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I'm reading the book $\textit{The Geometry of Schemes}$ and am a bit confused about the definition of the structure sheaf of an affine scheme.

For a ring $R$, we define the $\textit{distinguished open sets}$ of $X=\text{Spec}R$ to be the sets $$X_f= \text{Spec}R-V(f)=\{p \in \text{Spec}R : f \not\in p \}.$$

To define a sheaf $\mathcal{O}$ on $X$, we assign to each $X_f$ the ring $R_f$, i.e. the localization of $R$ at the element $f$.

Now the book says that if $X_g \subseteq X_f$, we have $g^n \in (f)$, for some integer $n$. Why is that?

Also, for the morphisms, the book says to define $$\text{res}_{X_f, X_g}: R_f \to R_{gf}=R_g$$ as the localization map. What does this mean? I'm mostly confused about the terminology here (what is the localization map?).

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2 Answers 2

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If $X_g \subset X_f$ then $V(f) \subset V(g)$. So, every prime ideal that contains $f$ also contains $g$, hence $g \in \sqrt{(f)}$. It means there exist some $n$ such that $g^n \in (f)$.

(Edit: leibnewtz has pointed out a typo in the definition of the map)

Let's say $g^n = fh$, for some $h \in R$. The restriction map is just the map $R_f \to R_g$ that sends $\frac{a}{f^k} \in R_f$ to $\frac{ah^k}{g^{nk}} \in R_g$.

It's quite straightforward to check that this map is in fact well defined and satisfies the composition requirement.

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  • $\begingroup$ I know I accepted your answer already, but I realized I don't understand it as well as I thought. Firstly, why is $R_{fg}=R_g$? Also, in what sense is the restriction map natural enough that no book seems to define it explicitly? I would also appreciate a hint for showing that it is well-defined $\endgroup$
    – Exit path
    Apr 26, 2016 at 20:45
  • $\begingroup$ Shouldn't the map send $a/f^k$ to $ah^k/g^{nk}$? That's the only way I can see that the map would be well-defined: $a/f^k=ah^k/(fh)^k=ah^k/g^{nk}$. $\endgroup$
    – Exit path
    Apr 26, 2016 at 21:03
  • $\begingroup$ @leibnewtz First of all, I don't think $R_{fg} = R_g$. I suppose it's a typo. And you're right, the map I've defined is wrong. The correct is the one you wrote. Sorry, my mistake. I had the same feeling as you when studied that, cause nobody defined this map, but it's not so difficut to guess what it should be. $\endgroup$
    – remath7
    Apr 27, 2016 at 1:27
  • $\begingroup$ To show it's well defined, suppose that $\frac{a}{f^k} = \frac{b}{f^l}$. Then, $f^r(af^l - bf^k) = 0$, for some $r\in \mathbb{N}$. Now multiply this expression by $h^{k+l+r}$. Can you see it now? $\endgroup$
    – remath7
    Apr 27, 2016 at 1:30
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    $\begingroup$ Well, I was thinking about the equality $R_{fg} = R_g$. My guess is that the author meant $R_{fg}$ is the localization of the ring $R_f $ in the element $g$. Now, it is true that $(R_f)_g = R_g$, since $g^n = fh$ $\endgroup$
    – remath7
    Apr 27, 2016 at 2:26
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For your first question: $X_g\subseteq X_f$ means every prime ideal containing $f$ contains $g$. However, the intersection of the prime ideal containing $f$ is the radical of the ideal generated by $f$. In particular, $g$ is in this radical. So by definition of radical, $g^n\in (f)$ for some $n$.

For your second question: The term localisation map refers to the localisation map of the structural sheaf $\mathcal{O}$. The sheaf is a contravariant functor from open subsets of $X$, so for any open subsets $U_1\subseteq U_2\subseteq X$, there should be a map from $\mathcal{O}(U_2)$ to $\mathcal{O}(U_1)$.

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  • $\begingroup$ I'm aware of the requirements the functor is supposed to satisfy. My question was how the restriction maps in this specific case are defined $\endgroup$
    – Exit path
    Apr 26, 2016 at 19:22

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