TITLE: Splitting field and polynomial of minimal degree
QUESTION [5 upvotes]: Let's assume that we have a splitting field $F$ over $Q$ that is a finite extension. Let $p(x)$ be the polynomial in $Q[x]$ that has $F$ as a splitting field and is of minimal degree. Is it correct that if $Gal(F,Q)$ is isomorphic to $S_n$ then $deg \ p(x)=n$? If yes, is it true that, more generally, $deg \ p(x)$ equals $n$ where $n$ is the minimal natural number such that $Gal(F,Q)$ is contained in $S_n$? 
Thanks a lot in advance.

REPLY [6 votes]: Yes and yes. In one direction, if $p(x)$ is a polynomial with splitting field $F$, then $G = \text{Gal}(F/\mathbb{Q})$ acts faithfully on the roots of $p(x)$, and this gives an embedding of $G$ into $S_n$ where $n = \deg p$. 
In the other direction, the data of an embedding of $G$ into $S_n$ is the data of a faithful action of $G$ on a set with $n$ elements. Decompose this action into its orbits, and let $H_1, H_2, \dots H_k$ be the corresponding stabilizers. The condition that $G$ embeds into $S_n$ is precisely the condition that the intersection of the conjugates of all of the $H_i$ is trivial. Let
$$F_i = F^{H_i}$$
be the subextensions of $F$ corresponding to the $H_i$. Now the condition that $G$ embeds into $S_n$ is precisely the condition that the $F_i$ have Galois closure $F$. Let $p_i(x)$ be the minimal polynomial of a primitive element of $F_i$, and let
$$p(x) = \prod p_i(x).$$
Then $p(x)$ is a polynomial of degree $\sum |G/H_i| = n$ with splitting field $F$.