# Related Forms for the Riemann Hypothesis over Function Fields

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. Got no answer there in the duration of 8 months.)

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.

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.

• Dear Will, thank you for the detailed answer (and for the corrections!). I understand now the first three equivalences - the "right" way to look at the first three zeta functions is by using the Euler product formula. Regarding the abelian extension that you have constructed - where have you seen it? I am somewhat familiar with function-field cyclotomic extensions (which give the desired Galois group) which involve adjoining the roots of function-field analogues of the usual cyclotomic polynomials. The construction you show seem to be different (but perhaps it is, magically, equivalent to it). – Ofir Gorodetsky Oct 27 '16 at 14:17
• An abelian extension of $\mathbf F_q(T)$ with Galois group $(\mathbf F_q[T]/(f))^\times$ in which the Frobenius at monic irreducible $\pi(T)$ not dividing $f(T)$ is the coset $\pi(T) \bmod f(T)$ can be made without class field theory. Use the splitting field over $\mathbf F_q(T)$ of the Carlitz polynomial in $\mathbf F_q[T][X]$ that is associated to $f(T)$. See Theorems 2.17 and 5.2 of math.uconn.edu/~kconrad/blurbs/gradnumthy/carlitz.pdf. There the constant field is $\mathbf F_p$, but it is straightforward to carry out all the arguments over a general finite field (see Section 9). – KConrad Oct 27 '16 at 14:21
• @OfirGorodetsky When the group of polynomials modulo $f(X)$ is replaced with a general algebraic group, this is known as the Lang isogeny. It is a natural generalisation of Artin-Schreier coverings. For any algebraic group $G$, it produces a covering of $G$ whose Galois group is $G(\mathbb F_q)$. Then if you have a map $\mathbb A^1 \to G$ you obtain a covering of $\mathbb A^1$ whose Galois group is (a subgroup of) $G(\mathbb F_q)$. One just has to line up the map to have the right Frobenius elements. – Will Sawin Oct 27 '16 at 15:59
• @OfirGorodetsky I think you're thinking of the Carlitz module, which Keith brought up. If you look at the definition of the Carlitz module and the definition I gave, I think you will see the equivalence is not so magical - they're both defined by an explicit equation involving Frobenius and modding out by the polynomial. I'm going to try writing down the precise equivalence between the two definitions. – Will Sawin Oct 27 '16 at 16:14
• @KConrad I'm pretty sure this is just the same cover, expressed using slightly different equations. I've written down a map from one to the other, and one can show it is a bijection using the fact that they are each connected coves of the same degree, but I'm guessing there is an explicit inverse as well. I think it's just a matter of preference. – Will Sawin Oct 27 '16 at 20:55