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Let $f(x)\in \mathbb{Z}[x]$ be an irreducible quartic polynomial with $S_4$ as Galois Group. Let $\theta$ be a root of $f(x)$ and set $K=\mathbb{Q}(\theta)$. Now, the question is:

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

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

As I have adjoined a root of an irreducible quartic, I can see that $K$ is of degree $4$ over $\mathbb{Q}$.

But, why does there is no proper subfield of $K$ containing $\mathbb{Q}$?

Suppose $L$ is proper subfield of $K$, then $L$ has to be of degree $2$ over $\mathbb{Q}$. So, $L$ is Galois over $\mathbb{Q}$, i.e., $L$ is normal. So the corresponding subgroup of Galois group has to be normal.

I tried working in this way but could not able to conclude anything from this.

Any help/suggestion would be appreciated.

Thank You

user26857
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2 Answers2

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Convert the question to a problem in permutation groups.

Let $F$ be the splitting field of $$f(x)=(x-\theta_1)(x-\theta_2)(x-\theta_3)(x-\theta_4)$$ with $\theta=\theta_1$.

We were given that the Galois group realizes all the 24 permutations of the roots $\theta_i,i=1,2,3,4.$ Therefore $$ \operatorname{Gal}(F/\mathbb{Q}(\theta))=\operatorname{Sym}(\{\theta_2,\theta_3,\theta_4\}) $$ contains automorphisms realizing all the six permutations of the other roots.

Galois correspondence then means that the claim is equivalent to:

There are no subgroups $H$ properly between $\operatorname{Sym}(\{\theta_2,\theta_3,\theta_4\})$ and $\operatorname{Sym}(\{\theta_1,\theta_2,\theta_3,\theta_4\})$.

In other words, this is equivalent to proving that the obvious copy of $S_3$ inside $S_4$ is a maximal subgroup. Have you seen that? If not, can you prove it?

Jyrki Lahtonen
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  • I do not know, but for some reason, i am not able to view maps as elements of permutation group :( I have to work on that i guess –  Aug 10 '13 at 07:36
  • No,the order of $S_{3}$ is $3! = 6.$ – Geoff Robinson Aug 10 '13 at 08:04
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    @Praphulla: If you know that $A_4$ is the only subgroup of index two in $S_4$, then you can use that simply by observing that $A_4$ does not have $S_3$ as a subgroup, because the latter contains odd permutations. – Jyrki Lahtonen Aug 10 '13 at 08:21
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    More generally, isn't it true for all $n$ that $S_{n-1}$ is a maximal subgroup of $S_n$? – Gerry Myerson Aug 11 '13 at 00:25
  • @Gerry, sure. Assume that $S_{n-1}$ is the point stabilizer of $n$, and $S_{n-1}\subset H\le S_n$. Then $H$ moves the point $n$, and is thus transitive. But it also has full point stabilizer, so then has to be all of $S_n$. I didn't want to assume that all the readers have sufficient familiarity with group actions for that to be all clear. May be I worried too much (given that I identify a Galois group with its action on the roots anyway)? – Jyrki Lahtonen Aug 11 '13 at 06:21
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    Nothing wrong with what you posted, and thanks for confirming maximality of $S_{n-1}$. So this means that if $f$ is irreducible of degree $n$ with Galois group $S_n$, and $\alpha$ is a root of $f$ in some extension of the rationals, then the field generated by $\alpha$ has degree $n$ and no subfields but itself and the rationals. That's good to know. – Gerry Myerson Aug 11 '13 at 09:55
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As has been remarked, the non-existence of intermediate fields is equivalent to $S_{3}$ being a maximal subroup of $S_{4}.$ If not, then there is a subgroup $H$ of $S_{4}$ with $[S_{4}:H] = [H:S_{3}] = 2.$ Now $S_{3} \lhd H$ and $S_{3}$ contains all Sylow $3$-subgroup of $H.$ But $S_{3}$ has a unique Sylow $3$-subgroup, which is therefore normal in $H.$ Hence $H$ contains all Sylow $3$-subgroups of $S_{4}$ as $H \lhd S_{4}$ and $[S_{4}:H] = 2.$ Then since $H$ only has one Sylow $3$-subgroup, $S_{4}$ has only one Sylow $3$-subgroup, a contradiction (for example, $\langle (123) \rangle$ and $\langle (124) \rangle$ are different Sylow $3$-subgroups of $S_{4}$).

Geoff Robinson
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