Let $K$ be a Galois extension of $\mathbb{Q}$ whose Galois group is isomorphic to $S_5$. Prove that $K$ is the splitting field of some polynomial of degree $5$ over $\mathbb{Q}$.
Since $K$ is a finite Galios extension over $\mathbb{Q}$ we know that $K$ is the splitting field of a separable polynomial $f$ over $\mathbb{Q}$. Let $n$ be the degree of this separable polynomial. Since the Galois group acts on the roots $f$ via permutation we know that the Galois group is isomorphic to a subgroup of $S_n$ and hence $n \geq 5$. Let $\alpha$ be a root of $f$. Since since $|K : \mathbb{Q}| = |K :\mathbb{Q}(\alpha)| |\mathbb{Q}(\alpha):\mathbb{Q}| = n|K :\mathbb{Q}(\alpha)|$, we have that $n|5! = 120$. Therefore $n \in \{5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120\}$
Any help would be appreciated.
So I think this is it.
Since there are only finitely may subgroups of $S_5$, there are only finitely many intermediate fields of $K|\mathbb{Q}$, and so each intermediate field is a simple extension of $\mathbb{Q}$. Since $S_4$ is a subgroup of $S_5$, there exists an intermediate field $\mathbb{Q}(b)$ such that $S_4 = \mathrm{Gal}(K/\mathbb{Q}(b))$. Since $24 = |K : \mathbb{Q}(b)|$ we have $5 = |\mathbb{Q}(b): \mathbb{Q}|$ and hence $\deg \min(b,\mathbb{Q}) = 5$.
Let $g =\min (b,\mathbb{Q})$. Now since $K$ is Galois over $\mathbb{Q}$, we know that $g$ splits in $K$, and furthermore the splitting field $F$ of $g$ over $\mathbb{Q}$ contained in $K$, is a Galois extension of $g$ and so Gal$(K/F)$ must be a normal subgroup of $S_5$ that is contained in $S_4$. Since the only such subgroup is the trivial subgroup is the trial subgroup, it follows that $K = F$, and hence $K$ is the splitting field of $g$.