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:
- It's irreducible
- 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!
Best 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 $\Phi_p(x+1)=\frac{(x+1)^p-1}{x}$.