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Of course, Euler proved that the Riemann zeta function can be defined as the analytic continuation of a product over all primes.

$$\zeta(s) = \prod_{p \in \mathbb{P}}\frac1{1-p^{-s}}$$

It is well known (but not something I understand) that the positions of zeros of the zeta function allows one to make inferences about the asymptotic behavior of primes. Is this a general phenomenon? Does Euler's transform generalize to products over other subsets of the natural numbers $\mathbb{A}$?

$$\alpha(s) = \prod_{a \in (\mathbb{A} \subset \mathbb{N})}\frac1{1-a^{-s}}$$

Can one then reverse Euler's transform and derive the generating subset $\mathbb{A}$ completely from the new function's zero set? More generally, how do properties of the derived function's zeros translate to properties of the generating subset?

And, specifically for the standard Riemann zeta function, if it was shown that exactly one single zero existed off the critical line, what would its position say about the distribution of primes?

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Is this question on Polar Density of a Set of Primes of interest to you? –  draks ... Apr 8 '12 at 19:53
    
... it was recently answered! –  draks ... Jul 19 '12 at 21:17

3 Answers 3

I'll just answer your last question for now. It is not possible for the $\zeta$ function to have a single zero off the critical line. The functional equation for the Riemann $\zeta$ function immediately shows, that the zeros in the critical strip are symmetric about the line $\text{Re}(z)=1/2$. Hence, if there is one zero, there must be two. Plus, $\overline{\zeta(\bar s)}=\zeta(s)$. So, if you have one zero, you have four.

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For the related question of finitely-many vs. infinitely-many exceptions to RH, see this MO question, especially Frank Thorne's and David Hansen's answers: mathoverflow.net/questions/50186/… –  B R Feb 27 '12 at 16:01

As you noted, to even talk about the "zero set" of the Riemann zeta function, one needs to have an analytic continuation to the left of the half-plane $\{\Re(s)>1\}$ in which the Euler product converges. For a generic set $A\subset{\mathbb N}$, it's not even clear that the corresponding Euler product has an anlytic continuation at all. In this case, questions concerning the "zero set" of the corresponding function are ill-posed.

You can search the relevant literature with the terms "Beurling primes" or "generalized primes".

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Let me answer the main question, concerning "An inverse for Euler's $\zeta$ function product formula":

In fact my answer to your question was part of one of my own questions. Provided with the answers given there, you can get the generating set or primes $\mathbb{P}$ with the following iterative procedure:

Assume the following process:

  1. Let's start with the set of primes $\{p_k\}$
  2. Then we use the Euler product being equivalent to Riemann's Zeta function $$ \prod_{p \text{ prime}} \frac{1}{1-p^{-s}} = \sum_{n=1}^\infty\frac{1}{n^s} = \zeta(s). $$
  3. Now the roots $\rho$ of $\zeta(s)$ contribute to the Prime Counting Function $\pi(x)$ in the follwing way $$\pi(x) = \operatorname{R}(x) - \sum_{\rho}\operatorname{R}(x^{\rho}) - \frac1{\ln x} + \frac1\pi \arctan \frac\pi{\ln x} , $$ with $ \operatorname{R}(x) = \sum_{n=1}^{\infty} \frac{ \mu (n)}{n} \operatorname{li}(x^{1/n})$. (Very nice demonstration can be found here.)
  4. The $k$th prime $p_k$ can now be calutated by using $\pi(p_k)=k$.
  5. So we get back to where we started: (1.) the set of primes $\{p_k\}$ and we now could start again.

So it's not a direct inverse, but a kind of cyclic process. Unfortunately, I have no idea for the general case.

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