# Primes in arithmetic progressions in number fields

My general question is how does one prove equi-distribution results for primes in arithmetic progressions in number fields? I am interested in the equi-distribution of prime elements of the ring of integers in cosets coprime to a fixed ideal (in some reasonable subset of the canonical embedding).

In particular, for the integers, one has the pleasant identity $$\sum_{n \leq x , n \equiv a \pmod{q}} \Lambda(n) = \frac{1}{\phi(q)} \sum_{\chi} \bar{\chi}(a) \sum_{n \leq x} \Lambda(n) \chi(n).$$ Then one can use classical tools from analytic number theory, such as zero-free regions for $L$-functions, to estimate the sum on the right.

For number fields, it seems like there are significant obstacles to the above approach. Is there a reference that addresses this? (the friendlier, the better!).

Edit: Allow me to be more precise (in response to KConrad). Fix $K$ to be a number field and $O_K$ the ring of integers. Let $I \leq O_K$ be an ideal and $R$ be some bounded region in the canonical embedding, which we shall denote $i$. Let $a \in O_K$ such that $((a) , I) = 1$. I am interested in $$\sum_{\substack{\alpha \in R \\ \alpha - i(a) \in I}} 1_P(\alpha),$$ where $1_P(\alpha)$ is one if $(\alpha)$ is a prime ideal and zero otherwise.

• What is this supposed to mean when the number field is not a UFD? – KConrad May 9 '16 at 21:14
• Remember that no one knows that there are infinitely many primes of the form $x^2+1$ so, if your $K = \mathbb{Q}[i]$ and $R$ is $\{ x+iy : 1/2<x<3/2 \}$, this is open. You need to somehow require $R$ to be "big" to have any hope. – David E Speyer May 9 '16 at 23:20
• @DavidSpeyer Good point, thanks. I do not mind if one makes some strong assumptions on $R$, I just do not want to be the person to make such assumptions! – George Shakan May 9 '16 at 23:26

Theorem 6 in Chapter XV, $\S$5 of Lang Algebraic Number Theory is a result of the sort you want. Lang formulates it using ideles, but he gives an application in more classical language in Example 3 on the following page. I first quote what he shows in Example 3, then I'll try to rewrite it to sound a bit more like what you want.

Lang's Example 3 Let $k$ be a number field of class number $1$. Let $k_{\infty}$ be the product of the archimedean completions of $k$, so $k_{\infty} \cong \mathbb{R}^s \times \mathbb{C}^t$ if $k$ has $s$ real and $2t$ imaginary embeddings. Let $k_{\infty}^{\ast}$ be the group of invertible elements of $k_{\infty}$, so $k_{\infty}^{\ast} \cong \mathbb{R}^{s+t} \times (\mathbb{Z}/2)^s \times (S^1)^t$, and let $k_{\infty}^{\ast,1}$ be the subgroup of elements with norm $1$. Let $U$ be the unit group of $k$, so $k_{\infty}^{\ast} / U \cong \mathbb{R} \times (S^1)^{s+2t-1} \times \mathbb{Z}/2^j$ for some $j$ and $k_{\infty}^{\ast,1} / U \cong (S^1)^{s+2t-1} \times \mathbb{Z}/2^j$. Let $\sigma : k_{\infty}^{\ast} / U \to S^1$ be a continuous homomorphism whose restriction to $k_{\infty}^{\ast,1} / U$ is surjective. Then $\sigma(\pi)$ is equidistributed in $S^1$, as $\pi$ ranges over generators of prime ideals, each prime ideal taken once.

Okay, but you also want to be allowed to impose congruence conditions on your ideals and work with other class numbers. Lang's Theorem 6 does that, except he only states it idelically. I believe the following is the classical version.

Let $\mathfrak{m}$ be a (nonzero) ideal of $k$ and let $H$ be the $\mathfrak{m}$-ray class group, meaning ideals relatively prime to $\mathfrak{m}$ modulo principal ideals whose generators are $1 \bmod \mathfrak{m}$. By Cebatorov, prime ideals are equidistributed in the finite group $H$, so we can consider a single class of $H$ and ask about the archimedean behavior of primes in that class.

Fix $I$ an ideal in the desired class, so an ideal is in this class if and only if it is of the form $\alpha I$ for some $\alpha \in I^{-1}$ with $\alpha = 1 \bmod \mathfrak{m}$. Let $P \subset I^{-1}$ be the set of $\alpha$ which are $1 \bmod \mathfrak{m}$ for which $\alpha I$ is prime. We have $P \subset I^{-1} \subset k \otimes \mathbb{R} = k_{\infty}$. Roughly, we are going to claim that $P$ is equidistributed.

Let $U$ be the group of units which are $1 \bmod \mathfrak{m}$. Choose a continuous group homomorphism $\sigma$ retracting $k^{\ast}_{\infty}/U$ onto $k^{\ast,1}_{\infty}/U$. Then $\sigma(P)$ is equidistributed in the compact group $k^{\ast,1}_{\infty}/U$.

If this is too abstract, note that a concrete case is equidistribution of Gaussian primes in wedge shaped regions of $\mathbb{C}$. See here, which is where I learned of the Lang reference.

The classical way to do this is to reformulate "primes in arithmetic progressions" as telling you about how the image of Frobenius is equidistributed. More precisely in this case, look at Frob$_p$ in $\text{Gal}(\mathbb Q(\zeta_q)/\mathbb Q)\cong(\mathbb Z/q\mathbb Z)^*$, and Dirichlet's theorem says that as $p$ varies, the images are equidistributed among the congruence classes in $(\mathbb Z/q\mathbb Z)^*$. Of course, the image of Frob$_p$ is exactly $p\bmod q$. Now go to a number field $K$, replace $\mathbb Q(\zeta_q)$ by any Galois extension $L/K$, and for primes $\mathfrak p$ in $O_K$, look at the conjugacy class of Frob$_{\mathfrak p}$ in Gal$(L/K)$. The Chebotarev density theorem says that these images are equidistributed, where one has to weight by the size of the conjugacy class. The proof is a generalization of Dirichlet's and uses zero-free regions, etc. You can find this explained in many standard texts.

• Thank you for your response, I am familiar with Chebotarev's density theorem. This is not the generalization I had in mind; I edited my question appropriately. – George Shakan May 9 '16 at 23:01
• Okay, thanks for the clarification. The answer to your edited question is going to depend how the region $R$ interacts with the action of the unit group and the class group, I think. So in general it will be very complicated. OTOH, if you want to take a region $R$ that has some nice symmetry and convexity properties, you might be able to do something. I'd suggest (but you've probably already thought of this) starting with quadratic imaginary fields and/or quadratic real fields of class number 1. – Joe Silverman May 9 '16 at 23:05

Very good responses by David Speyer and Joe Silverman. I just want to add a related reference that I recently came across. It contains a version of Linnik's theorem for number fields, it is nice and useful (see especially Theorem 5.2): Weiss - The least prime ideal.

• Indeed this is very useful! – George Shakan Jul 27 '16 at 20:40