How can the Zeta function be zero?

Question

How can the Zeta function be zero?

If the zeta function is the Euler product:

$$\zeta(s)=\prod_p \frac{1}{1-p^{-s}}$$

Then being a product my first thought was that it could only be zero if one or more of its terms were zero.

This would require $\frac{1}{1-p^{-s}}$ to be zero for some prime $p$

So there would have to be some prime $p$ for which $p^{-s}$ is infinite.

Clearly I'm misunderstanding something. Are the zeroes where the terms $(1-p^{-s})$ diverge?

Answer 1

• The Euler product only converges for $\Re(s) > 1$.

• The Riemann hypothesis is that $$\phi(s) = \prod_{k=1}^\infty \frac{1-(k \log k)^{-s}}{1-p_k^{-s}}$$ converges for $\Re(s) > 1/2$ (it is easy to show it converges for $\Re(s) > 1$ and the prime number theorem is that it converges for $\Re(s) \ge 1$)

• That it converges means to an analytic function, ie. the series $\sum_{k=1}^\infty \log \frac{1-(k \log k)^{-s}}{1-p_k^{-s}}$ converges to an analytic function so that $\log \phi(s)$ is finite and $\phi(s)$ has no zeros.

  Since the analytic continuation of $\psi(s) = \sum_{k=1}^\infty \log (1-(k \log k)^{-s})$ is known to have no zeros for $\Re(s) > 0$, it implies $\log \zeta(s)-\psi(s)$ and hence $\log \zeta(s)$ have no zeros for $\Re(s) > 1/2$

• That $\phi(s)$ converges for $\Re(s) > 1/2$ is equivalent to the number theoretic statement $$\pi(x) -\underbrace{ \sum_{2 \le k \le x} \frac{1}{\log k}}_{ = \ \text{Li}(x)+\mathcal{O}(1)} = \mathcal{O}(x^{1/2+\epsilon})$$

Answer 2

The Euler product only converges for $\mathrm{Re} (s) > 1$. For $\mathrm{Re} (s) \leq 1$, you need to consider a different representation of the zeta function.

The different representation comes from the "analytic continuation" of the zeta function. This has been written about extensively on this site. See for instance

1. Riemann zeta function's analytic continuation
2. What exactly is the Riemann zeta function?
3. How are zeta function values computed in the critical strip (which is where all the interesting zeroes are).