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Let $f(x) \in \mathbb Q[x]$ be irreducible of degree 4. Let $\alpha$ be a root of $f(x)$. Let $E = \mathbb Q(\alpha)$ and let $K$ be the splitting field of $f(x)$ over $\mathbb Q$. Prove that there is no field properly between $\mathbb Q$ and $E$ if and only if $G(K/\mathbb Q) \cong A_4$ or $S_4$.

At this point in the course we've just gone through the Fundamental Theorem of Galois Theory. I'm fairly lost, so any hints, sketches of a proof would help. Thanks!

Robert Cardona
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    Robert, are you able to sucessfully use the fundamental theorem to turn this into a group theory problem? – Alex Youcis Apr 09 '13 at 05:30
  • Not yet. I'm trying the converse first because it looks straight forward. I suppose that $G(K/\mathbb Q) \cong A_4$ or $S_4$. It would suffice to pick one of these and show that there is no proper subfield between $\mathbb Q$ and $E$. I'm picking $A_4$ because it has smaller order and I'm hoping I can get a contradiction to supposing that there is a proper subgroup between $\mathbb Q$ and $E$. – Robert Cardona Apr 09 '13 at 05:36

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Yes, start with the converse.

If $G = G(K/\mathbb Q) \cong A_4$ or $S_4$, we have that $G$ is $2$-transitive, hence primitive, on the roots. It follows that the stabilizer $G_{\alpha}$ (which is respectively $A_3$ or $S_3$) is a maximal subgroup of $G$. (You can also verify directly that those stabilizers are maximal.)

Now the fixed field of $G_{\alpha}$ is $E = \Bbb{Q}(\alpha)$: use the Galois correspondence.

For the converse, see which subgroups of $S_4$ other than $A_4$ and $S_4$ can arise (hint: they should act transitively on the roots) and check that the stabilizers are not maximal in those cases.

Andreas Caranti
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