Suppose $f(x)\in \mathbb{Z}[x]$ be an irreducible Quartic polynomial with Galois Group as $S_4$. 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 degree $\mathbb{Q}$ of degree 4 which has no proper Subfields?

Are there any Galois Extensions of $\mathbb{Q}$ of degree 4 with no proper sub fields.

As i have adjoined a root of 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 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

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    $\begingroup$ You're on the way. The group of the splitting field over $K$ has to be normal of order 6 in the group of the splitting field over $L$, which has to be normal of order 12 in $S_4$. Now, what do you know about subgroups of $S_4$? $\endgroup$ – Gerry Myerson Aug 10 '13 at 7:21
  • $\begingroup$ yes, so, corresponding subgroup of order 12 is $A_4$ but, it is normal. So i do not get any contradiction with this :( $\endgroup$ – user87543 Aug 10 '13 at 7:26
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    $\begingroup$ Keep going. You haven't used the "normal of order 6" part. $\endgroup$ – Gerry Myerson Aug 10 '13 at 7:28
  • $\begingroup$ oh yes, but $A_4$ has no group of order 6, leave about being normal :) :) :) $\endgroup$ – user87543 Aug 10 '13 at 7:31
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    $\begingroup$ There you go. Now, you can write it up and post is as an answer. $\endgroup$ – Gerry Myerson Aug 10 '13 at 7:33

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?

  • $\begingroup$ 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 $\endgroup$ – user87543 Aug 10 '13 at 7:36
  • $\begingroup$ No,the order of $S_{3}$ is $3! = 6.$ $\endgroup$ – Geoff Robinson Aug 10 '13 at 8:04
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    $\begingroup$ @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. $\endgroup$ – Jyrki Lahtonen Aug 10 '13 at 8:21
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    $\begingroup$ More generally, isn't it true for all $n$ that $S_{n-1}$ is a maximal subgroup of $S_n$? $\endgroup$ – Gerry Myerson Aug 11 '13 at 0:25
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    $\begingroup$ 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. $\endgroup$ – Gerry Myerson Aug 11 '13 at 9:55

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 $S_{4}$).

  • $\begingroup$ More generally, isn't it true for all $n$ that $S_{n-1}$ is a maximal subgroup of $S_n$? $\endgroup$ – Gerry Myerson Aug 11 '13 at 0:24
  • $\begingroup$ That is true of course $\endgroup$ – Geoff Robinson Aug 13 '13 at 1:44

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