is unramified in K ′/K and we have
dK′/K = (α) b2, hence dL/K = (α) c2, where b and c are ideals of K.
We suppose in the sequel that dL/K is odd; thus we can choose, modulo
K×2, an odd α, which implies that c is odd. We then have to compute
NK/Q (c)2 and NK/Q (α).
3. Computation of NK/Q (c)2.
From class field theory over Q we get NK/Q (c) ∈ AK , the Artin group of
K which is that of Kab, where Kab is the maximal abelian subextension of
K.
So we see that to obtain nontrivial congruences modulo a power of 2 we
must suppose that this Artin group is roughly a ray group mudulo a power
of 2 in the following way.
Let Q(µ2∞) be the field generated by all roots of unity of order a power
of 2. The best hypothesis is that K does contain a subfield k of Q(µ2∞) of
degree 2m, m ≥ 0.
For m = 0 we get k = Q (which is also Q(0) in the description below)
and for any m ≥ 1, the field k is equal to one of the following three fields,
for which we indicate its Artin group as a subgroup of AQ := {uZ, u ∈
Q×, u odd } (see e.g. [Gr, II.5.5.2]):
• k = Q(m) is the subfield, of degree 2m, of the cyclotomic Z2-extension
of Q; its Artin group is:
AQ(m) = {uZ, u ∈ Q
×, u > 0, u ≡ ±1 mod (4 · 2m)} ;
• k = Q′(m), m ≥ 1, is the subfield of Q(µ4·2m) of relative degree 2,
distinct from Q(µ4·2m−1) and from Q(m); its Artin group is
AQ′(m) = {uZ, u ∈ Q
×, u > 0, u ≡ 1 or − 1 + 4 · 2m−1 mod (4 · 2m)} ;
Stickelberger’s congruences for absolute norms of relative discriminants 399
• k = Q(µ4·2m−1), m ≥ 1; its Artin group is
AQ(µ4·2m−1 ) = {uZ, u ∈ Q
×, u > 0, u ≡ 1 mod (4 · 2m−1)} .
So this yields
NK/Q (c)2 ∈ {uZ, u ∈ Q×, u > 0, u ≡ 1 mod (4 · 2m)} ,
except if k = Q(m), in which case
NK/Q (c)2 ∈ {uZ, u ∈ Q×, u > 0, u ≡ 1 mod (4 · 2m+1)} ;
in other words, taking absolute norms:
NK/Q (c)2 ≡ 1 mod (4 · 2m), if k = Q′(m) or Q(µ4·2m−1), m ≥ 1,
NK/Q (c)2 ≡ 1 mod (4 · 2m+1), if k = Q(m), m ≥ 0.
4. Computation of NK/Q (α).
The best way is to use local class field theory by computing NK/Q (α) =∏
v|2
NKv/Q2 (α), where Kv is the completion of K at the place v | 2 of K and
NKv/Q2 the local norm.
The result is given by the following generalization of the result of [Pi].
Let p ≥ 2 be a prime number, let Kv be the completion of the number
field K at v | p (or any finite extension of Qp); we suppose that Kv contains
µp and a subfield k(v) of Qp(µp∞) of degree pm over Qp(µp),m ≥ 0 (if p = 2,
the context is that of the previous section for which the hypothesis are
satisfied for all v | 2, with k(v) = kv := Q2 k and [k(v) : Q2] = [k : Q] = 2m,
independently of v | 2, since k/Q is totally ramified at 2).1
The local norm group of k(v), restricted to the norms of units, is the
following subgroup of Z×p = µp−1 ⊕ (1 + pZp) for p 6= 2 or of
Z×2 = 〈−1 〉 ⊕ (1 + 4 Z2) for p = 2:
• p 6= 2, k(v) = Qp(µpm+1), m ≥ 0; the norm group is 1 + pm+1 Zp;
• p = 2, k(v) = Q
(m)
2 , m ≥ 0; the norm group is 〈−1 〉⊕ (1 + 4 ·2m Z2);
• p = 2, k(v) = Q′2(m), m ≥ 1; the norm group is 〈−1 + 4 · 2m−1 〉Z2 ;
• p = 2, k(v) = Q2(µ4·2m−1), m ≥ 1; the norm group is 1 + 4 · 2m−1 Z2.
We then have:
1Take care that if k = K ∩ Q(µ2∞ ), kv may not be equal to Kv ∩ Q2(µ2∞ ) (for instance
K = Q(
√
−17 ) for which k = Q, kv = Q2, m = 0); but Theorem 1 applies to k(v) = Kv ∩
Q2(µ2∞ ) = Q2(
√
−1 ) with m = 1.
400 Georges Gras
Theorem 1. Let Kv be the completion of a number field K at v | p; sup-
pose that Kv contains µp and a subfield k(v) of Qp(µp∞) of degree pm over
Qp(µp), m ≥ 0. Let Kv( p
√
α ), α ∈ K×v , be an unramified Kummer extension
of Kv (modulo K×pv we can suppose that α is a local unit).
Then we have NKv/Qp (α) ≡ 1 mod (pm+2).
Moreover, if p = 2 and k(v) = Q
(m)
2 , m ≥ 0, and if at least one of the
following two conditions holds:
(i) α ∈ K×2v ,
(ii) the index of ramification ev(Kv/k(v)) of Kv/k(v) is even,
we then have NKv/Q2 (α) ≡ 1 mod (4 · 2m+1).
Proof. We consider the following diagram:
KnrvKv( p
√
α)
.
Kv
knr(v)k(v) .
Fv Fv( p
√
α′ )Qp(µp)
pm
Qp(µp)nr
where for any field L, Lnr is the maximal unramified pro-extension of L;
we know that we have for instance Lnr = LQnrp (see e.g. [Gr, II.1.1.5]).
Put Fv := Kv ∩Qp(µp)nr. All horizontal extensions are unramified and all
vertical extensions are totally ramified.
Consider the intersection Kv( p
√
α)∩Qp(µp)nr as a Kummer extension of
Fv; thus there exists a suitable local unit α′ ∈ F×v such that α = α′ xp with
x ∈ K×v .
Then NKv/Qp (α) = NKv/Qp (α′) · NKv/Qp (x)p; since Fv(
p
√
α′ )/Fv is
unramified we have (see e.g. [Gr, I.6.3, (ii)]):
α′ = x′p (1 + p (1− ζ) y′) ,
x′, y′ ∈ F×v , v(y′) ≥ 0, where ζ is a primitive pth root of unity (for p = 2
we have p (1− ζ) = 4). Then:
NKv/Qp (α) = NKv/Qp (1 + p (1− ζ) y
′) ·NKv/Qp (xx
′)p ;
but (see the above list of norm groups of k(v)), we have
NKv/Qp (xx
′)p ≡ 1 mod (pm+2) ,
and in the particular case when p = 2 and k(v) = Q
(m)
2 ,
NKv/Q2 (xx
′)2 ≡ 1 mod (4 · 2m+1) ,
Stickelberger’s congruences for absolute norms of relative discriminants 401
and
NKv/Qp (1 + p (1− ζ) y
′) = NFv/Qp (1 + p (1− ζ) y
′)[Kv :Fv ] ≡ 1 mod (pm+2)
since [Kv : Fv] is a multiple of pm, which implies first that
(1 + p (1− ζ) y′)[Kv :Fv ] = 1 + pm+1 (1− ζ) y′′ ,
and then that
NFv/Qp (1 + p
m+1 (1− ζ) y′′) ∈ 1 + pm+2 Zp ;
hence we have the congruence
NKv/Qp (α) ≡ 1 mod (p
m+2) .
(i) If p = 2, k(v) = Q
(m)
2 , m ≥ 0, and α ∈ K×2v , then we obtain
NKv/Q2 (α) ≡ 1 mod (4 · 2m+1).
(ii) If p = 2, ev(Kv/k(v)) is even, the above computation yields
NFv/Q2 (1 + 4 y′)[Kv :Fv ] ≡ 1 mod (4 · 2m+1); if moreover k(v) = Q
(m)
2 , since
NKv/Q2 (xx′)2 ≡ 1 mod (4 · 2m+1) in that case, we obtain
NKv/Q2 (α) ≡ 1 mod (4 · 2
m+1) .
This completes the proof of the theorem. 
5. Statement of the main result
We return to the case p = 2. In Sections 3 and 4, we have computed
NK/Q (dL/K), making use of NK/Q (c)2 (absolute norm) and of NK/Q (α)
(arithmetic norm) from the NKv/Q2 (α), taking into account that the con-
gruence NKv/Q2 (α) ≡ 1 mod (4 · 2m) is independent of v with the choice of
k(v) := kv = Q2 k for all v | 2.
To determine NK/Q (α) we note that NK/Q (α) = (−1)ρNK/Q (α), where
ρ is the number of conjugates of α which are negative in the real embeddings
of K; from [Ma, § 3], the numbers ρ and c have same parity.
Thus we have obtained in general:
Theorem 2. Let L/K be a finite extension of number fields, unramified
at 2. Denote by dL/K the discriminant of L/K, by c the number of complex
places of L which lie above a real place of K, and by NK/Q (dL/K) the
absolute norm of dL/K . Let k be the maximal subfield of Q(µ2∞) contained
in K and put [k : Q] =: 2m, m ≥ 0.
Then we have the congruence (−1)cNK/Q (dL/K) ≡ 1 mod (4 · 2m).
Remark 1. We have the following improvement in two particular circum-
stances: if k = Q(m), m ≥ 0, then under at least one of the following two
conditions:
402 Georges Gras
(i) 2 splits totally in K(
√
α )/K, 2
(ii) the indices of ramification of v | 2 in K/k are all even,
we obtain the congruence (−1)c .NK/Q (dL/K) ≡ 1 mod (4 · 2m+1). 3
References
[Gr] G. Gras, Class Field Theory: from theory to practice. SMM, Springer-Verlag, 2003; second
corrected printing: 2005.
[Ma] J. Martinet, Les discriminants quadratiques et la congruence de Stickelberger. Sém.
Théorie des Nombres, Bordeaux 1 (1989), 197–204.
[Pi] S. Pisolkar, Absolute norms of p-primary units. Jour. de Théorie des Nombres de Bordeaux
21 (2009), 733–740.
Georges Gras
Villa la Gardette chemin Château Gagnière
F-38520 Le Bourg d’Oisans
E-mail: g.mn.gras@wanadoo.fr
URL: http://maths.g.mn.gras.monsite-orange.fr/
2If α = x2 (1 + 4 y), y ∈ K×, y 2-integer, this condition is equivalent to TrFv/F2 (y) = 0 for all
v | 2, where TrFv/F2 is the absolute trace from the residue field Fv of K (see [Gr, I.6.3, Lemma]).
3Use the computation of NK/Q (c)2 at the end of Sections 3, then Theorem 1 for the compu-
tation of NKv/Q2 (α) for v | 2 in these particular cases.


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