Why is \(\Phi_p(x) = \frac{x^p - 1}{x - 1} = x^{p-1} + x^{p-2} + \cdots + x + 1\) irreducible over \(\mathbb{Q}\) for \(p\) prime?
Solution
Set \[ g(y)=\Phi_p(y+1)=\frac{(y+1)^p-1}{y} =\sum_{m=1}^p\binom pm y^{m-1}. \] Its leading coefficient is \(1\). For \(1\le m\le p-1\), the prime \(p\) divides \(\binom pm\), while the constant term is \(\binom p1=p\), which is not divisible by \(p^2\). Thus \(g(y)\) is Eisenstein at \(p\), hence irreducible over \(\mathbb Q\).
The substitution \(x\mapsto y+1\) is an automorphism of \(\mathbb Q[x]\), so reducibility of \(\Phi_p(x)\) would imply reducibility of \(\Phi_p(y+1)=g(y)\). Therefore \[ \Phi_p(x)=1+x+\cdots+x^{p-1} \] is irreducible over \(\mathbb Q\).