Consider a scheme $X$; for every $x\in X$ with residue field $k(x)$, we have the canonical surjection $\mathcal O_{X,x}\longrightarrow k(x)$ that induces the morphism of affine schemes $\operatorname{Spec}(k(x))\longrightarrow\operatorname{Spec}(\mathcal O_{X,x})$. Now if $\operatorname{Spec} A=U\subseteq X$ is an affine open containing $x$, then we have a morphism $$A=\mathcal O_X(U)\longrightarrow \mathcal O_{U,x}=\mathcal O_{X,x} $$

that induces a morphism of schemes $\operatorname{Spec}(\mathcal O_{X,x})\longrightarrow U$.

Reassuming, by composing with the immersion of $U$ in $X$ we obtain a morphism of schemes $\operatorname{Spec}(k(x))\longrightarrow X$. Many texts say that this morphism is independent from the choice of the open affine $U$, but I don't understand why.


1 Answer 1


There are multiple ways of attacking this:

Method 1:

Show that if $V\subseteq U$ is another affine containing $x$, then the maps $\text{Spec}(\mathcal{O}_{X,x})\to V$ and $\text{Spec}(\mathcal{O}_{X,x})\to U$ are the same (this isn't that hard) (EDIT: As Asal Beag Dubh points out below, this means that $\text{Spec}(\mathcal{O}_{X,x})\to U$ factors as $\text{Spec}(\mathcal{O}_{X,x})\to V\to U$). Then, for any two affines $W,U$ pass to some affine open $V\subseteq W\cap U$.

Method 2:

Let $Z\subseteq X$ be the subset of $X$ consisting of points which generalize $x$. Consider the topological map $i:Z\hookrightarrow X$. Define $\mathcal{O}_Z:=i^{-1}\mathcal{O}_X$. Shown then that $(Z,\mathcal{O}_Z)$ is a scheme, and that for any choice of affine $x\in U$ one has that $Z\to X$ is isomorphic to $\mathcal{O}_{X,x}$ (i.e. that $Z\cong \mathcal{O}_{X,x}$ in a way compatible with these mappings).

  • $\begingroup$ Dear Alex, does the first sentence of Method 1 mean that one map factors through the other? I am not sure I can make sense of the assertion that they are "the same". $\endgroup$
    – user64687
    Commented Feb 27, 2014 at 9:34
  • $\begingroup$ @AsalBeagDubh I guess what I technically mean is that the maps $\text{Spec}(\mathcal{O}_{X,x})\to V\to U$ and $\text{Spec}(\mathcal{O}_{X,x})\to U$ are the same--so yes, the one factors into the other. Does that clarify? :) $\endgroup$ Commented Feb 27, 2014 at 9:36
  • 1
    $\begingroup$ Dear Alex: yes, I knew what you meant. An edit might make the point a bit clearer, but that's up to you. $\endgroup$
    – user64687
    Commented Feb 27, 2014 at 9:40
  • $\begingroup$ @AsalBeagDubh Done, thanks! $\endgroup$ Commented Feb 27, 2014 at 9:44

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