This is something more like a small doubt than some problem that I need help with.

I'm doing and exercise that is asking me to find subextensions of a given extension of all posible orders, and find the exact number of them.

To place the question in concrete terms, I've been given the field extension $\mathbb{Q}(\sqrt[6]{6},\sqrt{3}i)\vert \mathbb{Q}$ which I know is galois. I've found its degree, which is $12$, and I know that the Galois group $G$ is the dihedral group of order 12, $D_6$, and now I need to find subextensions of all posible orders. I know that the Fundamental Theorem of Galois theory says that the intermediate fields are in bijection to the subgroups of the Galois group, but I'm not completely sure that I've understood it correctly, so the first question is

I need to find field extensions whose degree is a divisor of the order of the Galois group (order $12$), don't I?

Or in more general terms

If the Galois group of a field extension $E\vert K$ has order $n$, do all subextensions of $E \vert K$ need to be of degree $d$, being $d$ such that $d \vert n$?

Now, the next section asks me to find the number of extensions of degree $3$ and $4$, so my last question would be

Should I seek for subgroups of the Galois group of orders $4$ and $3$ to find the number of subextensions of orders $3$ and $4$ respectively?

Or in more general terms

Does the number of subextensions of degre $d$ equal to the order of the Galois group, $G$, divided by the number of subgroups of $G$ of order $d$?

Thanks in advance for you help.

  • $\begingroup$ First at all you have to know if (I don't know at first sight) the Galois group has exactly $12$ elements. Maybe it has more. $\endgroup$ – Piquito Jun 14 '16 at 18:45
  • $\begingroup$ @Piquito of that I'm pretty sure. The extension is galois (it's the splitting field of $t^6-2$) so the order of the Galois group equals the degree of the extension. $\endgroup$ – user313212 Jun 14 '16 at 19:05
  • $\begingroup$ You need to know what group of order $\;12\;$ is $\;Gal(E/K)\;$ , otherwise how are you going to know what subgroups it has? There are five such groups up to isomorphism, two of them abelian, and one of them ( the alternating $\;A_4$ ) without subgroups of order six. Observe that this already counts this group out (why?) . $\endgroup$ – DonAntonio Jun 14 '16 at 19:11
  • $\begingroup$ @Joanpemo Sorry, I've edited the question. The Galois group is in fact the dihedral group $D_6$. $\endgroup$ – user313212 Jun 14 '16 at 19:17
  • $\begingroup$ Well, then it is going to be interesting as this group has $\;14\;$ proper non-trivial subgroups... $\endgroup$ – DonAntonio Jun 14 '16 at 19:24

Your Answer

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

Browse other questions tagged or ask your own question.