# 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 $\mathbb{Q}(\xi_4,\sqrt[4]{2})$. I'm pretty unsure because maybe its $\mathbb{Q}(\xi_4,\sqrt[4]{-2})$ but $\sqrt[4]{-2}$=$i\sqrt[4]{2}$ 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

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.