$F$ is perfect, $E=F(\beta)$, where $\beta^n=a\in F$, then $E$ is a Galois extension over $F$ if and only if the field contains the n-th primitive root $\omega$, where $\omega^n=1$
Question: I am confused on this word the field, does this refers to $F$ or $F(\beta)$?
My work: Perfect fields are separable, hence to show it is Galois, we need to show it is a splitting field. All roots for $x^n-a=0$ are in the form of $\beta \omega^k, k=0,1,..., n-1$. Hence if $F(\beta)$ is the splitting field, then $F(\beta)$ must contains all these roots, hence contains the quotient of these roots, namely $$\omega\in F(\beta)$$
So it seems the field refers to $F(\beta)$, but I want to find an example, such that $\omega\in F(\beta)\land \omega\notin F$?
Here is an example. Let $F=Q$, and $\beta=1+i$ is a root of $f(x)=x^4-4=0$, so $n=4$ and $\beta^4=4=a\in Q$ in this case, and $\omega=i$ is the primitive root of $\omega^4=1$
The splitting field of $f(x)$ is $K=Q(i+1,i-1,-i+1,-i-1)=Q(i)$ with degree $2$, and
$E=F(\beta)=Q(i+1)=Q(i)$, hence $E=K$ is the splitting field of $f(x)$. Namely, $E=F(\beta)$ is the Galois extension of $F$. However, $\omega=i\in F(\beta)\land \omega\notin F$.