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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 $F$.

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Do you know some about Galois correspondance ? – Lierre Feb 12 '12 at 13:10
@Lierre: the Galois correspondence is not necessary here. You can prove it just starting from the definition of $[K:F]$. – Damian Sobota Feb 12 '12 at 13:32
@DamianSobota: Maybe (although I don't see any immediate solution without it) ! But if it is an exercise, we should guess what is the expected proof. That's why I asked. – Lierre Feb 12 '12 at 13:37
@Damian: really? This is quite amazing: could you please elaborate? – Georges Elencwajg Feb 12 '12 at 17:19

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 $K$.

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
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.

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Can you give me ab example where extension has composite order and still has no sub-extensions? – Swapnil Tripathi Oct 6 '14 at 22:20
@Swapnil: for any composite $n$ part 2) of the answer gives an example. – Georges Elencwajg Oct 7 '14 at 10:54

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