# Showing that the splitting fields of $x^{2^m}+1$ and $x^{2^{m+1}}-1$ over $\mathbb{F}_p$ are isomorphic.

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.

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