Facts about the inverse of a maximal ideal in an algebraic integer ring

Let K be an algebraic number field with ring of integers \mathcal{O}. Fix an ideal A \subseteq \mathcal{O}. Recall that, for an ideal A \subseteq \mathcal{O}, A^{-1} = \{r \in K \ |\ rA \subseteq \mathcal{O}\}. Prove the following: (1) A^{-1} is an \mathcal{O}-submodule of K, (2) If B \subseteq A is an ideal, then A^{-1}B \subseteq \mathcal{O} is an ideal, and (3) If C \subseteq B \subseteq A are ideals, then A^{-1}C \subseteq A^{-1}B.

Let x,y \in A^{-1} and let r \in \mathcal{O}. Now for all k \in A, we have (x+ry)k = xk + ryk \in \mathcal{O}, since xk \in \mathcal{O} and y(rk) \in \mathcal{O} (because A is an ideal). Since 0 \in A^{-1}, by the submodule criterion, A^{-1} is an \mathcal{O}-submodule of K.

Suppose C \subseteq B \subseteq A. Note that if x \in A^{-1}C, then x = \sum r_ic_i where r_i \in A^{-1} and c_i \in C. Certainly then x \in A^{-1}B, as desired in (3).

Suppose B \subseteq A; by the previous argument, we have A^{-1}B \subseteq A^{-1}A \subseteq \mathcal{O}. Suppose x,y \in A^{-1}B and r \in \mathcal{O}, with x = \sum r_ib_i and y = \sum s_ic_i. Certainly then since B is an ideal we have x+ry = \sum r_ib_i + \sum s_irc_i \in A^{-1}B. Since 0 \in A^{-1}B, by the submodule criterion, A^{-1}B is an ideal in \mathcal{O}.

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