$F_{q^d}/F_q$ a finite field extension, a question about the homomorphism: $Aut_{F_q}(F_{q^d})\to Aut(F^{\times}_{q^d})$.

Question

I want to find the range of such homomorphism and to what Frobenius endomorphism of $q$, i.e $Fr_q$, is sent to? what is the kernel?

Can you help me with hints?

Thanks! Even keywords to use for better google searching.

Answer 1

Here the group of field automorphisms $Aut_{\Bbb{F}_q}(\Bbb{F}_{q^d})$ is known to be cyclic of order $d$, and generated by the Frobenius $F$.

The latter group consists of automorphisms of the multiplicative group $G=\Bbb{F}_{q^d}^*$. The group $G$ is known to be cyclic of order $q^d-1$. If $g\in G$ is a generator, a homomorphism $\phi: G\to G$ is uniquely determined by where it sends $g$. So we known that $\phi(g)=g^j$ for some natural number $j$. Call this homomorphism $\phi_j$. For it to be an automorphism of $G$ we need it to be surjective (because $G$ is finite surjectivitiy implies bijectivity). This happens if and only if $\phi_j(g)=g^j$ is another generater of $G$ if and only if $\gcd(j,q^d-1)=1$. As $\phi_j\circ\phi_k=\phi_{jk}$ with the product $jk$ calculated modulo $q^d-1$ we arrive at the description $$Aut(G)\simeq \Bbb{Z}_{q^d-1}^*.$$

So any homomorphism from $Aut_{\Bbb{F}_q}(\Bbb{F}_{q^d})$ to $Aut(G)$ is really a homomorphism $\psi$ from $C_d$ to $\Bbb{Z}_{q^d-1}^*$. Again, $\psi$ is uniquely determined by where it maps $F$. Because $F^d=id$ we get $\psi(F)^d=\phi_1$ as the only constraing. So whenever $a^d\equiv1\pmod{q^d-1}$, we get a well defined homomorphism $\psi_a$ uniquely determined by $\psi_a(F)=\phi_a$.

If the order of $a$ in $\Bbb{Z}_{q^d-1}^*$ is $m$, then we must have $m\mid d$. In that case $\operatorname{Im}(\psi_a)=\langle \phi_a\rangle\simeq C_m$, and $\operatorname{Ker}(\psi_a)=\langle F^m\rangle\simeq C_{d/m}$. It may be worth observing that this is in line with the first isomorphism theorem.

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However, there is a very natural homomorphism that simply maps $F$ to its restriction to the multiplicative group. Really, $F=\phi_q$. That special homomorphism is the homomorphism $\psi_q$ from above. It is not unnatural to think that wherever this question came from, this restriction (we really just exclude zero from the domain and the range) is the intended homomorphism. However, it is impossible to say without seeing the source. Should this be the intended interpretation, then the order of $a=q$ is $m=d$, and the homomorphism $\psi=\psi_q$ is injective. Hence its kernel is trivial and the image consists of powers of the Frobenius.