I meet an exercise when I study Galois Theory: Suppose $E/F$ is a finite Galois extension. Suppose $a\in E$ $\\$ is an element fixed only by the identity automorphism of E, i.e.$\{\phi\in \rm{Aut}(E):\phi(a)=a\}=\{id_E\}$. Then do we have $E=F(a)$?
My idea is: We can easily have $E/F$ is a simple extension by primitive element theorem, i.e.$E=F(b)$ for some $b\in E\setminus F$.
My question is: I do not know how to get $b=a$. Furthermore, if we take $\forall x\in E\setminus F$, can we have $E=F(x)$? I think it might be wrong, can we make it right after adding some condition on $x$? I think it might be right if we ask $x$ is algebraic over F or $x$ is separable.
Appreciating for any idea and suggestion.
$\DeclareMathOperator{\Gal}{Gal}$ Let $n = [E : F] = \lvert\Gal(E/F)\rvert$.
Since the only element of $\Gal(E/F)$ fixing $a$ is the identity, we see that there are $n$ distinct elements in the orbit of $a$ under $\Gal(E/F)$. This is the same as the degree of the extension $E/F$. What can you conclude?
Edit: A further hint: Recall that given elements $\alpha, \beta \in E$, there exists $\sigma \in \Gal(E/F)$ with $\sigma(\alpha) = \beta$ only if $\alpha$ and $\beta$ have the same minimal polynomial over $F$.