I am looking for a degree $4$ extension of $\mathbb {Q}$ with no intermediate field. I know such extension is not Galois (equivalently not normal). So I was trying to adjoin a root of an irreducible quartic. But I got stuck. Any hint/idea/solution?
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2You could start by searching for Galois extensions of $\mathbb{Q}$ of degree divisible by $4$ and Galois groups which have no subgroups of index $2$... – Sebastian Schoennenbeck Nov 12 '14 at 14:23
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1See Example 4.16 and Remark 4.18 in http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/galoisaspermgp.pdf – KCd Nov 12 '14 at 14:43
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See also this question. Close to being a duplicate. – Jyrki Lahtonen Nov 12 '14 at 21:21
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With regard to Sebastian Schoennenbeck's comment, an extension of $\mathbb{Q}$ with Galois group $A_4$ (alternating group on 4 points) will do the trick. Such an extension certainly exists, in fact all alternating groups are Galois groups over $\mathbb{Q}$.
Oliver Braun
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5Presumably you wanted to do the following. Let $L/\Bbb{Q}$ be Galois with Galois group $A_4$. Then let $K\subset L$ be the fixed field of $\langle (123)\rangle$. Then $[K:\Bbb{Q}]=12/3=4$, and there are no intermediate fields, because $A_3$ is a maximal subgroup of $A_4$. – Jyrki Lahtonen Nov 12 '14 at 21:23
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Without going for the maximality concept we can also use the fact that if there is a intermediate field say $F$ then $[F:\mathbb{Q}]$ has to be 2 and hence ${F}/{\mathbb{Q}}$ will be a Galois extension. Which implies that we have a normal subgroup of order $12/2=6$ in $A_4$ by the F.T. of G.T. – nkh99 Oct 31 '23 at 13:15