rational integers in ideals

Any non-zero ideal of an algebraic number field $K$, i.e. of the maximal order ${𝒪}_K$ of $K$, contains positive rational integers.

Proof.  Let  $𝔞\ne (0)$  be any ideal of ${𝒪}_K$.  Take a nonzero element $\alpha$ of$𝔞$.  The norm (http://planetmath.org/NormInNumberField) of $\alpha$ is the product

$$\text{N}(\alpha )={\alpha }^{(1)}\underset{\gamma }{\underbrace{{\alpha }^{(2)}\mathrm{\cdots }{\alpha }^{(n)}}}$$

where $n$ is the degree of the number field and ${\alpha }^{(1)},{\alpha }^{(2)},\mathrm{\cdots },{\alpha }^{(n)}$ is the set of the http://planetmath.org/node/12046$\mathrm{K}$-conjugates of  $\alpha ={\alpha }^{(1)}$.  The number

$$\gamma =\frac{\text{N}(\alpha )}{\alpha }$$

belongs to the field $K$ and it is an algebraic integer, since ${\alpha }^{(2)},\mathrm{\cdots },{\alpha }^{(n)}$ are, as algebraic conjugates of $\alpha$, also algebraic integers.  Thus  $\gamma \in {𝒪}_K$.  Consequently, the non-zero integer

$$\text{N}(\alpha )=\alpha \gamma$$

belongs to the ideal $𝔞$, and similarly its opposite number.  So, $𝔞$ contains positive integers, in fact infinitely many.