 The Eisenstein prime does not divide the power-basis index

Let $f=X^n+a_{n-1}{X}^{n-1}+\cdots +a_0\in \mathbb{Z}[X]$ satisfy Eisenstein's criterion at a prime number $p$, and let $\theta$ be a root. ThenIndeed, ${\theta }^n/p$ is an algebraic integer, whereas for $p\nmid r$ and $0\le i<n$ the field norm of $r{\theta }^i/p$ has $p$-adic valuation $i-n<0$, so that element is not integral. If $\gamma =p^{-1}{\sum }_{i=0}^{n-1}{r}_i{\theta }^i$ is integral, multiply it by ${\theta }^{n-1}$. Every term except $r_0{\theta }^{n-1}/p$ is integral; thus $p\mid r_0$. Subtract $r_0/p$, multiply the remainder by ${\theta }^{n-2}$, and continue to obtain $p\mid r_i$ for every $i$. Hence there is no element of order $p$ in the finite group ${\mathcal{O}}_{\mathbb{Q}(\theta )}/\mathbb{Z}[\theta ]$, proving the claim. The result concerns the Eisenstein prime only: other primes may still divide the index.