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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?

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    $\begingroup$ To the person who downvoted: I would be glad to hear your comments on how to improve this question. If you think that I could have anwered this question by myself, I must say the following. If an author of a book denotes some kind of objects by $R_f$, then it is (usually) assumed that $R_p$ stands for the same kind of objects just with $f$ replaced by $p$. How am I supposed to know that for $f=p$, $R_f$, i.e., $R_p$ means completely different object? Not to mention that Mumford writes $R_P$ for the localization w.r.t. $P$ two lines below. (I have Second edition.) $\endgroup$
    – user557
    Oct 26, 2016 at 15:15

2 Answers 2

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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 $\alpha: R_{f}\to R_{\mathfrak{p}}$.

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  • $\begingroup$ It is also known that this unique map is given by $\alpha(b/f^n)=\varphi(b)/\varphi(f^n)=\frac{b}{1}/\frac{f^n}{1}=\frac{b}{f^n}$, so this map still appears to be the identity map? $\endgroup$
    – user557
    Oct 26, 2016 at 15:58
  • $\begingroup$ It is not quite the identity map, since the word identity map specifically refers to the map $\text{id}: X\to X$ given by $x\mapsto x$. But in this case, the map $\alpha: R_{f} \to R_{\mathfrak{p}}$ is between different spaces! If $R$ is an integral domain, then $R_{\mathfrak{p}}$ is actually "bigger" than $R_{f}$. The map $\alpha$ is injective in this case. $\endgroup$
    – Prism
    Oct 26, 2016 at 19:06
  • $\begingroup$ But if it is incorrect (in general) to say that this map is given by $\frac{b}{f^n}\mapsto \frac{b}{f^n}$, then how does this fact agree with the fact that the unique map $\alpha$ in general case (replace $R_p$ with an arbitrary ring $R'$, $S$ with an arbitrary multiplicative set, and $R_f$ with $S^{-1}R$) is given by $\alpha({\frac{r}{s}})=\frac{\varphi (r)}{\varphi (s)}$? And is it possible to define it by a concrete formula in a proper way? $\endgroup$
    – user557
    Oct 26, 2016 at 19:17
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    $\begingroup$ You are right. It is formulaically given by $b/f^n \mapsto b/f^{n}$. The think to keep in mind is that these two expressions are in two different rings! And the respective equivalence relations on $R_f$ and $R_{\mathfrak{p}}$ are different! (I mean the equivalence relation that declares $b/f^{n} \sim c/f^{m}$ if there exists $s(bf^{m}-cf^{n})=0$ for some $s\in S$). $\endgroup$
    – Prism
    Oct 26, 2016 at 19:31
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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 $R_{\mathfrak p}$)}$.

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  • $\begingroup$ What do you mean by dots in $\dots \frac{x}{f^n}$? $\endgroup$
    – user557
    Oct 26, 2016 at 14:20
  • $\begingroup$ They just mean ‘ Guess what?’. It's metamathematic. $\endgroup$
    – Bernard
    Oct 26, 2016 at 14:23
  • $\begingroup$ So $\dots=1$? This is well-defined. $\endgroup$
    – user557
    Oct 26, 2016 at 14:36
  • $\begingroup$ You mean the identity map? Not exactly. Looks like a canonical injection, except it's not necessarily injective $\endgroup$
    – Bernard
    Oct 26, 2016 at 14:38
  • $\begingroup$ No other ideas come to me... $\endgroup$
    – user557
    Oct 26, 2016 at 14:51

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