Galois action on algebraic K-theory of finite fields

Question

This might be well-known to experts. I was just teaching a course where we went through some parts of Quillen's theorem computing the higher algebraic K-theory of finite fields. Denote by $\mathbb F_q$ the field with $q$ elements.

Theorem (Quillen) Let $q=p^k$ for some prime $p$ and $k \in \mathbb N_{\geq 1}$. $$K_{2i-1}(\mathbb F_q) = \mathbb Z/(q^i-1)\mathbb Z \quad \mbox{and} \quad K_{2i}(\mathbb F_q)= 0.$$

Canonically, $K_1(\mathbb F_q) = \mathbb F_q^\times$, so that the action of ${\rm Aut}(\mathbb F_q)$ on $K_1(\mathbb F_q)$ is clear. Since higher Milnor K-theory of finite fields vanishes, this does not help to understand the action on the higher K-groups.

Question: How does ${\rm Aut}(\mathbb F_q)$ act on $K_{2i-1}(\mathbb F_q)$?

Quillen's proof relies on the choice of an embedding $\bar {\mathbb F}_q^\times \to S^1$, which makes the isomorphism somewhat non-canonical, as far as I understand. Anyway, my guess would be that the Frobenius $F(x) = x^p$ acts as $a \mapsto p^i a$ on $\mathbb Z/(q^i-1)\mathbb Z$, but I do not know how to prove this.

Answer 1

$K(\overline{\mathbb{F}_p})$ is, after completion at any prime $\ell \neq p$, equivalent to $ku^\wedge_\ell$. It's true that the equivalence relies on a non-canonical embedding, but the conclusion that the homotopy groups of the $\ell$-completion are a polynomial ring, is independent of the concrete isomorphism. Since the homotopy groups of $K(\overline{\mathbb{F}_p})^\wedge_\ell$ are $\ell$-divisible and concentrated in odd degrees, we get $$ \pi_{2n} K(\overline{\mathbb{F}_p})^\wedge_\ell = \operatorname{Hom}(\mathbb{Q}_\ell/\mathbb{Z}_\ell, \pi_{2n-1}K(\overline{\mathbb{F}_p})). $$ We know the Frobenius acts by multiplication by $p$ on $\pi_1K(\overline{\mathbb{F}_p})\cong \overline{\mathbb{F}_p}^\times$, so it acts by multiplication by $p$ on $\pi_2 K^\wedge_\ell$, hence by multiplication by $p^n$ on $\pi_{2n} K^\wedge_\ell$, hence by multiplication by $p^n$ on $\pi_{2n-1} K(\overline{\mathbb{F}_p})$ as you expected (by going over all $\ell\neq p$). The finite field case follows from this by comparison along the canonical map between them.