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Algebraic Number Theory – Lecture 12
Lee Butler
To see a World in a Grain of Sand And a Heaven in a Wild Flower, Hold Infinity in the palm of your hand And Eternity in an hour.
- Some algebra
Let K be a number field and consider the sum
(1)
a
N (a)s
where the sum is over all ideals a ⊆ OK. Recall that the norm of an ideal is
N (a) = |OK /a|.
One of the questions you might ask about the series (1) is: does it converge? Well, consider the partial sums (^) ∑
N (a) 6 x
N (a)s^
If these are bounded then the series must converge. Write s = σ + iτ , then, since the ideals form a UFD, ∑
N (a) 6 x
N (a)σ^
N (pi) 6 x ei> 0
(N (p 1 )e^1 · · · )σ
N (p) 6 x
N (p)σ^
N (p)^2 σ^
N (p) 6 x
N (p)σ
Andrew assured us that for each prime ideal p there is a unique prime number p such that N (p) = pf for some f ∈ N. How many prime ideals can correspond to the same prime number p? Well N (p) = pf if and only if pOK = pepe 22 · · · pe rr
and f is the inertia degree of p over p. So from Andrew’s lecture on Hilbert ramification and the fundamental identity therein, at most [K : Q] prime ideals can correspond to the same prime number p. So ∑
N (a) 6 x
N (a)σ^
p 6 x
pσ
)−[K:Q]
The product on the right converges for σ > 1 and so the series (1) converges. 1
Since we bounded a finite sum by an infinite one above we lost equality. But consider the infinite product ∏
p
N (p)s^
N (p)^2 s^
N (p 1 )s^
N (p 2 )s^
N (p 1 )^2 s^
(N (p 1 )N (p 2 ))s^
N (p 1 )^3 s^
(N (p 1 )^2 N (p 2 ))s^
N (p 1 )s^
N (p 2 )s^
N (p^21 )s^
N (p 1 p 2 )s^
N (p^31 )s^
N (p^21 p 2 )s^
Since the ideals in OK form a UFD each denominator is a unique ideal in OK , and each ideal in OK appears as a unique denominator. So
∑
a
N (a)s^
p
N (p)s
We call this function on s the Dedekind zeta function of K, ζK (s).
Example. The Dedekind zeta function on Q is
∑
n∈N
N (n)s^
∑^ ∞
n=
ns^
= ζ(s).
- Analytic stuff
The Dedekind zeta function of K contains a vast amount of information about K. Hecke proved in 1917 that
lim s→ 1 (s − 1)ζK (s) =
2 r^1 (2π)r^2 hK RK w
|dK |
where
- r 1 , r 2 are respectively the number of real and pairs of complex embeddings K ↪→ C,
- hK is the class number of K,
- dK is the discriminant of K,
- w is the number of roots of unity in K, and
- RK is the regulator of K, which we’ll define another time^1.
We’re not going to prove this formula. Instead we’ll prove a special case of it, modulo knowing about the regulator and roots of unity. We need some analysis first, though.
Lemma 1. Let (am)m∈N be a sequence in C and suppose that
A(x) =
m 6 x
am = O(xδ^ )
for some δ > 0. Then ∑∞
m=
am ms
converges for <(s) > δ and in this half plane we have
∑^ ∞
m=
am ms^
= s
1
A(x) xs+^
dx.
(^1) It’s det(log |σi(εj )|) where ε 1 ,... , εr 1 are fundamental units and σ 1 ,... , σr 1 are the real embeddings.
with A(x) =
n 6 x an. By lemma 2 we know A(x) =
πx 2
D
+ O(
x).
So
ζK (s) = s
1
πx 2
Dxs+^
E(x) xs+
dx
πs 2
D(s − 1)
1
E(x) xs+^
dx,
where E(x) = O(
x). This integral converges for <(s) > 12 , giving us our analytic continuation:
(s − 1)ζK (s) = πs 2
D
1
E(x) xs+^
dx.
From this we get
lim s→ 1 (s − 1)ζK (s) =
π 2
D
From Sandro’s lecture, dK = − 4 D, so 2
D =
|dK |, as required.
Assuming the class number formula, and since we know that r 1 = 0, r 2 = 1, and hK = 1, we get that w = 2RK. Thus to calculate RK , which is a rather unpleasant determinant, we need only check if i ∈ K.
