Question about the lying over theorem. I have some questions about the proof of the Lying over theorem in the book Introduction to commutative algebra by Atiyah and Macdonald. 
(1) In the proof of Theorem 5.10 of Page 62, is the map $\alpha: A \to A_{\mathfrak{p}}$ the natural map which sends $a$ to $a/1$ for $a \in A$? 
(2) How to show that $\mathfrak{q}\cap A=\alpha^{-1}(\mathfrak{m})$?
(3) How to show that $\alpha^{-1}(\mathfrak{m})=\mathfrak{p}$? 
Thank you very much.
For (2), let $x \in \mathfrak{q} \cap A$. Then $\alpha(x) \in A_{\mathfrak{p}}$. Since the diagram commutes and the horizontal maps are injective, $\alpha(x)=\beta(x)\in \mathfrak{n}$. Therefore $x \in \alpha^{-1}(\mathfrak{n} \cap A_{\mathfrak{p}})=\alpha^{-1}(\mathfrak{m})$. On the other hand, suppose that $x \in \alpha^{-1}(\mathfrak{n} \cap A_{\mathfrak{p}})=\alpha^{-1}(\mathfrak{m})$. Then $\alpha(x) \in \mathfrak{n} \cap A_{\mathfrak{p}}$. Therefore $x \in A$ and $x \in \alpha^{-1}(\mathfrak{n}) = \beta^{-1}(\mathfrak{n})=\mathfrak{q}$. Hence (2) is true. Is this correct?

 A: 1) Yes.
2) Let the top horizontal arrow be $\phi: A \to B$ and the bottom one $\psi:A_{\mathfrak{p}} \to  B_{\mathfrak{p}}$. By commutativity, $\beta \phi = \psi \alpha$, so $(\beta \phi)^{-1}(E) = (\psi \alpha)^{-1}(E)$ for any subset $E$ (remember $(\beta \phi)^{-1}(E) = \phi^{-1}(\beta^{-1}(E))$). $(\beta \phi)^{-1}(\mathfrak{n}) = \phi^{-1}(\mathfrak{q})=A\cap \mathfrak{q}$. For the other side $(\psi \alpha)^{-1}(\mathfrak{n}) = \alpha^{-1}(\mathfrak{m})$. (I didn't carefully read your solution, but I imagine that it proves $(\beta \phi)^{-1}(E) = \phi^{-1}(\beta^{-1}(E))$ in more detail.)
3) The map $\alpha^{-1}$ provides an order preserving bijection between the primes in $A_{\mathfrak{p}}$ and the primes in $A$ contained in $\mathfrak{p}$ (see prop 3.11). $\mathfrak{m}\subset A_{\mathfrak{p}}$ is maximal with respect to the prime ideals in $A_{\mathfrak{p}}$, so the corresponding prime $\alpha^{-1}(\mathfrak{m})$ in $A$ is maximal among the set of primes contained in $\mathfrak{p}$. This maximal element is $\mathfrak{p}$. (other idea: show that $\alpha^{-1}(\mathfrak{p}A_{\mathfrak{p}}) = \mathfrak{p}$ directly, and note that $\mathfrak{m} = \mathfrak{p}A_{\mathfrak{p}}$ since both are maximal in the local ring $A_{\mathfrak{p}}$.)
