Let $R$ be an integral domain. Let $Y$ be a multiplicatively closed subset of $R$ which contains $1$ but not $0$.
Define $S=RY^{-1}=\lbrace ry^{-1} : r \in R, y \in Y \rbrace$ as well as $\mathcal{I}(R)$ and $\mathcal{I}(S)$ to be the sets of ideals of $R$ and $S$ respectively.
The maps $c:\mathcal{I}(S) \to \mathcal{I}(R)$ and $e:\mathcal{I}(R) \to \mathcal{I}(S)$ are given by $c(J)=J \cap R$ and $e(I)=IS$.
I'm trying to show that these maps respect inclusions, intersections and sums.
I'm confused about the non-trivial ones, i.e. sums for both maps and also intersections in the case of $e$. I feel like this is all very much just about notations and manipulation but I can't get my head around it. Can anyone give me a hand?
EDIT: manipulations give $(I+J)Y^{-1}=IY^{-1}+JY^{-1}$ for any two ideals $I$ and $J$ of $R$ so $(I+J)S=(I+J)RY^{-1}=IRY^{-1}+JRY^{-1}=IS+JS$ so $e$ respects sums.
For $S^{-1}I\cap S^{-1}J\subseteq S^{-1}(I\cap J)$, let $x\in S^{-1}I\cap S^{-1}J$. Then $x=\frac as=\frac bt$ with $a\in I$ and $b\in J$. Then there is $u\in S$ such that $uta=usb\in I\cap J$. Now write $x=\frac{uta}{uts}\in S^{-1}(I\cap J)$.
Let $R=K[X,Y]$, $f=X+Y$, $S=\{1,f,f^2,\dots\}$, $I=(X)$, and $J=(Y)$. Then $S^{-1}I\cap R=I$ and $S^{-1}J\cap R=J$ (since $I,J$ are prime ideals and $I\cap S=J\cap S=\emptyset$), while $(S^{-1}I+S^{-1}J)\cap R=S^{-1}(I+J)\cap R=S^{-1}R\cap R=R$ (since $(I+J)\cap S\ne\emptyset$).