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Suppose $f(x)\in \mathbb{Z}[x]$ is an irreducible quartic whose splitting field has Galois group $S_4$ over $\mathbb{Q}$. Let $\theta$ be a root of $f(x)$ and set $K=\mathbb{Q}(\theta)$.

a) Prove that $K$ is an extension of $\mathbb{Q}$ of degree $4$ which has no proper subfields.

b) Are there any Galois extensions of $\mathbb{Q}$ of degree $4$ with no proper subfields?

I have already solved part (a). But I don't understand what I have to do for part (b). Do I have to find an example? Can anyone please help me?

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  • $\begingroup$ Find an example or show there are no examples. Use Galois theory to think about properties of such an example as in (b) in terms of group theory. $\endgroup$ – KCd Nov 26 '14 at 4:11
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It looks like this problem is leading up to the fundamental theorem of Galois theory, which states that there is a one-to-one correspondence between the subfields of a Galois extension and subgroups of its Galois group. So our answer will be, of course, that no such Galois extension exists (which is why it's difficult to find an example).


First, note that if $K$ is a Galois extension of degree $4$ over $\mathbb{Q}$, then it must be the splitting field of a quartic polynomial. Either that polynomial is irreducible, or it can break into a product of irreducible quadratics. Of course, you'll want convince yourself of these cases and that the cases are indeed exhaustive.

If the polynomial breaks into a product of two irreducible quadratics, then I claim that it's rather easy to show that $K$ must have a subfield.

On the other hand, if the polynomial is irreducible, then $K = \mathbb{Q}[\alpha]$, where $\alpha$ is one of its roots. Since the degree of the Galois extension is $4$, we also know the Galois group has order $4$. As such, the Galois group must have an element $\phi$ of order $2$ that sends $\alpha \mapsto \beta$, where $\beta$ is another root of the polynomial.

Claim: $\mathbb{Q}[\alpha\beta] \subsetneq \mathbb{Q}[\alpha]$

To show this, suppose for contradiction they were actually the same field. Then both would be of degree $4$ over $\mathbb{Q}$, and the former would have a basis $\{1, \alpha\beta, (\alpha\beta)^2, (\alpha\beta)^3\}$. Because $\alpha$ would be an element of both fields, we could write:

$$\alpha = c_1 + c_2(\alpha\beta) + c_3(\alpha\beta)^2 + c_4(\alpha\beta)^3$$

What happens when the automorphism $\phi$ acts on both sides of this expression? What can we conclude?

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    $\begingroup$ I think you're working too hard. If it's Galois of degree 4, then its Galois group is a group of 4 elements. There are only two such groups (up to isomorphism), and each has a subgroup of index 2, and by Galois Theory the fixed field of such a subgroup will have degree 2, done. $\endgroup$ – Gerry Myerson Nov 26 '14 at 5:50
  • $\begingroup$ @GerryMyerson, that's precisely the path I would've liked to take. However, it seemed that the problem was assuming no prior knowledge of the fundamental theorem. $\endgroup$ – Kaj Hansen Nov 26 '14 at 5:54
  • $\begingroup$ Fair enough. That possibility did not occur to me. $\endgroup$ – Gerry Myerson Nov 26 '14 at 5:56

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