Solution
= Solution
Since $\beta$ is <integral>, $R[\beta]$ is generated over $R$ by finitely many powers, hence is fractional. It is invertible by (c). Because it is a <ring>, $I^2=I$; multiplying by $I^{-1}$ gives $I=R$. Thus every $\beta\in\mathcal O_K$ lies in $R$, so $R=\mathcal O_K$.