Let $K$ be a field containing $\mathbb{Q}$. What does it mean to say that an element of $K$ is algebraic? Show that if $\alpha \in K$ is algebraic and non-zero, then there exists $\beta \in \mathbb{Z}[\alpha]$ such that $\alpha \beta$ is a non-zero (rational) integer.

Now let $K$ be a number field, with ring of integers $\mathcal{O}_{K}$. Let $R$ be a subring of $\mathcal{O}_{K}$ whose field of fractions equals $K$. Show that every element of $K$ can be written as $r / m$, where $r \in R$ and $m$ is a positive integer.

Prove that $R$ is a free abelian group of $\operatorname{rank}[K: \mathbb{Q}]$, and that $R$ has finite index in $\mathcal{O}_{K}$. Show also that for every nonzero ideal $I$ of $R$, the index $(R: I)$ of $I$ in $R$ is finite, and that for some positive integer $m, m \mathcal{O}_{K}$ is an ideal of $R$.

Suppose that for every pair of non-zero ideals $I, J \subset R$, we have

$$(R: I J)=(R: I)(R: J) .$$

Show that $R=\mathcal{O}_{K}$.

[You may assume without proof that $\mathcal{O}_{K}$ is a free abelian group of rank $[K: \mathbb{Q}]$ ] ]