Let $L/K$ be an extension of number fields of degree $n$. Their integral rings being $\mathcal{O}_K$ and $\mathcal{O}_L$. Let $\mathfrak{p}$ be a prime ideal of $\mathcal{O}_K$. We let $\mathcal{O}_{K,\mathfrak{p}}$ and $\mathcal{O}_{L,\mathfrak{p}}$ denote the localizations of $\mathcal{O}_K$ and $\mathcal{O}_L$ at $\mathfrak{p}$. Here we must note that $\mathcal{O}_{L,\mathfrak{p}}=\{\frac{a}{b}\mid a\in \mathcal{O}_L, b\in \mathcal{O}_K-\mathfrak{p}\}$. I want to show that:
$\mathfrak{p}\mathcal{O}_{L,\mathfrak{p}}\cap \mathcal{O}_L=\mathfrak{p}\mathcal{O}_L$
I know that $\supseteq$ holds but don't know how to show the other direction. I even wonder whether this is true. This equality comes from Tom Weston's lecture notes of Algebraic Number Theory. Can you help me to prove this equalty. Thanks very much!!
This is a general property of localized rings. Let $A$ be a commutative domain, $S$ a subset of $A$ stable under multiplication, not containing $0$ nor $1$, $A'=S^{-1}A$ the localized ring w.r.t. $S$ (your particular case corresponds to $S$ = the complementary of a prime ideal $P$ of $A$). For any prime ideal $Q'$ of $A'$ (resp. $Q$ of $A$), denote $Q=Q'\cap A$ (resp. $Q'=QA'$). Then these two "restriction-extension" maps are mutually inverse bijections between the set of prime ideals of $A'$ and the set of prime ideals of $A$ not intersecting $S$. Note that we don't need $A$ to be a Dedekind domain. See e.g. P. Samuel's ANT, chap.5, §5.1.