Is this function part of the Galois group? Let $\Bbb K$ be a Galois extension of $\Bbb F_p$, where $p$ is prime.  Let $\Phi$ be a function on $\Bbb K$.
If $\Phi$ is an automorphism of $\Bbb F_p$ that permutes the roots of the minimal polynomial generating $\Bbb K$, can we conclude that $\Phi$ is a member of Gal$(\Bbb K/\Bbb F_p)$?
 A: Let $K/F$ be a normal extension where $K$ is the splitting field of $m(x)\in F[x]$. One can easily define a function $\Phi$ which is a given permutation on $m^{-1}(0)$ and acts trivially on $F$ (assuming $m^{-1}(0)\cap F$ is empty) but then one can choose how $\Phi$ acts on the rest of $K\setminus(F\cup m^{-1}(0))$ however one wants; we at least want our $\Phi$ to be $F$-linear. In general a permutation on $m^{-1}(0)$ does not even lift to an $F$-linear map on $K$ as an $F$-space, let alone an $F$-automorphism of it as a field. This is because proper subsets of the roots can be $F$-linearly dependent. For example in $\Bbb Q(\zeta_7)/\Bbb Q$ we have $\zeta+\zeta^{-1}\in F$ but $\zeta+\zeta^2\not\in F$ (here the polynomial is $\frac{x^7-1}{x-1}$) so a permutation that fixes $\zeta$ but exchanges $\zeta^{-1}$ for $\zeta^2$ would break this relation and hence can't be extended to an $F$-linear map. Moreoever the roots may not span $K$ (we may need powers of them too) which means a permutation isn't enough to determine a linear map, and the extra wiggle room would allow us to make non-automorphisms surely.
The only way a permutation on $m^{-1}(0)$ lifts to an $F$-linear automorphism is if ${\rm Aut}_FK$'s image in ${\rm Perm}(m^{-1}(0))$ is not a proper subgroup, else we could pick permutations outside it with no lift.
In particular with $F=\Bbb F_p$, $p$ odd and $K$ the splitting field of $X^p-X-a$ it is not enough. Observe that if $\alpha$ is a root of this polynomial then $\alpha+\Bbb F_p$ is the set of all roots so $[K:F]=p$ and the Frobenius map generates the Galois group which is cyclic, $C_p$. There are $p$ elements in this Galois group but $p!$ elements in the set of permutations on $\alpha+\Bbb F_p$, so most permutations don't lift to automorphisms. However $p=2$ does work out (in fact we don't even have to stipulate $F$-linear, as $K=F\cup m^{-1}(0)$ then).
