# Proving there are no subfields

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!

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Look at the subgroup structure of $S_4$ and apply the fundamental theorem of Galois Theory. Don's answer below is the way to go. – user38268 Jul 22 '12 at 10:39

If $F$ were an intermediate field of $K/\mathbb Q$, then $[K:F]=[F:\mathbb Q]=2$. Therefore the minimal polynomial of $\theta$ over $F$ is of the form $g(X) = X^2 + aX + b$.

But then $f(X) = g(X) h(X)$ with $g, h \in F[X]$ of degree $2$. Adjoining $\theta$ as a root of $g$ and a root $\beta$ of $h$, we get an extension $F(\theta, \beta)$ in which $g$ and $h$ split, hence $f$ splits in $F(\theta, \beta)$.

Now on the one hand $$[F(\theta, \beta):F] = [F(\theta, \beta):F(\beta)]\cdot[F(\beta):F]\le 4$$ implies $[F(\theta, \beta):\mathbb Q] \le 8< 24 = 4!$.

On the other hand, $F(\theta, \beta)$ contains the splitting field of $f$. This is clearly not possible if the splitting field has Galois group $S_4$ of order $4!$.

Btw: I don't think that it is necessarily true that $\theta = \sqrt{b}$ for some $b\in F$. The general form is rather $\theta = a \pm \sqrt{b}$ for $a,b\in F$, I believe.

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I like this solution very much since no knoledge of $S_4$ is needed (exept the size, of course). thank you! – Belgi Jul 23 '12 at 1:18

An idea: $\,S_4\,$ has a unique subgroup of order $\,4\,$ which is then normal and contained in $\,A_4\,$ , as can be swiftly checked.

If there was a subextension of order two this would be mean $\,A_4\,$ has a subgroup of order $\,6\,$ (of or index $\,2\,$, however you want to attack this), which is does not, as we know.

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