When $A\subset B$ is an extension of number rings, I want to show that $B_{\mathfrak{p}}=(A-\mathfrak{p})^{-1}B$ is semi local for every prime $\mathfrak{p}\subset A$.
So written differently, $B_{\mathfrak{p}}=\left\{\frac{a}{b}\mid a\in B,\;b\in A-\mathfrak{p}\right\}$.
Can anyone give some type of hint? I'm really lost.
Let $A\subset B$ be two integral domains and $P$ a maximal ideal $A$.
For $q$ a prime ideal of $(A-P)^{-1} B$ then $p= q \cap A$ is a prime ideal of $A$. If $p \not\subset P$ there exists $a\in p, a\not \in P$, thus $a^{-1} \in (A-p)^{-1} B$ so that $a\in q \implies a a^{-1}=1 \in q$, a contradiction.
Thus the only prime ideals of $(A-P)^{-1} B$ are those above prime ideals $\subset P$.
If $Frac(A),Frac(B)$ are finite extensions of $\Bbb{Q}$ then any non-zero prime ideal is maximal and any quotient of $(A-P)^{-1}B/(c)$ is a finite ring thus there are only finitely many ideals above $P$ and there are finitely many prime ideals in $(A-P)^{-1}B/(c)$.