Let $p$ be a prime and $m$ be a positive integer. Prove that the splitting fields of $x^{2^m} + 1$ and $x^{2^{m+1}} - 1$ over $\mathbb{F}_p$ are isomorphic.

Any help appreciated!


Hint: Write $$x^{2^{m+1}}-1 = (x^{2^m})^2 - 1^2 = (x^{2^m}-1)(x^{2^m}+1).$$

Then, note that if $\alpha$ is a root of $x^{2^m}+1 = 0$, then $\alpha^2$ is a root of $x^{2^m} - 1 = 0$.

Expanding upon the hint.

We see that $f(x) = x^{2^m} + 1$ is a factor of $g(x) = x^{2^{m+1}} - 1$. So, if $K$ is a splitting field of $g$ over $\mathbb{F}_p$, then $K$ contains an isomorphic copy of a splitting field $L$ of $f$ over $\mathbb{F}_p$. We want to show that $L = K$. This follows from the claim that $$S = \{ \pm \alpha^2 : f(\alpha) = 0 \}$$ is precisely the set of roots of $h(x) = x^{2^m} - 1$. If $p \neq 2$, then $f$ is separable, so it has $2^m$ distinct roots. Hence, $S$ has $2^m$ distinct elements, and we can verify that they are all roots of $h$. If $p = 2$, then $f = h$, so $g = f^2$, so the splitting fields coincide anyway.

  • $\begingroup$ I did - I even factorized x^(2^m) - 1 further, but I do not see how to continue. $\endgroup$ – DesmondMiles Oct 11 '18 at 16:46
  • $\begingroup$ @DesmondMiles I'll expand upon the hint then, just a minute. $\endgroup$ – Brahadeesh Oct 11 '18 at 16:46
  • $\begingroup$ @DesmondMiles I've updated the answer, please let me know if it works for you. $\endgroup$ – Brahadeesh Oct 11 '18 at 16:59
  • $\begingroup$ Yes, I see the key now, thank you! $\endgroup$ – DesmondMiles Oct 11 '18 at 17:00

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