4
$\begingroup$

The book I am using for my Abstract Algebra course is Contemporary Abstract Algebra by Joseph A. Gallian.

Let $E/F$ be a Galois extension with Galois group isomorphic to $\mathbb{Z}_2\oplus\mathbb{Z}_4$. Determine the subfield lattice for $E/F$.

I was wondering if someone could guide me into the right direction on how to solve this problem.

From what I do understand, Gal$(E/F)\cong \mathbb{Z}_2\oplus\mathbb{Z}_4$. Galois Theory, in a nutshell, is used to find how many subfields are between $E$ and $F$. If $E/F$ is "nice" then there is a one-to-one correspondence between the set of subfields and the set of subgroups.

I was able to find subgroups of the direct product $\mathbb{Z}_2\oplus\mathbb{Z}_4$. I don't know if I have found all subgroups to be honest with you. I did them by hand and wasn't sure if there was a theorem to verify if I got the right number of subgroups. The list is as follows.

  • $S_0=\{(0,0)\}$, the trivial subgroup.
  • $S_1=\{(0,0),(0,2)\}$
  • $S_2=\{(0,0),(1,2)\}$
  • $S_3=\{(0,0),(1,0)\}$
  • $S_4=\{(0,0),(0,1),(0,2),(0,3)\}$
  • $S_5=\{(0,0),(1,1),(0,2),(1,3)\}$
  • $S_6=\{(0,0),(0,2),(1,0),(1,2)\}$
  • $\mathbb{Z}_2\oplus\mathbb{Z}_4=\{(0,0),(0,1),(0,2),(0,3),(1,0),(1,1),(1,2),(1,3)\}$

From the information I got, we can see that $\mathbb{Z}_2\oplus\mathbb{Z}_4$ has 3 subgroups of order 4 and 3 groups of order 2. Hence we have 6 subfields between $E$ and $F$. This result in the following lattice.

enter image description here

Am I on the right track? Would the approach be similar if the Galois group was isomorphic to $\mathbb{Z}_3\oplus\mathbb{Z}_4$? Would there be a faster way to find the subgroups of direct products?


Sorry for the rather long read. I sincerely thank you for taking the time to read this post. I greatly appreciate any assistance you may provide.

$\endgroup$
2
$\begingroup$

Seems correct, based on a cursory reading of the subgroups, I will assume you have them correct, and refer you to this question for a complete description of the subgroups of a direct product. Subgroups of a direct product

I feel that I should point out that you have written $G$ at the top of your diagram for the fields rather than $E$. Also while this is a correct drawing of the lattice by the symmetry of the lattice itself, I feel that I should make certain that you are aware that $K_1$ does not correspond to $S_1$, but rather $S_6$, and $F$ corresponds to $\mathbb{Z}_2\oplus \mathbb{Z}_4$ rather than $S_0$ as is suggested by the way the diagrams have been drawn in parallel. If this was intentional, I mean no offense, just wished to point it out in case it wasn't.

$\endgroup$
  • $\begingroup$ thank you for pointing these things out. If you don't mind me asking, why does $K_1$ correspond to $S_6$? And why does $F$ correspond to $\mathbb{Z}_2\times\mathbb{Z}_4$? Small details like this still confuse me here and there. Thank you once again for your help. It helps me a lot. $\endgroup$ – Kevin_H Apr 28 '15 at 6:25
  • 1
    $\begingroup$ The subfield you get corresponds to the fixed field of the subgroup of the Galois group. So since $F$ is fixed by all the automorphisms in the Galois group, (and since the extension is Galois), the fixed field of the Galois group is $F$ and so the whole group $\mathbb{Z}_2\times\mathbb{Z}_4$ corresponds to $F$. In general small groups correspond to big fields, and vice versa. Since $K_1$ has dimension two over $F$, it corresponds to a group with index $2$ or size $4$ in the Galois group. Then since the diagram has to line up still, it must correspond to $S_6$. $\endgroup$ – jgon Apr 28 '15 at 6:31
  • 1
    $\begingroup$ The reason the index should be 2, is that the Galois group of an intermediate field extension is the quotient of the Galois group $G$ of $K/F$ by the automorphisms that fix the intermediate field $L$. $\endgroup$ – jgon Apr 28 '15 at 6:32

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.