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Let $K$ be a finite and normal extension of $F$ and suppose that the extension $K/F$ has no proper intermediate extension. Show that the degree $[K:F]$ is a prime number.

Do we consider the Galois group $\text{Gal}(K/F)$ and that $[K:F]=|\text{Gal}(K/F)|$ ?

Since the extension $K/F$ has no proper intermediate extension do we get that the Galois group has no proper subgroup? And that would mean that there is no $n$ such that $n\mid |\text{Gal}(K/F)|$ with $n\neq 1$ an $n\neq |\text{Gal}(K/F)|$ ?

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    $\begingroup$ Yes ${}{}{}{}{}{}{}{}{}{}{}{}$ $\endgroup$ – DonAntonio Nov 24 '18 at 22:16
  • $\begingroup$ Can we consider the Gaois group because the extension is normal? @DonAntonio $\endgroup$ – Mary Star Nov 24 '18 at 22:17
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    $\begingroup$ Yes, but I notice now that you didn't say anything about separable... But assuming you meant also separable, then yes: the extension is Galois and we can use directly the Galois group. $\endgroup$ – DonAntonio Nov 24 '18 at 22:24
  • $\begingroup$ At the exercise statement it is not mentioned that it is separable. Can we just assume that? Or do we have to take cases? If the extension is not separable does the stamenet also hold? @DonAntonio $\endgroup$ – Mary Star Nov 24 '18 at 22:47
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    $\begingroup$ @MaryStar Are you normally dealing with characteristic zero? $\endgroup$ – Display Name Nov 24 '18 at 23:45
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As the commenters pointed out, the possibility of inseparability adds a few extra twists.

If the extension $K/F$ is also assumed to be separable, then it is Galois. Furthermore, lack of intermediate fields is then equivalent to the Galois group not having non-trivial subgroups. That is easily seen to be equivalent to the Galois group being cyclic of prime order proving the claim in that case.

If the extension $K/F$ is not known to be separable, then we can proceed as follows. Let $\alpha\in K\setminus F$ be arbitrary. Because there are no intermediate fields, you can deduce that $K=F(\alpha)$. Let $m(x)\in F[x]$ be the minimal polynomial of $\alpha$ over $F$. If $m(x)$ is separable, then so is $K/F$ and we are back in the first case. Therefore we can deduce that $m(x)$ has multiple zeros, and the usual argument then shows that we must be in positive characteristic, say $p$, and that $m'(x)=0$ implying that $$ m(x)=x^{np}+a_{n-1}x^{(n-1)p}+\cdots+a_1x^p+a_0. $$ In other words $m(x)=f(x^p)$ for the polynomial $f(x)=x^n+\sum_{i=0}^{n-1}a_ix^i\in F[x]$.

Here obviously $f(\alpha^p)=m(\alpha)=0$. Therefore the minimal polynomial of $\alpha^p$ is a factor of $f(x)$. But $f(x)$ is necessarily irreducible for if $f(x)=g(x)h(x)$ non-trivially then $m(x)=g(x^p)h(x^p)$ in violation of irreducibility of $m(x)$. So $f(x)$ must be the minimal polynomial of $\alpha^p$.

Because $\deg f(x)<\deg m(x)=[K:F]$ we can deduce that $F(\alpha^p)$ is a proper subfield of $K$. The lack of intermediate fields means that $F(\alpha^p)=F$. In other words, $\alpha^p$ is an element of $F$. But this implies that the minimal polynomial of $\alpha^p$ over $F$ is the linear polynomial $f(x)=x-\alpha^p$. Therefore $$m(x)=f(x^p)=x^p-\alpha^p.$$ Hence $[K:F]=\deg m(x)=p$. QED.

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  • $\begingroup$ For the sake of completeness: in this thread we discuss the need for the normality assumption. As explained there, without normality there are field extensions of non-prime degree such that there are no intermediate fields. $\endgroup$ – Jyrki Lahtonen Nov 25 '18 at 9:14

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