Equality of ideals using GCD I am having trouble with a line in a proof of Marcus's Number Fields of the following result: If $I$ and $J$ are ideals in a number field $R$, then $\|IJ\|=\|I\|\|J\|$, where $\|I\|=|R/I|$.
Most of the proof I'm fine with, except for the following statement about ideals. We have that for a prime ideal $P$ in $R$ a chain $R\supset P\supset P^2\supset \cdots\supset P^m$. Let $\alpha\in P^k-P^{k+1}$. The claim is that $(\alpha R)\cap P^{k+1}=\alpha P$. 
Now, note that $(\alpha R)\cap P^{k+1}=\operatorname{lcm}(\alpha R,P^{k+1})$, and that the highest power of $P$ dividing $\alpha R$ is $P^k$. 
Then if $\alpha R=P^kQ_1\cdots Q_s$ for prime ideals $Q_i$, it would seems we would have $\operatorname{lcm}(\alpha R, P^{k+1})=Q_1\cdots Q_s P^{k+1}$, but I'm not sure how this implies the result.
What am I missing?
 A: As Marcus remarked $3$ pages prior (page $42$), gcd and lcm  are definable in the familiar way from the  unique prime ideal factorization, which immediately yields the familiar formula
$\quad\begin{align} {\rm lcm}(A,B) &= AB/\gcd(A,B)\\[.3em]
\Longrightarrow\ (\alpha)\cap P^{k+1}\! = {\rm lcm}((\alpha),P^{k+1}) &= (\alpha)\color{#c00}{P^{k+1}}/\!\!\underbrace{\gcd((\alpha),P^{k+1})}_{\textstyle \color{#c00}{P^k}\ \rm by\ hypothesis}\!\!\! = (\alpha)\color{#c00}P
\end{align}$
A: It seems that this equality of sets can be proved in a "lowbrow" way, just by talking about elements.
Suppose $x\in\alpha P$. Then certainly $x\in\alpha R$ since $P\subset R$; and also $x\in P^{k+1}$ since $\alpha\in P^k$. Therefore $\alpha P \subset \alpha R \cap P^{k+1}$.
Conversely, suppose $x\in \alpha R \cap P^{k+1}$. Since $x\in \alpha R$, write $x\alpha r$ for some $r\in R$; it suffices to show that $r\in P$. But if $r\notin P$, then we have $\alpha\notin P^{k+1}$ and $r\notin P$, which implies $\alpha r\notin P^{k+1}$ since $P$ is prime. Therefore we do in fact have $r\in P$. We conclude that $\alpha R \cap P^{k+1} \subset \alpha P$, which completes the proof of the desired set equality.
