# Primitive characters over function fields

Definitions:

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?

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

$$A/(f)^\times=A/(\prod_j f_j^{e_j})^\times\cong \prod_j A/(f_j^{e_j})^\times$$,
$$d_j=\deg(f_j)$$, $$D=\deg(f)$$,
$$A/(f_j)^\times$$ 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 $$A/(\prod_j f_j)^\times$$.
• Cool, this is what I was thinking too. For your penultimate line, we can make do with $\gcd(q,l)=1$ right? Commented Feb 7, 2020 at 2:55
• Yes ${}{}{}{}{}$ Commented Feb 7, 2020 at 2:56