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ON GAUSS SUMS, CYCLOTOMIC NUMBER FIELDSAND STICKELBERGER’S
THEOREM
KONSTANTIN DRAGOMIRETSKIY
01 June 2010
Undergraduate Honors Thesis
Advisor: Professor Harold Stark
University of California, San Diego
Department of Mathematics
0
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Contents
1. Introduction 2
2. Classical Gauss Sum 3
3. Quadratic Reciprocity 7
4. Gauss Sums 11
5. Quadratic Gauss Sums 15
6. Stickelberger’s Theorem 19
7. Brumer-Stark Conjecture 25
References 26
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1. Introduction
This undergraduate thesis will survey Gauss sums, starting from
theclassical Gauss sum analyzed by C. F. Gauss, and continuing to
moregeneral character sums. Along the way, the classical Gauss sum
willbe utilized to prove the famous law of quadratic reciprocity,
whichGauss refered to as the ”fundamental theorem” in the
DisquisitionesArithmeticae. Various decompositions and recursions
of Gauss sumswill be developed. Prime ideal factorizations will be
developed throughStickelberg’s theorem. Finally, a closing remarks
on the Stickelbergerelement annihilating the cyclotomic class group
and its generalization,the Brumer-Stark conjecture.
AcknowledgementFirst and foremost, I would like to thank Dr.
Harold Stark, the advisorfor my undergraduate honors thesis and my
concurrent algebraic num-ber theory course professor. I would have
prefered meeting ProfessorStark as a graduate or postdoctoral
student to be able to have muchdeeper dialogues in mathematics,
nevertheless I feel very fortunate thatour paths crossed and that I
had the privilege of working with him. Ithank him for his time
invested into this thesis and wish him good luckwith his book.
I thank Dr. Cristian Popescu for encouraging number theory to
meduring my undergraduate career and rea�ring my aspirations to
be-come a mathematician.
I finally thank Dr. Lance Small, Dr. Audrey Terras, and Dr.
CristianPopescu for teaching me mathematics and for recommending me
forthe mathematics doctorate program at the University of
California,Los Angeles.
Konstantin Vitalyevich Dragomiretskiy
2
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2. Classical Gauss Sum
The Classical Gauss Sum was studied extensively by Carl
FriedrichGauss.
This section is series of exercises from Abstract Algebra by
Dummitand Foote.
Put Kp = Q(⇣p), where ⇣p = exp(2⇡ip ) and p is an odd prime.A
finite dimensional Q-vector space is called a number field.
Kp is the pth cyclotomic number field. This is because the
monic,irreducible polynomial over Q is the pth cyclotomic
polynomial:�p(x) =
xp� 1
x� 1= 1 + x + . . . + xp�1 and Q[x]/h�p(x)i ⇠= Kp
Put Gp = Gal(Kp/Q) = {� 2 Aut(Kp/Q)| �(q) = q 8q 2 Q}
8a 2 Kp, a =�(p)X
i=1
qi⇣i
p=
p�1X
i=1
qi⇣i
pfor qi 2 Q where {⇣ ip}
p�1i=1 is a
power basis for Kp over Q. Hence ⌧ 2 Gp is completely determined
by⌧(⇣p) = ⇣↵p where (↵, p) = 1.
This observation elucidates the well-known isomorphism Gp ⇠=
Z/pZ⇥.
Definition 2.1. ⌘0 =X
⌧2H
⌧(⇣p) and ⌘1 =X
⌧2�H
⌧(⇣p)
where H is the unique normal subgroup of index 2 and the coset
�H =G \ H. ⌘0 and ⌘1 are called the two periods of ⇣p with respect
to H.
By the aformentioned isomorphism, Gp is cyclic, and hence the
follow-ing one-to-one correspondence can be observed:
h⌧1i = {⌧i 2 G |⌧i(⇣p) = ⇣gi
p} ! {1, g, . . . , gp�1} = hgi = Gp.
Remark. h⌧1i is cyclic by the action of composition.
Lemma 2.1. ⌧1(⌘0) = ⌘1 and ⌧1(⌘1) = ⌘0
Proof. ⌧1(⌘0) = ⌧1(X
⌧2H
⌧(⇣p)) =X
⌧2H
⌧1(⌧(⇣p)) =
p�12X
i=1
⌧1(⌧2i(⇣p)) =
p�12X
i=1
⌧2i�1(⇣p) =
X
⌧2�H
⌧(⇣p) = ⌘1.
3
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⌧1(⌘1) = ⌧1(X
⌧2�H
⌧(⇣p)) =X
⌧2�H
⌧1(⌧(⇣p)) =
p�12X
i=1
⌧1(⌧2i�1(⇣p)) =
p�12X
i=1
⌧2i(⇣p) =
X
⌧2H
⌧(⇣p) = ⌘0. ⇤
Lemma 2.2. ⌘0 =X
a⌘square⇣
a
pand ⌘1 =
X
b6⌘square
⇣b
p
Proof. a 2 H , a = g2k for some k 2 N , a = (gk)2 = g2k, a is
a
square, hence a is a square (modulo p) if and only if a 2 H.
X
a⌘square⇣
a
p=
p�12X
i=1
⇣g2i
p=
p�12X
i=1
⌧2i(⇣p) =X
⌧2H
⌧(⇣p) = ⌘0
If b is not a square then b 62 H and hence b = g2i�1 for some i
2 NX
b6⌘square
⇣b
p=
p�12X
i=1
⇣g2i�1
p=
p�12X
i=1
⌧2i�1(⇣p) =
p�12X
i=1
⌧1(⌧2i(⇣p)) = ⌧1(
p�12X
i=1
⌧2i(⇣p)) =
⌧1(⌘0) = ⌘1 ⇤
Definition 2.2. Let k be a field and K a cyclic extension with
Galoisgroup G of order n, given ↵ 2 K, the Lagrange resolvent
is
(↵, ⇣) = ↵+ ⇣�(↵) + . . . + ⇣n�1�n�1(↵) =n�1X
i=0
⇣i�
i(↵)
where ⇣ is an nth root of unity and � is a generator of G.
Lemma 2.3. ⌘0 + ⌘1 = (⇣p, 1) = �1 and ⌘0 � ⌘1 = (⇣p,�1)
Proof. ⌘0 + ⌘1 =X
a⌘square⇣
a
p+
X
b6⌘square
⇣b
p=
p�12X
i=1
⌧2i(⇣p) +
p�12X
i=1
⌧2i�1(⇣p) =
p�1X
i=1
⌧i(⇣p) = ⌧1(⇣p) + ⌧2(⇣p) + . . . + ⌧p�1(⇣p) = ⌧1(⇣p) + ⌧21 (⇣p)
+ . . . +
⌧p�21 (⇣p) + ⇣p = ⇣p + ⌧1(⇣p) + . . . + ⌧
�(p)�11 (⇣p) = (⇣p, 1).
⌘0 + ⌘1 =p�1X
i=1
⌧i(⇣p) =p�1X
i=1
⇣i
pvia rearrangement, and hence ⌘0 + ⌘1 =
�p(⇣p)� 1 = �1, where �p(x) is the pth cyclotomic
polynomial.
4
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⌘0 � ⌘1 = (⇣g2
p+ ⇣g
4
p+ . . . + ⇣g
p�1p
) � (⇣g1
p+ ⇣g
3
p+ . . . + ⇣g
p�2p
) =p�1X
i=1
(�1)i⌧ i1(⇣p) =p�2X
i=0
(�1)i⌧ i1(⇣p) = (⇣p,�1). ⇤
Lemma 2.4.p�1X
i=1
⇣i2
p= (⇣p,�1)
Proof. Observe that (p� k)2 ⌘ p2 � 2pk + k2 ⌘ k2(mod p).
Put � =p�1X
i=0
⇣i2
p. � = ⇣0
p+ ⇣1
p+ ⇣2
2
p+ . . . + ⇣(p�2)
2
p + ⇣(p�1)2p =
1 + 2(⇣1p
+ ⇣22
p+ . . . + ⇣
( p�12 )2
p ). Rewritting the residues 1, 2, . . . ,p�12
as powers of g, � = 1 + 2(⇣(ga1 )2
p + ⇣(ga2 )2p + . . . + ⇣
(ga p�1
2 )2
p ) = 1 +
2(⇣g2a1
p+ ⇣g
2a2
p+ . . .+ ⇣g
2a p�12
p) = 1+2⌘0 since each exponent of g in the
parenthesized sum is even and unique mod �(p).� = 1 + 2⌘0 = 2⌘0
� (�1) = 2⌘0 � (⇣p, 1) = 2⌘0 � (⌘0 + ⌘1) = ⌘0 � ⌘1 =(⇣p,�1). ⇤Lemma
2.5. ⌧(�) = � if ⌧ 2 H and ⌧(�) = �� if ⌧ 62 H
Proof. ⌧ 2 H ) ⌧ = ⌧2i for some i, 1 i p�12 ) ⌧(�) = ⌧(⌘0� ⌘1)
=
⌧(⌘0) � ⌧(⌘1) = ⌧2i(⌘0) � ⌧2i(⌘1) = ⌘0 � ⌘1 = � since by Lemma
2.1,⌧2 = ⌧1 � ⌧1 is the identity on both ⌘0 and ⌘1.⌧ 62 H ) ⌧ =
⌧2i+1 for some i, 0 i <
p�12 ) ⌧(�) = ⌧(⌘0 � ⌘1) =
⌧(⌘0) � ⌧(⌘1) = ⌧2i+1(⌘0) � ⌧2i+1(⌘1) = ⌘1 � ⌘0 = �� since ⌧2i+1
=⌧ � ⌧2i. ⇤Lemma 2.6. �̄ = � when p ⌘ 1 (mod 4) and �̄ = �� when p
⌘ 3(mod 4) where �̄ is the complex conjugate of �.
Proof. By Lemma 2.5, � is fixed by H, and thus Q ✓ Q(�) ✓ KH .
Itis easily verifiable that � 62 Q and thus Q(�) = KH
p, the subfield of Kp
fixed by H.
Gal(Q(�)/Q) ⇠=Gal(Kp/Q)
Gal(Kp/Q(�))= G/H = h��1i ⇠= Z/2Z where ��1 is
complex conjugation.
A simple observation of orders gives us ��1 = ⌧p�12
1 = ⌧ p�12
.
p ⌘ 1 (mod 4) ) p�12 ⌘ 0 (mod 2) ) ⌧ p�122 H ) �̄ = ��1(�) =
⌧ p�12
(�) = �.5
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p ⌘ 3 (mod 4) ) p�12 ⌘ 1 (mod 2) ) ⌧ p�1262 H ) �̄ = ��1(�)
=
⌧ p�12
(�) = ��. ⇤Lemma 2.7. ��̄ = p
Proof. �̄ =p�2X
i=0
(�1)i⌧ i1(⇣p) =p�2X
i=0
(�1)i⌧�i1 (⇣p))
��̄ = (p�2X
i=0
(�1)i⌧ i1(⇣p))(p�2X
j=0
(�1)j⌧�j1 (⇣p)) =p�2X
i=0
p�2X
j�0
(�1)i�j⌧ j1 (⌧
i�j1 (⇣p)
⇣p)
=p�2X
k=0
(�1)kp�2X
j=0
⌧j
1 (⌧
k
1 (⇣p)
⇣p) where k = i� j.
k = 0)⌧
k
1 (⇣p)
⇣p= 1) ⌧ j1 (
⌧k
1 (⇣p)
⇣p) = 1.
k 6= 0)⌧
k
1 (⇣p)
⇣p= ⇣g
k�1p
= ⇣↵kp
= ⌧↵k1 (⇣p) where gcd(↵k, p) = 1.
��̄ = (�1)0(p�1)+p�2X
k=1
(�1)kp�2X
j=0
⌧j
1 (⌧
k
1 (⇣p)
⇣p) = (p�1)+
p�2X
k=1
(�1)kp�2X
j=0
⌧j
1 (⇣g
k�1p
) =
(p�1))+p�2X
k=1
(�1)kp�2X
j=0
⌧j
1 (⌧↵k1 (⇣p)) = (p�1)+
p�2X
k=1
(�1)k⌧↵k1 (p�2X
j=0
⌧j
1 (⇣p)) =
(p�1)+p�2X
k=1
(�1)k⌧↵k1 [(⇣p, 1)] = (p�1)+p�2X
k=1
(�1)k⌧↵k1 (�1) = (p�1)+
p�2X
k=1
(�1)k+1 = (p� 1) + 1� 1 + . . . + 1� 1 + 1 = (p� 1) + 1 = p.
⇤
Theorem 2.1. �2 = (�1)p�12 p
Proof. This follows directly from Lemma 2.6 and Lemma 2.7. ⇤
We conclude that � = ±q
(�1)p�12 p with [Q(�) : Q] = 2.
Remark. By Galois correspondence, Q(�) is the unique quadratic
sub-field of Kp.
6
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3. Quadratic Reciprocity
We will use the results from the previous section to prove one
of Gauss’favorite theorems, the law of quadratic reciprocity.
First we will define the Legendre symbol and develop lemmas and
sup-plementary laws to assist the proof.
Definition 3.1. For odd prime p,
✓a
p
◆=
8><
>:
1 if a is a quadratic residue (mod p) and a 6⌘ 0 (mod p)
�1 if a is a quadratic non-residue (mod p)
0 if a ⌘ 0 (mod p).
this is called the Legendre symbol.
Lemma 3.1. 8a, b 2 Z/pZ⇥, (a + b)p ⌘ ap + bp (mod p)
Proof. (a + b)p =pX
i=0
✓p
i
◆a
ibp�i where
✓p
i
◆=
p!
(p� i)!i!)
For 1 i p� 1, i! 6⌘ 0 (mod p) and (p� i)! 6⌘ 0 (mod p) since
bothi and p� i are both strictly less than p.✓
p
i
◆(p� i)!i! = p! ⌘ 0 (mod p))
✓p
i
◆⌘ 0 (mod p)
(a + b)p = ap +p�1X
i=1
✓p
i
◆a
ibp�i + bp ⌘ ap + bp (mod p). ⇤
Lemma 3.2. Let p be an odd prime,
✓v
p
◆⌘ (v)
p�12 (mod p)
Proof. Let p be an odd prime. Z/pZ⇥ is cyclic, with generator
g.If v is a square modulo p, then v = g2k for some k, 0 k
p� 1
2.
vp�12 ⌘ g
(p�1)k⌘ 1k ⌘ 1 ⌘
✓v
p
◆.
Notice that this accounts for the case where v ⌘ 0 (mod p).
If v is not a square modulo p, then v = g2k+1 for some k, 0 k
<p� 1
2.
vp�12 ⌘ g
(p�1)k+ p�12 ⌘ gp�12 ⌘ �1 ⌘
✓v
p
◆. ⇤
Lemma 3.3. (Supplementary Law 1)
✓�1
p
◆=
(1 if p ⌘ 1 (mod 4)
�1 if p ⌘ 3 (mod 4)7
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Proof. Put v = �1 in Lemma 3.2 ⇤
Lemma 3.4. (Supplementary Law 2)
✓2
p
◆=
(1 if p ⌘ ±1 (mod 8)
�1 if p ⌘ ±3 (mod 8)
Proof. Set ⇣8 = exp(2⇡i8 ) and � = ⇣8 + ⇣
�18 = 2 cos
2⇡8 =p
2.�
2 = (⇣8 + ⇣�18 )
2 = ⇣28 + 2 + ⇣�28 = 2.
By Lemma 3.2:
✓2
p
◆⌘ 2
p�12 ⌘ (�2)
p�12 ⌘ �
p�1 (mod p).
Multiplying by � on both sides yields: �
✓2
p
◆⌘ �
p (mod p)
By Lemma 3.1: �p = (⇣8+⇣�18 )
p⌘ ⇣
p
8 +⇣�p8 ⌘ (⇣8+⇣
�18 )
✓2
p
◆(mod p).
p ⌘ ±1 (mod 8)) ⇣p8 + ⇣�p8 = ⇣8 + ⇣
�18 = � )
✓2
p
◆⌘ 1 (mod p).
p ⌘ ±3 (mod 8) ) ⇣p8 + ⇣�p8 = ⇣
38 + ⇣
�38 = (⇣8 + ⇣8)(⇣
28 � 1 + ⇣
�18 ) =
�(⇣28 + 2 + ⇣�18 � 3) = �(�
2� 3) = �� )
✓2
p
◆⌘ �1 (mod p) ⇤
Theorem 3.1. Let p, q be two distinct positive prime numbers,
then
✓p
q
◆=
8>><
>>:
✓q
p
◆if p or q ⌘ 1 (mod 4)
�
✓q
p
◆if p ⌘ q ⌘ 3 (mod 4)
Proof. From the previous section, we analyzed � =p�1X
i=1
⇣i2
p= ⌘0 � ⌘1.
� is also expressible as � =p�1X
v=1
✓v
p
◆⇣
v
p.
By Theorem 2.1, �q = �(�2)q�12 = �((�1)
p�12 p)
q�12 = �(�1)
p�12
q�12 p
q�12
By Lemma 3.2, �q ⌘ �(�1)p�12
q�12
✓p
q
◆(mod q).
By Lemma 3.1, �q =
p�1X
v=1
✓v
p
◆⇣
v
p
!q⌘
p�1X
v=1
✓v
p
◆q⇣
qv
p⌘
p�1X
v=1
✓v
p
◆⇣
qv
p⌘
p�1X
v=1
✓q
p
◆✓q
p
◆✓v
p
◆⇣
qv
p⌘
✓q
p
◆ p�1X
v=1
✓qv
p
◆⇣
qv
p⌘
✓q
p
◆� (mod q).
The last congruence is observable via variable change v !
qv8
-
By equating both congruences of �q (mod q),
�(�1)p�12
q�12
✓p
q
◆⌘
✓q
p
◆� ) �
2(�1)p�12
q�12
✓p
q
◆⌘
✓q
p
◆�
2)
(±p)(�1)p�12
q�12
✓p
q
◆⌘
✓q
p
◆(±p)) (�1)
p�12
q�12
✓p
q
◆⌘
✓q
p
◆(mod q)
±p is coprime to q, and thus has an inverse modulo q.
Because (�1)n,
✓p
q
◆, and
✓q
p
◆take values in {±1}, the congruence
can be strengthened to an equality.
p or q ⌘ 1 (mod 4)) (p�1) or (q�1) ⌘ 0 (mod 4))p� 1
2
q � 1
2⌘ 0
(mod 2))
✓p
q
◆=
✓q
p
◆
p ⌘ q ⌘ 3 (mod 4))p� 1
2
q � 1
2⌘ 1 (mod 2))
✓p
q
◆= �
✓q
p
◆⇤
Lemmas 3.3 and 3.4 and Theorem 3.1 give the full statement of
qua-dratic reciprocity. The Legendre symbol being multiplicative
(as wewill see in the next section, the Legendre symbol is a
quadratic mul-tiplicative character), together with the lemmas and
theorem, allow✓
n
p
◆to be computed 8n 2 Z.
Example.
✓541
7919
◆=
✓7919
541
◆=
✓345
541
◆=
✓3
541
◆✓5
541
◆✓23
541
◆=
✓541
3
◆✓541
5
◆✓541
23
◆=
✓1
3
◆✓1
5
◆✓12
23
◆=
✓4
23
◆✓3
23
◆=
✓3
23
◆=
�
✓23
3
◆= �
✓2
3
◆= 1.
The last equality agrees with both lemmas when 2 is viewed as
-1(mod 3) as well.
Definition 3.2. The generalization of the Legendre symbol is the
Ja-cobi. Without restriction to composite n > 0, if n = p↵11
p
↵22 . . . p
↵kk
,
then the Jacobi symbol⇣
a
n
⌘=
✓a
p1
◆↵1 ✓a
p2
◆↵2. . .
✓a
pk
◆↵k. This is
a product of Legendre symbols. It shares some properties common
tothe Legendre symbol, but loses a valuable property as well.
Remark. Both Supplementary Laws and Theorem 3.1 still hold
whendistinct primes p and q are replaced by composite n and m
whereboth n, m are positive and odd. Unfortunately, there is
significantloss of information. Specifically, the Jacobi symbol
does not share the
9
-
property:⇣
a
n
⌘= 1 ) a is a square modulo n. This is due to the
indiscernability between an even parity of quadratic nonresidues
in theproduct of the Jacobi symbol and the necessary condition of a
beinga quadratic residue modulo all primes dividing n. Despite
this, Jacobisymbols are still very useful in various algorithms
such as in primalitytesting.
We will now state the law of cubic reciprocity, without proof,
for theinterest of the reader.
Definition 3.3. An Eisenstein integer is an element of
Z[⇣3].Definition 3.4. Let ⇡ be and Eistenstein prime with N(⇡) 6=
3, and
let ↵ 2 Z[⇣3]. If ⇡|↵, set⇣↵
⇡
⌘
3= 0. If ⇡ 6 |↵, let
⇣↵
⇡
⌘
3be the unique
power of ⇣3 defined by
↵(N(⇡)�1)/3
⌘
⇣↵
⇡
⌘
3(mod ⇡)
This symbol is multiplicative like the Legendre symbol. Lastly,
if � is
a unit of Z[⇣3], define⇣↵
⇡
⌘
3= 1 for every ↵ 2 Z[⇣3]
Definition 3.5. Let ↵ 2 Z[⇣3]. ↵ is called primary if ↵ ⌘ ±1
(mod 3)Theorem 3.2. (Cubic Reciprocity). Let ↵ and � be coprime
primaryEisenstein integers. Then
✓↵
�
◆
3
=
✓�
↵
◆
3
For more information about cubic reciprocity see [3].
10
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4. Gauss Sums
Definition 4.1. Given a finite abelian group G, the
multiplicativehomomorphism � : G ,! S1 ⇢ C⇥ is called a character
of G.g 2 G, g = ga11 . . . g
ann
. �(g) = exp(2⇡iX
j
aj
fj) where ordG(gj) = fj.
More generally, �b1,...,bn(g) = exp(2⇡iX
j
bjaj
fj) =
Y
j
exp(2⇡ibjaj
fj) =
Y
j
�b1,...,bn(gj)ajbj The group of characters is denoted by Ĝ.
The definition of � used in this paper will be �(g) =
�1,...,1(g), omittingthe trivial weights.
Let �0 denote the trivial character, with �0(g) = 1 8g 2 G.
The finite abelian group on which � is defined will be a Galois
group ofa number field, typically Z/mZ⇥ corresponding to the mth
cyclotomicnumber field.
Definition 4.2. G = Z/mZ⇥. � can be extended by defining �(n)
=�(n̄) where n̄ is the residue of n modulo m. Setting �(n) = 0 for
nwith gcd(n,m)>1, extends � to the integers with periodicity,
such that�(n) = �(n+m) 8n,2 Z. We say that � is a character to the
modulusm. This extended character � is a Dirichlet character.
Definition 4.3. G = Z/mZ⇥. A character � has a minimum modulusof
definition, called the conductor, denoted f(�). A character is
calledprimitive if f(�) = m, and imprimitive otherwise, in which
case f(�) <m.
Example. Let m = 8, then Z/mZ⇥ = {1, 3, 5, 7}. Every
non-trivialcharacter is quadratic. By the multiplicative
relationships, it is easilyverifiable that {3,5} generate the
group, and hence � is defined by thevalues it takes on 3 and 5.
This is the character table:
� 0 1 2 3 4 5 6 7�1 0 1 0 1 0 1 0 1�2 0 1 0 -1 0 1 0 -1�3 0 1 0
1 0 -1 0 -1�4 0 1 0 -1 0 -1 0 1
f(�1) = 1 < 8 and f(�2) = 4 < 8, hence �1 and �2 are
imprimitive(�1 is obviously imprimitive because it is the trivial
character!)
11
-
�3 and �4 are both primitive because they cannot be defined on
asmaller modulus.
Remark. Unless stated otherwise, � will be a primitive
character.
Definition 4.4. The Gauss sum of Dirichlet character � modulo N
is
GN(�, a) =NX
n=1
�(n)⇣anN
where ⇣N = exp(2⇡iN
) and � is a (primitive)
character of (Galois) group G (= Z/NZ⇥). The notation GN(�)
=GN(�, 1) will be used if there is no ambiguity.
Remark. The Gauss sum generalizes the Langrange resolvent,
definedin the previous section.
When � is the trivial character, Gp(�0, 1) = �p(⇣p)� 1 = �1.
When � is the quadratic character, Gp(�2, 1) =pX
n=1
�2(n)⇣n =
pX
n=1
✓n
p
◆⇣
n
where
✓n
p
◆is the Legendre symbol.
The Classical Gauss sum from the Section 2 can now be viewed as
theGauss sum Gp(�2) where �2 is the unique quadratic character of
theGalois group.
Lemma 4.1. Let � and be characters modulo m. If gcd(f(�), f( ))
=1, then f(� ) = f(�)f( )
Proof. Apply the Chinese Remainder Theorem. ⇤Remark. Lemma 4.1
allows us to factor characters modulo m. Ifm = ab, and gcd(a, b) =
1 then we can factor a character � modulo minto a product of
characters �a and �b modulo a and b, respectively.
Lemma 4.2. gcd(a, m) = 1) Gm(�, a) = �(a)Gm(�, 1)
Proof. Gm(�, a) =mX
n=1
�(n)⇣anm
=mX
n=1
�(a)�(a)�(n)⇣anm
=mX
n=1
�(a)�(an)⇣anm
=
�(a)mX
n=1
�(an)⇣anm
= �(a)Gm(�, 1). The last equality holds because
gcd(a, m) = 1) n! an is a bijection on Z/mZ⇥. ⇤Lemma 4.3. gcd(a,
m) > 1) Gm(�, a) = 0
12
-
Proof. gcd(a, m) = d > 1) m = rd and a = bd and gcd(r, b) =
1There exists a c, such that �(c) 6= 1 and c ⌘ 1 (mod r). The
existenceof such a c arises from � being non-trivial on the kernel
of the mappingh : Z/mZ⇥ �! Z/rZ⇥. If it were trivial on the kernel,
then � woulddetermine a character of Im(h) ✓ Z/rZ⇥, extending to a
character �0of Z/mZ⇥, and �0 would induce �. This contradicts �
being primitive.It is easy to show that a ⌘ ca mod m, and thus
⇣a
m= ⇣ca
m
�(c)Gm(�, a) =mX
n=1
�(cn)⇣anm
=mX
n=1
�(cn)⇣canr
= Gm(�, a) by the vari-
able change cn �! n, since gcd(c, m) = 1.(�(c)� 1)Gm(�, a) = 0)
Gm(�, a) = 0, since �(c) 6= 1 ⇤
We will now apply Lemmas 4.2 and 4.3 to generalize Theorem
2.1.
Theorem 4.1. For primitive character � with conductor m, if
gcd(a, m) =1 then |Gm(�, a)| =
pm.
Proof. |Gm(�, a)|2 = Gm(�, a)Gm(�, a) =X
x
X
y
�(x)�(y)⇣(x�y)p
where
x and y run over the residues modulo m.x ⌘ y (mod m))
X
n (mod m)
⇣n(x�y)m
=X
n (mod m)
1 = m
x 6⌘ y (mod m))X
n (mod m)
⇣n(x�y)m
= �m(⇣m) = 0
By Lemma 4.3, gcd(x, m) > 1 ) �(x) = 0X
n (mod m)
|Gm(�, n)|2 =
X
(n,m)=1
|Gm(�, n)|2 =
X
(n,m)=1
X
x
X
y
�(x)�(y)⇣(x�y)m
=X
(n,m)=1
X
x (mod m)
�(x)�(x) =
X
(n,m)=1
|�(x)|m = mX
(n,m)=1
1 = m�(m).
By Lemma 4.2, for n coprime to m, Gm(�, n) = �(n)Gm(�, 1)
)|Gm(�, n)|2 = |�(n)|2|Gm(�, 1)|2.
Appealing to Lemmas 4.2 and 4.3, |Gm(�, n)|2 =
(|Gm(�, 1)|2 if gcd(n, m) = 1
0 if gcd(n, m) > 1X
n (mod m)
|Gm(�, n)|2 =
X
(n.m)=1
|Gm(�, 1)|2 = �(m)|Gm(�, 1)|
2.
Equating the two yields �(m)|Gm(�, 1)|2 = m�(m)) |Gm(�, 1)|2 =
m
gcd(a, m) = 1) |Gm(�, a)|2 = |Gm(�, 1)|2 ) |Gm(�, a)| =p
m ⇤13
-
Remark. Clearly, by Lemma 4.3, gcd(a, m) > 1) |Gm(�, a)| =
0
We will now develop a decomposition of a Gauss sum of
compositemodulus into simpler Gauss sums.
Theorem 4.2. Let m = ab with gcd(a,b)=1. Let � = �a�b be a
charac-
ter modulo m. Gm(�) = �a(b�1a )Ga(�a)�b(a�1b
)Gb(�b) where f(�a) = aand f(�b) = b.
Proof. For any n, 1 n m, we would like to write n ⌘ n1 (mod
a)and n ⌘ n2 (mod b). By the Chinese Remainder Theorem,
n = ↵n1 + �n2 where ↵ ⌘
(1 (mod a)
0 (mod b)and � ⌘
(0 (mod a)
1 (mod b)
Let ba denote the residue of b (mod a) and ab the residue of a
(mod b).↵ = bab�1a = bb
�1a
and � = aba�1b
= aa�1b
where b�1a
and a�1b
are integerrepresentatives of the residues between 1 and their
respecitve moduli.�(n) = �m(n) = �a(n)�b(n) =
�a(↵n1+�n2)�b(↵n1+�n2) = �a(n1)�b(n2)
Gm(�) =mX
n=1
�m(n)⇣n
m=X
n1,n2
�a(n1)�b(n2)⇣↵n1+�n2m
=X
n1,n2
�a(n1)�b(n2)⇣↵n1m
⇣�n2m
⇣↵n1m
= exp(2⇡i↵n1
m) = exp(
2⇡ibb�1a
n1
ab) = exp(
2⇡ib�1a
n1
a) = ⇣b
�1a
a
⇣�n2m
= exp(2⇡i�n2
m) = exp(
2⇡iaa�1b
n2
ab) = exp(
2⇡ia�1b
n2
b) = ⇣
a�1b
b
Gm(�) =X
n1,n2
�a(n1)�b(n2)⇣b�1a n1
a⇣
a�1b n2
b=X
n1
�a(n1)⇣b�1a n1
a
X
n2
�b(n2)⇣a�1b n2
b.
Be Lemma 4.2,X
n1
�a(n1)⇣b�1a n1
a= �a(b�1a )
X
n1
�a(n1)⇣n1a
= �a(b�1a )Ga(�a)
andX
n2
�b(n2)⇣a�1b n2
b= �b(a
�1b
)X
n2
�b(n2)⇣n2b
= �b(a�1b
)Gb(�b).
Substituting, we get Gm(�) = �(b�1a )Ga(�a)�(a�1b
)Gb(�b) ⇤Remark. The process in Theorem 4.2 can be repeated
until m is fac-
tored into prime powers. m =kY
i=1
p↵ii) Gm(�) = µ
Y
i
Gp
↵ii
(�p
↵ii
)
where µ is the appropritate root of unity.
14
-
5. Quadratic Gauss Sums
We will now focus on quadratic Gauss sums and their
decomposition.
Definition 5.1. The quadratic Gauss sum g(a, N) =NX
n=1
⇣an
2
N.
The quadratic Gauss sum can be decomposed into simpler
quadraticGauss sums, and ultimately factored analogously to the
Gauss sum inTheorem 4.2.
Lemma 5.1. For odd prime p, g(a, p) =
✓a
p
◆g(1, p).
Proof. Appealing to Lemma 2.4 and Lemma 4.2, we see that gcd(a,
p) =
1 ) g(a, p) =pX
n=1
⇣an
2
p=
pX
n=1
�2(a)⇣an
p= G(�2, a) = �2(a)G(�2, 1) =
�2(a)G(�2, 1) = �2(a)pX
n=1
�(n)⇣np
= �2(a)pX
n=1
⇣n
2
p= �2(a)g(1, p), where
�2 is the Legendre symbol. Likewise, if gcd(a, p) = p) �2(a) =
0. ⇤Remark. By induction, Lemma 5.1 can be extended through the
Ja-cobi symbol so that the restriction on p is lifted from having
to beprime to having to only be odd and positive. Even the
restriction ofbeing odd can be lifted, but the proof requires
analytic computation.
We will now develop some recursion formulae, useful in the
process offactorizing quadratic Gauss sums.
Lemma 5.2. For odd prime p, and j � 2, g(1, pj) = pg(1,
pj�2).
Proof. Writing n p-adically, n = r+spj�1 with r a residue modulo
pj�1
and s a residue modulo p. n2 = (r + spj�1)2 = r2 + 2rspj�1 +
sp2j�2 ⌘r2 + 2rspj�1 (mod pj).
g(1, pj) =p
jX
n=1
⇣n
2
pj=X
r
X
s
⇣r2+2rspj�1
pj=X
r
X
s
⇣r2
pj⇣
2rsp
= pX
r,p|r
⇣r2
pj.
gcd(p, r) = p ) ⇣2rsp
= exp(2⇡is) = 1, so the inner summation is p ass runs through
the residues modulo p.gcd(p, r) = 1) s �! 2rs is a bijection on
Z/pZ and as s runs throughthe residues modulo p, each pth root of
unity appears once, so the innersummation is 0.
15
-
g(1, pj) = pX
r,p|r
⇣r2
pj= p(⇣0
pj+⇣p
2
pj+⇣4p
2
pj+. . .+⇣p
2(j�1)2pj
) = p(⇣0pj�2+⇣
1pj�2+
⇣4pj�2 + . . . + ⇣
(j�2)2�1pj�2 ) = p
pj�2�1X
n=0
⇣n
2
pj�2 = pp
j�2X
n=1
⇣n
2
pj�2 = pg(1, pj�2) ⇤
Lemma 5.3. g(1, 2j) = 2g(1, 2j�2) for j � 4.
Proof. Writing n 2-adically, n = r+s2j�2 with r a residue modulo
2j�2
and s a residue modulo 2. n2 = (r + s2j�2)2 = r2 + 2rs2j�2 +
sp2j�4 ⌘r2 + rs2j�1 (mod 2j).
g(1, 2j) =2j�1X
n=0
⇣n
2
2j =X
r
X
s
⇣r2+rs2j�1
2j =X
r
X
s
⇣r2
2j (�1)rs = 2
2(j�1)�1X
2|r,r=0
⇣r2
2j .
The last equality holds by the same arguement as in Lemma
5.2.
g(1, 2j) = 22(j�1)�1X
2|r,r=0
⇣r2
2j = 42(j�2)�1X
2|r,r=0
⇣r2
2j = 4(⇣02j +⇣
22
2j +⇣24
2j +. . .+⇣2(j�3)
2
2j ) =
4(⇣02j�2 +⇣12j�2 +⇣
22
2j�2 + . . .+⇣2(j�4)
2
2j�2 ) = 42(j�3)�1X
n=0
⇣n
2
2j�2 = 22(j�2)�1X
n=0
⇣n
2
2j�2 =
2g(1, 2j�2). The second and sixth equalities are due to
symmetry. ⇤
Lemma 5.4. g(1, pj) =
(p
j2 p
j⌘ 1 (mod 4)
ipj2 p
j⌘ 3 (mod 4)
Proof. g(1, pj) = pg(1, pj�2) = . . . = pkg(1, pj�2k) after k
iterations.
j ⌘ 0 (mod 2)) g(1, pj) = . . . = pj�22 g(1, p2) = p
j2 g(1, 1) = p
j2 .
j ⌘ 1 (mod 2) ) g(1, pj) = . . . = pj�12 g(1, p) = p
j�12
q(�1)
p�12 p =
q(�1)
p�12 p
j2 =
(p
j2 if p ⌘ 1 (mod 4), pj ⌘ 1 (mod 4)
ipj2 if p ⌘ 3 (mod 4), pj ⌘ 3 (mod 4)
⇤
Lemma 5.5. g(1, 2j) =
((1 + i)2
j2 2j ⌘ 0 (mod 4)
0 2j ⌘ 2 (mod 4)
Proof. g(1, 2j) = 2g(1, 2j�2) = . . . = 2kg(1, 2j�2k) after k
iterations.
2j ⌘ 0 (mod 4) ) g(1, 2j) = . . . = 2j�22 g(1, 4) = 2
j�22 (i1 + i4 + i9 +
i16) = 2(1 + i)2
j�22 = (1 + i)2
j2 .
2j ⌘ 2 (mod 4)) g(1, 2j) = . . . = 2j�12 g(1, 2) = 2
j�12 (1� 1) = 0 ⇤
16
-
Lemma 5.6. Let a be odd. Then g(a, 2r) = ✏(a)
✓�2r
a
◆g(1, 2r) where
✏(a) =
(1 if a ⌘ 1 (mod 4)
i if a ⌘ 3 (mod 4)
Proof. Let �a : ⇣2r �! ⇣a2r . Then g(a, 2r) = �a(1, 2r) =
�a(1+i)�a(2
r2 ).
�a(1 + i) = 1 + ia =
(1 + i if a ⌘ 1 (mod 4)
1� i if a ⌘ 3 (mod 4)=
✓�1
a
◆✏(a)(1 + i)
�a(2r2 ) =
✓2
a
◆r2
r2 =
✓2r
a
◆2
r2 ) g(a, 2r) = ✏(a)
✓�2r
a
◆g(1, 2r) ⇤
Analogous to Theorem 4.2, there is a way to decompose a
quadraticGauss sum of composite modulus.
Theorem 5.1. Let m = ab with gcd(a, b) = 1. g(1, m) = µg(1,
a)g(1, b)where µ is an appropriate root of unity.
Proof. For any n, 1 n m, we would like to write n ⌘ n1 (mod
a)and n ⌘ n2 (mod b). By the Chinese Remainder Theorem,
n = ↵n1 + �n2 where ↵ ⌘
(1 (mod a)
0 (mod b)and � ⌘
(0 (mod a)
1 (mod b)
Let ba denote the residue of b (mod a) and ab the residue of a
(mod b).↵ = bab�1a = bb
�1a
and � = aba�1b
= aa�1b
where b�1a
and a�1b
are integerrepresentatives of the residues between 1 and their
respecitve moduli.
n2 = (↵n1 + �n2)2 = ↵2n21 + 2↵�n1n2 + �
2n
22 = b
2ab�2a
n21 + a
2ba�2b
+
2mb�1a
a�1b⌘ b
2ab�2a
n21 + a
2ba�2b
(mod m) ) ⇣n2
m= ⇣
(b2ab�2a )n21+(a
2ba�2b )n
22
m =
⇣(b2ab
�2a )n21
m ⇣(a2ba
�2b )n
22
m = ⇣(bb�2a )n21a ⇣
(aa�2b )n22
b) g(1, m) = g(bb�2
a, a)g(aa�2
b, b).
m = 2jk where k is odd and j � 0) g(1, m) = g(kk�22j ,
2j)g(2j2j
�2k
, k).Appealing to Lemma 5.1, the subsequent remark, and Lemma
5.6,g(1, m) = g(kk�22j , 2
j)g(2j2j�2k
, k) =
✏(kk�22j )
✓�2j
kk�22j
◆g(1, 2j)
2j2j
�2k
k
!g(1, k) =
"✏(kk�22j )
✓�2j
kk�22j
◆ 2j2j
�2k
k
!#g(1, 2j)g(1, k).
k =Y
piodd
paii
, so further decomposition of g(1, k) is possible but the
computation of the root of unity becomes rather cumbersome.
⇤17
-
Remark. m =Y
piprime
paii
. Using Theorem 5.1 and great diligence,
g(1, m) = µY
piprime
g(1, paii
) where µ, the appropriate root of unity, can
be computed through methods in the proof of Theorem 5.1.
We conclude this section with an elegant equation for evaluating
qua-dratic Gauss sums due to Dirichlet. The proof of the theorem
usesanalytic methods and is beyond the scope of this paper.
Theorem 5.2. For any positive integer b, g(1, b) =1 + i�b
1 + i�1p
b.
Proof. See Algebraic Number Theory by S. Lang, pp. 88-90. [5]
⇤Remark. Theorems 5.1 and 5.2 allow g(a, b) to be computed for
anypositive integers a, b.
18
-
6. Stickelberger’s Theorem
To pursue the prime factorization of the Gauss sum, we will look
atthe Gauss sum as an ideal (or divisor) in the appropriate number
field.
We must first introduce the Teichmüller character, which will
be acanonical character dependent on a prime ideal that will be the
basisfor expressing all other characters. The existence of such a
characterwill be justified by Hensel’s lemma.
Theorem 6.1. (Trivial case of Hensel’s lemma). Let K be a
numberfield and let p be a prime ideal in the ring of integers OK.
Let Kp be thecompletion of K at the finite place p and let Op be
the ring of integersin Kp. Let f(x) be a polynomial with coe�cients
in Op and supposethere exists ↵0 2 Op such that
f(↵0) ⌘ 0 mod p, f0(↵0) 6= 0 mod p
Then there exist a root ↵ 2 Kp of f(x), i.e. f(↵) = 0.
Proof. See http://planetmath.org/encyclopedia/HenselsLemma.html.
⇤Corollary. Let p be a prime number. The ring of p-adic integers
Zpcontains exactly p � 1 distinct (p � 1)th roots of unity.
Furthermore,every (p� 1)th root of unity is distinct modulo p.
Proof. Notice that Qp, the p-adic rationals, is a field.
Therefore f(x) =x
p�1� 1 has at most p� 1 roots in Qp. Moreover, if we let a 2 Z
with
1 a p � 1 then f(a) = ap�1 � 1 ⌘ 0 mod p by Fermat’s
littletheorem. Since f 0(a) = (p � 1)ap�2 is non-zero modulo p, the
trivialcase of Hensel’s lemma implies that there exist a root of
xp�1� 1 in Zpwhich is congruent to a modulo p. Hence, there are at
least p� 1 rootsin Zp, and we can conclude that there are exactly
p� 1 roots. ⇤Definition 6.1. The Teichmüller character (or
Teichmüller lift) is theunique character ! of F⇥
psatisfying !(a) ⌘ a (mod p).
Lemma 6.1. The Teichmüller character generates the character
groupof F⇥
p.
Proof. F⇥p
is cyclic and !(⇣p�1+p) = ⇣p�1, and so ! has order p�1. ⇤
Remark. Every character of F⇥p
is expressable as !�k. We will thus
use !(a) ⌘ a�1 (mod p) as our generator to avoid negative
powers.From this point on, we will refer to ! as this
generator.
19
-
The following lemmas will explain this diagram.
Pp�11 . . . Pp�1r
Kp�1,p
pppppp
ppppp
LLLL
LLLL
LLL
p1 . . . p�(p�1) Kp�1
NNNNNN
NNNNNN
NKp
qqqq
qqqq
qqqq
q ⇡p�1
Q
p
Put Kn = Q(⇣n) and let Km,n be the composite of Km and Kn.
It is well known in algebraic number theory that in a number
field K,an ideal factors into a product of prime ideals, the
algebraic integers
of K. n =rY
i=1
piei
Lemma 6.2. Let n be an ideal of OK. n =rY
i=1
peii.
rX
i=1
eifi = [K : Q]. Where fi = [OK/pi : Z/n]
fi is called the inertial degree of pi.
Proof. n[K:Q] = NK/Q(n) =
rY
i=1
NK/Q(pi)ei =rY
i=1
neifi = n
Pri=1 eifi ⇤
Remark. If K is a Galois extension, then ei = e and fi = f for1
i r As a consequence, efr = [K : Q].
Kp and Kp�1 are both Galois extensions because ⇣p and ⇣p�1
generatethe all of the pth and (p� 1)st roots of unity in Kp and
Kp�1, respec-tively, thus the fields are normal, and because all
number fields areseparable.
Theorem 6.2. Write m in the form pkn, p 6 | n. Then e = �(pk)
andf is the multiplicative order of p modulo n.
20
-
Proof. See Number Fields by D. Marcus pp. 76-78. [7] ⇤Corollary.
If p 6 | n, then p splits in �(m)/f distinct primes ideals inOKp�1,
where f is the order of p mod m.
Lemma 6.3. p totally splits in Kp�1.
Proof. Appealing to the corollary when m = p�1 gives us that p
splitsinto �(p� 1) distinct prime ideals because f = 1. ⇤Lemma 6.4.
p totally ramifies in Kp.
Proof. ⇡ = (1 � ⇣p)|(1 � ⇣ ip) for 1 i p � 1 ) ⇡p�1
|
p�1Y
i=1
(1� ⇣ ip) =
�(1) = p. For ⇡, (p � 1)fr = efr = [Kp : Q] = �(p) = p � 1 ) f
=r = 1. ⇡ has full ramification index, and must be prime, otherwise
itwould factor causing efr > [Kp : Q]. ⇤
Lemma 6.5. p =Y
Pp�11 . . . Pp�1�(p�1) in OKp�1,p.
Proof. Appealing to Lemmas 6.2, 6.3 and 6.4,p = p1 . . . p�(p�1)
in OKp�1 ) r � �(p� 1). p = ⇡
p�1) e � (p� 1).
ef � �(p� 1)�(p) = [Kp�1,p : Q] = efr � ef . ⇤
Now we know the structure of the factorization of p, and hence
theGauss sum with absolute value p. Stickelberger’s theorem will
explainthe factorization of the Gauss sum in terms of prime ideals
in Kp�1,p. In full generality, q = pn ⌘ 1 (mod m). The following
theorems willstate the general case, but the content of this paper
will only analyzethe case of n = 1 and m = p� 1
Definition 6.2. For integer k, looking at k (mod q� 1), we write
thep-adic expansion k = k0 + k1p + . . . + kn�1pn�1 with 0 ki p�
1.
s(k) =n�1X
i=0
ki and �(k) =n�1Y
i=0
ki!
Both s(k) and �(k) are periodic q � 1.
Theorem 6.3. (Stickelberger’s Theorem). For any integer k, we
havethe congruence
G(!k, ⇣Trp
)
(⇣p � 1)s(k)⌘�1
�(k)(mod P)
21
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In particular,
ordPG(!k, ⇣
Tr
p) = s(k)
Proof. See Cyclotomic Fields I,II by S. Lang, pp. 7-9. [6] ⇤
Corollary. In the case of n = 1, q = pn = p and Tr(x) = xp0
= x.s(k) = k0 ⌘ k (mod q) and �(k) = k0!In this case,
Stickelberger’s Theorem states: for 0 k p� 1
G(!k, ⇣p)
(⇣p � 1)k⌘�1
k!(mod P)
in particular,
ordPG(!k, ⇣p) = k
Proof. Let p be a prime ideal over p. !(n) ⌘ n�1 (mod p) = (mod
Pp�1).
Gp(!) =p�1X
n=1
!(n)⇣np⌘
p�1X
n=1
n�1(1� ⇡)n (mod Pp�1)
Gp(!) ⌘p�1X
n=1
n�1(1� n⇡) (mod P2) since P||⇡.
Gp(!) ⌘p�1X
n=1
n�1 + Gp(!) ⌘
p�1X
n=1
⇡ ⌘ 0� (p� 1)⇡ ⌘ ⇡ (mod P2)
Hence we have that, P||Gp(!). Similarily, for 1 k p� 2,
Gp(!k) =p�1X
n=1
!(n)k⇣np⌘
p�1X
n=1
n�k(1� ⇡)n (mod Pp�1)
Gp(!k) ⌘p�1X
n=1
n�k
nX
j=0
(�1)j✓
n
j
◆⇡
j⌘
kX
j=0
p�1X
n=1
n�k(1 + ↵1n⇡ + ↵2n
2⇡
2 + . . . + (�1)k1
k!n
k⇡
k) (mod Pk+1)
For 1 j k � 1,p�1X
n=1
nj�k
↵j⇡j⌘ ↵j⇡
j
p�1X
n=1
nj�k⌘ 0 (mod Pk+1)
Gp(!k) ⌘p�1X
n=1
(�1)1
k!⇡
k⌘ (�1)k+1
1
k!⇡
k6⌘ 0 (mod Pk+1).
Hence we have that, Pk||Gp(!k)) ordPG(!k) = k. ⇤
Stickelberger’s Theorem gives us a correspondence between kth
powersof a particular prime ideal P exactly dividing the Gauss sums
Gp(!k).
22
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Using the result of the theorem, we can understand the full
factorizationof a particular Gauss sum Gp(!a)
Take �a = �a,1 2 Gal(Kp�1,p/Q) such that �a(⇣p) = ⇣p and �a(!) =
!a.
Gp(!)�a =p�1X
n=1
!(n)a⇣np
= Gp(!a).
Pa||Gp(!a)) (P��1a )
a
||Gp(!a)��1a ) (P�
�1a )
a
||Gp(!)
We now have the full factorization of Gp(!) =Y
(a,p�1)=1
(P��1a )
a
.
Via conjugation, we know the full factorization of Gp(!k) =
Gp(!)�k =Y
(a,p�1)=1
(P��1a �k)
a
for any 1 k p� 2 with gcd(k, p� 1) = 1.
The following well known theorem in algebraic number theory
providesa method for computing the primes p and P lying over p.
Lemma 6.6. If � has order m, then Gp(�)m 2 Km.
Proof. Let ⌧ 2 Gal(Km,p/Q) such that ⌧(⇣p) = ⇣vp and ⌧(⇣m) = ⇣m.
Infact, ⌧ 2 Gal(Km,p/Km).
⌧(Gp(�)m) = ⌧(X
n (mod q)
�(n)⇣np) =
X
n (mod q)
�(n)⇣vpn = �(v)
m
Gp(�)m.
⇤Theorem 6.4. Let p be a prime ideal (or divisor) in the number
fieldk. If Ok[✓] = OK, then there is a prime ideal (or divisor)
Pi|p ofK/k of residue class degree (inertial degree) f = deg(hi(x))
given byP = hi(✓) + pOK (or P = (hi(✓), p) as a divisor), where
h(x) is
the (monic) irreducible polynomial of ✓ and h(x) =Y
hi(x) is the
projection of h(x) in Fq[x] ⇠= Ok/p[x]
Remark. It is necessary that Ok[✓] = OK . For ✓ = ⇣n this
holds.
Remark. The proof is elementary but relatively long and can be
foundin many texts. The result of the theorem will be used for
computationalexamples, hence the proof is excluded.
Proof. See Number Fields by D. Marcus pp.79-82. [7] or Analytic
The-ory of Algebraic Numbers by H. Stark Theorem, 8.1.3. [9]
⇤Example. Let k = Q, p = 7 and K = K6 where ✓ = ⇣6. h(x) =�6(x) =
x2 � x + 1 ⌘ (x � 3)(x � 5) (mod 7). There are exactly twoprimes
p1, p2 dividing p, both of inertial degree 1. p1 = (⇣6 � 3, 7)
and
23
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p2 = (⇣6 � 5, 7). K6 is a quadratic extension, hence p = p1p2.To
verify, 7 = (2 +
p�3)(2�
p�3) = (1 + 2⇣6)(3� 2⇣6) =
[1 · 7 + 2(⇣6 � 3)][(�1) · 7� 2(⇣6 � 5)].Using our previous
results, p = p1p2 = P61P
62 in K6,7.
Definition 6.3. The Stickelberger element is
✓(k, p) =X
c2Z/mZ⇥h
kc
q � 1i�
�1c2 Q[G]
where hti ⌘ t (mod Z) and 0 hti 1 and m is the order of
thecharacter !k (mod p). !k will be defined contextually.
Lemma 6.7. For any integer k, we have s(k) = (p� 1)n�1X
i=0
hkp
i
q � 1i.
Proof. We may assume 1 k q � 1 since both sides of the
equationare (q � 1)-periodic in k. k = k0 + k1p + . . . + kn�1pn�1
)kp
i = kn�i+kn�(i�1)p+. . .+kn�1pi�1+k0pi+k1pi+1+. . .+kn�(i+1)pn�1
)
hkp
i
q � 1i =
kpi
q � 1)
n�1X
i=0
hkp
i
q � 1i =
n�1X
i=0
kpi
q � 1=
s(k)(1 + p + . . . + pn�1)
q � 1=
s(k) q�1p�1
q � 1=
s(k)
p� 1⇤
Theorem 6.5. Gp(!k, ⇣Trp ) = P(p�1)✓(k,p) = p✓(k,p) as ideals in
Kp�1,p.
Proof. ord��1c P
G(!k, ⇣Trp
) = ordP�cG(!k, ⇣Trp ) = ordPG(!ck
, ⇣Tr
p) =
s(kc) by Theorem 6.3. {�pi} fixes Gp(!k⇣
T
pr) and hence fixes p (this is
the decomposition group of p) ) in the ideal p✓(k,p), ��1c
p occurs with
multiplicityn�1X
i=0
hkcp
i
q � 1i =
s(kc)
p� 1, by Lemma 6.6. Hence in the ideal
P(p�1)✓(k,p), ��1c
P occurs with multiplicity (p� 1)s(kc)
p� 1= s(kc). ⇤
24
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Example. n = 1 and q = p = 7. The only primes above p are P1P2in
K6,7, and thus are the only primes dividing Gp(!k, ⇣p)k Gp(!k,
⇣p)
1 P(p�1)✓(k,p) = P6(h16 i�
�11 +h
56 i�
�15 ) = P1P52
2 P(p�1)✓(k,p) = P6(h26 i�
�11 +h
106 i�
�15 ) = P21P
42
3 P(p�1)✓(k,p) = P6(h36 i�
�11 +h
156 i�
�15 ) = P31P
32
4 P(p�1)✓(k,p) = P6(h46 i�
�11 +h
206 i�
�15 ) = P41P
22
5 P(p�1)✓(k,p) = P6(h56 i�
�11 +h
256 i�
�15 ) = P51P
12
Theorem 6.6. Let C be the ideal class group of Q(⇣m). Then for
all bprime to m,
(b� �b)✓(m)
annihilates C
Proof. (Sketch:) �p = !N(p�1)/m and put ✓(m) =X
c2Z(m)⇥h
c
mi�
�1c
. As a
special case of Theorem 6.5, we obtain the factorization
Gp(�p, ⇣Tr
p) = p✓(m))
Therefore, if b is an integer prime to m, then
Gp(�p, ⇣Tr
p)b��b = p✓(m)(b��b)
✓(m)(b� �b) lies in Z[G]. This gives us a factorization of the
✓(m)(b��b)th power of the Gauss sum in terms of p and its
conjugates in Q(⇣m).In every ideal class there exists an ideal
prime to m (Hilbert).p✓(m)(b��b) is principal for every p 6 | m
⇤
7. Brumer-Stark Conjecture
The Brumer-Stark conjecture is the generalization of the
Stickelbergerideal annihilating the ideal class group of
Q(µm).Conjecture. (Brumer-Stark). Every ideal a of K has the
followingproperty: There exists an element ↵ 2 K satisfying |↵|v =
1 for everyArchimedian place v of K such that a!✓S,K/k = (↵) and
such that theextension K( !
p↵) is Abelian.
This conjecture has recently been proven by Dr. Cristian D.
Popescu.In his seminar’s closing remarks, he mentioned that he is
working on afurther generalization of the Brumer-Stark conjecture
for non-Abelianextensions.
25
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References
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[2] H. Davenport Multiplicative Number Theory, 3rd ed.,
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[4] E. Landau Elementary Number Theory, 2nd ed., Chelsea
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[5] S. Lang Algebraic Number Theory, Addison-Wesley, Reading,
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[7] D. Marcus Number Fields, Springer-Verlag, New York, 1977,
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[8] T. Ono An Introduction to Algebraic Number Theory, Plenum
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26
