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 $\alpha$.

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 5:50Algebraic Number Theoryincludes 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