DIRICHLET CHARACTERS AND PRIMES IN ARITHMETIC
PROGRESSIONS
We plan to prove the following.
Theorem 1 (Dirichlet’s Theorem). Let (a, k) = 1. Then the arithmetic progression
{a + nk : n = 0, 1, 2, . . .} = {m : m ≡ a mod k}
contains infinitely many primes.
Euler gave an analytic proof of the infinitude of primes by showing that
X1
→ ∞.
p
p≤x
In doing so he used properties of the zeta function
−1
∞
X
Y
1
1
ζ(s) =
=
1− s
.
s
n
p
n=1
p
We will attempt something similar and show that
X 1
→∞
p
p≤x
p≡a mod k
P
(for technical reasons we will in fact demonstrate the divergence of p≡a mod k logp p ).
P
Note that the above sum can be written as
p fa,k (p)/p where fa,k is the characteristic function of the property p ≡ a mod k. For example, one could take
fa,k (p) = b| cos(π(p − a)/k)|c but this is superficial and no good to anybody. It
would be nice if our function, or some slight variant, made sense as the coefficients of
a Dirichlet series and allowed for an Euler product. This would already imply more
depth than our previous example for fa,k .
1. Dirichlet characters
Definition 1. A Dirichlet Character modulo k is an arithmetic function χ : N → C
satisfying
(1) χ(n + k) = χ(n) ∀n ∈ N
(2) χ(mn) = χ(m)χ(n) ∀m, n ∈ N
(3) χ(n) 6= 0 ⇐⇒ (n, k) = 1.
1
DIRICHLET CHARACTERS AND PRIMES IN ARITHMETIC PROGRESSIONS
2
Definition 2. The principal character modulo k is the unique Dirichlet character χ1
such that χ1 (n) = 1 ⇐⇒ (n, k) = 1
Firstly, note that χ(1) = 1 since for (n, k) = 1 we have 0 6= χ(n) = χ(1 · n) =
χ(1)χ(n). Also, note that χ’s non-zero values are φ(k)th-roots of unity since for
(n, k) = 1, we have nφ(k) ≡ 1 mod k and hence χ(n)φ(k) = χ(nφ(k) ) = χ(1 + mk) =
χ(1) = 1.
Examples
(1) k = 1: One character, χ(n) = 1 ∀n.
(2) k = 2: One character, χ(n) = 1 for n odd, χ(n) = 0 for n even.
(3) k = 3: We have χ1 . Suppose χ 6= χ1 . Then χ(2) 6= 1 and χ(2)2 = χ(4) =
χ(1) = 1 hence χ(2) = −1.
(4) k = 4: Similar to k = 3.
Note that by periodicity we only need consider one representive n from a given
residue class n̂ mod k. Also, we may restrict attention to those representatives
n with (n, k) = 1 since χ(n) = 0 otherwise. In other words, we can restrict our
attention to the group (Z/kZ)× of units modulo k. By their multiplicative property,
we see that any given Dirichlet character χ : N → C induces a homomorphism
χ∗ : (Z/kZ)× → C× = GL1 (C), n̂ 7→ χ(n). Conversely, given a homomorphism
f : (Z/kZ)× → C× we can acquire a Dirichlet character by setting χ(n) = f (n̂)
when (n, k) = 1 and χ(n) = 0 when (n, k) > 1.
If we view the set of Dirichlet characters mod k as homomorphisms from (Z/kZ)× →
×
C then they form a group when equipped with the pointwise multiplication defined
by (χ · χ0 )(n) = χ(n)χ0 (n). The identity element is given by χ1 and the inverse of an
element χ is given by χ since (χ · χ)(n) = |χ(n)|2 = 1 = χ1 (n). This means that the
Dirichlet characters modulo k can be thought of as the elements of something called
the dual group Ĝ of the group G = (Z/kZ)× . A famous result of harmonic analysis
(Pontryagin Duality) implies that G is isomorphic to Ĝ, and hence we have the first
part of the following theorem.
Theorem 2. There are exactly φ(k) Dirichlet characters modulo k. Also, for any
given n coprime to k with n 6≡ 1 mod k there exists a χ such that χ(n) 6= 1.
Theorem 3 (Orthogonality). Let k be a positive integer and let χ be a Dirichlet
character modulo k. Then
(
k
X
φ(k) ifχ = χ1
(1)
χ(n) =
.
0
otherwise
n=1
DIRICHLET CHARACTERS AND PRIMES IN ARITHMETIC PROGRESSIONS
3
Let n be a positive integer. Then
(
φ(k)
χ(n) =
0
mod k
X
(2)
χ
if n ≡ 1 mod k
,
otherwise
where the summation is over all Dirichlet characters mod k. Let χ, χ0 be Dirichlet
characters mod k. Then
(
k
X
φ(k) ifχ = χ0
0
(3)
χ(n)χ (n) =
0
otherwise.
n=1
For positive integers n1 , n2 we have
(
X
φ(k)
(4)
χ(n1 )χ(n2 ) =
0
χ mod k
if n1 ≡ n2 mod k and (n1 , k) = 1
otherwise.
Proof. For the first sum the result is obvious if χ = χ1 . If χ 6= χ1 then pick an m
with (m, k) = 1 such that χ(m) 6= 1. Then as n runs through a reduced residue
system mod k so does mn. Therefore,
χ(m)
k
X
χ(n) =
k
X
χ(mn) =
χ(n)
n=1
n=1
n=1
k
X
and the result follows.
For the second sum note that if (n, k) > 1 then all terms in the sum are zero. If
n ≡ 1 mod k then χ(n) = χ(1) = 1 for all characters and so the sum simply equals
the number of characters mod k, which is φ(k) by Theorem 2. So now suppose that
(n, k) = 1 and that n 6≡ 1 mod k. Pick a χ0 such that χ0 (n) 6= 1. Then as χ runs
over the characters mod k so does χ0 χ, since they form a group. Therefore,
X
X
X
X
χ(n)
(χ0 · χ)(n) =
χ0 (n)χ(n) =
χ(n) =
χ0 (n)
χ
χ
χ
χ
and the sum is therefore equal to zero in this case.
For the third sum take χ = χχ0 in (1). Finally, for the fourth sum note that if
either (n1 , k) or (n2 , k) is > 1 then the sum is zero. So suppose (n1 , k) = (n2 , k) = 1
and let n2 denote the inverse of n2 modulo k i.e. the number satisfying n2 n2 ≡ 1
mod k. We now apply (2) with n = n1 n2 . The sum in question is given by
X
X
X
χ(n) =
χ(n1 )χ(n2 ) =
χ(n1 )χ(n2 ).
χ
χ
χ
This last equality follows on noting that χ(n1 )χ(n2 ) = χ(n2 n2 ) = χ(1) = 1 and hence
χ(n2 ) = χ(n2 ) since χ maps onto the unit circle for integers coprime to k. If n1 ≡ n2
DIRICHLET CHARACTERS AND PRIMES IN ARITHMETIC PROGRESSIONS
4
mod k then n ≡ 1 mod k and the above sum equals φ(k) by (2). If n1 6≡ n2 mod k
then n ≡
6 1 mod k and the remaining case follows.
2. Dirichlet L-functions
Let χ be a Dirichlet character mod k. Then the Dirichlet L-function associated to
χ is given by
∞
X
χ(n)
(5)
L(s, χ) =
, s = σ + it
s
n
n=1
Proposition 2.1. The series for L(s, χ) converges absolutely for σ > 1 and for fixed
δ > 0 it converges uniformly for σ ≥ 1 + δ. It is therefore analytic in the region
σ > 1.
Proof. Postponed.
The derivative of L(s, χ) is given by the series
∞
X
χ(n) log n
0
.
(6)
L (s, χ) = −
ns
n=1
Proposition 2.2. If χ 6= χ1 then L(s, χ) and L0 (s, χ) converge (conditionally) for
σ > 0. In particular, their values at s = 1 are defined.
Proof. If χ 6= χ1 then by partial summation we have
Z xX
X χ(n)
1 X
(7)
= s
χ(n) + s
χ(n) t−s−1 dt.
s
n
x
1
n≤x
n≤x
n≤t
P
By (1) the sums n≤x χ(n) are bounded. Therefore
Z x
X χ(n)
−σ
x +s
t−σ−1 dt.
(8)
s
n
1
n≤x
This last expression is O(1) for σ > 0 and so the result follows on letting x → ∞.
For the result involving the derivative just use log t t in the above.
The following is not absolutely necessary for our purposes but it does give more
meaning to some of our results.
Theorem 4 (Euler product). For σ > 1 we have the absolutely convergent product
−1
Y
χ(p)
.
(9)
L(s, χ) =
1− s
p
p
DIRICHLET CHARACTERS AND PRIMES IN ARITHMETIC PROGRESSIONS
5
Also, for σ > 1 we have
∞
(10)
X µ(n)χ(n)
1
=
L(s, χ) n=1
ns
and the sum is absolutely convergent in this region.
Proof. Note the factors in the product are the sums of geometric series:
−1 X
∞
χ(p)
χ(pm )
1− s
=
.
p
pms
m=0
Taking the product over primes less than a given x gives
mj
∞
1
X χ(n)
YX
X χ(pm
χ(pm )
1 · · · pj )
=
=
mj s
1
pms
ns
(pm
1 · · · pj )
p≤x m=0
p ,...,p ≤x
1
j
n∈A(x)
m1 ,...,mj ≥0
where A(x) = {n ∈ N : p|n =⇒ p ≤ x}. Clearly, limx→∞ A(x) = N. Now,
X χ(n)
X χ(n)
→0
L(s, χ) −
s
ns
n
n>x
n∈A(x)
P∞
−s
as x → ∞ since the series n=1 χ(n)n is convergent for σ > 1.
Since an absolutely convergent product of non-zero terms is non-zero we see that
L(s, χ) 6= 0 for σ > 1. Taking the reciprocal gives
Y
1
χ(p)
=
1− s
.
L(s, χ)
p
p
Expanding this product we see that the resultant series has coefficients µ(n)χ(n) and
so (10) follows. If the rigour police show up then just use the previous argument.
Absolute convergence of the series follows on applying the integral test after using
|µ(n)χ(n)| ≤ 1 and |ns | = nσ .
3. Dirichlet’s Theorem
We are now in a position to prove Dirichlet’s Theorem. We shall in fact prove the
following
Theorem 5. Suppose (a, k) = 1. Then
X log p
1
(11)
=
log x + O(1).
p
φ(k)
p≤x
p≡a mod k
The starting point is to use (4) as a suitable characteristic function.
DIRICHLET CHARACTERS AND PRIMES IN ARITHMETIC PROGRESSIONS
6
Lemma 3.1. Let (a, k) = 1. Then
X χ(p) log p
X log p
1
1 X
=
log x +
χ(a)
+ O(1)
(12)
p
φ(k)
φ(k) χ6=χ
p
p≤x
p≤x
1
p≡a mod k
Proof. By(4) we have
X log p X log p 1 X
=
χ(a)χ(p)
p
p φ(k) χ
p≤x
p≤x
p≡a mod k
X
1 X log p
=
χ1 (p) +
χ(a)χ(p)
φ(k) p≤x p
χ6=χ
(13)
1
X χ(p) log p
1 X χ1 (p) log p
1 X
=
+
χ(a)
.
φ(k) p≤x
p
φ(k) χ6=χ
p
p≤x
1
Now,
X χ1 (p) log p
p≤x
p
X log p X X log p
=
=
−
p
p
p≤x
p≤x
p≤x
(p,k)=1
(14)
=
p|k
X log p
p≤x
p
+ O(1)
= log x + O(1).
The above Lemma implies that if we can show
X χ(p) log p
= O(1)
p
p≤x
for χ 6= χ1 , then Dirichlet’s Theorem will follow. This is now our main focus.
Informal Discussion: As we have seen before, incorporating prime powers into
a sums involving log p usually just adds an O(1) term. So we can expect
X χ(p) log p X χ(n)Λ(n)
=
+ O(1)
p
n
p≤x
n≤x
where Λ(n) = log p if n = pm and Λ(n) = 0 otherwise. Why are we always adding in
prime power terms if it doesn’t really change anything? Well, this latter sum admits
a more concise description as the partial sums of the series for −L0 (1, χ)/L(1, χ).
DIRICHLET CHARACTERS AND PRIMES IN ARITHMETIC PROGRESSIONS
Indeed, on logarithmic differentiation of the Euler product: L(s, χ) =
χ(p)p−s )−1 , we have
∞
(15)
−
7
Q
p (1
−
∞
XX
X χ(n)Λ(n)
L0 (s, χ) X
log p
m −ms
=
=
(log
p)χ(p
)p
=
.
−s
s
L(s, χ)
1
−
χ(p)p
n
p
p m=0
n=1
From this we see that
X χ(p) log p
(16)
p
p≤x
≈−
L0 (1, χ)
L(1, χ)
and this is bounded as long as L(1, χ) 6= 0. We now attempt to make our discussion
into a more rigorous argument. Our first step is to demonstrate a relationship similar
to (16), but in a more quantitative form. We then show that this form is bounded if
L(1, χ) 6= 0 for χ 6= χ1 . Finally, we prove the latter.
Lemma 3.2. We have
X χ(p) log p X χ(n)Λ(n)
(17)
=
+ O(1).
p
n
p≤x
n≤x
Proof. We have
X χ(n)Λ(n)
n≤x
n
=
X χ(pm ) log p
pm
p,m
pm ≤x
(18)
=
X χ(p) log p
p≤x
p
+
X
p,m
pm ≤x,m≥2
χ(pm ) log p
.
pm
Now, the second sum in the above is than
X
log p
p
X log p
X 1
X log n
=
<
= O(1).
pm
p(p − 1)
n(n − 1)
p
n
m≥2
Lemma 3.3. For χ 6= χ1 we have
(19)
X χ(p) log p
p≤x
p
= −L0 (1, χ)
X µ(n)χ(n)
n≤x
n
+ O(1).
DIRICHLET CHARACTERS AND PRIMES IN ARITHMETIC PROGRESSIONS
Proof. By the previous Lemma and the fact that Λ(n) =
X χ(p) log p
(20)
p≤x
p
+ O(1) =
X χ(n)Λ(n)
n≤x
n
=
P
d|n
X χ(n) X
n≤x
n
8
µ(d) log(n/d) we have
µ(d) log(n/d)
d|n
X χ(d)µ(d) X χ(m) log m
=
d
m
d≤x
m≤x/d
after rearranging in terms of the divisors d and using the multiplicative properties
of χ. Now,
X X
∞
X
χ(m) log m
χ(m) log m
0
−L (1, χ) =
=
+
,
m
m
m=1
m≤Y
m>Y
and
X
0
Z ∞X
X χ(m) log m
log Y
log t
log Y
χ(n)
=−
−
.
χ(n)
dt m
Y
t
Y
Y
n≤t
m>Y
n≤Y
P
Here we have used the fact that χ 6= χ1 and hence the sum n≤Y χ(n) is bounded.
Therefore,
!
X χ(d)µ(d)
X 1 log x/d
X χ(n)Λ(n)
= − L0 (1, χ)
+O
n
d
d x/d
n≤x
d≤x
d≤x
(21)
X χ(d)µ(d)
= − L0 (1, χ)
+ O (1)
d
d≤x
since
X
(log x − log d) = bxc log x − x log x + O(x) = O(x).
d≤x
Note that by equation (10), the sum
X µ(n)χ(n)
n≤x
n
looks a lot like 1/L(1, χ) and so we have essentially established the relation (16).
Lemma 3.4. Suppose χ 6= χ1 . If L(1, χ) 6= 0 then
X µ(n)χ(n)
= O(1).
(22)
n
n≤x
DIRICHLET CHARACTERS AND PRIMES IN ARITHMETIC PROGRESSIONS
9
Proof. Note that
S :=
(23)
X χ(mn)µ(n)
X µ(n)χ(n) X χ(m)
=
n
m
mn
m,n
n≤x
m≤x/n
mn≤x
after rearranging. We now group together the terms mn, writing mn = ` say, and
rearrange to give
X χ(`) X
(24)
S=
µ(n) = 1
`
`≤x
n|`
after using
X
n|`
(
1 if ` = 1,
µ(n) =
0 otherwise.
On the other hand,
X µ(n)χ(n) (25)
X χ(m) S=
L(1, χ) −
n
m
n≤x
m>x/n
X µ(n)χ(n) =
L(1, χ) + O(n/x)
n
n≤x
=L(1, χ)
X µ(n)χ(n)
n≤x
=L(1, χ)
n
X µ(n)χ(n)
n≤x
n
+O
!
1X
χ(n)µ(n)
x n≤x
+ O (1) .
Since L(1, χ) 6= 0, we may divide through and the result follows.
4. The non-vanishing of L(1, χ) for non-principal characters.
As the section heading suggests, we plan to prove the following.
Theorem 6. If χ 6= χ1 then L(1, χ) 6= 0.
The proof is split into two cases: one where χ is real, and the other where χ is
complex. We will start with the real case.
Lemma 4.1. Suppose χ is a real Dirichlet character mod k. Let
X
(26)
A(n) =
χ(d).
d|n
DIRICHLET CHARACTERS AND PRIMES IN ARITHMETIC PROGRESSIONS
10
Then A(n) ≥ 0 for all n and A(n) ≥ 1 if n is a square.
Proof. First note A(n) is multiplicative since it is the convolution of two multiplicative functions, namely 1(n) := 1 ∀n and χ(n). Therefore, it is determined by its
values at prime powers. We have
χ(p) =0 =⇒ A(pα ) = 1,
χ(p) =1 =⇒ A(pα ) = d(pα ) = α + 1,
(
1 if α is even,
χ(p) = − 1 =⇒ A(pα ) =
0 if α is odd.
Q αp
Q
Writing n = p p we see A(n) = p A(pαp ) and hence A(n) ≥ 0 for all n. If n
is square then all the powers αp in the prime decomposition of n are even and so
A(n) ≥ 1.
Proposition 4.2. Suppose χ is a real non-principal Dirichlet character mod k. Let
X A(n)
.
(27)
B(x) =
1/2
n
n≤x
Then
(1) B(x) → ∞ as x → ∞.
(2) B(x) = 2x1/2 L(1, χ) + O(1) and hence L(1, χ) 6= 0.
Proof. For the first part note that by the Lemma
X 1
X 1
X A(n)
1
≥
=
∼
log x → ∞.
B(x) ≥
n1/2
n1/2
m
2
1/2
n≤x
n≤x
n=m2
n=m2
m≤x
For the second part we have
X 1 X
X χ(d) X 1
B(x) =
, after rearranging
χ(d)
=
n1/2
d1/2
m1/2
n≤x
d≤x
d|n
m≤x/d
X χ(d) x 1/2
=
2
using Euler summation
+ O(1) ,
d1/2
d
d≤x
X
X χ(d)
(28)
1/2
−1/2
=2x
+O
χ(d)d
d
d≤x
d≤x
X χ(d) =2x1/2 L(1, χ) −
+ O(1)
d
d>x
=2x1/2 L(1, χ) + O(1)
DIRICHLET CHARACTERS AND PRIMES IN ARITHMETIC PROGRESSIONS
11
where we have used
X χ(d)
d>x
d
1X
=−
χ(d) +
x d≤x
∞
Z
x
1 X
χ(d) dt = O(x−1 ).
2
t d≤t
We can now turn to the case where χ is complex. We will proceed towards a
contradiction.
Lemma 4.3. If χ 6= χ1 and L(1, χ) = 0 then
(29)
L0 (1, χ)
X µ(n)χ(n)
n
n≤x
= log x + O(1).
Proof. We have
S :=
X µ(n)χ(n) X χ(m) log(x/mn)
n
m
n≤x
m≤x/n
(30)
X µ(n)χ(mn) log(x/mn)
=
.
mn
m,n
mn≤x
Once again, we group together the terms mn, writing mn = ` say, and rearrange to
give
(31)
S=
X χ(`) log(x/`) X
`≤x
`
µ(n) = log x
n|`
after using
X
n|`
(
1 if ` = 1,
µ(n) =
0 otherwise.
DIRICHLET CHARACTERS AND PRIMES IN ARITHMETIC PROGRESSIONS
12
On the other hand,
X µ(n)χ(n) X χ(m)
(− log m + log(x/n))
S=
n
m
n≤x
m≤x/n
X χ(m) log m
X χ(m) X µ(n)χ(n) −
+ log(x/n)
=
n
m
m
n≤x
m≤x/n
m≤x/n
X µ(n)χ(n) X χ(m) log m 0
=
L (1, χ) +
n
m
n≤x
m>x/n
(32)
X χ(m) + log(x/n) L(1, χ) −
m
m>x/n
X µ(n)χ(n)
log(x/n)
L0 (1, χ) + O
=
n
x/n
n≤x
log(x/n)
+ log(x/n)L(1, χ) + O
.
x/n
Since we’re assuming L(1, χ) = 0 this equals
(33)
0
L (1, χ)
X µ(n)χ(n)
n≤x
n
X
1
+O
log(x/n) .
x n≤x
The result now follows on noting that the sum in ‘big O’ term is x, as has been
seen previously.
Let
(34)
S(k) = χ mod k : χ 6= χ1 , L(1, χ) = 0
and let
N (k) = |S(k)|.
(35)
Now, if L(1, χ) = 0 then L(1, χ) = 0 and so if χ ∈ S(k) then χ ∈ S(k) (note these
characters are distinct since χ 6= χ for complex characters). Therefore, N (k) is even.
Proposition 4.4. We have
X log p
1 − N (k)
(36)
=
log x + O(1).
p
φ(k)
p≤x
p≡1(k)
DIRICHLET CHARACTERS AND PRIMES IN ARITHMETIC PROGRESSIONS
13
This implies that N (k) = 0 since otherwise N (k) ≥ 2 and hence the right hand side
of the above would be negative for large enough x, contrary to the sum on the left
being made up solely of positive terms.
Proof. By Lemmas 3.1 and 3.3 we have
X log p
1
1 X X χ(p) log p
=
log x +
+ O(1)
p
φ(k)
φ(k)
p
p≤x
χ6=χ p≤x
1
(37)
p≡1(k)
=
X µ(n)χ(n)
1
1 X 0
log x −
+ O(1).
L (1, χ)
φ(k)
φ(k) χ6=χ
n
n≤x
1
Now, by Lemma 3.4 the sum over n ≤ x is O(1) if L(1, χ) 6= 0. If L(1, χ) = 0 then
the sum times L0 (1, χ) is log x + O(1) by Lemma 4.3. Therefore, after writing 1P for
the characteristic function of a given property P , we have
X log p
1
1 X
=
log x −
1L(1,χ)6=0 · O(1)
p
φ(k)
φ(k)
p≤x
χ6=χ
1
p≡1(k)
+ 1L(1,χ)=0 · log x + O(1) + O(1)
(38)
=
1
N (k)
log x −
log x + O(1).
φ(k)
φ(k)