Show that every finite extension of $F$ is cyclic. Let $k$ be a field, $k^a$ an algebraic closure, and $\sigma$ be an automorphism of $k^a$ leaving $k$ fixed. Let $F$ be the fixed field of $\sigma$. Show that every finite extension of $F$ is cyclic. 
I'm completely clueless, can anyone give some hint/solution ? 
 A: Let $E$ be such an extension. Can you find a relation between $Gal(E/F)$ and $Gal(k^a/F)$?

As the multiple downvoting shows, this hint-question appears to have been insufficient. Hence the following elaboration:
Let $a\in k^a$. Then $a$ is algebraic over $k$. Thus its orbit $\langle \sigma\rangle a$ under the operation of $\langle \sigma\rangle$ is contained in the set of conjugates of $a$, which is finite. Then the polynomial $$\prod_{b\in\langle \sigma\rangle a}(X-b)\in k^a[X]$$ is $\sigma$-invariant, hence is in fact $\in F[X]$, showing that $a$ is separable over $F$.
Being an algebraic closure, $k^a$ is also normal over $k$ and $F$, hence Galois.
Via the famous correspondence between subgroups of $\operatorname{Gal}(k^a/F)$ and intermediate fields between $F$ and $k^a$, both $\operatorname{Gal}(k^a/F)$ and $\langle \sigma\rangle$ correspond to $F$. Thus $\operatorname{Gal}(k^a/F)=\langle\sigma\rangle$.
Now let $E$ be a finite extension of $F$. As that makes $E$ algebraic over $F$, we can embed $E$ into $k^a$ and by abuse of language view $E$ as an intermediate field between $F$ and $k^a$.
Then $E$ corresponds via Galois correspondence to a subgroup $H\le \langle \sigma\rangle$. As any subgroups of a cyclic (or even abelian) groups is normal, we conclude that the extension $E/F$ is Galois and we know that $\operatorname{Gal}(E/F) =\operatorname{Gal}(k^a/F)/H$, which is cyclic.
A: I think we can do it as follows: Let $E$ be a finite extension of $F$. From above post we can see that $E/F$ is separable, hence simple. Let $E=F(a)$. Let $f_a(x)$ be the  minimal polynomial of $a$ over $F$. Now let $E'$ be the splitting field of $f_a$, i.e let $E'=F(a_1,a_2,..,a_k)$ where $a_i's$ are conjugates of $a$. So, $E'/F$ is normal and hence its is Galois, since again from above post we can say $E'/F$ is separable. Now using fundamental theorem of Galois theory we can say $Gal(E'/F)=\langle\sigma_1\rangle$ where $\sigma_1$ is restriction of $\sigma$ to $E'$ (since $E'/F$ is normal, hence $\sigma_1 \in Gal(E'/F)$). And now if $H=Gal(E'/E)$ then $Gal(E'/F)= \langle \sigma_1 \rangle /H$, and hence cyclic.
