Let $L/K$ be a geometric Galois extension of function fields over $\mathbb F\_q$. Let $\chi$ be a non-trivial irreducible character of $\text{Gal}(L/K)$. According to Michael Rosen's *Number Theory in Function Fields* p.131, Weil's proof of the Riemann Hypothesis for curves over finite fields shows that the Artin L-function associated to $\chi$ is a polynomial in the variable $u := q^{-s}$ of degree $$\chi(1)(2g\_K-2) + \deg\_K \mathfrak{f}(\chi),$$ where $g\_k$ is the genus of $K$ and $\mathfrak{f}(\chi)$ is an effective divisor of $K$ called the Artin conductor of $\chi$.



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> **Question** : I would like to have bounds on the previous degree, so for example on $\deg\_K \mathfrak{f}$, in terms of $\chi(1)$ and possibly other invariants related to $L/K$, like $\deg\_K D\_{L/K}$ where $D\_{L/K}$ is the discriminant of $L/K$.
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In the number field case, I know there are for example the Odlyzko bounds on Artin conductors, and I was wondering if such bounds could be found in the function field case. Thanks in advance for any answer.



Bounds on Artin conductors over function fields
It's not completely clear from your question if you want lower or upper bounds.


For lower bounds, the Odlyzko bounds come from analysis of the $L$-function. The lower bound we get by analyzing the $L$-function in the function field case is
$$ \deg\_K \mathfrak{f}(\chi) \geq (2 -2g\_K) \chi(1) $$ coming from the obvious fact that the degree of a polynomial is nonnegative. This is sharp, which we can see by, for instance, taking $f$ to be a generic polynomial of degree $d$, $K = \mathbb F\_q ( f(T))$, $L$ to be the Galois closure of $\mathbb F\_q(T)$ over $K$, $\chi$ to be the $d-1$-dimensional standard representation of the Galois group $S\_{d-1}$ of, in which case this $L$-function is indeed a polynomial of degree zero.


For upper bounds, the conductor-discriminant formula gives


$$ \deg\_K D\_{L/K} = \sum\_{\chi} \chi(1) \deg\_K \mathfrak{f}(\chi)$$ where the sum is over irreducible $\chi$ so $$\deg\_K \mathfrak{f}(\chi) \leq \frac{ \deg\_K D\_{L/K} }{\chi(1) } $$ because all these degrees are nonnegative.


This is sharp for a few Galois groups.




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From the definition of the Artin conductor, we can see that $$\deg \mathfrak{f} (\chi) = \frac{1}{|G|} \sum\_{g \in G} i(\sigma) ( \chi(e) - \chi(\sigma) ) =\frac{1}{|G|} \sum\_{g \in G} i(\sigma) \operatorname{Re}( \chi(e) - \chi(\sigma) ) $$ where $i(\sigma)$ is the maximum $i$ such that the $i$'th ramification group includes $\sigma$ and note that $Re( \chi(e) - \chi(\sigma)) > 0 $ because $\chi$ is a faithful representation of $g$. Combining this with the conductor-discriminant formula, we get


$$\deg\_K D\_{L/K} = \sum\_{ \sigma \in G, \sigma \neq e} i(\sigma)$$ and thus


$$ \frac{ \deg \mathfrak f(\chi) } { \deg D\_{L/K}}\geq \min\_{ \sigma \in G, \sigma \neq e} \frac{\operatorname{Re} ( \chi(e) -\chi(\sigma)) } {|G|} .$$


In fact we can do slightly better because we can average in the numerator over all conjugate $\sigma$, where $\sigma$ and $\sigma'$ are conjugate if $\sigma$ is a power of $\sigma'$ and vice versa. This has the effect of averaging over all Galois conjugates of $\chi(e) -\chi(\sigma)$, which upon consideration of roots of unity means that the numerator is at least $1/2$.


Now $|G| = [L:K]$ so this gives $$\deg \mathfrak f(\chi) \geq \frac{ \deg D\_{L/K}} { 2[L:K]}.$$


