Let $R$ be a ring, $X=\operatorname{Spec}R$, and write $X_f$ for $X\setminus V(f)$.
We know that $X_g\subset X_f$ iff $g\in \sqrt{(f)}$ iff $g^n=fh$ for some $n \in \mathbb{N}$ and $h\in R$. So there is a well-defined map $R_f\rightarrow R_g$ given by $\frac{a}{f^m}\mapsto\frac{ah^m}{g^{nm}}$ whenever $X_g\subset X_f$. Here $R_f$ stands for the localization of $R$ w.r.t. $\{f,f^2,\dots\}$.
Now I have a few questions about the proof of the following lemma from Mumford's "Red Book":
First, I don't understand how the equivalence of statement 4 with the other ones follows. (The first three statements are indeed clearly equivalent.)
Second, why $g$ goes to $0$ in $R_P$? Since Mumford appeals to the commutativity of the diagram, I need to trace where $g$ goes under the upper arrow and further under the right arrow. Since $X_{f_\alpha}\subset X_f$, it follows that $f^k_{\alpha}=fh$ for some $k\in \mathbb{N}$ and $h\in R$, and so under the upper arrow I have $\frac{b}{f^n}\mapsto\frac{bh^n}{f_{\alpha}^{nk}}$. To understand what is its image under the right arrow, I asked this (general) question, but I'm still not sure about this map. The image must be zero. Why is this so?
Thirdly, I'm not sure about the last sentence. Of course we arrive at a contradiction to the fact that $A\subset P$. However we started with assuming that $g\ne 0$, and to finish the proof we must show that there exists an index $\alpha_0$ such that the image of $g$ in $R_{f_{\alpha_0}}$ is nonzero. I guess this $\alpha_0$ equals $\alpha$, but I can't see why the image of $g$ in $R_{f_\alpha}$ equals zero.
First $\sqrt{A}=\bigcap\limits_{A\subseteq P, P\text{prime}} P$; this is a basic result in algebra (using AC).
Second, $g$ goes to zero by the commutative diagram, as it goes to zero in $R_{f_{\alpha}}$.
Third, the idea is $g\neq 0\Rightarrow \neg iv\Rightarrow \text{contradiction}$