I want to prove that if $\alpha$ is a root of $x^{p^{d}} - x$ and $n=d\cdot s$, then $\alpha$ is also a root of $x^{p^{n}} - x$.
So far I know that $x^{p^{d}-1} =1$ and I've tried this:
$$x^{p^{n} -1} = x^{p^{ds}-1}$$
I have also tried:
$$x^{p^{n}} = x^{p^{ds}} = (x^{p^{d}})^s=\alpha^s$$
but I can't conclude that $\alpha^s=\alpha$. Can someone help me?
Asked
Active
Viewed 20 times
0
kubo
- 1,918
-
1$x^{p^{ds}} \not= (x^{p^{d}})^s$ – ancient mathematician Nov 11 '20 at 12:32
-
Sometimes $\frac{x(x^k-1)}{x(x^m-1)}$ is a polynomial – reuns Nov 11 '20 at 13:12
-
$p^d-1$ is a factor of $p^{ds}-1$ (and the claim follows). See, for example here where we have a more general result. This is possibly more directly applicable. – Jyrki Lahtonen Nov 11 '20 at 13:19
-
Your specific question is also frequent within the tag [tag:finite-fields]. – Jyrki Lahtonen Nov 11 '20 at 13:22