The Class Number Formula
Version 1.0—22nd November, 2013 klokken 11:26.
A fundamental tool in the study of an algebraic number field is the Dedekind Zeta
function. It is a generating function encoding a lot of the arithmetic properties of the
field.
We fix a number field K whose ring of integers is denoted by A, and we let d =
[K : Q]. Then the Dedekind Zeta function of K is
X 1
ζK (s) =
(c)
s
N
(a)
a⊆ A
where a runs through all nontrivial, proper ideals of A. If we specialize K to be the
field Q of rational numbers, all ideals are principal with a unique positive generator
and so they are is in a one-one correspondence with the natural numbers, and we get
back the Riemann Zeta function.
We shall in this section establish the basic facts about the Dedekind Zeta function,
and the first question that arises is: For which s is it defined? It turns out that the series
in (c) converges for Re > 1, but one may show that the function can be analytically
continued to the whole complex plain, except at s = 1 where the zeta function has a
simple pole. For our modest needs it will be sufficient to know that the series converges
for all real s > 1.
Just like the Riemann Zeta function, the Dedekind Zeta function enjoys the property
of having an Euler product. In the Riemann case the Euler product is a consequence of
the fundamental theorem of arithmetic, stating that integers have a unique factorization
in primes. Unique factorization is true for ideals in Dedkind rings, and therefore the
Dedekind Zeta functions also has an Euler product:
Y
1
,
ζK (s) =
−s
1
−
N
(p)
p
where p runs through all prime ideals in A. The proof follows the same lines as for
the Riemann Zeta function, and we shall later on give it. There are several results and
many conjectures about the values of the Dedekind Zeta function at the integers, the
classic result being a formula for the residue of ζk (s) at s = 1 which involves the class
number of K and several other basic invariants of K. It reads
lim+ (s − 1)ζK (s) =
s→1
2r+t π t Rh
p
µ |∆|
(e)
Where r and t are the number of real and complex conjugate pairs of embeddings of
K, R is the regulator of K and µ the number of roots of unity contained in K, and of
course we have the old acquaintances, the class number h and the discriminant ∆.
The Class Number Formula
MAT4250 — Høst 2013
This formula is a main tool in computing the class number for many fields, the other
invariants occurring in the formula being more accessible, and therefore several call it
the class number formula. The first instance of the class number formula was found by
Dirichlet already in 1837. He proved a version for quadratic fields formulated in terms
of quadratic forms. The general formula as presented here was proven by Dedekind.
Let σ1 , . . . , σr+t be embeddings of K, chosen like we did in section xxx. That is
the first r are the real embeddings, in some order, and remaining t are chose one from
each pair of complex conjugate embeddings. Dirichlet’s unit theorem gives us a set
of fundamental units η1 , . . . , ηr+t−1 . Then the regulator is the absolute value any of
the minors (they are all equal up to sign since the row-sums of the matrix vanish, see
lemma 3 on page 11 below) of the (r + t) × (r + t − 1)-matrix
R• = (j log |σj (ηi )|)
(P)
The regulator plays a more conceptual role as the volume of the fundamental parallellotopes of the logarithmic unit lattice in the trace-zero hyperplane H in Rr+t−1 .
We closely follow the presentation in chapter 5 of Borevich and Shafarevisch [BorShaf].
Introducing the class group
The sumP(c) above defining the Dedekind Zeta function may be split into a sum
ζK (s) = c ζc (s) of functions ζc (s), each corresponding to a class c of the class group
CK . Indeed, one may take
X 1
ζc (s) =
N (a)s
a∈c
where the ideals a over which the sum is takes, are integral and, as indicated, confined
to the class c. The functions ζc (s) all have the same residue at s = 1. This explains
that the class number h appears as a factor in the class group formula, and the proof
of the class group formula reduces to a computation of the common residue at s = 1
of the functions ζc (s), the result being the formula
lim+ ζc (s) =
s→1
2r+t π t R
p
.
µ |∆|
Now, let us fix an ideal a0 ∈ c. Any other element a of the class c is of the form (f )a0
for some f ∈ K ∗ . For a to be an integral ideal a necessary and sufficient condition is
−1
that f belongs to a−1
0 (this is nothing else but the definition of a0 ). Introducing this
in the sum defining ζc (s), we find
ζc (s) =
X
a∈c
X
X
1
1
1
1
=
=
.
s
s
s
N (a)
N ((f )a0 )
N (a0 )
N ((f ))s
−1
−1
(f )⊆ a0
(f )⊆ a0
The two last sums are taken over all integral and principal ideals (f ) contained in a−1
0 .
0
0
Now, two principal ideals (f ) and (f ) are equal if and only if the elements f and f are
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The Class Number Formula
MAT4250 — Høst 2013
associate, that is, f = uf 0 for a unit u. So, letting S be a set of representatives of the
associates, i.e., S contains exactly one element from each class of associates in a−1
0 —or
phrased in a slightly different manner, S is a fundamental domain for the action of the
unit group U = A∗ on a−1
0 —one has
ζc (s) =
1
X
f ∈S
NK/Q (f )s
,
where we as well has replaced the counting norm N ((f )) by the the norm NK/Q (f )
(they are equal!).
The proof of this formula has two main ingredients. First we establish the formula
in
lims→1+ ζc (s) = γ/Γ where Γ is the volume of the Minkowski type lattice La−1
0
−1
K ⊗Q R = Rr ⊕ Ct associated to a0 , and γ is—with a friendly interpretation— the
volume of the part of the quotient a0 /U where the norm is at most one in absolute
value (the absolute value of the norm is a well defined function on the quotient since
N (u) = ±1 for units u). This is a special case of more general statement about counting
lattice elements belonging to subsets of a certain type.
We know the volume of the lattice La−1
. It was computed in xxx:
0
Γ = N (a0 )−1 2−t
p
|∆K |
The second ingredient of the proof is the computation of γ, and we shall find:
γ = 2r π t R/µ.
All together, this gives:
lim+ (s − 1)ζK (s) =N (a0 )−1
s→1
X
c∈CK
=N (a0 )−1 h
lim (s − 1)ζc (s) =
s→1+
γ
2r π t R/µ
2r+t π t RH
p
= p
.
= N (a0 )−1 h
Γ
N (a0 )−1 2−t |∆K |
µ |∆K |
The Minkowski setting
We recall the setting from the section on Minkowski’s geometry of numbers. Let
V = K ⊗Q R ' Rr ⊕ Ct ; the isomorphism being an isomorphism of algebras with
multiplication in Rr ⊕ Ct being defined componentwise. There is the embedding Σ of
K into V which is given by
Σ(α) = (σj (α))
where the σi -s are r + t chosen embeddings, chosen according to the usual rule that
the r first be the real embeddings and among the t next there should be one from each
pair of complex conjugate embeddings.
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The Class Number Formula
MAT4250 — Høst 2013
The logarithmic map. Recall the logarithmic map l : V0 → Rr+t given by
l(x) = (j log |xj |)
where x = (xj ) and the j are the weights corresponding to the embeddings σj ; with
j = 1 when σj is real, and j = 2 when σj is complex. The set V0 where the logarithm
is defined, consists of the elements in V = Rr ⊕Ct all of whose coordinates are non-zero.
Q
Norm and trace. The norm map N (x) on V is just the product xi i of the weighted
coordinates. It is multiplicative, and if ρx is the endomorphism of V given by multiplication by x, then det ρx = N (x).The trace map tr(y) on Rr+t is just the sum of the
coordinates of y. One has log |N (x)| = tr(l(x)).
The group of units. The group U of units is completely described by Dirichlet’s
unit theorem. It decomposes in a product U ' µK × Zr+t−1 where µK ⊆ K ∗ is the
group of roots of unity in K. We choose a basis η1 , . . . , ηr+t−1 for the free part, or as
ones says, the units η1 , . . . , ηr+t−1 form a fundamental set of units. We have shown
(and that was the hard part of proof of Dirichlet’s unit theorem) that the vectors
e1 = l(η1 ), . . . , er+t−1 = l(ηr+t−1 ) are linearly independent in Rr+t ; indeed, we showed
they form a basis for the trace-zero hyperplane H.
Introducing the vector e0 = (j ) = (1, . . . , 1, 2, . . . , 2) we thus have a basis e0 , . . . , er+t−1
for Rr+t . For any x ∈ V0 the coordinates of l(x) with respect to this will be denoted
by ξ0 , . . . , ξr+t−1 ; that is one has
l(x) = ξ0 e0 +
X
ξj ej .
1≤j≤r+t−1
The action of the unit group
The group of units U = A∗ acts on K by multiplication, of course, and when this
action is transported to V —where it shows up as multiplication by Σ(u)—it will still
be denoted by ux (where u ∈ u and x ∈ V ). This action is isometric since det ρu =
NK/Q (u) = ±1. The action is free on the set V0 where N (x) 6= 0, that is the set
of elements all whose coordinates are non-zero. Indeed if u is unit, clearly all the
σj (u) 6= 0, and for x ∈ V0 the equality ux = x therefore implies that u = 1.
The fundamental domain. We are in need of an explicit description of a fundamental domain for this action of U on V0 , that is a subset X⊆ V0 such each orbit
{ uv | u ∈ U } contains exactly one point from X.
The subset X we are looking for is defined by the following three constraints:
N (x) 6= 0
l(x) = ξ0 e0 +
P
1≤j≤r+t−1 ξj ej
where 0 ≤ ξj < 1 for 1 ≤ j ≤ r + t − 1
0 ≤ arg x1 < 2π/m.
There are two remarks to be made. First, in the second condition there is no
constraint on the coordinate ξ0 , Secondly, in case K has a real embedding, the last
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The Class Number Formula
MAT4250 — Høst 2013
condition reduces to the requirement that x1 be positive. Indeed, in that case µK =
{±1}, and m = 2.
Lemma 1 The set X described above is a fundamental domain for the action of U on
V0 .
Proof: Let x ∈ V0 and write
l(x) = ξ0 e0 +
X
ξj ej = ξ0 e0 +
X
(nj + γj )ej
j
j
where
Qnj n=j [ξj ] is the integral part of ξj and γj = ξj − nj satisfies 0 ≤ γj < 1. Letting
u = j ηj , one sees that
l(u−1 x) = l(x) − l(u) = ξ0 eo +
X
ξj ej −
X
j
nj ej = ξ0 eo +
j
X
γj ej .
j
Hence u−1 x satisfies the second request above. The first is trivially satisfied since
N (u−1 x) = N (u)−1 N (x) = N (x). Since µm = µK ⊆ K, we clearly may find a root of
unity ζ in K such that 0 ≤ arg ζx01 < 2π/m where x01 is the first coordinate of u−1 x.
Hence ζu−1 x satisfies the third request, and it faithfully continues to satisfy the second
since ζ being in the kernel of the logarithmic map l, gives l(ζu−1 x) = l(u−1 x).
We have established that every U -orbit hitsQ
X. To see that it hits just in one point,
0
0
assume that ux = x for x, x ∈ X and u = ζ j η nj . Writing out the coordinates of
l(x0 ) and l(ux) we get
ξ0 e0 +
X
ξj0 ej = ξ0 eo +
j
X
(nj + ξj )ej
j
It follows that nj = 0 since all of the ξi ’s and the ξi0 ’s lie between 0 and 1.
To finish off the proof, if both x1 and ζx1 has an argument in the interval [0, 2π/m),
the same holds for ζ, and being a m-th root of unity, it follows that ζ = 1.
o
The fundamental domain is a cone. We shall at a certain point need this fact.
So let t be a positive real number and let x ∈ X. Multiplication by t only changes
the component
Pof l(x) along e0 and does not affect the other components. Indeed, if
l(x) = ξ0 e0 + j ξj ej one finds
l(tx) = (j log |txj |) = (log |t| j ) + l(x) = (ξ0 + log |t|)e0 +
X
ξj ej ,
j
so we see that tx continues to satisfy the second request. For the third request, since t
is real and positive it certainly holds true that arg tx1 = arg x1 , and of course N (tx) =
td N (x), so N (tx) 6= 0 whenever N (x) 6= 0, and the first request is satisfied as well.
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The Class Number Formula
MAT4250 — Høst 2013
Riemann’s Zeta function
As everybody knows Riemann’s Zeta function is defined by
ζ(s) =
X 1
ns
n≥1
(X)
For our modest needs we assume that s is a real number lying in the interval (0, ∞),
although one may consider complex s in the half space Re s > 1. We shall not use ζ(s)
extensively, and only the following property:
Proposition 1 The sum (X) converges for s > 1, and lims→0+ (s − 1)ζ(s) = 1.
Proof: Elementary calculus gives
Z n+1
n
1
≤ s ≤
n
Z
n
n−1
dx
xs
where the first inequality holds for n ≥ 1 and the second for n ≥ 2. Hence, summing
over n one obtains for s > 1
Z ∞
Z ∞
1
dx
dx
1
=
≤ ζ(s) ≤ 1 +
=1+
.
s
s
s−1
x
x
s−1
1
1
o
The proof is divide into a few parts:
Volumes vs the residue of the zeta function
This paragraph explains the connection between the Dedekind Zeta function and certain volumes in the space Rr ⊕Cs . To be precise, the volume of the lattice La−1
and the
0
volume of the part of fundamental domain X where the norm takes values no greater
than one.
We shall work in more general situation, thus setting the points of the computation
in a clearer relief.
The setting is a real vector space V of dimension d equipped with an inner product
and a full lattice L⊆ V . The volume of L is denoted by Γ (recall, this is the volume of
a fundamental parallelotope of L).
Further we are given a cone X in V , i.e., a subset such that if x ∈ X and t > 0 a
real number, then tx ∈ X; and the last player in the plot, is a non-negative real-valued
function F (x) defined on V , taking positive values on X and satisfying the following
two conditions
F is homogeneous of degree d, i.e., F (tx) = td F (x)
The set T = { x ∈ V | F (x) ≤ 1 } is bounded and has a finite volume γ.
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The Class Number Formula
MAT4250 — Høst 2013
Then we have
Proposition 2 For s > 1 the series f (s) =
1
x∈L∩X F (x)s
P
lim+ (s − 1)f (x) =
s→1
converges, and one has
γ
.
Γ
Proof: There are three points in the proof.
Lattice points vs volumes. The first point of the proof is built on the intuitively
obvious statement, that the number of lattice points in a bounded set W , is approximately equal to the volume of the set divided by the volume of the fundamental
parallelotope of the lattice.
To be precise, let r > 0. The scaled down lattice r−1 L has a fundamental domain of
volume Γ/rd . We let N (r) be the number of lattice points from r−1 L lying in T , that
is N (r) = #T ∩ r−1 L. We may then cover T with N (r) translates of the fundamental
domain of scaled lattice r−1 L, and in that way obtain “tiling of T ” of total volume
N (r)Γ/rd . By letting r grow, one finds (that this holds is in fact the precise content
of the second assumption above)
γ = lim N (r)
r→∞
or
lim
r→∞
N (r)
Γ
= Γ lim
,
d
r→∞
r
rd
N (r)
γ
= .
d
r
Γ
(Z)
Volumes vs the function F . The second point of the proof is to link the volume
fraction γ/Γ to the behavior of F (x) for big values of x. Instead of making the lattice
small, one may make the set T big. Clearly N (r) = rT ∩L, and since F is a homogenous
function of x and X a cone, one sees that
rT = { rx | x ∈ X and F (x) ≤ 1 } = { y | y ∈ X and F (y) ≤ rd }.
Hence
N (r) = #{ y ∈ L ∩ X | F (y) ≤ rd }.
The set X ∩ L is countable, and we may order its the points according to increasing
values of F , that is X ∩ L = {x1 , x2 , . . . } with
0 < F (x1 ) ≤ F (x2 ) ≤ F (x3 ) ≤, . . . . . . ,
and we let rk be the real numbers such that rkn = F (xk ). The second point of the proof
is to establish the equality:
γ
k
= .
(c)
lim
k→∞ F (xk )
Γ
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The Class Number Formula
MAT4250 — Høst 2013
To that end, we have for all > 0
N (rk − ) ≤ k ≤ N (rk ),
hence as rkd = F (xk ) this gives
N (rk − ) (rk − )d
k
N (rk )
≤
≤
d
d
(rk − )
F (rk )
rk
rkd
Letting k → ∞ one obtains (c) in view of (Z).
The function
f (s) vs Riemann zeta.P The last point of the proof is to compare
P
1
1
the series n≥1 ns with the series f (s) = s∈L∩X F (x)
s.
For k >> 0, say k > k0 , one has
(
hence
(
γ
1
1
γ
1
− ) <
< ( − ) ,
Γ
k
F (xk )
Γ
k
1
γ
γ
1
1
− )s s <
< ( − )s s .
s
Γ
k
F (xk )
Γ
k
Hence summing for k ≥ k0 we get
(
X
X 1
X 1
1
γ
γ
s
− )s
≤
≤
(
−
)
Γ
k s k≥k F (xk )s
Γ
ks
k≥k
k≥k
0
this shows that thew series
s → 0+ , we obtain
(
P
1
k≥1 ks
0
0
converges, and multiplying by s − 1 and letting
γ
γ
− ) ≤ lim inf
(s
−
1)f
(s)
≤
lim
sup
(s
−
1)f
(s)
≤
(
− ),
s→1+
Γ
Γ
s→1+
from which it follows that lims→1+ (s − 1)f (s) = γ/Γ since lims→1+ (s − 1)ζ(s) = 1. o
The final volume computation
The only missing ingredient of our computation now, is the evaluation of the volume
of T . Recall that T is the subset of the fundamental domain X consisting of elements
of norm no greater than one. It is described by the following constraints:
0 < |N (x)| ≤ 1
P
l(x) = ξ0 e0 + 1≤j≤r+t−1 ξj ej where 0 ≤ ξj < 1 for 1 ≤ j ≤ r + t − 1
0 ≤ arg x1 < 2π/m.
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The Class Number Formula
MAT4250 — Høst 2013
Proposition 3 The volume γ of T is given as
vol(T ) =
2r π t R
µ
The whole of this paragraph is devoted to proof of this proposition. The proof is not
very deep, but kind of technical. Basically, the point is to find a good parametrization
of T . One uses polar coordinates for the complex coordinates, and observes that,
except for the third constraint, the angles can vary freely. One might say that the
“moral” content of the proof is that up to the action of roots of unity, and forgetting
the arguments of the complex part, T corresponds to a logarithmic cone over the
fundamental parallelotope of the unit lattice in Rr+t . The set T itself is bounded, but
its image l(T ) is not, it is stretch logarithmically in a direction away from the trace-zero
hyperplane. The precise content of these somehow vague descriptions is expressed in
the next lemma:
Lemma 2 For x ∈ V0 , one has the following expression for the logarithmic map
l(x) =
X
1
log |N (x)| e0 +
ξj ej
d
j
Proof:
For j > 0 the ej ’s live
P
P in the trace-zero hyperplane, so writing
P l(x) = ξe0 +
ξ
e
gives
tr(l(x))
=
dξ
=
log
|x
|
=
log
|N
(x)|,
since
tr(e
)
=
j
0
j j j
j j
j j = r+2t =
d.
o
The formula in the lemma inspire us to write down a parametrization of T , in fact
is a composition of two coordinate changes. Firstly, we use polar coordinates for the
complex coordinates and let ri denote the radius
Q vector, i.e., ri = |x|i . The Jacobian
of this coordinate change is the product J1 = r+1≤i≤r+t ri .
The arguments of these complex numbers do not appear in the two first constraints
defining T , and if we for a moment ignore the last constraint, they can be chosen freely.
The same applies to the signs of the real coordinates, and, as well, for the moment
we ignore the signs and assume the real coordinates xj for j ≤ r are positive, then
ri = xi for all i. This means that we shall give a parametrization of the set T0 of
points from X satisfying the first constraints above and whose real coordinates are all
positive. The parametrization is—in addition to keeping track of the angles— given
by the expression
X
i
ξj i log |σi (ηj )|
i log ri = log ξ0 +
d
j
where rj is given implicitly as a function of the ξj ’s. One observes that N (x) = ξ0 , and
hence the two first constraints defining T0 translate into 0 < ξj < 1. The domain of
the parameters is therefore (0, 1)r+t × [0, 2π)t .
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The Class Number Formula
MAT4250 — Høst 2013
The remaining step is to compute the Jacobian of this parametrization, and the
partials are given by:

i
if j = 0
i ∂ ri 
= dξ0
ri ∂ ξj i log |σi (ηk )| if j 6= 0 .
Hence the Jacobian determinant appears as
1 · · ·
·
·
·
·
·
·
.
.
..
Q ..
1
i ri Q
· · · i log |σi (ηj )| · · · J=
dξ0 i i .
..
..
.
···
··· r+t · · ·
Q
t
The factor
in
front
of
the
determinant
simplifies
to
1/(
r+1≤i≤r+t ri )2 d since ξ0 =
Q i
N (x) = i ri . The rightmost (r + t − 1) × (r + t)-minor is nothing else but the
transposed of the regulator matrix R• as defined in formula (P) on page 2, hence
evoking lemma 3 below we see that the Jacobian is given as
J=
R
2t
Q
r+1≤i≤r+t ri
.
Computing the integral one uses the product of the two Jacobians J1 and J, the product
of the ri ’s cancels, and integrand reduces to R/2t ; hence the volume of T0 is given as
vol T0 = π t R.
Coping with the third constraint. Let T1 be the subsets of X where only the two
first constraints are in force. We may divide T1 into m disjoints parts of equal volume,
of which T is one (in fact, the parts will be isometrically equivalent). To see this, let
ζ ∈ µK be a primitive root of unity. The multiplication map x 7→ S
ζx is an isometry of
V (its determinant is one, being the norm of ζ), and clearly T1 = 0≤k<m ζ k T ; indeed,
arg ζ k x1 = arg x1 + 2kπ/m. Replacing T by T1 , we have γ = vol(T1 )/m.
Coping with the signs of the real coordinates The two first constraints are
not sensitive to the sign of the real coordinates, so T1 is invariant under changing those
signs. And again, altering a sign is an isometry; indeed it is realized as multiplication
with an element of norm −1 all whose coordinates are 1 except one which equals −1.
This allows us to replace T1 by the set T0 consisting of those elements in T1 having
all their real coordinates positive. Clearly vol(T1 ) = 2r vol(T0 ), and finally we arrive
where we wanted to be:
2r π t R
γ = vol(T ) = 2 /m vol(T0 ) =
.
m
r
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The Class Number Formula
MAT4250 — Høst 2013
A simple lemma on determinants
The following lemma use a few times when the regulator enters the scene:
Lemma 3 Let A = (aij ) be an m × m-matrix whose column sums satisfy
(
X
a for j = 1
aij =
0 for 2 ≤ j ≤ m
i
that is, except for the first, the column sums all vanish. Then letting M be any the
minor of A obtained by deleting the first column and any row, we have det A = aM
Proof: In det A, one may replace the elements in k-th row by the corresponding
column sum, transforming the row into one with zeros everywhere except at the first
entry where there is an a. Developing the determinant along that row gives the lemma.
o
Hence if a (m − 1) × m-matrix has all column sums equal to zero, all the minors
are up to sign the same; we just add a column with all entries equal to 1/m and use
the lemma.
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The Class Number Formula
MAT4250 — Høst 2013
References
[BorShaf] Z. I. Borvich and I. R. Shafarevich Number theory, Academic Press 1966.
— 12 —