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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!

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  • $\begingroup$ 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 $[\Bbb{Q}(\root p\of 2,\zeta_p):\Bbb{Q}(\root p\of2)]$. $\endgroup$ – Jyrki Lahtonen Feb 11 at 6:31
  • $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Feb 11 at 6:33
  • $\begingroup$ 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). $\endgroup$ – Jyrki Lahtonen Feb 11 at 6:40
  • $\begingroup$ 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. $\endgroup$ – Servaes Feb 11 at 6:58
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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 $\Phi_p(x+1)=\frac{(x+1)^p-1}{x}$.

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