# Linear characters of algebraic closed fields

This question is a bit of a follow up to this question.

Let us consider the finite field $\mathbb{F}_q$ and its algebraic closure $\mathbb{F}$, viewed as an additive abelian group. Its group of linear characters, ${\rm Hom}(\mathbb{F},\mathbb{C}^*)$, consists of the group homomorphisms from the (additive) abelian group $\mathbb{F}$ to the multiplicative group of $\mathbb{C}\setminus \{0\}$. If $\lambda$ is linear character that is fixed by the Frobenius morphism ($x\mapsto x^q$), then is it true that $\lambda$ trivial? (Possibly because the Lang map is surjective?)

And is there a good description of the linear characters? And a good relation with ${\rm Hom}(\mathbb{F}_q,\mathbb{C}^*)$ commuting with the trace maps?

• I’m pretty sure you don’t mean $\text{Hom}(\mathbb{F},\mathbb{C})$ ...? – Jeremy Rickard Oct 22 '17 at 21:08
• Do you want just arbitrary group homomorphisms? Or are you assuming some sort of continuity? – Johannes Hahn Oct 22 '17 at 23:10
• I'm assuming arbitrary group morphisms. – João Dias Oct 22 '17 at 23:15


Since all elements in $\overline{\IF_p}$ have order $p$, any group homomorphism $\overline{\IF_p}\to\IC^\times$ has either image contained in the group of $p$-th roots of unity $\mu_p$. In other words we can equivalently ask for homomorphisms of additive groups $\overline{\IF_p}\to\mathbb{Z}/p$, i.e. $\IF_p$-vector space homomorphisms $\overline{\IF_p}\to\IF_p$.

Let $F$ be the Frobenius automorphism. Since $F-1: \overline{\IF_p}\to\overline{\IF_p}$ is surjective, we have $\phi\circ F=\phi \iff 0=\phi\circ(F-1) \iff \phi = 0$.

The next best thing is the normalised galois trace $\tau: \bigcup_{(p,m)=1} \IF_{p^m} \to \IF_p$. That is the unique (up to multiplication with a constant) $F$-invariant linear form on that field.

Every finite field has a one-dimensional space of $F$-invariant linear forms, namely the dual space of $\IF_{p^m} / im(F_{|\IF_{p^m}}-1)$. That is one-dimensional because $\dim im(F-1) = m - \dim ker(F-1) = m-1$. The problem of course is that there is no canonical one and these linear forms do not extend to $F$-invariant linear forms on field extensions of $\IF_{p^m}$ in general (they extend iff the degree of the extension is not divisible by $p$).

Together with $\tau$ you can build an $F$-invariant linear form on $\IF_{p^a} \otimes \bigcup_{(p,m)=1} \IF_{p^m}$ but that is as far as you can go.

• But if the normalized galois trace is $F-$invariant does not mean that is zero? – João Dias Oct 23 '17 at 1:04
• No, of course not. At the very least because $\tau(1)=1$. But I see what you mean: $\tau$ is not defined on $\overline{\mathbb{F}_p}$ (you can't normalise for extensions of degree divisible by $p$), but on the other field I mentioned. I will correct that. – Johannes Hahn Oct 23 '17 at 11:00