I was reading this question: When does a polynomial divide $x^k - 1$ for some $k$?
I followed the procedure given by Bill Dubuque in his answer (the "Graeffe" method) for the polynomial $f(x) = x+1$.
First you compute $g(x^2)$ and $xh(x^2)$, where $g(x^2)$ is the even part of $f(x)$ and $xh(x^2)$ is the odd part of $f(x)$.
So $$f(x) = x+1 = g(x^2)+xh(x^2) = 1+x*1$$
$$g(x) = 1, h(x) = 1$$
Then you compute $g(x)^2-xh(x)^2$ and normalize it: $$f_1(x) = ||g(x)^2-xh(x)^2||= ||1-x|| = x-1$$
Now if you follow the algorithm:
- $f_1(x) \neq f(x)$
- $f_1(x) \neq f(-x)$
- $f_1(x)$ is irreducible
So by the algorithm there $f(x)$ is not a cyclotomic polynomial. Am I missing something obvious?
