1

Despite it might be silly question, I'm eager to know it

Let the $ SF(x^n-a/\mathbb{Q})$ and $SF(x^n+a/\mathbb{Q}) $ be the splitting field of the $x^n-a$ and $x^n+a$ over the $\mathbb{Q}$ respectively. The $n $ is a natural number.

say $\omega = e^{ \frac{2\pi i }{n}}$, $\alpha = (-a)^{ \frac{1}{n}} $ And say $\beta =a^{ \frac{1}{n}}$

Then, $ SF(x^n-a/\mathbb{Q})= \mathbb{Q}( \beta , \omega ) $

on the other hand, $ SF(x^n+a/\mathbb{Q})= \mathbb{Q}( \alpha , \omega ) $

I've considered the $n$ is a odd case. Since the $(-1)^{ \frac{1}{n}} = -1 $ in either $\mathbb{Q}$ or $\mathbb{R}$ so $\alpha = -\beta $ in splitting fields over $\mathbb{Q}$.

Therefore my conclusion is If the $n$ is odd number then, $ SF(x^n-a/\mathbb{Q}) = SF(x^n+a/\mathbb{Q}) $.

Is my guess and conclusion right? Are there any other conditions for claiming $ SF(x^n-a/\mathbb{Q}) = SF(x^n+a/\mathbb{Q}) $ generally ?

Thanks for reading my post

se-hyuck yang
  • 2,236
  • 9
  • 22
  • See for example here for $a=1$. The splitting field of $x^n+1$ is $\Bbb Q (\zeta_{2n})$, and the one of $x^n-1$ is $\Bbb Q(\zeta_n)$. Then use this post. Or just see that for $n$ odd we have $\Phi_{2n}(X)=\Phi_n(-X)$. – Dietrich Burde May 23 '21 at 08:34
  • For general $a$ see here and this post. From these posts you should be able to answer your question. I am hesitating to write an answer, since repeating known answers is frowned upon. – Dietrich Burde May 23 '21 at 08:38

0 Answers0