I am trying to solve Q11 at pg. 582 from the book Abstract algebra by Dummit and Foote, the question is:
Let $f\in\mathbb{Z}[x]$ be an irreducible quartic whose splitting field has Galois group $S_{4}$over $\mathbb{Q}$. Let $\theta$ be a root of $f$ and denote $K=\mathbb{Q}(\theta)$ prove that $[K:\mathbb{Q}]=4$ which has no proper subfield.
My efforts: $f$ is irreducible over $\mathbb{Q}$ otherwise the splitting field of $f$ would of been of degree $\leq3!$ , hence $[K:\mathbb{Q}]=4$.
I can't figure out the second part, if such subfield exist then it is of degree $2$ over $\mathbb{Q}$ and is of the form $\mathbb{Q}(\sqrt{a})$ where $a\in\mathbb{Q}$. since $[K:\mathbb{Q}(\sqrt{a})]=2$ it also holds that $\theta=\sqrt{b}$ where $b\in\mathbb{Q}(\sqrt{a})$ . this is where I am stuck.
How do I prove that $K$ has no proper subfield ? help is appriciated!