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 $m$?
Best Answer
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).