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Let $X$ be a scheme, and $x \in X.$ Let $U=\text{Spec}(A)$ be an open affine subset containing $x,$ then we have the natural morphism $\mathcal{O}_X(U) \to \mathcal{O}_{X,x}$ inducing a morphism $ \text{Spec} \;\mathcal{O}_{X,x} \to U$ and by composing it with the open immersion $U \hookrightarrow X$ we get a morphism $f: \text{Spec} \;\mathcal{O}_{X,x} \to X.$

  1. Why this definition does not depend on the choice of $U?$ and

  2. What is the image of $f?$

Sorry if these are stupid questions!

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Do you just mean the set-theoretic image? Think about it in the affine case. What are the primes of $A_\mathfrak p$? What do they correspond to, topologically? –  Dylan Moreland Jul 22 '12 at 2:51
The map doesn't depend on the choice of open affine because we've identified the overlaps of an affine cover to define $X$ (the gluing construction). –  Andrew Jul 22 '12 at 2:57

2 Answers 2

up vote 5 down vote accepted

(1) Let $U$ and $V$ be open affine subsets of the scheme $X$ such that $x\in U\cap V$. Choose an open affine subset $x\in W\subseteq U\cap V$. Prove that the compositions $\text{Spec}(O_{X,x})\to W\to U\to U\cup V$ and $\text{Spec}(O_{X,x})\to W\to V\to U\cup V$ are equal to the composition $\text{Spec}(O_{X,x})\to W\to U\cap V\to U\cup V$. (Hint: recall that $U,V,W$ are affine open subsets of $X$ and you understand affine schemes by commutative algebra!)

(2) If $A$ is a commutative ring and if $p$ is a prime ideal, then the spectrum of the localization homomorphism $A\to A_{p}$ is the map $\text{Spec}(A_p)\to \text{Spec}(A)$. The image of this map equals the set of all prime ideals of $A$ contained in $p$ (prove this fact from commutative algebra if it is not obvious).

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Let me add to Amitesh's absolutely correct answer a few words describing the image $I\stackrel {\text {def}}{=}Im(f)$ of the canonical morphism $f:\text{Spec} (\mathcal O_{X,x}) \to X$ .

a) The set $I$ is exactly the intersection of all neighbourhoodz of $x$ in $X$: it is a kind of microgerm of $X$ at $x$.
Beware that $I$ is not a subscheme of $X$ since it is not locally closed.

b) More geometrically (and thus more interestingly!) consider the irreducible subvariety $V=\overline {\lbrace x\rbrace}\subset X$ whose generic point is $x$.
Let $Y\subset X$ be a closed irreducible subscheme on which $V$ lies: $V\subset Y$ and let $\eta_Y$ be the generic point of $Y$.
Then our subset $I$ is exactly the set of all those generic points $\eta_Y$. We say that $I$ is the set of generizations of $x$.

c) Two examples:
1) If $X$ is an irreducible scheme with generic point $\eta$, then for $x=\eta$ we have $I={\lbrace \eta\rbrace}$.
2) If $X=\mathbb A^2_\mathbb C=\text {Spec}(\mathbb C[x,y])$ and $x=(a,b)$ (more accurately $x$ is the maximal ideal $\mathfrak m= (x-a,x-b)$) , then $I$ is the set consisting in $x$, the generic point of $\mathbb A^2_\mathbb C$ and the generic points of all irreducible curves going through $x$, like for example the curve $(y-b)^2-(x-a)^3=0$.

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Dear @Georges thank you for the clarification. –  Ehsan M. Kermani Jul 22 '12 at 17:44
Dear Georges, do you know any reference for these very interesting remarks? –  Adeel Jul 23 '12 at 5:14
Dear Adeel, you can look at Mumford's The red book of varieties and schemes, page 74, Example F. You can also look at Grothendieck-Dieudonné's EGA, Chapitre 1, Section 2.4, pages 101-102 . –  Georges Elencwajg Jul 23 '12 at 8:30

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