Characterize ideals in a number ring The following question is taken from D. A. Marcus' Number Fields Chapter 3 Exercise 9(c). There is a question on this already but I do not understand the answer there provided by @AdamHughes (Algebraic number theory, Marcus, Chapter 3, Question 9)

Let $K$ and $L$ be the number field such that $K\subset L$ and let $R,S$ be their algebraic integers, respectively.
a) Let $I$ and $J$ be ideals in $R$, and suppose $IS|JS$. Showw that $I|J$.
b) Show that for each ideal $I$ in $R$, we $I=IS\cap R$.
c) Characterize those ideals $I$ of $S$ such that $I=(I\cap R)S$.

Any hints/explanations would be appreciated for the special case when $I=Q^{m}$ or the general case.
Edit: I have attempted a proof and answered this question on my own. Feel free to comment any improvements/shorter approaches.
 A: This is an answer to (c) only.
We first consider the case when $I=\mathfrak{P}_{1}^{a_{1}}\cdots\mathfrak{P}_{k}^{a_{k}}$ and suppose that $\mathfrak{P}_{j}$ all lie over $\mathfrak{p}$. Since $\mathfrak{p}$ is a prime and hence maximal ideal in $R$, then $\mathfrak{p}S\subsetneq S$. This gives $\mathfrak{p}S|I$ and we can write
\begin{equation*}
\mathfrak{p}S=\mathfrak{P}_{1}^{\alpha_{1}}\cdots\mathfrak{P}_{k}^{\alpha_{k}}
\end{equation*}
where $\alpha_{i}\leq a_{i}$ for all $i$. Let $\mathfrak{p}^{\ell}S$ be the largest power of $\mathfrak{p}S$ dividing $I$, so that $\mathfrak{p}^{\ell+1}S\nmid I$. Now let
\begin{equation*}
I\cap R=\mathfrak{p}^{\ell}J
\end{equation*}
and note that $\mathfrak{p}\not\supseteq J$ by assumption on $\ell$. Suppose that $J$ is a proper ideal of $R$ and so $J\subseteq\mathfrak{q}$ for some prime ideal $\mathfrak{q}$ in $R$. Then
\begin{equation*}
\mathfrak{P}_{1}^{a_{1}}\cdots\mathfrak{P}_{k}^{a_{k}}=I=\left(I\cap R\right)S=\mathfrak{p}^{\ell}JS\subseteq\mathfrak{p}^{\ell}\mathfrak{q}S
\end{equation*}
so that $\mathfrak{q}S=\mathfrak{P}_{1}^{b_{1}}\cdots\mathfrak{P}_{k}^{b_{k}}$. Now,
\begin{equation*}
\mathfrak{q}=\mathfrak{q}S\cap R=\mathfrak{P}_{1}^{b_{1}}\cdots\mathfrak{P}_{k}^{b_{k}}\cap R\subseteq\mathfrak{P}_{1}\cap R=\mathfrak{p}
\end{equation*}
so that $\mathfrak{p}=\mathfrak{q}$, a contradiction. Therefore, $J=R$ is an improper ideal. Therefore, $I=\left(I\cap R\right)S=\mathfrak{p}^{\ell}S$. Next, we consider the case when $I=\mathfrak{Q}_{1}^{b_{1}}\mathfrak{Q}_{2}^{b_{2}}$, where $\mathfrak{Q}_{1}\cap R=\mathfrak{q}_{1}$ and $\mathfrak{Q}_{2}\cap R=\mathfrak{q}_{2}$ and $\mathfrak{q}_{1}+\mathfrak{q}_{2}=1$. If $\mathfrak{Q}_{2}|\mathfrak{q}_{1}S$, which gives us $\mathfrak{q}_{1}=\mathfrak{q}_{2}$, a contradiction. Since $I\subseteq\mathfrak{Q}_{1}$, then $I\cap R\subseteq\mathfrak{Q}_{1}\cap R=\mathfrak{q}_{1}$ which implies that $\mathfrak{q}_{1}|I\cap R$ and hence $\mathfrak{q}_{1}S|\left(I\cap R\right)S=I$. This together with $\mathfrak{Q_{2}}\nmid\mathfrak{q}_{1}S$ shows that $\mathfrak{q}_{1}S=\mathfrak{Q}_{1}^{\beta_{1}}$ where $\beta_{1}\leq b_{1}$. Similarly, $\mathfrak{q}_{2}S=\mathfrak{Q}_{2}^{\beta_{2}}$ where $\beta_{2}\leq b_{2}$. Let $\mathfrak{q}_{1}^{\ell_{1}}S$ and $\mathfrak{q}_{2}^{\ell_{2}}S$ be the largest power of $\mathfrak{q}_{1}S$ and $\mathfrak{q}_{2}S$ dividing $\mathfrak{Q}_{1}^{b_{1}}$ and $\mathfrak{Q}_{2}^{b_{2}}$ respectively. This implies that
\begin{equation*}
I\cap R=\mathfrak{Q}_{1}^{b_{1}}\cap R\cap\mathfrak{Q}_{2}^{b_{2}}\cap R\subseteq\mathfrak{q}_{1}^{\ell_{1}}\cap\mathfrak{q}_{2}^{\ell_{2}}=\mathfrak{q}_{1}^{\ell_{1}}\mathfrak{q}_{2}^{\ell_{2}}
\end{equation*}
so that we can write $I\cap R=\mathfrak{q}_{1}^{\ell_{1}}\mathfrak{q}_{2}^{\ell_{2}}J$ for some ideal $J\subseteq R$. If $J$ is a proper ideal of $R$, then $J\subseteq\mathfrak{m}$ for some maximal ideal $\mathfrak{m}$ in $R$. Then
\begin{equation*}
I=\left(I\cap R\right)S=\mathfrak{q}_{1}^{\ell_{1}}\mathfrak{q}_{2}^{\ell_{2}}JS\subseteq\mathfrak{q}_{1}^{\ell_{1}}\mathfrak{q}_{2}^{\ell_{2}}\mathfrak{m}S\subseteq\mathfrak{m}S
\end{equation*}
so that $\mathfrak{m}S|I$. Write $\mathfrak{m}S=\mathfrak{Q}_{1}^{\gamma_{1}}\mathfrak{Q}_{2}^{\gamma_{2}}$. This implies that by (b),
\begin{equation*}
\mathfrak{m}=\mathfrak{m}S\cap R=\mathfrak{Q}_{1}^{\gamma_{1}}\mathfrak{Q}_{2}^{\gamma_{2}}\cap R\subseteq\mathfrak{Q}_{1}\cap R=\mathfrak{q}_{1}
\end{equation*}
so that $\mathfrak{m}=\mathfrak{q}_{1}$ and $\mathfrak{q}_{1}^{\ell_{1}+1}|\mathfrak{Q}_{1}^{b_{1}}$, contradicting the choice of $\ell_{1}$. Therefore, $J=R$ and so $I\cap R=\mathfrak{q}_{1}^{\ell_{1}}\mathfrak{q}_{2}^{\ell_{2}}$ which implies that $I=\mathfrak{q}_{1}^{\ell_{1}}\mathfrak{q}_{2}^{\ell_{2}}S$. Lastly, we consider the case when $I=I_{1}I_{2}\cdots I_{k}$ where the primes in $I_{j}$ lies over $\mathfrak{p}_{j}$. This implies that $\mathfrak{p}_{j}S|I$. If $\mathfrak{p}_{j}S|I_{w}$ for some $w\neq j$, then letting $\mathfrak{P}_{j,1}$ be a prime dividing $\mathfrak{p}_{j}S$, we see that $\mathfrak{P}_{j,1}$ divides $I_{w}$, so that
\begin{equation*}
I_{w}\cap R=\mathfrak{P}_{j,1}^{\beta_{j,1}}J\cap R\subseteq \mathfrak{P}_{j,1}\cap R=\mathfrak{p}_{j}
\end{equation*}
contradicting that the primes in $I_{w}$ lies over $\mathfrak{p}_{w}$. Since $\mathfrak{p}_{j}S|I$ and $\mathfrak{p}_{j}S\nmid I_{w}$ for all $w\neq j$ and moreover, if $\mathfrak{Q}_{w}$ is any prime in $I_{w}$, $\mathfrak{Q}_{w}\nmid\mathfrak{p}_{j}S$, so that $\mathfrak{p}_{j}S|I_{j}$ for each $j$. Let $\mathfrak{p}_{j}^{\ell_{j}}S$ be the largest power of $\mathfrak{p}_{j}S$ dividing $I_{j}$. Since
\begin{equation*}
I\cap R=\bigcap_{i=1}^{k}\left(I_{i}\cap R\right)\subseteq\bigcap_{i=1}^{k}\mathfrak{p}_{i}^{\ell_{i}}=\mathfrak{p}_{1}^{\ell_{1}}\cdots\mathfrak{p}_{k}^{\ell_{k}}
\end{equation*}
so that we can write $I\cap R=\mathfrak{p}_{1}^{\ell_{1}}\cdots\mathfrak{p}_{k}^{\ell_{k}}U$ for some ideal $U$ of $R$. If $U\subseteq\mathfrak{M}$ for some maximal ideal $\mathfrak{M}$ of $R$, then
\begin{equation*}
I=\left(I\cap R\right)S\subseteq\prod_{i=1}^{k}\mathfrak{p}_{i}^{\ell_{i}}\mathfrak{M}S\subseteq\mathfrak{M}S.
\end{equation*}
Write $\mathfrak{M}S=\mathfrak{I}_{1}\cdots\mathfrak{I}_{k}$ where $\mathfrak{I}_{j}$ is some product of primes where each prime is in the factorisation of $I_{j}$ in $S$. This implies that
\begin{equation*}
\mathfrak{M}=\mathfrak{M}S\cap R\subseteq\mathfrak{I}_{j}\cap R\subseteq\mathfrak{p}_{j}
\end{equation*}
so that $\mathfrak{M}=\mathfrak{p}_{j}$, contradicting the choice of $\ell_{j}$. Therefore, $U=R$, and so $I\cap R=\mathfrak{p}_{1}^{\ell_{1}}\cdots\mathfrak{p}_{k}^{\ell_{k}}$. This implies that $I=\left(I\cap R\right)S=\mathfrak{p}_{1}^{\ell_{1}}\cdots\mathfrak{p}_{k}^{\ell_{k}}S$.
