Journal de Théorie des Nombres de Bordeaux

Oscar Ordaz

Andreas Philipp

Irene Santos

Wolfgang A. Schmid

On the Olson and the Strong Davenport constants

Alfred Geroldinger

David J. Grynkiewicz

The catenary degree of Krull monoids I

Dominik J. Leitner

Dominik J. LEITNER
Two exponential diophantine equations
Tome 23, no 2 (2011), p. 479-487.
<http://jtnb.cedram.org/item?id=JTNB_2011__23_2_479_0>
© Société Arithmétique de Bordeaux, 2011, tous droits réservés.
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Journal de Théorie des Nombres
de Bordeaux 23 (2011), 479-487
Two exponential diophantine equations
par Dominik J. LEITNER
Résumé. L’équation 3a + 5b − 7c = 1, dont les inconnues a, b, c
sont des entiers positifs, a été mentionnée par Masser comme un
exemple pour lequel il n’y a pas d’algorithme permettant une réso-
lution complète. Malgré cela, nous trouvons ici toutes les solutions.
L’équation y2 = 3a +2b +1, dont les inconnues a, b sont des entiers
positifs et y est un entier, a été mentionnée par Corvaja et Zannier
comme un exemple dont on ignore si le nombre de solutions est
fini. Mais nous trouvons également ici toutes les solutions ; il n’y
en a en fait que six.
Abstract. The equation 3a + 5b − 7c = 1, to be solved in non-
negative rational integers a, b, c, has been mentioned by Masser
as an example for which there is still no algorithm to solve com-
pletely. Despite this, we find here all the solutions. The equation
y2 = 3a + 2b + 1, to be solved in non-negative rational integers
a, b and a rational integer y, has been mentioned by Corvaja and
Zannier as an example for which the number of solutions is not
yet known even to be finite. But we find here all the solutions too;
there are in fact only six.
1. Introduction
In this note we find all solutions of the equation
(1.1) 3a + 5b − 7c = 1
in non-negative integers a, b, c, and also all solutions of the equation
(1.2) y2 = 3a + 2b + 1.
in integers y and non-negative integers a, b.
The equation (1.1) has been mentioned by Masser [M] (p.203) as an
example for which there is still no algorithm to solve completely. It can
be interpreted as a special case of an S-unit equation, or, in a broader
context, an equation of the type covered by the classical results of Mordell-
Lang type. The structure of the solution set can be determined using the
Subspace Theorem applied to the more general S-unit equation
(1.3) x0 + x1 + · · ·+ xn = 0
Manuscrit reçu le 13 janvier 2010.
480 Dominik J. Leitner
in non-zero rational integers x0, x1, . . . , xn with no common factor. When
these integers are composed of primes from a fixed finite set, the conse-
quence is that (1.3) has at most finitely many solutions satisfying
(1.4)
∑
i∈I
xi 6= 0
for all non-empty subsets I of {1, . . . , n}. This (1.4) in our case (1.1)
is hardly any restriction, and one finds at once that the solution set of
(1.1) is at most finite. The general theory also provides an explicit esti-
mate for the number of solutions. But the recent Theorem 1 (p.808) of
the paper [ESS] of Evertse, Schlickewei and Schmidt gives only the upper
bound exp(4.189) ≈ 10344585380964, which is little use in actually finding
the solutions. The same can be said even for the very recent improvement
241944 ≈ 102683 by Amoroso and Viada [AV]. And it is notorious that in
general there are no effective estimates at all for the sizes of the solutions of
(1.3). Here we will use a relatively simple method of congruences to show
that the only solutions are in fact a = b = c = 0 and a = b = c = 1.
The equation (1.2) has been mentioned by Zannier [Z1] (pp.61,62) and
[Z2] (p.1) and Corvaja and Zannier [CZ2] (p.296), [CZ3] (pp.168,169) (see
also [Z3] (p.434), [CZ1] and [C] (p.130)) in the context of the Lang-Vojta
Conjecture (see for example [HS] (p.486)). Here the term y2 prevents the
use of the Subspace Theorem as above. And indeed they remark that it
is not even known whether the solution set is finite or not, unless one
assumes such a conjecture. One can also assume a version for (1.3) which
was formulated in elementary terms by Vojta [V] (p.7). Namely, for every
λ > 1 there is a constant C and a non-zero homogeneous polynomial F ,
each depending only on n and λ, such that all solutions of (1.3) in coprime
integers satisfy
(1.5) max{|x0|, |x1|, . . . , |xn|} ≤ CP λ
where P is the product of all the primes dividing the x0, x1, . . . , xn; however
(1.4) now has to be replaced by
(1.6) F (x1, . . . , xn) 6= 0.
For n = 2 this is of course the intractable abc-conjecture.
With (1.2) we get at once y2 ≤ C(6|y|)λ in (1.5) and so it suffices to fix
λ < 2. Now the failure of (1.6) is not so trivial; but (with x0 = y2) it would
lead to a point (x1, x2) = (3a, 2b) on one of a finite set of fixed curves.
Now since 3 and 2 are multiplicatively independent a well-known result of
Liardet (see for example Theorem 7.3 (p.207) of [L]) implies that there are
at most finitely many such points unless x1 or x2 is constant on one of the
curves. But when a or b is constant in (1.2) then it is easy to establish the
finiteness, for example with n = 2 in Vojta’s Conjecture.
Two exponential diophantine equations 481
Thus a fortiori there is no algorithm for the complete solution. Never-
theless we will use the same congruence method to show that the set is
indeed finite and in fact that the only solutions are y = ±2, a = 0, b = 1
and y = ±6, a = 1, b = 5 and y = ±6, a = b = 3.
Because both equations do actually have solutions, it may seem impossi-
ble that we can use congruences to prove the finiteness. And indeed it would
be impossible for equations that are polynomial in all the variables. Here
we have exponential terms like 3a. The values of this for example modulo
12 at a = 0, 1, 2, 3, 4, . . . are 1, 3, 9, 3, 9, . . .; of course eventually periodic
but not at once. So if we can show that 3a must be 1 modulo 12, then
we deduce a = 0 and not just a congruence for a. It is this principle that
we shall exploit, for various moduli the largest of which is 1820. In fact
the various moduli could be taken together to show that we get no more
solutions of (1.1) modulo 27927900 (and even 20475); however this kind of
simplification seems not to be possible for (1.2).
Of course our method is far too special to be considered as a contribution
to the theory of either the S-unit equation or the Vojta Conjecture. See also
the remark in the footnote of [Z1] (p.57). But its success with the fairly
natural equations (1.1) and (1.2) perhaps gives hope that it can be applied
to other interesting equations of the same sort. This is certainly true of
y2 = 10a + 6b + 1 also mentioned in [Z1] (p.60), for example; and already
there the same is noted for the two-variable equation y2 = 5a + 2a + 7.
I am grateful to David Masser for advice on the preparation of this note.
After it was submitted for publication, Michael Bennett kindly drew my
attention to the article [BF] of Brenner and Foster; it turns out that they
had already proved our Theorem 2.1 about (1.1). However they considered
nothing like (1.2).
2. The equation 3a + 5b − 7c = 1
In this section we prove the following result.
Theorem 2.1. Let a, b, c in N0 = N ∪ {0} satisfy (1.1); then either a =
b = c = 0 or a = b = c = 1.
Proof. We need one simple observation.
Lemma 2.1. Let a, b, c be in N0 with (1.1) and abc = 0; then a = b = c = 0.
Proof. At first let a = 0. Then (1.1) appears as 5b = 7c which forces
b = c = 0. Similarly if we start with b = 0. Finally, c = 0 leads to 3a+5b = 2
and so again a = b = c = 0, which completes the proof of the present
lemma. 
Lemma 2.1 shows that either a = b = c = 0 or a, b, c ∈ N and hence in
the following we may assume a, b, c ∈ N.
482 Dominik J. Leitner
Let us consider the following table, where we calculate values of 3n, 5n, 7n
modulo 1820
n 1 2 3 4 5 6 7
3n (mod 1820) 3 9 27 81 243 729 367
5n (mod 1820) 5 25 125 625 1305 1065 1685
7n (mod 1820) 7 49 343 581 427 1169 903
n 8 9 10 11 12 13
3n (mod 1820) 1101 1483 809 607 1 3
5n (mod 1820) 1145 265 1325 1165 365 5
7n (mod 1820) 861 567 329 483 1561 7
Here we get the same values for n = 1 and n = 13, hence we see a period
of length 12 when we calculate the table above for all n in N.
Now, for m, k in N0 we define {m}k = m+kN0. Then the values of n for
which the triple (3n, 5n, 7n) lies in various congruence classes modulo 1820
form subsets {1}12, . . . , {11}12 of N.
Perhaps with the help of a computer we now look for (a, b, c) with 1 ≤
a, b, c ≤ 12 such that
3a + 5b − 7c ≡ 1 (mod 1820).
In fact we find that (a, b, c) = (1, 1, 1) is the only triple as required and this
proves that a, b, c lie in the set {1}12, which means that
(2.1) a ≡ b ≡ c ≡ 1 (mod 12).
However, we rerun the procedure above modulo 341. Due to (2.1) we just
consider values with n ≡ 1 (mod 12) and get the table
n 1 13 25 37 49 61
3n (mod 341) 3 148 254 141 136 3
5n (mod 341) 5 191 67 36 284 5
7n (mod 341) 7 112 87 28 107 7
Here we see a period of length 60 and, as before, we find that (1, 1, 1) is the
only solution of (1.1) modulo 341 from the table above. This implies that
a ≡ b ≡ c ≡ 1 (mod 60).
Let us continue with the following table modulo 50
n 1 61 121
3n (mod 50) 3 3 3
5n (mod 50) 5 25 25
7n (mod 50) 7 7 7
Two exponential diophantine equations 483
Here we have only the two classes {1}0 = {1} and {61}60, in which the first
class is finite because the sequence 5n (mod 50) is not periodic but only
eventually so. Now looking for solutions of (1.1) modulo 50 forces b = 1.
Thus with (1.1) we get the new equation
(2.2) 7c − 3a = 4.
And here
n 1 61 121
7n (mod 9) 7 7 7
3n (mod 9) 3 0 0
which forces in a similar way a = 1. Now (2.2) implies that c = 1 and this
completes the proof of Theorem 2.1. 
3. The equation y2 = 3a + 2b + 1
In this section we prove the following result.
Theorem 3.1. Let y in Z and a, b in N0 satisfy (1.2); then either y = ±2
and a = 0, b = 1 or y = ±6 and a = 1, b = 5 or a = b = 3.
At first we note that y 6= 0 and hence we may assume y ∈ N without loss
of generality.
Lemma 3.1. Let y in N and a, b in N0 with ab = 0 satisfy (1.2); then
y = 2 and a = 0, b = 1.
Proof. At first let a = 0. Then b 6= 0 because 3 is not a square. Further
b = 1 leads to y2 = 4 and so y = 2. If now b ≥ 2 then 4 | 2b and so
y2 ≡ 2 (mod 4),
impossible because y2 ≡ 0, 1 (mod 4).
Otherwise we have a in N and b = 0 which leads to
y2 ≡ 2 (mod 3),
impossible because y2 ≡ 0, 1 (mod 3). This completes the proof. 
Therefore we may assume that a, b are in N.
Lemma 3.2. Let y, a, b in N satisfy (1.2); then 6 divides y.
Proof. We calculate (1.2) modulo 2. Then
y2 ≡ 3a + 2b + 1 ≡ 1 + 0 + 1 ≡ 0 (mod 2),
hence y2 is even and so is y.
Similarly we consider (1.2) modulo 3. Here
(y + 1)(y − 1) ≡ y2 − 1 ≡ 3a + 2b ≡ 2b 6≡ 0 (mod 3).
484 Dominik J. Leitner
So neither y + 1 nor y − 1 is divisible by 3 and hence 3 divides y, which
completes the proof of the present lemma. 
Lemma 3.2 shows that y = 6x for some x in N and thus (1.2) appears as
(3.1) 36x2 = 3a + 2b + 1.
Lemma 3.3. Let x, a, b in N satisfy (3.1). Then exactly one of the following
holds:
(1) x = 1 and either a = 1, b = 5 or a = b = 3,
(2) x is odd, a ≡ 1 (mod 8), and b ≡ 3 (mod 6) with a 6= 1, b 6= 3.
Proof. Considering (3.1) modulo 36 we get
(3.2) 0 ≡ 3a + 2b + 1 (mod 36).
Further we calculate in the table below 3n and 2n modulo 36 for 1 ≤ n ≤ 8.
n 1 2 3 4 5 6 7 8
3n (mod 36) 3 9 27 9 27 9 27 9
2n (mod 36) 2 4 8 16 32 28 20 4
Here we note that we get the same values for n = 2 and n = 8. Hence we
see a period of length 6 when we calculate the table above for all n in N or
rather we get a partition of N with the classes
(3.3) {1}0 = {1}, {2}6 = 2 + 6N0, . . . , {7}6 = 7 + 6N0.
We now look for all (a, b) with 1 ≤ a, b ≤ 7 satisfying (3.2) and with the
table above we find
(a, b) = (1, 5), (3, 3), (5, 3), (7, 3).
Together with (3.3) this implies that for a, b ∈ N we have either a = 1 and
b ∈ {5}6 or a is in one of the sets {3}6, {5}6, {7}6 and b ∈ {3}6.
Consider first a = 1. Then (3.1) appears as
36x2 = 2b + 4
and hence
(6x+ 2)(6x− 2) = 2b.
Thus 6x+ 2 and 6x− 2 are powers of 2 and (6x+ 2)− (6x− 2) = 4 yields
6x+ 2 = 8 and 6x− 2 = 4 respectively; so x = 1 and b = 5.
Otherwise a 6= 1 is odd and b ≡ 3 (mod 6). Now b = 3 in (3.1) leads to
36x2 = 3a + 9
and similar to above we see that 6x + 3 and 6x − 3 are powers of 3 with
(6x + 3) −(6x − 3) = 6; hence x = 1 and a = 3, which completes the first
part of the lemma.
Two exponential diophantine equations 485
Let now b 6= 3. Since a is odd (3.1) yields
36x2 ≡ 3 + 0 + 1 ≡ 4 (mod 8)
and thus x is odd. So x = 2z + 1 and now
36x2 = 288z(z + 1)2 + 36 ≡ 4 (mod 32).
Therefore (3.1) leads to
3 ≡ 3a (mod 32).
For the values of 3n (mod 32) we consider the following table
n 1 3 5 7 9
3n (mod 32) 3 27 19 11 3
Here we see a period of length 8 and hence that a ≡ 1 (mod 8). Therefore
the present lemma is proved. 
Lemma 3.4. There are no x, a, b in N with (3.1) which satisfy the condi-
tions of the second part of Lemma 3.3.
Proof. Suppose that we have such x, a, b in N. We then consider (3.1) mod-
ulo 120. Therefore we use the table
n 3 5 7
3n (mod 120) 27 3 27
2n (mod 120) 8 32 8
Similar to above we see a period of length 4 and a ≡ 1 (mod 8) with a 6= 1
implies that a is in the set {5}4 = 5 + 4N0. Further we note that
36x2 ≡

36 (mod 120), x ≡ ±1 (mod 10),
84 (mod 120), x ≡ ±3 (mod 10),
60 (mod 120), x ≡ 5 (mod 10).
Therefore we find that b ∈ {5}4 as well because 36x2 6≡ 12 (mod 120). So
the left side of (3.1) is 36 (mod 120) and hence we have x ≡ ±1 (mod 10).
Further b ∈ {5}4 implies that b ≡ 9 (mod 12) because b ≡ 3 (mod 6).
Next we consider (3.1) modulo 560. Therefore we use the following table
n 5 9 13 17
3n (mod 560) 243 83 3 243
2n (mod 560) 32 512 352 32
Thus we see a period of length 12. Now a ≡ 1 (mod 8) with a 6= 1 means
that a is in one of the sets {5}12, {9}12, {13}12 and b ≡ 9 (mod 12) shows
486 Dominik J. Leitner
b ∈ {9}12. Further, x ≡ ±1 (mod 10) and we get
36x2 ≡

36 (mod 560), x ≡ ±1,±29 (mod 70),
116 (mod 560), x ≡ ±9,±19 (mod 70),
436 (mod 560), x ≡ ±11,±31 (mod 70),
196 (mod 560), x ≡ ±21 (mod 70);
and thus we see that a is not in the set {13}12.
Finally, let us consider (3.1) modulo 208. Here we have a table
n 5 9 13 17
3n (mod 208) 35 131 3 35
2n (mod 208) 32 96 80 32
Again we see a period of length 12 and as above it follows that a is in one
of the sets {5}12, {9}12, {13}12 and b ∈ {9}12. And x ∈ N is odd so we find
36x2 ≡

36 (mod 208), x ≡ ±1 (mod 26),
116 (mod 208), x ≡ ±3 (mod 26),
68 (mod 208), x ≡ ±5 (mod 26),
100 (mod 208), x ≡ ±7 (mod 26),
4 (mod 208), x ≡ ±9 (mod 26),
196 (mod 208), x ≡ ±11 (mod 26),
52 (mod 208), x ≡ 13 (mod 26).
But this forces a ∈ {13}12, which is a contradiction to above; and this
completes the proof of the present lemma. 
Now the proof of Theorem 3.1 follows directly from the lemmas above.
References
[AV] F. Amoroso and E. Viada, Small points on subvarieties of a torus. Duke Mathematical
Journal 150(3) (2009), 407–442.
[BF] J. L. Brenner and L. L. Foster, Exponential diophantine equations. Pacific Journal of
Mathematics 101 (1982), 263–301.
[C] P. Corvaja, Problems and results on integral points on rational surfaces. In Diophantine
geometry (ed. U. Zannier), Edizioni della Normale 2007, 123–141.
[CZ1] P. Corvaja and U. Zannier, On the diophantine equation f(am, y) = bn. Acta Arith-
metica 94 (2000), 25–40.
[CZ2] P. Corvaja and U. Zannier, Some cases of Vojta’s conjecture on integral points over
function fields. Journal of Algebraic Geometry 17 (2008), 295–333.
[CZ3] P. Corvaja and U. Zannier, Applications of the subspace theorem to certain diophan-
tine problems. In Diophantine approximation, Developments in Mathematics 18 (eds H.P.
Schlickewei, K. Schmidt, R.F. Tichy), Springer 2008, 161–174.
[ESS] J.-H. Evertse, H. P. Schlickewei and W. M. Schmidt, Linear equations in variables
which lie in a multiplicative group. Annals of Mathematics 155 (2002), 807–836.
[HS] M. Hindry and J. H. Silverman, Diophantine Geometry. Springer, 2000.
[L] S. Lang, Fundamentals of Diophantine Geometry. Springer, 1983.
[M] D. W. Masser, Mixing and linear equations over groups in positive characteristic. Israel
Journal of Mathematics 142 (2004), 189–204.
Two exponential diophantine equations 487
[V] P. Vojta, A more general abc conjecture. International Mathematics Research Notices 21
(1998), 1103–1116.
[Z1] U. Zannier, Some applications of diophantine approximation to diophantine equations.
Forum Editrice Universitaria Udinese S.r.l. (2003).
[Z2] U. Zannier, Polynomial squares of the form aXm+b(1−X)n+c. Rendiconti del Seminario
Matematico della Università di Padova 112 (2004), 1–9.
[Z3] U. Zannier, Diophantine equations with linear recurrences. An overview of some recent
progress. Journal de Théorie des Nombres de Bordeaux 17 (2005), 432–435.
Dominik J. Leitner
Mathematisches Institut
Universität Basel
Rheinsprung 21
4051 Basel, Switzerland
E-mail: dominik.leitner@unibas.ch


Two exponential diophantine equations

Alain Couvreur

Differential approach for the study of duals of algebraic-geometric codes on surfaces

Peter Bruin

Peter BRUIN
The Tate pairing for Abelian varieties over finite fields
Tome 23, no 2 (2011), p. 323-328.
<http://jtnb.cedram.org/item?id=JTNB_2011__23_2_323_0>
© Société Arithmétique de Bordeaux, 2011, 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 23 (2011), 323-328
The Tate pairing for Abelian varieties
over finite fields
par Peter BRUIN
Résumé. Nous décrivons un accouplement arithmétique associé
à une isogenie entre variétés abéliennes sur un corps fini. Nous
montrons qu’il généralise l’accouplement de Frey et Rück, donnant
ainsi une démonstration brève de la perfection de ce dernier.
Abstract. In this expository note, we describe an arithmetic
pairing associated to an isogeny between Abelian varieties over
a finite field. We show that it generalises the Frey–Rück pairing,
thereby giving a short proof of the perfectness of the latter.
1. Introduction
Throughout this note, k denotes a finite field of q elements, and k̄ denotes
an algebraic closure of k. If n is a positive integer, then µn denotes the group
of n-th roots of unity in k̄×.
Let C be a complete, smooth, geometrically connected curve over k, let
J be the Jacobian variety of C, and let n be a divisor of q − 1. In [1], Frey
and Rück defined a perfect pairing
{ , }n : J [n](k)× J(k)/nJ(k) −→ µn(k)
as follows: if D and E are divisors on C with disjoint supports and f is a
non-zero rational function with divisor nD, then
{[D], [E] mod nJ(k)}n = f(E)(q−1)/n,
where
f(E) =
∏
x∈C(k̄)
f(x)nx if E =
∑
x∈C(k̄)
nxx.
Remark. We have composed the map as defined in [1] with the isomor-
phism k×/(k×)n → µn(k) that raises elements to the power (q − 1)/n.
Manuscrit reçu le 25 mai 2010.
I am grateful to Hendrik Lenstra and Ed Schaefer for discussions on this subject, and to
Andreas Enge for the suggestion to publish this note. I thank the two referees for useful remarks.
324 Peter Bruin
To prove the perfectness of { , }n, Frey and Rück used a pairing intro-
duced by Tate [10], which relates certain cohomology groups of an Abelian
variety over a p-adic field, and an alternative description of this pairing
given by Lichtenbaum [6] in the case of the Jacobian of a curve over a
p-adic field. The names ‘Tate pairing’ and ‘Tate–Lichtenbaum pairing’ are
therefore often used for what we call the Frey–Rück pairing.
There are now several proofs of the non-degeneracy of the Frey–Rück
pairing that no longer use the cited results of Tate and Lichtenbaum; see
Heß [2] and Schaefer [8]. In this note, we consider a separable isogeny
φ : A → B between Abelian varieties over k such that kerφ is annihilated
by q − 1. We define a pairing
[ , ]φ : ker φ̂(k)× coker(φ(k)) −→ k×
between the groups
ker φ̂(k) = {b ∈ B(k) | φ̂(b) = 0}
and
coker(φ(k)) = B(k)/φ(A(k)).
We show that this pairing is perfect in the sense that it induces isomor-
phisms
ker φ̂(k) ∼−→ Hom(coker(φ(k)), k×)
and
coker(φ(k)) ∼−→ Hom(ker φ̂(k), k×).
Furthermore, we show that the Frey–Rück pairing is the special case of
multiplication by n on the Jacobian of a curve over k. The more general
pairing appears to be known, but has not to my knowledge appeared in
the literature. In any case, I do not make any claim to originality. It seems
appropriate to call [ , ]φ the Tate pairing associated to φ.
Remark. The condition that kerφ be annihilated by q − 1 can be relaxed
somewhat. Let m be the exponent of kerφ and let d be the order of q in
(Z/mZ)×. If m and d are coprime, then one can pass to an extension of
degree d of k and consider appropriate eigenspaces for the Frobenius action.
This reproduces a result of Frey and Rück [1, Proposition 2.5].
2. Preliminaries
By a pairing we mean a bilinear map
A×B → C
between Abelian groups. It is called perfect if the induced group homomor-
phisms
A→ Hom(B,C) and B → Hom(A,C)
are isomorphisms.
The Tate pairing for Abelian varieties over finite fields 325
The Galois group Gk = Gal(k̄/k) is canonically isomorphic to the profi-
nite group Ẑ, with the element 1 of Ẑ corresponding to the q-power Frobe-
nius automorphism σ. The only actions of Gk that we consider are con-
tinuous actions on Abelian groups with the discrete topology. An Abelian
group equipped with such an action of Gk is called a Gk-module.
The fact that Gk is generated as a topological group by σ implies that for
any Gk-module M , the Galois cohomology group H1(Gk,M) has an easy
description via the isomorphism
(2.1)
H1(Gk,M)
∼−→M/(σ − 1)M
[c] 7−→ c(σ) mod (σ − 1)M
of Abelian groups, where [c] denotes the class of a cocycle c : Gk →M ; see
Serre [9, XIII, § 1, proposition 1].
If F is a finite Gk-module, the Cartier dual of F is the Abelian group
F∨ = Hom(F, k̄×)
with the Gk-action given by
(σa)(x) = σ(a(σ−1x)) for all a ∈ F∨, σ ∈ Gk, x ∈ F .
The subgroup F∨(k) ofGk-invariants consists of the elements of Hom(F, k̄×)
that are homomorphisms of Gk-modules.
Lemma 2.1. Let F be a finite Gk-module annihilated by q− 1, and let F∨
be its Cartier dual. There is a perfect pairing
F∨(k)×H1(Gk, F ) −→ k×
sending (a, [c]) to a(c(σ)) for every a ∈ F∨(k) and every cocycle c : Gk → F .
Proof. By assumption, we can write
F∨(k) = Hom(F, k̄×)Gk
= Hom(F, k×)Gk
= Hom(F/(σ − 1)F, k×).
Applying the isomorphism (2.1) with M = F , we get an isomorphism
F∨(k) ∼−→ Hom(H1(Gk, F ), k×)
such that the induced bilinear map F∨(k)× H1(Gk, F ) −→ k× is given by
the formula in the statement of the lemma. Since H1(Gk, F ) is annihilated
by the order of k×, it is (non-canonically) isomorphic to F∨(k). This implies
that the map
H1(Gk, F ) −→ Hom(F∨(k), k×)
is also an isomorphism, so the pairing is perfect. 
326 Peter Bruin
3. Definition of the Tate pairing
Let φ : A → B be a separable isogeny between Abelian varieties over k
such that kerφ is annihilated by q−1. We write φ(k) for the homomorphism
A(k)→ B(k) induced by φ. We note that
(kerφ)(k) = ker(φ(k)),
and we denote this group from now on by kerφ(k). We also note that there
is no similar equality for cokernels.
We start with the short exact sequence
0 −→ kerφ −→ A −→ B −→ 0
of commutative group varieties over k. Since A, B and φ are defined over k,
taking k̄-points in this sequence gives a short exact sequence of Gk-modules.
Taking Galois cohomology, we obtain a long exact sequence
0 −→ kerφ(k) −→ A(k) φ(k)−→ B(k) δ−→ H1(Gk, kerφ(k̄)) −→ H1(Gk, A(k̄)).
The connecting homomorphism δ is given by
δ(b) = [τ 7→ τ(a)− a],
where a is any element of A(k̄) such that φ(a) = b. A theorem of Lang on
Abelian varieties over finite fields [4, Theorem 3] implies that H1(Gk, A(k̄))
vanishes. This means that δ is surjective and hence gives an isomorphism
B(k)/φ(A(k)) ∼−→ H1(Gk, kerφ(k̄)).
The pairing from Lemma 2.1 becomes a perfect pairing
(3.1) (kerφ)∨(k)×B(k)/φ(A(k)) −→ k×
sending (f, b mod φ(A(k))) to f(σ(a)−a), with a ∈ A(k̄) such that φ(a) = b.
Now let φ̂ : B̂ → Â denote the isogeny dual to φ. The Cartier duality
theorem for isogenies of Abelian varieties (see for example Mumford [7,
§ 15, Theorem 1]) gives a canonical isomorphism
εφ : ker φ̂
∼−→ (kerφ)∨
of Gk-modules. Combining this with the pairing (3.1) gives a perfect pairing
(3.2)
[ , ]φ : ker φ̂(k)× coker(φ(k)) −→ k×
(x, y) 7−→ (εφx)(σa− a),
where a is any element of A(k̄) with (φ(a) mod φ(A(k))) = y. We call this
pairing the Tate pairing associated to φ.
The Tate pairing for Abelian varieties over finite fields 327
4. Comparison to the Frey–Rück pairing
We now take A = B = J , where J is the Jacobian of a complete, smooth,
geometrically connected curve C over k, and we take φ to be the multipli-
cation by a positive integer n dividing q − 1. We identify J with its dual
Abelian variety using the canonical principal polarisation. Then the Tate
pairing (3.2) becomes a perfect pairing
[ , ]n : J [n](k)× J(k)/nJ(k) −→ k×.
Moreover, from the Cartier duality isomorphism εn : Ĵ [n]
∼−→ J [n]∨ and
the identification of J [n] with Ĵ [n] we obtain a perfect pairing
en : J [n]× J [n] −→ µn
(x, y) 7−→ (εny)(x) for all x, y ∈ J [n](k̄),
called the Weil pairing. It can be computed as follows. Let D and E be
two divisors of degree 0 on Ck̄ with disjoint supports and such that their
classes [D] and [E] in J(k̄) are n-torsion points. Then there exist non-zero
rational functions f and g on Ck̄ such that
nD = div f and nE = div g,
and we have
(4.1) en([D], [E]) =
g(D)
f(E) ;
see Howe [3]. The fact that this depends only on the linear equivalence
classes of D and E comes down to Weil reciprocity: if f and g are two
rational functions on Ck̄ whose divisors have disjoint supports, then
f(div g) = g(div f).
For a proof, we refer to Lang [5, Chapter VI, corollary to Theorem 10].
Theorem 4.1. The pairing [ , ]n equals the Frey–Rück pairing { , }n.
Proof. Consider elements of J [n](k) and J(k)/nJ(k), represented by (k-
rational) divisors D and E, respectively, such that the supports of D and E
are disjoint. We choose a divisor E0 on Ck̄ such that nE0 is linearly equiv-
alent to E and rational functions f on C and g on Ck̄ such that
nD = div f and nE0 − E = div g.
Then we have
div(σg/g) = n(σE0 − E0).
328 Peter Bruin
Using the definitions of the pairing [ , ]n and of the Weil pairing en, the
formula (4.1), the fact that σ is the q-power map on k̄×, and Weil reci-
procity, we can compute [ , ]n as follows:
[[D], [E] mod nJ(k)]n = (εn[D])([σE0 − E0])
= en([σE0 − E0], [D])
= f(σE0 − E0)(σg/g)(D)
= σ(f(E0))/f(E0)
σg(D)/g(D)
=
(
f(E0)/g(D)
)q−1
=
(
f(nE0)/g(nD)
)(q−1)/n
=
(
f(nE0)/f(nE0 − E)
)(q−1)/n
= f(E)(q−1)/n.
This equality is what we had to prove. 
References
[1] G. Frey and H.-G. Rück, A remark concerning m-divisibility and the discrete logarithm
in class groups of curves. Mathematics of Computation 62 (1994), 865–874.
[2] F. Heß, A note on the Tate pairing of curves over finite fields. Archiv der Mathematik 82
(2004), no. 1, 28-32.
[3] E. W. Howe, The Weil pairing and the Hilbert symbol. Mathematische Annalen 305 (1996),
387–392.
[4] S. Lang, Abelian varieties over finite fields. Proceedings of the National Academy of Sciences
of the U.S.A. 41 (1955), no. 3, 174–176.
[5] S. Lang, Abelian Varieties. Interscience, New York, 1959.
[6] S. Lichtenbaum, Duality theorems for curves over p-adic fields. Inventiones Mathematicae
7 (1969), 120–136.
[7] D. Mumford, Abelian Varieties. Tata Institute of Fundamental Research, Bombay, 1970.
[8] E. F. Schaefer, A new proof for the non-degeneracy of the Frey–Rück pairing and a con-
nection to isogenies over the base field. In: T. Shaska (editor), Computational Aspects of
Algebraic Curves (Conference held at the University of Idaho, 2005), 1–12. Lecture Notes
Series in Computing 13. World Scientific Publishing, Hackensack, NJ, 2005.
[9] J-P. Serre, Corps locaux. Hermann, Paris, 1962.
[10] J. Tate, WC-groups over p-adic fields. Séminaire Bourbaki, exposé 156. Secretariat mathé-
matique, Paris, 1957.
Peter Bruin
Université Paris-Sud 11
Département de Mathématiques
Bâtiment 425
91405 Orsay cedex
France
E-mail: Peter.Bruin@math.u-psud.fr


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Daniel BARSKY et Bénali BENZAGHOU
Erratum à l’article Nombres de Bell et somme de factorielles
Tome 23, no 2 (2011), p. 527-527.
<http://jtnb.cedram.org/item?id=JTNB_2011__23_2_527_0>
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Journal de Théorie des Nombres
de Bordeaux 23 (2011), 527
Erratum à l’article Nombres de Bell et somme de
factorielles
par Daniel BARSKY et Bénali BENZAGHOU
Farid Bencherif (Université des Sciences et des Techniques Houari Bou-
medienne) et Joseph Oesterlé (Université Paris 6-Pierre et Marie Curie)
nous ont signalé des erreurs de calculs irréparables dans la preuve du
théorème 3 de l’article Nombres de Bell et somme de factorielles publié
dans le Journal de Théorie des Nombres de Bordeaux 16 (2004), 1–17.
Le théorème 3 et sa preuve (conjecture de Kurepa) sont retirés de l’ar-
ticle, la conjecture de Kurepa (0!+1!+ · · ·+(p−1)! 6≡ 0 mod p pour p ≥ 3,
premier) n’est pas démontrée, par contre tous les résultats précédents jus-
qu’au lemme 10 inclus restent valides.
Daniel Barsky
7 rue La Condamine
75017 Paris
E-mail: barsky@math.univ-paris13.fr
Bénali Benzaghou
USTHB
Institut de Mathématiques
El Alia BP 32 Bab Ezzouar
1611 ALGER, Algérie


Erratum à l’article Nombres de Bell et somme de factorielles

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