0
$\begingroup$

The classical prime number theorem states that the prime counting function $$\pi(X) := \# \{ p \leq X \ | \ \text{$p$ prime} \}$$ is asymptotically equal to $X/\log(X)$.

It is also known (and much simpler to prove) that there is a finite function field analogue $$\#\{ f \in \mathbb{F}_q[T] \ | \ \text{$f$ monic, irreducible}, \ N(f) \leq X\} \sim X/\log_q(X)$$ where we consider the norm $N(f) := \#(\mathbb{f}_q[T]/(f)) = q^{\deg(f)}$.

Furthermore, we also have the Landau prime ideal theorem, which states that in a number field $K$, have $$\#\{ \mathfrak{p} \subseteq \mathcal{O}_K \ | \ \text{$\mathfrak{p}$ prime ideal}, \ N(\mathfrak{p}) \leq X \} \sim X/\log(X)$$

The similarity of these expressions is striking, and since $\mathcal{O}_K$ and $\mathbb{F}_q[T]$ (and finite extensions thereof) are the "classical" examples for Dedekind domains, it seems natural to conjecture an analogue for all/many Dedekind domains $R$.

So assuming we have a Dedekind domain $R$ such that $R/\mathfrak{p}$ is finite for all primes $\mathfrak{p}$, we might have $$\#\{ \mathfrak{p} \subseteq R \ | \ \text{$\mathfrak{p}$ prime ideal}, \ \#(R/\mathfrak{p}) \leq X \} \sim c_R X / \log(X)$$ for some constant $c_R$.

My basic question is now: Is this true/proven/conjectured?

To go futher, I would also be very interested in how this looks for more general rings? In particular, is there a class of integral rings $R$ (with $R/\mathfrak{p}$ finite for all primes $\mathfrak{p}$) that have "more" small prime ideals than the classical rings above, if we again measure the size of prime ideals via $\#(R/\mathfrak{p})$?

I have found this question, but while starting with the PNT, it is more interested in general connections of number fields and function fields.

Edit As Wojowu pointed out, it is clear that we can arbitrarily decrease the number of small primes by localization. Nevertheless, I am still very interested in my additional question: Are there rings that have asymptotically more than $X/\log(X)$ primes $\mathfrak{p}$ with $\#(R/\mathfrak{p}) \leq X$?

Improve this question
$\endgroup$
3
  • 3
    $\begingroup$ Very false in general. Consider some localizations of $\mathbb Z$, for instance $\endgroup$
    – Wojowu
    Commented Jul 28, 2023 at 11:10
  • $\begingroup$ These localizations should again have less small primes - is there some ring where we can find more primes than usual? $\endgroup$
    – Simon Pohmann
    Commented Jul 28, 2023 at 13:22
  • 1
    $\begingroup$ Something that is likely relevant for you is Abstract Analytic Number Theory. Essentially, for a broad class of normed commutative monoids $G$, if $N_G(x)$ is the number of elements with norm less than $x$ and $N_G(x) = Ax^\delta + O(x^\nu)$ as $x\to\infty$ for some $\nu < \delta$, then the number of "prime" elements is $\sim \frac{x^\delta}{\delta\log x}$. This is to say that one can mutually generalize the two PNT's you mention, but it isn't by realizing them both as dedekind domains. $\endgroup$
    – Mark Schultz-Wu
    Commented Sep 10 at 20:03

0

Reset to default

You must log in to answer this question.