## 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 $\mathcal{O}$.