How can the Zeta function be zero? 
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?
 A: 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


*

*Riemann zeta function's analytic continuation

*What exactly is the Riemann zeta function?

*How are zeta function values computed in the critical strip (which is where all the interesting zeroes are).

A: *

*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 $$\pi(x) -\underbrace{ \sum_{2 \le k \le x} \frac{1}{\log k}}_{ = \ \text{Li}(x)+\mathcal{O}(1)} = \mathcal{O}(x^{1/2+\epsilon})$$
