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.


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 Feb 27 '14 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? :) $\endgroup$ – Alex Youcis Feb 27 '14 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 Feb 27 '14 at 9:40
  • $\begingroup$ @AsalBeagDubh Done, thanks! $\endgroup$ – Alex Youcis Feb 27 '14 at 9:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.