I'm working on the following problem with no luck.
Let $A$ be a domain. If $0 \neq f \in A$, then prove that $A_{f} = \bigcap_{P \in\operatorname{Spec}(A), f \not\in P} { A_{P}}$ and conclude that $A_{f}$ depends only on the open subset of $\operatorname{Spec}(A)$ that contains $f$.
I'm thinking about the fact that the primes of $A_{f}$ are in correspondence with the primes of $A$ that do not contain $p$, however I don't see it... probably it's more complicated than that...
Help please, thank you!