Journal de Théorie des Nombres de Bordeaux

Jeremy Lovejoy

Robert Osburn

$M_2$-rank differences for partitions without repeated odd parts

John Voight

Computing fundamental domains  for Fuchsian groups

Supriya Pisolkar

Supriya PISOLKAR
Absolute norms of p-primary units
Tome 21, no 3 (2009), p. 735-742.
<http://jtnb.cedram.org/item?id=JTNB_2009__21_3_735_0>
© Université Bordeaux 1, 2009, tous droits réservés.
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Article mis en ligne dans le cadre du
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Journal de Théorie des Nombres
de Bordeaux 21 (2009), 735-742
Absolute norms of p-primary units
par Supriya PISOLKAR
Résumé. Nous prouvons un analogue local d’un théorème de
J. Martinet sur la norme absolue du discrimant relatif d’une ex-
tension de corps de nombres. Ce résultat peut être vu comme un
énoncé sur les unités 2-primaires. Nous prouvons également un
résultat similaire pour la norme absolue des unités p-primaires,
pour tout p premier.
Abstract. We prove a local analogue of a theorem of J. Mar-
tinet about the absolute norm of the relative discriminant ideal of
an extension of number fields. The result can be seen as a state-
ment about 2-primary units. We also prove a similar statement
about the absolute norms of p-primary units, for all primes p.
1. Introduction
Let K be a p-adic field containing a primitive p-th root of unity ζp. A
unit α ∈ UK is called p-primary if the extension K(α1/p) is an unramified
extension of K. For example, the discriminant of an integral basis of an
unramified extension L/K of a 2-adic field is a 2-primary unit since it has a
square root in L (see proof of Cor.1.1). It is interesting to observe that for
an extension L/K the norm of a p-primary unit in L is again a p-primary
unit in K (see Lemma 2.2). In this paper we prove a result about the
absolute norm of a p-primary unit which is motivated by the following
theorem of J. Martinet.
Theorem 1.1. ([4]; 1.4) Let L/K be a quadratic extension of number fields
such that the absolute norm of the relative discriminant ideal ∂L/K is odd.
If K contains the 2m-th roots of unity for some m ≥ 2, then NK/Q(∂L/K) ≡
1(mod 2m+1).
Although the above result has been stated only for a quadratic extension,
arguments in [4] are sufficient to obtain the same congruence for any finite
extension of number fields satisfying the above hypothesis, by reducing to
the quadratic case. The above result is about the behaviour of the absolute
norm of the discriminant ideal at the prime 2. Thus one may ask if there
is a local analogue of Martinet’s theorem.
Manuscrit reçu le 9 juillet 2008.
736 Supriya Pisolkar
The main contribution of this paper is to prove a local analogue (see
corollary 1.1) of Theorem 1.1 which turns out to be a statement about
2-primary units. In fact, we prove the following theorem about p-primary
units for all primes p.
Theorem 1.2. Let K be a finite extension of Qp containing a primitive
pm-th root of unity where m ≥ 1. Let α be a p-primary unit of UK. Then
NK/Qp(α) ≡ 1(mod p
m+1)
Remark. The absolute norm of a p-primary unit may not satisfy a better
congruence. Moreover, this congruence may not hold for arbitrary units.
See example 3.
Corollary 1.1. Let K be a finite extension of Q2 containing a primitive
2m-th root of unity for some m ≥ 1. Let L be an unramified extension of
K and let {α1, α2, . . . , αn} be an integral basis of L/K.Then,
NK/Q2(dL/K(α1, α2, . . . , αn)) ≡ 1(mod 2
m+1)
Proof. K(
√
dL/K(α1, . . . , αn) ) is a subextension of L/K and thus an un-
ramified extension of K. Therefore dL/K(α1, . . . , αn) is a 2-primary unit of
K and the result follows from the above theorem. 
Using the above local statement, one would now like to recover Theorem
1.1. We are able to do this in the case when L/K has an integral basis (see
corollary 1.3). We first prove the following.
Corollary 1.2. Let K be a number field containing the 2m-th roots of unity
where m ≥ 1. Let L/K be a finite extension, let ∂L/K be the relative dis-
criminant of L/K, and suppose that the absolute norm NK/Q(∂L/K) is odd.
Assume that α = {α1, ..., αn} is a basis of L/K such that
(i) α is contained in OL.
(ii) α forms a S−1OK-basis of S−1OL where S = Z\(2).
Let dL/K(α) denote the discriminant of the basis α. Then
NK/Q(dL/K(α)) ≡ 1(mod 2m+1)
Proof of Corollary 1.2. The hypothesis that the norm of the discriminant
ideal is odd is equivalent to saying that all primes of OK lying above the
prime ideal (2) are unramified in L. Fix a prime p of OK lying above 2. Let
{qj}rj=1 be the prime ideals of OL lying above p. Choose an integral basis
βj = {βj1, βj2, ...} of Lj/Kp. Here Lj and Kp denote the completion of L and
K at qj and p respectively. Let A = L ⊗K Kp. Since A = ⊕Lj , β =
r⋃
j=1
βj
forms an integral basis for the Kp algebra A. Here by integral basis of A
Absolute norms of p-primary units 737
we mean an OKp-basis of ⊕jOLj . Let dA/Kp(β) denote the discriminant of
A of β. Note that α is also an integral basis of A and the p-adic image
(dL/K(α))p is equal to dA/Kp(α). Therefore
(dL/K(α))p = dA/Kp(β) · u
2
for some unit u ∈ UKp . Since dA/Kp(β) =
r∏
j=1
dLj/Kp(βj), by Corollary 1.1
we have,
NKp/Q2dA/Kp(β) =
r∏
j=1
NKp/Q2(dLj/Kp(βj)) ≡ 1(mod 2
m+1)
As Q2(ζ2m) ⊂ Kp, by theorem 2.4 we have, NKp/Q2(u) ∈ Um,Q2 . This
implies that NKp/Q2(u)2 ∈ Um+1,Q2 . Thus
NKp/Q2(dL/K(α))p ≡ 1(mod 2
m+1)
By using ([6],II,3.2) we get
NK/Q(dL/K(α)) =
∏
p|2
NKp/Q2(dL/K(α))p ≡ 1(mod 2
m+1)

Corollary 1.3. Let K be as above with m ≥ 2. Let L/K be a finite ex-
tension having an integral basis α. Assume that the absolute norm NK/Q
of the discriminant ideal ∂L/K is odd. Then NK/Q(dL/K(α)) is the positive
generator of NK/Q(∂L/K). In particular
NK/Q(∂L/K) ≡ 1(mod 2m+1)
Proof. Since NK/Q(dL/K(α)) is a generator of NK/Q(∂L/K), it suffices to
show that it is positive. As K is purely imaginary, the norm NK/Q of any
nonzero element is positive. 
Theorem 1.2 will be proved in section 3. In section 2, we recall some
classical results on the behaviour of the norm map of a totally ramified
cyclic extension of local fields. See for example [2].
Acknowledgement : I am thankful to Prof. C. S. Dalawat for his help.
This problem was suggested to me by him. I am indebted to Joël Riou for
painstakingly going through the first draft of this paper and for important
suggestions. I am grateful to Amit Hogadi for his interest and stimulating
discussions. I thank Prof. Loïc Merel for his useful comments.
738 Supriya Pisolkar
2. Cyclic ramified extensions and the norm map.
In this section we state some results about the norm map of a totally
ramified cyclic extension of local fields. These results will be crucially used
in the proof of the theorem 1.2.
Notation. Let K be a finite extension of Qp. Let k be the residue field of
K. We denote by Ui,K the subgroup of UK given by
Ui,K = {x ∈ UK|vK(1− x) ≥ i}
We denote by eK the absolute ramification index of K and eK = eK/p− 1.
Theorem 2.1. ([6], p. 212) Let K be a finite extension of Qp. For all
n > eK, the map ( )p : Un → Un+eK is a bĳection.
Theorem 2.2. ([2], III.1.4) Let L/K be a totally ramified Galois exten-
sion of degree p. Let πL be a uniformiser of L. Let σ be a generator of
Gal(L/K). Then σ(πL)/πL ∈ U1,L. Further, if s is the largest integer such
that σ(πL)/πL ∈ Us,L, then s is independent of the uniformiser πL.
Thus the integer s, defined above depends only on the extension L/K.
Therefore we will denote it by s(L/K). Note that s(L/K) is the unique
ramification break of Gal(L/K).
Theorem 2.3. ([2],III.2.3) s(L/K) ≤ peK.
Example. Let K = Q2. Then K has six ramified quadratic extensions
namely, K(
√
−1),K(
√
−5),K(
√
±2),K(
√
±10). For the first two extensions
s = 1 and for the remaining extensions s = 2.
We will now state some results about the Hasse-Herbrand function, which
is an important tool in understanding the behaviour of the norm map in
wildly ramified extension.
Proposition 2.1. ([2], III, prop. 3.1) Let L/F be a finite Galois extension
of local fields and N = NL/F : L∗ → K∗ be the norm map. Let kF be
infinite. Then there exists a unique function
h = hL/F : N→ N
such that h(0) = 0 and
NUh(i),L ⊂ Ui,F, NUh(i),L 6⊂ Ui+1,F, NUh(i)+1,L ⊂ Ui+1,F.
(1) For L/F a totally tamely ramified extension,
h(i) = [L : F]i,
(2) If L/F is totally ramified extension of degree p = char(kF) then,
h(i) =
{
i, if i < s
s(1− p) + pi if i ≥ s,
Absolute norms of p-primary units 739
If we have the tower of field extension L ⊂ M ⊂ F then,
hL/F = hL/M ◦ hM/F.
To treat the case of local fields with the finite residue fields we have the
following.
Lemma 2.1. ([2], III, 3.2) Let L/F be a finite separable totally ramified
extension of local fields. Then for an element α ∈ L we get
NL/F(α) = NL̂ur/F̂ur .
where F̂ur is the completion of Fur, L̂ur = LF̂ur.
This lemma shows that, if kF is finite then, for a finite Galois extension
L/F, hL/F = hL̂ur/L̂ur .
We will be using the following theorem from class field theory,
Theorem 2.4. ([5], p. 45) The norm map carries units of Qp(ζpm) into
Um,Qp; i.e, NQp(ζpm )|Qp
(
UQp(ζpm )
)
= Um,Qp.
Next we recall some results discussed in [1]. Let K be a finite extension
of Qp. We know that K∗ has a filtration (Un)n∈N. Put U0 = UK and
Un = Un/Un ∩Up0. We thus a get a filtration on K∗/K∗
p by Fp subspaces,
· · · ⊂ Un ⊂ · · · ⊂ U1 ⊂ U0 ⊂ K∗/K∗p
From ([1], Prop. 33), if K∗ contains an element of order p then Upek is
an Fp-line in K∗/K∗p. Further, Fp-lines in K∗/K∗p are in bĳection with
the cyclic degree p extensions of K. The line which corresponds to the
unramified extension is given by following proposition.
Proposition 2.2. ([1], Prop. 16) The Fp-line in K∗/K∗p which gives the
unramified (Z/pZ)-extension of K upon adjoining p-th roots is UpeK.
We thus get the following important corollary.
Corollary 2.1. A unit α in UK is p-primary if and only if the image of α
in K∗/K∗p belongs to UpeK.
In the earlier version of this paper, the statement of the following lemma
was hidden in the proof of Theorem 1.2. I thank Joël Riou for his suggestion
of stating it as a separate lemma and also providing an elegant proof.
Lemma 2.2. Let E/F be an extension of p-adic fields containing a prim-
itive p-th root of unity. Then the norm map NE/F : E∗ → F∗ takes a
p-primary unit to a p-primary unit.
740 Supriya Pisolkar
Proof. It suffices to prove this result in two cases: E/F is a totally ramified
extension and E/F is an unramified extension. Let α ∈ E be a p-primary
unit. By definition of a p-primary unit, there exists an unramified extension
E′/E and β ∈ E′ such that α = βp.
Suppose that E/F is a totally ramified extension. Then the extension
E′/E corresponds to the extension of residue field of E. The residue field
of E is the same as that of F. Thus there exists an unramified extension
F′/F such that E′ = F′E. It is easy to see that NE/F(α) = NE′/F′(α) and
thus NE/F(α) = NE′/F′(β)p.
Suppose now E/F is an unramified extension. E′/E is also an unram-
ified extension. Since these extensions are Galois, NE/F(α) is a product
of all σ(α) where σ ∈ Gal(E/F). For each σ ∈ Gal(E/F), we choose
σ̃ ∈ Gal(E′/F) which extends σ. Then NE/F(α) is a p-th power of a product
σ̃(β). This product lies in an unramified extension of F namely E′ and thus
NE/F(α) is a p-primary unit. 
3. Proof of Theorem 1.2
Proof of Theorem 1.2 . By hypothesis, K contains the pm-th roots of unity
where m ≥ 1. By above Lemma 2.2, the norm NK/Qp(ζpm ) of a p-primary
unit in K is a p-primary unit in Qp(ζpm). Thus it is enough to prove the
result in the special case K = Qp(ζpm).
Suppose that K = Qp(ζpm). By corollary 2.1, a unit α ∈ K is p-primary
if and only if
α = u · wp
where u ∈ UpeK,K and w ∈ UK. By theorem 2.4, NK/Qp(u) and NK/Qp(w)
belong to Um,QP . Thus by theorem 2.1, NK/Qp(w)p ∈ Um+1,Qp . To prove
the theorem it now remains to show that NK/Qp(u) ∈ Um+1,Qp . We are
going to show this by using the Hasse-Herbrand function. In fact, we will
show that hK/Qp(m) = pm − 1. Then, by using the property of Hasse-
Herbrand function we will get that, NK/Qp(Uh(m)+1,K) = NK/Qp(Upm,K) ⊂
Um+1,Qp .
Consider the tower of field extensions K = Km ⊃ Km−1 ⊃ · · · ⊃ K1
where Ki = Qp(ζpi). Note that for each 2 ≤ i ≤ m, Ki/Ki−1 is a wildly
ramified cyclic extension of degree p, and K1/Qp is tamely ramified cyclic
extension of degree p − 1. Let vKi be the surjective valuation of Ki. By
([6], IV, Lemma 1(c)), sKi/Ki−1 = pi−1 − 1. Indeed, vKi(σ(ζP i) − ζpi) =
vKi(ζpζpi − ζpi) = vKi(ζp − 1) = pi−1.
Henceforth for simplicity of notation we write si for s(Ki/Ki−1).
Step 1: For m = 1, we want to prove that hK1/K(1) = p−1. Since K1/K is
a tamely ramified extension of the degree p− 1, this follows by the formula
of Hasse-Herbrand function for tamely ramified extensions. See (1).
Absolute norms of p-primary units 741
Step 2: For m ≥ 2, we want to show that hKm/Qp(m) = pm − 1. We will
use the transitivity of the Hasse-Harbrand function thorough the tower of
field extensions. For simplicity of notation, we will write hK1/Qp = h1,
hKi/Ki−1 = hi, and h = hm ◦ · · · ◦ h1.
(i) We know from (1) that h1(m) = m(p−1). To compute h2(m(p−1)),
first observe that s2 = p − 1 and m(p − 1) > s2. Now applying
formula (2),
h2(m(p− 1)) = (p− 1)(1− p) + p(m(p− 1)) = (m− 1)p2 − (m− 2)p− 1.
(ii) For 1 ≤ n < m, let us assume that
(∗) hn ◦ hn−1 ◦ · · · ◦ h1(m) = (m− (n− 1))pn − (m− n)pn−1 − 1
It is easy to check that
(m− (n− 1))pn − (m− n)pn−1 − 1 ≥ sn+1 = pn − 1.
Now we can apply the formula (2).
hn+1 ◦ (∗) = hn+1((m− (n− 1))pn − (m− n)pn−1 − 1)
= (pn − 1)(1− p) + p[(m− (n− 1))pn − (m− n)pn−1 − 1]
= (m− n)pn+1 − (m− (n+ 1))pn − 1.
Thus, for n+ 1 = m, we get hm ◦ · · · ◦ h1(m) = h(m) = pm − 1.
This proves that NK/Qp(Uh(m)+1,K) = NK/Qp(Upm,K) ⊂ Um+1,Qp and
thus
NK/Qp(u) ∈ Um+1,Qp .
This completes the proof of the Theorem 1.2.

Remark. The result NK/Qp(Upm,K) ⊂ Um+1,Qp can also be derived without
the explicit use of the function hK/Qp . This can be achieved by the repeated
application of the fact ([2], III, 1.5) that, for a totally ramified cyclic degree-
p extension L/K, we have
(1) NL/K(Us+pi,L) ⊆ Us+i,K ∀ i > 0
(2) NL/K(Us+i,L) = NL/K(Us+i+1,L) for i > 0, p - i.
and for totally tamely ramified Galois extension of degree-n ([2], III, 1.3),
(1) NL/K(Uni,L) ⊆ Ui,K
(2) NL/K(Ui,L) = NL/K(Ui+1,L) if n - i.
In fact, in the notation Km = Qp(ζpm), these formulae imply
NKi/Ki−1(Uri,Ki) ⊂ Uri−1,Ki−1 , for 2 ≤ i ≤ m.
where for 1 ≤ i ≤ m, ri = (m− i+ 1)pi− (m− i)pi−1. It is easy to see that
NK1/Qp(Ur1,K1) ⊂ Um+1,Qp .
742 Supriya Pisolkar
Example. Let us show that a p-primary unit may not satisfy a congruence
better than that of Theorem1.2. Suppose that K = Q3(ζ34). Let α ∈
U2,Q3\U3,Q3 . Then α ∈ U34,K i.e α ∈ U3eK,K. Therefore α is a 3-primary
unit by using 2.2. We claim that NK/Q3(α) 6≡ 1(mod 36). Now NK/Q3(α) =
α[K:Q3] = α54. Since U1,Q3
( )2−→ U1,Q3 is an isomorphism which preserves all
filtration levels, α2 ∈ U2,Q3\U3,Q3 . By using Prop.2.1,
U2,Q3/U3,Q3
( )27−→ U5,Q3/U6,Q3
is an isomorphism. This implies that NK/Q3(α) 6≡ 1(mod 36).
Now we will show that a general unit which is not a 3-primary unit of K
may not satisfy the congruence as in the Theorem 1.2. Let α ∈ U1,Q3\U2,Q3 .
Thus α ∈ U2.33,K\U2.33+1,K and hence α is not a 3-primary unit of K. As
above, NK/Qp(α) = α54 and by Prop. 2.1, α54 ∈ U4,Q3\U5,Q3 .
References
[1] Dalawat C.S., Local discriminants, kummerian extensions, and abelian curves.
ArXiv:0711.3878.
[2] Fesenko, I. B., Vostokov, S. V., Local fields and their extensions. Second edition. Trans-
lations of Mathematical Monographs 121, AMS Providence, RI, 2001.
[3] Hasse H., Number theory. Classics in Mathematics. Springer-Verlag, Berlin, 2002.
[4] Martinet, J., Les discriminants quadratiques et la congruence de Stickelberger. Sém. Théor.
Nombres Bordeaux (2) 1 (1989), no. 1, 197–204.
[5] Neukirch, J., Class field theory. Grund. der Math. Wiss. 280. Springer-Verlag, Berlin,
1986.
[6] Jean-Pierre Serre, Local Fields. GTM 67, Springer-Verlag, 1979.
Supriya Pisolkar
Harish-Chandra Research Institute
Chhatnag Road, Jhunsi
Allahabad -211019, India.
E-mail: supriya@mri.ernet.in


Absolute norms of $p$-primary units

Kiran S. Kedlaya

Some new directions in $p$-adic Hodge theory

Ben Kane

CM liftings of supersingular elliptic curves

Karin Halupczok

Karin HALUPCZOK
On the ternary Goldbach problem with primes in arithmetic progressions
having a common modulus
Tome 21, no 1 (2009), p. 203-213.
<http://jtnb.cedram.org/item?id=JTNB_2009__21_1_203_0>
© Université Bordeaux 1, 2009, 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 21 (2009), 203-213
On the ternary Goldbach problem with primes in
arithmetic progressions having a common
modulus
par Karin HALUPCZOK
Résumé. Pour ε > 0 et n impair suffisamment grand, nous mon-
trons que, pour presque tout k ≤ R := n1/5−ε, il existe une re-
présentation n = p1 + p2 + p3 avec des nombres premiers pi ≡ bi
modulo k pour presque tout triplet admissible b1, b2, b3 de résidus
modulo k.
Abstract. For ε > 0 and any sufficiently large odd n we show
that for almost all k ≤ R := n1/5−ε there exists a representation
n = p1+p2+p3 with primes pi ≡ bi mod k for almost all admissible
triplets b1, b2, b3 of reduced residues mod k.
1. Introduction and results
Let n be a sufficiently large integer, consider an integer k and let b1, b2, b3
be integers that are relatively prime to k ≥ 1, with 0 ≤ bi < k, i = 1, 2, 3.
We consider the ternary Goldbach problem of writing n as
n = p1 + p2 + p3
with primes p1, p2 and p3 satisfying the three congruences
pi ≡ bi mod k, i = 1, 2, 3
for the common modulus k. One is interested in the solvability of this
question for all sufficiently large n with the modulus k being as large as
some power of n. This problem has been studied intensely by many authors.
For an overview, see for example [2].
An obvious necessary condition for solvability is
n ≡ b1 + b2 + b3 mod k,
as otherwise no such representation of n is possible.
We call such a triplet b1, b2, b3 of reduced residues mod k admissible, and
a pair b1, b2 of reduced residues admissible, if (n − b1 − b2, k) = 1. For a
given b1 we call b2 admissible, if b1, b2 is an admissible pair. Let us denote
the number of these admissible pairs respectively triplets by A(k).
204 Karin Halupczok
We make our consideration of this strengthened ternary Goldbach prob-
lem more precise in the following way. Let
J3(n) := Jk,b1,b2,b3(n) :=
∑
m1,m2,m3≤n
m1+m2+m3=n
mi≡bi (k),
i=1,2,3
Λ(m1) Λ(m2) Λ(m3),
where Λ is von Mangoldt’s function. The function J3(n) corresponds closely
to the number of representations of n of the kind we are interested in.
In this paper we prove that the deviation of J3(n) from its expected main
term is uniformly small for large moduli, namely in the following sense.
Theorem 1.1. For every A, ε > 0, every sufficiently large n and for D ≤
n1/5−ε we have
E :=
∑
D<k≤2D
k
ϕ(k)
∑
(b1,k)=1
1
ϕ(k)
∑
b2 adm.
∣∣∣∣∣J3(n)− n2k2S(n, k)
∣∣∣∣∣  n2(log n)A .
Here S(n, k) denotes the singular series for this special Goldbach prob-
lem. In depends on k, while J3(n) depends on k, b1, b2; residue b3 is simply
b3 ≡ n− b1 − b2 (k). Namely, see [4], for odd n we have
S(n, k) = C(k)
∏
p|k
p3
(p− 1)3 + 1
∏
p|n
p-k
(p− 1)((p− 1)2 − 1)
(p− 1)3 + 1
∏
p>2
(
1+
1
(p− 1)3
)
,
where p > 2 throughout, C(k) = 2 for odd k and C(k) = 8 for even k.
As a consequence of Theorem 1.1, we prove in section 2 the following
result.
Theorem 1.2. Let A, ε > 0 and let n ∈ N be odd and sufficiently large.
Then for all k ≤ R := n1/5−ε with at most  R(log n)−A exceptions there
exists a representation n = p1 + p2 + p3 with primes pi ≡ bi (k) for all but
 A(k)(log n)−A many admissible triplets b1, b2, b3.
So there are few exceptions for k, and also the number of exceptions of
admissible triplets is small compared with the number A(k) of all admissible
triplets.
Let us compare this Theorem 1.2 with the result of J. Liu and T. Zhang
in [4] who show the assertion for R := n1/8−ε and all admissible triplets. In
another paper [3], Z. Cui improved this to R := n1/6−ε. Further C. Bauer
and Y. Wang showed in [2] the assertion for R := n5/48−ε, but with only
 (log n)B many exceptions.
Here we improved the bound for R again, but at the cost of possibly a
few exceptions of admissible triplets.
On the ternary Goldbach problem 205
2. Proof of Theorem 1.2
First of all we give a lower bound for A(k):
Lemma 2.1. For odd n we have A(k)  ϕ(k)
2
(log k)3
. More precisely, for every
reduced residue b1 mod k there are  ϕ(k)(log k)3 many reduced residues b2 mod
k with (n− b1 − b2, k) = 1.
Proof. Fix a reduced residue b1 mod k. Now count the b2 with (b2, k) =
(n − b1 − b2, k) = 1. So b2 is to be chosen such that for all prime divisors
p > 2 of k we have b2 6≡ 0 (p) and b2 6≡ n − b1 (p). This gives ≥ p − 2
possibilities for b2 mod p, and ≥ pl−1(p− 2) possiblilities for b2 mod pl. If
p = 2 for even k we have an odd b1, so n − b1 is even and therefore one
can take b2 ≡ 1(2), so there are 2ν2(k)−1 possibilities for b2 mod 2ν2(k), if
2ν2(k)||k.
Therefore the number of b2 is at least
2max{0,ν2(k)−1}
∏
pl||k
p6=2
pl−1(p− 2) = ϕ(k)
∏
p|k
p6=2
p− 2
p− 1
with ∏
p|k
p6=2
p− 1
p− 2
=
∏
p|k
p6=2
(
1 +
1
p− 2
)
≤
∏
p|k
(
1 +
2
p− 1
)
≤
k∑
q=1
µ(q)2 2ω(q)
ϕ(q)

k∑
q=1
τ(q)
q
log k  (log k)3.

Now we deduce Theorem 1.2 as a corollary of Theorem 1.1.
Fix A, ε > 0 and let n be odd and sufficiently large. Consider
R3(n) :=
∑
p1,p2,p3
p1+p2+p3=n
pi≡bi(k),
i=1,2,3
log p1 log p2 log p3 and r3(n) :=
∑
p1,p2,p3
p1+p2+p3=n
pi≡bi(k),
i=1,2,3
1.
Let D < k ≤ 2D with D ≤ R := n1/5−ε. For any admissible triplet
b1, b2, b3 mod k we have
|R3(n)− J3(n)| ≤ (log n)3W3,
where W3 denotes the number of solutions of pl + qj + rm = n with p, q, r
prime, and where l, j or m are at least 2 such that pl ≡ b1 (k), qj ≡ b2 (k)
and rm ≡ b3 (k).
206 Karin Halupczok
Now we prove that
∑
D<k≤2D
k max
b1,b2,b3
admissible
W3 
n2
(log n)A+3
.
For this, we split the number W3 according to whether at least two of
the exponents l, j,m are ≥ 2 or only one is, and for this we write W3 =
W1 + W2. There are at most
√
n prime powers ≤ n with exponent ≥ 2, so
in the first case we have W1  n, and the sum with W1 replacing W3 is
 D2W1  D2n  n
2
(log n)A+3
.
In the second case, if only one exponent is ≥ 2, we have W2 
√
n · nk =
n3/2
k , and so the sum with W2 replacing W3 is  Dn
3/2  n2
(log n)A+3
.
So for D ≤ n1/5−ε it follows from Theorem 1.1 that
∑
D<k≤2D
k
ϕ(k)
∑
(b1,k)=1
1
ϕ(k)
∑
b2 adm.
∣∣∣∣R3(n)− n2k2S(n, k)
∣∣∣∣
≤
∑
D<k≤2D
k
ϕ(k)
∑
(b1,k)=1
1
ϕ(k)
∑
b2 adm.
∣∣∣∣R3(n)− J3(n)∣∣∣∣
+
∑
D<k≤2D
k
ϕ(k)
∑
(b1,k)=1
1
ϕ(k)
∑
b2 adm.
∣∣∣∣J3(n)− n2k2S(n, k)
∣∣∣∣
 (log n)3
∑
D<k≤2D
k max
b1,b2,b3
admissible
W3 +
n2
(log n)A
 n
2
(log n)A
.
So the formula of Theorem 1.1 holds also with R3(n) replacing J3(n).
Now for D < k ≤ 2D we have A(k) := #{b1, b2 admissible mod k}. Let
T (k) := #{b1, b2 admissible mod k; R3(n) = 0}, consider the set
KD := {k; D < k ≤ 2D,T (k) ≥ A(k)(log n)−A}
and let KD be its cardinality.
On the ternary Goldbach problem 207
Since S(n, k)  1 if it is positive, which is the case for admissible triplets
and odd n (see its formula above as an Euler product), we have
KD ·
n2
D

∑
D<k≤2D
k∈KD
k
T (k)
∑
b1,b2 adm.
R3(n)=0
∣∣∣∣n2k2S(n, k)
∣∣∣∣

∑
D<k≤2D
k
A(k)
∑
b1,b2 adm.
(log n)A
∣∣∣∣R3(n)− n2k2S(n, k)
∣∣∣∣
 (log n)A+3
∑
D<k≤2D
k
ϕ(k)2
∑
b1,b2
adm.
∣∣∣∣R3(n)− n2k2S(n, k)
∣∣∣∣  n2(log n)A ,
using Lemma 2.1 and the above. Therefore it follows that KD  D(log n)−A,
so for all k 6∈ KD we have R3(n) > 0 for all but  A(k)(log n)−A many
admissible triplets b1, b2, b3, and then r3(n)  R3(n)(log n)−3 is positive,
too. This shows Theorem 1.2, since the overall number of exceptions is

∑
D=2i≤R
KD  (log n) ·
R
(log n)A+1
=
R
(log n)A
.

3. Proof of Theorem 1.1
Our proof is in two steps using the circle method.
Let A, ε, θ > 0, B ≥ 2A + 1 and D ≤ n1/4(log n)−θ.
We define major arcs M ⊆ R by
M :=
⋃
q≤D(log n)B
⋃
0<a<q
(a,q)=1
]
a
q
− D(log n)
B
qn
,
a
q
+
D(log n)B
qn
[
and minor arcs by
m :=
]
−D(log n)
B
n
, 1− D(log n)
B
n
[
\ M.
For α ∈ R and some residue b mod k denote
Sb(α) := Sb,k(α) :=
∑
m≤n
m≡b (k)
Λ(m) e(αm).
From the orthogonal relations for e(αm) it follows that
J3(n) =
∫ 1
0
Sb1(α)Sb2(α)Sb3(α) e(−nα) dα.
208 Karin Halupczok
By
JM3 (n) :=
∫
M
Sb1(α)Sb2(α)Sb3(α) e(−nα) dα
we denote the value of the integral for J3(n) on the major arcs M and by
Jm3 (n) := J3(n)− JM3 (n)
its value on the minor arcs m.
We first consider the major arcs.
Theorem 3.1. For D ≤ n1/5−ε we have
EM :=
∑
D<k≤2D
k max
b1,b2,b3
admissible
∣∣∣∣∣JM3 (n)− n2k2S(n, k)
∣∣∣∣∣  n2(log n)A .
We can give a very short proof of Theorem 3.1 simply by adapting the
result of J. Liu and T. Zhang in [4] for the major arcs used here. In fact,
by following their proof we see that for P := D(log n)B and Q := n
D(log n)B
and any U ≤ P , we have to choose D such that the conditions
U ≤ n1/3(log n)−E , (UQ)−1 ≤ U−3(log n)−E
DU ≤ D1/3−δn1/3(log n)−E , (UQ)−1 ≤ D1−δ(DU)−3(log n)−E
are satisfied for any E > 0 and small δ > 0. The optimal choice of D is
therefore given by D ≤ n1/5−ε. This proves Theorem 3.1. The improvement
in this paper comes from the different intervals used as major and minor
arcs such that working on the minor arcs with mean values over b1, b2 is
still possible.
Specifically for an estimate on the minor arcs, we show the following in
the next section.
Theorem 3.2. For D ≤ n1/4(log n)−θ we have
Em :=
∑
D<k≤2D
k
ϕ(k)
∑
(b1,k)=1
1
ϕ(k)
∑
(b2,k)=1
adm.
|Jm3 (n)| 
n2
(log n)A
.
Theorem 1.1 is then a corollary of Theorems 3.1 and 3.2 since E ≤
EM + Em.
This Theorem is the interesting part of Theorem 1.1, where we can attain
a higher power of n for the bound of D by considering the mean value
over b1, b2 instead of the maximum. But because of this we have to allow
exceptions of admissible triplets in Theorem 1.2, as we have seen in its
proof.
In both Theorems 3.1 and 3.2 the resulting bound for D is the optimum
with the given method, so these bounds cannot be balanced to get a larger
On the ternary Goldbach problem 209
range than n1/5. Also the cited method for the major arcs cannot be im-
proved by using mean values over b1, b2 since the character sum estimates
used are independent of b1, b2.
4. Proof of Theorem 3.2: the estimate on the minor arcs
Let D ≤ n1/4(log n)−θ and let
∑∗ denote a sum over reduced residues
mod k. We have
Em 
∑
D<k≤2D
k
ϕ(k)2
∑∗
b1,b2
|Jm3 (n)|
≤
∑
D<k≤2D
k
ϕ(k)2
∑∗
b1,b2
∫
m
|Sb1(α)Sb2(α)Sn−b1−b2(α)| dα
=
∑
D<k≤2D
k
ϕ(k)
∑∗
b1
∫
m
|Sb1(α)| ·
1
ϕ(k)
∑∗
b2
|Sb2(α)Sn−b1−b2(α)|dα
≤
∑
D<k≤2D
k
ϕ(k)
∫
m
∑∗
b1
|Sb1(α)|
· 1
ϕ(k)
( ∑
b2 mod k
|Sb2(α)|2
)1/2( ∑
b2 mod k
|Sn−b1−b2(α)|2
)1/2
dα
≤
∑
D<k≤2D
k
ϕ(k)
max
α∈m
∑∗
b1
|Sb1(α)|
1
ϕ(k)
∑
b2 mod k
∫ 1
0
|Sb2(β)|2dβ
n(log n)3
∑
D<k≤2D
1
ϕ(k)
max
α∈m
∑∗
b1
|Sb1(α)|
≤n(log n)3
∑
D<k≤2D
max
α∈m
(
1
ϕ(k)
∑∗
b1
|Sb1(α)|2
)1/2
n(log n)3
∑
D<k≤2D
(
n2
D2(log n)2A+6
)1/2
.
In the last step we use Lemma 4.1 below, valid for D ≤ n1/4(log n)−θ and
suitable chosen θ, B > 0 depending just on A > 0.
Now the above is  n(log n)3D n
D(log n)A+3
= n
2
(log n)A
as was to be shown
for the minor arcs. 
So what is left to show is the following.
Lemma 4.1. For all A > 0 and B ≥ 2A+1, θ ≥ B/2 let D ≤ n1/4(log n)−θ
and α ∈ R with ||α − uv || <
1
v2
for some integers u, v with (u, v) = 1 and
210 Karin Halupczok
D(log n)B ≤ v ≤ n
D(log n)B
. Then for D < d ≤ 2D we have
1
ϕ(d)
∑
c,(c,d)=1
|Sc,d(α)|2 
n2
D2(log n)A
.
We remark that for α ∈ m there exist u, v with (u, v) = 1, v ≤ n
D(log n)B
and ||α − uv || <
D(log n)B
vn ≤
1
v2
by Dirichlet’s approximation theorem, so
v ≥ D(log n)B since α ∈ m, and therefore the conditions of Lemma 4.1 are
satisfied.
For the proof we need the following auxiliary Lemma. (See [5], Lemma
4.11 and 4.10.)
Lemma 4.2. Let ||α− uv || ≤
1
v2
, (u, v) = 1. Then
(a)
∑
m≤X
min(Y, ||αm||−1)  XY
v
+ Y + (X + v)(log v),
(b)
∑
m≤X
min
(
Y
m
, ||αm||−1
)

(
Y
v
+ X + v
)
log(2vX)
Proof of Lemma 4.1. Fix n large and D ≤ n1/4(log n)−θ, and let α, u and
v be as given in Lemma 4.1.
We apply Vaughan’s identity on the exponential sum Sc,d(α), see for
example A. Balog in [1], where a similar Lemma is given (Lemma 2 there).
From that it follows that it suffices to show for any complex coefficients
|am|, |bk| ≤ 1 and any M ∈ N with
I : M ≤ V 2, if bk = 1 for all k,
II : V ≤ M ≤ n
V
else, where V := D(log n)B,
we have
∑
(c,d)=1
∣∣∣∣∣ ∑
m∼M
∑
k≤n/m
km≡c(d)
ambke(αmk)
∣∣∣∣∣
2
 n
2
D(log n)A
.
Here m ∼ M means M < m ≤ M ′ for some M ′ ≤ 2M .
On the ternary Goldbach problem 211
We consider first case II: Then the left hand side becomes (where m∗
denotes the inverse of m mod d):
II :=
∑
(c,d)=1
∣∣∣∣ ∑
m∼M
(m,d)=1
am
∑
k≤n/m
km≡c(d)
bk e(αmk)
∣∣∣∣2
≤
∑
(c,d)=1
M
∑
m∼M
(m,d)=1
∣∣∣∣ ∑
k≤n/m
k≡cm∗(d)
bk e(αmk)
∣∣∣∣2
= M
∑
m∼M
(m,d)=1
∑
(c,d)=1
∣∣∣∣ ∑
k≤n/m
k≡c(d)
bk e(αmk)
∣∣∣∣2
= M
∑
m∼M
∑
(c,d)=1
∑
k≤n/m
k≡c(d)
bk
∑
k′≤n/m
k′≡k(d)
bk′ e(αm(k − k′))
= M
∑
m∼M
∑
k≤n/m
(k,d)=1
bk
∑
k′≤n/m
k′≡k(d)
bk′ e(αm(k − k′))
= M
∑
m∼M
∑
k≤n/m
(k,d)=1
bk
∑
l≥(k−n/m)/d
l≤(n/m−1)/d
bk−ld e(αmld)
≤ M
∑
k≤n/M
∑
|l|≤n/Md
∣∣∣∣ ∑
m∼M
m≤n/k
m≤n/ max{k−ld,ld+1}
e(αmld)
∣∣∣∣.
Now the absolute value of the exponential sum is  min(M, ||αld||−1), so
II  M n
M
∑
|l|≤n/Md
min(M, ||αld||−1)
 n
∑
L≤n/M
d|L
min(M, ||αL||−1) + nM
≤ n
( ∑
L≤n/M
d|L
12
)1/2( ∑
L≤n/M
M min(M, ||αL||−1)
)1/2
+ nM
 n
(
n
Md
)1/2
M1/2
(
n
v
+ M +
(
n
M
+ v
)
(log n)
)1/2
+ nM,
212 Karin Halupczok
because of the auxiliary Lemma 4.2 (a). So expression II is n2
D(log n)A
since
we have D(log n)B = V ≤ M ≤ n/V in case II, and since D(log n)B 
v  n
D(log n)B
for B ≥ 2A + 1.
Now consider case I: Then the left hand side becomes (again m∗ denotes
the inverse of m mod d):
I :=
∑
(c,d)=1
∣∣∣∣ ∑
m∼M
(m,d)=1
am
∑
k≤n/m
km≡c(d)
e(αmk)
∣∣∣∣2
≤
∑
(c,d)=1
M
∑
m∼M
(m,d)=1
∣∣∣∣ ∑
k≤n/m
k≡cm∗(d)
e(αmk)
∣∣∣∣2
≤M
∑
m∼M
∑
(c,d)=1
∣∣∣∣ ∑
k≤n/m
k≡c(d)
e(αmk)
∣∣∣∣2
=M
∑
m∼M
∑
(c,d)=1
∑
k≤n/m
k≡c(d)
e(αmk)
∑
k′≤n/m
k≡k′(d)
e(−αmk′)
=M
∑
m∼M
∑
k≤n/m
(k,d)=1
∑
k′≤n/m
k≡k′(d)
e(αm(k − k′))
≤M
∑
m∼M
∑
k≤n/m
∣∣∣∣ ∑
l≥(k−n/m)/d
l≤(n/m−1)/d
e(αmdl)
∣∣∣∣
M
∑
m∼M
∑
k≤n/M
(
min
(
n
md
, ||αmd||−1
)
+ 1
)
n
∑
m∼M
min
(
n
md
, ||αmd||−1
)
+ Mn
n
∑
L∼Md
min
(
n
L
, ||αL||−1
)
+ Mn
n
(
n
v
+ Md + v
)
(log n) + Mn,
using auxiliary Lemma 4.2 (b). Now we get I  n2
D(log n)A
since D(log n)B 
v  n
D(log n)B
with B ≥ A + 1 and since Md  V 2d  D3(log n)B 
n
D(log n)B
for D ≤ n1/4(log n)−θ and θ ≥ B/2. So Lemma 4.1 is shown. 
On the ternary Goldbach problem 213
Final remark. As was kindly pointed out to me by Z. Cui, it is possible
to improve the statement on the major arcs such that Theorems 1.1, 3.1
and 3.2 hold for the improved exponent 1/4 instead of 1/5. The idea for
this major arc improvement comes from Z. Cui in [3].
References
[1] A. Balog, The Prime k-Tuplets Conjecture on Average. Analytic number theory, Proc.
Conf. in Honor of Paul T. Bateman, Urbana/IL (USA), 1989, Prog. Math. 85 (1990), 47–
75.
[2] C. Bauer, Y. Wang, On the Goldbach conjecture in arithmetic progressions. Rocky Moun-
tain J. Math. 36 (1) (2006), 35–66.
[3] Z. Cui, The ternary Goldbach problem in arithmetic progression II. Acta Math. Sinica (Chin.
Ser.) 49 (1) (2006), 129–138.
[4] J. Liu, T. Zhang, The ternary Goldbach problem in arithmetic progressions. Acta Arith.
82 (3) (1997), 197–227.
[5] M. B. Nathanson, Additive Number Theory: The Classical Bases. Graduate texts in Math-
ematics 164, Springer-Verlag, 1996.
Karin Halupczok
Albert-Ludwigs-Universität Freiburg
Eckerstr. 1
D-79104 Freiburg, Allemagne
E-mail : karin.halupczok@math.uni-freiburg.de
URL: http://home.mathematik.uni-freiburg.de/halupczok/


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