## Every nonzero ideal in a ring of algebraic integers contains infinitely many rational integers

Let $\mathcal{O}$ be the ring of integers in a fixed finite extension $K = \mathbb{Q}(\theta)$, and let $A$ be a nonzero ideal in $\mathcal{O}$. Prove that $A$ contains infinitely many rational integers.

We begin with a lemma.

Lemma: Let $A \subseteq \mathcal{O}$ be an ideal. If $\alpha \in A$, then $N(\alpha) \in A$, where $N$ denotes the field norm. Proof: Note that $N(\alpha) \in K$ since $N(\alpha)$ is a rational integer. So $N(\alpha)/\alpha \in K$. Moreover, by definition, $N(\alpha)/\alpha$ is an algebraic integer, being the product of the (certainly integral) conjugates of $\alpha$ for $K$. So $N(\alpha)/\alpha \in \mathcal{O}$. Since $A$ is an ideal, $\alpha(N(\alpha)/\alpha) = N(\alpha) \in A$. $\square$

Now if $A \neq 0$, then there exists a nonzero element $\alpha \in A$. We have $N(\alpha) \in A$ nonzero, and thus $kN(\alpha) \in A$ for all $k \in \mathbb{Z}$. In particular, $A$ contains all of the $\mathbb{Z}$-multiples of $N(\alpha)$.