 Solution

An element $\alpha$ is algebraic over $\mathbb{Q}$ when it is a root of a nonzero polynomial in $\mathbb{Q}[X]$. Clearing denominators givesFor $\alpha \ne 0$, take $\beta =-(a_1+a_2\alpha +\cdots +a_d{\alpha }^{d-1})\in \mathbb{Z}[\alpha ]$; then $\alpha \beta =a_0\ne 0$.

Since $K=\mathrm{Frac}R$, write $x=a/b$ with $a,b\in R$ and $b\ne 0$. Applying the first result to $b$ gives $b\beta =m\in \mathbb{Z}\setminus \{0\}$, whence $x=a\beta /m$. Changing signs makes $m>0$.

Choose a $\mathbb{Z}$-basis $e_1,\dots ,e_d$ of ${\mathcal{O}}_K$. Write each $e_i=r_i/m_i$ as above and take a common multiple $m$. ThenThus $R$ is a free abelian group of rank $d=[K:\mathbb{Q}]$ and has finite index in ${\mathcal{O}}_K$. If $0\ne a\in I$, then $aR\subseteq I$ and multiplication by $a$ has nonzero determinant on the rank-$d$ lattice $R$, so $(R:aR)$, and hence $(R:I)$, is finite. Also $m{\mathcal{O}}_K\subseteq R$ is an ideal of $R$.

Put $h=({\mathcal{O}}_K:R)$ and $I=m{\mathcal{O}}_K$. ThenThe assumed multiplicativity of indices givesso $h=1$. Therefore $\boxed{R={\mathcal{O}}_K}$.