Let $w\in\mathbb{C}$ be a primitive 10th root of unity. What is $[\mathbb{Q}(w):\mathbb{Q}]$?

Question

Let $w\in\mathbb{C}$ be a primitive 10th root of unity. What is $[\mathbb{Q}(w):\mathbb{Q}]$? Diagram the lattice of subfields of $\mathbb{Q}(w)$ giving generators for each.

So $w$ being a primitive 10th root of unity implies that $w^{10}=1$. We can rewrite the equation to be $$ w^{10}-1=(x^5-1)(x^5+1)=(x-1)(x^4+x^3+x^2+x+1)(x+1)(x^4-x^3+x^2-x+1). $$ By Eisenstein's Criterion, we know that the factors of degree 4 are irreducible. So how do I use this to determine the index $[\mathbb{Q}(w):\mathbb{Q}]$?

I know for the lattice diagram I need to have the factors so is it enough to say, for example $\mathbb{Q}/(x^4+x^3+x^2+1)$, as a group of the diagram?

Answer 1

The cyclotomic polynomial defined as $X^n-1=\prod_{m\mid n}\Phi_m(X)$ is always irreducible over $\mathbb{Q}$.

In this case, $\Phi_{10}(X)\Phi_5(X)\Phi_2(X)\Phi_1(X)=X^{10}-1$.

We know by this definition that $\Phi_5(X)\Phi_1(X)=X^5-1$ and $X^2-1=\Phi_2(X)\Phi_1(X)$ so $\Phi_2(X)=X+1$. Thus, $\Phi_5(X)(X+1)(X^5-1)=(X^{10}-1),$ which gives $\Phi_5(X)=(X^5+1)/(X+1)=X^4-X^3+X^2-X+1$ (You can show directly that this is irreducible by a transformation $Y=X+1$ and applying Eisenstein's Criterion.)

And, in general, $\deg(\Phi_m)=\phi(m),$ where $\phi$ is the Euler totient function which counts the number of positive integers less than $m$ which are coprime to $m$.

(Indeed, $\phi(10)=(5-1)(2-1)=4$ as we can see.)

Answer 2

Continuing your approach, from your observation that
$$ x^{10}-1=(x-1)(x+1)(x^4+x^3+x^2+x+1)(x^4-x^3+x^2-x+1) $$ Let the quartic factors be $f(x)$ and $g(x)$. Then $g(x)=f(-x)$. Therefore, they have the same splitting field, which is the splitting field of $x^{10}-1$. Since they are irreducible, this splitting field has degree $4$.