0

Que: Let $ \omega_p = e^{\frac{2\pi \iota}{p}} $ be the $p$-th primitive root of unity, with $p$ some prime. Prove $\mathbb Q(\sqrt[p]2, \omega_p) = \mathbb Q(\sqrt[p]2+ \omega_p)$.

Sol: To prove the equality we will show one is a subset of the other. Proving $\mathbb Q(\sqrt[p]2, \omega_p) \subset \mathbb Q(\sqrt[p]2+ \omega_p)$ is obvious because $\sqrt[p]2, \omega_p \in Q(\sqrt[p]2, \omega_p) \Rightarrow \sqrt[p]2 + \omega_p \in Q(\sqrt[p]2, \omega_p) \Rightarrow Q(\sqrt[p]2 + \omega_p) \subset Q(\sqrt[p]2, \omega_p)$. How to prove the other way? Can we work on powers of $\sqrt[p]2 + \omega_p$ to get $\sqrt[p]2$ and $\omega_p$? If there is any other way please tell.

Suzu Hirose
  • 11,660
Lalbahadur Sahu
  • 99
  • 1
    Try following the ideas of these examples: https://en.m.wikipedia.org/wiki/Primitive_element_theorem#The_theorems – Arkady Jun 11 '22 at 04:52
  • Are you familiar with the basic Galois correspondence? This field is the splitting field of $x^p-2$ (with well known Galois group), so Galois theory works like charm. – Jyrki Lahtonen Jun 11 '22 at 04:55
  • If you don't know Galois theory yet, you can mimick the proof of the primitive element theorem as Arkady commented. – Jyrki Lahtonen Jun 11 '22 at 05:05
  • Thank you @JyrkiLahtonen. Please have a look at my arguments. Since $[\mathbb Q(\sqrt[p]2, \omega_p):Q] = p(p-1)$ so $ \mathbb Q(\sqrt[p]2, \omega_p)$ is a finite and separable extenstion of $x^p-2$ which means that $Q(\sqrt[p]2, \omega_p) $ is simple by Primitive Element Theorem. Being simple it will have finitely many intermediate subfields and so by mimicking the proof of primitve element theorem $Q(\sqrt[p]2+c\omega_p) = Q(\sqrt[p]2+d\omega_p) $ which will lead us to get $\sqrt[p]2,\omega_p \in Q(\sqrt[p]2+c\omega_p)$ – Lalbahadur Sahu Jun 11 '22 at 07:58

0 Answers0