I have a few questions about proving the following identities:
$$\bigcap_{p \in SpecA}A_p = A \ \ \ \ \bigcup_{p \in SpecA}A_p = K$$
Here $A$ is an integral domain, $K$ is its field of fractions. For any prime ideal $p$ we can treat $A_p$ as a subring of $K$.
My questions are:
Why does it suffice to prove the first equality only for maximal ideals? I know that maximal ideals are prime, but that doesn't help me at the moment. Of course, the inclusion $\leftarrow$ is quite obvious in case of prime ideals.
In the second equality $\subset$ is easy, because $A_p$ is a subring of $K$. A localization of a domain $A$ with respect to some multiplicatively closed subset $S$ of $A$ is a subring of $K$ consisting of the elements $m/s$, with $m \in A$ and $s \in S$.
The opposite direction is troublesome.
Could you help me with that?
Regarding the first equality: Note that if $p\subset m$ then by definition $A_m\subset A_p$ and $A_p\cap A_m=A_m$.
Regarding the second equality: Let $s/t\in K$. Since $A$ is an integral domain, there is a prime ideal $p$ such that $t\not\in p$ and so $s/t\in A_p$.