1. Let $l$ be an odd prime and $q=p^n$ be a prime power. Set $A=\mathbb F_q[t]$.
  2. A Dirichlet character with modulus $f\in A$ is a group homomorphism $\chi:(A/fA)^*\to \mathbb C^* $.
  3. The character $\chi:(A/f)^*\to \mathbb C^*$ is called primitive if it does not factor through modulus $(A/f')^*$ for another $f'|f$ with $\deg(f')<\deg(f)$.
  4. The minimal possible modulus of $\chi$ is called its conductor.
  5. The order of $\chi$ is the smallest $k\geq0$ such that $\chi^k$ is the trivial character.

Does the following claim hold:

If $q\equiv 1\pmod l$, every $l$-th order character has squarefree conductor. What if we only have $\gcd ( l , p )=1$?

My argument is as follows. Let $\chi$ be a primitive character of conductor $f=\prod P_i^{e_i}$ for primes $P_i\in A$. Then $(A/f)^*=\prod (A/P_i^{e_i})^*$ The character $\chi$ is primitive so it's restricition to any piece $A/P_i^{e_i}$ is non-trivial. But the group $A/P_i^{e_i}$ admits a decomposition $$0 \to G_i \to (A/P_i^{e_i})^* \to (A/P_i)^* \to 0 $$

If $e_i>1$, the first piece of the exact sequence is a $p$-group (Rosen, Number theory in function fields, Chapter 1, prop 1.6). Since $q\equiv 1\pmod l$, we get that the character is trivial on the first piece. Thus the character factors through $(A/P_i)^*$ and thus $e_i=1$. In fact all we need is that $\gcd(l,p)=1$ for this argument to go through.

This seems to say that the only way you can get characters of non-square free conductor is if the order of the character has to somehow interact non-trivially with the characteristic of the base field. Does this sound right?

  • $\begingroup$ $l\equiv 1\pmod l$ seems to be a typo. $\endgroup$ Commented Feb 7, 2020 at 2:14
  • $\begingroup$ @KemonoChen Fixed! $\endgroup$
    – Arkady
    Commented Feb 7, 2020 at 2:21

1 Answer 1


$A/(f)^\times=A/(\prod_j f_j^{e_j})^\times\cong \prod_j A/(f_j^{e_j})^\times$,

$d_j=\deg(f_j)$, $D=\deg(f)$,

$A/(f_j)^\times$ has $q^{d_j}-1$ elements, $ A/(f_j^{e_j})^\times$ has $(q^{d_j}-1) q^{d_j(e_j-1)}$ elements, $\ker(A/(f_j^{e_j})^\times\to A/(f_j)^\times)$ has $q^{d_j(e_j-1)}$ elements and hence it is in the $q^D$-torsion of $A/(f)^\times$.

$\chi$ has order $l$ and $q\equiv 1\bmod l$ means that $\chi(a^{q^D})=\chi(a)^{q^D}=\chi(a)$

Thus $ \ker(\chi)$ contains the $q^D$ torsion of $A/(f)^\times$ and hence $\chi$ factors through $A/(\prod_j f_j)^\times$.

  • $\begingroup$ Cool, this is what I was thinking too. For your penultimate line, we can make do with $\gcd(q,l)=1$ right? $\endgroup$
    – Arkady
    Commented Feb 7, 2020 at 2:55
  • $\begingroup$ Yes ${}{}{}{}{}$ $\endgroup$
    – reuns
    Commented Feb 7, 2020 at 2:56

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