ZETA FUNCTIONS AND L-SERIES FOR A NUMBER FIELD

CARL MILLER

1. Introduction

Every number field K has an associated zeta function ζ_{K}(z), whose analytic properties are closely tied to number theoretic information about K. Zeta functions are also a subset of a larger class of analytic functions called L-series, which come from one-dimensional representations of subgroups of the ideal class group. These functions are useful for proving many results in class field theory.

We are concerned specifically with generalized Dirichlet L-series, which are associated to the ray class group of K for a chosen modulus 𝔪. We will define an L-series and prove convergence for Re(z) > 1. Since we are primarily interested in the values of the L-series near z = 1, we will need to show that the L-series has an analytic continuation to a neighborhood of 1. Our proof of the continuation relies on an important theorem about the distribution of norms in a generalized ideal class. In Section 6 we will give a summary proof of this theorem, referring the reader to Lang for some technical lemmas. Then, as an example application, we use L-series to prove a result in class field theory known as the Universal Norm Index Inequality.

2. Notation and Definitions

Throughout, let K/ℚ be a number field, let A denote its ring of integers, and let I and P denote the group of fractional ideals of K and principal fractional ideals of K, respectively. Let 𝔪 = 𝔪_{0}𝔪_{∞} be a formal product of places of K, with 𝔪_{0} the finite part and 𝔪_{∞} the infinite part. If v | 𝔪_{∞} is a valuation, let σ_{v} : K → k_{v} denote the corresponding embedding, with k_{v} = ℝ or ℂ. For convenience we will write x ≡ 1 mod 𝔪 to mean that

x ≡ 1 mod 𝔭^{m}

for all 𝔭^{m} | 𝔪_{0}, and

σ_{v}(x) > 0

for all real v | 𝔪_{∞}. Let I^{𝔪} denote the group of fractional ideals that are coprime to 𝔪_{0}, let P^{𝔪} ⊆ I^{𝔪} denote the subgroup generated by principal ideals (x) with x ≡ 1 mod 𝔪, and let Cl_{𝔪}(K) = I^{𝔪}/P^{𝔪} denote the ray class group of 𝔪. If J ⊆ A is an ideal, let N(J) denote the norm of J.

3. Dirichlet Characters

Let Ξ_{𝔪} be the group of complex characters χ : Cl_{𝔪}(K) → ℂ^{∗}. We note a few properties of this group. Recall that the ray class group is always finite (see [2], pp. 112). Proposition 3.1. The order of Ξ_{𝔪} is the same as the order of the group Cl_{𝔪}(K).

Proof. Write Cl_{𝔪}(K) as a direct product of finite cyclic groups,

Cl_{𝔪}(K) ≅ ℤ/n_{1}ℤ × … × ℤ/n_{r}ℤ

and let g_{i} ∈ Cl_{𝔪}(K) be their respective generators. A character χ is determined by the independent choice of mappings of each g_{i} to an n_{i}-th root of unity. There are n_{1}n_{2} ⋯ n_{r} = |Cl_{𝔪}(K)| possible choices. □

Proposition 3.2. Let χ_{0} ∈ Ξ_{𝔪} denote the trivial character. Any character χ : Cl_{𝔪}(K) → ℂ^{∗} satisfies

∑_{g∈Cl_{𝔪}(K)} χ(g) = \{ \begin{matrix}|Cl_{𝔪}(K)| & if χ = χ_{0}, \\ 0 & if χ ≠ χ_{0}. \end{matrix}

Proof. Note that for any h ∈ Cl_{𝔪}(K),

∑_{g∈Cl_{𝔪}(K)} χ(g) = ∑_{gh∈Cl_{𝔪}(K)} χ(gh) = χ(h) ∑_{g∈Cl_{𝔪}(K)} χ(g).

So either ∑_{g∈Cl_{𝔪}(K)} χ(g) = 0, or χ(h) = 1 for all h ∈ Cl_{𝔪}(K), and hence ∑_{g∈Cl_{𝔪}(K)} = |Cl_{𝔪}(K)|, as desired. □

By a similar trick, we have Proposition 3.3. For any ideal class g ∈ Cl_{𝔪}(K),

∑_{χ∈Ξ_{𝔪}} χ(g) = \{ \begin{matrix}|Ξ_{𝔪}| = |Cl_{𝔪}(K)| & if g = P^{𝔪}, \\ 0 & otherwise. \end{matrix} □

4. Zeta Functions and L-series

For a generalized Dirichlet character χ : Cl_{𝔪}(K) → ℂ^{∗}, the L-series L_{𝔪}(z, χ) is given by

L_{𝔪}(z, χ) = ∑_{(J,𝔪)=1} \frac{χ(J)}{N(J)^{z}}

where the sum is over integral ideals J that are coprime to 𝔪. Note that if 𝔪 = 1 and χ : Cl(K) → ℂ^{∗} is trivial, we obtain the zeta function of the number field K:

ζ_{K}(z) = ∑_{J⊆A} \frac{1}{N(J)^{z}} .

It will be useful when working with L-series to refer to the zeta function of an ideal class 𝔍 ∈ Cl_{𝔪}(K), given by

ζ(z, 𝔍) = ∑_{J∈𝔍} \frac{1}{N(J)^{z}} .

Note that

L_{𝔪}(z, χ) = ∑_{𝔍∈Cl_{𝔪}(K)} χ(𝔍)ζ(z, 𝔍).

We show first that these series converge when Re(z) > 1. Note that for p prime, the number of ideals 𝔭 satisfying N(𝔭) = p is less than or equal to d = [K : ℚ], and secondly, the number of prime factors (counted with multiplicity) for a given n is less than or equal to log_{2} n. Combining these facts shows that the number of ideals J satisfying N(J) = n is less than or equal to (log_{2} n)^{d}. Thus all the series above are dominated (in absolute value) by

∑_{n=1}^{∞} \frac{(log_{2} n)^{d}}{n^{Re(z)}} ,

which converges provided that Re(z) > 1.

The following product form of the L-series will be useful in later proofs. Proposition 4.1. For z ∈ ℂ with Re(z) > 1,

L_{𝔪}(z, χ) = ∏_{𝔭∤𝔪} \frac{1}{1 − \frac{χ(𝔭)}{N(𝔭)^{z}}} .

Proof. Taking the formal logarithm of the series gives

∑_{𝔭∤𝔪} ∑_{n=1}^{∞} \frac{χ(𝔭)^{n}}{nN(𝔭)^{nz}}

which is dominated by

∑_{p∈ℕ} ∑_{n=1}^{∞} \frac{d}{np^{nz}} < ∑_{m∈ℕ} \frac{d}{m^{z}} ,

and this converges for Re(z) > 1. Thus we may safely expand and multiply out the series:

∏_{𝔭∤𝔪} \frac{1}{1 − \frac{χ(𝔭)}{N(𝔭)^{z}}} = ∏_{𝔭∤𝔪} (1 + \frac{χ(𝔭)}{N(𝔭)^{z}} + \frac{χ(𝔭)^{2}}{N(𝔭)^{2z}} + ⋯)

The function χ(·)/N(·)^{z} is multiplicative, thus

∏_{𝔭∤𝔪} (1 + \frac{χ(𝔭)}{N(𝔭)^{z}} + \frac{χ(𝔭)^{2}}{N(𝔭)^{2z}} + ⋯) = ∑_{(J,𝔪)=1} \frac{χ(J)}{N(J)^{z}}

as desired. □

Already we have seen a connection between the L-series and the number theoretic properties of K: the multiplicative expression for L_{𝔪}(z, χ) is immediately equivalent to unique factorization for ideals in A.

5. The Riemann Zeta Function

We have seen that the series for L_{𝔪}(z, χ) converges to an analytic function for Re(z) > 1. However since we are interested in the values of L_{𝔪}(z, χ) near z = 1, we need to prove the existence of a larger analytic continuation. We begin by considering the Riemann zeta function,

ζ_{ℚ}(z) = ∑_{p∈ℕ} \frac{1}{p^{z}} .

The convergence theorem for the Riemann zeta function will be useful as an intermediate step in proving a similar theorem for general L-series. Theorem 5.1. There exists an analytic continuation of ζ_{ℚ}(z) to the domain {z : Re(z) > 0, z ≠ 1}.

Proof. Consider the alternating series

ζ_{2}(z) = 1 − \frac{1}{2^{z}} + \frac{1}{3^{z}} − \frac{1}{4^{z}} + ….

Let σ = Re(z); suppose σ > 0. Then

|ζ_{2}(z)| = |z ∫_{1}^{2} \frac{ds}{s^{z+1}} + z ∫_{3}^{4} \frac{ds}{s^{z+1}} + z ∫_{5}^{6} \frac{ds}{s^{z+1}} + …|
≤ \frac{|z|}{1^{σ+1}} + \frac{|z|}{3^{σ+1}} + \frac{|z|}{5^{σ+1}} + …

which is convergent. Thus ζ_{2}(z) is analytic on {z : Re(z) > 0}. Now note that

ζ_{ℚ}(z) − (\frac{2}{2^{z}} ) ζ_{ℚ}(z) = ζ_{2}(z),

for Re(z) > 1, hence the formula

ζ_{ℚ}(z) = (1 − \frac{1}{2^{z−1}} )^{−1} ζ_{2}(z)

gives an analytic continuation of ζ_{ℚ}(z) to the region {z : Re(z) > 0} minus the zeroes of the function (1 − 1/2^{z−1}). We can take care of all but one of these exceptional points by considering another alternating zeta series,

ζ_{3}(z) = ∑_{n=1}^{∞} (\frac{1}{(3n + 1)^{z}} + \frac{1}{(3n + 2)^{z}} − \frac{2}{(3n + 3)^{z}} ) .

By similar reasoning as above, ζ_{3}(z) converges for Re(z) > 0, and

ζ_{ℚ}(z) = (1 − \frac{1}{3^{z−1}} )^{−1} ζ_{3}(z)

Thus we have an analytic continuation of ζ_{ℚ}(z) to {z : Re(z) > 0} excluding the zeroes of (1 − 1/3^{z−1}). Now suppose z is a zero of both (1 − 1/2^{z−1}) and (1 − 1/3^{z−1}). Then 2^{z−1} = 3^{z−1} = 1, hence z = 1 + θi and

θ log 2 ≡ θ log 3 ≡ 0 mod 2π

But log 2/ log 3 is irrational, so θ = 0. Thus z = 1 is the only excluded point. □

Proposition 5.2. The zeta function ζ_{ℚ}(z) has a simple pole at z = 1 with residue 1.

Proof. Using the expression from the last proof,

ζ_{ℚ}(z) = (1 − \frac{1}{2^{z−1}} )^{−1} ζ_{2}(z).

Now ζ_{2}(1) is the power series expression for log 2, while the residue of (1 − 1/2^{z−1})^{−1} at 1 is 1/ log 2. Thus Res_{z=1} ζ_{ℚ}(z) = 1. □

In fact any zeta function may be extended to ℂ − {1}, but it is not necessary for us to prove that here (see [3], Ch. VII). All “nontrivial” zeroes of the Riemann zeta function lie in the strip {z : 0 ≤ z ≤ 1} ⊆ ℂ, so what we have proved above is sufficient to state the famous Riemann hypothesis. Conjecture 5.3. All nontrivial zeroes of ζ_{ℚ}(z) lie on the line {z : Re(z) = 1/2}.

6. The Distribution of Norms in an Ideal Class

Let 𝔍 ∈ Cl_{𝔪}(K) be an ideal class. In order to evaluate the zeta function ζ(z, 𝔍), we need an asymptotic estimate for the function

j(𝔍, t) = |{J ∈ 𝔍 : J ⊆ A, N(J) ≤ t}| .

Choose an (integral) ideal B from the inverse class 𝔍^{−1}. The one-to-one map J ↦ JB takes an integral ideal in 𝔍 to a principal ideal; the image of this map is precisely the set of ideals in P^{𝔪} that are generated by elements of B. Also, N(J) ≤ t if and only if N(JB) ≤ tN(B). So it suffices to count the number of ideals (x) ∈ P^{𝔪} such that x ∈ B and N(x) ≤ tN(B).

To put this another way: Lemma 6.1. The function j(𝔍, t) is equal to the number of elements x ∈ B satisfying

x ≡ 1 mod 𝔪 and N(x) ≤ tN(B)

modulo the unit group U^{𝔪}. □

We will describe a geometric interpretation for the set of such elements. Let S_{∞} denote the set of infinite primes of K. The embeddings {σ_{v}}_{v∈S_{∞}} give an embedding

K → ∏_{v∈S_{∞}} k_{v} ≅ ℝ^{n}.

The vector space ∏_{v∈S_{∞}} k_{v} has a norm

N(x) = ∏_{v∈S_{∞}} v(x_{v})^{N_{v}}

(here N_{v} denotes the multiplicity of v in 𝔪_{∞}). Thus we define the unit ball B^{1} ⊆ ∏_{v∈S_{∞}} k_{v}. Note that the image of an ideal under this map is a lattice.

We wish to make use of the following idea: if a region D ⊆ ℝ^{n} is sufficiently nice, then we can estimate its volume to an arbitrary degree of accuracy by counting lattice points inside D for sufficiently fine lattices. This is made precise in the theorem below. Theorem 6.2. Suppose D ⊆ ℝ^{n} is a subset whose boundary is (n − 1)-Lipshitz parametrizable, and suppose L ⊆ ℝ^{n} is a lattice with fundamental parallelotope P. Let λ(t, D, L) be the number of lattice points in tD. Then

λ(t, D, L) = \frac{Vol(D)}{Vol(P)}t^{n} + O(t^{n−1}).

Proof. See [1], pp. 128. □

We will show that j(𝔍, t) is equal to λ(t, D, L) for appropriately chosen D and L. First note that the group of units U^{𝔪} acts on ∏_{v∈S_{∞}} k_{v} via u(x) = (q(u) · x). This action becomes fixed-point free if we restrict to the subset

Q(S_{∞}) = {(q_{v})_{v∈S_{∞}} : q_{v} ≠ 0 for all v, and q_{v} > 0 for real v | 𝔪_{∞}} .

Let V ⊆ U^{𝔪} be a maximal free subgroup. Lemma 6.3. There exists a fundamental domain F ⊆ ℝ^{n} for the action of V on Q(S_{∞}) such that tF = F for all t > 0. Also, F ∩ B^{1} has (N − 1)-parametrizable boundary.

Proof. See [1], pp. 131. □

Now let D = F ∩ B^{1} and let L ⊆ ℝ^{n} be the image of the B𝔪_{0}.

Choose (by the Chinese Remainder Theorem) an element y ∈ B such that y ≡ 1 mod 𝔪_{0}. Consider the translated lattice y + L. Every element of (y + L) ∩ F is the image of an element x ∈ B such that (x) ∈ P^{𝔪} (recall Lemma 6.1). Conversely, every such principal ideal (x) has exactly |U^{𝔪}/V | representatives in (y + L) ∩ F, since F is a fundamental domain for the action V ⊆ U^{𝔪}. Let w_{𝔪} = |U^{𝔪}/V | (this is the number of roots of unity of K, which is finite). We have

w_{𝔪}j(𝔍, t^{n}) = |{x ∈ ℝ^{n} : x ∈ F, N(x) ≤ N(B)t^{n}}|
= λ(tN(B), D, y + L).

The translation by y is asymptotically irrelevant; by Theorem 6.2,

w_{𝔪}j(𝔍, t^{n}) = \frac{N(B) Vol(D)}{Vol(P)}t^{n} + O(t^{n−1}).

where P denotes the fundamental parallelotope of L. Therefore,

j(𝔍, t) = \frac{N(B) Vol(D)}{w_{𝔪} Vol(P)}t + O(t^{1−1/n}).

Now, the volume of the fundamental parallelotope corresponding to B is linearly related to the norm of B (see [1], pp. 115):

\frac{N(B)}{Vol(P)} = \frac{2^{r}}{\sqrt{|D_{K}|}}

where r is the number of complex primes of K and D_{K} is the discriminant of K. Thus let

ρ_{𝔪} = \frac{N(B) Vol(D)}{w_{𝔪} Vol(P)} = \frac{2^{r} Vol(D)}{w_{𝔪}\sqrt{|D_{K}|}}.

We have proven Theorem 6.4. There exist constants ρ_{𝔪} and n > 1 such that for any 𝔍 ∈ Cl_{𝔪}(K), the function j(𝔍, t) = |{J ∈ 𝔍 : N(J) ≤ t}| satisfies

j(𝔍, t) = ρ_{m}t + O(t^{1−1/n}).

Thus we see that the distribution of the norm function on an ideal class is asymptotically linear, and moreover, the linear coefficient is the same for any ideal class in Cl_{m}(K).

We note that there is a nice formula for Vol(D) in terms of the invariants of the number field K (but we won’t need to prove that here). This leads to a formula for the residue of ζ_{K}(z) at 1 called the Class Number Formula (see [3], pp. 467).

7. L-series Convergence

Consider the ideal class zeta function,

ζ(z, 𝔍) = ∑_{J∈𝔍} \frac{1}{N(J)^{z}} .

Let a_{m} = |{J ∈ 𝔍 : N(J) = m}|. We have

ζ(z, 𝔍) = ∑_{m=1}^{∞} \frac{a_{m}}{m^{z}} ,

and by Theorem 6.4,

∑_{m=1}^{t} a_{m} = ρ_{𝔪}t + O(t^{1−1/n}).

We will demonstrate an analytic continuation of ζ(z, 𝔍) via comparison with the Riemann zeta function. Let s_{t} = ∑_{m=1}^{t}(a_{m} − ρ_{𝔪}). Observe:

| ∑_{m=1}^{t} \frac{a_{m} − ρ_{𝔪}}{m^{z}} | = | ∑_{m=1}^{t} \frac{s_{m}}{m^{z}} − ∑_{m=0}^{t−1} \frac{s_{m}}{(m + 1)^{z}} |

= |\frac{s_{t}}{t^{z}} + ∑_{m=1}^{t−1} s_{m} (\frac{1}{m^{z}} − \frac{1}{(m + 1)^{z}} )|

≤ |\frac{s_{t}}{t^{z}} | + | ∑_{m=1}^{t−1} s_{m}z ∫_{m}^{m+1} \frac{ds}{s^{z+1}} |

≤ |\frac{s_{t}}{t^{z}} | + ∑_{m=1}^{t−1} \frac{|s_{m}z|}{m^{1−1/n}} ∫_{m}^{m+1} \frac{ds}{|s^{z+1/n}|}

Let σ = Re(z) and suppose σ > 1 − 1/n. s_{t} = O(t^{1−1/n}), so let C be an upper bound for |s_{t}/t^{1−1/n}|. We have

| ∑_{m=1}^{t} \frac{a_{m} − ρ_{𝔪}}{m^{z}} | = C + C|z| ∑_{m=1}^{t−1} ∫_{m}^{m+1} \frac{ds}{s^{δ+1/n}}

= C + C|z| ∫_{1}^{t} \frac{ds}{s^{δ+1/n}}

which is bounded. Thus,

∑_{m=1}^{t} \frac{a_{m} − ρ_{𝔪}}{m^{z}}

converges to an analytic function for Re(z) > 1 − 1/n. Using the results of Section 5,

ζ(z, 𝔍) = ( ∑_{m=1}^{t} \frac{a_{m} − ρ_{𝔪}}{m^{z}} ) + ρ_{𝔪}ζ_{ℚ}(z)

is analytic on {z : Re(z) > 1 − 1/n}, except at z = 1 where it has a pole with residue ρ_{𝔪}. The zeta function for the number field K satisfies

ζ_{K}(z) = ∑_{𝔍∈Cl(K)} ζ(z, 𝔍),

thus ζ_{K}(z) is analytic on {z : Re(z) > 1 − 1/n} except for a pole at 1 with residue ρ_{1} |Cl(K)|. Finally, if χ : Cl_{𝔪} → ℂ^{∗} is a nontrivial character, then

L_{𝔪}(z, χ) = ∑_{𝔍∈Cl(K)} χ(𝔍)ζ(z, 𝔍)

= ∑_{𝔍∈Cl(K)} χ(𝔍) ( ∑_{m=1}^{t} \frac{a_{m}(𝔍) − ρ_{𝔪}}{m^{z}} ) + ( ∑_{𝔍∈Cl(K)} χ(𝔍)) ρ_{𝔪}ζ_{ℚ}(z)

= ∑_{𝔍∈Cl(K)} χ(𝔍) ( ∑_{m=1}^{t} \frac{a_{m}(𝔍) − ρ_{𝔪}}{m^{z}} )

using Proposition 3.2. Thus L_{𝔪}(z, χ) is analytic on all of the region {z : Re(z) > 1 − 1/n}.

8. The Universal Norm Index Inequality

An inequality that is useful in class field theory can be proven from the information about ζ_{K}(z) and L_{𝔪}(z, χ) near z = 1 that we developed above. We will need some asymptotic notation: for analytic functions f and g, write f ∼ g (z → 1) to mean that f/g can be continued analytically to 1. Write f ≾ g (z → 1) and f ≿ g (z → 1) to mean that f/g is bounded above near 1, and bounded below near 1, respectively. Recall the series for the logarithm of L_{𝔪}:

L_{𝔪}(z, χ) = exp (∑_{𝔭∤𝔪} ∑_{m} \frac{χ(𝔭)^{m}}{mN(𝔭)^{mz}} ) .

The sum of the second- and higher-order terms in the series converges for z near 1, thus

L_{𝔪}(z, χ) ∼ exp (∑_{𝔭∤𝔪} \frac{χ(𝔭)}{N(𝔭)^{z}} ) (z → 1).

Now, suppose L/K is a Galois extension, and B ⊆ L the ring of integers. Let 𝔑_{L}^{𝔪} ⊆ K be the group of norms of fractional ideals in L that are coprime to 𝔪. We are interested in the subgroup of the ray class group,

P^{𝔪} ⊆ 𝔑_{L}^{𝔪}P^{𝔪} ⊆ I^{𝔪}.

Theorem 8.1. If 𝔪 is divisible by all the ramified primes of L/K, then

(I^{𝔪} : P^{𝔪}𝔑_{L}^{𝔪}) ≤ [L : K].

(The degree (I^{𝔪} : P^{𝔪}𝔑_{L}^{𝔪}) is called the norm index of K.)

Proof. Write H = P^{𝔪}𝔑_{L}^{𝔪} for convenience. Consider the product

∏_{χ:I^{𝔪}/H→ℂ^{∗}} L_{𝔪}(z, χ).

Recall that L_{𝔪}(z, χ) is analytic at 1 unless χ is trivial, in which case L_{𝔪}(z, χ) has a simple pole at 1. Thus,

\frac{1}{z − 1} ≿ ∏_{χ} L_{𝔪}(z, χ) (z → 1).

Replacing L_{𝔪}(z, χ) with the approximation given above,

\frac{1}{z − 1} ≿ exp (∑_{χ} ∑_{𝔭∤𝔪} \frac{χ(𝔭)}{N(𝔭)^{z}} )

= exp [ ∑_{𝔍∈I^{𝔪}/H} (∑_{χ} χ(𝔍)) ∑_{𝔭∈𝔍} \frac{1}{N(𝔭)^{z}} ]

= exp [(I^{𝔪} : H) ∑_{𝔭∈H} \frac{1}{N(𝔭)^{z}} ]

by Proposition 3.3.

Now, if 𝔭 is a prime in K, and {𝔮_{i}}_{i=1}^{k} are the primes lying above 𝔭 with residue field extension degrees deg 𝔮_{i}, then N(𝔮) = N(𝔭)^{deg 𝔮_{i}}, so

∑_{j=1}^{k} \frac{1}{N(𝔮_{j})^{z}} = ∑_{deg 𝔮_{𝔧}=1} \frac{1}{N(𝔭)^{z}} + \{terms with exponent ≥ 2z\}

≤ \frac{[L : K]}{N(𝔭)^{z}} + \{terms with exponent ≥ 2z\}

And every prime 𝔮 in B lies over a prime in H by definition. Thus, by ignoring terms with exponent ≥ 2z (which are asymptotically irrelevant near z = 1) we find

\frac{1}{z − 1} ≿ exp ((I^{𝔪} : H) ∑_{𝔭∈H} \frac{1}{N(𝔭)^{z}} )

≿ exp (\frac{(I^{𝔪} : H)}{[L : K]} ∑_{𝔮⊆B} \frac{1}{N(𝔮)^{z}} )

= ζ_{L}(z)^{\frac{(I^{m}:H)}{[L:K]}}

≿ (\frac{1}{z − 1})^{\frac{(I^{m}:H)}{[L:K]}}

Thus (I^{m} : H) ≤ [L : K], as desired. □

References

[1] S. Lang: Algebraic Number Theory, Addison-Wesley, 1970.

[2] G. Janusz: Algebraic Number Fields, Acadmic Press, Inc., 1973.

[3] J. Neukirch: Algebraic Number Theory, Springer-Verlag, 1999.