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.

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