The Dirichlet L-functions are an extremely important tool in studying primes in arithmetic progressions – their zeros “control” distribution of primes in arithmetic progressions in the same manner as zeros of Riemann zeta function control the overall distribution of primes. The first and the most elementary result involving these zeros, which is the key result in most proofs of Dirichlet’s theorem, is that there is never a zero at point $$s=1$$. This post will present a proof of this fact using results from algebraic number theory.

This post is based solely on the content of Marcus’s Number Fields. The prerequisities for it are basic results about ideals in number fields and a minute amount of complex analysis. No background in analytic number theory is necessary.

## Characters and Dirichlet characters

Definition: A Dirichlet character modulo $$m$$ (or just a character) is a function $$\chi:\mathbb Z\rightarrow\mathbb C$$ such that:

• $$\chi$$ is periodic with period $$m$$,
• $$\chi(n)=0$$ iff $$\gcd(n,m)>1$$,
• $$\chi(kl)=\chi(k)\chi(l)$$ for all $$k,l$$.

It is easily seen that a Dirichlet character can be equivalently seen as a homomorphism $$(\mathbb Z/m\mathbb Z)^\times\rightarrow\mathbb C^\times$$, later extended to a function on $$\mathbb Z/m\mathbb Z$$ and, by periodicity, to $$\mathbb Z$$.

First we count the number of characters modulo $$m$$. First express $$(\mathbb Z/m\mathbb Z)^\times$$ as a product of cyclic groups $$G_1\times\dots\times G_n$$, say $$G_i$$ has a generator $$g_i$$ of order $$c_i$$, so that $$c_1\dots c_n=|(\mathbb Z/m\mathbb Z)^\times|=\varphi(m)$$. Then $$\chi$$ is determined by its values on $$g_i$$. Because $$\chi(g_i)^{c_i}=\chi(g_i^{c_i})=\chi(1)=1$$, $$\chi(g_i)$$ must be a $$c_i$$th root of unity. We have exactly $$c_i$$ of these, and we easily see that any choice of $$\chi(g_i)$$ to be a $$c_i$$th root of unity for $$i=1,\dots,n$$ gives us a well-defined character. Thus we have shown the following:

Lemma 1: There are exactly $$c_1\dots c_n=\varphi(m)$$  characters modulo $$m$$.

It is also useful to know how many characters take given value at given point $$g$$ of order $$c$$. First we will deal with taking value $$1$$. Note that $$\chi$$ is $$1$$ on $$g$$ iff it’s $$1$$ on whole subgroup $$\langle g\rangle$$ of size $$c$$, iff it induces a well-defined function on $$(\mathbb Z/m\mathbb Z)^\times/\langle g\rangle$$. Such a function is a homomorphism $$(\mathbb Z/m\mathbb Z)^\times/\langle g\rangle\rightarrow\mathbb C^\times$$. As above, we can show that there is exactly $$|(\mathbb Z/m\mathbb Z)^\times/\langle g\rangle|=\frac{\varphi(m)}{c}$$ of them. We can deduce exactly $$\frac{\varphi(m)}{c}$$ characters take value $$1$$ on $$c$$. Now for two characters $$\chi_1,\chi_2$$, they take the same value on $$g$$ iff $$\chi_1^{-1}\chi_2$$, which is also a character, takes value $$1$$ on $$g$$. So every value is taken the same number of times. Since there are only $$c$$ possible values, we get:

Lemma 2: For any $$c$$th root of unity $$\zeta$$, $$\chi(g)=\zeta$$ for precisely $$\frac{\varphi(m)}{c}$$ characters.

[Edit: here is a somewhat more straightforward proof of lemma 2, without needing to pass to the quotient group. We may assume that groups $$G_i$$ have prime power order, say $$c_i$$ is a power of $$p_i$$ (these primes needn’t be distinct). Write $$g=g_1^{a_1}\dots g_n^{a_n}$$. The order of $$G$$ is equal to the least common multiple of $$\frac{c_i}{\gcd(a_i,c_i)}$$, which are powers of respective $$p_i$$. For each prime $$p\mid |G|$$ let $$i_p$$ be such that $$\frac{c_{i_p}}{\gcd(a_{i_p},c_{i_p})}$$ is the greatest power of $$p$$ possible. The product of these numbers is $$c$$. Consider $$\chi$$ defined such that $$\chi(g_{i_p})=\zeta_{c_{i_p}}$$ and $$\chi(g_i)=1$$ otherwise. Then $$\chi(g_{i_p}^{a_{i_p}})$$ is a primitive $$\frac{c_{i_p}}{\gcd(a_{i_p},c_{i_p})}$$th root of unity, and now we easily see that $$\chi(g)$$ is a primitive $$c$$th root of unity. By taking $$\chi^k$$, which are also characters, we get what we’ve desired.]

Out of all Dirichlet characters modulo $$m$$, exactly one of them always takes value $$1$$ on $$(\mathbb Z/m\mathbb Z)^\times$$. We denote this character by $$\chi_0$$ and call it the principal character modulo $$m$$.

Proposition 1: $$\displaystyle\sum_{i\in(\mathbb Z/m\mathbb Z)^\times}\chi(i)=\begin{cases} \varphi(m) & \text{if }\chi=\chi_0,\\ 0 & \text{otherwise.} \end{cases}$$

Proof: This is clear for $$\chi=\chi_0$$, since we get a sum over zeroes. Suppose $$\chi\neq\chi_0$$. Then $$\chi(k)\neq 1$$ for some $$k\in(\mathbb Z/m\mathbb Z)^\times$$. But then we have

$$\displaystyle\sum_{i\in(\mathbb Z/m\mathbb Z)^\times}\chi(i)=\sum_{i\in(\mathbb Z/m\mathbb Z)^\times}\chi(ki)=\chi(k)\sum_{i\in(\mathbb Z/m\mathbb Z)^\times}\chi(i)$$

so $$\sum_{i\in(\mathbb Z/m\mathbb Z)^\times}\chi(i)=0$$. $$\square$$

## Dirichlet L-functions

Definition: For a character $$\chi$$ we define the associated Dirichlet L-function $$L(s,\chi)$$ as

$$L(s,\chi)=\displaystyle\sum_{n=1}^\infty\frac{\chi(n)}{n^s}$$.

This is treated as a function of a complex variable $$s\in\mathbb C$$. By $$x,y$$ we will denote, respectively, the real and imaginary part of $$s$$. We will establish a general result involving convergence of more general Dirichlet series, which will be also useful later.

Proposition 2: Let $$a_n,n\in\mathbb Z_+$$ be a sequence of complex numbers such that, for some fixed $$r\in\mathbb R$$, $$\displaystyle\left|\sum_{n=1}^k a_n\right|=O(n^r)$$ (using big O notation). Then the series

$$f(s)=\displaystyle\sum_{n=1}^\infty\frac{a_n}{n^s}$$

converges uniformly on any bounded subset of the halfplane $$x>r+\varepsilon$$ for any $$\varepsilon>0$$, and therefore defines an analytic function on that halfplane.

Proof: We will use Cauchy’s uniform convergence criterion. Let $$A_n=\displaystyle\sum_{i=1}^n a_n$$. Consider

$$\displaystyle\sum_{n=m}^M\frac{a_n}{n^s}= \sum_{n=m}^M\frac{A_n-A_{n-1}}{n^s}= \sum_{n=m}^M\frac{A_n}{n^s}-\sum_{n=m}^M\frac{A_{n-1}}{n^s}= \sum_{n=m}^M\frac{A_n}{n^s}-\sum_{n=m-1}^{M-1}\frac{A_n}{(n+1)^s}=\frac{A_M}{M^s}-\frac{A_{m-1}}{(m-1)^s}+\sum_{n=m}^{M-1}A_n\left(\frac{1}{n^s}-\frac{1}{(n+1)^s}\right)$$

We assumed there is a constant $$C$$ such that $$|A_n|<Cn^r$$. Additionally, $$|n^s|=n^x$$ and $$\displaystyle\left|\frac{1}{n^s}-\frac{1}{(n+1)^s}\right|=\left|s\int_n^{n+1}\frac{dt}{t^{s+1}}\right|\leq|s|\int_n^{n+1}\frac{dt}{t^{x+1}}\leq\frac{|s|}{n^{x+1}}$$, so we get

$$\displaystyle\left|\sum_{n=m}^M\frac{a_n}{n^s}\right|\leq CM^{r-x}+C(m-1)^{r-x}+|s|\sum_{n=m}^{M-1}Cn^{r-x-1}$$.

It is now straightforward to see that, for $$x>r+\varepsilon$$ this expression can be uniformly bounded (recall we assume $$|s|$$ is bounded), letting us use Cauchy’s criterion. Latter part of the proposition follows from standard facts from complex analysis. $$\square$$

Using proposition 2 we can get:

Corollary: $$L(s,\chi)$$ is well-defined and analytic for $$x>1$$, and the defining series converges absolutely. If $$\chi\neq\chi_0$$, it is well-defined and analytic for $$x>0$$.

In particular, for $$\chi\neq\chi_0$$, $$L(1,\chi)$$ is a well-defined complex number. The goal of this post is to show that it is nonzero.

A very important Dirichlet series is the Riemann zeta function, defined by $$\zeta(s)=\displaystyle\sum_{n=1}^\infty\frac{1}{n^s}$$. It can be seen as the Dirichlet L-function for the unique character modulo $$m=1$$. Right now we show that $$\zeta(s)$$ has a simple pole at $$s=1$$.

To see that, multiply $$\zeta(s)$$ by $$1-2^{1-s}$$, which has a simple zero at $$s=1$$. We get the series $$\displaystyle\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^s}$$, which by proposition 2 is analytic around $$s=1$$ and clearly takes nonzero value there. Hence $$\zeta(s)$$ indeed has a simple pole at that point.

Because of the absolute convergence of Dirichlet L-functions and multiplicativity of characters, for $$x>1$$ we can write

$$L(s,\chi)=\displaystyle\prod_p\left(\sum_{i=0}^\infty\frac{\chi(p^k)}{(p^k)^s}\right)=\prod_p\left(\sum_{i=0}^\infty\left(\frac{\chi(p)}{p^s}\right)^k\right)=\prod_p\left(1-\frac{\chi(p)}{p^s}\right)^{-1}$$

with the products running over all primes $$p$$. The idea here is, if we were to expand the product, then, thanks to unique factorization, each term $$\frac{\chi(n)}{n^s}$$ will appear exactly once. In particular, for Riemann zeta function, $$\zeta(s)=\displaystyle\prod_p\left(1-\frac{1}{p^s}\right)^{-1}$$ and for principal characters modulo $$m$$, $$L(s,\chi_0)=\displaystyle\prod_{p\nmid m}\left(1-\frac{1}{p^s}\right)^{-1}=\zeta(s)\prod_{p\mid m}\left(1-\frac{1}{p^s}\right)$$. Since the latter product has finitely many terms, it’s clearly analytic and nonzero in the neighbourhood of $$s=1$$, so we see $$L(s,\chi_0)$$ also has a simple pole at this point.

## Dedekind zeta functions

Definition: Given a number field $$K$$, denote by $$j_n$$ number of ideals in $$\mathcal O_K$$ which have norm $$n$$. We then define the Dedekind zeta function of $$K$$ as

$$\zeta_K(s)=\displaystyle\sum_{n=1}^\infty\frac{j_n}{n^s}$$.

In the particular case of $$K=\mathbb Q$$ we have $$j_n=1$$ and $$\zeta_\mathbb{Q}(s)=\zeta(s)$$.

The following result is rather technical and will not be proven here. It involves carefully counting elements of bounded norm in an ideal and avoiding counting unit multiples.

Theorem 1: For any number field $$K$$ there is a constant $$\kappa>0$$ such that $$\sum j_i=\kappa n+O(n^{1-\varepsilon})$$ for $$\varepsilon=\frac{1}{[K:\mathbb Q]}$$.

By proposition 2 the series is absolutely convergent for $$x>1$$. Because for such $$s$$ the order of summation will not matter, we can write

$$\zeta_K(s)=\displaystyle\sum_\frak{a}\frac{1}{N\frak a^s}$$,

where the sum runs over all nonzero ideals $$\frak a$$ in $$\mathcal O_K$$ and $$N$$ denotes the ideal norm. Using unique factorization of ideals we can prove a product formula like in the case of Dirichlet L-functions: we have

$$\zeta_K(s)=\displaystyle\prod_\frak{p}\left(1-\frac{1}{N\frak p^s}\right)^{-1}$$,

where the product runs over all prime ideals $$\frak p$$.

Corollary: $$\zeta_K(s)$$ has a simple pole at $$s=1$$.

Proof of the corollary: Define $$f(s)=\zeta_K(s)-\kappa\zeta(s)=\displaystyle\sum_{n=1}^\infty\frac{j_n-\kappa}{n^s}$$. By theorem 1 $$\displaystyle\sum_{i=1}^n(j_n-\kappa)=-\kappa n+\sum_{i=1}^nj_n=O(n^{1-\varepsilon})$$, so in particular $$f(s)$$ is analytic at $$s=1$$. Hence $$\zeta_K(s)=f(s)+\kappa\zeta(s)$$ has a simple pole, since $$\kappa\neq 0$$. $$\square$$

Assume now $$K$$ is a Galois extension, so that all primes lying above a given prime have the same ramification degree and inertia index. By grouping together terms which correspond to the same prime in $$\mathbb Z$$, we can further write

$$\zeta_K(s)=\displaystyle F(s)\prod_\frak{p}\left(1-\frac{1}{(p^{f(p)})^s}\right)^{-[K:Q]/f(p)}$$,

where $$F(s)$$ is the finite product coming from ramified primes.

Finally, letting $$K=\mathbb Q(\zeta_m)$$, where $$m$$ is a primitive $$m$$th root of unity, we have

Proposition 3: If $$p\nmid m$$, $$f(p)=c(p)$$, where $$c(p)$$ is the order of $$p$$ modulo $$m$$.

Let’s first see why this will give us what we want. Consider the product of all L-functions of characters modulo $$m$$. If any of them had a zero at $$s=1$$, this zero would cancel the simple pole of $$L(s,\chi_0)$$, hence this product would be analytic at $$s=1$$. On the other hand, for $$x>1$$ (so that the products converge absolutely),

$$\displaystyle\prod_\chi L(s,\chi)=\prod_\chi\prod_{p\nmid m}\left(1-\frac{\chi(p)}{p^s}\right)^{-1}=\prod_{p\nmid m}\prod_\chi\left(1-\frac{\chi(p)}{p^s}\right)^{-1}$$

Lemma 2 and proposition 3 imply that the inner product is equal to

$$\displaystyle\left(\prod_{i=0}^{c(p)}\left(1-\frac{\zeta_{c(p)}^i}{p^s}\right)\right)^{-\varphi(m)/c(p)}=\left(1-\frac{1}{(p^{c(p)})^s}\right)^{-\varphi(m)/c(p)}=\left(1-\frac{1}{(p^{f(p)})^s}\right)^{-\varphi(m)/f(p)}$$,

where $$\zeta_{c(p)}$$ is a primitive $$c(p)$$th root of unity and the equality follows from factoring $$x^{c(p)}-\frac{1}{(p^s)^{c(p)}}$$ and plugging $$x=1$$. Therefore

$$\displaystyle\prod_\chi L(s,\chi)=\prod_{p\nmid m}\left(1-\frac{1}{(p^{f(p)})^s}\right)^{-\varphi(m)/f(p)}=G(s)\zeta_K(s)$$,

where $$G(s)$$ is a finite product coming from ramified primes and primes dividing $$m$$. $$G(s)$$ clearly doesn’t vanish, so $$G(s)\zeta_K(s)$$ still has a simple pole. This is a contradiction since, as we said before, the product of L-functions is analytic at $$s=1$$ if one of them is zero there. Thus we have proven:

Theorem: If $$\chi$$ is a nonprincipal character modulo $$m$$, then $$L(1,\chi)\neq 0$$.

Proof of proposition 3: Let $$\frak p$$ be a prime lying above $$p$$ and let $$\sigma$$ be the automorphism of $$\mathcal O_K/\frak p$$ given by $$\sigma(a)=a^p$$ (the Frobenius automorphism). $$\sigma$$ generates the group of automorphisms of this field, so $$f(p)$$ is equal to the order of this automorphism. Since $$\overline{\zeta_m}$$ generates this field, $$f(p)$$ is the least positive integer $$k$$ such that $$\zeta_m^{p^k}\equiv\zeta_m\pmod{\frak p}$$. We will be done if we show that $$\zeta_m^l\equiv\zeta_m\pmod{\frak p}$$ implies $$l\equiv 1\pmod m$$. To see why this is true, note that this congruence implies $$\frak p\mid 1-\zeta_m^{l-1}$$ so, unless $$l\equiv 1\pmod m$$, this implies $$\frak p\mid\displaystyle\prod_{i=1}^{m-1}(1-\zeta_m^i)=m$$, because the product is value at $$x=1$$ of $$\displaystyle\prod_{i=1}^{m-1}(1-\zeta_m^i)=\frac{x^m-1}{x-1}=\sum_{i=0}^{m-1}x^i$$. Since $$\frak p$$ lies over $$p$$, this would imply $$p\mid m$$, which we assumed is not the case. Hence necessarily $$l\equiv 1\pmod m$$. $$\square$$

## Summary

First I’d like to remark that, although the above proof seems to use a lot of complex analysis, most of the steps can be done using only functions of real variable. Many proofs would unfortunately become more technical.

Here is a rough sketch of the idea of the above proof. It is often instructive to summarize an argument by recalling the main points, so that, if necessary, one could reconstruct the proof by filling in (sometimes purely technical, but sometimes requiring more sophisticated ideas) the details.

1. The L-functions are convergent on halfplane $$x>0$$, except for principal character one, which has a simple pole at $$1$$.
2. Assuming one of the L-functions vanishes, their product must be analytic at $$1$$.
3. Using the asymptotics on number on the number of ideals of given norm, Dedekind zeta function of a cyclotomic field has a simple pole at $$1$$ (it’s true in general; we only need it for cyclotopmic fields).
4. Both the Dedeking zeta function and the product of L-functions can be expressed as a product over primes, one involving inertia degrees, the other involving multiplicative orders.
5. Using Frobenius automorphism, inertia degree and multiplicative order of a prime are equal.
6. Therefore, the two functions are equal up to a factor which is analytic at $$1$$.
7. We get a contradiction, since one of the functions has a simple pole, while the other one doesn’t.