Chapter 7
The Prime number theorem for
arithmetic progressions
7.1
The Prime number theorem
Denote by π(x) the number of primes 6 x. We prove the Prime Number Theorem.
x
as x → ∞.
Theorem 7.1. We have π(x) ∼
log x
Before giving the detailed proof, we outline our strategy. Define the functions
X
X
X
Λ(n),
log p, ψ(x) :=
log p =
θ(x) :=
p6x
k,p: pk 6x
n6x
where Λ is the von Mangoldt function, given by Λ(n) = log p if n = pk for some
prime p and some k > 1, and Λ(n) = 0 otherwise.
By Lemma 3.14 from Chapter 3 we have
∞
X
n=1
Λ(n)n−s = −
ζ 0 (s)
for Re s > 1.
ζ(s)
• By applying Theorem 6.3 (the Tauberian theorem for Dirichlet series) to the latter
we obtain
ψ(x)
1X
=
Λ(n) → 1 as x → ∞.
x
x n6x
107
• We prove that ψ(x) − θ(x) is small. This gives θ(x)/x → 1 as x → ∞.
• Using partial summation, we deduce π(x) log x/x → 1 as x → ∞.
We first verify the conditions of the Tauberian theorem.
P
−s
Lemma 7.2. ∞
can be extended to a function analytic on an open set
n=1 Λ(n)n
containing {s ∈ C , Re s > 1, s 6= 1}, with a simple pole with residue 1 at s = 1.
Proof. Let A := {s ∈ C , Re s > 1, s 6= 1}. By Theorem 5.2, ζ(s) is analytic on an
open set containing A with a simple pole at s = 1. Further, by Corollary 5.4 and
Theorem 5.5, ζ(s) 6= 0 on A, and hence also ζ(s) 6= 0 on an open set containing A.
So by Lemma 2.17, ζ 0 (s)/ζ(s) is analytic on an open set containing A, with a simple
pole with residue 1 at s = 1. This proves Lemma 7.2.
Lemma 7.3. (i) θ(x) = O(x) as x → ∞.
√
(ii) ψ(x) = θ(x) + O( x) as x → ∞.
(iii) ψ(x) = O(x) as x → ∞.
Q
Proof. (i) By homework exercise 3a, we have p6x p 6 4x for x > 2. This implies
X
θ(x) =
log p 6 x log 4 = O(x) as x → ∞.
p6x
(ii) We have
ψ(x) =
X
log p =
X
log p +
p6x
p,k: pk 6x
X
log p +
p2 6x
X
log p + · · ·
p3 6x
= θ(x) + θ(x1/2 ) + θ(x1/3 ) + · · ·
Notice that θ(t) = 0 if t < 2. So θ(x1/k ) = 0 if x1/k < 2, that is, if k > log x/ log 2.
Hence
[log x/ log 2]
ψ(x) − θ(x) =
X
θ(x
1/k
√
) 6 θ( x) +
k=2
[log x/ log 2]
X
√
θ( 3 x)
k=3
log x
√
√
√
√
6 θ( x) +
− 3 θ( 3 x) = O x + 3 x · log x
log 2
√
= O( x) as x → ∞.
(iii) Combine (i) and (ii). This follows also from homework exercise 6, but (ii) will
be needed anyhow.
108
Proof of Theorem 7.1. Lemmas 7.2 and 7.3 (iii) imply that LΛ (s) =
satisfies all conditions of Theorem 6.3, with α = 1. Hence
P∞
n=1
Λ(n)n−s
1X
ψ(x)
=
Λ(n) → 1 as x → ∞.
x
x n6x
From Lemma 7.3 (ii) we infer
√
θ(x)
ψ(x) + O( x)
ψ(x)
=
=
+ O(x−1/2 ) → 1 as x → ∞.
x
x
x
We now apply partial summation to obtain our result for π(x). Thus,
Z x
1 0
X
X
1
1
θ(t) ·
π(x) =
1=
log p ·
= θ(x)
−
dt
log
p
log
x
log
t
2
p6x
p6x
Z x
θ(t)
θ(x)
+
=
2 dt.
log x
2 t log t
By Lemma 7.3 (i) there is a constant C > 0 such that θ(t) 6 Ct for all t > 2.
Together with homework exercise 1, this implies
Z x
Z x
x θ(t)
dt
as x → ∞.
2 · dt 6 C ·
2 = O
log2 x
2 t log t
2 log t
Hence
log x
θ(x)
x π(x) log x
=
+O
·
x
x
x
log2 x
θ(x)
1
+O
→ 1 as x → ∞.
=
x
log x
7.2
The Prime number theorem for arithmetic
progressions
Let q, a be integers with q > 2, gcd(a, q) = 1. Define
π(x; q, a) := number of primes p 6 x with p ≡ a (mod q).
109
Theorem 7.4. We have π(x; q, a) ∼
x
1
·
as x → ∞.
ϕ(q) log x
The proof is very similar to that of the Prime number theorem. Define the
quantities
X
θ(x; q, a) :=
log p,
p6x, p≡a (mod q)
X
ψ(x; q, a) :=
p,k, pk 6x, pk ≡a (mod q)
Let
∞
X
F (s) :=
X
log p =
Λ(n).
n6x, n≡a (mod q)
Λ(n)n−s .
n=1, n≡a (mod q)
Let G(q) be the group of characters modulo q.
Lemma 7.5. For s ∈ C with Re s > 1 we have
F (s) = −
X
1
L0 (s, χ)
·
.
χ(a) ·
ϕ(q)
L(s, χ)
χ∈G(q)
Proof. By homework exercise 7a we have for χ ∈ G(q) and for s ∈ C with Re s >
1, since χ ∈ G(q) is a strongly multiplicative arithmetical function and L(s, χ)
converges absolutely,
∞
X
L0 (s, χ)
=−
χ(n)Λ(n)n−s .
L(s, χ)
n=1
Using Theorem 4.9 (ii) (one of the orthogonality relations for characters) we obtain
for s ∈ C with Re s > 1,
X
χ∈G(q)
∞
X
X
L0 (s, χ)
χ(a) ·
χ(a)
χ(n)Λ(n)n−s
=−
L(s, χ)
n=1
=−
χ∈G(q)
∞
X


X

n=1
χ(a)χ(n) Λ(n)n−s = −ϕ(q) ·
χ∈G(q)
∞
X
n=1, n≡a (mod q)
110
Λ(n)n−s .
Lemma 7.6. The function F (s) can be continued to a function analytic on an open
set containing {s ∈ C : Re s > 1, s 6= 1}, with a simple pole with residue ϕ(q)−1 at
s = 1.
Proof. Let A := {s ∈ C : Re s > 1, s 6= 1}, B := {s ∈ C : Re s > 1}.
(q)
By Theorem 5.3 (iii), L(s, χ0 ) is analytic on an open set containing A, with a
(q)
simple pole at s = 1. Further, by Corollary 5.4 and Theorem 5.5, L(s, χ0 ) 6= 0 for
(q)
(q)
s ∈ A, and hence for s in an open set containing A. Therefore, L0 (s, χ0 )/L(s, χ0 )
is analytic on an open set containing A. Further, by Lemma 2.17, it has a simple
pole with residue 1 at s = 1.
(q)
Let χ be a character mod q with χ 6= χ0 . By Theorem 5.3 (ii), L(s, χ) is
analytic on an open set containing B, and by Corollary 5.4 and Theorem 5.5, it is
non-zero on B, hence on an open set containing B. Therefore, L0 (s, χ)/L(s, χ) is
analytic on an open set containing B.
Now by Lemma 7.5, F (s) is analytic on an open set containing A, with a simple
(q)
pole with residue χ0 (a)/ϕ(q) = ϕ(q)−1 at s = 1.
Lemma 7.7. (i) θ(x; q, a) = O(x) as x → ∞.
√
(ii) ψ(x; q, a) − θ(x; q, a) = O( x) as x → ∞.
(iii) ψ(x; q, a) = O(x) as x → ∞.
Proof. (i) We have θ(x; q, a) 6 θ(x) = O(x) as x → ∞.
(ii) We have
X
ψ(x; q, a) − θ(x; q, a) =
log p
k,p, k>2,pk 6x, pk ≡a (mod q)
6
X
√
log p = ψ(x) − θ(x) = O( x) as x → ∞.
k,p, k>2,pk 6x
(iii) Obvious.
Proof of Theorem 7.4. Notice that F (s) satisfies the conditions of Theorem 6.3, with
α = ϕ(q)−1 . Hence
ψ(x; q, a)
1
→
as x → ∞,
x
ϕ(q)
111
and then by Lemma 7.7 (ii),
√
ψ(x; q, a) + O( x)
1
θ(x; q, a)
=
→
as x → ∞.
x
x
ϕ(q)
By partial summation we have
π(x; q, a) =
X
26p6x, p≡a (mod q)
Now
Z
06
2
x
θ(t; q, a)
· dt 6
t log2 t
1
θ(x; q, a)
log p ·
=
+
log p
log x
Z
2
x
Z
2
x
θ(t; q, a)
· dt.
t log2 t
x θ(t)
·
dt
=
O
as x → ∞,
t log2 t
log2 x
using the estimate from the proof of Theorem 7.1. So
1 π(x; q, a) log x
θ(x; q, a)
1
=
+O
→
as x → ∞.
x
x
log x
ϕ(q)
This completes our proof.
In Chapter 1 we mentioned that there are sharper versions of the Prime Number
Theorem, with an estimate for the error |π(x) − Li(x)|. Similarly, there are refimenents of the Prime number theorem for arithmetic progressions with an estimate
for the error |π(x; q, a) − Li(x)/ϕ(q)|. The simplest case is when we fix q and let
x → ∞, but for applications it is important to have also versions where q is allowed
to move in some range when we let x → ∞.
By an ‘absolute constant’ we mean a positive constant that does not depend
on anything. The following result was proved by Walfisz in 1936, with important
preliminary work by Landau and Siegel.
Theorem 7.8. There are an absolute constant C1 , and for every real A > 0 there
is a number C2 (A) depending on A, such that the following holds.
For every real x > 3, every integer q with 2 6 q 6 (log x)A and every integer a with
gcd(q, a) = 1, we have
√
π(x; q, a) − 1 Li(x) 6 C1 xe−C2 (A) log x .
ϕ(q)
112
The constants C1 , C2 (A) are ineffective, this means that by going through the
proof of the theorem one cannot compute the constants, but only show that they
exist.
As we mentioned in Chapter 1, there is an intricate connection between the
Rx
zero-free region of ζ(s) and estimates for |π(x) − Li(x)|, where Li(x) = 2 dt/ log t.
Similarly, there is a connection between zero-free regions of L-functions and estimates for |π(x; q, a) − Li(x)/ϕ(q)|. We recall the following, rather complicated,
result of Landau (1921) on the zero-free region of L-functions.
Theorem 7.9. There is an absolute constant c > 0 such that for every integer q > 2
the following holds. Among all characters χ modulo q, there is at most one such
that L(s, χ) has a zero in the region
c
R(q) := s ∈ C : Re s > 1 −
.
log(q(1 + |Im s|))
If such a character χ exists, it has no more than one zero in R(q) and moreover,
(q)
χ 6= χ0 , χ is a real character and the zero is real.
Any character χ modulo q having a zero in R(q) is called an exceptional character
mod q, and the zero of L(s, χ) in R(q) is called an exceptional zero. It is conjectured
that exceptional characters do not exist.
In order to obtain Theorem 7.8, one needs an estimate for the real part of a
possible exceptional zero of an L-function. The following result was proved by
Siegel (1935).
Theorem 7.10. For every ε > 0 there is a number c(ε) > 0 such that for every
integer q > 2 the following holds: if χ is an exceptional character modulo q and β
an exceptional zero of L(s, χ), then Re β < 1 − c(ε)q −ε .
Theorems 7.9 and 7.10 imply (after a lot of work) Theorem 7.8. Proofs of Theorems 7.8–7.10 may be found in H. Davenport, Multiplicative Number Theory, Graduate texts in mathematics 74, Springer Verlag, 2nd ed., 1980.
Knowing that an arithmetic progression contains infinitely many primes, one
would like to know when the first prime in such a progression occurs, i.e., the
smallest x such that π(x; q, a) > 0. The following estimate is due to Linnik (1944).
113
Theorem 7.11. Denote by P (q, a) the smallest prime number p with p ≡ a (mod q).
There are absolute constants c, L such that for every integer q > 2 and every integer
a with gcd(a, q) = 1 we have P (q, a) 6 cq L .
The exponent L is known as ’Linnik’s constant.’ Since the appearance of Linnik’s
paper, various people have tried to estimate it. The present record is L = 5.18, due
to Xylouris (2011).
114