Let $g(x)$ of degree 4 be an irreducible polynomial in $\Bbb Q$. Let $\alpha \in \Bbb C$ be a root of $g(x)$, and let $R=\Bbb Q(\alpha)$. We also let $K$ be the splitting field of $g(x)$ over $R$. Further, assume that there exists a quadratic field $L$ such that $\Bbb Q \subset L \subset R $, and that $[L:\Bbb Q]=2$, and assume that $K$ properly contains $R$ (i.e. $R \neq K$).

Prove that $[K:\Bbb Q]=8$ and find $Gal(K/\Bbb Q)$.

I have found that $Gal(K/\Bbb Q)$ needs to be a transitive subgroup of $S_4$, so then the options are $S_4, A_4, D_4$, Klein-Four subgroup, $\Bbb Z/4\Bbb Z$.

At this point, I would want to show that $[K:L]=4$ implying the first result. The second result most likely comes from using the resolvent cubic, but I don't see how to use it.

Any hints or help is appreciated, thanks

  • $\begingroup$ I think you mean $L$ is quadratic, not quartic and that $K$ properly contains $R$ i.e. $K\ne R$. $\endgroup$ – Adam Hughes Mar 29 '16 at 1:11
  • $\begingroup$ @AdamHughes I definitely meant that, thanks! $\endgroup$ – user218512 Mar 29 '16 at 1:15

First note that the Galois group has order equal to the degree of the splitting field, so really your only option is $D_4$ since it has order $8$, and Sylow's theorem says that all Sylow $p$ subgroups of a group are isomorphic (even conjugate). So as long as you can prove the degree is $8$, you're good to go.

For that you see that $4||\operatorname{Gal}(K/\Bbb Q)$ since $[R:\Bbb Q]=4$.

(Groups of order 4) We can rule out any group of order $4$--i.e. $\Bbb Z/4\Bbb Z$ and the Klein-$4$ group--since $[K:\Bbb Q]>[R:\Bbb Q]=4$.

($A_4$) We can rule out $A_4$ because $A_4$ has no subgroup of order $6$ (a basic result from group theory) and since $[L:\Bbb Q]=2$ is the index of $\operatorname{Gal}(K/L)$ in $\operatorname{Gal}(K/\Bbb Q)$.

($S_4$) Finally, to rule out $S_4$ we note that since $L\subset R\subset K$ means that we would have $L=K^{A_4}$ necessarily since $A_4$ is the unique subgroup of index $2$ in $S_4$, and by the FTGT we'd have that $\operatorname{Gal}(K/R)\le \operatorname{Gal}(K/L)$ where the order of $\operatorname{Gal}(K/R)$ is $[K:R]$ and if this is $6$--i.e. if $[K:\Bbb Q]=24$ as we are assuming--again we'd have a subgroup of $A_4$ of order $6$. Hence $S_4$ is ruled out.

This only leaves $D_4$, which must be the Galois group.

  • $\begingroup$ How does $A_4$ having no subgroup of order $6$ make it impossible for it to be the Galois group for $K$ in this scenario? I understand what your argument for ruling out $S_4$ is, but I do not understand how you ruled out $A_4$. $\endgroup$ – Noble Mushtak Mar 29 '16 at 1:39
  • $\begingroup$ @NobleMushtak Because the FTGT says that $\operatorname{Gal}(K/L)$ has index $2$ in the big group, but index $2$ is order $6$ by Lagrange's theorem. $\endgroup$ – Adam Hughes Mar 29 '16 at 1:41
  • $\begingroup$ Oh, that makes sense! Thanks for the explanation! $\endgroup$ – Noble Mushtak Mar 29 '16 at 1:42
  • $\begingroup$ I am simply someone who answers question on Math StackExchange a lot, saw your answer, and wanted to understand it. I am not the person who asked the question. $\endgroup$ – Noble Mushtak Mar 29 '16 at 19:05

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.