TITLE: $f \in K[X]$ of degree $n$ with Galois group $S_n$, why are there no non-trivial intermediate fields of $K \subset K(a)$ with $a$ a root of $f$?
QUESTION [11 upvotes]: Let $K$ be a field and $f \in K[X]$ of degree $n$ with Galois group $S_n$. Let $a$ be a root of $f$, $L = K(a)$, and let $E$ be a intermediate field of the extension $K \subset L$. Prove that $E = K$ or $E = L$. 
Any suggestions on where to start? Thanks.

REPLY [7 votes]: Let $M$ be the splitting field of $f$ and let $a_1,\ldots,a_n$ be the roots of $f$ where $a_1 = a$. Every automorphism of $M|K$ permutes the set $\{a_1,\ldots,a_n\}$ and the fact that the Galois group of $f$ is $S_n$ means that every permutation $\sigma \in S_n$ defines such an automorphism by sending $a_i$ to $a_{\sigma(i)}$. An automorphism $\sigma \in G(M|K)$ fixes $L$ if and only if $\sigma(a) = a$, i.e. $\sigma(1) = 1$ when considered as an element of $S_n$. That means that $G(M|L)$ as a subgroup of $S_n$ is the subgroup $S_{\{2,\ldots,n\}}$ of permutations fixing $1$.
The intermediate extensions $K \subseteq E \subseteq L$ now correspond to the subgroups $G$ of $S_n$ containing $S_{\{2,\ldots,n\}}$, so we have to show that every such subgroup is either $S_{\{2,\ldots,n\}}$ or $S_n$.
If $G$ contains an element $\sigma$ with $\sigma(1) \neq 1$, then it contains all transpositions of the form
$$(1\; i) = \sigma^{-1} ( \sigma(1) \; \sigma(i)) \sigma,$$
(unless $\sigma(i) = 1$) and also all transpositions $(i\; j)$ with $i,j \neq 1$. Since $S_n$ is generated by the transpositions it follows that $G = S_n$.
Edit: To show that $(1\; i) \in G$ where $\sigma(i) = 1$ we may assume $n \geq 3$ (the case $n \leq 2$ is trivial). Take any $j$ with $j \neq 1$ and $j \neq i$. Then $(1\;i) = (i\;j)(1\;j)(i\;j) \in G$.