I'm trying to see how $$ [\mathbb Q(\alpha):\mathbb Q]=4,$$ where $$ \alpha^5 = 1,\ \alpha \neq 1 .$$
I don't understand why it's 4 and not 5.
From my knowledge the basis could be
$$ {1, \sqrt[5]{1}, \sqrt[5]{1^2}, \sqrt[5]{1^3}, \sqrt[5]{1^4}} $$
Why isn't this the case?
Help would be much appreciated, thanks
The element $\alpha$ clearly satisfies $f(x) = x^5-1$. However, this polynomial factors over $\mathbb{Q}$: $$f(x) = x^5-1 = (x-1)(x^4+x^3+x^2+x+1).$$ Since $\alpha\neq 1$, then $\alpha$ must in fact be a root of $$\Phi_5(x) = x^4+x^3+x^2+x+1.$$ This already proves that $[\mathbb{Q}(\alpha):\mathbb{Q}]\leq 4$. The question is whether $\Phi_5$ is irreducible. You may already know it is (there is a standard proof using Eistenstein’s Criterion). This fits into the general theory of Cyclotomic extensions of $\mathbb{Q}$, as noted by Sunyata.