Let $K\subseteq L$ be number fields and let $R \subseteq S$ be the corresponding rings of integers.
Question. a) If $I$ and $J$ are integral ideals of $R$ such that $IS \mid JS,$ then $I \mid J.$
b) If $I$ is an integral ideal of $R,$ then $R\cap IS = I.$
c) Characterize the integral ideals $\mathfrak{A}$ of $S$ such that $(\mathfrak{A}\cap R)S=\mathfrak{A}.$
(This is Exercise 9 from Chapter 3 of "Number Fields" by Marcus).
I think I've managed to solve (a) and (b) but would be very grateful if someone could confirm (or refute!) and I'm just a bit stuck on (c)...
Attempted solution. a) For any maximal ideal $\mathfrak{P}$ of $S,$ we have $$v_\mathfrak{P}(IS)=v_\mathfrak{p}(I) \times e(\mathfrak{P}|\mathfrak{p})$$ $$v_\mathfrak{P}(JS)=v_\mathfrak{p}(J) \times e(\mathfrak{P}|\mathfrak{p})$$ where $\mathfrak{p}$ is the unique maximal ideal of $R$ below $\mathfrak{P}\;$ (i.e. $\mathfrak{p}=\mathfrak{P}\cap R$).
As such, we have the following chain of implications (in fact, they are equivalences, but whatever): $$IS\mid JS \; \Rightarrow \; v_\mathfrak{P}(IS) \leq v_\mathfrak{P}(JS) \;\; \forall \mathfrak{P} \; \Rightarrow \; v_\mathfrak{p}(I)\leq v_\mathfrak{p}(J) \;\; \forall\mathfrak{p} \; \Rightarrow \; I\mid J.$$ The second implication uses the fact that every max ideal of $R$ lies below some max ideal of $S.$
b) Let $J=R\cap IS.$ It is clear that $I\subseteq J.$
Moreover, it is clear that $JS=(R\cap IS)S \subseteq IS$ and so, by part (a), we get $J \subseteq I.$
c) I'm not really sure what the author expects here, I mean it's easy to see that $$(\mathfrak{A}\cap R)S=\mathfrak{A} \; \Leftrightarrow \; \mathfrak{A} = IS \text{ for some integral ideal } I \text { of } R$$ but I don't know what else to say... Any hints?
Many thanks!