Automorphism Groups of Finite Fields

Question

In the category Rng, if $F_1,F_2,...,F_n$ are finite fields, can the automorphism group $Aut(F_1 \times F_2 \times \cdots \times F_n)$ be expressed in terms of the automorphism groups $Aut(F_i)$.

Answer 1

$\newcommand{\Size}[1]{\left\lvert #1 \right\rvert}$$\DeclareMathOperator{\Aut}{Aut}$This is an attempt at a solution, very much in the direction indicated in the comments. I am using the fact that automorphisms preserve annihilators, sizes, etc.

Note that for each $i$, we have that the annihilator of $F_{i}$ in the rng $$ R = F_1 \times F_2 \times \cdots \times F_n $$ is of course $$ F_1 \times \cdots \times \widehat{F_{i}} \times \cdots \times F_n $$ (where the hat denotes omission), which is a rng with $2^{n-1}$ idempotents.

Let $G$ be an automorphism, and consider $F_{i}^{g}$, for some $i$. (Sorry, I have the bad habit of writing the action of automorphisms as exponents.)

If $F_{i}^{g}$ is not contained in a single component $F_{j}$, then there are elements of $F_{i}^{g}$ with non-zero components in $F_{h}$ and $F_{k}$, for some $h < k$, so that the annihilator of $F_{i}^{g}$ is contained in $$ F_1 \times \cdots \times \widehat{F_{h}} \times \cdots \widehat{F_{k}} \times \cdots\times F_n, $$ and thus will have at most $2^{n-2}$ idempotents. Thus $$F_{i}^{g} \subseteq F_{j}$$ for some $j$.

However the annihilator of $F_{i}$ has size $\Size{R}/\Size{F_{i}}$, whereas, $F_{j}$ being a field, the annihilator of $F_{i}^{g}$ has size $\Size{R}/\Size{F_{j}}$. It follows that $F_{i}^{g} = F_{j}$, and thus $F_{i}$ and $F_{j}$ are isomorphic.

It follows that $g$ induces a permutation of the $F_{i}$, which yields a homomorphism from the automorphism group $$A = \Aut(F_1 \times F_2 \times \cdots \times F_n)$$ to $S_{n}$. Let $H$ be image of this homomorphism. Clearly $H$ splits as the direct product of smaller symmetric groups, which permute among themselves the $F_{i}$ which are isomorphic. The kernel of this homomorphism is clearly $$ K = \Aut(F_{1}) \times \cdots \times \Aut(F_{n}), $$ and then $A$ is the semidirect product of $K$ by $H$.