# (Degree of) Splitting Field for $f(x) = x^p - 2$ over $\mathbb{Q}$, p prime

Here is the work I've done so far:

• $$\sqrt[p]{2}$$ is a real root of $$f(x)$$
• Any $$(\zeta \sqrt[p]{2})$$ where $$\zeta$$ is a $$p^{th}$$ root of unity is also a root of $$f(x)$$
• Since $$p$$ is prime, all its roots of unity are primitive, and since primitive roots of unity will generate the rest under repeated multiplication, it's the case that $$\mathbb{Q}\left( \sqrt[p]{2}, \zeta_p \right)$$ is a splitting field of $$f(x)$$ where $$\zeta_p$$ is a $$p^{th}$$ primitive root of unity (i.e. any root of unity)
• To find $$[\mathbb{Q}(\sqrt[p]{2}, \zeta_p) : \mathbb{Q}]$$ I want to use the tower theorem, i.e. find $$[\mathbb{Q}(\sqrt[p]{2}, \zeta_p) : \mathbb{Q}(\sqrt[p]{2})][\mathbb{Q}(\sqrt[p]{2}) : \mathbb{Q}]$$
• $$f(x)$$ is irreducible over $$\mathbb{Q}$$ by Eisenstein with $$p=2$$ so I can say that $$[\mathbb{Q}(\sqrt[p]{2}) : \mathbb{Q}] = p$$

What I'm stuck on is $$[\mathbb{Q}(\sqrt[p]{2}, \zeta_p) : \mathbb{Q}(\sqrt[p]{2})]$$. I'm thinking it has something to do with the polynomial $$p(x) = x^{p-1} + x^{p-2} +\dots + 1$$ being the minimal polynomial (just based on the fact I've seen it around ) but I've no idea how to go on about proving that:

1. It's irreducible
2. Whatever chosen $$p^{th}$$ primitive root of unity is a root of $$p(x)$$

Any help/hints to get me over this hurdle are appreciated. Thanks for reading!

• The tower formula is the way to go. You also need the facts that $[\Bbb{Q}(\root p\of2):\Bbb{Q}]=p$ (Eisenstein) and $[\Bbb{Q}(\zeta_p):\Bbb{Q}]=p-1$ (Eisenstein again, but needs a substitution, standard though). The tower theorem then says that $n=[\Bbb{Q}(\zeta_p,\root p\of2):\Bbb{Q}]$ must be a multiple of both $p$ and $p-1$ implying... Also, use the polynomial $p(x)$ to get an upper bound for $[\Bbb{Q}(\root p\of 2,\zeta_p):\Bbb{Q}(\root p\of2)]$. – Jyrki Lahtonen Feb 11 at 6:31
• Not posting the details as an answer because I'm fairly sure we have done this already on this site. I may have done it myself, but that may also have been about the related Galois group. – Jyrki Lahtonen Feb 11 at 6:33
• For details, see here. The accepted answer has them (if you strip away discussion of the Galois group). My post is here, but that came at a point, where this extension degree was already known (IIRC). – Jyrki Lahtonen Feb 11 at 6:40
• Your second bullet point is off; if $\zeta$ is a $p$-th root of unity then $(\zeta\sqrt[p]{2})^p=2$ is not a root of $f$, but $\zeta\sqrt[p]{2}$ is. – Servaes Feb 11 at 6:58

## 1 Answer

Your polynomial $$x^{p-1}+x^{p-2}+\cdots+1$$ is the $$p$$-th cyclotomic polynomial and is often denoted $$\Phi_p$$. It satisfies $$\Phi_p=\frac{x^p-1}{x-1},$$ which immediately shows that every primitive $$p$$-th root of unity is a root of $$\Phi_p$$. As there are $$p-1$$ primitive $$p$$-th roots of unity and the degree of $$\Phi_p$$ is $$p-1$$, it follows that these are all roots of $$\Phi_p$$, so $$\Phi_p=\prod_{i=1}^{p-1}(x-\zeta^p),$$ where $$\zeta$$ is any primitive $$p$$-th root of unity.

To see that $$\Phi_p$$ is irreducible, apply Eisensteins criterion to $$\Phi_p(x+1)=\frac{(x+1)^p-1}{x}$$.