Let $A$ be a ring with fraction field $K$. Let $L$ be a finite degree extension of $K$. Let $B$ be the integral closure of $A$ in $L$. Let $P$ be a prime ideal of $A$. Let $Q$ be a prime ideal of $B$, such that $Q\cap A=P$. Consider the localizations $A_P$ and $B_Q$.
It is easy to see that $$A_P\subseteq B_Q\cap K$$
How to prove that $$A_P=B_Q\cap K$$
I need this theorem when $A$ and $B$ are Dedekind domains. But I think it is true, regardless of whether or not $A$ and $B$ are Dedekind domains.
I have an idea when $A$ and $B$ are Dedekind domains. Correct me if I am wrong.
Consider $\frac{b}{t}\in B_Q\cap K$, where $b\in B$ and $t\in B-Q$. As $\frac{b}{t}\in K$, we have $\frac{b}{t}=\frac{a}{a'}$ for some $a,a'\in A$. Without loss of generality we can chose $a,a'\in A$ such that the ideals $\langle a\rangle$ and $\langle a'\rangle$ are coprime. If $a'\notin P$, then we are done. So assume that $a'\in P$. As $\langle a\rangle$ and $\langle a'\rangle$ are coprime, we have $P\nmid \langle a \rangle$. From $\frac{b}{t}=\frac{a}{a'}$, we get $a'b=at$. So in $B$ we have the followring equalty of ideals $$B\langle a'\rangle\langle b\rangle=B\langle a\rangle\langle t\rangle$$ Since, $P\mid\langle a'\rangle$, we have $Q\mid B\langle a'\rangle$. As, $Q$ divides the LHS, $Q$ must divide one of $B\langle a \rangle$ or $\langle t\rangle$. Since, $P\nmid\langle a\rangle$, we have $Q\nmid B\langle a\rangle$. Hence, $Q\mid \langle t\rangle$. But this is a contradiction, as $t\notin Q$. So, we must have $a'\notin P$.
Please tell me if my answer is correct or not.