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If a normal $K/F$ has no intermediate extensions, then $\[K : F\]$ is prime

I want an example of: K is a finite extension of F.There is no proper intermediate extensions of K/F. [K:F] is not prime. I do not know about 'discriminant' how to write comments here?.

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marked as duplicate by Qiaochu Yuan Feb 21 '12 at 18:01

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You can't write comments until you have enough points. You can edit your question, instead. –  Gerry Myerson Feb 17 '12 at 3:38
    
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3 Answers 3

For any positive integer $n$, you can find a finite Galois extension $K/F$ with Galois group $S_n$. Let $H$ be the subgroup of $S_n$ consisting of all permutations fixing a particular point -- say $1$ if you identify $S_n$ with the group of bijections of $(1,\ldots,n)$. Then $H$ has index $n$ in $S_n$ and is a maximal subgroup. It follows that $K/K^H$ is a degree $n$ extension with no proper intermediate extensions.

If you don't know this much Galois theory, you might just want to look at a specific $S_4$-extension of $K/\mathbb{Q}$ and try to find the index four subfield $K^H$ by hand. This will take some calculation, but it may be educational...

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Pete, it's a very basic result, but do you know who proved the claim in your first sentence? Does it date back to Galois himself? –  Charles Feb 16 '12 at 18:27
    
@Charles: no, it's too old for me to know about, I'm afraid. Famously Hilbert knew how to prove this, but also much stronger results. Perhaps Galois knew it in some sense, but I doubt he wrote down a proof. –  Pete L. Clark Feb 16 '12 at 18:55
    
Anyway, the modern proof is just to observe that $S_n$ naturally acts by field automorphisms of $\mathbb{C}(t_1,\ldots,t_n)$ and apply Artin's Theorem to get that $\mathbb{C}(t_1,\ldots,t_n)/\mathbb{C}(t_1,\ldots,t_n)^{S_n}$ must then be Galois with automorphism group $S_n$. –  Pete L. Clark Feb 16 '12 at 18:57
    
Well, at least that bounds the years. –  Charles Feb 16 '12 at 19:20
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All we need is to find a Galois extension with Galois group that has a maximal subgroup whose index is not prime.

Then taking the fixed field of the subgroup will give the example.

For example, let $L$ be the splitting field of $x^4-x-1$ over $F=\mathbb{Q}$. Then $\mathrm{Gal}(L/F) \cong S_4$. Let $\alpha_1,\alpha_2,\alpha_3,\alpha_4$ be the four roots of $x^4-x-1$, and let $H$ be the subgroup of $\mathrm{Gal}(L/F)$ that fixes $\alpha_4$. Then $H\cong S_3$, so $[S_4:H]=4$. However, $H$ is maximal in $S_4$, since the only subgroup of $S_4$ of index $2$ is $A_4$. Let $K$ be the fixed field of $H$ (i.e., $K=\mathbb{Q}(\alpha_4)$). The intermediate fields of $K/F$ are in one-to-one, inclusion reversing correspondence with the subgroups of $S_4$ that contain $H$, hence $K/F$ has no intermediate extensions. And $[K:F]=[S_4:H]=4$.

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Here's how to find one: Let $E/F$ be a Galois extension with Galois group $G$, let $H$ be a maximal subgroup of $G$ whose index is not prime, and let $K$ be the fixed field of $H$.

The smallest group I know with a maximal subgroup of non-prime order is $S_4$, which has maximal $S_3$ subgroups of index $4$. This gives us the following example:

Let $E$ be the field $\mathbb{Q}(x_1,x_2,x_3,x_4)$ of rational functions in four variables with rational coefficients. The group $S_4$ acts on $E$ by permutation of variables. Let $F$ be the fixed field of this action, i.e. the symmetric rational functions. Let $S_3$ denote the subgroup of $S_4$ consisting of permutations that fix $x_4$, and let $K$ be the fixed field of $S_3$ (i.e. rational functions that are symmetric between $x_1$, $x_2$, and $x_3$). Then $K/F$ is a degree four extension with no proper intermediate extensions.

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@Arturo Good point! I have edited my answer. –  Jim Belk Feb 16 '12 at 17:42
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