Order of splitting field of $x^5+8$

Question

I am trying to calculate the order of the splitting field of $f(x):=x^5+8$.

I started considering that all the roots of $f$ are $\zeta^i\cdot\sqrt[5]{-8}$ for $i=0,\dots,4$ where $\zeta$ is the 5th primitive root of unity. Let $K$ be the splitting field of $f(x)$, then $K$ contains $\sqrt[5]{-8}$ and $\zeta\cdot\sqrt[5]{-8}$, so it contains $\zeta$ as well. Thus, $K\supseteq\mathbb{Q}(\zeta, \sqrt[5]{-8})$, but every field that contains $\zeta$ and $\sqrt[5]{-8}$ must be a splitting field for $f(x)$, therefore $K=\mathbb{Q}(\zeta, \sqrt[5]{-8})$.

Is my proof allright still here?

Now I would like to prove that $K$ has order $20$.

• The minimal polynomial of $\sqrt[5]{-8}$ on $\mathbb{Q}$ is $x^5+8$, so $|\mathbb{Q}(\sqrt[5]{-8}):\mathbb{Q}|=5$.
• The minimal polynomial of $\zeta$ on $\mathbb{Q}$ is $\Phi_5(x)=x^4+x^3+x^2+x+1$, so $|\mathbb{Q}(\zeta):\mathbb{Q}|=4$.

But I'm stucked here, I'm not able to proof nor that $x^5+8$ is irreducible on $\mathbb{Q}(\zeta)$, nor that $\Phi_5(x)$ is irreducible on $\mathbb{Q}(\sqrt[5]{-8})$.

Thanks for your help.

Answer 1

Degrees of subfields must divide the degree of the field, so your field extension must be of degree a multiple of $4$ and $5$, i.e. a multiple of $20$.