## 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 $\mathcal{O}$.