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This is a problem from a comprehensive exam at my university. Let $f(x)$ be an irreducible quartic polynomial over $\mathbb{Q}$. Let $L$ be the splitting field of $f(x)$. Suppose $\mathbb{Q}(\alpha) \cap \mathbb{Q}(\beta)=\mathbb{Q}$ for any two distinct roots $\alpha$ and $\beta$ of $f(x)$. Now I want to calculate the Galois group.

My Try:

Firstly since it is irreducible it has to be a subgroup of $S_4$. Also it has to be a transitive subgroup of order greater than $4$. So the only options are $S_4, A_4$ and $D_8$. Now $\mathbb{Q}(\alpha) \cap \mathbb{Q}(\beta)=\mathbb{Q}$ implies the lattice for the group should contain index $4$ groups such that the only group containing any two is the whole group. This rules out $D_8$. Now I am not sure how to proceed. Any hints or suggestions are welcome. Thank you.

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  • $\begingroup$ I don't think you can say anything more. Won't the way you used Lord Shark the Unknown's sketch imply that when the Galois group is either $A_4$ or $S_4$ the condition $\Bbb{Q}(\alpha)\cap\Bbb{Q}(\beta)=\Bbb{Q}$ is automatically satisfied? That must be the answer. $\endgroup$ – Jyrki Lahtonen Aug 17 '17 at 5:15
  • $\begingroup$ @JyrkiLahtonen Yeah you are right. Thank you. $\endgroup$ – happymath Aug 17 '17 at 13:15
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The groups $S_4$ and $A_4$ are both $2$-transitive on the zeros of $f$. So if $\Bbb Q(\alpha)\cap\Bbb Q(\beta)=\Bbb Q$ holds for any pair of distinct zeroes it holds for all of them. By the Galois correspondence, $\text{Gal}(L/(\Bbb Q(\alpha)\cap\Bbb Q(\beta)))=\left<H_1,H_2\right>$ where $L$ is the splitting field and $H_1$ and $H_2$ are the subgroups of $G=\text{Gal}(L/\Bbb Q)$ corresponding to $\Bbb Q(\alpha)$ and $\Bbb Q(\beta)$. We can identify $G$ with a group of permutations of $\{1,2,3,4\}$ and $H_1$ and $H_2$ as the stabilisers of $1$ and $2$ respectively. So the question is now: what is $\left<H_1,H_2\right>$ when $G=S_4$, and when $G=A_4$?

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    $\begingroup$ I looked at the lattice diagrams for both $S_4$ and $A_4$ and I think $<H_1,H_2>$ is the whole group in both the cases $\endgroup$ – happymath Aug 17 '17 at 4:04

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