If A is an ideal in an algebraic integer ring and a an element of A, then there is an ideal B such that AB = (a)

Let \mathcal{O} be the ring of integers in an algebraic number field K. Suppose A \subseteq \mathcal{O} is an ideal and \alpha \in A an element. Show that there exists an ideal B \subseteq \mathcal{O} such that AB = (\alpha).


Write A = \prod_{i=1}^n P_i as a product of maximal ideals, and let A^{-1} = \prod_{i=n}^1 P_i. Note that as subsets of \mathcal{O}, we have (A^{-1}(\alpha))A = A^{-1}A(\alpha) = \mathcal{O}(\alpha) = (\alpha). (Using Lemma 8.23 in TAN.) On the other hand, since (\alpha) \subseteq A, A^{-1}(\alpha) is an ideal of \mathcal{O}, as we argue. By definition, A^{-1}(\alpha) \subseteq \mathcal{O}. If x,y \in A^{-1}(\alpha) and r \in \mathcal{O}, where x = \sum h_ia_i\alpha and y = \sum k_ib_i\alpha, then x+ry = \sum h_ia_i\alpha + \sum k_irb_i\alpha \in A^{-1}(\alpha). Since 0 \in A^{-1}(\alpha), by the submodule criterion A^{-1}(\alpha) is an ideal of \mathcal{O}.

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