If A is an ideal of integers in an algebraic number field and if bA is contained in A, then b is an integer

Let K be an algebraic number field with ring of integers \mathcal{O}, and let A \subseteq \mathcal{O} be an ideal. Suppose \beta \in K such that \beta A \subseteq A. Prove that \beta \in \mathcal{O}.

Let \{\omega_i\} be a basis for A over \mathbb{Z}. (This exists by Theorem 7.10). In particular, for each i, there exist rational integers c_{i,j} such that \beta\omega_i = \sum_i c_{i,j}\omega_j. Rearranging, we have 0 = \sum_j (c_{i,j} - \delta_{i,j}\beta)\omega_j for each i, where \delta_{i,j} is the Kronecker delta. In particular, [\omega_1 \cdots \omega_n]^\mathsf{T} is a nontrivial solution to the matrix equation Mx = 0, where M = [c_{i,j} - \delta_{i,j}\beta]. Thus \mathsf{det}(M) = 0. On the other hand, by the Leibniz expansion for determinants, \mathsf{det}(M) is a polynomial in \beta having coefficients in \mathbb{Z}. Thus \beta is an algebraic integer, and more specifically \beta \in \mathcal{O}.

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