If a normal finite extension $K/F$ has no intermediate extensions, then $[K : F]$ is prime Let $K$ be a finite normal extension of $F$ such that there are no proper intermediate extensions of $K/F$. Show that $[K:F]$ is
prime. Give a conterexample if $K$ is not normal over $F$.
 A: 1) If $K/F$ is Galois and there is no strictly intermediate extension, Galois theory tells us that $Gal(K/F)$ has  no non-trivial subgroup and thus is of order a prime $p$. Hence $K/F=p$ is prime too.  
2) Let's exhibit   for every $n\gt 1$ a field extension $K/F$ of degree $[K:F]=n$ without  any strictly intermediate subfield. 
a) Take  any Galois extension $L/F$ with Galois group the full symmetric group $S_n$.
[This is easy to find: for example,  take $k(T_1,...,T_n)/k(s_1,...,s_n)$ where the $T_1,...,T_n$ are indeterminates over an arbitrary field $k$ and the $s_i$'s are the elementatary symmetric  in these indeterminates. There are  examples with $F=\mathbb Q$ too, but that is more difficult] 
b) Take  $K=L^{ S_{n-1}}$, the fixed field under the subgroup $S_{n-1}\subset S_n$.
Since there is no subgroup strictly between $S_{n-1}$ and $S_n$, Galois theory implies that there is no strictly intermediate field between $F$ and $K$.  
Edit
Answer 1) remains true under br69's  weaker hypothesis that the extension   only be finite and normal (but not necessarily Galois):  
1') If the finite normal extension $K/F$ has no strictly intermediate extension, then $[K:F]$ is prime
Proof:
We start from the tower $F\subset K_{sep }\subset K$. The no-intermediate-field hypothesis ensures that one of the following two possibilities holds:
i) $K_{sep }= K$. Then the extension $F\subset K$ is Galois and we are back to the Galois case.
ii)   $K_{sep }= F$. Then the extension $F\subset K$ is purely inseparable, hence we are in characteristic $p\gt 0$ and $[K:F]=p^r$.
Now for every $b\in K\setminus F$ there exists some power $b^{p^s}=c$ with $c\notin F$ but $c^p\in F$ .
The inclusions $F\subsetneq F(c)\subset K$ and the no-intermediate-field hypothesis force $F(c)=K$ and thus $[K:F]=[F(c):F]=p$ as desired.
A: I'll give another proof of (1) under the weaker hypothesis that $K/F$ is not necessarily Galois.
Let $K\subseteq E\subseteq F$ the fixed field of $Aut_FK$. Then it has to be $E=K$ or $E=F$.

*

*If $E=F$ then $K/F$ is Galois $\checkmark$


*If $E=K$ then $Aut_FK=1$.



*

*If $F$ is perfect then every irreducible polynomial is seperable. $K/F$ is normal hence it is the splitting field of a collection of irreducible polynomials over $F$. If $f$ is one of these polynomials and $a$ is a root of $f$ which does not belong to $F$ then it is $F(a)=K$. So $K$ is the splitting field of $m_{a,F}$ hence $K/F$ is Galois $\checkmark$


*If $F$ is not perfect and $char(F)=p$ and $a$ is a root like before it is $|K:F|=deg(m_{a,F})$. Then $a$ is the one and only root of $f$ which does not belong to $F$ since if $b$ was another one then $F(a)=F(b)=K$ and we would have and automorphism of $K$ fixing $F$ and sending $a\to b$ which is a contradiction since $Aut_FK=1$. In $K$ it is $$m_{a,F}(x)=(x-a)(x-a_1)...(x-a_{n-1})$$ and $m_{a,F}$ irreducible with $a_i\not\in F\Rightarrow a_i=a \Rightarrow m_{a,F}(x)=(x-a)^n\in F[x]\Rightarrow a^n\in F$. If $a^k\in F$ for some $k<n$ then $m_{a,F}(x)|x^k-a^k$ which is a contradiction. Since $$(x-a)^n=\sum_{k=0}^n \dbinom{n}{k}x^ka^{n-k}$$ it has to be $p| \dbinom{n}{k},\ 1\leq k<n$. So $p|n$ hence we have $F\subseteq F(a^p)\subseteq F(a)\subseteq K\Rightarrow F=F(a^p)=F(a)=K\Rightarrow |K:F|=p$
