1

Let $m(X) := \frac{X^p-1}{X-1}$ and $n(X) := m(X^p)$. I have shown that $m$ is irr. over $\mathbb Q$. Now I want to show that this is true for $n(X)$, too. I know that $$ n(X+1)((X+1)^p-1)= (X+1)^{p^2}-1 $$ Working modulo $p$ I get that $$ n(X+1)X^p \equiv X^{p^2} \mod p $$ which means that $p$ divives each coefficient of $n(X+1)$ and not the leading coefficient. But I do not know how to show that $p^2 \nmid a_0$ where $a_0$ is the constant term of $n(X+1)$.

Pleas help :)

1 Answers1

1

Turning my comment into an answer: write $m(X) = \sum_{i=0}^{p-1}X^i = 1+X+X^2+\ldots+X^{p-1}$. Then $n(X+1) = m\left((X+1)^p\right) = 1+(X+1)^p+(X+1)^{2p}+\ldots+(X+1)^{p(p-1)}$. Now just use the binomial theorem to find the constant term of each of the terms of this sum.

Steven Stadnicki
  • 51,746
  • Ok thanks. So easy :D. I will accept this in a few minutes. –  Jun 18 '13 at 16:37
  • One little question: If I prove that $n(X+1)$ is irreducible then $n(X)$ is reducible because assume not .Then $n(x) = g(x)h(x)$ so $n(X+1) = g(X+1)h(X+1)$ but $g,h$ are no units so $g(X+1)$ and $h(X+1)$ are no units s.t. $n(X+1)$ would be reducible ? –  Jun 18 '13 at 16:43
  • 1
    @André precisely - in fact, it's the standard means for applying the Eisenstein Criterion to cyclotomic polynomials, as you've discovered. – Steven Stadnicki Jun 18 '13 at 16:51
  • Great. Thanks for your help. –  Jun 18 '13 at 19:14