In the set of ideals of a ring of algebraic integers in some fixed finite extension of QQ, if AB = A then B = (1)

Let A,B \subseteq \mathcal{O}_K be ideals in the ring of integers in K = \mathbb{Q}(\theta), with A \neq 0. Show that if AB = A then B = (1), using only the existence of a basis for A.

Let T = \{\omega_i\}_{i=1}^n be a basis for A over \mathbb{Z}. (T exists by Theorem 7.10 since A \neq 0.) Since AB = A, we have \omega_i = \sum_j \gamma_{i,j} \omega_j for each i, where \gamma_{i,j} \in B, using the definition of the ideal product and collecting terms in \omega_j as necessary. Rearranging, we have 0 = \sum_j (\gamma_{i,j} - \delta_{i,j}) \omega_j for each i, where \delta_{i,j} is the Kronecker delta. That is, [\omega_1\ \cdots\ \omega_n]^\mathsf{T} is a nontrivial solution to the matrix equation Mx = 0, where M = [\gamma_{i,j} - \delta_{i,j}]. Thinking of \mathcal{O} as embedded in the field K, we have \mathsf{det}(M) = 0. On the other hand, via the Leibniz expansion of \mathsf{det}(M), we have \mathsf{det}(M) = \beta - 1 for some \beta \in B. So -1 \in B, and we have B = (1).

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