I know how to do this in the case of $\mathbb{Z}/3\mathbb{Z}$, by looking at subfields of cyclotomic extensions. Specifically, consider the extension $\mathbb{Q}(\zeta_7)$ where $\zeta_7$ is a primitive $7^{th}$ root of unity. The Galois group is the cyclic group on 6 elements. We know that $\zeta_7+\zeta_7^{-1}$ is fixed by complex conjugation, and since complex conjugation generates a subgroup of index 3, we may simply take the fixed field of this subgroup. Note that since the Galois group is cyclic, it is abelian, thus every subgroup is normal so there are no issues here. It also turns out that complex conjugation is the only automorphism fixing $\zeta_7+\zeta_7^{-1}$, so our fixed field is precisely $\mathbb{Q}(\zeta_7+\zeta_7^{-1})$. My question is, can we find an analogue for $\mathbb{Z}/3\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z}$?

  • 1
    $\begingroup$ If you just want to find an example, there are many in the LMFDB. Here are the search results. $\endgroup$ Mar 29 '20 at 19:11
  • $\begingroup$ Thanks, I never knew that site existed. However, I would like to see a method of constructing such a field extension without using an unwieldy 9th degree polynomial. $\endgroup$ Mar 29 '20 at 19:16
  • 4
    $\begingroup$ I still think the information I linked can help you do that. For instance, take this field. If you scroll down to Intermediate fields, you'll see that it has subfields $K := \mathbb{Q}(\zeta_7 + \zeta_7^{-1})$ and $L := \mathbb{Q}(\zeta_9 + \zeta_9^{-1})$. If you can show that it is the compositum $KL$, I think you should be done. $\endgroup$ Mar 29 '20 at 19:45


Let $\mu_n$ denote the set of $n$-th roots of unity. If $n,m$ are coprime, then $\mathrm{Gal}(\mathbb{Q}(\mu_{nm})/\mathbb{Q})\cong\mathrm{Gal}(\mathbb{Q}(\mu_n)/\mathbb{Q})\times\mathrm{Gal}(\mathbb{Q}(\mu_m)/\mathbb{Q})$ via restrictions. Using this, you can find a Galois extension with Galois group $\mathbb{Z}/6\mathbb{Z}\times\mathbb{Z}/6\mathbb{Z}$ and then you can conclude via the fundamental theorem of Galois theory. (It may be worthwhile to note that if you know the sturcture theorem for finite abelian groups and a special case of Dirichlet's theorem on arithmetic progressions, then you can prove that any abelian group can be realized as Galois group over $\mathbb{Q}$ using this method).

  • 2
    $\begingroup$ So essentially do the same thing I did with $\zeta_7$ but instead use $\zeta_9$, then take their compositum? $\endgroup$ Mar 29 '20 at 21:01
  • $\begingroup$ Yes, that works. And indeed $\mathbb{Q}(\mu_n)\mathbb{Q}(\mu_m)=\mathbb{Q}(\mu_{nm})$. $\endgroup$
    – Thorgott
    Mar 29 '20 at 23:13

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.