Dirichlet's theorem for number fields

I'd like to see a formulation of Dirichlet's theorem for number fields, i.e. some analogue of the assertion:

The number of primes less than $N$ congruent to $a \pmod{m}$ where $(a,m)=1$ is

$\frac{1}{\phi(m)}\pi(N) + o(\pi(N))$

I am really only specifically concerned with such a theorem for the Eisenstein integers $\mathbb{Z}[\omega]$, but I am curious about the general case as well. Is this true, and if so can somebody provide me with a reference? Thanks!

• It's called Chebotarev's Density Theorem. – Xandi Tuni Jun 23 '10 at 5:13
• Also Dirichlet's theorem (in the usual sense) only states that there are infinitely many primes in a given (nontrivial) arithmetic progression. You're asking about a number field equivalent of a stronger statement, namely the Prime Number Theorem for arithmetic progressions (the analogue of which is of course Chebotarev's Density Theorem). – Peter Humphries Jun 23 '10 at 5:22
• Note that the analogue of Dirichlet's theorem in $Z[\omega]$ says "there are infinitely many primes congruent to $a$ mod $I$" with $I$ an ideal in $Z[\omega]$ and $a$ in $(Z[\omega]/I)^\times$, and not "there are infinitely many primes in the set $\{a,a+x,a+2x,a+3x,\ldots\}$", the latter (when $a$ and $x$ are coprime) being an open problem! Indeed, if we knew there were infinitely many primes in $Z[i]$ in the set $\{i,1+i,2+i,3+i,\ldots\}$ we would get infinitely many rational primes of the form $n^2+1$, which is already open. – Kevin Buzzard Jun 23 '10 at 5:50
• Since you asked for a reference, Lang's Algebraic Number Theory includes proofs of some of this material (though I believe he only proves the Chebotarev theorem for the Dirichlet density). – Akhil Mathew Jun 23 '10 at 13:12

To expand on the excellent comments a bit, one needs both a bit more and a bit less than Chebotarev's density theorem. :-)

Let's take a number field $K$, and a nonzero ideal $\mathfrak{N}$ of the ring $\mathfrak{O}$ of $K$. We divide the nonzero ideals coprime to $\mathfrak{N}$ into a finite number of classes depending on $\mathfrak{N}$. There is an equivalence relation on ideals defined by $\mathfrak{a}\sim\mathfrak{b}$ if $b\mathfrak{a}=a\mathfrak{b}$ where $a$, $b\in\mathfrak{O}$, $a\equiv b\equiv1$ (mod $\mathfrak{N}$) and $a$ and $b$ are totally positive (positive in each embedding of $K$ in $\mathbb{R}$). The equivalence classes are called ray classes modulo $\mathfrak{N}$.

The analogue of Dirichlet's theorem for $K$ is that the prime ideals of $\mathfrak{O}$ are equidistibuted amongst the ray classes. The strong form states that if $\pi_{\mathfrak{a}}(N)$ is the number of prime ideals of norm $\le N$ in the ray class of $\mathfrak{a}$ and $\pi_{\mathfrak{O}}(N)$ is the number of prime ideals of norm $\le N$ in $\mathfrak{O}$ then $$\lim_{N\to\infty}\frac{\pi_{\mathfrak{a}}(N)}{\pi_{\mathfrak{O}}(N)} =\frac1m$$ where $m$ is the number of ray classes modulo $\mathfrak{N}$.

To prove this we need less than Chebotarev's theorem, as we need to apply that only to an abelian extension of $K$, but we need more, namely a suitable extension $L/K$ to apply it to. This extension $L/K$ has abelian Galois group $G$ which is in natural correspondence with the set of ray classes. In detail the Frobenius element attached to a prime ideal $\mathfrak{p}$ is the element of $G$ corresponding to the ray class of $\mathfrak{p}$. This extension exists by the existence theorem of class field theory, quite a deep result.

Let's consider particular examples. For $K=\mathbb{Q}$ all ideals are principal so take $\mathfrak{N}=(N)$ for a positive integer $N$. Then for positive integers $r$ and $s$, the ideals $(r)\sim(s)$ iff there are positive integers $a$ and $b$ congruent to $1$ modulo $N$ with $br=as$. This condition is equivalent to $r\equiv s$ (mod $N$). So ray classes correspond to congruence classes and so we recover Dirichlet's theorem.

Now let $K=\mathbb{Q}(\omega)$. In this case all ideals are principal. Take an ideal $\mathfrak{N}=(\nu)$ of $\mathfrak{O}$. and we find this time that $(\alpha)\sim(\beta)$ if $\alpha\equiv\eta\beta$ (mod $\nu$) where $\eta=\pm \omega^j$ is a unit in $\mathfrak{O}$. (As $K$ has no real embeddings, the condition of total positivity is redundant.) So the analogue of Dirichlet here is that for $\alpha$ coprime to $\nu$ the density of prime ideals $\pi$ with $\pi\equiv\pm\omega^j\alpha$ (mod $\nu$) is indepdendent of the choice of $\alpha$.

• Dear Robin, Why do we need to apply Cebotarev density at all? Isn't it more direct, and closer to Dirichlet's original argument, to work directly with characters of the ray class group and to study the associated Hecke $L$-functions? – Emerton Jun 23 '10 at 14:09
• Matthew, you are quite right; the abelian case of Chebotarev follows the same lines as Dirichlet's theorem if you know (as we do in this case directly, and in general by CFT) that the characters of the Galois group arise from ray class characters. – Robin Chapman Jun 23 '10 at 14:31
• Thinking again, one can prove this density result by looking at ray class characters, but one needs to show the nonvanishing of their L-functions on the line $\mathrm{Re}s=1$. This is easy except for showing that $L(1,\chi)\ne0$ for quadratic Hecke characters $\chi$. The usual way to prove this would be to note that the L-functions are factors of the Dedekind zeta function of the ray class field extension; for this one needs CFT. Perhaps though one only needs appropriate quadratic extensions of $K$? – Robin Chapman Jun 23 '10 at 14:43
• @Robin: Can one derive a formula for $\pi_D(N)$ with Siegel-zero term, and error term of the form $O(x e^{-c\sqrt{\log x}})$? If so, do you know the reference? – i707107 Jan 9 '13 at 22:56
• I meant $\pi_a(N)$. – i707107 Jan 9 '13 at 22:57