> [!NOTE] Corollary > Let $p$ be a [[Prime numbers|prime]]. Then the $p$th [[Cyclotomic Polynomial|cyclotomic polynomial]] given by $\Phi_{p} (x) = \frac{x^{p}-1}{x-1} = x^{p-{1}} + x^{p-2} +\dots + x+1$is [[Irreducible Polynomial|irreducible]] over $\mathbb{Q}.$ **Proof**: Let $f(x)=\Phi_{p}(x+1)= \frac{(x+1)^{p}-1}{x} = x^{p-1}+ {p\choose{ 1}} x^{p-2} + {p \choose 2} x^{p-3} +\dots +{p\choose p-1} $Then, since every coefficient except that of $x^{p-1}$ is divisible by $p$ and the constant term is not divisible by $p^{2},$ by [[Schönemann-Eisenstein Criterion|Eisenstein's criterion]], $f$ is irreducible over $Q.$ Suppose $\Phi_{p}(x)=g(x)h(x)$ is a nontrivial factorization of $\Phi_{p}(x)$ over $\mathbb{Q},$ then $f(x)=\Phi_{p}(x+1)=g(x+1)\cdot h(x+1)$ is a nontrivial factorization of $f$ over $\mathbb{Q},$ a contradiction. Thus $\Phi_{p}$ is irreducible over $\mathbb{Q}.$