Right now I'm just trying to get a better grip on my understanding for prelims. The question is as follows:
Suppose we have the polynomial $x^6 -5$,
- Prove this is irreducible over $\mathbb{Q}$
- What is the degree over $Q$?
- Is this Galois?
- Find an intermediate field
- What is its Galois group?
Currently I am trying to prove the second part. I believe the first part follows from applying Gauss' Lemma after observing that our polynomial is irreducible in $\mathbb{Z}[x]$.
My work for the second part goes like this:
Observe first, that $x^6-5$ has roots $\sqrt[6]{5}, \omega \sqrt[6]{5}, \omega ^2 \sqrt[6]{5}, - \sqrt[6]{5}, - \omega \sqrt[6]{5}, - \omega^2 \sqrt[6]{5}$. Next I realized that we can recover $\omega$ by dividing $\omega^2 \sqrt[6]{5}$ by $\sqrt[6]{5}$.
So what does this mean? What's stopping me from saying now that the splitting field of this polynomial is $\mathbb{Q}[\sqrt[6]{5},\omega]$? Certainly I can get all the other roots in this field. However, I don't think I necessarily understand our degree over $\mathbb{Q}$.
My bad intuition says that degree will be 36. Since, naively, we have to raise $\sqrt[6]{5}, \omega$ to the sixth power respectively. In the back of my mind, I have the tower
$ [\mathbb{Q}[\sqrt[6]{5},\omega] : \mathbb{Q}] = [\mathbb{Q}[\sqrt[6]{5}, \omega] : \mathbb{Q}[\sqrt[6]{5}] ] \times [\mathbb{Q}[\sqrt[6]{5}] : \mathbb{Q} ]$
Where the extensions on the right are both clearly (to me) degree 6 extensions.
Am I missing something here in this argument?
As you have already noted, the roots of $x^6 - 5$ are those that satisfy $x^6 = {5}$, i.e. $\{\zeta_6^n \sqrt[6]{5} \mid n \in \mathbb{N} \} = \{\sqrt[6]{5}, \zeta_6 \sqrt[6]{5}, \zeta_6^2 \sqrt[6]{5}, \zeta_6^3 \sqrt[6]{5}, \zeta_6^4 \sqrt[6]{5}, \zeta_6^5 \sqrt[6]{5} \}$, where $\zeta_6$ is a primitive sixth root of unity (e.g. $\zeta_6 = e^{2\pi/6 i}$). Then, the splitting field is just formed by adjoining all of these roots, which as you have said can be summarised as $\mathbb{Q}(\sqrt[6]{5}, \zeta_6)$.
Your use of the tower law is a good way to determine the degree of this extension over $\mathbb{Q}$, that is: $$[\mathbb{Q}(\sqrt[6]{5}, \zeta_6): \mathbb{Q}] = [\mathbb{Q}(\sqrt[6]{5}, \zeta_6):\mathbb{Q}(\sqrt[6]{5})][\mathbb{Q}(\sqrt[6]{5}): \mathbb{Q}]$$
Where you went wrong is the degree of $\mathbb{Q}(\sqrt[6]{5}, \zeta_6) / \mathbb{Q}(\sqrt[6]{5})$, which is not 6. Your intuition is correct in saying that "we [can] raise $\zeta_6$ to the sixth power" to get an element of $\mathbb{Q}(\sqrt[6]{5})$, but this only tells us that the degree of $\zeta_6$ is at most 6, since the minimal polynomial could have (and in fact has) smaller degree.
The minimal polynomial over $\mathbb{Q}$ of an $n$-th primitive root of unity is the $n$-th cyclotomic polynomial $\Phi_n$. In this case, $\Phi_6 = X^2 - X + 1$. So in fact $[\mathbb{Q}(\sqrt[6]{5}, \zeta_6):\mathbb{Q}(\sqrt[6]{5})] = 2$, giving an overall degree of 12.
Here is the reason $\Phi_n$ has degree less than $n$ in general. Consider the polynomial $p = x^n - 1$, whose roots are the $n$-th roots of unity by definition. When computing the minimal polynomial of a primitive $n$-th root of unity $\zeta_n$, we want the monic polynomial having $\zeta_n$ as a root with least degree, so we want to remove any superfluous factors from the polynomial $x^n - 1$. For instance, we can remove the factor $x - 1$ straight away by noticing the trivial root $x = 1$ of $p$ (which is not a primitive $n$-th root of unity): we have $x^n - 1 = (x - 1)(x^{n - 1} + \cdots + x + 1)$, hence $\zeta_n$ is a root of $x^{n - 1} + \cdots + x + 1$. So in fact $\deg \Phi_n \le n - 1$. Generalising, for any $d$ that divides $n$, any primitive $d$-th root of unity $\zeta_d$ will be a root of $p$, because $\zeta_d^n = \zeta_d^{kd} = (\zeta_d^d)^k = 1^k = 1$. Conversely, every $n$-th root of unity is a primitive $d$-th root for some $d \mid n$. So we have to remove all the factors involving $d$-th roots of unit with $d < n$.
More precisely, since the minimal polynomial of an element divides all polynomials having that element as a root, we have that $\Phi_d$ divides $x^n - 1$ for all $d \mid n$, and in fact since this accounts for all possible $n$-th roots, we conclude: $$\prod_{d \mid n} \Phi_d = x^n - 1$$ and hence: $$\Phi_n = \dfrac{x^n - 1}{\displaystyle \prod_{\substack{d \mid n \\ d < n}} \Phi_d}$$ This allows us to compute cyclotomic polynomials inductively: $$\Phi_1 = x - 1$$ $$\Phi_2 = (x^2 - 1)/\Phi_1 = (x + 1)$$ $$\Phi_3 = (x^3 - 1)/\Phi_1 = (x^2 + x + 1)$$ $$\Phi_4 = (x^4 - 1)/\Phi_1/\Phi_2 = (x^3 + x^2 + x + 1)/\Phi_2 = x^2 + 1$$ And so on.