mona123 asked 9 months ago

Please help me to answer c) of the following problem:
Let $F=\mathbb{F}_3(t)$.
Let $f(x)=x^6+x^4+x^2-t\in F[x]$. You may assume that $f(x)$ is irreducible in $F[x]$. Let $E$ be splitting field of $f(x)$ over $F$.
>a) Show that $f(x)=f(-x)$ and $f(x+1)=f(x)$
>b) Determine $Gal(E/F)$ Hint: why is part a) there?
>c) For each non trivial $H< \text{Gal}(E/F)$ determine $\text{Fix}(H)$. Express the answer in form of $F(\beta)$ whith the minimal polynomial of $\beta$ over $F$ specified.
I proved $a)$ For $b)$, this is what i wrote:

Part (a) has the consequence that, if $r$ is a root of $f$ in an extension field, also $r+1$, $r+2$, $-r$, $-r+1$ and $-r+2$ are roots.
We note also that these are pairwise distinct:
* $r\ne r+1$, $r\ne r+2$, $r\ne -r$, $r\ne -r+1$, $r\ne -r+2$;
* $r+1\ne r+2$, $r+1\ne -r$, $r+1\ne -r+1$, $r+1\ne -r+2$;
* $r+2\ne -r$, $r+2\ne -r+1$, $r+2\ne -r+2$;
* $-r\ne -r+1$, $-r\ne -r+2$;
* $-r+1\ne -r+2$.
From what that, all the roots are contained in $F(r)$, thus $E=F(r)$, so $[E:F]=[F(r):F]=6$, so $Gal(E/F)$ has order $6$. As $Gal(E/F)$ acts transitively on the roots (due to the irreducibility of f), there are $\sigma, \tau \in Gal(E/F)$, such that $\sigma(r)=r+1$ and $\tau(r)=-r$.
Then we have $\sigma(\tau(r))=\sigma(-r)=-\sigma(r)=-r-1$ and $\tau(\sigma(r))=\tau(r+1)=\tau(r)+1=-r+1$ As $-1 \neq 1$, we see that $\sigma$ and $\tau$ don’t commute. Thus $Gal(E/F)$ is non-Abelian and has order $6$, thus $Gal(E/F)=S_3$.