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.

  • $\begingroup$ 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

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 $G=A_4$?

  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.