What is the most direct proof of the Riemann hypothesis for Dirichlet L-functions over function fields? 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 $m$?  
 A: 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).
A: 
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:  

Lemma:
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.
