What are $R_p$ and "a natural map" $R_f\rightarrow R_p$ in Mumford's Red Book? 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. $\{f,f^2,\dots\}$.
In his Red book, Mumford writes the following:

If $[P]\in X_f$, then $f\notin P$ and there is a natural map
  $R_f\rightarrow R_p$, since the multiplicative system $R-P$ contains
  the multiplicative system $\{f,f^2,\dots\}$.

I don't get what this small $p$ (or more generally $R_p$) stands for. First I naively thought that $R_p$ is just $R_P$, the localization of $R$ at the prime ideal $P$, but then realized that this makes no sense because the the existence of a map $R_f\rightarrow R_p$ assumes that $X_p\subset X_f$, which is of course true, but if $p=P$, then on trying to verify that this map is well-defined, one has to use $p\in\sqrt{(f)}$, which is nonsense in the case $p=P$.
So the question is what are $p$, $R_p$, and what "natural map" does Mumford mean? 
 A: Suppose that $f\notin\mathfrak{p}$. Consider the canonical map $\varphi: R\rightarrow R_{\mathfrak{p}}$ (this is the map $a\mapsto a/1$). Since $f\notin\mathfrak{p}$, it must be the case that $\varphi(f)=f/1$ is invertible in $R_{\mathfrak{p}}$. By the same reasoning, if we let $S=\{1, f, f^2, ... \}$, then $\varphi(S)\subset (R_{\mathfrak{p}})^{*}$, which is the group of units in $R_{\mathfrak{p}}$ (i.e. the invertible elements of ring). Now using the universal property of the localization (with respect to the multiplicative set $S$), we get a unique map $\alpha: R_{f} \to R_{\mathfrak{p}}$ such that $\varphi = \alpha\circ \psi$ where $\psi: R\rightarrow R_f$ is the canonical map into the localization. So the map Mumford talks about is the map $\alpha: R_{f}\to R_{\mathfrak{p}}$. 
A: Your first naive thought was correct: the prime ideal $\mathfrak p\in D(f)$ if $f\notin \mathfrak p$,, hence the set $\{1,f,f^2,\dots,f^n,\dots\}$ is a sub-multiplicative system of the multiplicative system $R\setminus\mathfrak p$, and the map
$R_f\to R_{\mathfrak p}$ is perfectly well defined. It is the map $\dfrac x{f^n}\;\text{(calculated in $R_f$)}\mapsto\dotsm\dfrac x{f^n}\;\text{(calculated in $R_{\mathfrak p}$)}$.
