## In an algebraic integer ring, the smallest positive integer contained in a given ideal divides the norm of the ideal

Let $K$ be an algebraic number field with ring of integers $\mathcal{O}$, and let $A \subseteq \mathcal{O}$ be and ideal. Suppose $m$ is the smallest positive integer contained in $A$. Show that $m|N(A)$. Suppose now that $N(A)$ is a rational prime; show that $\mathcal{O}/A = \{\overline{t}\ |\ 0 \leq t < p\}$.

By Theorem 9.16, $N(A) \in A$. Now $A \cap \mathbb{Z} = (m)$ is an ideal of $\mathbb{Z}$, which is principal since $\mathbb{Z}$ is a principal ideal domain. Certainly then $m$ is the smallest positive integer in $A$. Thus $m|N(A)$.

Now suppose $N(A)$ is a rational prime. Suppose there is a (minimal) integer $t \in A$ such that $0 < t < N(A)$. By part (1), $t|N(A)$, a contradiction. So none of the integers in $(0,N(A))$ are in $A$, and thus $t \not\equiv 0$ mod $A$ for all $t \in (0,N(A))$. Moreover, we have $t \not\equiv s$ mod $A$ for all $t,s \in (0,N(A))$ with $s \neq t$, since otherwise (assuming $s > t$) we have $s-t \equiv 0$ mod $A$. Since $N(A) = p$, the cosets $\{\overline{t}\ |\ t \in [0,N(A))\}$ exhaust $\mathcal{O}$ and are mutually exclusive.