Prime number theorem for Dedekind domains

Let $\mathscr P\subseteq \mathbb N$ be the set of prime numbers. The prime number theorem tells us that if $\pi(x)=|\{p\in\mathscr P\colon p\leq x\}|$ then $\pi(x)\sim \frac{x}{\log x}$. Now one could think about the fact that $\mathbb Z$ is a Dedekind domain and $\mathscr P$ can be seen as the set of prime ideals in $\mathbb Z$. So the function $\pi(x)$ can be thought as a way of counting prime ideals whose norm is not greater than $x$.

If $K$ is a number field, $\mathcal O_K$ its number ring and $\mathscr P_K$ the set of its prime ideals, we are in a very similar situation: we have a Dedekind domain and for each prime ideal $\mathfrak p\in \mathscr P_K$ the norm $N(\mathfrak p)$ of $\mathfrak p$ is well defined, being $|\mathcal O_K/\mathfrak p|$. Moreover, for each $n\in \mathbb N$ there is a finite number of (prime) ideals of norm $n$. So the question is: what can I say asymptotically about the function $\pi_K$ defined as $\pi_K(x)=|\{\mathfrak p\in \mathscr P_K\colon N(\mathfrak p)\leq x \}|$? For example if $K$ is a quadratic field, then there are $\pi(x)$ rational primes which are $\leq x$. Half of them remain prime in $K$ so they have norm $p^2$, so I expect them to contribute for $\pi(\sqrt{x})/2$ to $\pi_K(x)$, while the other half splits in $K$ in $2$ prime ideals each of which of norm $p$, so I am expecting this other half to contribute for $2\cdot \pi(x)/2=\pi(x)$ to $\pi_K(x)$. Would that be correct?

And finally, if $A$ is any Dedekind domain such that there are only finitely many prime ideals of a fixed norm (is this always the case?), is there any kind of general result about $\pi_A(x)=|\{\mathfrak p\in \mathscr P_A\colon N(\mathfrak p)\leq x\}|$?

• Landau's Prime Ideal Theorem says that $\pi_K(x) \sim x/\log x$, just as in the classical case $K = \mathbf Q$. If you want to try pushing this type of result into "very general Dedekind domains" then you might want to look at Knopfmacher's book "Abstract Analytic Number Theory". – KCd Feb 8 '14 at 4:08
• Notice that you need to assume that the quotients $A/p$ are finite for any prime ideal $p \subset A$ (in order to define the norm of $p$). This is not true for all Dedekind domains, as $\Bbb C[X]$ shows. – Watson May 21 '18 at 18:24