by
Jerson Leonardo Caro Reyes
Thesis advisor
Ph.D Guillermo Mantilla-Soler
A thesis presented for the degree of Master of Science in Mathematics
Departamento de Matem´aticas Facultad de Ciencias Universidad de los Andes
Colombia 2016
I would like to thank my advisor Guillermo Mantilla-Soler. First of all, because his office was always open whenever I needed advice and guidance in solving all the problems that arose while writting this thesis. Second for the trust that he gave me, consistently he allowed this thesis to be my own work, but pointed me in the right the direction whenever he thought I needed it.
I would also like to thank my friends and classmates: Daniel ´Avila, Santiago Pinz´on, Hernan Garc´ıa, Julian Forero, Andres Galindo, Nicolas Walteros, Gustavo Chaparro, Edison Lopez, Nicolas Esco-bar, Jorge Mu˜noz, Cesar Venegas, Carolina Herrera and Yeini Montes, who gave me a support in many opportunities when I wanted to put aside mathematics for personal problems.
Finally, I must express my very profound gratitude to my mother and my brother for providing me with unfailing support and continuous encouragement throughout my years of study and through the process of creation of this thesis. This accomplishment would not have been possible without them.
Thank you.
An important objective in Algebraic number theory is the study of number fields and their ring of algebraic integers. One of the crucial arithmetic invariants associated with a number field K is its Dedekind zeta functionζK(s) = Q1−kPk1 −s, where the product runs over all prime ideals
P 6= 0in the ring of algebraic integers OK, andkPk := |OK/P|. This function is the natural
generalization of the Riemann zeta function and gives us arithmetic information about the number field. For example, if we compute its residue at the isolated singularity 1, we get a formula for the order of the class group, in the case of non real quadratic fields.
Then a natural question is: What are some necessary and sufficient conditions over two number fields so that these fields have the same Dedekind zeta function? In this sense, Robert Perlis [Per77] proved that two number fields K and K0 are arithmetically equivalent (i.e., they have the same Dedekind zeta function) if and only if the subgroups H := Gal(N/K) and H0 := Gal(N/K0) ofG := Gal(N/Q)areGassmann equivalents, that is,cG∩H
=
cG∩H0
, for allc ∈ Gand
cG={gcg−1:g∈G}andN the normal closure ofKK0. To show this, he uses an ad hoc process that uses complex analysis. Motivated by this, it arose the idea of addressing the problem via Artin’s L-functions of specific Galois representations.
A good example of the power of Algebraic Number Theory can be seen when it is used to find solutions to the Diophantine equationx2+y2 =z2, withxyz6= 0and pairwise coprimes. For this purpose, we express this equation inZ[i]as(x+yi)(x−yi) = z2. Assuming that this ring is a
Unique Factorization Domain and since(x+yi, x−yi) = 1, we get thatx+yiis a square inZ[i],
that is,x+iy= (p+iq)2, and we obtain
x=p2−q2, y= 2pq, z=p2+q2.
In this sense, and with a similar process, Gabriel Lam´e in March of 1847 gave a “proof” of Fermat’s last theorem using the factorization inZ[ζ], whereζ is an-primitive root of the unity,
xn+yn= (x+y)(x+ζy)(x+ζ2y)· · ·(x+ζn−1y).
However, he assumed that this ring was a UFD. A few years before this, Kummer had already discovered that such unique factorization properties did not necessarily hold in these rings. He in-troduced the notion of ideals in an attempt to salvage the lack of unique factorization. Furthermore, he introduced the class number, which is the order of the quotient of all fractional ideals by the principal ideals, and an analytic formula describing it from the Dedekind zeta function ζK(s)in
cases likeQ(i).
On the other hand, the Dirichlet theorem on arithmetic progressions is also obtained by studying ζK(s)(see Chapter 2 and Chapter 3) and more generally the Tchebotarev density theorem, since
Dirichlet density is defined using the fact that this zeta function has a singularity at1.
Then a natural question is: What are some necessary and sufficient conditions over two number fields so that these fields have the same Dedekind zeta function? In 1977, Robert Perlis showed a characterization of the number fields via the Dedekind zeta function which he calls arithmetic equivalent [Per77]. The proof uses some results in complex analysis and group theory. It begins with two number fieldsKandK0and a Galois extensionN/QwithK, K0 ⊂N, thus, he considers
the subgroupsH:= Gal(N/K)andH0 := Gal(N/K0)ofG:= Gal(N/Q). This situation can be
visualized in the following diagram:
N
H
G H0
K K0
Q
He proves that this fields are arithmetic equivalents if and only if the groupsHandH0have a special property as subgroups ofG, called Gassmann equivalence i.e.,cG∩H
=
cG∩H0
, for allc∈G
andcG ={gcg−1 :g∈G}. In order to show that, he uses a intermediate step using the ramification of the primes inQ.
In 2016 Mantilla-Soler [MS16] improves Perlis’ result using Artin L-series of Galois Representa-tions. In order to achieve this, he weakens the hypothesis in the intermediate step of Perlis’ proof. The objectives of this thesis are the following: First, we show an alternative proof of [MS16], so that it can be understood by people with less expertise in Galois Representations. Second, we give a proof of another characterization of arithmetic equivalence, using the tools stated above.
In this thesis we assume that the reader has a basic knowledge of topics usually covered in a first course in Algebraic Number Theory, but we will recall basic concepts for the reader’s convenience.
The first chapter gives a brief motivation for studying of Dedekind zeta functions. Furthermore, we define some important invariants in basic algebraic number theory and general topics such as Dirichlet series, Artin L-functions, Galois representations, among others.
In the second chapter, it is proved the Class number formula, that is, the formula for the residue at1 of the Dedekind zeta function. For this purpose, we will use lattices to compute the residue of a partial Dedekind zeta function i.e., the sum restricted to all integral ideals in a fixed class in the Class group; this will be useful since such residue does not depend on the class, so, we obtain the class number formula multiplying this residue for the class number hk. Also, we give some
applications of this formula to compute the Class number of a complex quadratic field. Thanks to this result, we can define the Dirichlet density, which is a measure for the prime numbers.
The third chapter uses the Dirichlet density to “count” in a Galois extensionL/Khow many primes in K have their Frobenius automorphism in a fixed conjugacy class of Gal(L/K). This is the Tchebotarev density theorem. We begin by reducing the general case of the theorem to the cyclic case while simultaneously proving it for Cyclotomic extensions. Finally we will prove the theorem for cyclic extension using a weak form of the Dirichlet density theorem.
In the last chapter we sketch the proof of Perlis’ theorem and we give another one using the tools stated above. In the same way we prove a theorem which gives other characterization of arithmetic equivalence, based in the number of primes over a rational prime. This theorem was shown by Stuart, Donna and Perlis, Robert in 1995 [SP95].
1 Preliminaries 6
1.1 Basic Algebraic Number Theory . . . 6
1.2 Invariants of the Dirichlet series . . . 9
1.3 Artin’s L-functions . . . 12
1.3.1 Zeta and L-functions for number fields . . . 13
1.3.2 Artin’s L-functions . . . 15
1.3.3 The conductor . . . 18
1.4 Absolute Galois group and Galois representation . . . 20
2 The Class Number Formula 22 2.1 Class number Formula . . . 23
2.2 Applications to Quadratic Number Fields . . . 29
2.3 Natural and Dirichlet Density . . . 32
3 Tchebotarev Density Theorem 35 3.1 Reduction To The Cyclic Case . . . 35
3.2 Cyclotomic Extension Case . . . 37
3.3 Cyclic Case . . . 39
3.4 Dirichlet density theorem . . . 42
4 Arithmetic Equivalence through Galois representations 43 4.1 On the equationζK(s) =ζK0(s) . . . 46
4.2 A Galois theoretic characterization of arithmetic equivalence . . . 51
Preliminaries
1.1
Basic Algebraic Number Theory
We summarize here the most important notions and results used in this thesis. For basic notation and definitions, the reader can see [Neu13] Chapter I. We begin with a number fieldK, and its ring of integersOK, which is the set of allα ∈ K, such that its minimal monic polynomial belongs
to Z[x]. Since OK is a Dedekind Domain, i.e. is an integrally closed Noetherian domain with
Krull dimension one (every nonzero prime ideal is maximal), this ring has the unique factorization property for prime ideals. We denote bykAk:=|OK/P|, for any integral ideal ofK. So, ifp∈Z
is a prime, then
pOK =Pe11· · ·Pegg,
has a prime decomposition in OK with ramification indices e(Pi/p) := ei and residue degrees
f(Pi/p) = [FPi :Fp], whereFPi =OK/Pi. Then the followingfundamental equalityholds: g
X
i=1
e(Pi/p)f(Pi/p) = [K:Q].
Observation1. We classify some rational primes (i.e. the primes inZ) from their decomposition in
primes ofOK.
• We say thatpis totally split or splits completely ifg, the number of primes inOK over it, is
equal to[K :Q].
• The primepis called inert ifpOK is a prime ideal ofOK.
• The prime p is called ramified if some of these e(P/p) is greater than 1, and it is called unramified in any other case.
Ramified primes are characterized by the following fact: Letx1, . . . , xn∈OK be an integral basis
forK overQ. We define thediscriminantas follows: disc(OK) = det(Tr(xixj)), where Tr(x)
is the trace of the linear operator multiplication byx. Since Tr(OK) ⊂ Z, then Tr(xixj) ∈ Z, so
that disc(OK) ∈ Z. For more details and the proof that disc(OK)is independent of the integral
base, see [Neu13] chapter I, section 2. The prime ideals which ramify are exactly the primes that divide the discriminant. Hence there exist only a finite number of primes that ramify. For quadratic fields, we know the explicit value for the discriminant. LetK=Q(√d), withd6= 1an square free integer, then:
disc(OK) =
d if d≡41
4d if d≡4 3,2
Also the following theorem helps to find the decomposition of a rational prime in a number field.
Theorem 1 (Dedekind Criterion). Let K = Q(α) be a number field, f(x) the minimal monic
polynomial inZ[x]ofα, and take a primep∈Zsuch thatp-[OK :Z[α]]. LetQgi=1hi ei
(x)denote the monic irreducible factorization off(x)inFp[x]. Then the prime factorization ofpOK has the
form
pOK =pe11· · ·p
eg
g
wherepi =hp, hi(α)ifor anyhi ∈Z[x]liftinghi ∈Fp[x]. Moreover, there is an isomorphism of
residue fieldsFp[x]/hi →OK/piviax→αmodpi, so the residue field degreefi(pi/p)is equal to
deg(hi).
For a proof of this theorem the reader can see [Jan96] Theorem 7.4, page 37.
If the extension is Galois, we have some important results with respect to this decomposition. For example ifG:= Gal(K/Q)is its Galois group, thenGacts transitively in the prime ideals ofOK
over each rational primep, hence by unique factorization for each rational prime its ramification indices coincide, and the same occurs for the residue degrees.
Given a Galois extension of number fields L/K andpprime in L, we can define Dp(L/K) and Ip(L/K)the decomposition and inertia subgroups ofG, respectively,
Dp(L/K) :={σ ∈G:σ(p) =p},
and
Ip(L/K) :={σ ∈G:σ(x)≡x(modp)for allx∈OL}.
ClearlyIp(L/K)≤Dp(L/K). Whenever there is no possible ambiguity we denote these, withDp andIp, respectively, ande=|Ip|, whereeis the common ramification index.
Now, we can define the homomorphismπpfromDpontoGal(Fp/Fp)defined by the restriction of
σ ∈DptoOK. It can be shown that this homomorphism is surjective and its kernel isIp[see [Jan96] Chapter III, section 1]. HenceIpDpandDp/Ipis isomorphic toGal(Fp/Fp)which is cyclic. So,
whenpis unramifiedDp is cyclic; then we can define the Frobenius automorphismFrobp(L/K), which is a generator ofDp, and is the unique that satisfies the following equation
Frobp(L/K)(x)≡xkpk(modp), x∈OK.
Whenever there is no possible ambiguity we denote the Froenius element byFrobp.
Example 1. LetK = Q(ζ3). We can classify the prime ideals from its decomposition in three
different sets (by the fundamental equivalence) the ones that ramify, the ones that are inert and the ones that split completely, as follow:
(a) Since disc(OK) =−3then the only rational prime that ramifies is3, hence over3OKthere
is only one primeP, and from the fundamental equality, we have3OK =P2. Because the
norm is multiplicative, we havekPk= 3.
(b) Notice that a prime split completely if and only if its decomposition subgroup is the trivial group. Since the Frobenius automorphism generates this subgroup, we have that p splits completely if and only if
Frobp(L/K)ζ3 ≡ζ3p(modP),
and that happens, if and only ifp ≡ 1 mod 3. Finally, from the fundamental equality, we havepOK =P1P2, withkPik=pfor(i= 1,2).
(c) Finally a prime is inert if and only ifp≡2 mod 3, and we havekpOKk=p2.
On the other hand we can characterize the units in OK, denoted by (OK)∗ from the following
theorem.
Theorem2 (Dirichlet’s Unit Theorem). The group of units(OK)∗is the direct product of the torsion
subgroupµ(K)(the finite cyclic subgroup of all roots of unity inK) and a free abelian group of rankr+s−1.
In other words: there exist unitsξ1, . . . , ξr+s−1, called fundamental units, such that any other unit
ξcan be written uniquely as a product
ξ =ζξe1
1 · · ·ξ
er+s−1
r+s−1
withζ a root of unity and integersei.
For details of the proof the reader can see [Neu13], page 42, Theorem 7.4.
Finally, and also from the fact thatOK is a Dedekind Domain, we can considerIKthe group of all
fractional ideals ofK, andPKthe subgroup of all fractional principal ideals. SetCl(K) =IK/PK
the quotient group called the ideal class group. Cl(K) is finite from lattice theory [see [Neu13] chapter I, section 5], so, we denote byhKits order.
A Brief Motivation
One of first results in arithmetic is the infinitude of the primes. It is well known that Euclides gave a proof obtaining a new prime from a finite list of these. But Euler also gave a proof using the Riemann zeta function presented as a product, as follows: suppose that there are only a finitely many primes, then
ζ(s) =X
n≥0
1 ns =
Y
pprime
(1−p−s)−1, for Re(s)>1
the product on the right is finite and converge for anys6= 0and in the case ofs= 1the sum is the harmonic series, in which case does not converge.
The last proof shows an application of the zeta function. The following definition is a natural generalization of this function.
Definition1. A Dirichlet series is a function of the form
f(s) = ∞
X
n=1
a(n) ns ,
where thea(n)are complex numbers ands=σ+itis a complex variable.
Example2. (a) The Riemann zeta function is a Dirichlet serie takinga(n) = 1for alln.
(b) The function
ζK(s) = X
Aintegral
A6=0
1
kAks,
is a Dirichlet series takinga(n)the number of integral ideals of normn. In the Corollary 1 we will prove that this function is exactly the Dedekind zeta function.
The idea in the following section is to find the convergence range of a Dirichlet series and compute explicitly the residue at the isolated singularity 1 of the Riemann zeta function.
1.2
Invariants of the Dirichlet series
Before starting, we prove the following Lemma from [Jan96], which in particular states that for Dirichlet series such thatP
n≤xa(n)is bounded by a linear function forxlarge enough, this series
is analytic fors >1. This is the case of Dedekind zeta functions.
Lemma1. Letf(s)be a Dirichlet series and let
S(x) =X
n≤x
a(n).
Suppose there exist positive constantsaandbsuch that|S(x)| ≤axbfor allx≥rfor some positive r. Then the seriesf(s)is uniformly convergent in
D(b, δ, ε) :={s:Re(s)≥b+δ,|arg(s−b)| ≤π/2−ε},
for any positiveδ, ε.
Proof. Observe thata(n) =S(n)−S(n−1), so forv≥u+ 1we have
v X
n=u
a(n) ns = v X
n=u
S(n) ns −
v−1
X
n=u−1
S(n) (n+ 1)s
= S(v) vs −
S(u−1)
us +
v−1
X
n=u
S(n)
1 ns −
1 (n+ 1)s
≤ S(v) vs +
S(u−1) us +
v−1
X
n=u
|S(n)|
1 ns −
1 (n+ 1)s
Now notice that
1 ns −
1
(n+ 1)s =s Z n+1
n
dt ts+1.
Since|ns|=nσifs=σ+it,n >0and|S(x)| ≤axb(by hypothesis), we have
v X
n=u
a(n) ns ≤ a
vσ−b +
a uσ−b +
v−1
X
n=u
|s|(anb)
Z n+1
n
df ts+1
,
and note that
v−1
X
n=u
|s|(anb)
Z n+1
n
df ts+1
≤a|s|
Z ∞ n df ts+1−b
= a|s| (σ−b)uσ−b.
From the fact thatv > u >0, andσ−b >0, we set vσ−ba ≤ uσ−ba , then
v X
n=u
a(n) ns ≤ 2a
uσ−b +
a|s|
(σ−b)uσ−b,
now, let θ the argument of s−b, thencos(θ) = |σs−−bb|, and sinceσ −b = Re(s) −b ≥ δ by hypothesis, we get
|s|
σ−b ≤
|s−b|+b σ−b ≤
1 cos(θ) +
b δ,
sinces∈D(b, δ, ε), we have|arg(s−b)|=|θ| ≤π/2−ε, hence there exist a constantM ≥ 1 cos(θ).
Finally for anyε0, you can findnlarge enough such that
2a uσ−b +
a|s|
(σ−b)uσ−b ≤
2a+M+b/δ uσ−b ≤ε0.
Below we show that the Riemman Zeta function has residue 1 at its singularity 1.
Proposition1. Letζ(s)be the Riemman zeta function, then the residue at the isolated singularity 1, is equal to 1, i.e.
lim
s→1+(s−1)ζ(s) = 1.
Proof. Consider the series
ζ(s) 1−21−s=ζ(s)−21−sζ(s) [by lemma 1]
= ∞
X
n=1
1 ns −2
1−s
∞
X
n=1
1 ns =
∞
X
n=1
1 ns −
∞ X n=1 2 2ns = ∞ X n=1
(−1)n−1 ns .
Now we have
lim
s→1+(s−1)ζ(s) = lims→1+
s−1
1−21−s ·(1−2
1−s)ζ(s) = lim s→1+
s−1
1−21−s ·slim→1+(1−2
1−s)ζ(s)
= lim
s→1+
s−1
1−21−s ·slim→1+ ∞
X
n=1
(−1)n−1 ns .
Applying l’Hopital’s rule for the first limit (becauses−1and1−21−sare analytic), we obtain
lim
s→1+
s−1
1−21−s = lims→1+ 1 21−sln(2) =
1 ln(2).
Now Lemma 1 states that the Dirichlet series in the second limit is uniformly convergent inD(0, δ, ε), because the sum of the first n coefficients is 0 or 1. In particular it is analytic in the disk|s−1|< 12, i.e.
lim
s→1+ ∞
X
n=1
(−1)n−1
ns =
∞
X
n=1
(−1)n−1
n = ln(2).
Now, we can characterize the Dirichlet series which converge and agree forssufficiently large.
Lemma2. Let
f(s) = ∞
X
n=1
a(n)
ns andg(s) =
∞
X
n=1
b(n) ns ,
be two Dirichlet series, such that are uniformly convergent andf(s) =g(s)for Re(s)> bfor some b∈R. Thena(n) =b(n)for alln∈Z+.
Proof. Notice that
lim
s→∞f(s) =a(1)and slim→∞g(s) =b(1), thena(1) =b(1), hence
X
n≥2
a(n) ns =
X
n≥2
b(n) ns ,
Multiplying by2s, and lettings→ ∞we obtain thata(2) =b(2). Inductively we get thata(m) = b(m), for allm∈Z+.
Now, following the ideas in [FT93] we will show that the Dedekind zeta function converges for s >1. Before we need the following lemma.
Definition2. Let{bn}a sequence of complex numbers withbn 6= 1, for eachn, and consider the
productQ∞
n=1(1 +bn), it is said toconvergeif the partial products
Qm
n=1(1 +bn) converge to a
non-zero value. We call that this product converges absolutely, ifQ∞
n=1(1 +|bn|) converges. It is
well known that ifQ∞
Lemma3. Let{bn}be a sequence of complex numbers. The productQ∞n=1(1 +|bn|)converges
absolutely if and only if the seriesP∞
n=1bnconverges absolutely.
Proof. We can assume thatbn ≤0. SetPm =Qmn=1(1 +bn)andSn =Pmn=1bnand notice that
{Pn}and{Sn} are monotonic increasing sequences. SincePn ≥ 1 +Sn, we have thatP∞n=1bn
converges ifQ∞
n=1(1 +|bn|)does.
The Taylor expansion of the exponential function says that forb ≥ 0,eb and soeSn ≥ P
n, so, if P∞
n=1bnconverges, thenQ∞n=1(1 +|bn|)is upper bounded.
Lemma4. The Dedekind zeta function
ζK(s) = Y
p
1− kpk−s−1
,
converges fors >1.
Proof. Notice that
1 +kpk−s <1 + (kpks−1)−1
=(1− kpk−s)−1 ≤1 + 2kpk−s,
where the last inequality is obtained from kpk−s ≥ 2. Set (1− kpk−s)−1 = 1 +bp. By the fundamental equality we have that
X
p|p
bp =X p|p
(1− kpk−s)−1−1
≤X
p|p
2kpk−s≤2[K :Q]p−s,
sincekpk−s ≥pand the number the primes ofOKabovepare at most[K:Q], hence
X
p
bp ≤2[K:Q]
X
p
p−s≤2[K :Q]ζ(s),
then P
pbp converges, and for this reason
Q
p(1− kpk −s
)−1) converges because of the Lemma 3.
1.3
Artin’s L-functions
The idea in this section is to define the Artin’s L- functions. This will allow us to write some Dirichlet series like an Euler product. The Dedekind zeta functions are particular cases of Artin’s L-functions.
Definition3. A modulus forKis a formal product as follows
m=Y p
pn(p),
where the product runs over all finite primes (the integral prime ideals), infinite real primes (the real embeddings K ,→ R) and infinite complex primes (one of each pair of conjugate complex embeddingsK ,→ C). Each exponentn(p)is a nonnegative integer, andn(p)6= 0for only a finite
number ofp. In the case thatpis a real infinite prime thenn(p) = 0or1, andn(p) = 0in the case thatpis a complex infinite prime.
Also, we can considermas a productm=m0m∞, where
m0= Y pfinite
pn(p), m∞=
Y
preal pn(p).
For a modulus mofK, letIm be the multiplicative subgroup of fractional ideals ofK relatively prime tom0, i.e.,a ∈Im if and only if there is notp|a, withn(p) > 0. Also we defineUmbe the multiplicative subgroup of principal fractional ideals hαi, withα ∈ K∗ satisfying the conditions α≡1 mod pn(p), and forσa real embedding ofK dividingm,σ(α)>0. Then, we can consider Im/Um, which has finite order, in the same way that the Class Group (for details, see Chapter IV, section 2, Lemma (2.3) [Jan96]).
Example3. (a) Ifn(p) = 0for allp, thenImis exactlyIK, andUmare the principal idealsPK.
ObviouslyIm/Umis the class group.
(b) Letmbe an integer that is either odd or divisible by4, and letm= (mZ)·∞where∞denotes
the real prime ofQ. ThenImis the set of all ideals(a) = Q(p)r(p)where(m, p) = 1. And Um are the principal ideals generated by ` = a/b, with (a, m) = (b, m) = 1, ` > 0 and νp(a−1)≤rifpr|m, r >0. Therefore there is a well-defined map(b/c)→ b
m
·(cm)−1 : Im →(Z/mZ)∗, which one can show induces an isomorphismIm/Um →(Z/mZ)∗.
1.3.1 Zeta and L-functions for number fields
We begin, with a brief motivation of L-functions forQ, the rational numbers.
Example4. Consider the following series
X
n≥0
1
9n2+ 9n+ 2 =
X
n≥0
1 3n+ 1−
1 3n+ 2
.
This is a Dirichlet series; if we consider the nontrivial homomorphismχ: (Z/3Z)∗→ {±1}, then
X
n≥0
1
9n2+ 9n+ 2 =
∞
X
n=1
χ(n) ns ,
where by abuse of notation we denote the extension of χtoZby χ. Such extension is: χ(n) =
Definition 4. Ifk is a integer bigger than 1, then a functionχ(n) is called a Dirichlet character (modk) if it is completely multiplicative, periodic with periodk, and vanishes when(n, k)>1.
An example is the principal character (modk):
χ(n) =
1 if (n, k) = 1
0 otherwise
A Dirichlet characterχ(modk) is called primitive if for every proper divisordofkthere exists an integera≡d1, with(a, k) = 1andχ(a)≤1.
Definition5. Letχbe a nontrivial character of(Z/mZ)∗. We can extendχtoZ+, and by abuse of
notation we denote that extension withχ, as follows:χ(n) =χ(nm)if(n, m) = 1andχ(n) = 0,
otherwise. The Dirichlet series associated toχis defined as follows:
L(s, χ) = ∞
X
n=1
χ(n) ns .
It follows from the Lemma 1 thatL(s, χ)is analytic for Re(s)>0.
In general, letK be a number field,ma modulus forK andχa character of the groupsIm/Um. Notice that we can seeχlike a multiplicative homomorphism ofImwithUmin its kernel, and we defineχ(A)as a value ofχat the cosetAUm. The L-serie associated toχandmis
L(s, χ,m) = X A∈Im
Aintegral
A6=0
χ(A)
kAks,
where the sum runs over all integral ideals prime tom.
Theorem 3. For alls with Re(s) > 1 the function L(s, χ,m) can be expressed as a uniformly convergent product
L(s, χ,m) =Y p-m
1−χ(p) kpks
−1
,
where the product runs over all prime ideals not dividingm.
Proof. Notice that forpa prime ideal we have
1−χ(p) kpks
−1
= 1 + χ(p)
kpks +
χ(p2)
kp2ks +. . . ,
since the character and the norm are multiplicative and the uniqueness of prime decomposition. Let
p1, . . . ,pgall the prime ideals of norm less than or equal tonandpi|mfor alli= 1, . . . , g.
Now consider
Y
i
1− χ(pi) kpiks
−1
= X
ej≥0
χ(pe1
1 · · ·p
eg
g )
pe1
1 · · ·p
eg
g s
=X
A χ(A)
where the sum runs over all integral ideals divisible exactly for somepi’s. Then
L(s, χ,m)− Y
kpk≤n
1−χ(p) kpks
−1
≤ X
kAk>n
χ(A)
kAks
.
Since|χ(A)|= 1(because it is a root of the unity) and by Lemma 1 is uniformly convergent, so is 0whenn→ ∞.
Corollary1. In the case thatχis the trivial character,K =Qandm=∅,L(s, χ,m)is the Euler
product of the Riemann zeta function. In particular
ζK(s) = Y
p
1− kpk−s−1
=X
A
kAk−s.
1.3.2 Artin’s L-functions
In the last section we defined the L-series depending on a modulus and a character of the group Im/Um. This is important when we want to neglect the ramified primes. In this section, we will define the Artin’s L-functions. It is defined locally (at each prime) and the ramified primes are also included. An important result about Artin’s L-functions is that wheneverN/Kis a Galois extension of number fields, then ζN(s) = a·ζK(s), where adepends only on the nontrivial characters of
Gal(N/K).
We begin with a Galois extensionN/Kandψa character ofG= Gal(N/K), then we can define the Artin’s L-function, which is an Euler product:
L(s, ψ, N/K) = Y p∈PK
Lp(s, ψ, N/K),
where Lp (local Euler factor) is defined in the following sense: LetΨ : G → Gl(V) be a finite dimension complex representation ofGwith character ψ. Then ifp is unramified, we define the local factor, as:
Lp(s, ψ, N/K) :=
1
det(Id−Ψ(FrobP))kpk−s) .
In the case thatpis ramified, we takeP∈ PN be any prime overp. Letσ(P/p)be any Frobenius automorphism, i.e.σ(P/p)generatesGal(kP/kp), and define the Euler factor as follows:
Lp(s, ψ, N/K) :=
1
det[(Id−Ψ(σ(P/p))kpk−s)|
VIP]
.
where VIP denotes the vector subspace of V, fixed by the inertia group IP under the action of Ψ :G→Gl(V). In the case thatVIP = 0, we definedet[(Id−Ψ(σ(P/p))kpk−s)|
VIP] = 1.
Example5. (a) LetN/Kbe a Galois extension of number fields, andG := Gal(N/K). Con-sider the trivial character ofG, then
L(s, ψ, N/K) = Y p∈PK
1
(b) LetK =Q(i),G:= Gal(N/K), and consider theψnontrivial character ofGi.e.,ψ(Id) = 1
andψ(σ) =−1, whereσis the complex conjugacy. Then in the case thatpis unramified, we have
Lp(s, ψ, K/Q) = 1
det[(Id−Ψ(Frobp)p−s)]
= 1
1−ψ(Frobp)p−s
,
Notice thatFrobp =Idifpis completely split (p≡41) andFrobp =σifpis inert (p≡43).
On the other hand,2is the only prime that ramifies inK, because disc(OK) = 4, butVI2 = 0,
thenL2(s, ψ, K/Q) = 1, so:
L(s, ψ, K/Q) =
X
n≥0
(−1)n (2n+ 1)s.
(c) ConsiderK =Q(ζ3). By the example 1. If we consider the nontrivial character ofGal(K/Q),
like in the previous example, we have
X
n≥0
1
9n2+ 9n+ 2 =
∞
X
n=1
χ(n) ns .
whereχis the character like in Example 4.
Artin’s Formalism
In this section, we will define the Artin’s L-functions from the algebra of the representations, that is, the addition, restriction and induction.
Theorem4. LetN/Kbe a Galois extension of number fields with groupGandp∈ PK,
(a) For charactersψ1,ψ2ofGwe have:
Lp(s, ψ1+ψ2, N/K) =Lp(s, ψ1, N/K)·Lp(s, ψ2, N/K).
(b) LetN/L/K be a tower of fields withL/K Galois and π : Gal(N/K) → Gal(L/K) the natural projection. Then for any characterψofGal(L/K)and its liftingψ0 =ψ◦πtoGwe have:
Lp(s, ψ, L/K) =Lp(s, ψ0, N/K).
(c) LetN/L/K be a tower of fields,ψa character ofGal(N/L)andψGthe induced character ofG. Then we have:
Y
P∈PL P|p
Proof. For (a) takeΨ1 :G→ Gl(V1)andΨ2 :G→ Gl(V2)representations ofGwith characters
ψ1andψ2, respectively. Then
Lp(s, ψ1+ψ2, N/K) =
1
det(Id−Ψ1⊕Ψ2(F(P/p))kpk−s)
.
Now, we will compute the determinant
det(Id−Ψ1⊕Ψ2(F(P/p))kpk−s) =
IdV1⊕V2−
Ψ1(F(P/p)) 0
0 Ψ2(F(P/p)))
=
IdV1 −Ψ1(F(P/p)) 0
0 IdV2 −Ψ2(F(P/p)))
,
and we get the result.
The proof of (b) is obtained as follows: first, letP/p/pthe tower of primes, then ifpin is nonrami-fied inN is clear thatF(P/p)|L=F(P/p). In the case ofpramify inN, also we get this equality, because of the following exact sequence:
IP/p,−→IP/p→→Ip/p.
For a proof of item (c), see [Kli98].
Theorem5. LetN/Kbe a Galois extension of number fields andG:= Gal(N/K). Then we have:
ζN(s) = Y
ψ∈Gb
L(s, ψ, N/K)dimψ =ζK(s)· Y
ψ6=1G
L(s, ψ, N/K)dimψ.
Proof. Notice that
1G{1}(g) =X
x∈G
˙1{1}(x−1gx) =
|G| if x= 1
0 otherwise
Now by the orthogonality relation among the characters, we have for alla, b∈G
X
ψ∈Gb
ψ(a)ψ(b) =
|G| if a=b−1
0 otherwise
So,
1G{1}(x) = X
ψ∈Gb
ψ(1)ψ(x) = X
ψ∈Gb
Then
ζN(s) =(c)L(s,1{1}, N/N) =L(s,1G{1}, N/K)
=L(s,X
ψ∈Gb
dim(ψ)ψ(x), N/K) =(a)
Y
ψ
L(s, ψ, N/K)dimψ
=ζK(s)· Y
ψ6=1G
L(s, ψ, N/K)dimψ
Thanks to the above theorem we can describe the Dedekind zeta function in terms of the Riemann zeta function.
Example6. ConsiderK =Q(ζ3)and y Example 5 (c) and Theorem 5, we have the Dedekind zeta
function forK =Q(ζ3):
ζK(s) =ζ(s) Y
pprime
(1−χ(p)p−s)−1
Whereχis like in example 4, and by the same example, we have
ζK(s) =ζ(s) X
n≥0
1 (3n+ 1)s −
1 (3n+ 2)s
.
1.3.3 The conductor
In this section we will define the conductor from a representation of the group of a Galois extension, which is a generalization of the discriminant. This invariant is important because it can say when two extensions have the same discriminant. The definitions and facts given here may be consulted in [CF67] and [Ser13].
Definition6 (Ramification groups). LetL/Kbe a Galois extension of local fields with groupG:= Gal(L/K). We define the functioniG :G→ {Z∪ ∞}as follows. Forg∈G, letxbe a generator
ofOLas anOK-module. SetiG(g) =νL(g(x)−x).
Now defineGi for all positive integer numbersiby: g ∈ Gi if and only ifiG(g) ≤ i+ 1. The
groupsGiare called the ramification groups ofG.
Now, with the above conditions and χ be a character of G, then we can consider the following number called the Artin conductor:
f(χ) = ∞
X
i=0
gi
g0
(χ(1)−χ(Gi)),
wheregi =|Gi|andχ(Gi) = g1i Ps∈Giχ(s).
Definition 7 (Global Conductor). LetL/Qbe a finite Galois extension of number fields and let
global conductor ofχ as follows. Letp ∈ Zbe a prime and choose a prime ideal p inLwhich dividesp. LetGp = Gal(Lp/Qp)be the corresponding decomposition subgroup. Letf(χ, p) be
the Artin conductor of the restriction ofχtoGp as defined above. Then
f(χ) =Y
p
pf(χ,p).
The following lemma is a Corollary in [CF67] in page 159.
Lemma5. LetK/Qbe a finite extension and letN/Qthe normal closure ofK. SetH := Gal(/K)
andG:= Gal(N/Q), then:
fψIndG H{1H}
=disc(OK).
In the subsequent section we talk about Galois representation, because we will use the Artin’s L-functions with these representations, which holds the same properties that the representations of finite groups, such as the eigenvalues of each matrix in the image are roots of the unity, all the matrix in the image is diagonalizable, because has a finite image.
Observation2. LetK andK0 be two number fields, with the same normal closure N. SetG := Gal(N/Q),H := Gal(N/K)andH0 := Gal(N/K0). Suppose thatIndGH{1H} ∼= IndGH0{1H0}, then by Lemma 5, disc(OK) =disc(OK0).
Kronecker-Weber Theorem
Furthermore, Artin’s L-series allow us to consider problems with Galois extensions using the char-acters of its group. For example, the Kronecker-Weber theorem. This is one of the most important theorems in number theory of at the end of the 19th century and early 20th century, because it characterizes all abelian extensions.
Theorem6 (Kronecker-Weber Theorem). A number fieldLis an abelian extension ofQif and only
ifL⊂Q(ζm)forζmsomem-th root of unity.
Notice that Theorem 6 is equivalent to prove that Qab, is the extension of Qgenerates for all the
n-roots of the unity forn≥1whereQabis the maximal abelian extension, i.e. the field that contain
all abelian extensions ofQ.
From Artin’s L- series we can consider this theorem as follows:
Theorem7. LetK/Qbe an abelian extension with Galois groupG, letρ:G→C∗be a character
and L(ρ, s) its L-functionas as above. There exists a unique Dirichlet character χ modulo q for someq≥1such that
L(ρ, s, K/Q) =L(χ, s, K/Q).
Sketching the proof, we considerζK(s), like the product in Theorem 5. Then, we setmthe l.c.m.
of the conductors of the characters in such product. So, using the fact thatpis totally split inK, if is totally split inQ(ζm), and using Corollary 3, we getK ⊂ Q(ζm). Kronecker-Weber Theorem
1.4
Absolute Galois group and Galois representation
LetK/Qbe an algebraic infinite extension. We can define its Galois group as an inverse limit, as
follows
Gal(K/Q)∼= lim←−Gal(F/Q),
whereF runs over all the sub extensions ofK/QwithF/QGalois and finite. The isomorphism is
obtained identifying the automorphismσ ∈ Gal(K/Q)with the element(σ|F)F. For a definition
of inverse limit see [DDSMS03]. Now, we define a prime(pF)F ofK overp ∈Q, as a sequence
of primes, one for each finite sub extension ofK such that ifF0 ⊂ F thenpF is overpF0. In this sense, we can define the decomposition and inertia subgroups of a primepofK, as follows
Dp= lim
←−DpF, Ip = lim←−IpF.
WhenK =Qthe algebraic closure ofQwe call Gal(K/Q)the absolute Galois group. Now, we
define profinte groups because it facilitates the topology ofGal(Q/Q).
Definition 8. A profinite group is a compact Hausdorff topological group whose open subgroups form a base for the neighborhoods of the identity, that is, every open set containing the identity contains an open subgroup.
The following Proposition gives us a characterization of the inverse limit from profinite groups. For a proof the reader can see [DDSMS03] proposition 1.3, page 17.
Proposition2. IfGis a profinite group, thenGis (topologically) isomorphic to
lim
←−(G/N)N,
where the limit runs over all normal subgroups with finite index onG. Conversely, the inverse limit of any inverse system of finite groups is a profinite group.
Galois representation
Definition9. A Galois representation is a continuous homomorphism:
ρ: Gal(Q/Q)−→GLn(k),
wherekis an algebraically closed field (C,Qp,Fp). In the case thatk = Cwe call it an Artin’s
representation, ifk=Fpit is called discrete, and ifk=Qpis calledp-adic.
Now, we will prove that any Artin’s representation has finite image.
Lemma6. LetGbe a Lie group (for a definition see [DK12]), then there exists a neighborhoodU of the identity inGsuch thatU does not contain any non trivial subgroup ofG.
Proof. letgbe the Lie algebra ofG. We know thatT0(exp)(X) = XforX ∈g, that is,T0(exp)
is equal to the identityg → g. By the inverse mapping theorem, we know that: There is an open neighborhoodU0 (we may choose it with finite diameter) of0ingandV0 ofeinG, such that the
exponential map is aC1 diffeomorphism fromU0 toV0. SetV = exp(U0/2). We claim thatV is the desired neighborhood. Otherwise ifx ∈ V is not the identity, we have thatx = exp 12u
for someu∈U0. Then,
xn= exp
1 2u
· · ·exp
1 2u
(ntimes),
for any positive integern. Now, givenv, we can findN such that N2u ∈ U0\ 1 2U
0. Then,xN ∈
exp(V0)\exp(12U0) =V \V0.
Observation 3. Notice that Lemma 6 states that any profinite group G that is also Lie group is finite; indeed by definition, any neighborhood of the identity must contain an open subgroup ofG, so{e}is open. SinceGis a topological group,{x}is open for anyx∈G, henceGhas the discrete topology. And sinceGis compact, thenGis finite.
Theorem8. LetGbe a profinite group and letρ :G→GLn(C)be a continuous homomorphism
thenρhas a finite image.
Proof. Sinceρ:G→GLn(C)is continuous and the kernel ofρis the pre-image of{e}(a closed
subset becauseGLn(C)is Hausdorff),ker(ρ)is closed, so, the quotient is also a profinite group,
and we may assume ρ is injective. Nowρ : G → ρ(G)is a bijection. Since Gis compact and ρ(G)is Hausdorff with the subspace topology, thenρ : G → ρ(G)is an homeomorphism. Also, we know thatρ(G)is compact in the topology ofGLn(C)(because it is compact with the subspace
topology), hence is closed, since GLn(C) is Hausdorff. Then ρ(G) is a Lie subgroup with the
subspace topology. Thus, Ghas the same topology ofρ(G), hence it is a profinite group, and by Observation 1 we conclude that is finite.
The Class Number Formula
An important result in Algebraic number theory is the class number formula. It computes the residue at the isolated singularity 1 of a Dedekind zeta function. This residue is important because it gives a relation between the Dedekind zeta function ofKand some invariants of this field such as the order of the Class group ofOK, the determinant and other invariants. The proof that we will present in
this section was taken from [Jan96] and [Jar14].
We begin with the definition of a lattice, and its volume [Neu13], which measures in some sense the number of points in the lattice. Furthermore, we define the regulator, which is the volume of the units inOK.
Definition10. LetV be ann-dimensional R-vector space. AlatticeinV is an additive subgroup
of the form:
Γ =Zv1+· · ·+Zvm
withv1, . . . , vmlinearly independents inV. The set{v1, . . . , vm}is called abasisand the set
Ψ ={x1v1+· · ·+xmvm :xi ∈R,0≤xi <1}
thefundamental meshof the lattice. The lattice is calledcomplete, ifm=n, in whose case we can associate a volume toΓ, as follow:
vol(Γ) =vol(Ψ) =|detA|
whereAis the matrix of the basis change frome1, . . . , entov1, . . . , vn.
Definition11 (The regulator ofK). By Theorem 2, we know that(OK)∗ 'Zr+s−1×µ(K). Set
|µ(K)|=ωK. Then we can consider(OK)∗ like a complete lattice insideRr+s−1, as follows: let
ξ1, . . . , ξr+s−1 be the fundamental units. Remember that the norm of a real embedding is given by
kαki =|σi(α)|for1≤i≤r, and the norm of a complex embedding is given bykαkr+j =|τj(α)|2
for1≤j ≤s. Consider the following matrix
A=
logkξ1k1 logkξ2k1 · · · logkξr+s−1k1
logkξ1k2 logkξ2k2 · · · logkξr+s−1k2
..
. ... . .. ...
logkξ1kr+s logkξ2kr+s · · · logkξr+s−1kr+s
,
of order(r+s)×(r+s−1). We defineAito be the(r+s−1)×(r+s−1)minors obtained
by subtracting thei-th row of A. Then the regulator ofK, denoted reg(K), is given by|detAi|,
which is independent ofi. For details, see [Neu13], page 43, Proposition 7.5.
In other words, we considerΓ⊂Rr+sa lattice associated to(O
K)∗with base
vi= (logkξ1k1,logkξ2k1, . . . ,logkξr+s−1k1) [i= 1, . . . , r+s−1]
then if we take the image of Γ via thei-th natural projection ofRr+s toRr+s−1 then we have a
complete lattice inRr+s−1, and its volume is reg(K).
In the case thatr+s−1 = 0, i.e. K =Q(
√
d), withd < 0and squarefree orK =Q, we have
that reg(K) = 1.
2.1
Class number Formula
Let K be an algebraic number field, with r real embedding σ1, . . . , σr and s pairs of complex
embeddings τ1, τ1, . . . , τs, τs. For each nonzero integral ideal A(ideal of OK) we have defined
kAk =|OK/A|. The unique factorization of ideals and the fundamental equality, imply that there
is only a finite numberaK(n)of integral ideals with normn. Then, by Lemma 1 the Dedekind zeta
function converges fors >1we can also write this as follows
ζK(s) =
Y 1
1− kPk−s =
X
n≥1
aK(n)
ns .
We begin by defining theζ-function of a cosetkof the class groupCl(K)
ζK(s,k) = X
A∈k
Aintegral
1
kAks =
X
n≥1
a(n,k) ns ,
wherea(n,k)is the number of integral ideals inkhaving norm exactlyn. Notice that
ζK(s) = X
k∈Cl(K)
X
A∈k
Aintegral
1
kAks
= X
k∈Cl(K)
ζK(s,k),
and choosea∈ k−1 such thata ⊂ OK (it is well known that each class ofCl(K)has an integral
ideal [Jan96] page 146, Lemma 2.3), so thataAis principal, for allA∈k. Then multiplication by
agives a bijection between integral ideals inkand principal ideals divisible bya. Thus
ζK(s,k) =kaks X
α∈K∗
a|hαi 1
khαiks. (2.1)
First, we define the spaceLr,sto be the set of points(x
1, . . . , xr, xr+1, . . . , xr+s)where the firstr
coordinates are real and the remainingsare complex, which has dimensionr+ 2s=n. Notice that
Lr,sis a vector subspace of
Rn. With scalar multiplication as well as component wise addition and
multiplication of points, this forms a commutative ring and a linear space.
Finally, we define a norm on Lr,s as follow N(x) = |x
1· · ·xr| |xr+1|2· · · |xr+s|2. And we can
consider the injectionφ:K→ Lr,sdefined by
φ(α) = (σ1(α), . . . , σr(α);τ1(α), . . . , τs(α));
Clearlyφis a homomorphism of commutative rings, andN(φ(α)) =khαik.
Now, we need to rewrite the sum (2.1) in terms ofLr,s, as follows: LetAbe the set of theαsuch
that a| hαi, where from each possible set of associate values (i.e. allα that generates the same principal ideal) we select one. DefineΓ = φ(a) = {x ∈ Lr,s : x = φ(b)for someb ∈ a}, and
Θ ={x∈ Lr,s:x=φ(b)for someb∈ A}. Then
ζK(s,k) =kaks X
α∈Θ
1
N(α)s. (2.2)
So, the idea to compute the residue at the isolated singularity 1 of a Dedekind zeta function is to reform (2.2) in terms ofΓ. Then using Lemma 8 we will be able to compute such residue.
In order to achieve this, we denotelk(x) = log|xk|for1 ≤k ≤randlr+k(x) = log|xr+k|2 for
1 ≤ k ≤ s; we may then define for x ∈ Lr,s the vectorl(x) = (l
1(x), . . . , lr+s(x)). The set of
all points ofLr,swith nonzero components, form a group under componentwise multiplication, and
this mapping is a homomorphism onto the additive groupLr,s.
Ifα∈ K, then writel(α) = l(φ(α)). This geometric representationl(α)is called the logarithmic representation ofα, and the sum of its components is equal tolog|kαk|.
Letξ1, . . . , ξr+s−1 be the fundamental units. Defineλ= (1, . . . ,1; 2, . . . ,2). Then{λ, l(ξ1), . . . ,
l(ξr+s−1)}is a basis forRr+s, for more details the reader can see [Jan96], page 148. Forx∈ Lr+s,
we can writel(x)as follows:
l(x) =cλ+c1l(ξ1) +· · ·+cr+s−1l(ξr+s−1),
wherec= n1log(N(x)), since the sum of the components inl(x)islog(N(x)), and the sum of the components inl(ξj)islogkξjk= 0.
Now we need to give the definition of a cone:
Definition 12. LetV be an n-dimensional R-vector space. A set X ⊂ V is called a cone if for
x∈Xandα >0impliesαx∈X.
DefineXto be the cone consisting of allx∈ Lr,ssuch that:
(a) N(x)6= 0.
(b) 0≤ci <1fori= 1, . . . , r+s−1.
This is a cone because forα >0, we havel(αx) = (logα)λ+l(x), hence the coefficient ofl(ξi)
belongs to[0,1), andarg(αx1) = arg(x1).
Then the following Lemma states thatΘ = Γ∩X.
Lemma7. Let[α]⊂ OK be the set of all elements inOK which generatehαi. Then exactly one
member of[α]has image inX.
Proof. Since for allβ∈[α]there existsξ ∈Ok∗, such thatβ =ξα, then to prove this, is enough to show that giveny ∈ Rnwith nonzero norm,y can be written uniquely asx·φ(ξ), wherex ∈ X
(multiplication is componentwise) and ξ is a unit. Like above, we can put l(y) as the following sumcλ+c1l(ξ1) +...+cr+s−1l(ξr+s−1). For eachi, we can writeci equal tomi+µi , where
mi ∈ Z and0 ≤ µi < 1, and write u = ξ1m1· · ·ξrm+rs+−s−11. Then define z = y ·φ(u
−1), and
notice that the coefficients ofzfor eachl(ξi)are in the correct range. Now we can correctarg(z1);
let r be the unique integer such that0 ≤ arg(z1) − 2ωπrK < ω2Kπ, and choose a root of unityω
such that ρ1(ω) = e
2πi
ωK, where ρ1 gives the first component of the map φ. Then z·φ(ω−r) = y·φ(u−1)φ(ω−r) =: x ∈ X, theny = x·φ(uωr) as desired. The uniqueness follows from the construction.
We now use the result of Lemma 7 to rewrite:
ζK(s,k) =kaks X
x∈Γ∩X
1
N(x)s,
which we may evaluate from the following Lemma.
Lemma8. LetXbe a cone inRnandF :X →R>0be a function such thatF(ξx) =ξnF(x)for
x∈Xandξ ∈R>0, andF ={x ∈X :F(x)≤1}is bounded with vol(F) >0. LetΓ ⊂Rnbe
a complete lattice with volume vol(Γ). Suppose that
ζΓ(s) =
X
x∈Γ∩X
1 F(x)s,
converges on Re(s)>1, thenlims→1(s−1)ζΓ(s) = vol(
F)
vol(Γ).
Proof. Letrbe a positive real number, we know vol(1rΓ) = r1nvol(Γ).
The volume of F may be computed in the following way. For a real number r > 0 consider the points of 1rΓ∩ F. Using each such point as center, place an-dimensional parallelepiped equal to a translate of a fixed fundamental mesh of 1rΓ. LetT1(r)be the number of these parallelepipeds
contained inF. Their union is a polyhedron of volume r1nvol(Γ)T1(r)which serves to approximate
F from the inside. Thus, the volume ofF is the limit of r1nvol(Γ)T1(r)asr→ ∞.
Now we want to approximateFfrom the outside. LetT2(r)be the number of parallelepipeds with
center at some point of 1rΓand having nonempty intersection withF. The polyhedron consisting of these parallelepipeds containingF and the volume ofF is the limit ofr1nvol(Γ)T1(r)asr → ∞.
Finally letT(r)be the number of points of 1rΓ∩ F. ThenT1(r) < T(r) < T2(r)and it follows
that:
vol(F) =vol(Γ) lim
r→∞
1rΓ
∩ F
rn ,
then
vol(F)
vol(Γ) = limr→∞
1rΓ
∩ F
rn .
But by definition ofF,
1 rΓ ∩ F
=|{x∈Γ∩X:F(x)≤rn}|.
SinceF is bounded, for a fixrthe set 1rΓ
∩ F is finite. Therefore, we can list the points ofΓ∩X such that0 < F(x1)≤F(x2)≤. . .and definerk=F(xk)1/n. If we defineγ(r) =
1rΓ
∩ F,
then we have that forε >0,γ(rk−ε)< k≤γ(rk) :=|{i∈Z+:F(xi)≤F(xk)}|. Dividing by
rnk gives
γ(rk−ε)
(rk−ε)n
rk−ε
rk n
< k rnk ≤
γ(rk)
rnk .
Sincernk =F(xk), we havelimk→∞rkn k =
vol(F)
vol(Γ).
Now givenε >0, by the above inequality there existsk0such thatk≤k0implies
vol(F) vol(Γ) −ε
s
1 ks <
1 F(xk)s
<
vol(F) vol(Γ) +ε
s
1 ks.
Ifs→1+, in the sum over allk≥k0multiply by(s−1), and sincelims→1(s−1)ζ(s) = 1we get
lim
s→1
vol(F) vol(Γ) −ε
s
≤ lim
s→1(s−1)ζΓ(s)≤slim→1
vol(F) vol(Γ) +ε
s
,
and sinceεis arbitrary, we have
lim
s→1(s−1)ζΓ(s) =
vol(F) vol(Γ).
We thus need compute vol(F) =vol({x ∈X : kxk ≤ 1})and vol(Γ) = vol(φ(a)) = vol({x ∈ Lr,s :x=φ(b)for someb∈a}), because Lemma 8 states that
lim
s→1(s−1)ζK(s,k) =kak ·
vol(F) vol(Γ).
Now we will compute separately vol(F)and vol(Γ).
Proof. Sinceahas rankn, then is generated byα1, . . . , αn, so thatΓis generated byφ(α1),
· · · , φ(αn). LetB be the matrix with entries(σiαj), whereσi varies over all embeddings
(real and complex) of K. Then disc(a) = det(B)2 = kak2disc(OK). Also, let C be the
matrix consisting of n inner products (hφ(αi), φ(αj)i) = (Pnk=1σk(αi)σk(αj) = BTB.
Thus|detC|1/2 = |detB|, and since vol(Γ) = |detC|1/2 = disc(a), we have vol(Γ) =
kak |disc(OK)|1/2.
(b) vol(F) = 2r+sπωsreg(K)
k .
Proof. DefineFk for 0 ≤ k < ωK by applying the mapx 7→ e
2kπi
ωK x toF; since multi-plication by a n-th root of unity preserve volume, we have vol(F) = vol(Fk). LetFe =
SωK
k=0FK
∩ {(x1, . . . , xr;xr+1, . . . , xr+s) :x1, . . . , xr>0}. Multiplying any point inFe
by(±1, . . . ,±1; 1, . . . ,1)shows that vol(F) = ω2r
Kvol(Fe), and so we will compute vol(F) through multiple changes of variable.
First, we change from the (r +s)-dimensional complex space Lr,s to
Rn via the
trans-formation which maps a point (x1, . . . , xr;xr+1, . . . , xr+s) ∈ F to the real valued point
(ρ1, . . . , ρr, ρr+1, ϕr+1, . . . , ρr+s, ϕr+s), whereρj =|xj|andϕj = argxj for allj(we say
xj =yj+izj =ρjeiϕj). Now we compute the Jacobian of the transformationρs+1· · ·ρr+s.
ThenFeis given by the conditionsρ1, . . . , ρr+s >0;Qrj=1+sρ ej
j ≤1, whereej is thej-th
co-ordinate ofλ= (1, . . . ,1; 2, . . . ,2); and0≤γk <1in the formula for eachj-th coordinate
ofl(x):
logρej
j =
ej
n log
r+s Y
k=1
ρek
k !
+
r+s−1
X
k=1
γklj(ξk).
These conditions do not restrictϕjfor any valuer+ 1≤j≤r+s, so they take on all values
in [0,2π). We now change variables again, replacing ρ1, . . . , ρr+s with γ, γ1, . . . , γr+s−1
according to
logρej
j =
ej
n logγ+
r+s−1
X
k=1
γklj(ξk).
Since the sum of theej isn, andPrj=1+slj(ξk) = 0, we sum all the above equation and find
γ = Qrj=1+sρej
j . ThusFe is now defined by the conditions0 < γ ≤ 1and0 ≤ γk < 1for
1≤k≤r+s; clearly this set has positive volume now. This transformation has Jacobian
J = ρ1 nγ ρ1
e1l1(ξ1) · · ·
ρ1
e1l1(ξr+s−1)) ..
. ... . .. ...
ρr+s
nγ ρr+s
er+slr+s(ξ1) · · ·
ρr+s
er+slr+s(ξr+s−1))
= ρ1· · ·ρr+s nγ2s
e1 l1(ξ1) · · · l1(ξr+s−1))
..
. ... . .. ... er+s lr+s(ξ1) · · · lr+s(ξr+s−1))
= ρ1· · ·ρr+s nγ2s
n 0 · · · 0
e2 l2(ξ1) · · · l2(ξr+s−1))
..
. ... . .. ... er+s lr+s(ξ1) · · · lr+s(ξr+s−1))
.
This determinant is now exactlyn·reg(K), so
J = ρ1· · ·ρr+s
n(ρ1· · ·ρsρ2s+1· · ·ρ2r+s)2s
n·reg(K) = reg(K) (ρs+1· · ·ρr+s)2s
.
Now, we can compute the volume ofFe:
vol(Fe) = 2s Z
· · ·
Z
e
F
dx1· · ·dxrdyr+1dzr+1· · ·dyr+sdzr+s
= 2s
Z
· · ·
Z
e
F
ρr+1· · ·ρr+s·dρ1· · ·dρrdρr+1dϕr+1· · ·dρr+sdϕr+s
= 2s(2π)s
Z 1
0
· · ·
Z 1
0
ρr+1· · ·ρr+s· |J|dγdγ1· · ·dγr+s−1
= 2s(2π)sreg(K)
2s = (2π)
sreg(K),
Thus vol(F) = ω2r
Kvol(Fe) =
2r+sπsreg(K)
ωk as desired.
Using (a) and (b), we obtain the next result:
Theorem9 (The class number formula). LetζK(s)the Dedekind zeta function for the number field
K, then
lim
s→1+(s−1)ζK(s) =
2r+sπsreg(K) ωK(|disc(OK)|)1/2
hK,
whererandsare the numbers of real and complex primes ofKrespectively, reg(K)is the regulator of K (i.e. the volume of the(r+s−1)-dimensional lattice of units),ωK is the number of roots of
1 in K, disc(OK)is the discriminant ofK/Q, andhkis the class number.
More generally, but in the same sense, the residue of
ζK(s,k) = X
A∈k
Aintegral
1
wherekis a coset ofUmoverImwithma modulus forK. Is given for:
gm= 2
r+sπsreg(m)
ωmkm0k |disc(OK)|
,
where
reg(m) =vol((OK)∗∩Km,1)
ωm= number of unity in(OK)∗∩Km,1,
withα∈Km,1 if and only if satisfies the conditionsα≡1 modpn(p), and forσa real embedding
ofKdividingm,σ(α)>0.
Proposition3. Letχbe a character of the class groupIm/Um, then
lim
s→1(s−1)L(s, χ) =
hmgm if χ=χ0
0 otherwise
wherehmis the order of the class groupIm/Um.
Proof. Since
L(s, χ) =X
k
χ(k) X A∈k
Aintegral
kAk−s=X
k
χ(k)ζ(s,k),
hence
lim
s→1(s−1)L(s, χ) =
X
k
χ(k)gm,
and the result is obtained from the orthogonality relation of the characters.
2.2
Applications to Quadratic Number Fields
In this section we use the class number formula to compute the class number for quadratic extension. For instance, we know now that the only imaginary quadratic fieldsQ(
√
d),dsquarefree andd <0, which have class number1are those with
d=−1,−2,−3,−7,−11,−19,−43,−67,−163.
Example7. ConsiderK =Q(ζ3). In this case reg(K) = 1, the class number formula states
lim
s→1+(s−1)ζK(s) =
2r+sπsreg(K) ωK(|disc(OK)|)1/2
hK
= 2π
6(3)1/2hK =
π
By Example 6 we have
ζK(s) =ζ(s) X
n≥0
1 (3n+ 1)s −
1 (3n+ 2)s
. (2.4)
By (2.3) and (2.4), we obtain π
3√3hK = lims→1+(s−1)ζK(s)
= lim
s→1+(s−1)ζ(s)
X
n≥0
1 (3n+ 1)s −
1 (3n+ 2)s
.
Because this sum is a Dirichlet series and the sum of the first n-th partial sum is 0 or 1, by part (a) of Lemma 1, it is uniformly convergent inD(0;δ, ε). In particular, it is analytic in the disk|s−1|< 12. Then
π
3√3hK = lims→1+(s−1)ζ(s)
X
n≥0
1 (3n+ 1)s −
1 (3n+ 2)s
= lim
s→1+(s−1)ζ(s) lims→1+
X
n≥0
1 (3n+ 1)s −
1 (3n+ 2)s
=X
n≥0
1 3n+ 1−
1 3n+ 2
.
Consider
1
x2+x+ 1 =
1−x
1−x3 = (1−x)
X
n≥0
x3n=X
n≥0
x3n−x3n+1,
then 2 √ 3tan −1
2x+ 1
√
3
=
Z dx
x2+x+ 1 =
Z X
n≥0
x3n−x3n+1
dx
=X
n≥0
x3n+1
3n+ 1− x3n+2
3n+ 2
.
Now if we takex→1, we have that
X
n≥0
1 3n+ 1−
1 3n+ 2
= √2
3tan 3 √ 3 = π 3√3,
thenhK = 1, so,OKis a PID.
Example8. ConsiderK =Q(i). In this case reg(K) = 1, the class number formula tells us
lim
s→1+(s−1)ζK(s) =
2r+sπsreg(K) ωK(|disc(OK)|)1/2
hK
= 2π
4(4)1/2hK=
π
By Example 5 (b) and Theorem 5, we have the Dedekind zeta function forK =Q(i):
ζK(s) =ζ(s) X
n≥0
(−1)n
(2n+ 1)s. (2.6)
By (2.5) and (2.6), we obtain
π 4hK =
X
n≥0
(−1)n (2n+ 1)s.
And is well known that
X
n≥0
(−1)n (2n+ 1)s =
π 4,
thenhK = 1, soOKis a PID.
In the same way, we can generalize this process to quadratic extensions (different to the above examples), as follow
hK =−
1
|disc(OK)|
|disc(OK)|−1
X
n=1
disc(OK)
n
·n,
wheredisc(OK)
n
is the extended Jacobi symbol. For example, ifK =Q(
√
−163), since disc(OK) =
−163we have
hK =−
1 163
162
X
n=1
−
163 n
·n,
and implementing the following code in MAGMA, to compute the sum:
sum:=0;
for i in [1..162] do
sum:=sum+KroneckerSymbol(i, -163)*i; sum;
end for;
we obtain that this sum is equal to−163, so,hK = 1. In the case thatd≡4 3,2we use the code:
sum:=0;
for i in [1..|d|-1] do
sum:=sum+KroneckerSymbol(i, 4d-i)*i; sum;
end for;
d disc(OK) sum
−2 −8 −8
−7 −7 −7
−11 −11 −11
−19 −19 −19
−43 −43 −43
−67 −67 −67
Table 2.1: Quadratic number fields withhK= 1
2.3
Natural and Dirichlet Density
In this section the idea is to define a measure for the prime ideals of a number fieldK. We begin with the definition of Natural density and the Dirichlet density, which coincide when the natural density exists. So, we use the word “density”, for referring to Dirichlet density.
This concept is important because the Tchebotarev Density Theorem, which computes the density of primes in a number field K, whose Frobenius automorphism are in a fixed conjugacy class in Gal(L/K), whereL/K is a Galois extension of number fields.
Definition13 (Natural density). LetKbe any number field andAany set of prime idealsP6={0}
inOK. Then its Natural densityδN at(A)is defined as the following limit (if it exists)
δN at(A) = lim x→∞
|{P∈ A:kPk ≤x}| |{P∈ PK :kPk ≤x}|.
Example9. A nice example is an infinite subset of prime numbers with natural density0.
Setpnthen-th prime number, and consider the setS={pi|i=n2, n∈Z+}. By definition
δN at(S) = lim n→∞
|Sn|
|Pn|
,
wherePnare the primes less or equal thann, andSnare the primes inSwith the same condition.
So
lim
n→∞
|Sn|
|Pn|
= lim
n→∞
p
|Pn|
|Pn|
= 0.
Now, we define an important mathematical object which gives us information about the proportion of a subset of the prime numbers.
Notice that
ln(ζK(s)) = ln
Y
P∈PK
(1− kPk−s)−1
=− X
P
ln(1− kPk−s) =X P
X
m≥1
1 mkPksm.
And the last summation whenm >1converges fors= 1, hence
ln(ζK(s))∼ X
P
for the same reason we can neglect those prime idealsPwithkPk=pf ≥p2and get
ln(ζK(s))∼ X
P∈PK
f(P/p)=1
kPk−s,
and since from the class number formula we have
ζK(s)∼
2r+sπsreg(K) ωK(|disc(OK)|)1/2
hK·
1 s−1,
then
X
P∈PK
f(P/p)=1
kPk−s∼ln(ζK(s))∼ln
1 s−1
.
This allow us to define the Dirichlet density.
Definition14 (Dirichlet density). LetKbe any number field andAany set of prime idealsP6={0}
inOK. Then its Dirichlet densityδ(A)is defined as the number that grants the following asymptotic
equivalence (if it exists)
δ(A) X P∈A
f(P/p)=1
kPk−s∼δ(A)X P∈A
kPk−s∼ X
P∈PK
kPk−s∼log
1 s−1
.
It is well known that the existence ofδN at(M) implies the existence of δ(M), and that one has
δN at(M) = δ(M), see [Neu13], page 543. So that when we will mention the density of a set of
prime ideals, we assume that it is the Dirichlet density.
Lemma9. LetL/K be a Galois extension of number fields. Then the density of the primes inK that split completely inLis1/[L:K].
Proof. ConsiderSpl(L/K)the set of all prime ideals in PK such that split completely in L and
notice that over eachp∈Spl(L/K)there are exactly[L:K]primes. LetAbe the set of all primes ofPL, above some prime inSplS(L/K). Then:
ln
1 s−1
∼ X
P∈PK
f(P/p)=1
kPk−s= [L:K]X p∈A
kpk−s.
Thenδ(Spl(L/K)) = 1/[L:K].
Corollary2. Givenm∈Z+There exist infinitely many prime numberspsuch thatp≡1 (modm).
Proof. TakeK =Q(ζm), thenpis totaly split inK if and only ifFrobp is the identity. Since the
Frobenius automorphism is characterized by the equation xp ≡ Frobp(x) (modp), with p|p and
x∈OK, thenpis totaly split if and only ifp≡1(modm). So, the set of all primes satisfying that
Corollary3. Suppose thatL1andL2are Galois extensions ofK, andSis a finite subset of primes
inK. Then L1 = L2 if and only ifSplS(L1/K) = SplS(L2/K), whereSplS(Li/K) = {p ∈
PK:pis completely decomposed inLiandp /∈S}.
Proof. Is clear that ifL1 = L2 thenSplS(L1/K) = SplS(L2/K). On the other hand, consider
the Galois extensionL = L1L2, and the injective homomorphism Gal(L/K) → Gal(L1/K)×
Gal(L2/K), sendingσ ∈Gal(L/K)to(resL1(σ),resL2(σ))∈Gal(L1/K)×Gal(L2/K). Now, letPa prime inL1L2unramified inK, and letPi =P∩Liwithi= 1,2and notice that the
above homomorphism sendsFrobP(L/K)toFrobP1(L1/K)×FrobP2(L2/K). Suppose thatP
split completely, then the decomposition subgroupDPis trivial, so we have
SplS(L/K) =SplS(L1/K)∩SplS(L2/K) =SplS(L1/K) =SplS(L2/K).
By Lemma 9, we have thatδ(SplS(L/K)) = 1/[L:K], then
1 [L:K] =
1 [L1 :K]
= 1
[L2 :K]
⇒[L:K] = [L1 :K] = [L2 :K],
henceL1=L2.
An important result in basic algebraic number theory is:
Theorem10. LetL/Kbe a finite (not necessarily Galois) extension of number fields andN/Kthe normal closure ofL/K. Then the prime idealpis totally split inK if and only if is totally split in N.
For a proof of this theorem the reader can see [Neu13], page 58. Now, we can prove the following corollary.
Corollary4. Suppose thatL1andL2are finite extensions ofK, andSis a finite subset of primes
inK. ThenL1andL2have the same normal closure if and only ifSplS(L1/K) =SplS(L2/K).
Proof. IfL1andL2have the same normal closureN, then for a prime idealpofOKis totally split
inL1if and only if it is so inNby 10, and this happens if and only ifpis totally split inL2. On the
other hand, letN1andN2 be the normal closures ofL1 andL2, respectively. Then by Theorem 10
Tchebotarev Density Theorem
Theorem11 (Tchebotarev Density Theorem). LetE/Kbe a Galois extension of number fields with Galois groupGand letCbe a conjugacy class inG. Then the set
S :={p∈ PK : FrobP(E/K)∈ Cfor someP|p}
={p∈ PK :pis not ramified inE,Frobp(E/K) =C},
has densityc/|G|, where|C|=candFrobp(E/K) ={FrobP(E/K) :P|p}.
3.1
Reduction To The Cyclic Case
The idea of the proof is to reduced to the case of cyclic extensions, next we prove the theorem for cyclotomic extension, and finally using the last case we prove it in the cyclic case.
Proof. (of Theorem 11). Suppose the theorem holds for cyclic extensions. Letσ ∈GandLbe the field fixed byhσi. HenceE/Lis a cyclic extension of degree|σ|. By assumption the set
Sσ :={p∈ PL: FrobP(E/L) =σ for someP|p},
has density 1/|σ|. Now, notice that the set S0 := {p∈Sσ :f(p/p∩K) = 1} also has density
1/|σ|. We may defineSfromS0 as follows:
S=
p∈ PK:p∈S0for somep∈ PLsuch thatp|pnonramified , (3.1)
First, we show that for P ∈ PE it holds thatFrobP(E/K) = σ if and only if FrobP(E/L) = σandf(P∩L/p) = 1.
The left implications is clear because
FrobP(E/K)f(P∩L/p)= FrobP(E/L). (3.2)
On the other hand,σgeneratesGal (FP/Fp). And sinceσ∈Gal(E/L), this is possible if and only
if Fp = Fp, becauseσ also generatesGal (Fp/Fp), hence f(P∩L/p) = [Fp : Fp]. And again
using (3.2) we obtain the result. Now, notice that
p∈S⇐⇒phas a prime factorPinEwith FrobP(E/K) =σ
⇐⇒FrobP(E/L) =σandf(P∩L/p) = 1withP∈ PE,P|p
⇐⇒phas a prime factorp∈L, withp∈S0,
hence, we have (3.1).
Now, definerp =|{p∈S0 :p|p}|and notice that
δ(S0) = X p∈S0
kpk−s =X
p∈S X
p∈S0
kpk−s=X
p∈S
rp·p−s. (3.3)
Now we prove thatrp= c|·|Gσ||for allp∈S.
Notice thatrp = |{P∈ PE :P |p, FrobP(E/K) =σ}|, because for any p ∈ S0 and for some prime divisorP|p, we have thatFrobP(E/L) = σ, hencef(P/p) = |σ| = [E :L], becauseσ generatesGal(FP/Fp). ThusPis the only prime ofEoverp.
By the new definition ofS, rp ≤ 0 for allp ∈ S, then we can choosePinE overp, such that
FrobP(E/K) =σ. Then we can definerpas follows:
rp ={τP:τ ∈G, FrobτP(E/K)}
=
τP:τ ∈G, τ στ−1=σ .
Because any element inOE is of the formτ−1(x), withx∈OE, then
FrobP(E/K)τ−1(x)≡τ−1(x)
q
(modP),
composing withτ
τFrobP(E/K)τ−1(x)≡xq(mod τ(P)),
hence rp = |{τP:τ ∈CG(σ)}| = [CG(σ) : CG(σ)∩DP]whereCG(σ)is a centralizer of σ.
SinceFrobP(E/K) =σ, we haveDP =hσi ⊂CG(σ), then
rp=
|CG(σ)|
|σ| = |G|
|σ| ·c. (3.4)
Then by (3.3) and (3.4), we have
1
|σ| =δ(S
0) =r
pδ(S) =
|G| |σ| ·cδ(S),
hence
δ(S) = c
