Find the the splitting field of $X^5-2$ over $\mathbb{Q}$ and find it's degree.
My approach: The roots of $X^5-2$ are $\{\sqrt[5]{2},\sqrt[5]{2}\omega,\sqrt[5]{2}\omega^2, \sqrt[5]{2}\omega^3, \sqrt[5]{2}\omega^4\}$ where $\omega=e^{2\pi i/5}$.
It's quite easy to show that splitting field of $X^5-2$ over $\mathbb{Q}$ is $\mathbb{Q}(\sqrt[5]{2},\omega)$.
Let's find the value of $[\mathbb{Q}(\sqrt[5]{2},\omega):\mathbb{Q}]$.
By tower's Theorem $[\mathbb{Q}(\sqrt[5]{2},\omega):\mathbb{Q}]=[\mathbb{Q}(\sqrt[5]{2},\omega):\mathbb{Q}(\sqrt[5]{2})][\mathbb{Q}(\sqrt[5]{2}):\mathbb{Q}]$ and it's obvious that $[\mathbb{Q}(\sqrt[5]{2}):\mathbb{Q}]=5$.
$\omega$ is the root of polynomial $X^4+X^3+X^2+X+1$ which shows that $[\mathbb{Q}(\sqrt[5]{2},\omega):\mathbb{Q}(\sqrt[5]{2})]\leq 4$.
How to show that polynomial $X^4+X^3+X^2+X+1$ is irreducible over $\mathbb{Q}(\sqrt[5]{2})$?
I was trying in that way: since $\omega \notin \mathbb{Q}(\sqrt[5]{2})$ then it factors as a product of quadratic polynomials $$X^4+X^3+X^2+X+1=(X^2+AX+B)(X^2+CX+D),$$ where $A,B,C,D\in \mathbb{Q}(\sqrt[5]{2})$.
How to get contradiction?
I would be very thankful if anyone can show how to complete this reasoning?
Best Answer
It has degree $20$. It has subfields $\Bbb Q(\sqrt[5]2)$ of degree $5$ (Eisenstein) and $\Bbb Q(\omega)$ of degree $4$ (cyclotomy, or Eisenstein again). So its degree is a multiple of $4$ and of $5$, so is s multiple of $20$. But clearly the degree is at most $20$ also.