# Splitting fields and their degrees

I'm having some trouble with splitting fields and finding their degrees (the concept is relatively straight forward, but given a polynomial I'm not sure how to proceed).

Say I have the polynomial $x^8-1$. I know that the splitting field over $\mathbb{Q}$ is $\mathbb{Q}(\xi_{8})$ by an example in my book ($\xi_8$ being the 8th roots of unity). But how do I find the degree of this splitting field over $\mathbb{Q}$?

Even more confusing for me is something like $x^4+2$ over $\mathbb{Q}$. I think the splitting field is $\mathbb{Q}(\xi_4,\sqrt{2})$. I'm pretty unsure because maybe its $\mathbb{Q}(\xi_4,\sqrt{-2})$ but $\sqrt{-2}$=$i\sqrt{2}$ and $i \in \xi_4$. So how do I go about determining what the splitting field is? And in either case I run into the problem again of finding the degree of $\mathbb{Q}(\xi_n)$ for $n \in \mathbb{Z}^{+}$

I think once I see how to go about it it'll click (I often find that with the "concrete" problems in maths I am initially over complicating and later find that I was being silly). Sorry if I'm incoherent it's late and I have the exam in 2 days.

• Finding the extension degree of a splitting field can become quite taxing. There are standard techniques that you can absorb from seeing several examples worked out, but finding all the possible non-obvious relations among the zeros is not straight forward. For the degree of $\Bbb{Q}(\xi_n)$ there is a theorem stating that the degree is given by the Euler totient function $\phi(n)$. A pitfall related to the examples you list is that (one of) the eighth root(s) of unity is $\xi_8=(1+i)/\sqrt2$. So if you already need $\sqrt2$ in your field, then adjoining $\xi_4$ gives you $\xi_8$ too. – Jyrki Lahtonen Apr 29 '15 at 5:05

## 1 Answer

When you have $F(\alpha)/F$ for some $\alpha$, the degree is the degree of the minimal polynomial for $\alpha$. In the case of $\zeta_n$, the minimal polynomial is $\Phi_n$, the $n$th cyclotomic polynomial. If you only want to know the degree of the extension, it's worth noting that the degree of $\Phi_n$ is $\varphi(n)$ where $\varphi$ is Euler's totient function, so $\mathbb{Q}(\zeta_8)/\mathbb{Q}$ has degree $\varphi(8)=4$.

To see the fact about the degree of $\Phi_n$, note that $X^n-1=\prod_{d|n}\Phi_d(X)$, so $\deg(X^n-1)=n=\sum_{d|n} \deg \Phi_d$. Mobius inverting, we see $\deg \Phi_n = (\textrm{id} * \mu)(n) = \varphi(n)$

I'm not personally aware of a general method for finding splitting fields, it has always seemed to be a case by case thing. In that particular case, you could do it by listing all the roots, since the splitting field is $\mathbb{Q}(\alpha_1,\alpha_2,\alpha_3,\alpha_4)$ where the $\alpha_i$ are the roots of the polynomial. To find the degree of that extension, you could find a minimal set of these roots that will give you the rest of the roots.

Also what you were doing seemed fine to me already, so don't take this as saying your method was wrong.

• Is that property with the Euler totient function a well known thing (just so I don't pull some random lemma from thin air)? In any case thanks! – Sally Mander Apr 29 '15 at 5:01
• Well, we discussed it in my introductory number theory course, so I would imagine it's pretty common. – jgon Apr 29 '15 at 5:23