Does the Riemann Hypothesis for finite fields imply the original RH? Let $E$ be an elliptic curve over a finite field $\mathbb{F}_p$ where $p$ is a prime. The zeta function, $\zeta(E, s)$ for $E$ is defined as 
$\zeta(E,s) = \dfrac{(1-\alpha p^{-s})(1-\beta p^{-s})}{(1-p^{-s})(1-p^{1-s})}$.
where $\alpha$ and $\beta$ are certain algebraic integers whose defintion is not of much concern here. The Riemann Hypothesis (now a theorem) is the statement that  $\zeta(E,s)$ vanishes if and only if $\Re(s) = 1/2$.
Taking the product for $\zeta(E,s)$ over all primes, we have 
$\prod_p \zeta(E,s) = \dfrac{\zeta(s)\zeta(s-1)}{L(s)}$
where $\zeta(s)$ is the Riemann zeta function and $L(s) = \prod_p \dfrac{1}{(1-\alpha p^{-s})(1-\beta p^{-s})}$ is an $L$-series that is known to converge for $\Re(s)>3/2$.
Rearranging the above, we have 
$\dfrac{\zeta(s)}{\zeta(E,s)} = \dfrac{L(s)}{\zeta(s-1)}$
Since $\zeta(E,s)$ tends to $1$ rapidly as $\Re(s) \to \infty$, the Riemann Hypothesis for elliptic curves over finite fields can be stated in the form that:
                                                                         $1/\zeta(E,s)$ is convergent for $\Re(s) > 1/2$.
Therefore, $1/\zeta(E,s)$, $\zeta(s)$ and $L(s)$ are all convergent in the region $\Re(s) > 3/2$. This requires $1/\zeta(s-1)$ to be also convergent for $\Re(s) > 3/2$, which implies the convergence of $1/\zeta(s)$ for $\Re(s) > 1/2$, a condition that is known to be necessary and sufficient for the original Riemann Hypothesis.
EDIT: From the comments below, it seems the convergence of $1/\prod_p \zeta(E,s)$ for $\Re(s) > 1/2$, needs an explanation:
Take the above definition of $1/\zeta(E,s)$ with $\mid \alpha \mid = \mid \beta \mid = \sqrt p $. From here, one can quickly verify that $\mid \zeta (E,s) \mid \neq 0$, which implies the convergence of $1/\prod_p \zeta(E,s)$.
 A: The mistake you've made is the statement

The Riemann hypothesis for elliptic curves over finite fields can be stated in the form that $\zeta(E,s)$ is convergent for $\Re(s) > 1/2$.

I'm not sure why you think this is the case. I think one reason might be notation; you ought to define
\[\zeta_p(E,s) = \frac{(1 - \alpha_p p^{-s})(1 - \beta_p p^{-s})}{(1 - p^{-s})(1 - p^{1-s})}\]
to be the local zeta function of $E$, and then set
\[\zeta(E,s) = \prod_p \zeta_p(E,s).\]
Now the mistake becomes clear. Each $\zeta_p(E,s)$ defines a meromorphic function on $\mathbb{C}$ with poles whenever $p^{-s}$ or $p^{1-s}$ are $1$, and zeroes on the line $\Re(s) = 1/2$. But the Euler product
\[\zeta(E,s) = \prod_p \zeta_p(E,s)\]
does NOT share this property! This is an infinite product of meromorphic functions, and such an infinite product need not be as nicely behaved as each individual term in the product.
In particular, it is most definitely NOT known whether the function
\[\frac{1}{\zeta(E,s)}\]
is convergent (or rather, holomorphic) for $\Re(s) > 3/2$.
For example, the function
\[1 - p^{-s}\]
is holomorphic for all $s \in \mathbb{C}$, but this does not tell us anything about the behaviour of the function
\[\prod_p (1 - p^{-s}) = \frac{1}{\zeta(s)}\]
whenever $\Re(s) < 1$, because this Euler product does not converge there.
In fact, all we know about $\zeta(E,s)$ is that it is holomorphic for $\Re(s) > 3/2$, because $L(s)$, $\zeta(s - 1)$, and $\zeta(s)$ are all holomorphic there, and also that $\zeta(E,s)$ is nonvanishing for $\Re(s) > 2$, because $L(s)$, $\zeta(s - 1)$, and $\zeta(s)$ can all be expressed in terms of Euler products there.
