If $ f _p(x)=1+x+x^2+ \cdots +x^{p-1} $ where $p$ is prime number, $e$ is any positive integer.

Question

Show that $f_p (x^{p^{e-1}})$ is irreducible over $\mathbb{Q}$. I know it is obviously true for $ p=2$. Please suggest me any hints

Answer 1

The proof follows from:

• The cyclotomic polynomials, $\Phi_n(x)$, are irreducible over $\mathbb Q$.
• $f_p(x^{p^{e-1}})=\Phi_{p^e}(x).$
  Proof:
  Let $\zeta_n$ be the $n^{th}$ primitive root of unity, $\zeta_n=\exp(\frac{2\pi i}{n})$ and let $0\leq m \leq p-1 $.
  The ${p^{e-1}}^{th}$ roots of $\zeta_p^{m}\in\mathbb C$ are $\exp{\left(\frac{2k\pi i}{p^{e-1}}+\frac{2m \pi i}{p^e}\right)}, \ k=0,1,2,\ldots p^{e-1}-1$ or equivalently $\zeta^{kp+m}_{p^e}, \ k=0,1,\ldots,p^{e-1}-1$. Therefore \begin{align}x^{p^{e-1}}-\zeta_p^{m}=(x-\zeta^m_{p^e})(x-\zeta^{p+m}_{p^e})(x-\zeta^{2p+m}_{p^e})\ldots (x-\zeta^{(p^{e-1}-1)p+m}_{p^e}) \ \ \ \ \ \ \ \ \ .\end{align} Multiplying the previous by parts for $m=0,1,\ldots,p-1$ we obtain $$\Phi_p(x^{p^{e-1}})=f_p(x^{p^{e-1}})=\Phi_{p^e}(x).$$