Past exam of the number theory course of the University of Oxford / 2011 / B / 4 / b / Solution

The polynomial is irreducible by Eisenstein's criterion at $q$, so $[K:\mathbb{Q}]=n$. Its constant coefficient gives the field normThe defining equation givesIf $q\nmid r$ and $0\le i<n$, thenhas $q$-adic valuation $i-n<0$. It is not an integer. The norm of an algebraic integer is an integer, so $r{\theta }^i/q$ is not integral.

Suppose now that $\gamma$ is integral. Multiply it by ${\theta }^{n-1}$. All terms except $r_0{\theta }^{n-1}/q$ are integral, since their powers of $\theta$ are at least $n$. Thus this remaining term is integral, and the norm obstruction forces $q\mid r_0$. Subtract the integer $r_0/q$ from $\gamma$. Multiply the remainder by ${\theta }^{n-2}$; the only term of exponent below $n$ is $r_1{\theta }^{n-1}/q$, so $q\mid r_1$. Continuing in this way proves $q\mid r_j$ for every $j$. ThereforeFor $n=1$ the same conclusion follows directly from the norm obstruction with $i=0$.