Example of finite extension that is not prime with no proper intermediate extension. I want an example of a finite extension K of F such that there are no proper intermediate extensions of K/F, and
[K:F] is not prime.
I do not know about 'discriminant'.
 A: 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...
A: 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 $[K:F]=[S_4:H]=4$.
A: 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.
