There are several formulations and consequences of the Riemann Hypothesis over Function Fields (RH, from now on). I am interested in the logical implications between those, and in proofs\references for those implications (which are surely known to the experts). 


Personally, I would be most satisfied with proofs which are explicit and which only use the knowledge that was available to Weil at the time of his proof of the RH.


I will state below the various formulations\consequences. Let $\mathbb{F}\_q$ be the finite field with $q$ elements. 


1. Curves: Let $C/\overline{\mathbb{F}\_q}$ be a smooth, projective algebraic curve defined over $\mathbb{F}\_q$. Let $\zeta\_C(s)$ be the zeta function of $C$, defined as
$$\zeta\_C(s) = \exp(\sum\_{n \ge 1} \frac{N\_m}{m}q^{-ms}),$$
where $N\_m$ is the number of points of $C$ defined over the degree $m$ extension $\mathbb{F}\_{q^m}$ of $\mathbb{F}\_q$. **RH**: All the zeros of $\zeta\_C(s)$ lie on the line $\Re(s)=\frac{1}{2}$.
2. Field Extensions: Let $K/\mathbb{F}\_q$ be a function field with constant field $\mathbb{F}\_q$. Let $\zeta\_K(s)$ be the zeta function of $K$, defined as follows:
$$\zeta\_K(s) = \sum\_{A \ge 0} (NA)^{-s},$$
where the summation is over all effective divisors $A$ of $K$, and $NA=q^{\deg A}$.
**RH** implies: All the zeros of $\zeta\_K(s)$ lie on the line $\Re(s)=\frac{1}{2}$.
3. Rings of Integers (Dedekind zeta functions): Let $K/\mathbb{F}\_q(T)$ be a field extension of finite degree. Let $O\_K$ be the integral closure of $\mathbb{F}\_q[T]$ in $K$. Let $\zeta\_{O\_K}(s)$ be the zeta function of $O\_K$, defined as follows:
$$\zeta\_{O\_K}(s) = \sum\_{I \ge 0} (NI)^{-s},$$
where the summation is over all ideals $I$ of $O\_K$, and $NI=|O\_K/I|$.
**RH** implies: All the zeros of $\zeta\_{O\_K}(s)$ lie on the line $\Re(s)=\frac{1}{2}$.
4. Characters (L-functions): Let $\chi:\mathbb{F}\_q[T] \to \mathbb{C}$ be a Dirichlet character. Let $$L(s,\chi) = \sum\_{f \in \mathbb{F}\_q[T], f \text{ monic}} \chi(f) |f|^{-s}$$
be its L-function, where $|f|=|\mathbb{F}\_q[T]/f|$. **RH** implies: All the zeros of $L(s,\chi)$ lie on the line $\Re(s) = \frac{1}{2}$.


I believe the implications are $1 \leftrightarrow 2 \leftrightarrow 3 \implies 4$, although I cannot show this rigorously. Are these the only implications, and are they correct?


Below are some hand-waving arguments that partially explain the implications. They are not proofs, and I am *not* satisfied with them.


* I am most comfortable with Variant 2, mostly because there is an elementary proof for it, due to Stepanov-Bombieri, found in the appendix "Number Theory in Function Fields" by Michael Rosen.
* Morally, Variant 1 and Variant 2 are equivalent, since one can associate a function field to any curve, and vice versa.
* The only difference between Variant 2 and Variant 3 seems to be the contribution of the prime at infinity, which only contributes a pole ($s=1$).
* Variant 3 implies Variant 4, as follows: Note $L(s,\chi)$ is a polynomial. Construct an abelian extension $K$ of $\mathbb{F}\_q(T)$ whose set $S$ of "associated Dirichlet characters" contains $\chi$. In that case, $\zeta\_{O\_K}(s) = \prod\_{\chi' \in S} L(s,\chi')$. $RH$ for $\zeta\_{O\_K}(s)$ implies $RH$ for $L(s,\chi)$. The hard part, which I am not sure how to do, is the construction of the abelian extension - Is there a simple procedure for producing such an extension?


(Cross-posted from [MSE](https://math.stackexchange.com/questions/1665778/related-forms-for-the-riemann-hypothesis-over-finite-fields). Got no answer there in the duration of 8 months.)



Related Forms for the Riemann Hypothesis over Function Fields
1 is not just morally equivalent with 2 - the equivalence is really easy. As you say, there is a correspondence between smooth, projective, geometrically connected algebraic curves over $\mathbb F\_q$ and function fields with constant field $\mathbb F\_q$. The equivalence sends the set effective divisors on the curve (=formal sums of closed points) to the set of effective divisors of the field. By the definition of effective divisors, one writes the zeta function as a product over the closed points $P$ of $1/(1- q^{-d\_P s})$ where $P$ is the degree of $d$. Because a closed point of degree $d$ corresponds to $d$ $\mathbb F\_{q^m}$ points when $m$ is a multiple of $d$ and $0$ otherwise, these zeta functions are identical.


3 as stated is wrong, because there may be extra primes at $\infty$, which create zeroes off the line, and the prime at $\infty$ may be inert, creating the same effect. It is only valid when the extension is totally ramified at $\infty$.


4 is also wrong as stated when $\chi$ is trivial on $\mathbb F\_q^\times$. Indeed one can see this explicitly, as by the polynomial description, $L(0,\chi)$ is equal to the sum over monic polynomials $f$ of bounded degree of $\chi(f)$, thus $1/(q-1)$ times the sum over all polynomials of bounded degree of $\chi(f)$, which vanishes when the bound is large enough.


The abelian extension may be constructed by class field theory. This can be done explicitly. If $\chi$ is a character modulo a polynomial $f(T)$ of degree $d$, consider the extension generated by variables $a\_0,\dots,a\_{d-1}$ satisfying the system of equations $$(a\_0^q+ a\_1^q X + \dots + a\_{d-1}^q X^{d-1}) \equiv (a\_0+ a\_1 X + \dots + a\_{d-1} X^{d-1}) (T-X) \mod f(X)$$


where $X$ is a formal variable. The left side and right side, after applying Euclids algorithm to mod out by $f(X)$, are polynomials of degree $d-1$ in $X$, so this is really $d$ equations. Also we should demand $(a\_0+ a\_1 X + \dots + a\_{d-1} X^{d-1})$ be invertible modulo $f(X)$.


Now the group $(\mathbb F\_q[X]/f(X))^\times$ acts on this equation via its action by multiplication on $(a\_0+ a\_1 X + \dots + a\_{d-1} X^{d-1})$ modulo $f(X)$. (Because the coefficients are $\mathbb F\_q$, they are equal to their own $q$th powers, and cancel on both sides).


In fact these equations define a field extension with Galois group $(\mathbb F\_q[X]/f(X))^\times$. I don't remember how easy it is to prove this. The claim is that this extension's zeta function has the factorisation you seek. The idea is that it is sufficient to calculate how the Frobenius elements of various primes lie in the Galois group, and this can be done easily because, by the equation we have written down, the Frobenius acts on $(a\_0+ a\_1 X + \dots + a\_{d-1} X^{d-1})$ by multiplication by $X-T$.


If you do this calculation you will see an extra factor appearing at $\infty$ when $\chi$ is trivial on $\mathbb F\_q^\times$.


Finally, you are correct that 4 does not imply 3 because not all function fields are of this form, or are subfields of fields of this form.




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Because it came up in Keith's and Ofir's comments, let me explain how this relates to the Carlitz module. If we let $\mathbb F\_q[Y]$ act on $\overline{\mathbb F\_q[T]}$ in the unique linear way such that $Y (g) = g^p + Tg$, then $\overline{\mathbb F\_q[T]}$ is called the Carlitz module and the Carlitz torsion - the elements sent to $0$ by the action of $f(Y)$ - generate an abelian extension, which I claim is equivalent to this one (modulo sign changes). Indeed let me adjust this slightly so $Y(g) = T - g^p$, which should be the same thing.


The system of equations


$$(a\_0^q+ a\_1^q X + \dots + a\_{d-1}^q X^{d-1}) \equiv (a\_0+ a\_1 X + \dots + a\_{d-1} X^{d-1}) (T-X) \mod f(X)$$


may be rewritten


$$ X (a\_0+ a\_1 X + \dots + a\_{d-1} X^{d-1}) \equiv (Ta\_0-a\_0^q+ (Ta\_1-a\_1^q) X + \dots + (Ta\_{d-1}-a\_{d-1}^q) X^{d-1}) \mod f(X)$$


So the action of $X$ by multiplication is the same as the action of $Y$. Since the action of X on the left side clearly satisfies $f(X)=0$, it follows that the action of $Y$ satisfies $f(Y)(a\_i)=0$. So the elements $a\_0, \dots, a\_{d-1}$ are all elements of the Carlitz $f$-torsion. 


