# Determining A Splitting Field

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?

• Splitting fields over $\mathbb{Q}$ are always Galois. – Cocopuffs Aug 19 '13 at 23:32
• @Cocopuffs: If the polynomial is irreducible. – Clayton Aug 19 '13 at 23:33
• 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. – Cocopuffs Aug 19 '13 at 23:36
• @Cocopuffs: Sorry, you're right. I'm getting ahead of myself... – Clayton Aug 19 '13 at 23:47
• @Cocopuffs: More generally, splitting fields for any family polynomial over characteristic $0$ will be Galois, right? – Clayton Aug 20 '13 at 0:37

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 $$[\mathbb{Q}(\zeta_{1155}, \sqrt[15]{5}) : \mathbb{Q}] = 480 \cdot 15 = 7200.$$
• +$1$: I think this answers my question very well. Thank you! – Clayton Aug 20 '13 at 0:38
• 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. – Clayton Aug 20 '13 at 17:53
• @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 $(\mathbb{Z}/15\mathbb{Z})^{\times}$. – Cocopuffs Aug 20 '13 at 18:03
• @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 – Cocopuffs Aug 20 '13 at 23:28