$IS \mid JS$ implies that $I\mid J$ where $I$ and $J$ are ideals of a number ring contained in another number ring $S$ $K$ and $L$ are number fields and $K \subset L$ . Now, if $I$ and $J$ are two ideals in number ring of $K$ and  $IS \mid JS$ then we have to show that $I \mid J$.     Here, $S$ is number ring of $L$.
The book suggests to factor $I$ and $J$ into primes of number ring of $K$ and then consider what happens in $S$. 
This question has been asked here before: 
Algebraic number theory, Marcus, Chapter 3, Question 9
once but it doesn't use the hint the book gives and instead part $(b)$ of the same problem to prove this result. 
Can anybody explain what the author means by 'considering what happens in $S$' or give hints but no full solution please

Let $I = \mathfrak p_1^{e_1}\dots\mathfrak p_s^{e_s}$, $J = \mathfrak q_1^{k_1}\dots\mathfrak q_r^{k_r}$ in $R$, the ring of integers of $K$. Let $P_1$ be a prime ideal in the factorization of $\mathfrak p_1S$
Then we have $P_1 \mid \mathfrak p_1S \mid IS \mid JS$. So as $P_1$ is a prime ideal and the factorzation in prime ideals is unique, $P_1$ must appear in the ideal factorization of $JS$. On the other side the factorization of $JS$ is completely determined by the factorization of $J$ in $R$. In other words we first factorize each $\mathfrak q_iS$ and just multiply them to get the factorization of $JS$. Therefore $P_1$ must come from a factorization of one of the prime ideal factors of $J$, WLOG let it be $\mathfrak q_1S$.
However we know that a prime ideal lies above a unique prime ideal. In particular, $\mathfrak p_1S \subseteq P_1 \implies P_1 \cap R = \mathfrak p_1$. From above we get $\mathfrak p_1 = P_1 \cap R = \mathfrak q_1$. Furthermore from $IS \mid JS$ it's not hard to conclude that $e_1 \le k_1$. Now repeat the same for all $\mathfrak p_i$
