Dummit and Foote, Exercise 14.2.9 Give an example of fields $F_1,F_2,F_3$ with $\mathbb{Q}\subset F_1\subset F_2 \subset F_3$,
$[F_3:\mathbb{Q}]=8$ and each field is galois over all its sub fields with the exception of $F_2$ not galois over $\mathbb{Q}$
As $[F_3:\mathbb{Q}]=8$, most possibly i should add some 4th root of an element in Q and 4th root of unity $i$. So, $F_3=\mathbb{Q}(\sqrt[4]{2},i)$ with out loss of generality $[F_3:F_2]=[F_2:F_1]=[F_1:\mathbb{Q}]=2$
I want $F_1$ to be galois over $\mathbb{Q}$ and $[F_1:\mathbb{Q}]=2$ So, $F_1$ could be $ \mathbb{Q}(\sqrt{2})$.
I want $F_2$ to be galois over $F_1$ and $[F_2:F_1]=2$ So, $F_2$ has to be $\mathbb{Q}(\sqrt[4]{2})$.
$F_3$ is already chosen, and we have $F_3$ galois over $F_2, F_1, \mathbb{Q}$.
$F_2$ is not galois over $\mathbb{Q}$.
The reason i am not happy about this is that this came to my thought just by chance.
Though i have written a big justification i am not so sure if every extension should be like this????
Does any other extension should be of this type???
I am hoping for a more simpler example.
Any hint/suggestion would be appreciated :)
 A: Hint:
Start with the Galois group $G$ of $F_3$ over $\mathbb Q$. The conditions on $F_3$ can be translated into conditions on the Galois group, and it will turn out that up to isomorphism there is a single group $G$ satisfying these conditions. (Update: Since $G$ must have a subgroup of order $2$ which is not normal, the only possibility is the dihedral group of order $8$, see the above discussion with Jyrki Lahtonen).
Now you have to find a field extension $\mathbb Q \subseteq F_3$ such that its Galois group equals $G$.
The remaining fields $F_2$ and $F_1$ are now determined by suitable subgroups of $G$ via the Galois correspondence.
A: I don't understand the request for a simpler example --- what could possibly be simpler than adjoining $\sqrt2$, then $\root4\of2$, then $i$? 
Anyway, to build an answer from the bottom up, $F_1$ is to have degree 2 over the rationals, so it has to be ${\bf Q}(\sqrt d)$ for some nonsquare integer $d$; that's what extensions of degree 2 look like. Then $F_2$ has to be $F_1(\sqrt{\alpha})$ for some $\alpha$ in $F_1$. $F_2=F_1(\sqrt e)$ for some rational $e$ won't work, since that would make $F_2$ Galois over the rationals. So $\alpha$ has to be something that's in $F_1$ but not rational, and what simpler choice is there than $\alpha=\sqrt d$? So, $F_2={\bf Q}(\root4\of d)$ 
Now you have to get $F_3$ by adjoining $\sqrt{\beta}$ to $F_2$ for some $\beta$ in $F_2$. Since $F_3$ contains $\root4\of d$ and is Galois, it must contain $i\root4\of d$, and thus it must contain $i$. So what could be simpler than letting $F_3=F_2(i)$? 
So I would suggest that you didn't just come to you by chance --- it came to you because it's the most logical & intelligent thing to do, if you are looking to construct the simplest possible answer from the rationals up. And if you analyze the process carefully, you'll see whether you might be able to modify it to get other constructions of examples. 
