Dirichlet's Theorem on Primes in Arithmetic Progression

Published August 27, 2026Last updated August 28, 2026

Abstract

The fact that there are infinitely many primes is one of the oldest and most fundamental results in number theory. This result extends to arithmetic progressions of the form $a, a + q, a + 2q, \ldots$ , where $a$ and $q$ are coprime. Dirichlet’s theorem states that any such progression contains infinitely many primes. In this talk, we discuss tools from analytic number theory, including Dirichlet characters and $L$ -functions, and sketch a proof of Dirichlet’s theorem.

Introduction

Dirichlet’s theorem on primes in arithmetic progressions (roughly) states:

Dirichlet’s theorem (initial statement). Let $a, q$ be natural numbers such that $(a, q) = 1$ . Then the sequence $a, a + q, a + 2q, \ldots$ contains infinitely many primes.

Here the notation $(a,b)$ is the same as $\gcd(a,b)$ . This is a remarkable result! However, to prove it, we will need to state it in a slightly different manner. Consider the following:

Let $a,q$ be natural numbers which are coprime. Then

\sum_{p \equiv a\,(\mathrm{mod}\,q)} \frac{\log p}{p} = \infty.

By convention, we only allow $p$ to be a prime in analytic number theory. So this theorem states that the sum of $\log(p)/p$ over all primes which are congruent to $a$ modulo $q$ , i.e., are part of the sequence $a, a + q, a + 2q, \ldots$ , is infinite.

This directly implies that the number of such primes is infinite, since if the number of primes were finite then the sum would be over a finite set and thus would trivially be finite. This is the formulation of Dirichlet’s theorem for which we will sketch a proof.

We introduce a few preliminaries, define Dirichlet characters and $L$ -functions and prove some facts about them, and finally sketch the proof of Dirichlet’s theorem.

Important functions

Write $s=\sigma+it$ , where $\sigma,t\in\mathbb{R}$ . The Riemann zeta function is defined on the half-plane $\sigma>1$ as

\zeta(s) = \sum_{n=1}^{\infty} n^{-s}.

This sum converges absolutely. It can also be expressed as the Euler product

\zeta(s) = \prod_{p\text{ prime}} \left(1 - \frac{1}{p^s}\right)^{-1}.

Definition 2 (von Mangoldt function). The von Mangoldt function is the arithmetic function $\Lambda(n)$ defined as

\Lambda(n) = \begin{cases} \log p & \text{if } n=p^k \text{ for some prime } p, \\ 0 & \text{otherwise}. \end{cases}

Dirichlet characters

We now begin the study of the actual objects involved in analytic number theory which lead to the proof of Dirichlet’s theorem.

Definition 3 (Dirichlet character modulo $m$ ). For any positive integer $m$ , a Dirichlet character modulo $m$ is a function $\chi:\mathbb{Z}\to\mathbb{C}$ which satisfies the following conditions.

$\chi$ is completely multiplicative, i.e., $\chi(ab)=\chi(a)\chi(b)$ for every $a,b\in\mathbb{Z}$ ,
$\chi(a)=0$ if and only if $(a,m)>1$ , and
$\chi$ is periodic with period $m$ , i.e., $\chi(a)=\chi(a+m)$ for every $a\in\mathbb{Z}$ .

One observes that the values of $\chi$ are determined uniquely by those integers which are between $0$ and $m$ and are coprime to $m$ ; this gives rise to an alternative characterization of Dirichlet characters of which the above is a special case.

Definition 4 (Character on a group). Let $G$ be a finite abelian group. Then a character of the group $G$ is a group homomorphism $\chi:G\to\mathbb{C}^{\times}$ .

Clearly, we see that a character modulo $m$ can be viewed as a special case of this construction by letting $G=(\mathbb{Z}/m\mathbb{Z})^{\times}$ and, letting $\chi$ be the group character, defining $\chi':\mathbb{Z}\to\mathbb{C}$ to be

\chi'(n) = \begin{cases} \chi_G(\overline{n}) & \text{if } (n,m)=1, \\ 0 & \text{otherwise}, \end{cases}

where $\overline{n}$ is the equivalence class mod $m$ to which $n$ maps. It is not hard to see that this relationship defines a one-to-one correspondence between the set of group characters of $(\mathbb{Z}/m\mathbb{Z})^{\times}$ and the set of Dirichlet characters mod $m$ .

Remark. Recall that $(\mathbb{Z}/m\mathbb{Z})^{\times}$ , sometimes denoted $U_m$ , is the multiplicative group of the numbers between $0$ and $m$ which are coprime to $m$ . It has order $\phi(m)$ , where $\phi$ is Euler’s totient function.

Remark. Notice that if $G$ has order $n$ , then for every $x\in G$ and $\chi\in\widehat{G}$ , it holds that $\chi(x)$ is an $n$ -th root of unity, since $1=\chi(1)=\chi(x^n)=(\chi(x))^n$ .

Definition 5. For any group $G$ (likewise any modulus $m$ ), we always have a character which maps all values of the group to $1$ (likewise maps all values coprime to $m$ to $1$ and all other values to $0$ ). This character, known as the trivial character, is denoted $\chi_0$ . The set of all characters of a group $G$ is denoted as $\widehat{G}$ .

We can define a multiplication on $\widehat{G}$ by letting

(\chi_a\chi_b)(x)=\chi_a(x)\chi_b(x).

Observe that this makes $\widehat{G}$ an abelian group—commutativity and associativity follow because $\mathbb{C}^{\times}$ has those properties. The trivial character is an identity element, and given a character $\chi$ , we can set its inverse $\overline{\chi}$ such that

\overline{\chi}(x)=\overline{\chi(x)},

where $\overline{z}$ is the complex conjugate of $z$ . It is not hard to show that $\overline{\chi}$ is also a character, and showing that it is the inverse of $\chi$ can be done by exploiting the fact that multiplicative inverses of roots of unity are their complex conjugates.

Let $\chi$ be a character of a group $G$ which is cyclic with order $n$ and generator $a$ . Then the elements of $\widehat{G}$ can be identified with values of $k$ for $k\in\{0,1,2,\ldots,n-1\}$ such that

\chi_k(a^m)=e(km/n)=e^{(2\pi i k m)/n}.

Moreover, in this case $G\cong\widehat{G}$ .

Consider an arbitrary character $\chi$ . We know $a$ generates $G$ , and that $\chi(a)$ is an $n$ -th root of unity. So we know it is of the form $e(k/n)$ for some $k$ . But since $\chi$ is a group homomorphism we know for every $m$ that $\chi(a^m)=e(k/n)^m=e(km/n)$ . Hence every character is of this form, and since we see that $k$ wholly describes the value $\chi$ takes at every point, the elements of $\widehat{G}$ correspond to the set $\{0,1,\ldots,n-1\}$ .

Further, notice that, for arbitrary $\alpha,\beta\in\{0,1,\ldots,n-1\}$ ,

\begin{aligned} (\chi_\alpha\chi_\beta)(a^m) &=e(\alpha m/n)\cdot e(\beta m/n) \\ &=e((\alpha+\beta)m/n) \\ &=\chi_{\alpha+\beta}(a^m). \end{aligned}

From this it follows that $\widehat{G}\cong((\mathbb{Z}/n\mathbb{Z}),+)$ , which is cyclic with order $n$ . Immediately it follows that $\widehat{G}\cong G$ since there is only one cyclic group of any given order (up to isomorphism).

Let $G$ be a finite abelian group which is the direct product of two groups, i.e., $G=G_1\times G_2$ . Then, for every character $\chi_1$ of $G_1$ and $\chi_2$ of $G_2$ , we can define a character of $G$ as $\chi((x_1,x_2))=\chi_1(x_1)\chi_2(x_2)$ . Any character $\chi$ of $G$ can likewise be decomposed into unique characters of $G_1$ and $G_2$ .

This is Lemma 4.3 of [MV06]. The proof is omitted here for brevity.

Corollary 3.1. Let $G$ be a cyclic group of order $n$ . Then the following identities hold. For every character $\chi\in\widehat{G}$ ,

\sum_{x\in G}\chi(x)=\begin{cases} n & \text{if } \chi=\chi_0, \\ 0 & \text{otherwise}, \end{cases}

and for every $x\in G$ ,

\sum_{\chi\in\widehat{G}}\chi(x)=\begin{cases} n & \text{if } x=e, \\ 0 & \text{otherwise}, \end{cases}

where $e$ is the identity in $G$ . These identities follow from the enumeration of characters using the set $\{0,\ldots,n-1\}$ .

Since every finite abelian group can be written as the direct product of some cyclic groups and due to Theorem 3, we can extend the results above to noncyclic groups (alternatively, we can use the Chinese remainder theorem). In particular, we know that $G\cong\widehat{G}$ for every finite abelian group. Further, we know the following.

Consider the group $(\mathbb{Z}/m\mathbb{Z})^{\times}$ , which has order $\phi(m)$ . Then, for every character $\chi\in\widehat{(\mathbb{Z}/m\mathbb{Z})^{\times}}$ ,

\sum_{x\in(\mathbb{Z}/m\mathbb{Z})^{\times}}\chi(x)=\begin{cases} \phi(m) & \text{if } \chi=\chi_0, \\ 0 & \text{otherwise}, \end{cases}

and for every $x\in(\mathbb{Z}/m\mathbb{Z})^{\times}$ ,

\sum_{\chi\in\widehat{(\mathbb{Z}/m\mathbb{Z})^{\times}}}\chi(x)=\begin{cases} \phi(m) & \text{if } x=1, \\ 0 & \text{otherwise}. \end{cases}

Dirichlet $L$ -functions

Definition 6 (Dirichlet $L$ -function). Let $\chi$ be a character modulo $q$ . Then the Dirichlet $L$ -function of $\chi$ is defined in the half-plane $\sigma>1$ as

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

We know that this sum converges using a $p$ -series test since $\chi(n)$ is bounded (in particular it has absolute value at most $1$ ) and the real part of $s$ is strictly greater than $1$ . In particular, we know it converges absolutely, so we are free to rearrange terms.

Notice that any $n$ over which we sum in the definition of the $L$ -function has a prime factorization. Furthermore, we know that $\chi$ is completely multiplicative, so we can, in a sense, swap multiplication (of prime factors to make $n$ ) and addition (of prime powers) in the given equation.

L(s,\chi) =\prod_{p\text{ prime}}\left(\sum_{k=0}^{\infty}\frac{\chi(p^k)}{p^{ks}}\right) =\prod_{p\text{ prime}}\left(\sum_{k=0}^{\infty}\left(\frac{\chi(p)}{p^s}\right)^k\right).

Now notice that the product on the inside in the above equation is nothing but a geometric series. By using the formula for a geometric series, we thence arrive at the following formula for $L(s,\chi)$ , known as the Euler product.

\therefore L(s,\chi)=\prod_{p\text{ prime}}\left(1-\frac{\chi(p)}{p^s}\right)^{-1}.

Proof of Dirichlet’s theorem

We now use these tools to sketch the proof of Dirichlet’s theorem.

Fix some $q$ and consider the $L$ -function of the principal character mod $q$ . We have

L(s,\chi_0)=\sum_{n=1}^{\infty}\chi_0(n)n^{-s}=\prod_p\left(1-\frac{\chi_0(p)}{p^s}\right)^{-1}.

But since $\chi_0(n)$ is one if $n$ and $q$ are coprime and zero otherwise, we can split sums and products according to whether their indices are coprime to $q$ .

\therefore L(s,\chi_0)=\sum_{\substack{n=1\\(n,q)=1}}^{\infty}n^{-s}.

And similarly,

\begin{aligned} \therefore L(s,\chi_0) &=\left(\prod_{p\mid q}\left(1-\frac{\chi(p)}{p^s}\right)^{-1}\right) \left(\prod_{p\nmid q}\left(1-\frac{\chi(p)}{p^s}\right)^{-1}\right) \\ &=\left(\prod_{p\mid q}\left(1-\frac{0}{p^s}\right)^{-1}\right) \left(\prod_{p\nmid q}\left(1-\frac{1}{p^s}\right)^{-1}\right) \\ &=\prod_{p\nmid q}\left(1-\frac{1}{p^s}\right)^{-1}. \end{aligned}

This is a much nicer form to work with, but we are still not quite done. Recall that

\zeta(s)=\prod_p\left(1-\frac{1}{p^s}\right)^{-1},

and thus

L(s,\chi_0)=\zeta(s)\prod_{p\mid q}(1-p^{-s}).

Now let $\chi\neq\chi_0$ . Let $x$ be arbitrary and let $k,r$ be the quotient and remainder respectively when dividing $x$ by $q$ . Then

\begin{aligned} \sum_{1\leq n\leq x}\chi(n) &=\sum_{1\leq n\leq kq}\chi(n)+\sum_{kq<n\leq kq+r}\chi(n) \\ &=0+\sum_{1\leq n\leq r}\chi(n), \\ \therefore\left|\sum_{1\leq n\leq x}\chi(n)\right| &=\left|\sum_{1\leq n\leq r}\chi(n)\right| \\ &\leq\sum_{1\leq n\leq r}|\chi(n)| \\ &\leq\sum_{1\leq n\leq r}1<q, \end{aligned}

where the last inequality follows because $0\leq r<q$ .

Now, consider the logarithm of $L(s,\chi)$ .

\begin{aligned} \log(L(s,\chi)) &=\log\left(\prod_p\left(1-\frac{\chi(p)}{p^s}\right)^{-1}\right) \\ &=\sum_p-\log\left(1-\frac{\chi(p)}{p^s}\right) \\ &=\sum_p\sum_{k=1}^{\infty}\frac{\chi(p)^k}{kp^{ks}}. \end{aligned}

Here the last equality arises from the Taylor series $-\log(1-x)=\sum_{k=1}^{\infty}x^k/k$ for $|x|<1$ .

Now, notice that we are essentially iterating over prime powers—this is a key area where the Von Mangoldt function is useful. In fact, we see that

\begin{aligned} \log(L(s,\chi)) &=\sum_p\sum_{k=1}^{\infty}\frac{\chi(p)^k}{kp^{ks}} \\ &=\sum_p\sum_{k=1}^{\infty}\frac{\log p\cdot\chi(p)^k}{\log p\cdot kp^{ks}} \\ &=\sum_p\sum_{k=1}^{\infty}\frac{\Lambda(p^k)\cdot\chi(p^k)}{\log(p^k)\cdot(p^k)^s}. \end{aligned}

If we instead choose to sum over $n:=p^k$ , we immediately see that

\log(L(s,\chi))=\sum_{n=2}^{\infty}\frac{\Lambda(n)\chi(n)}{n^s\log(n)},

and

\frac{\mathrm{d}}{\mathrm{d}s}\log(L(s,\chi)) =\frac{L'}{L}(s,\chi) =-\sum_{n=1}^{\infty}\Lambda(n)\chi(n)n^{-s}.

Both formulae are valid for $\sigma>1$ . The bounded partial sums above, together with partial summation, show that when $\chi\neq\chi_0$ , the Dirichlet series for $L(s,\chi)$ extends analytically to $\sigma>0$ . For the principal character,

L(s,\chi_0)=\zeta(s)\prod_{p\mid q}(1-p^{-s})

has a simple pole at $s=1$ with residue $\phi(q)/q$ .

Fix some $a,q$ which are coprime. Then

\frac{1}{\phi(q)}\sum_{\chi\in\widehat{(\mathbb{Z}/q\mathbb{Z})^{\times}}}\overline{\chi}(a)\chi(n) =\begin{cases} 1 & \text{if } n\equiv a\pmod q, \\ 0 & \text{otherwise}. \end{cases}

If $(n,q)>1$ , every character modulo $q$ vanishes at $n$ , while $n\not\equiv a\pmod q$ , so both sides are zero. Suppose now that $(n,q)=1$ , and define $t:=na^{-1}\pmod q$ . Then $t\in(\mathbb{Z}/q\mathbb{Z})^\times$ , and

\begin{aligned} \frac{1}{\phi(q)}\sum_{\chi\in\widehat{(\mathbb{Z}/q\mathbb{Z})^{\times}}}\overline{\chi}(a)\chi(n) &=\frac{1}{\phi(q)}\sum_{\chi\in\widehat{(\mathbb{Z}/q\mathbb{Z})^{\times}}}\overline{\chi}(a)\chi(at) \\ &=\frac{1}{\phi(q)}\sum_{\chi\in\widehat{(\mathbb{Z}/q\mathbb{Z})^{\times}}}\chi(t). \end{aligned}

If $n$ and $a$ lie in the same residue class modulo $q$ , then $t=1$ ; if not, then $t\neq1$ . The result now follows from Theorem 4.

Let us use this identity to advance our goal. We use it as a “filter” in the following, combining several pieces from the previous steps:

\begin{aligned} \sum_{n\equiv a\,(\mathrm{mod}\,q)}\Lambda(n)n^{-s} &=\sum_{n=1}^{\infty}\Lambda(n)n^{-s}\left( \frac{1}{\phi(q)}\sum_{\chi\in\widehat{(\mathbb{Z}/q\mathbb{Z})^{\times}}}\overline{\chi}(a)\chi(n)\right) \\ &=\frac{1}{\phi(q)}\sum_{\chi\in\widehat{(\mathbb{Z}/q\mathbb{Z})^{\times}}}\overline{\chi}(a) \sum_{n=1}^{\infty}\Lambda(n)n^{-s}\chi(n) \\ &=-\frac{1}{\phi(q)}\sum_{\chi\in\widehat{(\mathbb{Z}/q\mathbb{Z})^{\times}}}\overline{\chi}(a)\frac{L'}{L}(s,\chi). \end{aligned}

Consider the term of this sum where $\chi=\chi_0$ . Since $L(s,\chi_0)$ has a simple pole at $s=1$ , its logarithmic derivative $L'/L$ has a simple pole there with residue $-1$ . Hence the $\chi_0$ term can be estimated as

-\frac{1}{\phi(q)}\overline{\chi_0}(a)\frac{L'}{L}(s,\chi_0) =\frac{1}{\phi(q)(s-1)}+O(1)

when $s$ approaches $1^+$ . Now, notice that for every other character, $L(s,\chi)$ is analytic at $1$ , so

\lim_{s\to1^+}L(s,\chi)=L(1,\chi).

A central nonvanishing theorem, whose proof we omit here, states that $L(1,\chi)\neq0$ whenever $\chi$ is nonprincipal; see [MV06, Apo76]. Thus $L'/L(s,\chi)=O(1)$ as $s\to1^+$ for every nonprincipal $\chi$ , and consequently

\sum_{n\equiv a\,(\mathrm{mod}\,q)}\frac{\Lambda(n)}{n^s} =\frac{1}{\phi(q)(s-1)}+O(1)\longrightarrow\infty.

Because all the terms are nonnegative, letting $s\to1^+$ allows us to conclude that

\sum_{n\equiv a\,(\mathrm{mod}\,q)}\frac{\Lambda(n)}{n}=\infty,

which is nothing but the sum over prime powers

\sum_{\substack{n\equiv a\,(\mathrm{mod}\,q)\\n=p^k}}\frac{\log p}{p^k}=\infty.

Consider now the prime powers which contribute to this sum. Clearly

\begin{aligned} \sum_{\substack{n\equiv a\,(\mathrm{mod}\,q)\\n=p^k,\,k\geq2}}\frac{\log p}{p^k} &\leq\sum_{\substack{n=p^k\\k\geq2}}\frac{\log p}{p^k} \\ &\leq\sum_p\sum_{k\geq2}\frac{\log p}{p^k} \\ &=\sum_p\log p\left(\sum_{k\geq2}p^{-k}\right). \end{aligned}

Observe that the inner sum is a geometric series with common ratio $1/p$ and first term $1/p^2$ . Hence, the total sum will be given as $1/(p(p-1))$ .

\begin{aligned} \therefore\sum_{\substack{n\equiv a\,(\mathrm{mod}\,q)\\n=p^k,\,k\geq2}}\frac{\log p}{p^k} &\leq\sum_p\frac{\log p}{p(p-1)} \\ &\leq\sum_{n\geq2}\frac{\log n}{n(n-1)}. \end{aligned}

The last sum converges by comparison with $\sum_{n\geq2}(\log n)/n^2$ . Thus the total contribution from higher prime powers, those with $k\geq2$ , is finite. Hence,

\therefore\sum_{p\equiv a\,(\mathrm{mod}\,q)}\frac{\log p}{p}=\infty,

completing the proof of the theorem.