How do I show $[\mathbb{Q}(\alpha):\mathbb{Q}]=4$.

Question

Let $\alpha\in\Bbb C\setminus\{-1\}$ be a root of the polynomial $x^5+1$. Show that $[\Bbb Q(\alpha):\Bbb Q]=4$.

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My attempt:

Let $p(x)=x^5+1=(x+1)(x^4-x^3+x^2-x+1)$. Now assume that $q(x)=(x^4-x^3+x^2-x+1)$.

Claim: The polynomial $q(x)$ is irreducible over $\Bbb Q$.

Replace $x$ with $x-1$, then $q(x-1)=x^4-5x^3+10x^2-10x+5$. By Eisenstein criteria, this polynomial is irreducible over $\Bbb Q$ for $p=5$. So, there exist a field extension $\Bbb Q(\alpha)$, and $[\Bbb Q(\alpha):\Bbb Q]=$degree of minimal polynomial.

In this case, $[\Bbb Q(\alpha):\Bbb Q]=4$.

Can anyone suggest me some improvement or mistakes in this solution?

Answer 1

Yes, it is correct. You showed that $q(x)$ is the minimal polynomial of $\alpha$ over $\mathbb{Q}$ (since $q(\alpha) = 0$ and $q$ is irreducible). Thus

$$|\mathbb{Q}(\alpha): \mathbb{Q}| = \deg q = 4$$