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I am trying to determine the splitting fields of a bunch of polynomials. I'll ask one here and hope that a general enough technique can be described to find the rest of them.

Currently, I'm trying to find the splitting field of $(x^{15}-5)(x^{77}-1)$ over $\Bbb Q$, find the degree, and determine if it's a Galois extension.

Now, I know that the right polynomial is the cyclotomic polynomial, hence has degree $\varphi(77)=60$, over $\Bbb Q$. The left polynomial is irreducible by Eisenstein's Criterion, hence adjoining $\sqrt[15]{5}$ gives a degree 15 extension and as a separate extension, adjoining $\zeta_{15}$ (a primitive $15^{th}$ root of unity) gives a degree $8$ extension. Since $8$ and $15$ are relatively prime, I know that the degree of the extension for the splitting field of $x^{15}-5$ is $120$.

All of this seems well and good, but now I'm lost. The splitting field itself is obviously $\Bbb Q(\zeta_{77},\sqrt[15]{5},\zeta_{15})$, but how can I check the degree and determine if it is Galois?

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    $\begingroup$ Splitting fields over $\mathbb{Q}$ are always Galois. $\endgroup$
    – Cocopuffs
    Commented Aug 19, 2013 at 23:32
  • $\begingroup$ @Cocopuffs: If the polynomial is irreducible. $\endgroup$
    – Clayton
    Commented Aug 19, 2013 at 23:33
  • $\begingroup$ It doesn't matter whether it's irreducible or not. The splitting field of any family of polynomials is normal by definition, and Galois since $\mathbb{Q}$ is perfect. $\endgroup$
    – Cocopuffs
    Commented Aug 19, 2013 at 23:36
  • $\begingroup$ @Cocopuffs: Sorry, you're right. I'm getting ahead of myself... $\endgroup$
    – Clayton
    Commented Aug 19, 2013 at 23:47
  • $\begingroup$ @Cocopuffs: More generally, splitting fields for any family polynomial over characteristic $0$ will be Galois, right? $\endgroup$
    – Clayton
    Commented Aug 20, 2013 at 0:37

1 Answer 1

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First reduce the $\zeta$'s: you have $$\mathbb{Q}(\zeta_{77}, \zeta_{15}) = \mathbb{Q}(\zeta_{lcm(77,15)}) = \mathbb{Q}(\zeta_{1155}).$$ This is an extension of $\mathbb{Q}$ of degree $\varphi(1155) = \varphi(77)\cdot \varphi(15) = 480.$

What is the intersection $\mathbb{Q}(\zeta_{1155}) \cap \mathbb{Q}(\sqrt[15]{5})$?

Subfields of cyclotomic fields are abelian (the converse is also true), that is, they are Galois with abelian Galois groups. However, the nontrivial subfields of $\mathbb{Q}(\sqrt[15]{5})$ - $\mathbb{Q}(\sqrt[3]{5})$, $\mathbb{Q}(\sqrt[5]{5})$ and $\mathbb{Q}(\sqrt[15]{5})$ - are not Galois. So you have $\mathbb{Q}(\zeta_{1155}) \cap \mathbb{Q}(\sqrt[15]{5}) = \mathbb{Q}$ and $$[\mathbb{Q}(\zeta_{1155}, \sqrt[15]{5}) : \mathbb{Q}] = 480 \cdot 15 = 7200.$$

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  • $\begingroup$ +$1$: I think this answers my question very well. Thank you! $\endgroup$
    – Clayton
    Commented Aug 20, 2013 at 0:38
  • $\begingroup$ Cocopuffs, how do we know what the nontrivial subfields of $\Bbb Q(\sqrt[15]{5})$ are exactly? I certainly agree that the field is the compositum of the two others listed, so there isn't any quarrel about that. Just why any intermediate field has to be one of those two. $\endgroup$
    – Clayton
    Commented Aug 20, 2013 at 17:53
  • $\begingroup$ @Clayton You can use Galois theory in $\mathbb{Q}(\sqrt[15]{5}, \zeta_{15})$. The subfields of $\mathbb{Q}(\sqrt[15]{5})$ correspond to subgroups of $(\mathbb{Z}/15\mathbb{Z})^{\times} \rtimes \mathbb{Z}/15\mathbb{Z}$ containing $(\mathbb{Z}/15\mathbb{Z})^{\times}$. $\endgroup$
    – Cocopuffs
    Commented Aug 20, 2013 at 18:03
  • $\begingroup$ Is that group isomorphic to anything more elementary? My professor doesn't like semidirect products and this is one of her previous final exams I am using to study for my prelims. $\endgroup$
    – Clayton
    Commented Aug 20, 2013 at 18:22
  • $\begingroup$ @Clayton I don't know of it. But it winds up being that you look for subgroups of $\mathbb{Z}/15\mathbb{Z}$. There is probably an easier proof that I don't see $\endgroup$
    – Cocopuffs
    Commented Aug 20, 2013 at 23:28

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