Simple algebraic extension of fields

Question

Let $L/K$ be a finite extension of fields such that for every $a\in L\setminus K$, we have $[L:K]=[K(a):K]$. Can we say that $[L:K]$ is a prime number?

Answer 1

The answer is no: if $n>1$ is an integer, and the field $K$ is such that it admits a Galois extension with group $S_n$, then there exists a degree $n$ extension $L$ of $\mathbb{Q}$ such that there are no subextensions $K \subsetneq K' \subsetneq L$. It follows that, in that case, for any $a \in L$ with $a \notin K$ we have $K(a)=L$.

To show this, let $F/K$ be a Galois extension with group $S_n$, and let $L \subset F$ be the subfield which corresponds to the subgroup $S$ of permutations $\pi \in S_n$ which satisfy $\pi(1)=1$ (or which have any other fixed point, it doesn't matter which one). It is easy to convince yourself that there are no subgroups $S \subsetneq S' \subsetneq S_n$, hence by the main theorem of Galois theory there are no subextensions $K \subsetneq K' \subsetneq L$.

As an example, the field $\mathbb{Q}$ admits $S_n$ extensions for every $n$, and hence there are field extensions $\mathbb{Q} \subset L$ of every degree $n>1$ that have no proper subextensions.

For an explicit example: the polynomial $f = x^4+x+1$ has Galois group $S_4$ over $\mathbb{Q}$. If $F=\mathbb{Q}(\alpha_1,\alpha_2,\alpha_3,\alpha_4)$ is the splitting field of $f$, where the $\alpha_i$ are the zeros of $f$, then the subextension corresponding to the permutations which keep the $i$-th zero fixed is just $\mathbb{Q}(\alpha_i)$. So for any zero $\alpha$ of $x^4+x+1$, the field extension $\mathbb{Q}(\alpha)/\mathbb{Q}$ is an example.