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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 $1/\prod_p \zeta(E,s)$.

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    $\begingroup$ Hard to find the question. $\endgroup$ – Thomas Andrews Mar 25 '16 at 17:12
  • $\begingroup$ @Thomas, please check the edit. Thanks. $\endgroup$ – User1 Mar 25 '16 at 17:14
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    $\begingroup$ Why did you ask another question as opposed to editing the earlier version into a better shape? I am not quite ready to call it a duplicate, but it is close. And the problem with the convergence of the product of $\prod_p\zeta(E,s)$ is there $\endgroup$ – Jyrki Lahtonen Mar 25 '16 at 17:21
  • $\begingroup$ Its not really the same question: the earlier version concerned a zeta function over $\mathbb{Q}$, whilst this is over a finite field. In any case, asking another question only improves clarity, since the former seemed to be misunderstood/ not clear. $\endgroup$ – User1 Mar 25 '16 at 17:25
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    $\begingroup$ Just because your feelings were hurt doesn't mean I was being impolite. And it is still wrong. $\endgroup$ – Thomas Andrews Mar 25 '16 at 19:28
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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.

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    $\begingroup$ Why does the convergence surely hold? This has nothing to do with the fact that $\zeta(E,s) \to 1$ as $\Re(s) \to \infty$. $\endgroup$ – Peter Humphries Mar 25 '16 at 17:36
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    $\begingroup$ We do NOT know that all the zeroes of $\zeta(E,s)$ have real part $1/2$! Just because all the zeroes of $\zeta_p(E,s)$ have real part $1/2$ does NOT mean that $\zeta(E,s)$ has only zeroes there. $\endgroup$ – Peter Humphries Mar 25 '16 at 17:47
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    $\begingroup$ For example, $\zeta_p(s) = (1 - p^{-s})^{-1}$ has no zeroes at all! But the Riemann zeta function $\zeta(s)$ has infinitely many zeroes. When $\Re(s) > 1$, it is true that $\zeta(s) = \prod_p (1 - p^{-s})^{-1}$, but this infinite product does not converge otherwise. Instead, we have to define $\zeta(s)$ for $\Re(s) \leq 1$ via the process of analytic continuation. $\endgroup$ – Peter Humphries Mar 25 '16 at 17:49
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    $\begingroup$ Nope, it does not converge too. You need to look up the definition of convergence of an Euler product. See the closely related question here: mathoverflow.net/questions/63714/… $\endgroup$ – Peter Humphries Mar 25 '16 at 18:09
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    $\begingroup$ In particular, the product of holomorphic functions need not converge. This is the case with $\prod_p (1 - p^{-s})^{-1}$; each individual function $(1 - p^{-s})^{-1}$ defines a nonvanishing holomorphic function for $\Re(s) > 0$, but the infinite product of these convergent functions does not converge whenever $0 < \Re(s) < 1$. $\endgroup$ – Peter Humphries Mar 25 '16 at 18:11

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