Let $\mathbb{F}$ be a finite field of order $q$, let $m$ be an irreducible polynomial in the ring $\mathbb{F}[T]$, and let $\chi$ be a Dirichlet character modulo $m$. Define the function field Dirichlet $L$-function $$ L(s,\chi) := \sum'_f \frac{\chi(f)}{|f|^s}$$ where the sum is over monic polynomials in $\mathbb{F}[T]$, and $|f| = q^{\mathrm{deg}(f)}$ is the usual valuation. The Riemann hypothesis for this $L$-function asserts that the non-trivial zeroes of this $L$-function all lie on the line $\mathrm{Re}(s) = \frac{1}{2}$. An equivalent form of this result is that the error term in the prime number theorem for arithmetic progressions is of square root type in the function field setting.

The only way I know how to prove this is to show that such a Dirichlet $L$-function can be multiplied with other Dirichlet $L$-functions or zeta functions to create (up to some local factors) a Dedekind zeta function over some finite extension of $\mathbb{F}[T]$, which is essentially the local zeta function of some curve over $\mathbb{F}$, and at this point one can use any of the usual proofs of RH for such curves (Weil, Bombieri-Stepanov, etc.). But to get the finite extension I either need to appeal to some general theorem in class field theory (existence of ray class fields, which I understand to be a difficult result) or to explicitly construct the extension using Carlitz modules or something equivalent to such modules (the latter is discussed for instance in the answer to this other MathOverflow post).

My question is whether there is a more direct way to establish RH for Dirichlet L-functions over function fields without having to locate a suitable field extension (or whether there is some "soft" way to abstractly demonstrate the existence of such an extension without a huge amount of effort). For instance, is it possible to interpret the Dirichlet $L$-function directly as the zeta function of some $\ell$-adic sheaf? Or can the elementary methods of Stepanov type be adapted directly to the Dirichlet $L$-function (or perhaps the product of all $L$-functions of the given modulus $m$?

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    $\begingroup$ But the curve you need is $y^n = m(x)$ if your character has order $n$, so you don't need any fancy theory to write down the curve. You can then write Stepanov's proof directly in terms of polynomials in $x$. See Stark's paper on hyperelliptic curves for $n=2$. $\endgroup$ – Felipe Voloch May 14 '19 at 23:44
  • $\begingroup$ Felipe: for the $n=2$ case I can see the connection using quadratic reciprocity, but for higher $n$ what is the reciprocity relationship that links a Dirichlet character of modulus $m$ and order $n$ to the curve $y^n = m(x)$? I am a little surprised the answer is so simple given that the literature seems to tout the field extensions coming from Carlitz modules as the "correct" analogue of cyclotomic field extensions. (But perhaps this is more to handle the case when $n$ and $q$ share a common factor, which doesn't arise here?) $\endgroup$ – Terry Tao May 15 '19 at 1:19
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    $\begingroup$ There is a general statement (Weil reciprocity) that works but the $n$-th power reciprocity for rational function fields (see e.g. Hasse, Number Theory, Ch 5, at the very end) is easy to prove and is enough to give the result. You are correct in your comment about the Carlitz module not being necessary in the prime to $q$ case. $\endgroup$ – Felipe Voloch May 15 '19 at 2:42
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    $\begingroup$ Yes, to apply the results directly, we need $n|(q-1)$ and we can get around that by passing to an extension of some degree $d$. That changes the roots $\alpha$ of the $L$-functions to $\alpha^d$ (in the variable $q^{-s}$). $\endgroup$ – Felipe Voloch May 15 '19 at 17:52
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    $\begingroup$ Regardless of the details in my answer, did you have a plan for how getting a $\ell$-adic sheaf interpretation would give a more direct proof? Don't you then have to define $\ell$-adic cohomology, set up the basic formalism, and give Deligne's proof of the Riemann hypothesis? Even with the simplifications for the special case of curves, that can't be less than a book. $\endgroup$ – Will Sawin May 16 '19 at 2:58

Switching from comment to answer because the comment thread is getting too long.

Let $\# \mathbb{F}=q, t=q^{-s}$ and consider $L(t,\chi)$. Then (by taking the logarithmic derivative of the Euler product)

$$L(t,\chi) = \exp (\sum_{n=1}^{\infty} S_n t^n/n )$$

where $S_n = \sum_{\deg P | n} \chi(P)\deg P$

and $P$ runs through irreducible polynomials of $\mathbb{F}[T]$.

Then for any integer $d>0$,

$$\prod_{\zeta^d =1} L(\zeta t,\chi) = \exp (\sum_{n=1}^{\infty} S_{dn} t^{dn}/n )$$

Now, if $Q$ is an irreducible polynomial of degree $n$ over the field of $q^d$ elements, then $Q$ has $m$ (some $m|d$) conjugates $Q_i$ over $\mathbb{F}$ and the product of these conjugates is an irreducible polynomial $P$ in $\mathbb{F}[T]$ so (edit: fixed error pointed out in comments)

$$\sum_i \chi(Q_i)\deg Q_i = (\sum \chi(Q_i))\deg Q = \chi(P)m\deg(Q) = \chi(P)\deg(P).$$

Using this, one checks that the equivalent of $S_n$ over the field extension equals $S_{nd}$ and this gives $\exp (\sum_{n=1}^{\infty} S_{dn} t^n/n )$ is the $L$-function in the extension field, say $L_d(t,\chi)$. Another way of stating this is $\prod_{\zeta^d =1} L(\zeta t,\chi)= L_d(t^d,\chi)$.The relation with the zeros follows. (This is e.g. in Weil, Basic Number Theory, Appendix 5, lemma 4 in much more generality and fancier language).

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    $\begingroup$ You should mention that the character you call $\chi$ of $\mathbb F_{q^d} [T]$ is the original character $\chi$ composed with the norm map. $\endgroup$ – Will Sawin May 16 '19 at 3:00
  • $\begingroup$ @WillSawin yes. $\endgroup$ – Felipe Voloch May 16 '19 at 3:03
  • $\begingroup$ Sorry for being dense, but could you explain why one has $\sum \chi(Q_i)$ equal to $\chi(P)$? Even with Will's clarification, I don't see this. $\endgroup$ – Terry Tao May 16 '19 at 3:17
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    $\begingroup$ @TerryTao I would write $S_n$ as $\sum_{ P \in \mathbb F_q[T], \deg P = n} \chi(P) \Lambda(P)$ where $\Lambda(P)$ is $0$ if $P$ is not a prime power and the degree of that prime power otherwise. Then we need to check $\sum_{ P \in \mathbb F_q[T], \deg P = dn} = \sum_{ Q \in \mathbb F_{q^d} [T], \deg Q = n} \chi(\operatorname{Norm}(Q)) \Lambda(Q))$. For $Q$ a prime power in $\mathbb F_{q^d}[T]$, the norm of $Q$ is a prime power in $\mathbb F_q[T]$, and the prime of $\mathbb F_q^d[T]$ will lie over the prime of $\mathbb F_q[T]$. $\endgroup$ – Will Sawin May 16 '19 at 3:28
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    $\begingroup$ @TerryTao So breaking the sum into powers of primes lying over a given prime, we need to know that the total degree of the prime polynomials living over a given prime polynomial is the degree of that polynomial. This follows because their product is the base polynomial. $\endgroup$ – Will Sawin May 16 '19 at 3:31

Is it possible to interpret the Dirichlet $L$-function directly as the zeta function of some $\ell$-adic sheaf?

Yes and no. Yes in that it's possible to express the Dirichlet $L$-function directly as the zeta function of some $\ell$-adic sheaf. No in that doing so is essentially the same as constructing the relevant Galois extension, and so you're forced to again rely on an explicit construction or an abstract existence result. Indeed the relevant $\ell$-adic sheaf has rank one, so you get a homomorphism $\pi_1^{et}( \operatorname{Spec} \mathbb F [T , m^{-1} ] ) \to \overline{\mathbb Q}_\ell^\times$ but $\pi_1^{et}( \operatorname{Spec} \mathbb F [T , m^{-1} ] )$ is just a quotient of $\operatorname{Gal} (\mathbb F (T))$ so you can derive the abelian Galois extension from the $\ell$-adic sheaf.

whether there is some "soft" way to abstractly demonstrate the existence of such an extension without a huge amount of effort.

I don't think the explicit existence proof of the field extension is really that bad. I might regret saying this, but I don't think it's that much worse than the proof of power reciprocity.

An (I think) essentially self-contained proof using mainly elementary algebraic number theory is below:


Fix $m$ a polynomial of degree $d$ over $\mathbb F_q$.

Let $k$ be an algebraically closed field containing $\mathbb F_q$ and let $c_0,\dots, c_{d-1}$ be elements of $k$. If $\operatorname{Res} ( c_0 + \dots + c_{d-1} X^{d-1} , m(X) ) \neq 0$, then the solutions $a_0,\dots, a_{d-1} \in k$ with $$(a_0^q + \dots + a_{d-1}^{q} X^{d-1} ) = ( c_0 + \dots + c_{d-1} X^{d-1} ) ( a_0 + \dots + a_{d-1} X^{d-1} ) \mod m(X)$$ and $$\operatorname{Res} ( a_0 + \dots + a_{d-1} X^{d-1} , m(X) ) \neq 0$$ form exactly one orbit under the action of $(\mathbb F_q[X]/m(X))^\times$.

Proof: Multiplying by an element of $(\mathbb F_q[X]/m(X))^\times$ gives another solution, so the solutions form a union of orbits. The ratio between any two solutions is an element of $(k[X]/m(x))^\times$ with each coordinate equal to its own $q$th power, hence an element of $(\mathbb F_q[X]/m(X))^\times$, so there is at most one orbit. To check that there is at least one orbit, first note that if $a_0,\dots, a_{d-1}$ are independent transcendentals, then $c_0,\dots, c_{d-1}$ would be independent transcendentals (or otherwise there would be infinitely many solutions), so there exist solutions when $c_0,\dots, c_{d-1}$ are independent transcendentals. Given any tuple $c_0,\dots, c_{d-1}$ in $k$, take $c'_0,\dots, c'_{d-1}$ independent transcendentals in $k' = k(c_0',\dots, c_{d-1}'$, and then the coefficients of $$( c_0 + \dots + c_{d-1} X^{d-1})( c_0' + \dots + c_{d-1}' X^{d-1})\mod m$$ are themselves independent transcendentals, so dividing the solutions for these two, there are solutions in $k'$, and then because $k$ is algebraically closed, solutions in $k$. QED

Now adjoin to $\mathbb F_q(T)$ a root in $\overline{\mathbb F_q(T)}$ of the system of equations $a_0,\dots, a_{d-1} \in k$ with $$(a_0^q + \dots + a_{d-1}^{q} X^{d-1} ) = ( X-T ) ( a_0 + \dots + a_{d-1} X^{d-1} ) \mod m(X)$$ and $$\operatorname{Res} ( a_0 + \dots + a_{d-1} X^{d-1} , m(X) ) \neq 0$$

Because the set of roots forms an orbit under $(\mathbb F_q[X]/m(X) )^\times$, the Galois group is a subgroup of $(\mathbb F_q[X]/m(X) )^\times$. We want to check that the Frobenius element associated to a prime $\pi(T)$ not dividing $m(T)$ inside this Galois group is equal to the reduction of $m(X)$ mod $(X)$. It follows immediately from the definition of the zeta function of a global field that the zeta function of the function field generated by this root is a product of Dirichlet $L$-functions for characters of $(\mathbb F_q[X]/m(X) )^\times$.

The Frobenius element in the Galois group is the same as the Frobenius element in the local field $K_\pi$ for $K = \mathbb F_q(T)$. By Hensel's lemma, each solution in the algebraic closure of the residue field $\overline{k}_\pi$ lifts to a solution in an unramified extension of $K_\pi$ and thus to an element of the algebraic closure $\overline{K}_\pi$. Because $\overline{k}_\pi$ and $\overline{K}_\pi$ have the same number of solutions, all solutions in $\overline{K}_\pi$ arise this way, so the action of Frobenius on solutions over the local field $\overline{K}_\pi$ is equal to its action on solutions over the residue field $k_\pi = \mathbb F_q[T]/\pi(T)$.

But the Frobenius in $\overline{ \mathbb F_q[T]/\pi(T)}$ sends $a_i$ to $a_i^{q^{\deg \pi}}$, so it acts on $(a_0 + \dots + a_{d-1} X^d)$ by multiplication by $$(X-T) (X-T^q) \dots (X - T^{ q^{ \deg \pi -1 } } ) .$$ Because $T \in \mathbb F_q[T]/\pi(T)$ is a root of $\pi(T)$, $T^q,T^{q^2}, \dots$ must be the remaining roots, and so Frobenius acts by multiplication by $\pi(X)$, as desired.

  • $\begingroup$ Stepanov's method produces a function in $\mathbb{F}(x,y), y^n = m(x)$ but this function is invariant under the automorphism $y \mapsto \zeta y$ so it comes out as a polynomial in $x$, and you can set up the construction this way to begin with. Same for Artin-Schreier $y^p-y=f(x)$ and the automorphism $y \mapsto y+1$. The original papers treat only these cases in this way. I am definitely not talking about later extensions such as Bombieri's. $\endgroup$ – Felipe Voloch May 16 '19 at 2:58
  • $\begingroup$ @FelipeVoloch Ah, I didn't know that as I am too sophisticated to understand proofs that don't have Riemann-Roch in them. If the relationship between the classical case and general case is just working with invariant functions in the classical case, then surely there is a version of the classical argument that can be made for general Dirichlet characters using the explicit equations for this field extension avoiding some of the abstract steps. But I could be wrong about this as well. $\endgroup$ – Will Sawin May 16 '19 at 3:21
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    $\begingroup$ Thanks for this. While I am not yet proficient enough in the etale formalism to fully follow this more modern interpretation of the Carlitz construction, it at least does seem less ad hoc this way. $\endgroup$ – Terry Tao May 16 '19 at 3:44
  • $\begingroup$ @TerryTao One doesn't really need the etale formalism - it should all be doable in the language of algebraic number theory, which I have attempted to do above. $\endgroup$ – Will Sawin May 16 '19 at 14:06
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    $\begingroup$ @TerryTao Your comment that this is the more modern interpretation made me wonder by how much. It seems that this idea is Lang 1956 (doi.org/10.2307%2F2372673 "In the case the group $G$ is commutative than one can use it to derive the class field theory over a variety $V$ having a rational map to $G$") while the Carlitz module approach is Carlitz 1935. So about 20 years more modern. But the Carlitz module approach seems much better known, possibly because Lang didn't fully write it up in that paper ("The abelian class field theory is carried out in detail elsewhere") $\endgroup$ – Will Sawin May 16 '19 at 14:09

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