8Lfunctions, Dirichlet series*
II Number Fields
8 Lfunctions, Dirichlet series*
This section is nonexaminable.
We start by proving the exciting fact that there are infinitely many primes.
Theorem (Euclid). There are infinitely many primes.
Proof. Consider the function
Y
p primes
1 −
1
p
−1
=
Y
p prime
1 +
1
p
+
1
p
2
+ ···
=
X
n>0
1
n
.
This is since every
n
=
p
e
1
1
···p
e
r
r
factors uniquely as a product of primes, and
each such product appears exactly once in this. If there were finitely many
primes, as
P
1
p
n
converges to
1 −
1
p
−1
, the sum
X
n≥1
1
n
=
Y
p prime
1 −
1
p
must be finite. But the harmonic series diverges. This is a contradiction.
We all knew that. What we now want to prove is something more interesting.
Theorem
(Dirichlet’s theorem)
.
Let
a, q ∈ Z
be coprime. Then there exists
infinitely many primes in the sequence
a, a + q, a + 2q, ··· ,
i.e. there are infinitely many primes in any such arithmetic progression.
We want to imitate the Euler proof, but then that would amount to showing
that
Y
p≡a mod q
p prime
1 −
1
p
−1
is divergent, and there is no nice expression for this. So it will be a lot more
work.
To begin with, we define the Riemann zeta function.
Definition
(Riemann zeta function)
.
The Riemann zeta function is defined as
ζ(s) =
X
n≥1
n
−s
for s ∈ C.
There are some properties we will show (or assert):
Proposition.
(i) The Riemann zeta function ζ(s) converges for Re(s) > 1.
(ii) The function
ζ(s) −
1
s − 1
extends to a holomorphic function when Re(s) > 0.
In other words,
ζ
(
s
) extends to a meromorphic function on
Re
(
s
)
>
0 with
a simple pole at 1 with residue 1.
(iii) We have the expression
ζ(s) =
Y
p prime
1 −
1
p
s
−1
for
Re
(
s
)
>
1, and the product is absolutely convergent. This is the Euler
product.
The first part follows from the following general fact about Dirichlet series.
Definition
(Dirichlet series)
.
A Dirichlet series is a series of the form
P
a
n
n
−s
,
where a
1
, a
2
, ··· ∈ C.
Lemma. If there is a real number r ∈ R such that
a
1
+ ··· + a
N
= O(N
r
),
then
X
a
n
n
−s
converges for Re(s) > r, and is a holomorphic function there.
Then (i) is immediate by picking
r
= 1, since in the Riemann zeta function,
a
1
= a
2
= ··· = 1.
Recall that x
s
= e
s log x
has
x
s
 = x
Re(s)

if x ∈ R, x > 0.
Proof. This is just IA Analysis. Suppose Re(s) > r. Then we can write
N
X
n=1
a
n
n
−s
= a
1
(1
−s
− 2
−s
) + (a
1
+ a
2
)(2
−s
− 3
−s
) + ···
+ (a
1
+ ··· + a
N−1
)((N − 1)
−s
− N
−s
) + R
N
,
where
R
N
=
a
1
+ ··· + a
N
N
s
.
This is getting annoying, so let’s write
T (N) = a
1
+ ··· + a
N
.
We know
T (N)
N
s
=
T (N)
N
r
1
N
Re(s)−r
→ 0
as N → ∞, by assumption. Thus we have
X
n≥1
a
n
n
−s
=
X
n≥1
T (n)(n
−s
− (n + 1)
−s
)
if
Re
(
s
)
> r
. But again by assumption,
T
(
n
)
≤ B · n
r
for some constant
B
and
all n. So it is enough to show that
X
n
n
r
(n
−s
− (n + 1)
−s
)
converges. But
n
−s
− (n + 1)
−s
=
Z
n+1
n
s
x
s+1
dx,
and if x ∈ [n, n + 1], then n
r
≤ x
r
. So we have
n
r
(n
−s
− (n + 1)
−s
) ≤
Z
n+1
n
x
r
s
x
s+1
dx = s
Z
n+1
n
dx
x
s+1−r
.
It thus suffices to show that
Z
n
1
dx
x
s+1−r
converges, which it does (to
s
s−r
).
We omit the proof of (ii). The idea is to write
1
s − 1
=
∞
X
n=1
Z
n+1
n
dx
x
s
,
and show that
P
φ
n
is uniformly convergent when Re(s) > 0, where
φ
n
= n
−s
−
Z
n+1
n
dx
x
s
.
For (iii), consider the first r primes p
1
, ··· , p
r
, and
r
Y
i=1
(1 − p
−s
i
)
−1
=
X
n
−s
,
where the sum is over the positive integers
n
whose prime divisors are among
p
1
, ··· , p
r
. Notice that 1, ··· , r are certainly in the set.
So
ζ(s) −
r
Y
i=1
(1 − p
−s
i
)
−1
≤
X
n≥r
n
−s
 =
X
n≥r
n
−Re(s)
.
But
P
n≥r
n
−Re(s)
→
0 as
r → ∞
, proving the result, if we also show that it
converges absolutely. We omit this proof, but it follows from the fact that
X
p prime
p
−s
≤
X
n
n
−s
.
and the latter converges absolutely, plus the fact that
Q
(1
− a
n
) converges if
and only if
P
a
n
converges, by IA Analysis I.
This is good, but not what we want. Let’s mimic this definition for an
arbitrary number field!
Definition
(Zeta function)
.
Let
L ⊇ Q
be a number field, and [
L
:
Q
] =
n
. We
define the zeta function of L by
ζ
L
(s) =
X
aCO
L
N(a)
−s
.
It is clear that if
L
=
Q
and
O
L
=
Z
, then this is just the Riemann zeta
function.
Theorem.
(i) ζ
L
(s) converges to a holomorphic function if Re(s) > 1.
(ii)
Analytic class number formula:
ζ
L
(
s
) is a meromorphic function if
Re
(
s
)
>
1 −
1
n
and has a simple pole at s = 1 with residue
cl
L
2
r
(2π)
s
R
L
D
L

1/2
µ
L

,
where
cl
L
is the class group,
r
and
s
are the number of real and complex
embeddings, you know what
π
is,
R
L
is the regulator,
D
L
is the discriminant
and µ
L
is the roots of unity in L.
(iii)
ζ
L
(s) =
Y
pCO
L
prime ideal
(1 − N(p)
−s
)
−1
.
This is again known as the Euler product.
We will not prove this, but the proof does not actually require any new ideas.
Note that
X
aCO
L
N(a)
−s
=
Y
pCO
L
,p prime
(1 − N(p)
−s
)
−1
holds “formally”, as in the terms match up when you expand, as an immediate
consequence of the unique factorization of ideals into a product of prime ideals.
The issue is to study convergence of
P
N
(
a
)
−s
, and this comes down to estimating
the number of ideals of fixed norm geometrically, and that is where all the factors
in the pole come it.
Example.
We try to compute
ζ
L
(
s
), where
L
=
Q
(
√
d
). This has discriminant
D, which may be d or 4d. We first look at the prime ideals.
If
p
is a prime ideal in
O
L
, then
p  hpi
for a unique
p
. So let’s enumerate
the factors of η
L
controlled by p ∈ Z.
Now if
p  D
L

, then
hpi
=
p
2
ramifies, and
N
(
p
) =
p
. So this contributes a
factor of (1 − p
−s
)
−1
.
Now if p remains prime, then we have N(hpi) = p
2
. So we get a factor of
(1 − p
−2s
)
−1
= (1 − p
−s
)
−1
(1 + p
−s
)
−1
.
If p splits completely, then
hpi = p
1
p
2
.
So
N(p
i
) = p,
and so we get a factor of
(1 − p
−s
)
−1
(1 − p
−s
)
−1
.
So we find that
ζ
L
(s) = ζ(s)L(χ
D
, s),
where we define
Definition (Lfunction). We define the Lfunction by
L(χ, s) =
Y
p prime
(1 − χ(p)p
−s
)
−1
.
In our case, χ is given by
χ
D
(p) =
0 p  D
−1 p remains prime
1 p splits
=
(
D
p
p is odd
depends on d mod 8 p = 2
.
Example. If L = Q(
√
−1), then we know
−4
p
=
−1
p
= (−1)
p−1
2
if p 6= 2,
and χ
D
(2) = 0 as 2 ramifies. We then have
L(χ
D
, s) =
Y
p>2 prime
(1 − (−1)
p−1
2
p
−s
)
−1
= 1 −
1
3
s
+
1
5
s
−
1
7
s
+ ··· .
Note that
χ
D
was defined for primes only, but we can extend it to a function
χ
D
: Z → C by imposing
χ
D
(nm) = χ
D
(n)χ
D
(m),
i.e. we define
χ
D
(p
e
1
1
···p
e
r
r
) = χ
D
(p
1
)
e
1
···χ
D
(p
r
)
e
r
.
Example. Let L = Q(
√
−1). Then
χ
−4
(m) =
(
(−1)
m−1
2
m odd
0 m even.
It is an exercise to show that this is really the extension, i.e.
χ
−4
(mn) = χ
−4
(m)χ
−4
(n).
Notice that this has the property that
χ
−4
(m − 4) = χ
−4
(m).
We give these some special names
Definition
(Dirichlet character)
.
A function
χ
:
Z → C
is a Dirichlet character
of modulus D if there exists a group homomorphism
w :
Z
DZ
×
→ C
×
such that
χ(m) =
(
w(m mod D) gcd(m, D) = 1
0 otherwise
.
We say χ is nontrivial if ω is nontrivial.
Example. χ
−4
is a Dirichlet character of modulus 4.
Note that
χ(mn) = χ(m)χ(n)
for such Dirichlet characters, and so
L(χ, s) =
Y
p prime
(1 − χ(p)p
−s
)
−1
=
X
n≥1
χ(n)
n
s
for such χ.
Proposition. χ
D
, as defined for
L
=
Q
(
√
d
) is a Dirichlet character of modulus
D.
Note that this is a very special Dirichlet character, as it only takes values
0, ±1. We call this a quadratic Dirichlet character.
Proof. We must show that
χ
D
(p + Da) = χ
D
(p)
for all p, a.
(i) If d ≡ 3 (mod 4), then D = 4d. Then
χ
D
(2) = 0,
as (2) ramifies. So χ
D
(even) = 0. For p > 2, we have
χ
D
(p) =
D
p
=
d
p
=
p
d
(−1)
p−1
2
as
d−1
2
≡ 1 (mod 2), by quadratic reciprocity. So
χ
D
(p + Da) =
p + Da
d
(−1)
p−1
2
(−1)
4da/2
= χ
D
(p).
(ii) If d ≡ 1, 2 (mod 4), see example sheet.
Lemma.
Let
χ
be any nontrivial Dirichlet character. Then
L
(
χ, s
) is holomor
phic for Re(s) > 0.
Proof. By our lemma on convergence of Dirichlet series, we have to show that
N
X
i=1
χ(i) = O(1),
i.e. it is bounded. Recall from Representation Theory that distinct irreducible
characters of a finite group G are orthogonal, i.e.
1
G
X
g∈G
χ
1
(g)χ
2
(g) =
(
1 χ
1
= χ
2
0 otherwise
.
We apply this to
G
= (
Z/DZ
)
×
, where
χ
1
is trivial and
χ
2
=
χ
. So orthogonality
gives
X
aD<i≤(a+1)D
χ(i) =
X
i∈(Z/DZ)
×
χ(i) = 0,
using that χ(i) = 0 if i is not coprime to D. So we are done.
Corollary. For quadratic characters χ
D
, we have
L(χ
D
, 1) 6= 0.
For example, if D < 0, then
L(χ
D
, 1) =
2πcl
Q(
√
d)

D
1/2
µ
Q(
√
d)

.
Proof. We have shown that
ζ
Q(
√
d)
(s) = ζ
Q
(s)L(χ
D
, s).
Note that
ζ
Q(
√
d)
(
s
) and
ζ
Q
(
s
) have simple poles at
s
= 1, while
L
(
χ
D
, s
) is
holomorphic at s = 1.
Since the residue of
ζ
Q
(
s
) at
s
= 1 is 1, while the residue of
ζ
Q(
√
D)
at
s
= 1
is nonzero by the analytic class number formula. So
L
(
χ
D
,
1) is nonzero, and
given by the analytic class number formula.
Example. If L = Q(
√
−1), then
1 −
1
3
+
1
5
−
1
7
+ ··· =
2π ·1
2 · 4
=
π
4
.
In general, for any field whose class number we know, we can get a series
expansion for π. And it converges incredibly slow.
Note that this corollary required two things — the analytic input for the
analytic class number formula, and quadratic reciprocity (to show that
χ
D
is a
Dirichlet character).
More ambitiously, we now compute the zeta function of a cyclotomic field,
L
=
Q
(
ω
q
), where
ω
q
is the primitive
q
th root of unity and
q ∈ N
. We need to
know the following facts about cyclotomic extensions:
Proposition.
(i) We have [L : Q] = ϕ(q), where
ϕ(q) = (Z/qZ)
×
.
(ii) L ⊇ Q is a Galois extension, with
Gal(L/Q) = (Z/qZ)
×
,
where if
r ∈
(
Z/qZ
)
×
, then
r
acts on
Q
(
w
Q
) by sending
ω
q
7→ ω
r
q
. This is
what plays the role of quadratic reciprocity for cyclotomic fields.
(iii) The ring of integers is
O
L
= Z[ω
q
] = Z[x]/Φ
q
(x),
where
Φ
q
(x) =
x
q
− 1
Q
dq,d6=q
Φ
d
(x)
is the qth cyclotomic polynomial.
(iv)
Let
p
be a prime. Then
p
ramifies in
O
L
if and only if
p  D
L
, if and only
if
p  q
. So while
D
might be messy, the prime factors of
D
are the prime
factors of q.
(v)
Let
p
be a prime and
p  q
. Then
hpi
factors as a product of
ϕ
(
q
)
/f
distinct
prime ideals, each of norm p
f
, where f is the order of p in (Z/qZ)
×
.
Proof.
(i) In the Galois theory course.
(ii) In the Galois theory course.
(iii) In the example sheet.
(iv) In the example sheet.
(v) Requires proof, but is easy Galois theory, and is omitted.
Example. Let q = 8. Then
Φ
8
=
x
8
− 1
(x + 1)(x − 1)(x
2
+ 1)
=
x
8
− 1
x
4
− 1
= x
4
+ 1.
So given a prime p (that is not 2), we need to understand
O
L
/p =
F
p
[x]
Φ
8
,
i.e. how Φ
8
factors factors mod p (Dedekind’s criterion). We have
(Z/8)
×
= {1, 3, 5, 7} = {1, 3, −3, −1} = Z/2 × Z/2.
Then (v) says if p = 17, then x
4
factors into 4 linear factors, which it does.
If p = 3, then (v) says x
4
factors into 2 quadratic factors. Indeed, we have
(x
2
− x − 1)(x
2
+ x − 1) = (x
2
− 1)
2
− x
2
= x
4
+ 1.
Given all of these, let’s compute the zeta function! Recall that
ζ
Q(ω
q
)
(s) =
Y
p
(1 − N(p)
−s
)
−1
.
We consider the prime ideals
p
dividing
hpi
, where
p
is a fixed integer prime
number. If p  q, then (v) says this contributes a factor of
(1 − p
−fs
)
−ϕ(q)/f
,
to the zeta function, where
f
is the order of
p
in (
Z/qZ
)
×
. We observe that this
thing factors, since
1 − t
f
=
Y
γ∈µ
f
(1 − γt),
with
µ
f
= {γ ∈ C : γ
f
= 1},
and we can put t = p
−s
.
We let
ω
1
, ··· , ω
ϕ(q)
: (Z/qZ)
×
→ C
×
be the distinct irreducible (onedimensional) representations of (
Z/qZ
)
×
, with
ω
1
being the trivial representation, i.e. ω
1
(a) = 1 for all a ∈ (Z/qZ)
×
.
The claim is that
ω
1
(
p
)
, ··· , ω
ϕ(q)
(
p
) are
f
th roots of 1, each repeated
ϕ
(
q
)
/f
times. We either say this is obvious, or we can use some representation theory.
We know
p
generates a cyclic subgroup
hpi
of (
Z/qZ
)
×
of order
f
, by definition
of
f
. So this is equivalent to saying the restrictions of
ω
1
, ··· , ω
ϕ(q)
to
p
are the
f distinct irreducible characters of hpi
∼
=
Z/f, each repeated ϕ(q)/f times.
Equivalently, note that
Res
(Z/qZ)
×
hpi
(ω
1
⊕ ··· ⊕ ω
ϕ(q)
) = Res
(Z/qZ)
×
hpi
(regular representation of (Z/qZ)
×
).
So this claims that
Res
(Z/qZ)
×
hpi
(regular rep. of (Z/qZ)
×
) =
ϕ(q)
f
(regular rep. of Z/f ).
But this is true for any group, since
Res
G
H
CG = G/HCH,
as the character of both sides is Gδ
e
.
So we have
(1 − p
−fs
)
−ϕ(q)/f
=
ϕ(q)
Y
i=1
(1 − ω
i
(p)p
−s
)
−1
.
So we let
χ
i
(n) =
(
w
i
(n mod q) gcd(n, q) = 1
0 otherwise
be the corresponding Dirichlet characters. So we have just shown that
Proposition. We have
ζ
Q(ω
q
)
(s) =
ϕ(q)
Y
i=1
L(χ
i
, s) · (corr. factor) = ζ
Q
(s)
ϕ(q)
Y
i=2
L(χ
i
, s) · (corr. factor)
where the correction factor is a finite product coming from the primes that divide
q.
By defining the
L
functions in a slightly more clever way, we can hide the
correction factors into the
L
(
χ, s
), and then the
ζ
function is just the product of
these Lfunctions.
Proof.
Our analysis covered all primes
p  q
, and the correction factor is just to
include the terms with p  q. The second part is just saying that
ζ
Q
(s) = L(χ
1
, s)
Y
pq
(1 − p
−s
)
−1
.
This allows us to improve our result on the nonvanishing of
L
(
χ,
1) to all
Dirichlet characters, and not just quadratic Dirichlet characters.
Corollary. If χ is any nontrivial Dirichlet character, then L(χ, 1) 6= 0.
Proof.
By definition, Dirichlet characters come from representations of some
(
Z/qZ
)
×
, so they appear in the formula of the
ζ
function of some cyclotomic
extension.
Consider the formula
ζ
Q(ω
q
)
(s) = ζ
Q
(s)
ϕ(q)
Y
i=2
L(χ
i
, s) · (corr. factor)
at
s
= 1. We know that the
L
(
χ
i
, s
) are all holomorphic at
s
= 1. Moreover,
both
ζ
Q(ω
q
)
and
ζ
Q
have a simple pole at 0. Since the correction terms are finite,
it must be the case that all L(χ
i
, s) are nonzero.
Theorem
(Dirichlet, 1839)
.
Let
a, q ∈ N
be coprime, i.e.
gcd
(
a, q
) = 1. Then
there are infinitely many primes in the arithmetic progression
a, a + q, a + 2q, a + 3q, ··· .
Proof. As before, let
ω
1
, ··· , ω
ϕ(q)
: (Z/qZ)
×
→ C
×
be the irreducible characters, and let
χ
1
, ··· , χ
ϕ(q)
: Z → C
be the corresponding Dirichlet character, with ω
1
the trivial one.
Recall the orthogonality of columns of the character table, which says that if
gcd(p, q) = 1, then
1
ϕ(q)
X
i
ω
i
(a)ω
i
(p) =
(
1 a ≡ p (mod q)
0 otherwise
.
Hence we know
1
ϕ(q)
X
i
χ
i
(a)χ
i
(p) =
(
1 a ≡ p (mod q)
0 otherwise
,
even if gcd(p, q) 6= 1, as then χ
i
(p) = 0. So
X
p≡a mod q
p prime
p
−s
=
1
ϕ(q)
X
i
χ
i
(a)
X
all primes p
χ
i
(p)p
−s
. (‡)
We want to show this has a pole at s = 1, as in Euclid’s proof.
To do so, we show that
P
p
χ
i
(
p
)
p
−s
is “essentially”
log L
(
χ
i
, s
), up to some
bounded terms. We Taylor expand
log L(χ, s) = −
X
log(1 − χ(p)p
−s
) =
X
n≥1
p prime
χ(p)
n
np
ns
=
X
n≥1
p prime
χ(p
n
)
np
ns
.
What we care about is the n = 1 term. So we claim that
X
n≥2,p prime
χ(p
n
)
np
ns
converges at s = 1. This follows from the geometric sum
X
p
X
n≥2
χ(p
n
)
np
ns
≤
X
p
X
n≥2
p
−ns
=
X
p prime
1
p
s
(p
s
− 1)
≤
X
n≥2
1
n
s
(n
s
− 1)
< ∞.
Hence we know
log L(χ, s) =
X
p
χ
i
(p)p
−s
+ bounded stuff
near s = 1.
So at s = 1, we have
(‡) ∼
1
ϕ(q)
X
i
χ
i
(a) log L(χ
i
, s).
and we have to show that the right hand side has a pole at s = 1.
We know that for
i 6
= 1, i.e.
χ
i
nontrivial,
L
(
χ
i
, s
) is holomorphic and
nonzero at
s
= 1. So we just have to show that
log L
(
χ
1
, s
) has a pole. Note
that L(χ
1
, s) is essentially ζ
Q
(s). Precisely, we have
L(χ
1
, s) = ζ
Q
(s)
Y
pq
(1 − p
−s
).
Moreover, we already know that ζ
Q
(s) blows up at s = 1. We have
ζ
Q
(s) =
1
s − 1
+ holomorphic function
=
1
s − 1
(1 + (s − 1)(holomorphic function)).
So we know
log L(χ
1
, s) ∼ log ζ
Q
(s) ∼ log
1
s − 1
,
and this does blow up at s = 1.
So far, we have been working with abelian extensions over
Q
, i.e. extensions
L/Q
whose Galois group is abelian. By the Kronecker–Weber theorem, every
abelian extension of
Q
is contained within some cyclotomic extension. So in
some sense, we have considered the “most general” abelian extension.
Can we move on to consider more complicated number fields? In general,
suppose
L/Q
is Galois, and
G
=
Gal
(
L/Q
). We can still make sense of the
ζ
functions, and it turns out it always factors as
ζ
L
(s) =
Y
ρ
L(ρ, s)
dim ρ
,
where
ρ
ranges over all the irreducible representations of
G
, and
L
(
ρ, s
) is the
Artin
L
function. It takes some effort to define the Artin
L
function, and we
shall not do so here. However, it is worth noting that
L
(1
, s
) is just
ζ
Q
(
s
), and
for ρ 6= 1, we still have a factorization of the form
L(ρ, s) =
Y
p prime
L
p
(ρ, s).
This L
p
(ρ, s) is known as the Euler factor.
One can show that
L
(
ρ, s
) is always a meromorphic function of
s
, and is
conjectured to be holomorphic for all s (if ρ 6= 1, of course).
If
ρ
is onedimensional, then
L
(
ρ, s
) is a Dirichlet series
L
(
χ, s
) for some
χ
.
Recall that to establish this fact for quadratic fields, we had to use quadratic
reciprocity. In general given a
ρ
, finding
χ
is a higher version of “quadratic
reciprocity”. This area is known as class field theory. If dim ρ > 1, then this is
“nonabelian class field theory”, known as Langlands programme.