Journal de Théorie des Nombres de Bordeaux

Antonio Lei

Sarah Livia Zerbes

Signed Selmer groups over $p$-adic Lie extensions

Bianca Viray

Bianca VIRAY
Failure of the Hasse principle for Châtelet surfaces in characteristic 2
Tome 24, no 1 (2012), p. 231-236.
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Journal de Théorie des Nombres
de Bordeaux 24 (2012), 231-236
Failure of the Hasse principle for Châtelet
surfaces in characteristic 2
par Bianca VIRAY
Résumé. Soit k un corps global de caractéristique 2. On construit
une surface de Châtelet sur k possédant une obstruction de Brauer-
Manin au principe de Hasse. Cette construction étend un résultat
de Poonen à la caractéristique 2, en montrant ainsi que l’obs-
truction de Brauer-Manin après revêtement fini étale n’est pas
suffisante pour expliquer tous les contre-exemples au principe de
Hasse sur un corps global arbitraire.
Abstract. Given any global field k of characteristic 2, we con-
struct a Châtelet surface over k that fails to satisfy the Hasse
principle. This failure is due to a Brauer-Manin obstruction. This
construction extends a result of Poonen to characteristic 2, thereby
showing that the étale-Brauer obstruction is insufficient to explain
all failures of the Hasse principle over a global field of any char-
acteristic.
1. Introduction
Poonen recently showed that, for a global field k of characteristic different
from 2, the étale-Brauer obstruction is insufficient to explain failures of the
Hasse principle [6]. This result relied on the existence of a Châtelet surface
over k that violates the Hasse principle [5, Prop 5.1 and §11]. Poonen’s
construction fails in characteristic 2 due to the inseparability of y2 − az2.
Classically, Châtelet surfaces have only been studied over fields of char-
acteristic different from 2. In this paper, we define Châtelet surfaces over
fields of characteristic 2 and obtain a result analogous to [5, Prop 5.1].
Theorem 1.1. Let k be any global field of characteristic 2. There exists a
Châtelet surface X over k that violates the Hasse principle.
The only assumption on characteristic in [6] is in using [5, Prop 5.1] (all
other arguments go through exactly as stated after replacing any polyno-
mial of the form by2 + az2 by its Artin-Schreier analogue, by2 + byz+ az2).
Manuscrit reçu le 7 novembre 2010, révisé le 29 décembre 2010.
This research was supported by the Mentored Research Award from University of California,
Berkeley.
Mots clefs. Hasse principle, Brauer-Manin obstruction, Châtelet surface, rational points.
Classification math. 11G35, 14G05, 14G25, 14G40.
232 Bianca Viray
Therefore, Theorem 1.1 extends the main result of [6] to global fields of
characteristic 2, thereby showing that the étale-Brauer obstruction is in-
sufficient to explain all failures of the Hasse principle over a global field of
any characteristic.
The proof of Theorem 1.1 is constructive. The difficulty in the proof lies
in finding suitable equations so that the Brauer set is easy to compute and
empty.
2. Background
2.1. Brauer-Manin obstructions. The counterexamples to the Hasse
principle referred to in Theorem 1.1 are all explained by the Brauer-Manin
obstruction, which we recall here [1, Thm. 1]. Let k be a global field and let
Ak be the adèle ring of k. Recall that for a projective variety X, we have
the equality X(Ak) =
∏
v X(kv), where v runs over all nontrivial places of
k. The Brauer group of X, denoted BrX, is the group of equivalence classes
of Azumaya algebras on X. Let invv denote the morphism from class field
theory BrQv → Q/Z [7, XIII.3, Prop. 6]. For any A ∈ BrX, P ∈ X(S),
let evA(P ) be the image of A under the map P ∗ : BrX → BrS, which is
induced by P : S → X. Define
X(Ak)Br :=
{
(Pv)v ∈ X(Ak) :
∑
v
invv (evA(Pv)) = 0 for all A ∈ BrX
}
By class field theory we have
X(k) ⊆ X(Ak)Br ⊆ X(Ak).
Thus, if X(Ak)Br = ∅, then X has no k-points. We say there is a Brauer-
Manin obstruction to the Hasse principle if X(Ak) 6= ∅ but X(Ak)Br = ∅.
See [8, §5.2] for more details.
2.2. Châtelet surfaces in characteristic 2. A conic bundle X over
P1 is the zero-locus of a nowhere-vanishing global section s of Sym2(E) in
PE , for some rank 3 vector sheaf E on P1. Consider the special case where
E = O⊕O⊕O(2) and s = s1−s2 where s1 is a global section of Sym2(O⊕O)
and s2 is a global section of O(2)⊗2 = O(4). Take a ∈ k× and P (x) a
separable polynomial over k of degree 3 or 4. If s1 = y2 + yz + az2 and
s2 = w4P (x/w), thenX contains the affine variety defined by y2+yz+az2 =
P (x) as an open subset. In this case we sayX is the Châtelet surface defined
by
y2 + yz + az2 = P (x).
By the same basic argument used in [5, Lemma 3.1], we can show that X
is smooth. See [5, §3 and §5] for the construction of a Châtelet surface in
the case where the characteristic is different from 2.
Failure of the Hasse principle 233
3. Proof of Theorem 1.1
Let k denote a global field of characteristic 2. Let F denote its constant
field and let n denote the order of F×. Fix a prime p of k of odd degree (recall
that the degree of a prime p is the degree of the field extension Op/p over
F), and let S = {p}. Let Ok,S denote the ring of S-integers. Let γ ∈ F be
such that T 2 +T +γ is irreducible in F[T ]. By the Chebotarev density theo-
rem [4, Thm 13.4, p. 545] applied to the compositum of k[T ]/(T 2 + T + γ)
and the Hilbert class field of k, we can find an element b ∈ Ok,S that
generates a prime of odd degree. Similarly, by applying the Chebotarev
density theorem to the compositum of k[T ]/(T 2 + T + γ) and the ray class
field of modulus b2, we can find an element a ∈ Ok,S that generates a prime
of even degree and that is congruent to γ (mod b2Ok,S). These conditions
imply that vp(a) is even and negative and that vp(b) is odd and negative.
Define
f(x) = a−4nbx2 + x+ ab−1,
g(x) = a−8nb2x2 + a−4nbx+ a1−4n + γ.
Note that g(x) = a−4nbf(x) + γ. Let X be the Châtelet surface given by
(∗) y2 + yz + γz2 = f(x)g(x).
In Lemma 3.1 we show X(Ak) 6= ∅, and in Lemma 3.3 we show
X(Ak)Br = ∅. Together, these show that X has a Brauer-Manin obstruction
to the Hasse principle.
Lemma 3.1. The Châtelet surface X has a kv-point for every place v.
Proof. Suppose that v = va. Since a generates a prime of even degree,
the left-hand side of (∗) factors into two distinct linear factors in kv[y, z].
Therefore, we can change variables so that (∗) becomes w1w2 = f(x)g(x)
and hence there is a solution over kv.
Now suppose that v 6= va. Since y2 + yz + γz2 is a norm form for an
unramified extension of kv for all v, in order to prove the existence of a
kv-point, it suffices to find an x ∈ kv such that the valuation of the right-
hand side of (∗) is even.
Suppose further that v 6= vp, vb. Choose x such that v(x) = −1. Then
the right-hand side of (∗) has valuation −4 so there exists a kv-point.
Suppose that v = vp. Let π be a uniformizer for v and take x = πa2/b.
Then
f(x) = b−1a4−4nπ2 + a2b−1π + ab−1.
Since a has negative even valuation and n ≥ 1, we have v(f(x)) = v(a2b−1π)
which is even. Now let us consider
g(x) = a4−8nπ2 + a2−4nπ + a1−4n + γ.
234 Bianca Viray
By the same conditions mentioned above, all terms except for γ have posi-
tive valuation. Therefore v(g(x)) = 0.
Finally suppose that v = vb. Take x = 1b + 1. Then
f(x) = 1
b
(
a−4n + a+ 1 + b+ a−4nb2
)
.
Note that by the conditions imposed on a,
(
a−4n + a+ 1 + b+ a−4nb2
)
≡
γ + b (mod b2Ok,S). Thus v(f(x)) = −1. Now consider
g(x) = a−8n + a−8nb2 + a−4n + a−4nb+ a1−4n + γ
modulo b2Ok,S . By the conditions imposed on a, we have
g(x) ≡ 1 + 1 + b+ γ + γ ≡ b (mod b2Ok,S).
Thus v(g(x)) = 1, so v (f(x)g(x)) is even. 
Let L = k[T ]/(T 2 +T+γ) and let A denote the class of the cyclic algebra
(L/k, f(x))2 in Br k(X)(see [2, §2.5] for a detailed definition). Using the
defining equation of the surface, we can show that (L/k, g(x))2 is also a
representative for A. Since g(x) +a−4nbf(x) is a v-adic unit, g(x) and f(x)
have no common zeroes. Since A is the class of a cyclic algebra of order 2,
the algebra
(
L/k, f(x)/x2
)
2 is another representative for A. Note that for
any point P of X, there exists an open neighborhood U containing P such
that either f(x), g(x), or f(x)/x2 is a nowhere vanishing regular function
on U . Therefore, A is an element of BrX.
To show that X(Ak)Br = ∅, we use the continuity of the map evA. While
the result is well-known, it is difficult to find in the literature so we give a
proof for reader’s convenience.
Lemma 3.2. Let kv be a local field and let V be a smooth projective scheme
over kv. For any [B] ∈ BrV ,
evB : V (kv)→ Br kv
is locally constant.
Proof. To prove continuity, it suffices to show that ev−1B (B′) is open for any
B′ in the image of evB. By replacing [B] with [B] − [evB(x)], we reduce to
showing that ev−1B (0) is open.
Fix a representative B of the element [B] ∈ BrV . Let n2 denote the rank
of B and let fB : YB → V be the PGLn-torsor associated to B. Then we
observe that the set ev−1B (0) is equal to fB(YB(kv)) ⊂ V (kv). This set is
open by the implicit function theorem [3, Thm. 2.2.1]. 
Failure of the Hasse principle 235
Lemma 3.3. Let Pv ∈ X(kv). Then
invv(evA(Pv)) =
{
1/2 if v = vb,
0 otherwise.
Therefore X(Ak)Br = ∅.
Proof. The surface X contains an open affine subset that can be identified
with
V (y2 + yz + az2 − P (x)) ⊆ A3.
Let X0 denote this open subset. Since evA is continuous by Lemma 3.2 and
invv is an isomorphism onto its image, it suffices to prove that invv takes
the desired value on the v-adically dense subset X0(kv) ⊂ X(kv).
Since L/k is an unramified extension for all places v, evaluating the
invariant map reduces to computing the parity of the valuation of f(x) or
g(x).
Suppose that v 6= va, vb, vp. If v(x0) < 0, then by the strong triangle
inequality, v(f(x0)) = v(x20). Now suppose that v(x0) ≥ 0. Then both
f(x0) and g(x0) are v-adic integers, but since g(x)− a−4nbf(x) = γ either
f(x0) or g(x0) is a v-adic unit. Thus, for all Pv ∈ X0(kv), invv(A(Pv)) = 0.
Suppose that v = va. Since a generates a prime of even degree,
T 2 + T + γ splits in ka. Therefore, (L/k, h) is trivial for any h ∈ ka(V )×
and so invv(A(Pv)) = 0 for all Pv ∈ X0(kv).
Suppose that v = vp. We will use the representative (L/k, g(x)) of A. If
v(x0) < v(a4nb−1) then the quadratic term of g(x0) has even valuation and
dominates the other terms. If v(x0) > v(a4nb−1) then the constant term of
g(x0) has even valuation and dominates the other terms. Now assume that
x0 = a4nb−1u, where u is a v-adic unit. Then we have
g(x0) = u2 + u+ γ + a1−4n.
Since γ was chosen such that T 2 + T + γ is irreducible in F[T ] and p is a
prime of odd degree, T 2 +T +γ is irreducible in Fp[T ]. Thus, for any v-adic
unit u, u2 + u+ γ 6≡ 0 (mod p). Since a1−4n ≡ 0 mod p, this shows g(x0)
is a v-adic unit. Hence invv(A(Pv)) = 0 for all Pv ∈ X0(kv).
Finally suppose that v = vb. We will use the representative (L/k, f(x)) of
A. If v(x0) < −1 then the quadratic term has odd valuation and dominates
the other terms in f(x0). If v(x0) > −1 then the constant term has odd
valuation and dominates the other terms in f(x0). Now assume x0 = b−1u
where u is any v-adic unit. Then we have
f(x0) =
1
b
(
a−4nu2 + u+ a
)
.
236 Bianca Viray
It suffices to show that a−4nu2 + u + a 6≡ 0 (mod bOk,S). Since a ≡ γ
(mod bOk,S), we have
a−4nu2 + u+ a ≡ u2 + u+ γ.
Using the same argument as in the previous case, we see that a−4nu2 +u+
a 6≡ 0 (mod bOk,s) and thus v(g(x0)) = −1. Therefore invv(A(Pv)) = 12 for
all Pv ∈ X0(kv). 
Acknowledgements
I thank my advisor, Bjorn Poonen, for suggesting the problem and for
many helpful conversations. I thank Olivier Wittenberg for a sketch of the
proof of Lemma 3.2 and thank Daniel Erman and the referees for their
comments.
References
[1] Y. I. Manin, Le groupe de Brauer-Grothendieck en géométrie diophantienne. Actes du
Congrès International des Mathématiciens (Nice, 1970), Gauthier-Villars, Paris, 1971, 401–
411.
[2] Philippe Gille, Tamás Szamuely, Central simple algebras and Galois cohomology. Cam-
bridge Studies in Advanced Mathematics, vol. 101, Cambridge University Press, Cambridge,
2006.
[3] Jun-ichi Igusa, An introduction to the theory of local zeta functions. AMS/IP Studies in
Advanced Mathematics, vol. 14, American Mathematical Society, Providence, RI, 2000.
[4] Jürgen Neukirch, Algebraic number theory. Grundlehren der Mathematischen Wis-
senschaften [Fundamental Principles of Mathematical Sciences], vol. 322, Translated from
the 1992 German original and with a note by Norbert Schappacher; With a foreword by G.
Harder, Springer-Verlag, Berlin, 1999.
[5] Bjorn Poonen, Existence of rational points on smooth projective varieties. J. Eur. Math.
Soc. (JEMS) 11 (2009), no. 3, 529–543.
[6] Bjorn Poonen, Insufficiency of the Brauer-Manin obstruction applied to étale covers. Ann.
of Math. (2) 171 (2010), no. 3, 2157–2169.
[7] Jean-Pierre Serre, Local fields. Graduate Texts in Mathematics, vol. 67, Translated from
the French by Marvin Jay Greenberg, Springer-Verlag, New York, 1979.
[8] Alexei Skorobogatov, Torsors and rational points. Cambridge Tracts in Mathematics,
vol. 144, Cambridge University Press, Cambridge, 2001.
Bianca Viray
Mathematics Department
Box 1917
Brown University
Providence, RI 02912 USA
E-mail: bviray@math.brown.edu
URL: http://math.brown.edu/~bviray


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Signed Selmer groups over $p$ -adic Lie extensions

Failure of the Hasse principle for Châtelet surfaces in characteristic $2$

Weak multipliers for generalized van der Corput sequences

Explicit bounds for split reductions of simple abelian varieties

On the number of representations in the Waring-Goldbach problem with a prime variable in an arithmetic progression

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