In Chapter 7 of Marcus' Number Fields, he defines the polar density of a set $A$ of primes of a number field $K$ as follows:

Definition: If some $n$th power of the function $$\zeta_{K,A}(s) = \prod_{P \in A} \left(1-\frac{1}{\|P\|^s} \right)^{-1}$$ can be extended to a meromorphic function in a neighborhood of $s=1$, having a pole of order $m$ at $s=1$, then the polar density of $A$ is $\delta(A) = \dfrac{m}{n}$.

Are there any relatively easily describable sets of primes such that this density would not be defined?

  • $\begingroup$ Since this definition of density is always rational, would you be interested in a set of primes with irrational density? $\endgroup$ – Erick Wong Jul 12 '12 at 22:36
  • $\begingroup$ @ErickWong at least I would be interested in such a set. Go ahead, please... $\endgroup$ – draks ... Jul 14 '12 at 9:00

We might as well restrict to $K=\mathbb Q$, so that $\zeta_A(s) = \prod_{p\in A} (1 - p^{-s})^{-1}$. If the polar density of $A$ exists, then the Dirichlet density exists and is equal. In particular, if the natural density and polar density of $A$ both exist, then they must be the same.

Since the definition of polar density is only capable of producing rational numbers, it suffices to name a set of primes $A$ having a natural density that is irrational.

I can think of several natural sets of primes which should have irrational density, but I haven't thought of one that is both aesthetically pleasing and of provably irrational density.

Some incomplete examples:

  • The set of primes with $p-1$ squarefree has density equal to Artin's constant $\prod_p (1-\frac{1}{p(p-1)})$, see https://math.stackexchange.com/a/155117/30402. Artin's constant is presumed but not known to be irrational.
  • The set of primes for which $p-1$ has a prime factor greater than $\sqrt{p}$ is conjectured to be $\log 2$, which is the exact density of integers $n$ possessing a similar property.

Of course it is possible to describe less succinctly a set of primes having any density in $[0,1]$. I still wonder if there is a simple variation that will give an obviously irrational singular series constant like $6/\pi^2$.

Another approach is to take a sparse set of primes for which $\prod_p(1-1/p)^{-1}$ still diverges slowly (for instance, if the counting function is $x/(\log x \log \log x)$). This will have natural density $0$, but the singularity at $s=1$ prevents $\zeta_A(s)$ from being continued to a holomorphic function in a neighbourhood of $1$ (by a result of Landau).

The disjoint union of such a set with one already having polar density should also lack a polar density, even with a rational natural density.

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  • $\begingroup$ +1 and Thx for your answer. How is the sparse set of primes defined? Can you give an explicite example? $\endgroup$ – draks ... Jul 16 '12 at 9:17
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    $\begingroup$ For those who would be interested, the claims in your first paragraph are proved in Proposition 4.1, p. 190, Class Field Theory, Milne. $\endgroup$ – Watson Dec 29 '16 at 11:12

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