Let $R$ be a integral domain and let $\mathfrak U\subseteq\mathfrak B$ two ideals of $R$ such that $\mathfrak UR_\mathfrak p=\mathfrak BR_\mathfrak p$ for all maximal ideals. Then $\mathfrak U=\mathfrak B$.
Proof: Let $b\in \mathfrak B$. For each maximal ideal $\mathfrak p$, we have $bR_\mathfrak p\subseteq \mathfrak UR_\mathfrak p$. So there is some $a\in \mathfrak U$ and some $s\in R\setminus\mathfrak p$ such that $b=a/s$. The ideal of $R$ defined by
$$\{y\in R: by\in\mathfrak U\}$$
must contains $s$ and so it does not belong to $\mathfrak p$. This ideal is not contained in any maximal ideal so it must be all of $R$. Thus $b=1b\in\mathfrak U$ as required.
My question is: if the above proof holds if $\mathfrak U,\mathfrak B$ are fractional ideals of $R$.
I think so, I'm not entirely sure.
Thanks.